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What is the torque produced by 2 rotating bodies with a clutch I am trying to simulate a car engine etc, but I have failed to find any equations governing the torque created by $2$ different constant velocity shafts of different angular momenta joining together with some given slip or friction factor. I know $I_1w1 + I_2w_2 = I_3w_3$ which gives me the end angular velocities, but its not giving me the torque acting on each shaft at any moment in time whilst they are joining.
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A clutch has two sides and a degree of freedom between them. Let's call one side the engine side with known torque $\tau_E$ and unknown speed $\omega_E$. On the other side, the transmission side, the speed is known $\omega_T$, but the transmitted torque $\tau_T$ isn't.
The clutch itself has a critical torque $\tau_C$ which limits its ability to transfer torque. This limits, depends on the engagement of the clutch, and it varies between zero and some maximum value that is more than the maximum the engine can supply $\tau_{C_{\rm max}} >\tau_{E_{\rm max}}$.
So value of the maximum critical torque (torque rating) does not matter for the operation of the car. All it matters is that with 100% clutch in $\tau_C = 0$ and with 0% clutch in $\tau_C > \tau_{E_{\rm max}}$.
Here is a system diagram that will drive the equations
The components are shown in boxes, and the system variables connect the components.
First, the kinematics are valid regardless of how the clutch behaves
$$ \begin{aligned}
\omega_W & = \frac{v}{R} & \dot \omega_W & = \frac{\dot v}{R} \\
\omega_T & = \gamma\; \omega_W & \dot \omega_T & = \gamma\; \dot \omega_W
\end{aligned} \tag{1} $$
Where $gamma$ is the total gear ratio, and $R$ the wheel radius.
And thus the engine torque is always known if the engine speed $\omega_E$ is known $$ \tau_E = f(\omega_E) \tag{2}$$
Considering that each component has each own mass moment of inertia, as well as the combined mass of the car $m$ and any drag force $F_D$ applied, the equations of motion for the car, wheel and transmission are:
$$ \begin{aligned}
F & = m \dot v + F_D \\
\tau_W &= I_W \dot \omega_W + F\,R \\
\tau_T &= I_T \dot \omega_T + \tfrac{1}{\gamma} \tau_W
\end{aligned} \tag{3}$$
Notice that all the accelerations can be written in terms of $\dot v$, such as $\dot \omega_W = \tfrac{1}{R} \dot v$ and $\dot \omega_T = \tfrac{\gamma}{R} \dot v$.
Then the two scenarios for the clutch:
*
*Clutch engaged - engine speed matches transmission speed, $\omega_E = \omega_T$ as well as $\dot \omega_E = \dot \omega_T$
$$ \begin{aligned} \tau_E &= \tau_C + I_E \dot \omega_T \\
\tau_C & = I_C \dot \omega_T + \tau_T \end{aligned} \\ \tag{4a} $$
The above 5 equations in (3) & (4a) are solved for $\tau_C$, $\tau_T$, $\tau_W$, $F$ and finally $\dot{v}$
Notice that the engine speed is directly linked to the car speed $\omega_E = \omega_T = \tfrac{\gamma}{R} v $ and so engine torque is known $\tau_E = f(\omega_E)$.
*
*Clutch slipping - transmission torque limited to $\tau_C$
$$ \begin{aligned} \tau_E &= \tau_C + I_E \dot \omega_E \\
\tau_C & = I_C \dot \omega_T + \tau_T \end{aligned} \tag{4b} $$
The above 5 equations in (3) & (4b) are solved for $\omega_E$, $\tau_T$ $\tau_W$, $F$ and finally $\dot{v}$
Here the engine speed is found from the numeric integral $\Delta \omega_E = \int \dot \omega_E \,{\rm d}t $ and the car acceleration $\dot v$ does not depend on engine torque.
The conditions of slipping the clutch are $\tau_E > \tau_C$, and for engaging the clutch $| \omega_E - \omega_T | \approx 0$.
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Would the existence of a universal wave function automatically mean the Many Worlds Interpretation? If the Universal Wave Function definitely existed, would that mean the Many-Worlds Interpretation was automatically true or would it only imply that?
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The concept of "wavefunction"is inherent in the definition of Quantum Mechanics,in the same way the concept of mass is inherent in the definition of classical mechanics: it is a mathematical function which is used to model data and predict physical behavior of matter at small dimensions .
Present physics modeling assumes that the underlying microscopic interactions obey quantum mechanical equations, and it has been shown that the classical theories emerge from this underlying level. In this sense a universal wave function, assuming extreme measurement accuracies, exists and has been modeled with the density matrix formalism .
[The many-worlds interpretation] (MWI) is an interpretation of quantum mechanics that asserts that the universal wavefunction is objectively real, and that there is no wavefunction collapse. This implies that all possible outcomes of quantum measurements are physically realized in some "world" or universe
As in principle a universal wave function exists, the crucial word is "interpretation". It means that the proposal gives exactly the same numerical predictions for new studies , just a different way of interpreting the internal functions.
would that mean the Many-Worlds Interpretation was automatically true or would it only imply that?
The universal wavefunction exists, can be formulated mathematically . The mathematics is true, but the many-worlds interpretation of the mathematics is just a different optical angle,there are no new predictions to be able to distinguish if it is valid or not.
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Expansion Postulate Quantum Mechanics How does the expansion postulate allow predictions to be made about measurement outcomes?
I understand the postulate as:
$$
ψ =\sum_{n} a_n φ_n
$$
with coefficients calculated by:
$$
a_n =\int φ_n^*ψdτ.
$$
I think that:
$$
|a_n|^2
$$
is the probability of the system being in state φ, but I do not think that is the answer to the question.
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Yes, $\vert a_n \vert^2$ is not the probability of the system being in the state $\phi_n$. It is the probability that the system (which is with $100\%$ probability in the state $\psi$) will be found in a state $\phi_n$ upon a measurement of an operator whose eigenstates are $\{\phi_n\}$.
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Why exactly do we feel a shock when we place our hand into a conducting solution? I have a very naive question.
Suppose you have pure water in a flask, and you place two ends of a copper wire (which are connected to a battery) into the water.
If you were to place your hand into the water, you would not feel any shock, as pure water does not conduct electricity.
However, if you add an electrolyte like common salt to the same water, you would probably feel a shock.
Adding salt makes the solution conducting. When the two wires are placed in the solution, the ions are attracted to the end of the wire which has an opposite charge.
However, what does the movement of those ions have to do with whether or not your hand feels a shock? Shouldn't whether you feel a shock just depend on what resistance your hand offers?
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when you immerse your hand into the conducting water, your skin is connected to the conducting medium. The charge being transported through the water is then faced with a choice, as follows.
Your hand and the water surrounding it represent two resistors in parallel. If your hand is more conductive than the water, the electricity will flow preferentially through your hand and you will feel a shock.
If the water is more conductive than your hand, most of the charge transport will occur through the water, and not much will occur in your hand- and you will feel little or no shock.
Dry skin is an insulator, but it readily absorbs water and becomes a good conductor because of the salt it contains. This means you might not immediately feel a shock when you splunge your hand into the water, but you will as the outer layers of your skin get hydrated.
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Does it make sense to say that something is almost infinite? If yes, then why? I remember hearing someone say "almost infinite" in this YouTube video. At 1:23, he says that "almost infinite" pieces of vertical lines are placed along $X$ length.
As someone who hasn't studied very much math, "almost infinite" sounds like nonsense. Either something ends or it doesn't, there really isn't a spectrum of unending-ness.
Why not infinite?
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Almost infinite can make a lot of sense in physics. There isn't a precise definition but I would interpret it as the following: when something is 'almost infinite' the properties we are considering will barely change when we make the system actually infinite.
Examples:
*
*In thermodynamics the particle number is often of the order of Avogadro's number $N\approx 6.022\cdot10^{23}$. For most properties considered this is basically infinite.
*Let's say we have a gaussian distribution $f(x)=e^{-\pi x^2}$. The integral over the whole number line is $\int_{-\infty}^{\infty}e^{-\pi x^2}\text d x=1$, but most of the area is in a small portion centered around zero. If we take instead $\int_{-L}^{L}e^{-\pi x^2}\text d x$ then this will approximate 1 to many decimal places even if $L$ is as small as 5. If you take $L=100$ then, as far as $f$ is considered, $L$ is infinite. In quantum mechanics this $f$ could be the wavefunction of a particle for example.
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Continuity equation in QM I found this question in a quantum mechanics exam:
What is the physical interpretation of the continuity equation $\frac{\partial\rho}{\partial t}+\frac{\partial j}{\partial x}=0$? Here $\rho(x,t)$ is the probability density and $j(x,t)$ is the probability current.
I assume they want a one liner like "probability is conserved". But to be honest I do not understand this. Can any one help me here? What's the one liner they are looking for and why? Many thanks!
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You're actually right, it stems from the "conservation" of probability, or the fact that probability sums to 1. It is literally the equation that says if $\rho$ changes, then that must be due to $j$.
Consider the integral version of this equation. In 3D the space derivative is a divergence,
$$\int \left[\frac{\partial \rho}{\partial t} + \nabla\cdot j\right] dV$$
$$\frac{\partial P}{\partial t} = - \oint j \cdot dA$$
The time rate of change of probability in an area is equal to the amount of probability "leaving" the area in any direction (through the surface that defines the area).
In fact, this is the same as the differential form continuity equation in fluids, charge (electromagnetism), heat, etc.
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What is a Topological Twist? I have come across topological twists on numerous occasions but I have never actually seen them explained in an understandable way. So, I was wondering
*
*What does it physically mean to topologically twist a theory?
*What does it mathematically mean to topologically twist a theory?
*What is the motivation for talking about topological twists?
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So, I think I've found an answer to my own question. One can take a supersymmetric theory defined on $\mathbb{R}^n$ and topologically twist it by redefining the rotation group of the theory into a mixture of the (spacetime) rotation group and the R-symmetry group.
Physically, this is just making one sector of the theory "more apparent". We can then restrict to this sector of the theory to get a topological field theory. In general, we will lose the supersymmetry (or at least lose some of it) since we are only considering one sector of the theory.
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Could light from the Sun be used to power the Breakthrough Starshot microships? The Breakthrough Starshot initiative aims to accelerate a swarm of 16m$^2$-area solar sails to 15% of $c$ using Earth-based lasers in the order of 100GW power in 10 minute bursts. Considering loses from various medium densities and temperatures of the atmosphere they estimate the sails to receive 60 times the amount light that the Sun provides.
Sunlight in Space is abundant, 8% of its spectrum is ultra-violet light which is more energetic, and has an intensity of around 1000 W/m$^2$.
Could the gigawatt lasers be replaced entirely with an apparatus, in Space, that would be some arrangement of Fresnel lenses, lenses, mirrors and crystalline materials so as to forward and beam pumped sunlight to the sails, possibly for longer exposition times?
If yes, what would be the size and composition of such an apparatus, roughly?
Somewhat of a follow up question: what kind of heat would such a beam of sunlight-turned-coherent carry in relation to distance from the spacecraft?
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Because of its angular width (low spatial coherence), the sun's light cannot be focused to a small (meter-sized) spot at large distances. Any kind of lens arrangement can at best form an image of the sun.
However, large arrays of photovoltaic cells could certainly be used to power a laser array for Starshot. A laser beam with a large (kilometer-scale) aperture can be focused to the requisite meter-scale spot at interplanetary distances because it has very high spatial coherence.
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Describing forces in rolling Consider a wheel on a frictionless horizontal surface. If we apply a horizontal force (parallel to the surface and above the level of the center of mass), what happens to the wheel? Does it roll or slide forward or rotate only or does any other phenomenon happen? Please guide me. Also draw a free body diagram.
Note: This is a thought experiment. If the question is not satisfying, I am sorry for that and please guide me.
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Consider the Free Body Diagram:
where:
*
*$F$ is a driving force
*$mg$ the weight of the wheel
*$F_N$ a reactive force, called the Normal force
*$F_f$ a friction force
We can now establish some force/torque balances.
In the vertical ($y$) direction, there no motion because with $\text{N2L}$:
$$\Sigma F_y=F_N-mg=0 \Rightarrow F_N=mg$$
The friction force is usually modeled as:
$$F_f=\mu F_N=\mu mg$$
As long as no slipping occurs, $\mu$ is the static friction coefficient.
Now, looking at the balance of torques about the CoG (marked as $+$) we have a net torque balance $\tau$:
$$\tau=F\lambda-F_f R=F\lambda-\mu mg R$$
As per $\text{N2L}$ (applied for rotation) this causes angular acceleration $\alpha$, in the clockwise direction:
$$\tau= I_w \alpha \Rightarrow \alpha=\frac{\tau}{I_w}$$
where $I_w$ is the inertial moment of the wheel.
Note that $\alpha=\frac{\text{d}\omega}{\text{d}t}$.
Without slipping/sliding, we can write $v=\omega R$ and also:
$$a=\alpha R$$
Or:
$$a=\frac{F\lambda-\mu mg R}{I_w}R$$
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Why is total kinetic energy always equal to the sum of rotational and translational kinetic energies? My derivation is as follows.
*
*The total KE, $T_r$ for a rigid object purely rotating about an axis with angular velocity $\bf{ω}$ and with the $i$th particle rotating with velocity $ \textbf{v}_{(rot)i} = \textbf{r}_i \times \textbf{ω}$, (summing over the i'th particle) is $T_r = \frac{1}{2}m_i|\textbf{r}_i \times \textbf{ω}|^2$, as long as the origin passes through the axis of rotation.
*Let's decompose these using a coordinate system with 3 arbitrary orthogonal unit vectors, whose directions have subscript (1,2,3), and expand the brackets. The result can be shown to be $T_r = \frac{1}{2}I_{ij}ω_iω_j$ summing from $i,j = 1$ to $3$, and where $I_{ij}$ are elements of the moment/product of inertia tensor in the given coordinate system.
This seems like the standard expression for rotational kinetic energy. The only assumption was that the object has a constant rotation and that our chosen origin lies on the axis of rotation.
*Now let's boost the object with a velocity $\textbf{v}_o$. The total velocity is now $\textbf{v}_o + \textbf{v}_{(rot)i}$ so the total KE is $\frac{1}{2}Mv_o^2 + T_r + m_i \textbf{v}_o \cdot({\textbf{r}_i \times\textbf{ω})}$
It seems to me that the third term is not trivially zero. If it is, can anyone show this? If not, then why do we simply add rotation and translation energies in mechanics?
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You're right that the third term does not generically vanish. The key element in decomposing the kinetic energy into rotational and translational parts is that you calculate the rotational kinetic energy about the center of mass.
If the center of mass of the object is at the coordinate origin, and $\mathbf r_i$ is the position of the $i^{th}$ mass, then everything proceeds as you suggest. The velocity of the $i^{th}$ mass is $\mathbf v_i = \mathbf r_i \times \boldsymbol \omega$, and so the total kinetic energy is
$$T = T_r = \sum_i\frac{1}{2} m_i (\mathbf r_i\times \boldsymbol \omega)^2$$
If we perform a boost, then we would have that $\mathbf v_i = \mathbf (r_i-\mathbf R) \times \boldsymbol \omega_{CM} + \mathbf v_0$, where $\boldsymbol \omega_{CM}$ is the angular velocity about the center of mass and $\mathbf R$ is the position of the center of mass. This would give us
$$T=\sum_i\frac{1}{2}m_i\left([\mathbf r_i -\mathbf R]\times\boldsymbol \omega_{CM} + \mathbf v_0\right)^2$$
$$= \sum_i\left\{ \frac{1}{2}m_i\big([\mathbf r_i - \mathbf R]\times \boldsymbol \omega_{CM}\big)^2 + \frac{1}{2}m_i \mathbf v_0^2 + m_i \mathbf v_0\cdot [\mathbf r_i-\mathbf R]\times \boldsymbol \omega_{CM}\right\}$$
The first term is the rotational kinetic energy about the center of mass. The second term is the translational kinetic energy, calculated as though the entire mass $M$ were concentrated at the center of mass position. The third term vanishes because if we sum over masses,
$$\sum_i (m_i\mathbf r_i - m_i \mathbf R) = M\mathbf R - M\mathbf R = 0$$.
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What is happening when magnetic field lines snap or break? In discussions of sun spots and auroras on Earth, magnetic field lines are often described as "snapping" or "breaking", with the result of releasing charged particles very energetically.
My understanding is that field lines are just a visualization tool. I don't understand, intuitively, how a field line could snap or break, or why that would result in a release of energy.
I'm having trouble even framing this question because the concept of a field line breaking just doesn't make sense to me. What is happening when a magnetic field "snaps"?
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You are right: magnetic field lines can't snap or break because they are not physical objects. They are more analogous to elevation lines on a topographical map, or more precisely to lines perpendicular to elevation lines: to the fall lines on a ski slope. However, they do describe something physical, which is the magnetic field distribution. When the sources of the magnetic field rearrange, the "magnetic field lines" can change discontinuously, and it's the discontinuous change that is referred to as "snapping" or "breaking".
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Why use virtual displacement to make constraint forces vanish? Why do we use virtual displacement to vanish work done by constraint forces instead of the actual displacement?
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*
*In a nutshell, d'Alembert's principle states that a certain vector should be perpendicular to a constraint surface, i.e. perpendicular to all its tangent vectors, i.e. perpendicular to all infinitesimal virtual displacements. See also e.g. this related Phys.SE post.
*Note that at this stage, we are just trying to find the EOMs of the physical system under investigation. At a later stage we will then try to solve the EOMs. We cannot use the actual displacement because we don't know the solution yet.
*Btw, it should probably be stressed that the actual displacement never coincides with a virtual displacement, because the latter is frozen in time, cf. e.g. this & this related Phys.SE posts.
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What is the energy of a (conduction/valence) band? I'm trying to figure out what determines the energy of the bands, either conduction or valence band. Mostly I can read about the bandgap energy, which is mostly just the difference between $ E_C $ and $ E_V $ but there are also more concise expressions like $ E_g(T) = E_g(0) - \frac{ \alpha \cdot T^2 }{T + \beta} $. According to wiki, $ E_g(0) $ is "just" a material constant.
In short words: What determines $ E_V $ and $ E_C $ ?
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Really it depends how deep you want to go! I hopefully will keep it fairly simple.
If you consider an atom, the energy levels are determined by the orbital the electron is in, calculated from the Schrodinger wave equation. It is a similar concept for a material, albeit complicated by there being a large number more atoms and the interactions between them. It should be noted here that these energy levels also vary periodically throughout the lattice.
Essentially this means that Ev and Ec are determined by the solution to the electron wavefunction for the highest occupied and lowest unoccupied energy band in the material.
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Is it possible for two observers to observe different wavefunctions for one electron? Suppose there are 2 scientists who have decided to measure the location of an electron at a same fixed time. Is possible that while one observes the wavepacket localized at (position=x) while the other observes the wavepacket localized at (position=y). The condition however is position x is not equal to y.Please dont confuse about the degree of localization which can be quite varying depending upon which measurement-momentum or position is given priority
:( I have little to no experience with quantum superposition and gaussian wavepackets.......kindly manage with my rough knowledge
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I will confess that this I am not certain on this, as this is a quite odd question. but I have formulated what seems like a very reasonable answer. Also the very first thing to mention is that one cannot physically observe the wave-function, we can only predict how it evolves, and then measure a property of a particle, this collapsing the wave-function.
When we (in theory at least) make a measurement of position, the wave-function collapses to a Dirac-delta distribution (like a spike at where we measure the particle). Without going into the specifics, after we make the measurement, if we leave the system alone, the wave-function will spread out over time.
So lets consider what happens if we make two position measurements, at different times, but in the limit that the time goes to zero.
So if the first measurement gives $x_1$, then at some time $t$ after the measurement, we make another measurement. This measurement could be literally any value, but for a free particle will have a mean at $x_1$. And it will have some standard deviation (or uncertainty) that is greater than zero. However, as we take this $t$ to zero, that probability of the measurement being away from $x_1$ will decrease, and the standard deviation will decrease as well, until in the limit that $t=0$, the probability of the measurement being anywhere except $x_1$.
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'Why' is the Schrödinger equation non-relativistic? The transition amplitude for a particle currently in one spacetime point to appear up in another point doesn't respect causality which becomes one of the main reasons to abandon non-relativistic quantum mechanics. We impose the relativistic Hamiltonian $H=\sqrt{c^2p^2+m^2c^4}$ to get the Klein–Gordon equation or more correctly "add" special relativity after 2nd quantizing to fields, which shows how antiparticles crop up and help in preserving causality in this case. Apart from that, the equation is not even Lorentz covariant, which proves it to be non-relativistic.
But why does this occur? I mean, the Schrödinger equation is consistent with the de Broglie hypothesis and the latter is so much consistent with relativity, that some books even offer a "derivation" of the same by equating $E=h\nu$ and $E=mc^2$ probably resulting from a misinterpretation of de Broglie's Ph.D. paper. (A derivation isn't exactly possible though). So, the Schrödinger equation should include relativity in it, right? But it doesn't... How does relativity vanish from the Schrödinger equation or did the de-Broglie hypothesis ever not "include" relativity in any way?
My suspicion—The "derivation" is not possible, so the common $\lambda=h/mv $ with m as the rest mass, doesn't include relativity in any way. End of story. Is this the reason or is there something else?
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The Schrödinger equation is non-relativistic by construction. It follows from the nonrelativistic classical energy expression by applying De Broglie's idea to replace $(E,\vec p)$ by $-i\hbar (\partial_t, \vec \nabla)$.
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Isn't the disturbance in quantum field an interpretation of quantum wave function? In quantum mechanics, a particle is neither a wave nor a particle but is described by a wave function which does not have a physical interpretation. But in quantum field theory, a particle is a disturbance in the associated field. Isn't the disturbance in quantum field an interpretation of quantum wave function?
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In quantum mechanics, a particle is neither a wave nor a particle but is described by a wave function which does not have a physical interpretation
Note the bold a particle. The quantum mechanical equations , Dirac, Klein Gordon, quantized Maxwell, all describe a single particle.
The postulates of Quantum Field Theory are the postulates of Quantum mechanics, exactly. QFT was invented as a mathematical tool for the study of many particle systems in quantum mechanics, that allows for calculating crossections of interactions and decay probabilities.
Isn't the disturbance in quantum field an interpretation of quantum wave function?
If you mean by interpretation a different mathematical formulation of the wave function for interacting particles, the answer is yes.
Note that the fields on which the creation and annihilation operators work are the plane wave wavefunctions, (for example see here) i.e. the solution of the corresponding quantum mechanical equations for single particles. As a probability the $Ψ^*Ψ$ of a plane wave cannot describe the location of a particle as measured in the lab. For single tracks of particles the wavepacket solution has to be used .
For calculating crossections and decays, the Feynman diagrams using QFT are accurate and validated by experiments.
See also my answer here .
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What is the difference between a force and a net force? I read in Newton's first law, it states that an object will continue to have a constant velocity unless acted upon by a force whilst for other articles, it states "unless acted upon by a net force." Which one is correct? Are they both interchangeable? Is there any difference between these two concepts?
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Besides the answers already given, you can think of Newton's first law as a special case of Newtons second law. For Newton's second law the "force" is explicitly stated as a "net force", or
$$F_{net}=ma$$
Assuming constant mass, if the net external force acting on an object is zero, then the acceleration of the object is also zero, i.e., the object will either be at rest or moving at constant velocity with respect to an observer in any inertial frame per Newton's first law.
Hope this helps.
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How does the Brownian motion of air molecules compare to the threshold of human hearing as a function of frequency? This fantastic question essentially asks what is the noise floor of air? Both the answer given on that thread and the value stated by Microsoft are around -23 or -24 dBSPL.
However, overall loudness is only one metric. What does the amplitude of the noise in dBSPL look like when graphed out as a function of frequency in the audible range? How does the shape and level of the curve defined by that graph compare to the threshold of human hearing as described by the equal loudness contour?
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It depends.
*
*The threshold of human hearing is adaptive and varies depending on the preceding noise level (for a couple of hours). After a shift in an anechoic chamber, engineers start hearing how their own blood goes through veins, food being digested, etc. The threshold can not be assigned a "fixed" value. The "usual" -94dBPa was suggested as a median value in a certain experiment with a few dozen of people, tones, and headphones. Now, such an experiment would be deemed unscientific.
*People are very different. There can not be a common "threshold" because it means nothing. My hearing at ~25yo was 20dB better than "normal", and I know people who are even better than me. Some blind people show absolute wonders when it comes to their abilities to "see" with their ears.
*The methodic used in the MS chamber: The best existing mic, by B&K, is -2dBA. Thus, to obtain a lower figure of room noise, they compared the noise of a mic in total silence (inside a very heavy box) with the noise of the same mic in the room. Then they assume that the noise in the room is sqrt(self_nse^2+room_nse^2). Self mic noise is mostly the thermal noise of preamp (the diaphragm noise is high-freq and low), so the robustness of the methodic is not unquestionable.
I would find the Hollywood assumption that you hold your breath, then no living creature can hear you, quite ... inspiring.
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Is it always the case that the square root of a lagrangian gives the same equations of motion as the lagrangian itself? Inspired by the Phys.SE post Geodesic Equation from variation: Is the squared lagrangian equivalent? I was wondering if it is always the case that the square root of a lagrangian gives the same equations of motion as the lagrangian itself? Are there specific counterexamples, or is there any way to derive a set of conditions the lagrangian has to satisfy, for it to have this particular property?
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That is rarely$^1$ the case. Sufficient conditions and examples are given in this duplicate Math.SE post.
--
$^1$ An instructive example is perhaps this Phys.SE post: The coefficients of each squared term in the Lagrangian have been carefully fine-tuned to make it work. It is not just $(T-V)^2$.
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What does the uncertainty principle tell us about the harmonic oscillator? For the harmonic oscillator we have $\sigma_x \sigma_p = \hbar(n+1/2) $ and by the uncertainty principle $\sigma_x \sigma_p \geq \frac{\hbar}{2}$.
In one of the exercises I was doing I was asked to comment on this result. This makes me think that I am missing something special, as the most I can say is that
*
*there is some connection between energy and uncertainty as $\frac{E_n}{\omega}= \sigma_x \sigma_p = \hbar(n+1/2) $
*the groundstate is a minimum uncertainty state.
What's so special about the above result? Can anyone help me out?
Many thanks!
|
I cannot read the mind of the person who asked you to comment, but my guess is that they were looking for something like "In this case, the uncertainty principle doesn't tell me anything beyond what I already knew because it's obvious that $n+1/2\ge 1/2$."
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On the Application of the Einstein-Brillouin-Keller (EBK) Method Consider the Born-Sommerfeld quantization condition (modified) [see Einstein–Brillouin–Keller (EBK)]
$$I_{i} = \frac{1}{2\pi}S_{i} = \frac{1}{2\pi}\oint p_{i} dq_{i} = \hbar \left(n_{i} + \frac{\mu_{i}}{4} + \frac{b_{i}}{2} \right),\tag{1}$$
when applied to the semi-harmonic oscillator with potential,
$$V(x) = \frac{1}{2}m\omega^{2}x^{2}\text{ for }x>0\text{ and }V(x) = \infty\text{ otherwise.} \tag{2}$$
We here have a turning point at $x=x_{1}$, say $x_{1} = a$ and obtain the expression for this as
$$a = \frac{1}{\omega}\sqrt{\frac{2E}{m}}.\tag{3}$$
We can find that upon integration, $\int_{0}^{a}k(x) dx = \frac{\pi E}{2\hbar \omega}$. Now, in the quantization condition, the Maslov indices take the values $\mu = 1$ and $b = 1$ for the reasons that there is one turning point and since there is one reflection at the hard wall (and also since $\Psi(0) = 0$ holds) respectively. Upon doing this substitution, we get
$$I_{x} = \frac{1}{2\pi}S_{x} = \frac{1}{2\pi}\int p(x) dx = \left(n + \frac{3}{4} \right)\hbar,\tag{4}$$
and this, when equated to what was found previously doesn't give the correct expression for $E_{n}$. This seems to work only when we take $S_{x} = 2\int p(x) dx$, since this correctly gives $E_{n} = \left(2n+\frac{3}{2}\right)\hbar\omega$. How do we account for this factor of "$2$"? Is this to do with phase changes due to the reflections or that we take into consideration both possibilities of solutions with $E<V(x)$ and $E>V(x)$ at the turning point?
Links to other questions about EKB:
for an outline of general features:
1,
for derivation and additional references: 2, about the Maslov index in Bohr-Sommerfeld quantization condition: 3, for references: 4.
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Hint: Eq. (1) is for a closed orbit, i.e. the particle traverses the interval $[0,a]$ twice, i.e. forth & back. This explains the factor 2.
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Is a capacitor a dipole? A few more questions about understanding dipoles I recently learned about dipoles, according to its definition I was wondering if a capacitor can be considered also as a dipole?
Also I was wondering what is the physical meaning of the dipole moment $\vec{p}=qd$?
And my last question is what is the motivation of studying dipoles? What is so special about it? From what I have learned it is just 2 equal and opposite charges that are a distance $d$ apart and they create an electric field according to what expected.
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No capacitor is not a dipole it is combination of oppositely charged plates so that electric fireplace within then is zero . The value of electric for any other system is different from that a dipole . The basic motivation for it's study is to understand the change in values of electric field and potential when a system consists of two oppositely charged bodies.
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Error in equations for harmonic spherical wave? I am currently studying Optics, fifth edition, by Hecht. In chapter 2.9 Spherical Waves, the author says the following:
$$\dfrac{\partial^2}{\partial{r}^2}(r \psi) = \dfrac{1}{v^2} \dfrac{\partial^2}{\partial{t}^2} (r \psi) \tag{2.71}$$
Notice that this expression is now just the one-dimensional differential wave equation, Eq. (2.11), where the space variable is $r$ and the wavefunction is the product $(r \psi)$. The solution of Eq. (2.71) is then simply
$$r \psi(r, t) = f(r - vt)$$
or $$\psi(r, t) = \dfrac{f(r - vt)}{r} \tag{2.72}$$
This represents a spherical wave progressing radially outward from the origin, at a constant speed $v$, and having an arbitrary functional form $f$. Another solution is given by
$$\psi(r, t) = \dfrac{g(r + vt)}{r}$$
and in this case the wave is converging toward the origin. The fact that this expression blows up at $r = 0$ is of little practical concern.
A special case of the general solution
$$\psi(r, t) = C_1\dfrac{f(r - vt)}{r} + C_2 \dfrac{g(r + vt)}{r} \tag{2.73}$$
is the harmonic spherical wave
$$\psi(r, t) = \left( \dfrac{\mathcal{A}}{r} \right) \cos k(r \mp vt) \tag{2.74}$$
or $$\psi(r, t) = \left( \dfrac{\mathcal{A}}{r} \right) e^{ik(r \mp vt)} \tag{2.75}$$
wherein the constant $\mathcal{A}$ is called the source strength.
Now, see my question 1. here. It seems to me that the author has made the same error of writing $\cos k(r \mp vt)$ and $e^{ik(r \mp vt)}$, instead of $\cos (kr \mp vt)$ and $e^{i(kr \mp vt)}$, respectively. But this repeat of the error now makes me wonder: Is this actually an error on the part of the author, or am I misunderstanding something?
I would greatly appreciate it if people would please take the time to clarify this.
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It's no error. $k$ has dimensions of inverse length, $r$ has dimensions of length, $v$ has dimensions of length per time, and $t$ has dimensions of time.
What you propose is dimensionally incorrect, as $kr$ is dimensionless and $vt$ has dimensions of length. On the other hand, $k(r\mp vt)$ is a valid operation, and it gives us an overall dimensionless quantity that we need for the argument of an exponential function.
You might be getting mixed up with distributing that $k$ to get $kr\mp \omega t$, which is a typical way to denote such an argument. $\omega$ had dimensions of inverse time, and $v=\omega/k$
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Do atoms absorb the same amount of light? I'm currently working on a project on my own where I'm interested in finding information about an object based on a spectrum. Namely, I want to use the spectrum that I input into my program to be able to analyze what atoms are present in the analyzed object. (I know this is probably hard but it's a fun project). However, when I started to work on this my question arose:
Do atoms that are exposed to the same amount of light absorb the same amount as well? (Albeit different frequencies).
So, when the atoms are exposed to light (uniform over the EM spectrum), will two atoms that absorb different frequencies absorb the same amount of light? And if so, one could infer that the less light of a specific frequency that we can find, (the less compared to the maximum that would be emitted at that frequency) the more there is of the element that absorbs this specific frequency? (Though it would probably be useful to look at more than one "black line" in the spectrum)
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The amount of light is not a precisely defined quantity here. For your purposes it may be safe to assume that each atom may absorb only one photon, however these photons would have different frequencies. The number of photons of different frequencies absorbed is then proportional to the number of atoms absorbing this frequency. Thus, distinct peaks in a spectrum would correspond to different types of atoms, whereas the intensity of the peaks (proportional to their height or their area) would correspond to the number of atoms of each type. However, the above analysis of a spectrum is true only for the atoms in a gaesous phase - the absorption spectrum of a liquid or solid is rather different from the spectra of the atoms, of which it is made.
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Baryon number vs electromagnetic charge, what is the difference? What exactly is a Baryon number? I looked up definition from wikipedia and still struggle to understand this. And how does this differ than the
electromagnetic charge?
My textbook did the following computation:
It is calculating the electromagnetic charge right and not the Baryon number?
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Baryon number ($B$) and electric charge ($q$) are different quantities.
Both quantities are useful because both obey their own conservation law.
To make this more clear consider the proton ($p$), the neutron ($n$),
their constituent quarks ($u$ and $d$), and the electron ($e$).
Proton and neutron have the same baryon number,
but different electric charges.
$$
\begin{array}{c|c|c}
\text{particle} & B & q \\ \hline
p\ (uud) & 1 & +e \\ \hline
n\ (udd) & 1 & 0 \\ \hline
u & \frac 13 & +\frac 23 e \\ \hline
d & \frac 13 & -\frac 13 e \\ \hline
e & 0 & -e
\end{array}$$
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In metals, the conductivity decreases with increasing temperature? I am currently studying Principles of Optics: Electromagnetic Theory of Propagation, Interference and Diffraction of Light, 7th edition, by Max Born and Emil Wolf. Chapter 1.1.2 Material equations says the following:
Metals are very good conductors, but there are other classes of good conducting materials such as ionic solutions in liquids and also in solids. In metals the conductivity decreases with increasing temperature. However, in other classes of materials, known as semiconductors (e.g. germanium), conductivity increases with temperature over a wide range.
An increasing temperature means that, on average, there is greater mobility of the atoms that constitute the metal. And since conductivity is due to the movement of electrons in the material, shouldn't this mean that conductivity increases as temperature increases?
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In a metal, the number of charge carriers is almost unaffected by temperature. There is an enormous number of free electrons per unit volume (compared to a semiconductor like germanium), so Fermi Dirac statistics applies.
As the temperature increases, the scattering events of the electrons with the phonons increase because the number of phonon increases. These scattering events increase the resistivity of the material, hence the conductivity lowers.
Edit to answer the comment "Can you please clarify what you mean by "scattering events"?" : A scattering event in this context is an interaction between two quasiparticles, the electron (in a solid) and a phonon. They both carry energy and momentum, can interact and thus the electrons can have their momentum drastically changed and their energy somewhat changed, after an interaction with a phonon.
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Does a book get lighter if you rearrange the letters? I'm wondering if the information lost by rearranging the letters of a book is measurable as a difference in its initial and final mass.
Choose a long, random string over an alphabet, say $\{0,1\}$, of length $N$. It should be random in the sense that it is incompressible. You might also choose a big book, at random, and compress it.
Once you have the book, or have written down the string in a book, measure the book's mass $m_0$.
Convert the letters into a standard alphabet by using, say, the ASCII encoding scheme. The letters should be more or less distributed uniformly, unlike English which has a rank-frequency distribution for the letters. Rearrange the letters into the complete works of Shakespeare, or as much literature as you possibly can. Then apply the encoding to get a bitstring. This process can be represented by a 0/1 permutation matrix $\sigma_1$ which acts on the bitstring.
Finally, move all the 0's to the left, and 1's to the right. This can be represented as another permutation matrix $\sigma_2$. Measure the mass of the book to get $m_2$.
It appears that the information content of the book at the beginning is $S_0=N$ bits. The information content $S_1$ of the complete works of Shakespeare is around 1.98MB (less than really, size of zipped text file). The last state is very compressible, and $S_2 \approx 2\log_2(N/2)$.
Suppose $N$ is large, say Avagadros' number $N=N_{A}=6.02214076*10^{23}$, more than a zetta and less than a yotta. Then $\triangle S = S_0 - S_2 \approx N_A$. If 1 bit represents about $10^{-23} J/K$, then at $300K$ the information lost corresponds to $~20.1$ picograms.
I suppose the lost information is carried away by the matrices $\sigma_1$, $\sigma_2$ if no one watches or records the rearrangement as it occurs. Is that correct?
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Well if you compress the more ordered book into code and write the code in a new book. Yes, the new book will be lighter.
The original book will not be lighter because although you have ordered the letters according to a scheme that a given algorithm can compress well, how does the Universe know which algorithm you are thinking of? It will compress well according to one algorithm but badly according to almost every other.
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Difference between massless neutrino flavours The title pretty much sums up the the question, what's the difference between massless neutrino flavours?
I know that an electron neutrino interact with the electron and so on for the muon and the $\tau$. I also get the basic of how they enter the standard model Larangian where they are in $SU(2)$ doublets and from there I can guess that a $\nu_e$ interacts only with an electron and so on.
Are we giving this description the status of reality? I mean is that since we describe them in this way in the SM we conclude that there must be 3 of them?
I don't quite understand what's the difference, experimentally, from 3 particles or just one particle that interacts with an electron, a muon or a tauon with certain probability. Why can't it be only one neutrino?
I hope my question is clear.
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Experimentally, we can determine the "flavor" of a neutrino by colliding it with matter and seeing the products. Electron/anti-electron neutrinos collisions more commonly produce an imbalance in observable electron/anti-electron count, and similarly for muon/anti-muon and tau/anti-tau neutrinos.
On an aside, this is how Neutrino Oscillation (where neutrinos convert flavor!) has been discovered/quantified - one experiment for quantifying neutrino oscillation is putting neutrino detectors at different distances relative to a source and observing the distance-dependent (and therefore time-dependent) change in the detection rate of a particular kind of neutrino.
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Time evolution of the standard deviation of an operator How would I find the time evolution of the standard deviation of an operator? For example, how might I find the time evolution $\sigma_x (t)$ of the standard deviation
$\sigma_x = \sqrt{ \langle \hat{x}^2 \rangle - \langle \hat{x} \rangle^2}$
of the position operator $\hat{x}$ given a state $| 0 \rangle$ representing a particle in the ground state of a harmonic oscillator?
Can I multiply the initial value
$\sigma_x (0) = \sqrt{ \dfrac{\hbar}{2 m \omega} }$
by the generator of time translation
$\hat{U} = \large e^{\frac{-i \hat{H} t}{\hbar}}$? I've also tried switching to the Heisenberg picture and applying the Heisenberg equations of motion but have been unable to reach a conclusion. Any help would be greatly appreciated, thanks.
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In case you're planning to actually compute it, here's how:
Derivatives commute with operator averages in the Heisenberg picture (since your state is constant; alternatively in the Schrodinger picture, you can take derivatives of the state), so you can do something like this:
$$\frac{d}{dt}\sqrt{\langle x^2\rangle-\langle x\rangle^2}=\frac{1}{2\sqrt{\langle x^2\rangle-\langle x\rangle^2}}\frac{d}{dt}(\langle x^2\rangle-\langle x\rangle^2)$$
$$=\frac{1}{2\sqrt{\langle x^2\rangle-\langle x\rangle^2}}\left(\langle \frac{d}{dt} x^2\rangle-\frac{d}{dt}\langle x\rangle^2\right)$$
$$=\frac{1}{2\sqrt{\langle x^2\rangle-\langle x\rangle^2}}\left(\langle \frac{d}{dt} x^2\rangle-2\langle x\rangle\frac{d}{dt}\langle x\rangle\right)$$
Notice the product rule for operator derivatives is not commutative
$$=\frac{1}{2\sqrt{\langle x^2\rangle-\langle x\rangle^2}}\left(\langle \frac{dx}{dt} x\rangle+\langle x\frac{dx}{dt}\rangle-2\langle x\rangle\langle \frac{dx}{dt}\rangle\right).$$
Then plug in Heisenberg's equation $\frac{dx}{dt}=\frac{i}{\hbar}[H,x]$,
$$=\frac{1}{2\sqrt{\langle x^2\rangle-\langle x\rangle^2}}\frac{i}{\hbar}\left(\langle [H,x] x\rangle+\langle x[H,x]\rangle-2\langle x\rangle\langle[H,x]\rangle\right)$$
$$=\frac{1}{2\sqrt{\langle x^2\rangle-\langle x\rangle^2}}\frac{i}{\hbar}\left(\langle [H,x^2]\rangle-2\langle x\rangle\langle[H,x]\rangle\right)$$
And some further operator bashing will give you the answer. Alternatively, you can do this in the Schrodinger picture directly. Here, for example, you can evaluate derivatives like so:
$$i\hbar\frac{d}{dt}\langle\psi\rvert x^2\lvert \psi\rangle=i\hbar\left(\frac{d}{dt}\langle\psi\rvert\right) x^2\lvert \psi\rangle+i\hbar\langle\psi\rvert x^2\frac{d}{dt}\lvert \psi\rangle=i\hbar\left(\frac{d}{dt}\lvert \psi\rangle\right)^\dagger x^2\lvert \psi\rangle+\langle\psi\rvert x^2H\lvert \psi\rangle$$
$$=-\langle \psi\lvert H x^2\lvert \psi\rangle+\langle\psi\rvert x^2H\lvert \psi\rangle=-\langle \psi\lvert [H, x^2]\lvert \psi\rangle$$
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What is a wavepacket without a photon? There are a lot of questions on this site about photons and wavepackets.
Relation between radio waves and photons generated by a classical current
This one lists a lot of them as reference. None of them in detail specifically answers my question.
What happens when a photon hits a beamsplitter?
I have read this:
https://arxiv.org/ftp/arxiv/papers/1410/1410.3416.pdf
A wave.packet enters the
beam splitter and breaks into two smaller wave.packets. The incident particle,
shown as a dot in picture, emerges along path a accompanied by a wave packet. The
wave packet in path b is empty.
I am not asking about the photon traveling as a wave, or a particle, or why the photon is delocalized when traveling and why it is taking all possible paths. I am not asking what happens when a photon hits a beam splitter.
I am asking if it is really possible what the paper suggests, that there can exist a wavepacket, without a photon, that is, where the photon number is 0.
In physics, a wave packet (or wave train) is a short "burst" or "envelope" of localized wave action that travels as a unit.
https://en.wikipedia.org/wiki/Wave_packet
The best definition of a wavepacket I found on this site is saying that a wavepacket is basically a localized excitation of the EM field. This is a contradiction, because the photon itself is usually defined as the excitation of the EM field. How could there be a wavepacket (a localized excitation), without a photon (excitation of the EM field)?
Question:
*
*What is a wavepacket without a photon?
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A wave packet is just a localized superposition of waves. In the context of QM it describes the probability of detecting a particle like any other quantum wave function.
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How to infer about the volume of the body from measuring the mass of the object in 2 different liquids? While solving a problem on hydro statics subject I saw a statement that argued that if I know the weight of the object in the air and then I know the weight of the object in the water.so when I subtract between the weight I can conclude the volume of the objects.
I have numeric example:
weight in air is $740 gm$
weight in water is $690 gm$
so the difference is $$m = 740- 690= 50 gm$$ and somehow they concluded that the volume is $50 cm^3$
what is the explanation for that? is it about the Archimedes law?
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There can be two possible cases
Case 1
The statement in your book might be wrong (Though more likely possibility is that statement in the book might be correct but what you read /inferred from it is wrong).In that case you are right. We can only subtract two quantities with same dimensions. In your case it is mass.
We can't get a different quantity when we subtract two same quantity with same dimensions .
Case 2
What i think it is-
They meant volume of object submerged in the water is $50cm^3$. It is only because as density of water is 1 $gm \ cm^{-3}$. Now in case of water 50gm = 50 $cm^3$ but it still not actual volume of the object. It is the volume of object submerged in the liquid.If the object is fully submerged then the book is correct.
Infact, this the actual statement of Archimedes Principle - The volume of liquid displaced is equal to the the volume of object submerged in the liquid.
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Pseudo Force and Inertial and Non-Inertial frames In the figure given below is block placed on an incline $\theta$. Now the lift is accelerating upwards with an acceleration $a_0$. Now if we make our measurements from the lift frame we will have to apply a pseudo force $-ma_0$. Which will have two components one in the direction of $Mg\cos\theta$. And other in the direction of $Mg\sin\theta$. Now $Mg\sin\theta+Ma_0\sin\theta=Ma_\text{net}$. Where $a_\text{net}$ is the net acceleration in that direction.
Now let's observe it from the ground or an inertial frame here the object has a net upward acceleration which has a component opposite to $Mg\sin\theta$. Therefore $Mg\sin\theta=- Ma_0\sin\theta$. Now what I thought was that this is not possible and hence there is another force acting opposite to $Mg\sin\theta$, $Ma_\text{net}$. Now this doesn't make any sense to me if there is a force acting opposite to $Mg\sin\theta$, and the net is also in that direction, then won't the object move upwards on the incline. Now that doesn't make any sense. Can someone tell me what Is happening and from where is this $a_\text{net}$ coming from when observing in the inertial frame?
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Perhaps you can see it better with this figure.
to apply NEWTON second law , you have to calculate the components of the position vector to the mass in inertial system.
$$\vec{R}= \pm\begin{bmatrix}
x_0 \\
y_0 \\
\end{bmatrix}=\pm
\left[ \begin {array}{c} s\cos \left( \vartheta \right)
\\ s\sin \left( \vartheta \right) + y \left( \tau
\right) \end {array} \right]
\tag 1$$
where "+" from inertial system and "-" from lab system .
with equation (1) you can obtain the kinetic energy and with the potential energy $U=m\,g\,R_y$ you get the equation of motion:
$$M\,{\ddot{s}}\pm{M}\,g\sin \left( \vartheta \right) +{M}\,\sin \left( \vartheta
\right) \underbrace{{\frac {d^{2}}{d{\tau}^{2}}}y \left( \tau \right)}_{a_0}
=0$$
thus for "+" sign you get $$M\,a_{\text{net+}}=M\,g\,\sin(\vartheta)+{M}\,\sin(\vartheta)\,a_0\quad \Rightarrow\quad g\mapsto g+a_0$$
and for "-" sign you get:
$$M\,a_{\text{net-}}=-M\,g\,\sin(\vartheta)+{M}\,\sin(\vartheta)\,a_0\quad \Rightarrow\quad g\mapsto a_0-g$$
|
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Is excited electrons life-time = half-life? Everywhere I read something related to excited electrons in atoms I only see the word "lifetime".
For example most excited states have a lifetime of a few nano to micro seconds and metastable states have a lifetime of a few miliseconds as far as I understand.
Shouldn't we say "half-life" instead? Since we are dealing with probabilities..
I'm surprised because I don't see the word "half-life" anywhere when talking about excited states. Therefore I would love to set the record straight.
|
Half-life ($t_{1/2}$) and lifetime ($\tau$) are related to one another.
$$
t_{1/2} = \tau \ln(2)
$$
From the equivalent equations,
$$
N(t) = N_0 \left(\frac {1}{2}\right)^{\frac{t}{t_{1/2}}} \\
N(t) = N_0 e^{-\frac{t}{\tau}}
$$
It’s just convention; forms adopted by different fields. I would say the half life approach is more intuitively understandable, so I can see this being used in fields such a radioactivity where safety is a concern.
However, is optics, rate equations are a hugely important tool, and using lifetimes makes the mathematics slightly easier to understand and work with.
Source https://en.m.wikipedia.org/wiki/Half-life
|
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Has the ballistic motion of an electron in gravitational field ever been measured? Reading this question I thought of an argument that an electron's trajectory would bend in the gravitational field despite the electron's being incapable of strong interaction; this would then disprove the conjecture stated in that question. But then I couldn't find any references to actual experiments that have been done to measure this ballistic motion of an electron in gravitational field.
Has such motion actually been measured?
|
EDIT: see the answer by S. McGrew. Turns out there's some clever methods of releasing extremely low energy free electrons, which makes gravitational effects just measureable.
Let's think about the size of the effect you're looking for.
Suppose you sputter off an electron with a very low energy of around 1 eV. This corresponds to an electron velocity of about 600 km/s. If you set up a vacuum chamber 1 km long, the electron travels across it in around 1.5 ms. In that time, gravity on Earth's surface will have deflected it by under 30 microns - 0.03 mm! Actual electron emissions are much higher energy, especially for long-chamber situations - most particle accelerators operate with energies in the MeV range, which gives deflections a thousand times less. More or less, this is completely impractical to measure, since this will be many orders of magnitude smaller than your beam width.
|
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Which force is doing the work here? My text book (Fundamentals of Physics by Halliday, Resnick, and Walker) mentions the following about the work done in internal energy transfers:
An initially stationary ice-skater pushes away from a railing and then slides over the ice. Her kinetic energy increases because of an external force F on her from the rail. However, that force does not transfer energy from the rail to her. Thus, the force does no work on her. Rather, her kinetic energy increases as a result of internal transfers from the biochemical energy in her muscles.
This is confusing me a lot. The energy transfer is clearly internal but work must be done by the force as work done is defined as the (dot) product of force and displacement and the definition makes no reference to any transfer of energy.
I thought work done by a force just means that the force is causing a transfer of energy to (or from) an object, and gives no information about whether the energy is coming from the object exerting the force.
My confusion is not over whether work is being done or not but which force is doing the work which ends up causing the change in kinetic energy.
|
I think I agree there is something off in that quoted section.
The purpose of that section is not practical application, the purpose is to bring an abstract concept into focus. And that means the usual simplifications for practical purposes should not be used.
There is a physics joke that goes as follows: When Arnold Schwarzenegger does his push-ups, he is actually pushing the Earth away from himself.
In the case of the skater the standard for-practical-purposes simplification is: she is in effect pushing off against the entire Earth, so we treat the railing she is pushing off against as immovable.
But: that simplification violates the third law. It is more instructive to think of a setup where the third law is visibly involved. You can have the skater push off against a sled that is, say, twice her mass. And when you have a clear concept of the third-law-visibly-involved case you can take that to the limit of pushing off against something that is vastly more massive than the skater.
|
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Why does water keep dripping out of the bucket even after the faucet is turned off? So I just noticed that when I filled my bucket with water until it overflowed and then I turned the faucet off, the water kept dripping for like 20 seconds. Why does this happens? Shouldn't it have stopped dripping sooner after I turned the faucet off?
|
There are two possible scenarios here, which I don't know which it is unless I have more details.
One is where you fill slowly until overflow just begins to happen; this results in Adrian's answer, which I will elaborate on a bit:
The surface tension of water means that the water level can rise above the level of the bucket edge by forming a "bubble" above the lip; Laplace pressure, pressure due to the curvature of such a surface (surface geometry looks like the illustration here), counteracts gravitational pressure, preventing the water from spilling. Once you overfill enough, this pressure is no longer sufficient, the surface breaks, and water begins to spill; once it does, the vertical curvature at the overflow point reduces (to form the outflow stream), weakening the Laplace pressure, which results in the water continuing to spill out until the level lowers enough that the gravitational pressure is lower than the new, lowered Laplace pressure.
The other scenario is where you fill the water fast. Here, it's just due to the inertial and viscosity forces of the fluid; there must be a surface level imbalance for fluid with weight/viscosity to flow (think a river, flowing from high to low), which means there is a gradient in surface level from where you're filling the bucket to where it is spilling out (flow requires a pressure gradient, which results in a surface level gradient); once you stop the inflow, the fluid will leak until the surface is level, which takes some time. At low flows, the surface level gradient is rougly proportional to flow, which means that technically, without any other effects, the flow will only exponentially decay rather than actually stop; however, surface tension as mentioned above will eventually cause water to abruptly stop flowing.
|
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Why is internal resistance of battery considered outside the terminals although it is present between the terminals inside the battery In ideal battery the internal resistance is zero whereas in non-ideal battery there is some internal resistance now this internal resistance is due to the battery material (electrolyte) and is present inside the battery between the terminals then why do we represent and eventually do calculations by considering that internal resistance to be connected with battery terminals externally. I’m totally unable to get the point. Please help
|
Replacing a real battery by an ideal battery (voltage $V_0$)
and an internal resistance ($R$) is just a model.
You can apply Kirchhoff's voltage law to this model and derive
a relation between the real voltage $V$ and current $I$:
$$V=V_0 - R I. \tag{1}$$
On the other hand:
You can take a real battery, and measure
the real voltage $V$ and current $I$ for various
external loads connected to the battery.
You may get a $V$-$I$ plot like this:
(image from djb microtech ltd - internal resistance)
This plot is equivalent to the formula (1) above,
when you take $V_0=1.53$ V and $R=0.421$ Ohm.
So the ultimate justification for the simple theoretical model
(an ideal battery and internal resistance) is
that it matches the experimental observations
(measured voltage and current).
Of course, this doesn't mean that the ideal voltage $V_0$
and the internal resistance $R$ are separate entities inside a real battery.
From electrochemistry of galvanic cells we are aware these two things
are not separate. Both are caused by complex processes on the
electrodes and in the electrolyt of the real battery.
But for predicting the measured $V$-$I$ relation we don't
need to care about these details.
|
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Finding acceleration of center of mass in cart pole problem In this link about finding equations of motion of cart pole problem,
There is an equation about acceleration of center of mass of the pole.
Screenshots of them below.
I don't understand why they have more than two parts about angular acceleration - $\varepsilon \times r_p$ and $\omega \times (\omega \times r_p)$?
If I'm being right, first one is torque, and second one is acceleration of a point in circular movement.
I guess in some part I'm being incorrect, but I don't understand why they put two angular acceleration of it? It's copy of them, aren't they?
|
When you transfer velocity from one point to another in a rigid body you end up with an equation like
$$ \boldsymbol{v}_P = \boldsymbol{v}_C + \boldsymbol{\omega} \times \boldsymbol{r}_P $$
Acceleration is just the time derivative of the above with
$$ \begin{aligned}
\tfrac{\rm d}{{\rm d}t} \boldsymbol{v}_P &= \boldsymbol{a}_P \\
\tfrac{\rm d}{{\rm d}t}\boldsymbol{v}_C & = \boldsymbol{a}_C \\
\tfrac{\rm d}{{\rm d}t} \boldsymbol{\omega} &= \boldsymbol{\epsilon} \\
\tfrac{\rm d}{{\rm d}t} \boldsymbol{r}_P &= \boldsymbol{\omega} \times \boldsymbol{r}_P
\end{aligned} $$
The last part is because $\boldsymbol{r}_P$ is a fixed vector riding along with the rigid body.
The transformation of acceleration is thus
$$ \boldsymbol{a}_P = \boldsymbol{a}_C + (\tfrac{\rm d}{{\rm d}t}\boldsymbol{\omega} )\times \boldsymbol{r}_P + \boldsymbol{\omega} \times (\tfrac{\rm d}{{\rm d}t} \boldsymbol{r}_P) $$
$$\boldsymbol{a}_P = \boldsymbol{a}_C + \boldsymbol{\epsilon} \times \boldsymbol{r}_P + \boldsymbol{\omega} \times ( \boldsymbol{\omega} \times \boldsymbol{r}_P) $$
|
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Is $E^2-P^2=m^2$ true only for free particles? I'm studying Friedman and Susskind's Special Relativity and Classical Field Theory and follow them in using $c=1$.
They derive the above relation by first using Lagrangian of a free particle $\mathcal L=-m\sqrt{1-v^2}$ to show that conjugate momenta are given by $P^i = mU^i$ (where $U^\mu$ is the 4-velocity). Then they write out the Hamiltonian for the free particle using this Lagrangian and show that $H=mU^0$. Then writing $E$ for $H$ and since $U^\mu$ is a 4-vector and $m$ a scalar, they conclude that the quadruple $(E, P^1, P^2, P^3)$ forms a 4-vector. Then they use the invariance of the norm of the this 4-vector to show that $E^2 - P^2 = m^2$ (by observing the free particle in its rest frame, which is possible only since the particle was free to start with, otherwise there'd be no inertial frame in which the particle is seen at rest always).
Now the entire argument above was for a free particle. My question is whether the oft-used relation $E^2-P^2=m^2$ (or a modification thereof) also valid for a non-free particle (by which I mean a particle whole Lagrangian differs by that of a free particle).
Edit:
I realized (thanks to probably_someone in a comment below) that the original argument that the authors give is slightly different than mine. They multiply the equation $(U^0)^2-(U^i)^2 = 1$ with $m^2$ to yield $E^2-P^2=m^2$. Nevertheless, I still like my argument better. :)
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It does depend on what is meant by P, and E.
If $E^2=P^2-m^2$, even with forces, the particle would have a constant velocity.
I think that the canonical forms are more commonly used, so
$(E-V)^2=({\bf P}-q{\bf A})^2+(m+S)^2$, with $\bf P$ the canonical momentum,
and $(V,{\bf A})$and S vector and scalar potentials.
|
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The average velocity of a SHO is 4 times the maximum velocity? I am looking at simple harmonic oscillators. The maximum potential energy is equal to the maximum kinetic energy:
$k {x_{max}}^2 = m {v_{max}}^2 \rightarrow x_{max}=v_{max}\sqrt{\frac{m}{k}} = \frac{v_{max}}{\omega}$
Also, the time to find the average speed, calculate the total distance traveled over one oscillation by the total time it takes.
$v_{ave} = \frac{4 {x_{max}}}{\tau} = 4 {x_{max}} \omega$
Now if I plug in the first result into the second result I get:
$v_{ave} = 4 v_{max}$
But surely this is not right. The average speed is equal to 4 times the maximum? How can this be?
|
You made a simple mistake converting time period to angular frequency: $\omega = \frac{2\pi}{T}$, so $\frac{1}{T} = \frac{W}{2\pi}$.
I will not work it through for you, but you should get the answers you are looking for.
If you are interested in working out max velocity and average velocity in general, i.e. not just a mass on a spring, then you could use the general SHO wave equation:
$$y = A\cos(wt).$$
Where $y$ is the displacement and $A$ is the amplitude (your $x_\text{max}$).
We can differentiate to find the velocity,
$$\frac{dy}{dt} = -Aw\sin(wt).$$
From which we can calculate the average velocity and maximum velocity.
The maximum velocity is easiest as we know the largest value of $\sin(wt)$ is one, so
$$v_\text{max} = \text{max } \left|\frac{dy}{dt}\right| = Aw.$$
Which is what you got: $v_\text{max} = x_\text{max}w$.
The average velocity is a little harder. The average value of a function $g(x)$ in the interval $[a,b]$ is defined as
$$\frac{1}{b-a}\int_a^bg(x) \, dx.$$
So we could take the interval to be one time period (because any more and it just repeats itself!), but the problem with this is that we will just get zero for the average velocity ($\frac{dy}{dt}$). This is because the average value of $\sin$ from $0$ to $2\pi$ is zero and $w$ just stretches this graph.
The mistake we have made is that we actually want to find the average value of the speed, $\left|\frac{dy}{dt}\right|$, given by the integral
$$\frac{1}{T}\int_0^T\left|\frac{dy}{dt}\right|\, dt = \frac{Aw}{T}\int_0^T|\sin(wt)|\, dt$$
There are two ways I can see of evaluating this, either write $|\sin (wt)|$ as $\sqrt{\sin^2(wt)}$ and use a double-angle substitution, or we can use the symmetry of the sine function, seen below.
From this symmetry, we can see that
$$\int_0^{2\pi}|\sin x| \, dx = 2\int_0^{\pi}\sin x \, dx$$.
Or in our case,
\begin{align}
\frac{Aw}{T}\int_0^T|\sin(wt)|\, dt &= \frac{2Aw}{T}\int_0^{T/2}
\sin(wt)\, dt \\
&= \left. -\frac{2A}{T}\cos(wt)\right|_0^{T/2} \\
&= -\frac{2A}{T}\left[\cos\left(\frac{wT}{2}\right) - \cos(0)\right] \\
&= -\frac{2A}{T}\left[\cos(\pi) - \cos(0)\right] \\
&= -\frac{2A}{T}\left[(-1) - 1\right] \\
&= \frac{4A}{T} = v_\text{ave}
\end{align}
which is of course what you got with the simpler $\text{speed} = \frac{\text{distance}}{\text{time}}$ approach: $v_\text{ave} = \frac{4x_\text{max}}{T}$.
So although your inkling that the average speed shouldn't be greater than the max speed was correct and solved by the simple calculation error I pointed out at the start, I hope you can learn something extra from this answer in the form of an alternative, more mathematical approach, that you could have taken.
I will conclude with the final results.
\begin{align}
v_\text{max} = Aw && v_\text{ave} = \frac{4A}{T} = \frac{2Aw}{\pi}
\end{align}
|
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Work Equals Torque? Horsepower, Pulleys While reading one definition of torque, I saw its units are Newton-meter, which is the same as work. But sources usually make it a point to emphasize "even though both work and torque units are the same, they should not be confused, they are very different". One is like an object being pushed with force certain distance, the other force applied to a wrench etc. at certain length, applied around an axis of rotation.
But if we think of the pulley seen below,
Isn't radius and distance related here? If I rotate a wrench of length $r$ with force $F$ from top position to 90 degree position, isn't the same thing as pushing an object with force $F$ at a distance of $r$?
|
You can still do work by applying a torque, but that is not the same thing as torque being equivalent to work. An easy way to see this is that you can apply a constant torque, but the work done by the torque depends on the angle through which the torque rotates the object.
In your wrench example, assuming maximum torque for the given $F$ and $r$, your torque is then $\tau = rF$. If you rotate the wrench through some angle $\theta$, then the work you have done is $W=rF\theta$. This is equivalent to pushing an object a distance $r\theta$ with a constant force $F$, i.e., the "rotational work" is just the work over the arc length distance of the rotation.
|
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Energy of a rotating wheel Why does a rotating wheel have kinetic energy $K=\frac{1}{2}mv^2$ associated with its movement?
I will clarify.
Let's say I have a rotating wheel in an empty void which has an angular velocity $\omega$, then its total energy will be $E=\frac{1}{2} I \omega^2$, as the wheel is not moving. Now suppose I put this wheel on a surface. The wheel starts moving and, from what I understand, the total energy associated to the wheel is $E=\frac{1}{2} m v^2+ \frac{1}{2} I\omega^2$. How is that possible since energy cannot be created nor destroyed? The wheel is the same, so how can it have two different amounts of energy?
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Find the rotational and translational kinetic energy of a ball with rotational energy $\frac{1}{2} I\omega_0^2$ initially if put on a slant.
\begin{align*}
\frac{1}{2} m v^2+ \frac{1}{2} I\omega^2&=\frac{1}{2} I\omega_0^2\\
\frac{1}{2} m v^2+ \frac{1}{2} \alpha mR^2\omega^2&=\frac{1}{2} \alpha mR^2\omega_0^2&(\text{dimensions of }I\rightarrow [M][L]^2\Rightarrow \alpha\text{ is dimensionless})\\
\frac{1}{2} m v^2+ \frac{1}{2} \alpha mv^2&=\frac{1}{2} \alpha mR^2\omega_0^2&(\because v=wr\text{ if the ball is rolling without slipping})\\
\end{align*}
For $\alpha=1$, exactly half the rotational kinetic energy $\frac{1}{2} I\omega_0^2$ goes into translational kinetic energy in no rolling case.
|
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Does entanglement of a bipartite PPT state $\rho$ imply entanglement of $\rho + \rho^{\Gamma}$? Consider an entangled bipartite quantum state $\rho \in \mathcal{M}_d(\mathbb{C}) \otimes \mathcal{M}_{d'}(\mathbb{C})$ which is positive under partial transposition, i.e., $\rho^\Gamma \geq 0$. As separability of $\rho$ is equivalent to separability of its partial transpose $\rho^\Gamma$, we know that $\rho^\Gamma$ is entangled. Does this imply that the sum $\rho + \rho^\Gamma$ (ignoring trace normalization) is also entangled? If not, can we impose restrictions on $\rho$ which guarantee that the above proposition holds?
In the language of entanglement witnesses, the problem reduces to finding a common witness that detects both $\rho$ and $\rho^\Gamma$. Let $W$ be the entanglement witness detecting $\rho$, i.e., $\text{Tr} (W\rho) < 0$. Then $W$ is non-decomposable (as $\rho$ is PPT) and is of the canonical form $P+Q^\Gamma - \epsilon \mathbb{I}$, where $P, Q \geq 0$ are such that $\text{range}(P) \subseteq\text{ker}(\delta)$ and $\text{range}(Q) \subseteq \text{ker}(\delta^\Gamma)$ for some bipartite edge state $\delta$ (these are special states that violate the range criterion for separability in an extreme manner, see edge states) and $0 < \epsilon \leq \text{inf}_{|e,f\rangle} \langle e,f | P+Q^\Gamma | e,f \rangle$. If $\delta$ is such that $\text{ker}(\delta) \cap \text{ker}(\delta^\Gamma)$ is not empty, then we can choose $P=Q$ to be the orthogonal projector on $\text{ker}(\delta) \cap \text{ker}(\delta^\Gamma)$, in which case $W=W^\Gamma$ is the common witness. But is this always true? Can we use optimization of entanglement witness to ensure this condition?
Cross posted on math.SE
Cross posted on quantumcomputing.SE
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I believe this is not true, based on https://arxiv.org/abs/quant-ph/9903012, Eq. (8), where a given $\rho\in\mathcal{M}_2(\mathbb{C})\otimes\mathcal{M}_N(\mathbb{C})$ can be always written as
$$\rho=\frac{\rho+\rho^{T_A}}{2}+\frac{\rho-\rho^{T_A}}{2}=\rho_s+\sigma_y^A\otimes B,$$ i.e., a separable part $2\rho_s=\rho+\rho^{T_A}$ and a part that might be inseparable. Thus, if I am not mistaken, even if $\rho$ is PPT entangled, $\frac{\rho+\rho^{T_A}}{2}$ is separable.
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Are there any quantum effects which we can see in every day life? I am wondering if there are any natural phenomenon in every-day life that cannot be explained by classical physics but can only be explained by quantum mechanics. By classical physics, I mean Newtonian mechanics and Maxwell's electromagnetic theory.
I know that there are macro-scale quantum phenomena such as superconductivity, but that isn't something that we can see in ordinary life.
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The way the question is phrased, specifically: what is not explained by Newton or Maxwell, there are 2 obvious candidates:
*
*The stability of matter: atoms governed by Maxwell's equations and Newtonian mechanics decay immediately due to EM radiation.
*The lack of an Ultraviolet Catastrophe: thermal radiation has infinite power, which we do not observe.
however, one would not generally list those as "quantum effects".
|
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In the generation of X-Rays, why the incoming electron generated from anode knocks out the K shell electron rather than outer shell electrons? When the high energy beam of particles or photon hits the cathode, electrons from $K$ shell are knocked in the generation of characteristic x-rays. Why do inner electrons get knocked out?
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Well, The story begins by the physicist C. Barkla. He noticed that atoms appeared to emit two types of X-rays. The two types of X-rays differed in energy and Barkla originally called the higher energy X-ray type A and the lower energy X-ray type B. He later renamed these two types K and L since he realized that the highest-energy X-rays produced in his experiments might not be the highest-energy X-ray possible. He wanted to make certain that there was room to add more discoveries without ending up with an alphabetical list of X-rays whose energies were mixed up. As it turns out, the K type X-ray is the highest energy X-ray an atom can emit. It is produced when an electron in the innermost shell is knocked free and then recaptured. This innermost shell is now called the K-shell, after the label used for the X-ray. So, you can say that the question that why accelerating electron coming in didn't hit the outer shell, is not an appropriate statement. You must ask why, when the electron hits the innermost shell(K-shell) produces maximum energy X-ray.
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Right vs Left Derivatives Let $\theta$ be a fermionic quantity and $f(\theta)=f(0)+\theta\frac{\partial f}{\partial\theta}=f(0)+\frac{\partial_r f}{\partial\theta}\theta$. Under a variation $\theta\mapsto\theta+\delta\theta$ we have
$$f(\theta)\mapsto f(\theta)+\delta\theta\frac{\partial f}{\partial\theta},$$
using the first formula, or
$$f(\theta)\mapsto f(\theta)+\frac{\partial_r f}{\partial\theta}\delta\theta,$$
using the second one. However,
$$\delta\theta\frac{\partial f}{\partial\theta}=(-1)^{|\delta\theta|(|f|+|\theta|)}\frac{\partial f}{\partial\theta}\delta\theta=(-1)^{|\delta\theta|(|f|+|\theta|)+|\theta|(|f|+1)}\frac{\partial_rf}{\partial\theta}\delta\theta$$
which is different from $\frac{\partial_rf}{\partial\theta}\delta\theta$ in general. This yields a contradiction between both variations. Of course problems are avoided if $|\delta\theta|=|\theta|$ but I don't see how this affect the first two equations. I am very confused by this!
|
*
*Yes, by definition the Grassmann parity $|\delta z|$ of a variation $\delta z$ of a supernumber $z$ (of definite Grassmann parity) is the same as the Grassmann parity $|z|$ of the supernumber $z$ itself:
$$|\delta|~=~0.\tag{1}$$
*Perhaps OP is wondering about the following question.
Question: How does an infinitesimal variation $\delta$ relate to a left vector-field/linear derivation $X$ of Grassmann-parity $|X|$?
Answer: In order to relate $X$ to an infinitesimal variation$^1$ $$\delta~=~\epsilon X,\tag{2L}$$
one needs to introduce an infinitesimal parameter $\epsilon$ of the same Grassmann-parity $|\epsilon|=|X|$.
--
$^1$ For a right vector-field/linear derivation $X_R$, we instead have
$$\delta~=~X_R \epsilon ,\tag{2R}$$
with $|\epsilon|=|X_R|$.
|
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To what precision is the heat capacity an extensive quantity We know that heat capacity is an extensive quantity, basically meaning for double the amount of substance you need double the energy to increase temperature. To what extend is this actually true, like:
*
*Are there e.g. (measurable) surface effects?
*Have (precision) experiments been done on this?
|
Classically, the heat capacity of all substances would follow the law of Dupont and Petit with the same value per atom, so then there are no surface effects.
In the quantum regime, at low temperatures, one would expect a surface effect because the vibrational modes are different at surfaces. For measurements one would need a substance with a large surface area. But then a signal might be due to ordering or desorption of adsorbed gasses.
That is why one studied $c_v$ of things like very fluffy graphite: the 2-dimensional phase transitions. The difference between the surface and the bulk of graphite is not large and I would not expect that a difference in $c_v$ between graphite and fluffy graphite would have been studied. There are other things in such systems that are much more interesting than a small difference in the Debye temperature.
|
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|
Why are there no operators in classical mechanics? I have been wondering to find the answer of some fundamental questions in quantum mechanics and the answer to the above question will help me to clear doubts of quantum world
|
Yes, there are operators in classical mechanics; one can formulate classical mechanics in terms of Hilbert space and operators, the only difference with quantum operators is that classical operators is that the former commute, see https://en.wikipedia.org/wiki/Koopman%E2%80%93von_Neumann_classical_mechanics for more information.
|
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Locally flatness in general relativity My professor made following statement:
The spacetime of GR is curved in the presence of strong gravitational fields. The effects of curvature
manifest themselves at large distances. Locally, one can choose a flat Minkowskian metric.
I dont get it:
I thought, gravitation is expressed by curvature. If I sit on my chair, this will be due to gravitation. But this is not an effect shown on large distances (as e.g earth and sun).
|
I think now I understand my confusion:
One can choose riemann normal coordinates to get the canonical form of the metric (which can be the minkowski metrik). So locally we can choose a flat metric.
I thought, this implies that spacetime is locally flat without curvature.
But the curvature is described by the riemann tensor, which depends on the second derivative of the metric. Which is not vanishing by choosing riemann normal coordinated!
|
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Impulse operator on real wave function The impulse operator in quantum mechanics is given by
\begin{align}
\hat{p} = \frac{\hbar}{i}\nabla
\end{align}
As a Hermitian operator, the expected value of this operator $\langle{p}\rangle = \langle \psi|\hat{p}\psi\rangle$ should be real. However, for a real wave function $\psi(\vec
{r})\in \mathbb{R}$ (a valid solution to the Schrödinger equation) the resulting integral is imaginary:
\begin{align}
\langle{p}\rangle = \frac{\hbar}{i}\int d^3r \cdot \psi \nabla \psi
\end{align}
Is there an error in my thinking or is it impossible to calculate the expected value that way? An alternative approach would be to use the Fourier transform.
|
The possibility that is not accounted for in the question is that the integral may be zero. In fact, it can be shown that a wave function corresponding to a stationary state can always be chosen real, and the momentum of a stationary state is definitely zero.
Another insight may come from considering wave function
$$\phi_+(x) = \psi_k(x) + \psi_{-k}(x) = e^{ikx} + e^{-ikx} = 2\cos(kx).$$
The average momentum in this state is zero, as it is a sum of two states with opposite momenta, $\pm\hbar k$.
To conclude: your formula for the average momentum is correct, since it is obtained from general rule. And, since it would give an unphysical imaginary value for a real wave function, it means that all such wave functions correspond to states with zero momentum.
|
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|
In a Parallel Radioactive decay , what is the ratio of sum of stable nuclei of the two products? Here is the full Question
The question gives you two Rate constants and asks for what time will the ratio of "sum of stable nuclei of Ca:Ar" be equal to 99.But what is this ?If i think of number of nuclei in terms of concentration then the ratio of them is coming to be simply $$ \frac{k_1}{k_2}$$ and its value is already fixed , so either there must be an error in my working or there must be different understanding of "sum of stable nuclei of Ca:Ar"
Here is my attempt at doing this from start
Here is my notes on Parallel Reaction
|
You're misreading the problem.
Let the numbers of potassium, calcium, argon nuclei be $N_\text{K}$, $N_\text{Ca}$, $N_\text{Ar}$, all of which are functions of time.
You are correct that the ratio
$$\frac{N_\text{Ca}}{N_\text{Ar}}$$
is independent of time. But the question is about the ratio
$$\frac{N_\text{Ca}+N_\text{Ar}}{N_\text{K}}$$
which increases as the potassium decays away.
|
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If a pendulum of mass $m$ and length $l$ is considered quantum mechanically, what will be the approximate ground state energy? I have seen that the solution of the quantum pendulum is obtained by solving the Mathieu's equation form of the Schrodinger equation and it also depends on a parameter 'q' called the energy barrier.
But can we approximate the energy levels of the quantum pendulum to that of the quantum harmonic oscillator? Under what limit can we do this?
|
I could not find the reference I mentioned where they perform perturbation theory starting from both the quantum harmonic oscillator limit and the quantum rotor limit and connect their solutions and energies with the exact solutions of the quantum pendulum. Here I will just briefly scheme the basics of this procedure.
The Hamiltonian for a quantum pendulum can be written in terms of the angle with respect to the vertical $\phi$ with $-\pi \leq \phi < \pi$ producing the Schrödinger equation
\begin{equation}
0 = \left[ - \frac{\hbar^2}{2 I} \frac{\partial^2}{\partial \phi^2} + I \omega^2 \left( 1 - \cos \phi \right) - E \right] \psi\left( \phi \right) \,
\end{equation}
with $I = m R^2$ and $\omega = \sqrt{\frac{g}{R}}$. With a transformation like
\begin{equation}
z = \frac{\pi}{2} + \frac{\phi}{2} \, ,
\end{equation}
you can put it the form of Mathieu equation,
\begin{equation}
0 = \frac{\partial^2 y}{\partial z^2} + \left[ \lambda - 2 q \cos(2z) \right] y \,
\end{equation}
with
\begin{equation}
q = \frac{R^4}{L^4_{HO}}
\end{equation}
where
\begin{equation}
L_{HO} = \sqrt{\frac{\hbar}{2 m \omega}}
\end{equation}
is the standard factor of the harmonic oscillator position operator when written in terms of creation annihilation operators, $ x = \sqrt{\frac{\hbar}{2 m \omega}} \left( a + a^\dagger \right)$. Notice that the energy is in $\lambda$,
\begin{equation}
\lambda(q) = - 2 q + 4 \sqrt{q} \frac{E(q)}{\hbar \omega} \, ,
\end{equation}
or, in other words,
\begin{equation} \tag{E}\label{E}
\frac{E(q)}{\hbar \omega} = \frac{\sqrt{q}}{2} + \frac{1}{4 \sqrt{q}} \lambda(q) \, .
\end{equation}
Several properties are well known of the periodic solutions with period $\pi$ of the Mathieu equations, which are the ones of interest to us in this case. In particular, the value of $\lambda$ for these particular solutions is often expanded around two limits: $q \rightarrow \infty$ and $q \rightarrow 0$. Both are useful in their own way.
You can slowly reproduce the terms in the expansion of $\lambda$ around $q \rightarrow \infty$ if you use perturbation theory breaking your Schrödinger equation in the form
\begin{equation}
0 = \left[ - \frac{\hbar^2}{2 I} \frac{\partial^2}{\partial \phi^2} + \frac{1}{2}I \omega^2 \phi^2 - E \right] \psi\left( \phi \right) + \Delta H \psi\left( \phi \right) \,
\end{equation}
where we see the order-$0$ Hamiltonian is of the harmonic oscillator kind, and $\Delta H$ contains all the $\mathcal{O}(\phi^k)$, $k \geq 4$ terms of the expansion of the potential term around $\phi = 0$. The energies of the order-$0$ Hamiltonian are the well known
\begin{equation}
E^{(HO)}_k = \frac{1}{2} \left( 2k + 1 \right) \hbar \omega \, ,
\end{equation}
with $k \in \{0\} \cup \mathbf{Z^+}$. When putting the subsequent terms in the expansion, in turns out the term proportional to $\sqrt{q}$ in (\ref{E}) is ultimately cancelled, and the $E^{(HO)}_k$ contribution is followed by a $\mathcal{O}\left(q^{-1/2}\right)$ correction.
On the other hand, you can slowly reproduce the terms in the expansion of $\lambda$ around $q \rightarrow 0$ if you use perturbation theory breaking your Schrödinger equation in the form
\begin{equation}
0 = \left[ - \frac{\hbar^2}{2 I} \frac{\partial^2}{\partial \phi^2} - E \right] \psi\left( \phi \right) + \Delta \tilde{H} \psi\left( \phi \right) \,
\end{equation}
where we see the order-$0$ Hamiltonian is of the free rotor kind, and $\Delta \tilde{H}$ contains all the original potential term. The energies of the order-$0$ Hamiltonian are
\begin{equation}
E^{(R)}_n = \frac{n^2}{\sqrt{q}} \hbar \omega \, ,
\end{equation}
with $n \in \{0\} \cup \mathbf{Z^+}$. This unperturbed spectrum is also degenerate for $n \geq 1$. I have not fully checked this case, but I believe this order-$0$ energy is universally followed by the $\frac{\sqrt{q}}{2}$ term in (\ref{E}), followed then by higher $q$-power contributions which at some point should break the unperturbed degeneracy.
Turns out Mathematica has functions for the values of $\lambda$ that we need. Using the notation of Mathieu functions, their characteristic values for periodic solutions of period $\pi$ are labeled $a_{2l}(q)$ and $b_{2l+2}(q)$, for even and odd solutions respectively. The result for the exact values of the energy can be written
\begin{eqnarray}
\frac{E_r}{\hbar \omega} &=& \frac{\sqrt{q}}{2} + \frac{1}{4 \sqrt{q}} a_{r}(q) \, \, \, , \, \, \,\mathrm{for} \, \, r \, \, \mathrm{even} \, ,
\\
\frac{E_r}{\hbar \omega} &=& \frac{\sqrt{q}}{2} + \frac{1}{4 \sqrt{q}} b_{r+1}(q) \, \, \, , \, \, \, \mathrm{for} \, \, r \, \, \mathrm{odd} \, .
\end{eqnarray}
Plotting this allows us to visualize very nicely the two limits we have discussed: from the degenerate free rotor at small $q$ to the uniformly spaced spectrum of the harmonic oscillator for large $q$. We also see how, for the ground state, the energy changes continuously from $0$ to $\frac{\hbar \omega}{2}$:
|
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Why Einstein action is not Yang-Mills action for gauge theory of Poincaré algebra? It is well known, how to construct Einstein gravity as gauge theory of Poincare algebra. See for example General relativity as a gauge theory of the Poincaré algebra.
There are
*
*Construction of covariant derivative:
$$ \nabla_m = \partial_m -i e_m^{\;a}P_a -\frac{i}{2}\omega_m^{\;\;\;cd}M_{cd}.$$
*Impose covariant constraint on geometry:
$$
[\nabla_m, \nabla_n] = -i R_{mn}^{\;\;\;a}P_a -\frac{i}{2}R_{mn}^{\;\;\;ab}M_{ab}
$$
$$
R_{mn}^{\;\;\;a} = 0.
$$
From this equation, spin connection $ω^{\;\;\;cd}_m$ is expressed in terms of veilbein $e^{\;\;a}_m$.
*Now, one can easily construct Einstein-Hilbert action:
$$
S_{EH} = \int d^d x e \;R_{mn}^{\;\;\;ab} e_a^{\;m}e_b^{\;n}
$$
$e_a^{\;m}$ is inverse veilbein $e_a^{\;m} e_m^{\;b}= \delta_a^b $. Metric tensor:
$$
g_{mn} = e_m^{\;a}e_n^{\;b} \eta_{ab}.
$$
But one can modify second step and obtain another actions, with additional dynamical spin connection:
*
*$$
S_{EH} = \int d^d x e \;R_{mn}^{\;\;\;ab} e_a^{\;m}e_b^{\;n}.
$$
*$$
S_{YM} = \int d^d x e \left(\;R_{mn}^{\;\;\;ab} R_{kl}^{\;\;\;cd}g^{mk}g^{nl}\eta_{ad}\eta_{bc} + R_{mn}^{\;\;\;a} R_{kl}^{\;\;\;b}g^{mk}g^{nl}\eta_{ab}\right).
$$
So I have few questions:
What will standard Einstein-Hilbert action describe in this case?
What is Yang-Mills theory for Poincare group? Which properties have such theory?
Why Einstein action is not Yang-Mills theory for Poincare group?
|
The YM action for the Porcare group as you write down is perfectly allowable in the effective field theory framework, as long as you double check that pathological tachyons are absent. There are tons of papers devoted to the so called $f(R)$ and $f(T)$ theories with higher-order Lagrangian terms (like $R^2$, $T^2$).
The catch is that, comparing with the EH term, the YM term is suppressed by a factor of $O(p^2/M_p^2)$, where $M_p$ is the Planck mass. Therefore, the YM term is negligible, except in extreme situations, e.g. shortly after the Big Bang.
|
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Optics: mirrors While drawing ray diagrams for plane and spherical mirrors, what is generally taken as the point of observation?
Eg, If a concave mirror is presented with, say a wire turned into a triangle, placed from focus towards the mirror, the image can be obtained following rules of reflection but where would the eyes need to be for that particular image to be seen? Like if I were to move around, I'd see different parts of an object. What angle is the images drawn from?
|
We can think of every extended source as a collection of point sources. So, let us only think of point sources. When we draw ray diagrams we often say that an image of a point is formed where 2 rays meet. I was just as baffled as you are but then it clicked of what i think is corect.
Image forming means seeing. Our eyes have evolved to form images of point sources out of diverging rays. (This totally makes sense because you can only have divergent rays from a point source.) The mechanism of our eyes has cornea and lens which converge divergent rays.--so in short we have no problem focusing divergent rays onto our retina.
Here is the trick. If 2 rays are converging then they must be diverging on the other side.
So for a 2D scenario, the correct way to see an image is to put your eye behind the image such that the rays enter your eyes. These divergent rays will be converged onto your retina. Done
And for a 3D scenario, a light cone emerges out (for point sources again) so if you manage to get your head such that the cone enters your eyes you will see the image. But again for maximum clarity, put your head on the diverging part of the cone. :-)
Actually, a sharp image is formed if every point forms a point on retina and if every point falls on a region of retina instead of a point, a collection of such points-an extended source becomes blurred.
I cant be sure that my answer is correct as I didn't refer any source.. Require assessment from good physicists.
[![This is for the 2D scenario, apply the same to cone][1]][1]
The divergent rays will converge onto the retina by lens. Though only two rays have been shown many rays come out forming the complete image. FYI img is formed between 'C' and 'F'.
|
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Negative energy density There are some universe models where $\Lambda < 0$. In this case, the energy density of the dark-energy becomes negative. At this point, does it make sense to talk about "negative dark energy density"? Or is it possible to think of this energy as curvature on space-time? Such that, $\Lambda < 0$ will imply a negative curvature and $\Lambda > 0$ positive one.
For instance if we have a only matter with $\Omega_m = 0.3$ the universe will have negative curvature with $\Omega_{\kappa} = 0.7$.
When we add positive dark energy such that $\Omega_{\Lambda} = 0.7$ we would have $\Omega_{\kappa} = 0$.
If we add negative dark energy such that $\Omega_{\Lambda} = -0.7$, we would have $\Omega_{\kappa} = 1.4$
So adding positive energy density increases the curvature.
In other words is there a something call "negative" energy density ? or
Does all negative energy densities are thought in terms of the curvature effects on space-time?
|
Negative energy density may exist theoretical and the closest believed real phenomenon is the Casmir effect. If it exists, it does cause repulsion of ordinary matter, though how it happens is a little more indirect.
When there are two plates, that are very closely placed in an environment that has 0 energy density, the possible wavelengths that can exist in between the plates is limited. This leads to fewer wavelengths of electromagnetic waves, than is found in the surroundings. This causes an energy density lower than 0, which is basically negative. This is Casmir effect in simple words. The Wikipedia link given can give better insights.
The Wikipedia link here talks about something quite similar.
|
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Choice of representation $R$ in the $\rm SU(2)$ Yang-Mills action $\frac{1}{g^2} \mathrm{Tr}_{R} (F\wedge \star F)$ Usually we write the Yang-Mills theory with gauge group $G$ as
$$\frac{1}{g^2} \mathrm{Tr}_{R} (F\wedge \star F)$$
But here we need to choose what $R$ is. There are several cases one may expect:
*
*$R$ has to be the rep with minimal dimension. Different choice of $R$ gives different gauge theories. For example, for $\rm SU(2)$ $R$ is 2d rep where the three generators are represented by three Pauli matrices. For $\rm SO(3)$, $R$ is 3d rep.
*$R$ has to take all over all possible rep of the gauge group. ($d=2,3,...$ for $\rm SU(2)$.
*The action does not depend on what $R$ is.
My question is: which of the above is correct? and why?
|
The action should be invariant under the action of gauge group, therefore, you have to construct something, that when considering the action of gauge group remains unchanged.
Assume that the field stregth tensor $F$ transform under some representation of $G$:
$$
F \Rightarrow U^{i_1 \ldots i_n}_{j_1 \ldots j_m} F
$$
To make this more clear, let us list several simple examples:
$$
\phi^i \Rightarrow U_{j}^{i} \phi^j - \text{fundamental representation}
$$
$$
\phi_i \Rightarrow U_{i}^{j} \phi_j - \text{antifundamental representation}
$$
$$
\phi_i^{j} \Rightarrow U_{i}^{a} \phi_a^{b} U_{b}^{j} - \text{adjoint representation}
$$
After multiplying $F$ and $F^{*}$ and taking of the trace, there has to remain and object without free indices. The operation of multiplication and $\text{Tr}$ contracts equal number of lower and upper indices, so the only consistent way to make singlet is to have $m = n$.
$$
\text{Tr} \ F \wedge F^{*} = \text{Tr} \ U^{i_1 \ldots i_m}_{j_1 \ldots j_m} U^{j_1 \ldots j_m}_{k_1 \ldots k_m} \ F \wedge F^{*} = \text{Tr} \ U^{i_1 \ldots i_m}_{j_1 \ldots j_m} U^{j_1 \ldots j_m}_{i_1 \ldots i_m} \ F \wedge F^{*}
$$
The written by far expression by far doesn't correspond to irreducible representation in general - i.e, one can symmetrize (antisymmetrize) over $i_a$ or $i_j$. I may be wrong, but it seems that the general action of the aforementioned form is some irreducible representation of tensor product of several adjoint representations.
|
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|
Clarifying the relationship between pressure and temperature? From the ideal gas law, we are aware that PV = nRT, which seems to suggest a direct relationship between pressure and temperature, or that as temperature increases, pressure increases.
In my geography book, however, it is written that "The equator receives direct rays of the Sun, this causes the temperature to rise, hence causing an equatorial low-pressure region." Later the book writes that "Colder air causes a higher pressure." Not just this, when I plotted data between sea level pressure, and average temperature for a weather station in New York, the plot I got was this which seems totally counter-intuitive to the direct relationship suggested by the ideal gas law.
Can someone please help me fix this conundrum?
|
The ideal gas law takes no account of weather patterns over land and water; it takes no account of atmospheric circulation. So the connection you attempt to draw between the gas law and weather reporting is invalid.
Furthermore, the gas law's relationship between temperature and pressure you cite requires that the volume be held constant. No such rule obtains when describing air circulation patterns in the atmosphere.
|
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Classical Wave theory and the photoelectric effect I read that according to classical wave theory, light is viewed as a wave whose intensity is continuously variable. And for this reason, it is unable to explain the photoelectric effect. My questions are:
*
*What does "continuously variable" mean?
*How does the intensity of light being continuously variable make it unable to explain the photoelectric effect?
|
Electrons ejected from a sodium metal surface were measured as an electric current. Finding the opposing voltage it took to stop all the electrons gave a measure of the maximum kinetic energy of the electrons in electron volts.
What does intensity "continuously variable" mean?
It means that for any frequency one can make a continuous increase in the energy of the beam of light.
How does the intensity of light being continuously variable make it unable to explain the photoelectric effect?
As is seen in this particular experiment no electrons come out bellow a certain frequency even though the material is irradiated with it. This means that the photo electron energy cannot be built up by the intensity of the light shone on the material.
|
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Is the entropy of a rotating body largest when the axis of rotation passes through it's centre of mass? I am looking for an answer to the observation that a body always rotates about its centre of mass when freely tossed. It can be explained if the entropy is highest in the case when the axis passes through the com, however, I am unable to prove it.
I am doing this to be able to visualise the motion of a body in space, when struck tangentially.
|
This is easily explained by Newton's second law. If there is no net force applied to a body then the center of mass will not accelerate. it will either be stationary or move in a straight line. The only allowed motion is a rotation about the center of mass.
Changes in entropy, arise from the exchange of energy at some temperature and has nothing to do with the mechanics of rotational motion.
|
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|
Is energy required in generating magnetic field in simple resistance circuit? Consider a simple resistance circuit with a cell and a resistor. It is stated that energy stored in cell appears as heat in resistance as current flows in ideal circuit (neglecting EM radiation) as whole.
POWER/RATE OF HEAT GENERATION = POWER/RATE OF ENERGY CONSUMPTION in CELL = VI
However we also know that flowing current produces magnetic field.
So my questions are:
*
*Is energy needed to create magnetic field in general?
*Does the energy of cell also appears in the energy of the magnetic field?
*Is there any such thing as "energy of magnetic field"
*Any relevant information.
P.S. I am an undergrad. I do not know Special Relativity but I understand that feeling the effects of magnetic field depends on frame of reference.
|
1.Is energy needed to create magnetic field in general?
Yes. When you are using circuit theory the mechanism for dealing with magnetism is called inductance and is usually represented by the variable $L$. The inductance gives the total magnetic field so that you can still use a lumped element approach and do not need to know the details of the magnetic field strength at each location.
2.Does the energy of cell also appears in the energy of the magnetic field?
Yes. Power is given by $P=VI$. Power from the battery goes not only into the resistor but also any inductance or capacitance or any other circuit elements attached.
3.Is there any such thing as "energy of magnetic field"
Yes, for an inductor $V=L\frac{dI}{dt}$ so $P=IV=IL\frac{dI}{dt}$. Notice that when the circuit is at steady state $\frac{dI}{dt}=0$ so $P=0$. Meaning the energy is only briefly put into the magnetic field, not continuously dissipated.
The total energy in the magnetic field is given by the integral of the power over time. It is $E=\frac{1}{2}LI^2$
|
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Are there non-constant potentials that result in eigenstates of the Hamiltonian that are all plane waves? It is commonly known that the eigenstates to the Hamiltonian of a constant potential are plane waves, aka
$$
V(r) = V_0 \Rightarrow H\psi = n \text{ with } \psi = \exp\left(\frac{ip}{\hbar}x\right)\exp\left(-i\frac{E}{\hbar}t\right)
$$
But is there another $V(r)$ for which this holds?
I suppose the question can be simplified to: is a Hamiltonian uniquely defined by its eigenstates, because if this is the case then the answer should be no, there are no other hamiltonians with those eigenstates.
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No, there are no other potentials with the same eigenstates. (set $\hbar = m = 1$ for clarity)
Suppose there is another Hamiltonian $\hat{H}'$, such that the eigenstates of $\hat{H}$ and $\hat{H'}$ are the same, the complete, orthogonal set $| \psi \rangle$ (obtained via the spicy spectral theorem). Then, we know that $[H, H'] = 0$, since $\hat{H} \hat{H}' |\psi \rangle = E' \hat{H} |\psi \rangle = E' E |\psi \rangle$. Neat!
Then we can do this: (Note here that $V$ is an analytic function, and by "$V(\hat{x})$", I mean $\sum_{k=0}^\infty \frac{1}{k!} \frac{\partial V(0)}{\partial x} \hat{x}^k$
$$
0=[\hat{H},\hat{H}'] = [\frac{1}{2}\hat{p}^2 + V, \frac{1}{2}\hat{p}^2 + V'] = [\frac{1}{2}\hat{p}^2, V'] + [V, \frac{1}{2}\hat{p}^2]\\
\Rightarrow [\hat{p}^2, V(\hat{x})] = [\hat{p}^2, V'(\hat{x})]
$$
That's pretty big, since that translates (after a bit of algebra, involving expanding V in a Taylor series and using $[x, p] = i$ ) to
$$
-2i \frac{\partial V}{\partial x}\hat{p} - \frac{\partial^2 V}{\partial x^2}
= -2i \frac{\partial V'}{\partial x}\hat{p} - \frac{\partial^2 V'}{\partial x^2} $$
By substituting in e.g. momentum eigenstates, this condition begins to look like
$$\partial_xV'(x) = \partial_xV(x)$$
By simply integrating this here equation, we see that $V-V' = constant$.
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What is the point of a voltage divider if you can't drive anything with it? The voltage divider formula is only valid if there is no current drawn across the output voltage, so how could they be used practically? Since using the voltage for anything would require drawing current, that would invalidate the formula. So what's the point; how can they be used?
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The first and most obvious use of the voltage divider is to use your intended load as one of the resistors in the divider. This works well if the load is well approximated as a resistor.
That's how the in-cable volume regulator of the headphones work.
When the load is not linear, one can simply make it more linear by paralleling it with another resistor.
That's how most negative feedbacks work - the amplifier input is rarely linear and not always has exactly known V/A behaviour, but it is guaranteed that the impedance is more than a certain value. You parallel this input impedance with a known, much lower resistor value and use it for one of the divider legs.
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Exact Diagonalization of a tight-binding Hamiltonian with periodically alternating potential My question is, can we diagonalize a general Hamiltonian ,
$$H=-t\sum_i^N (c_i^{\dagger}c_{i+1}+h.c.)+\sum_i \mu_i c_i^{\dagger}c_i$$ where,
$$\mu_i=\begin{cases}
\mu_0, &\text{if mod}(i,p)=0 \\
0, &\text{otherwise}.
\end{cases}$$
Obviously, $p$ is the periodicity of the lattice and $c$ is Fermionic annihilation operator. I know $p=2,3,4$ will have an analytic solution but from Abel-Ruffini's theorem $p=5$ onwards may or may not have an analytic solution. Now I am sure because of periodicity there should be a certain degree of symmetry present in solutions, from Bloch's theorem, but I just cannot find a method to analytically solve the problem to get the eigenvalues and eigenvectors.
Numerically, I have found the solution, but any suggestions to solve it analytically?
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I think this should be treated as a tight-binding model with period $p$ and $p$ states in every site. One could do it by first introducing operators:
$$a_{l, \nu} = c_{pl +\nu}, \nu=0,...,p-1,$$
and then looking for plane wave solutions $\sim e^{ikl}$.
|
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Specific note that is not clear for me in and the derivation of maxwell equation $\oint \vec{B} \cdot d \vec{r}=\mu_{0} I_{e n c}$ I know this is not the full equation but right now in this path of the course that what we learned so far.
We studied that a wire along the $z$ axis produces magnetic field $\vec{B}=\frac{\mu_{0} I}{2 \pi \rho} \hat{\varphi}$
then for every closed loop that goes around the wire we can write $$\oint \vec{B} \cdot d \vec{r}=\int \frac{\mu_{0} I}{2 \pi \rho} \rho d \varphi=\mu_{0} I$$
He mentioned that, the critical point in the proof is that, we can say the magnetic field behaves like $\rho^{-1}$ which is not clear for me why is it so critical. What would happen if it did not behave like that?
In addition he wrote that if the loop that we take does not go along the wire, the circulation of $\vec{B}$ will be equal to zero. This is not clear for me physically or mathematically.
for example loop like that :
while the black dot is the wire in the origin.
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I don't quite understand what your instructor was telling you, but since the field due to the wire is $\propto 1/\rho$, when integrating along a hypothetical Ampèrean loop around the wire, the factor of $dr = \rho \, d\varphi$ cancels out the $1/\rho$ of the field, leaving the trivial integral $\int_\text{loop} \, d\varphi$ behind, along with some constant factors. As for the reason why $B \propto 1/\rho$, that has to do with the nature of magnetic fields in general (specifically $\nabla \times \textbf{B} = \mu_0 \textbf{J}),$ which allows one to derive Ampère's law in the first place. The law can then be used to obtain the field around simple geometries, such as the long current-carrying wire in your question.
It occurs to me that your instructor might have derived the field due to the wire using something other than Ampère's law and was trying to show that Ampère's law works, using the wire as an example. Indeed, you can derive the same using the phenomenological Biot-Savart law, which while different from Ampère's law, can be shown to be consistent with it. I hope this clarifies your doubts.
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Relative Velocity of two particles If two particle are neither approaching towards nor receding away from other then their relative velocity is non zero.
How is this possible??
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Turning my comment into an answer ...
The velocity can be non-zero because velocity is represented by a vector, so the vector changes when the direction changes, even if the magnitude ("speed" in this case) remains constant.
For example, consider the case of two objects in circular orbits around the barycenter. If I am riding on top of one of them their separation is constant, but the direction to the other one is changing, describing a circle. The direction changes, so velocity is not constant.
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Enthalpy during compression of water What happens to the temperature of water when compressed?
Enthalpy $H = U + PV$.
$H$ is conserved in a closed system. By which I mean adiabatic and negligible external work applied.
We compress a litre of water to 10 bar (say). This requires negligible work because water is almost incompressible. But $P$ goes up, $V$ hardly changes, so $PV$ goes up, so $U$ should go down. It should get quite a bit colder.
But is that correct? Because very little work was done, yet $PV$ changes quite a lot, and it would produce a lot of cooling.
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Enthalpy certainly isn't conserved in a system you do work on. However, entropy is if you do the work quasistatically and adiabatically. Thus, you seek $\left(\frac{\partial T}{\partial P}\right)_S$, the temperature rise upon isentropic compression. Applying the triple product rule and a Maxwell relation, we have
$$\left(\frac{\partial T}{\partial P}\right)_S=-\left(\frac{\partial S}{\partial T}\right)^{-1}_P\left(\frac{\partial S}{\partial P}\right)^{-1}_T=\left(\frac{\partial S}{\partial T}\right)^{-1}_P\left(\frac{\partial T}{\partial V}\right)^{-1}_P=\frac{\alpha TV}{C_P}=\frac{\alpha T}{\rho c_P},$$
where $\alpha$ is the volumetric thermal expansion coefficient, $c_P$ is the specific heat capacity, and $\rho$ is the density. For water at room temperature, we should therefore expect a temperature increase of about 0.015°C/bar.
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Doppler effect observed on pendulums I wanted to investigate the doppler effect on pendulums and set up an experiment where:
*
*The bob of a pendulum is a speaker which emits a specific sound frequency.
*Directly underneath the lowest point of the pendulum sits a microphone that observes the frequency being emitted by the pendulum.
In the experiment, I would release the speaker from various amplitudes and record the frequency that the microphone perceives as the speaker passes the equilibrium position.
I know that the standard doppler equation is:
$$f=f_0\left(\frac{c}{v\pm c}\right)$$
where $f$ is the observed frequency
$f_0$ is the initial frequency
$c$ is the speed of sound
and $v$ is the speed of the moving source
However this equation only applies to non-accelerating systems; even though the speed of the speaker at the bottom of the pendulum can be predicted with the equation
$$v=\sqrt{2gh}$$
substituting this velocity into the doppler equation will not give accurate results because of the acceleration of the pendulum.
So, my question is, how can one calculate the frequency observed by the microphone at the bottom of the pendulum given the various release heights that the speaker is released from?
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Your formula will give the velocity at any height on the swing down, and that can be used in the Doppler equation.
Recording the sound is not a problem, but measuring the (variable) frequency may be a challenge. If you can freeze (or video) the display on an oscilloscope, you might measure the period of each cycle of the sound wave.
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Why didn't heavier elements settle at the core of the solar system? As the solar system formed, why didn't all of the heavier elements such as iron, collect where the sun is leaving the lighter elements in the outer solar system?
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That is actually somewhat the case...
The planetesimals formed in the inner Solar System were rocky planets (our terrestrial planets evolved from these), whereas the outer planets (the gas giants) in the cooler zone evolved from lighter elements such has hydrogen and helium.
This is of course an incomplete answer. From exoplanets surveys we have found many systems with inner gas giants. There are many aspects we are yet to fully understand about the evolution of these systems. Such aspects could be planet migration, long term tidal effects (affecting orbital energy) and impact events.
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Insulator or conductor with different boundary conditions I'm studying the 1-D SSH model. It's a toy model for a topological insulator. Here's the reference I'm using. If the hopping amplitudes $v$ and $w$ are equal, then with periodic boundary conditions we have a conductor. If however, we take open boundary conditions, a plot of the eigenvalues shows that there is a band gap with the same values of $v$ and $w$. What is going on?
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Are you refering to fig 1.4 in the paper you cite as evidence for an enegy gap at $u=v=1$? If so, you need to realize that in the that plot the authors are using a very small (10 site) system. The gaps between the eigenvalues are therefore due to the finite-size discreteness of the eigenvalues $E_n=\pm 2\cos k$ with $k$ something like $k=(n+1/2)\pi/N$ for $N$ sites. There will only be an exact zero energy $E=2 \cos \pi/2$ state when $N$ is odd. $N=10$ is even in their plot. As $N$ becomes larger the spaces between the $N=1$ eigenvalues will fill in, and there will be no gap at zero energy in the $u=v$ case.
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What is the probability that two chemical species of binding energy $E$ will be bound? This may seem like a very simple question, but I've been agonising over it for days. What is the probability, $p$, that two chemical species with binding energy $E$, will be bound.
My first instinct is that
\begin{equation} \tag{1}\label{1} p = Ae^{-\beta E} \end{equation}
with A some normalisation factor. This would give the correct behaviour ($p\to1$ as $E\to -\infty$ and $p\to 0$ as $E\to 1$). But if this is the case, what is the form of $A$.
However, for
\begin{equation}\tag{2}\label{2} p = 1 - e^{\beta E} \end{equation}
we see the same behaviour.
I have had a look in some literature for this, and have become even more confused, seeing both \eqref{1} and \eqref{2} used.
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The answer is not a simple function of binding energy E. What you need to know is the equilibrium constant $[AB]/[A][B]=K=\exp (-\beta {{G}_{0}})$, where ${{G}_{0}}$ denotes the free energy difference at reference concentrations. Doubling the concentration of unbound A & B will quadruple the concentration of bound AB.
K is proportional (not equal) to $\exp (\beta E)$ as you suspected, but there is a further factor that pertains to entropy. In gas phase, the entropy difference can be calculated via the semi-classical Sackur-Tetrode equation, which drags in Planck’s constant, cubed. The situation in liquid phase is not conceptually different, but solvation energies complicate things.
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Derivation of Schwarzschild metric I was studying the book of Hartle on general relativity. In chapter 9, "The Geometry Outside
a Spherical Star", he suddenly introduces a metric named Schwarzschild metric and then goes on describing the geometry it produces. I did not quite get how exactly this was a metric generated by a spherical start. There must be some methodology of arriving at this metric. In non-relativistic Newtonian limit, I know how to show $g_{00} = 1+2\phi/c^2$, but for general case, I did not find anything useful.
What is the systematic logical sequence behind this result?
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A typical method is to make the ansatz that the spherically-symmetric metric has the form
$$ds^2=-A(r)dt^2+B(r)dr^2+r^2(d\theta^2+\sin^2{\theta}d\phi^2)$$
and determine which functions $A(r)$ and $B(r)$ make the metric satisfy the Einstein field equations, which in vacuum are $R_{\mu\nu}=0$ everywhere (except perhaps at some singularity).
This ansatz lets you reduce partial differential equations to ordinary differential equations.
You should try doing this yourself, by hand! Calculate the Christoffel symbols, the Riemann tensor, and the Ricci tensor in terms of $A$ and $B$ and their first and second $r$-derivatives. Then set Ricci to zero and solve for $A$ and $B$. You will have a great sense of satisfaction in solving Einstein’s field equations for a Schwarzschild black hole.
It turns out that this metric describes not just black holes but also the vacuum outside an uncollapsed and non-rotating star. But it is easiest to first think about the all-vacuum black hole case.
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What is the maximum deflection angle of a pendulum in a car, when the car, initially at rest, accelerates suddenly? I was doing Kleppner-D.-Kolenkow-R.J. and I came across the following problem:-
A pendulum is tied vertically to a car at rest, the car suddenly accelerates at a rate A. Find the maximum angle of deflection $\phi$ through which the weight swings.
MY TRY: I saw the solution of this problem in the book which uses car's frame of reference, which was fairly simple.
I tried to do it in the ground frame of reference.
Deflection of the pendulum will be maximum when the angular velocity of the mass hung to pendulum relative to the hanged point will be zero, hence the velocity of mass relative to the car, perpendicular to the string is zero. But the constraint of a taut string doesn't allow velocity of mass relative to the car along the string also. So, velocity of mass relative to car is zero at the point of maximum deflection.
I have the following two tools to solve the problem:-
*
*Apply Work energy theorem to the mass.
*Use the string constraint i.e. the acceleration of the mass and the topmost point along the string will be equal at any instant i.e. $T-mgcos(\theta)=masin(\theta)$
The tension force and gravity are only two forces acting on the mass. But, how could one find the work done by tension on the mass in the journey from $A$ to $B$. Any hint would be a great help!
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General relativity tells us that gravity is equivalent to acceleration. In this case the pendulum suddenly finds itself in a system where gravity has shifted backwards by an angle with a tangent of a/g. Being a pendulum it will swing toward this new equilibrium, and momentum will carry it an equal angle beyond.
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Lagrangian potential for Newtonian gravity In the Wikipedia site for Lagrangian (field theory) the Lagrangian density for Newtonian gravity is given by
$${\cal L}(\mathbf{x},t) = \frac{1}{2}\rho(\mathbf{x},t)\mathbf{v}^2 -\rho(\mathbf{x},t) \Phi(\mathbf{x},t) – \frac{1}{8\pi G}(\nabla\Phi(\mathbf{x},t))^2$$
I understand how variation of this Lagrangian leads to the correct Poisson equation $\nabla^2\Phi=4πG\rho$.
However, if ${\cal L}$ is simply the differential form of $T-V$, I would have thought that just the first two terms would be sufficient. Where does the third term in ${\cal L}$ come from?
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TL;DR: JoshuaTS' answer is exactly right: (Minus) the 3rd term ${\cal V}_3=\frac{1}{8\pi G}(\nabla\Phi)^2$ is the energy density of the gravitational field.
*
*In total OP's Lagrangian density contains 3 terms: ${\cal L}={\cal T}_1-{\cal V}_2-{\cal V}_3$. Here ${\cal T}_1$ is a kinetic term for matter, while ${\cal V}_2=\rho\Phi$ is an interaction/source term between matter and the gravitational field.
*The EL equation for the specific gravitational potential $\Phi$ is the Poisson eq. $\nabla^2\Phi=4\pi G \rho$, as OP already mentions.
*Next let us integrate out/eliminate the $\Phi$-field by using its EL equation. Then one may show that ${\cal V}_3=-\frac{1}{2}{\cal V}_2$, so the 3rd term is important for the correct normalization of the potential energy.
*In fact, it's a fun exercise to check that if the mass density $$\rho({\bf x},t)=\sum_i m_i \delta^3({\bf x}-{\bf x}_i(t))$$ is a sum of point masses, then the potential energy
$$V~=~\int_{\mathbb{R}^3}\!d^3x~({\cal V}_2+{\cal V}_3)~=~- \frac{G}{2}\sum_{i\neq j} \frac{m_im_j}{|{\bf x}_i(t)-{\bf x}_j(t)|}$$
becomes Newton's gravitational formula after discarding the singular self-interaction terms.
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Relativity without constancy of light speed Using homogeneity of space, isotropy of space and the principle of relativity (without the constancy of light speed), one can derive:
$$x' = \frac{x-vt}{\sqrt{1+\kappa v^2}}$$
$$t' = \frac{t+\kappa vx}{\sqrt{1+\kappa v^2}}$$
$\kappa = 0$ denotes Galilean and $\kappa < 0$ denotes Lorentz Transformation.
What does $\kappa > 0$ denote? Is it physically possible? I was told that it is self-inconsistent. Can somebody help me with a proof of this?
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$\kappa > 0$ represents Euclidean geometry, in which the time axis is equivalent to (and freely interchangeable with) the spatial ones. In other words it acts like a fourth spatial dimension.
So you can for example take a left turn into the time axis and go forward, then turn around and go back in time. Many consider this non-physical, leaving the choice beweeen $\kappa = 0$ (Galilean Transformation) and $\kappa < 0$ (Lorentz Transformation).
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How do I interpret uncertainty in velocity greater than the speed of light? I just studied Heisenberg's uncertainty principle in school and I came up with an interesting problem.
Assume an electron which is moving very slowly and we observe it with a distance uncertainty of say $\Delta x=1\times10^{-13} \text{ m}$ if we try finding uncertainty of velocity using the formula $$\Delta x \cdot \Delta v\ge \dfrac{h}{4\pi m}$$
$$\Delta v=578838179.9 \text{ m/s}$$ Which is clearly greater than the speed of light but that is not possible. How did physicists overcome this challenge?
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What you've discovered is that "normal" Quantum Mechanics is incompatible with relativity. As Valter Moretti pointed out, using a relativistic expression for momentum solves this problem. There are, however, more problems that cannot be solved by simply using relativistic expression for energy and momentum. For example,
*
*The relativistic equation $E=mc^2$ implies that it is possible for energy to be converted into new particles. The time-energy uncertainty principle $\left(\Delta E\cdot\Delta t\geq\hbar/2\right)$ implies that it is possible for particles to be created out of thin air, even when, from a classical viewpoint, there is not enough energy present.
*Even when single-particle quantum mechanics is modified to use a relativistic Hamiltonian, as in the Klein-Gordon equation, there is always a non-zero probability that a particle can teleport across a space-like interval (faster than the speed of light).
These problems are resolved by the introduction of Quantum Field Theory. Basically, instead of quantizing individual particles, we quantize fields. The particles are excitations of the fields, and new particles can appear out of thin air. Quantum field theories are designed to preserve causality so that they work well with relativity. The mathematics is all very complicated, but that's the basic idea.
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Need Help Understanding Why $\Delta \vec{v}$ is Perpendicular to $\vec{v}$ I am confused about the statement of how vector $\Delta \vec{v}$ is perpendicular to vector. I highlighted the statement in pink. I ended up copying the image of the right vector $\vec{v}$ in the velocity isosceles triangle and moved its tail to touch the tail of vector $\vec{v}(t)$. It does not look perpendicular so could someone clarify my misunderstanding?
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This is only true if ${\bf v}$ just changes direction but keeps the same length. If ${\bf v}$ and ${\bf v}+ \Delta {\bf v}$ have the same length, then
$$
{\bf v}\cdot {\bf v}= ({\bf v}+ \Delta {\bf v})\cdot ({\bf v}+ \Delta {\bf v})= {\bf v}\cdot {\bf v}+ 2{\bf v}\cdot \Delta{\bf v}+ (\Delta {\bf v})\cdot (\Delta {\bf v})
$$
so as $\Delta {\bf v}$ gets small we must have ${\bf v}\cdot \Delta{\bf v}=0$. i.e. they become perpendicular.
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Why can't photons cancel each other? The textbook argument against photons canceling each other draws upon the conservation of energy. Does this mean that energy conservation is a "stronger" principle than superposition? Waves in other media than the EM field, e.g., sound or water, do cancel out---presumably by passing on their energy to some other degree of freedom (e.g., heat). Could this imply that EM waves don't have any alternative channel to pass on the destructed energy and thus can't cancel out?
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We should be careful to distinguish interaction, correlation, annihilation and interference. Photons do not interfere. Any interference takes place at wave function level, so impacts the probability of finding a number of photons. Photons can annihilate but this requires two photons of a least 511 keV each in order to create an electron-positron pair. Photons can interact (scatter) via transient vacuum charge fluctuations. Finally photons can be correlated by Bose statistics.
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How to derive this $\dfrac{dT}{d\tau}$? I am studying the paper "Gravitational field of a particle falling in a Scharzschild geometry analyzed in tensor harmonics" by Zerilli. The author calculates the gravitational radiation emitted by a particle falling along a geodesic into a Schwarzschild black hole.
The stress energy tensor of the particle is:
$$T^{\mu\nu}=m_0 \dfrac{dT}{d\tau} \dfrac{dz^\mu}{dt} \dfrac{dz^\nu}{dt} \dfrac{1}{r^2}\delta(r-R(t)) \delta^{(2)}(\Omega-\Omega(t))$$
where the trajectory of the particle is $z^\mu=\left( T(\tau),R(\tau), \theta(\tau), \phi(\tau) \right)$.
In the appendices the author calculates the expression of the stress energy tensor of a particle falling radially into the black hole with the method of tensor harminics. For expample the 00 component is:
$$A_{lm}^{(0)} = m_0 \dfrac{dT}{d\tau} \left(1-\dfrac{2m}{r}\right)^2 \dfrac{1}{r^2} \delta(r-R(t)) Y_{lm}^*$$
In order to calculate the gravitational radiation emitted one must Fourier transform this expression: the author gives the procedure: 1) multiply by $\exp(i\omega t)dt$, the write $dR=dt/(dR/dt)$ so that the delta is simplified thanks to its properties. At the end of the calculations the author reports:
$$A_{lm}^{(0)}=\dfrac{m_0}{2\pi} \sqrt{\left(l+\dfrac{1}{2}\right) \dfrac{r}{2m}} \dfrac{1}{r^2} \exp(i\omega T(r))$$
The last two expressions give me troubles:
*
*Starting from the Fourier transform: where does the $\dfrac{dT}{d\tau}$ goes? Since it is the time component of the trajectory of the particle I thought that it could be derived by the Lagrangian as Wald does, i.e. from
$$-1=-\left( 1-\dfrac{2m}{r} \right) \dfrac{dT}{d\tau} + \left( 1- \dfrac{2m}{r} \right) \dfrac{dR}{d\tau}$$
from this I can derive the $\dfrac{dT}{d\tau}$ term, but it depends on $\dfrac{dR}{d\tau}$, whose dependence I do not know. Wald derives $\dfrac{dR}{d\tau}$ from the above Lagrangian by using the fact that (through Killing vecotrs) one has $E=\left(1-\dfrac{2m}{r} \right)\dfrac{dT}{d\tau}$, but then I need the energy $E$. From thecontour conditions of Zerilli problem the particle stars at infinity with 0 velocity, so at infinity $E=m_0c^2$, but this doesn't seem to be the way Zerilli calculated the Fourier transform, so I am lost;
*The second problem I have is with the the second equation I have reported: where does the $\left( 1-\dfrac{2m}{r} \right)^2$ term comes from?
|
The equations of motion for a geodesic starting at rest at infinity are very simple, and as usual the key is to start with the constants of motion:
$$ -1 = \frac{d z^\mu}{d\tau}\frac{d z^\nu}{d\tau} g_{\mu\nu} = -\left(1-\frac{2m}{r}\right)\left(\frac{d T}{d\tau}\right)^2 + \left(1-\frac{2m}{r}\right)^{-1}\left(\frac{d R}{d\tau}\right)^2$$
and
$$ 1=\mathcal{E} = -g_{t\mu}\frac{d z^\mu}{d\tau} = \left(1-\frac{2m}{r}\right)\frac{d T}{d\tau}$$
(The specific energy $\mathcal{E}$ is one because the particle starts at rest at infinity.)
These can be solved for $\frac{d T}{d\tau}$ and $\frac{d R}{d\tau}$
\begin{align}
\frac{d T}{d\tau} &= \left(1-\frac{2m}{r}\right)^{-1}\\
\frac{d R}{d\tau} &= -\sqrt{\frac{2m}{r}}
\end{align}
These can be combined to calculate.
$$\frac{d R}{dT} = \frac{\frac{d R}{d\tau}}{\frac{d R}{d\tau}} = -\sqrt{\frac{2m}{r}}\left(1-\frac{2m}{r}\right) $$
You now have all the ingredients you need to calculate $A_{LM}$. (Also note that you can make your life easy by placing the incoming particle on the pole. This implies that $A_{LM}=0$ for all $M$ except 0.)
|
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|
Does dark matter follow all principles of regular physics? Is dark matter bound by all the laws of regular physics?
i.e. laws of thermodynamics, speed of light, length contraction, mass-energy relation.
What about Newton's laws of motion (since all of Newton's laws assume an interaction between particles)?
|
Is dark matter bound by all the laws of regular physics?
We call "physics" our mathematical modelling of the world's behavior, so dark matter will be covered by its laws eventually. Its behavior diverges from the behavior we manage to observe with great accuracy for matter in much closer vicinity to us, so either dark matter consists of different particles than what we are used to, or we are overspecialising our theories about matter in our vicinity and overlooking a variable that appears fixed to us, in a similar vein to how there does not actually exist a discrepancy between non-relativistic and relativistic particles.
In that case, the problem is not that dark matter is not covered by what you call the laws of "regular" physics, but that no matter is. It's just that dark matter gives us the "Doh." about it much less subtly than matter closer to us does.
|
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|
Why are aerodynamic / streamlined shapes always stumpy at the front? I'm building an autonomous boat, to which I now add a keel below it with a weight at the bottom. I was wondering about the shape that weight should get. Most of the time aerodynamic shapes take some shape like this:
The usual explanation is that the long pointy tail prevents turbulence. I understand that, but I haven't found a reason why the front of the shape is so stumpy. I would expect a shape such as this to be way more aerodynamic:
Why then, are shapes that have good reason to be aero-/hydrodynamic/streamlined (wings/submarines/etc) always more or less shaped like a drop with a stumpy front?
|
Any speculation about what shape might be best is meaningless without specifying the flow conditions. For the keel on a boat, the main one is the Reynolds Number, a parameter that is proportional to the the length multiplied by the speed.
In most low-speed applications, a sharp leading-edge is not the best. With any incidence, the flow will tend to separate too readily, but even when going straight through the fluid there are velocity gradients that need to be considered. The flow increases and then decreases in speed as it moves along the keel, and the drag that this causes depends on the details of the viscous boundary layer development.
The figure below (which I generated using a readily available analysis code) shows some approximate optimizations of 2D foil sections at several different values of Reynolds Number. The best shapes at the highest speeds (at the top of the figure) have smaller leading-edge radii than the low-speed ones (at the bottom), which are extremely blunt.
|
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|
At which point in the universe history will inflation prevent galaxies feeding from intergalactic matter? It's my understanding that galaxies formed from accretion of intergalactic matter around supermassive black holes. As the universe expands the amount of matter entering a galaxy decreases, until at some point this process stops or becomes negligible.
At which point in the past or the future will matter accretion from the intergalactic medium cease? How is this point in time shifted by the filamentous organization of baryonic intergalactic matter?
|
Although galaxies are moving apart from one another faster and faster, it is only at the largest scales where the average energy density of matter is smaller than the dark energy density. Clusters and superclusters of galaxies are unlikely to dissolve because they are more energy-dense than dark energy, and they will only become more dense as they collapse from their own self-gravity.
|
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|
Why is maximal kinetic energy lost in a perfectly inelastic collision? A perfectly inelastic collision is one where both of the colliding objects stick together and move as one.
My question is, why, of all possible combinations of final velocities that conserve momentum, does this one lead to the greatest loss in kinetic energy?
One reasoning I found was that this is the only combination in which the total kinetic energy of the system becomes 0 in some frame of reference (com frame). But just because the KE is 0 in some frame doesn't mean that it is the least possible in every other frame, does it?
|
If I understand your question correctly, then the answer would simply be realizing what the C.O.M frame does to a two-body problem. If you recall a two body problem can be converted into a one-body problem in the C.O.M frame. This means that equations of motion for the now new one body are equivalent to the original scenario.
Since Kinetic Energy(at least classically) cannot be negative , minimum corresponds to it being zero (with max loss) , hence it cannot be any lower for any other frame.
Now what I mean by this is. Let's take this inelastic collision from some other frame where the K.E has some value. If this collision instead were elastic , computing K.E from the chosen frame$ will give value greater than that for the previous case. So basically getting the most minimum value possible (=0) for K.E. from C.O.M frame implies that if you had chosen any other frame where K.E. was not zero , an elastic collision observed from this frame will give a greater K.E and hence and lesser loss. The key is that for comparing elastic and inelastic collision you have to stick to one frame whatever that maybe.
|
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|
Is it possible the many-worlds hypothesis explains dark matter? To provide context for the title, is it possible that dark matter is just gravity from other 'branches' of the universal wave function that have split from ours, that are weakly interacting with the gravity in our branch? Is this an idea anyone has explored before?
|
Dark matter candidates arise frequently in theories that suggest physics beyond the Standard Model, such as supersymmetry and extra dimensions. One theory suggests the existence of a “Hidden Valley”, a parallel world made of dark matter having very little in common with matter we know. If one of these theories proved to be true, it could help scientists gain a better understanding of the composition of our universe and, in particular, how galaxies hold together.
In the google search "many worlds dark matter" there is this arxiv paper , and others that show people are researching other dimensions for explanations of dark matter. research is on going.
|
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|
Why is the electromotive force (EMF) highest in the loop when $\theta$ = $90$? My question is in regard of the following snippet provided by my textbook.
So why is the electromotive force (EMF) highest in the loop when $\theta$ = $90$ or $270$?
So the magnitude of the induced EMF will be determined by the rate at which the loop is rotating, according to Faraday's Law. EMF will be maximum when the rate of change of flux is at maximum.
But why does this means that the loop has moved to a position parallel to the magnetic field and the flux through the loop is zero? Since there is no magnetic field penetrates the area at that instant shouldn't there be no current? In turns, shouldn't there be no magnetic field?
|
The field mentioned in this discussion is from an external source. The maximum flux does occur when the normal vector representing the loop is parallel to the field (The field is passing through the loop), as indicated in the diagram. The maximum “rate of change” of the flux occurs when the plane of the loop is parallel with the field (and the normal vector is at 90 degrees.
|
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|
How is melting time affected by flow rate and temperature of surroundings? Suppose you have a solid sphere of m, where m is an element with freezing point of 0 degrees Celsius.
In one scenario, you place your sphere in a (“static”) 25 degree Celsius environment and measure time, t, until melting. The sphere is fixed and cannot be displaced.
In the other, you place your sphere in environment with temperature, T, and with constant flow rate, v. Again, you measure time, t, until melting.
What is the equation that would relate the two scenarios? In other words, at what temperature and flow rate would time required for melting in the second scenario equal time required in the first?
|
The answer to this is very subtle, and is the core subject of interest in convective heat transfer. In either case, you’ll find that most engineers would model either scenario using Newton’s law of cooling:
$$Q = hA(T-T_{\infty})$$
where $Q$ is the heat transfer rate, $A$ is the surface area of the object in contact with its surroundings, $T$ is the temperature of the object and $T_{\infty}$ is the (approximate) temperature of the surroundings. $h$ is a sort of catchall term called the “heat transfer coefficient”, which is affected by all sorts of things—in particular, by flow in the surroundings of the embedded object. Most engineers find this coefficient through empirical studies.
That being said, flow in general increases the amount of heat transfer, and so an object embedded in surroundings at a different temperature & a uniform flow will heat up/cool down to the surrounding temperature faster than without the flow.
In the case without flow, temperature gradients will actually cause flow themselves by changing the density of the fluid near the object with a different temperature, so there will still be some minor convective heat transfer—this is usually called natural convection.
|
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|
Inverse metric for 3+1 decomposition I was trying to read about 3+1 decomposition of spacetime from section 12.2 of Padmanabhan's book Gravitation Foundations and Frontiers. However, other sources can also provide the context for my question.
Once the coordinate system $(t,y^\alpha)$ has been adopted on the spacetime from the foliation, $x^a=x^a(t,y^\alpha)$, then we can write (also the book uses the convention $a=0,1,2,3$; $\alpha=1,2,3$ or latin indices represent spacetime and greek indices only space),
\begin{align}
dx^a&=\frac{\partial x^a}{\partial t}dt+\frac{\partial x^a}{\partial y^\alpha}dy^\alpha\nonumber\\
&=t^adt+e^a_\alpha dy^\alpha\nonumber\\
&=\left(Nn^a+N^\alpha e^a_\alpha\right)dt+e^a_\alpha dy^\alpha\nonumber\\
&=\left(Ndt\right)n^a+\left(N^\alpha dt+dy^\alpha\right)e^a_\alpha
\end{align}
Where we have used the fact that the tangent to the curves parametrized by $t$ is $t^a=\partial x^a/\partial t=Nn^a+N^\alpha e^a_\alpha$; and $N$ is called the lapse function and $N^\alpha$ is called the shift vector. $e^a_\alpha=\partial x^a/\partial y^\alpha$ are the tangent to the hypersurface called tetrads.
The line element (squared) now becomes,
\begin{align}
ds^2&=g_{mn}dx^mdx^n\nonumber\\
&=g_{mn}\left[\left(Ndt\right)n^m+\left(N^\alpha dt+dx^\alpha\right)e^m_\alpha\right]\left[\left(Ndt\right)n^n+\left(N^\beta dt+dx^\beta\right)e^n_\beta\right]\nonumber\\
&=-N^2dt^2+h_{\alpha\beta}\left(dx^\alpha+N^\alpha dt\right)\left(dx^\beta+N^\beta dt\right),
\end{align}
here,
\begin{align}
h_{\alpha\beta}=g_{mn}e^m_\alpha e^n_{\beta}=g_{\alpha\beta}.
\end{align}
The metric can be read out from the above line element,
\begin{align}
g_{00}=-N^2+N_\gamma N^\gamma,\quad g_{0\alpha}=N_\alpha,\quad g_{\alpha\beta}=h_{\alpha\beta}
\end{align}
In matrix form,
\begin{align}
g_{mn}=\begin{pmatrix}
-N^2+N_\gamma N^\gamma & N_\alpha\\
N_\alpha & h_{\alpha\beta}
\end{pmatrix}
\end{align}
My question is how to calculate the inverse of this metric?
I have tried to do that but not succeeded except for the component $g^{00}$ and I am not sure whether that derivation is correct. So let me describe the process in the following.
Now, as $\partial_a t=\delta^t_a=\delta^0_a$ in the coordinate system $(t,y^\alpha)$. Thus,
\begin{align}
g^{00}&=g^{ab}\partial_a t\partial_b t\nonumber\\
&=\frac{1}{N^2}g^{ab}n_an_b\nonumber\\
&=-N^{-2}.
\end{align}
Where I have used the fact that the normal vectors are defined as $n_a=-N\partial_a t$ and the normalization for spacelike hypersurfaces is such that $n^an_a=-1$.
In the book, the components for the inverse metric are given to be,
\begin{align}
g^{00}=-N^{-2},\quad g^{0\alpha}=N^{-2}N^{\alpha},\quad g^{\alpha\gamma}=h^{\alpha\gamma}-N^{-2}N^\alpha N^\gamma
\end{align}
Therefore the answer I am looking for is the step by step derivation of the inverse metric given the components of the metric and also one should verify whether my calculation for $g^{00}$ is correct. Thank you.
|
Well, maybe there is a more clear way to do this, without some guessing. I would start from the definition of an inverse matrix:
$$
g^{\mu \alpha} g_{\alpha \nu} = \delta_{\nu}^{\mu}
$$
Or more concretely:
$$
\begin{pmatrix}
-N^2+N_\gamma N^\gamma & N_\alpha\\
N_\alpha & h_{\alpha\beta}
\end{pmatrix}
\begin{pmatrix}
g^{00} & g^{0 \alpha}\\
g^{0 \alpha} & g^{\alpha \beta}
\end{pmatrix} =
\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}
$$
Written in components:
$$
\begin{align}
(-N^2+N_\gamma N^\gamma) g^{00} + N_\alpha g^{0 \alpha} = 1 \\
(-N^2+N_\gamma N^\gamma) g^{0\alpha} + N_\beta g^{\beta \alpha} = 0 \\
N_\alpha g^{0\beta} + h_{\alpha \gamma} g^{\gamma \beta } = \delta_\alpha^{\beta}
\end{align}
$$
Now, using the symmetry of $g_{\mu \nu}, h_{\mu \nu}$ under exchange of $\mu \leftrightarrow \nu$, one may see, that there are $ D(D+1) / 2$ linear equations on the same number of unknows, which can be in principle solved.
Doing these directly seems a tedious task, so there can be an educated guess. Supposing we knew that $g^{00}$ is $-N^2$, in general ansatz could be $\alpha N^2 + \beta N_\alpha N^{\alpha}$, then the first equation is immediately resolved by setting :
$$
g^{0 \alpha} = N^{-2} N^{\alpha}
$$
Then one may look on the second line. Here is also natural to assume, that $g^{\mu \nu} = h^{\mu \nu} + b^{\mu \nu}$, where $b^{\mu \nu}$ is also symmetric. This substitution gives:
$$
-N^{\alpha} - N^{-2} N_\gamma N^\gamma N^{\alpha} + N^{\alpha} + N_\beta b^{\beta \alpha} = 0
$$
Here also, one may see, that the $b^{\mu \nu} = -N^{-2} N^{\beta} N^{\alpha}$ does the job.
|
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What is the force required to tilt a rectangular object when exerted at 2 different locations?
I have a question to confirm if my "hunch" is correct. Assuming 2 scenarios of similar force but is exerted at different height of a rectangular object with a pivot point (shown in the pic below). Based on Torque (perpendicular force * distance from the pivot point), it should be easier to tilt the object from F1 location vs F2 location assuming a similar amount of Torque needed to lift the object.
Torque=d1 * F1 = d2 * F2 ). Based on that formula, then F1<F2 to get same torque.
However, when I use the 2nd class lever equation, I find F1 ==F2 based on the equation below:
Assuming the CoM is in the middle of the object, using the 2nd class lever formula, i.e. d1 * F1= mg(CoM)* 0.5 *d1) , and so F1 = (mg(C0M) * 0.5 * d1)/ d1. I used the same formula for F2 and I found out F1=F2. Is this the exact force to tilt the object and if so, why both F1=F2 when it should be F1 < F2, or am I confusing force and torque?
|
I see you have got the definition of torque wrong. It is not just force multiplied by distance from pivot point. Torque is distance from pivot point multiplied by the amount of force which is perpendicular to the direction of point of application from pivot point. It can also be stated as the full force multiplied by the perpendicular distance . So in our case h1f1= mglength of block/2=h2*f2 and we cannse f1 is not equal to f2 and ever which is bigger.
You can also try taking cross products of displacement vector and force vector.
|
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|
Lorentz boost expressed as Hyperbolic versors At this link https://en.wikipedia.org/wiki/Versor#Hyperbolic_versor it is claimed that an hyperbolic versor, defined as:
$$
\exp(a \mathbf{r})=\cosh a+\mathbf{r}\sinh a
$$
where $||\mathbf{r}||=1$ correspond to a Lorentz boost. But I cannot work out a proof. Can anyone help?
I assume one starts by applying the exponential to a 4-vector $\mathbf{s}$ as follows:
$$
\mathbf{s}'=\exp(\frac{a}{2} \mathbf{r})\mathbf{s}\exp(-\frac{a}{2} \mathbf{r})
$$
Then I get
$$
\begin{align}
\mathbf{s}'&= (\cosh a/2+\mathbf{r}\sinh a/2 )\mathbf{s} (\cosh a/2 -\mathbf{r}\sinh a/2)\\
&= (\cosh a/2+\mathbf{r}\sinh a/2 ) (\mathbf{s}\cosh a/2 -\mathbf{s}\mathbf{r}\sinh a/2)\\
&=\mathbf{s} \cosh^2a/2-\mathbf{s}\mathbf{r}\cosh a/2\sinh a/2+ \mathbf{r}\mathbf{s}\sinh a/2\cosh a/2 - \mathbf{r}\mathbf{s}\mathbf{r}\sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2+\cosh a/2\sinh a/2(-\mathbf{s}\mathbf{r}+ \mathbf{r}\mathbf{s}) - \mathbf{r}\mathbf{s}\mathbf{r}\sinh^2a/2
\end{align}
$$
edit (based on answer):
$$
\begin{align}
\mathbf{s}'&=\mathbf{s} \cosh^2a/2+ 2\mathbf{r}s_\perp \cosh a/2\sinh a/2 - (s_\parallel-s_\perp) \sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2+ \mathbf{r}s_\perp \sinh a - (s_\parallel-s_\perp) \sinh^2a/2
\end{align}
$$
edit2:
$$
\begin{align}
\mathbf{s}'&=\mathbf{s} \cosh^2a/2+ 2\mathbf{r}s_\perp \cosh a/2\sinh a/2 - (s_\parallel-s_\perp) \sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2+ \mathbf{r}s_\perp \sinh a - (s_\parallel-s_\perp) \sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2+ \mathbf{r}s_\perp \sinh a - (s_\parallel-s_\perp+2s_\perp-2s_\perp) \sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2+ \mathbf{r}s_\perp \sinh a - (s_\parallel+s_\perp-2s_\perp) \sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2+ \mathbf{r}s_\perp \sinh a - (\mathbf{s}-2s_\perp) \sinh^2a/2\\
&=\mathbf{s} \cosh^2a/2-\mathbf{s}\sinh^2a/2+ \mathbf{r}s_\perp \sinh a +2s_\perp \sinh^2a/2\\
&=\mathbf{s} + \mathbf{r}s_\perp \sinh a +2s_\perp \sinh^2a/2\\
&=\mathbf{s} + \mathbf{r}s_\perp \sinh a +s_\perp (\cosh a -1)\\
&=s_\parallel + ( \cosh a + \mathbf{r} \sinh a )s_\perp
\end{align}
$$
Is this a Lorentz boost? How do I show that it is?
|
You can write $\mathbf s$ as a sum of parallel and perpendicular parts which commute and anticommute respectively with $\mathbf r$. Then you have $\mathbf r\mathbf s-\mathbf s\mathbf r = 2\mathbf r\mathbf s_\perp$ and $\mathbf r\mathbf s\mathbf r = s_\|-s_\perp$ and I think you'll get the Lorentz transformation.
If you're interested in this then I'd encourage you to work with Clifford algebras instead of the special-case algebras. You can think of a Clifford algebra as having one imaginary unit for each element of an orthonormal basis of the space; the units square to $\pm1$ depending on the signature, and always anticommute with each other. Vectors represent reflections, and therefore products of even numbers of vectors represent rotations. All of the special-case algebras that were introduced historically are even subalgebras of a Clifford algebra. For example, $a+b\mathbf x\mathbf y$ is the complex numbers, $a+b\mathbf x\mathbf t$ is the split-complex numbers, etc. Clifford algebras also have a natural notation for the vectors that the rotations act on, which the special-case algebras don't.
|
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|
Rewriting the Laplacian on a curved manifold I guess there is a sense in which the following is true:
"The Laplacian written on a Riemannian manifold $(M,g)$ can be seen as adding a coordinate dependent mass field to the Laplacian on Euclidean space."
*
*Can someone kindly refer me to a place where this is exactly derived?
(or feel free to write in the derivation here if its not too long!)
Just so that we are on the same page :
For ``nice" real valued functions $f$ on $(M,g)$ we have for the square of the gradient of $f$,
$\Vert {\nabla_g f} \Vert ^2 = g(\nabla_g f,\nabla_g f) = \sum_{j=1}^n \sum_{i=1}^n g^{ij} \partial_i f \partial_j f$
and the Laplacian of $f$ being,
$\nabla_g^2 f := \frac{1}{\sqrt{\det(g)}} \sum_{i,j=1}^n \frac{\partial }{\partial x_i} \left ( \sqrt{\det(g)} g^{ij} \frac{f}{\partial x_j}\right )$
where we define the metric as $g = [g_{ij}] = g \left ( \partial_{x_i}, \partial_{x_j} \right )$ and $g^{-1} = [g^{ij}]$.
|
*
*Recall that the quantum Hamiltonian$^1$ $\hat{H}=-\frac{\hbar^2}{2}\Delta$ for a point particle is (in the Schrödinger representation) associated with the Laplacian $\Delta$.
*The corresponding classical Hamiltonian is $H=\frac{1}{2}\sum_{i,j=1}^np_i g^{ij} p_j$.
*In flat space the Hamiltonian for a free particles is $H=\sum_{i=1}^n\frac{p_ip_i}{2m_i}$.
*Comparing 2 & 3 we see that the inverse metric $g^{ij}$ plays the role of a position-dependent inverse mass matrix ${\rm diag}(1/m_1,\ldots, 1/m_n)$, which is essentially the statement quoted by OP.
For more details, see e.g. this related Phys.SE post.
--
$^1$ We ignore for simplicity potential terms $V$.
|
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Why rotating objects stops? Sorry for asking this simple question, but really I couldn't find a good document discuss what I need exactly.
I am implementing a flight simulation, but my question is related to physics rather than aerodynamics so I find to ask the question to physics experts.
Suppose that I am having a cuboid (Simple form of the plane) with the following dimension:
Length: 14.8m
Height: 4.8m
Depth: 10.0m
The coordinate system is X is right, y is Up and z depth (inside the paper).
I applied a torque on the Y axis, the rectangle begins to gain angular velocity and it rotates in the XZ plane around its center of gravity.
Every thing works fine for now, but after I remove the torque the cuboid should stops i.e. the angular velocity should be decreased till reaches zero. How this happens?
I think this should be due to the moment of inertia, as I am using 3D coordinate system the inertia should be inertia tensor and the right way to calculate the inertia tensor from these dimensions.
If what I thought is right so I need the equation for how the inertia tensor is affecting the angular velocity till the angular velocity reaches zero.
If I am not right, what is the force that affect the cuboid to stop rotating?
|
The answer to your question is that in real life any time an object moves in air there are surface forces developing due to the boundary layer of air.
The aerodynamics of rotating objects are very complex (see the magnus effect for example), but the end result is that there is net torque applied opposing the rotational motion, as well as translational forces (lift/drag etc) due to the motion.
Consider a rotating bar, and resolve the velocity $\vec{v} = \vec{\omega} \times \vec{r}$ of the object (relative to the air) into two components, $v_n$ for normal velocity and $v_t$ for tangential velocity.
The two opposing forces act on that surface element $F_n$ being the pressure drag, and $F_t$ being the surface friction. They are not proportional to each other since the latter depends on air viscosity and the first on density.
Add up all the combined effects all around the body to get an idea of what the net forces and torques are.
|
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|
When did the Big Bang happen? Did the Big Bang happen at a point? goes through the fact that the Big Bang happened everywhere at the same time. John Rennie's answer explains this as being a consequence of all points in space beings squished into a single point, so technically the Big Bang happened everywhere. But, when we talk about relativity, we also use a fourth dimension of time. So, at the Big Bang all points in space, as well as all points in time should be squished (for lack of better terms) into a single point. So did the Big Bang happen at happen 'every-time'?
Edit: The age of the universe goes into how the true age of the universe can be determined as time flows in different ways for different observers. I am asking whether the Big Bang happened at all points in time together.
|
The real answer is that the big bang singularity isn't part of modern cosmology, in part because of the horizon problem. The big bang model is only valid back to an early era when the scale factor was nonzero; before that, something else happened. Various inflationary models are the most popular, but there are others. What these models have in common is that the state they start with isn't the homogeneous, isotropic universe of big bang cosmology (because that homogeneous, isotropic state is what they aim to explain), so big bang cosmology can't say anything about any singularities that they may have.
If we pretend that's not the case, then the answer is that the scale factor only scales the spatial dimensions, not the time dimension, so it happens everywhere but not everywhen.
|
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|
Is the cosmic event horizon viewed from Andromeda different from the one viewed from Earth? If the most distant galaxies we see in deep space are in proximity of our cosmic event horizon does it mean that if we were on Andromeda galaxy and looking in the same direction we would see objects that are beyond the Earth cosmic event horizon?
|
If you're talking about the observable universe as bordered by the cosmic microwave background (which is not a true event horizon, just an opaque surface), then yes, the portion of the universe observable from Andromeda is slightly shifted relative to ours. They'd see different patterns in the cosmic microwave background. They might just barely be able to see a few extra galaxies.
If you're talking about the theoretical particle horizon (what we'd see if the CMBR and everything beyond it was transparent), then we don't know. It depends on what happened in the early universe.
If you're talking about the future cosmological horizon (the infinite-time limit of the particle horizon), that depends on the limit of your position as the time goes to infinity, not your position at any finite time. If someone from Andromeda and someone from the Milky Way stayed within a few billion light years of each other until the end of time, then they'd have the same future cosmological horizon.
|
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Frequency and intensity in photoelectric effect In the explanation of photoelectric effect it is written that intensity and frequency of radiation have different results i.e. higher intensity means greater number of emitted photoelectrons and higher frequency means greater kinetic energy of emitted photoelectrons.
But what we have learned about intensity is that it is the energy released per unit area per unit time.
So if we say about the intensity of light in terms of photons it would look like this :
I = nhf/area/time.
If we take area to be 1 unit and time to be 1 second then the equation will look like this :
I = nhf
The equations above show that with fixed intensity if frequency is increased then number of photons or say photoelectrons decreases and vice versa. But according to Einstein's theory number of photoelectrons is affected only by the intensity of light not its frequency.
What's the cause of this contradiction ?? Is it okay to define intensity only as number of photons ??
Am I wrong somewhere ?
I am just a beginner so please cooperate.
|
Whatever you have said is right but only one thing wrong. "But according to Einstein's theory number of photoelectrons is affected only by the intensity of light not its frequency." That's is not in Einstein's theory. You can check it in any book. It is written as "The photocurrent increases if the intensity of
the incident light is increased". Einstein have not said anything about when the intensity is fixed but the frequency is changed.
|
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|
Is my understanding of how a force is applied through a string correct? Let there be a situation where a force 'F' is acting on one end of an in extensible string which is connected to a box named 'A' resting on a friction less horizontal surface through the other end.
The Force F is transmitted through the string and acts on box 'A'. As a reaction to this box 'A' pulls the string toward itself with force 'F'.
In the FBDs we can see that the string is balanced by the forces but the box is not, so the box starts accelerating towards the right. As the box starts to accelerate, for a very small moment there is slack in the string which makes the tension force 0 for a brief moment.Then the force F being applied on the string is used to make the string taut again and tension forces again begin to accelerate box A.
This keeps repeating itself.
Is this how it works or there something wrong?
|
As the box starts to accelerate, for a very small moment there is slack in the string which makes the tension force 0 for a brief moment.
Wrong.
Tension is zero before the force is applied. Once the force starts acting and the string becomes taut, the block and string gain the same acceleration and the string doesn't slack again.
|
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What colour would neutronium be? Everything we learn about colour in relation to matter is based on "normal" matter that has electrons around it. Absorption and emission of electromagnetic radiation is explained in terms of electrons transitioning between quantum levels with different colours being caused by the energy difference between transitions.
In this thought experiment I have a piece of neutronium at room temperature. As there are no electrons to interact with light of any wavelength, what colour would it be? (You need to do the measurement quickly in the few attoseconds before it explodes).
I can only make 3 guesses but I can't think of any way of deciding which one is the least unlikely.
1 - Transparent.
2 - Perfect mirror
3 - Perfectly black.
My limited physics suggests #1 as being plausible as there are no electrons, no orbitals and therefore no interactions with light. However, a totally non-scientific gut feeling says that transparent is ridiculous. How can something with such insane density have no interaction with light and look like it's essentially invisible? Surely it has to either reflect light perfectly, or absorb it perfectly?
Then again, perhaps there would be the neutron matter equivalent of an absorption spectrum. That at relatively low photon energies (visible light), neutronium would be transparent, but at stupidly high energies (cosmic rays from matter falling into black holes,) it would absorb photons.
There's no great reason for asking, just intellectual curiosity. A mental itch that needs scratching.
|
Neutrons are composed of quarks and quarks do have electric charge and so clearly photons would interact with neutrons. Light interacts with all charged particles and not just electrons. Because of its nature, neutronuim
would behave like a black body and therefore would emit light in the form of black-body radiation. By definition, a black-body is "black" and so you would probably be right with answer "3. perfectly black".
|
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Reason for peaks in graph of binding energy per nucleon A similar question was asked before, but it asked for a different thing. My question here is: What is the reason for spikes in this graph? The graph initially has spikes and then shows a constant decrease. Is it related to something called magic numbers as it is seen in multiple of 4?
|
There are two effects that lead to the presence of the small jagged peaks and valleys in the binding energy per nucleon. (The main shape of the curve is given by the semi-empirical mass formula, derived from the liquid drop model of the nucleus. The model has a positive binding energy proportional to the number of adjacent nucleon-nucleon pairs, a Coulomb repulsion term related to the square of of the number of protons, and a term related to the proton-neutron imbalance.) On top of this, producing the zigzags, there are pairing effects and magic number effects.
The pairing effects come from the fact that the bound nucleons have slightly lower energies when they are correlated in in proton-proton or neutron-neutron pairs. That tends to make the binding energy per nucleon for an odd-odd nucleus like $^{10}$B less than the odd-even $^{11}$B. Or $^{17}$O is less tightly bound than $^{16}$O and $^{18}$O on either side of it, since $^{16}$O and $^{18}$O are both even-even.
The other effect is due to the presence of magic numbers, which are related to filled nuclear orbitals. Just like atomic electrons are most stable when they form a filled outer shell, nuclei are most stable when the protons and/or neutrons fill certain nuclear shells. For example, $^{4}$He is much more tightly bound than $^{3}$H or $^{3}$He, since the $^{4}$He has two protons and two neutrons, with each pair filling a $1$S shell. Another (double) magic nucleus is $^{16}$O, with the eight protons and neutrons each filling the $1$S and $1$P shells. (The notation for shells differs a bit here from that used with electrons. Nuclear shells are denoted by $n$S, $n$P, etc., where $n$ starts separately at one for each value of the angular momentum.) The $^{18}$O nucleus has to have its two extra neutrons shunted into the higher-energy $2$S shell, lowering the binding energy per nucleon. Another magic number occurs at 10, which is why $^{20}$Ca is especially stable; the ten protons and ten neutrons fill up the $1$S, $1$P, and $2$S shells. (The pattern of magic numbers gets a bit more complicated than this, because of the strong spin-orbit coupling in the nucleus, but this is a reasonable picture of the general behavior.)
|
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|
Energy Conservation in Rolling without Slipping Scenario
A solid ball with mass $M$ and radius $R$ is placed on a table and given a sharp impulse so that its center of mass initially moves with velocity $v_o$, with no rolling. The ball has a friction coefficient (both kinetic and static) $μ$ with the table. How far does the ball travel before it starts rolling without slipping?
The solution I found starts by setting up a conservation of energy and setting $v = rw$:
$$ \ \frac{1}{2}m v_o^2 = \frac{1}{2}m v^2 + \frac{1}{2}Iw^2 \to v_o^2 = \frac{7}{5}v^2 \quad{(1)}$$
It then goes on to say $W = \Delta K_{rotation}$ and solves for $D$ :
$$ \ \int_{0}^{D} F_{f} dx = μmgD= \frac{1}{2}Iw^2\quad{(2)}$$
There are a couple of things I do not understand about this approach. How does $(1)$ account for the loss of energy due to the friction force which is causing the rotation and the slipping that occurs before it starts rolling purely? Second, how does $(2)$ account for the change in center of mass velocity? Wouldn't $W = \Delta K_{rotation} + \Delta K_{transitional}$ ? I am most likely misunderstanding something and help is greatly appreciated.
|
I am not sure about this solution. I would set up the equations of motion as follows.
Translational motion:
there is only one force acting on the system, which is dynamic friction of modulus $F_d=\mu mg$. The motion is uniformly decelerated with acceleration $a = \mu g$. The (horizontal component of the) translational velocity then follows the equation
$$
v(t) = v_0 - at
$$
Rotational motion
The dynamical friction acts with a torque of modulus $\mu mg R$ on the system (taking the center of the ball as a pole), then the angular acceleration is given by the rotation dynamic equation $I \alpha = \mu mgR$, which gives $\alpha = 2\mu g/(5R)$. The angular velocity $\omega$ as a function of time reads
$$
\omega(t) = \alpha t
$$
Pure rolling condition is obtained setting $v(t) = \omega(t) R$, which gives you the time needed to reach pure rolling, which is $t = v_0/(a+\alpha R)$. Now you can use this time to compute the distance with the kinematics law of the translation $D = v_0 t - (1/2)a t^2$.
I think you should account for friction in the energy balance of the problem, so $W=\Delta K_{trasl}+\Delta K_{rot}$, but then you would have two unknown variables: $D$ and the final velocity, so you should use kinematics as stated above in this case.
|
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|
Inconsistency in applying work-energy theorem in the classical problem of a kink on an inclined wedge Consider the typical problems of mechanics where we had to find the velocity for some mass which reaches bottom of a wedge after metting some changes in the wedge angles (kinks). The following is a particluar type of problem.
The ball starts sliding from the top of the incline A and reaches the ground encountering a kink at C. We have to find the velocity of the mass as it reaches the point B somewhere on the ground. Assume every surface the ball encounters is frictionless and ignore the rotational motion of the ball.
Edit: Assume that the collision of the ball with the ground at C is perfectly inelastic.
We used energy conservation from A to C and momentum conservation along the horizontal at C to state that the velocity of the ball as it reaches the ground is $\sqrt{2gh}\space cos(\theta)$.
But, work energy theorem implies that the work done by all forces on the ball must be equal to the change in the kinetic energy of the particle. What work does the impulsive normal (The kink at C) do on the ball? There being no displacement at C (C being a point), shouldn't the work done by the impulsive normal be zero? If so, shouldn't the velocity of the ball at point B be $\sqrt{2gh}$?
I am sure that the velocity at B has to be $\sqrt{2gh}\space cos(\theta)$. But, I am unable to find out why is such inconsistency? I heard Work-energy theorem is universally applicable. Any insights to such problems will be highly be appreciated.
I am new to this site, so please ask for clarifications, in case of any discrepancies.
|
The work energy theorem is alright. The problem is that you are not taking $N_x$ into account in momentum conservation
|
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|
Lorentz Force on a Current Carrying Wire Does the Lorentz Force on a Current Carrying wire given by the equation
$$\mathbf{F} = I \int \text{d}\ell \times \mathbf{B}$$
constitute an action reaction pair? That is, if i have two arbitrarily shaped current carrying wires, is it true that force on any one of them due to the magnetic field of the other is equal to the force on the other due to the magnetic field of the first?
|
Newton's third law in modern terms states conservation of momentum. Electrostatic forced conserve $P_{kin}=\sum_i m_i p_i$ but magnetic forces do not , as argued by @Rohit. In the presence of electromagnetic fields the conserved momentum is $P = P_{kin} + P_{pot}$, where $P_{pot} = - \sum_i q_i \vec A_i$.
|
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|
Elastic moduli interrelations for homogeneous isotropic materials The classical equation for the Young modulus in elasticity theory for a homogeneous isotropic material in one-dimension is commonly given in the formulation
$$ E = \frac{\sigma}{\epsilon} \quad,$$
with $\sigma$ as the uniaxial stress, and $\epsilon$ as the dimensionless strain parameter.
However, then I also discovered that $E$ can be rewritten in a very different form in terms of the Lamé constants $\mu$ and $\lambda$,
$$ E = \frac{\mu \left(3\lambda + 2\mu\right)}{\lambda + \mu} $$ (vide, e.g. wikipedia).
However, in an unpublished manuscript on stellar astrophysics, a variant of the bulk modulus $K$, which I know now decribes the volumetric elasticity, i.e. elasticity in three-dimensions, is simply stated without derivation as
\begin{equation}
K = \rho \cdot\frac{\partial P}{\partial \rho} \quad,
\end{equation}
with $\rho$ as mass density and $P$ as pressure.
I was wondering how the equation for $K$ above, for a spherical, isotropic and homogenous body can be derived and if it can be directly linked or expressed in terms of the one-dimensional Young modulus $E$?
|
The Bulk modulus in terms of the Lame parameters is
$$
\kappa =\lambda +\frac 23 \mu.
$$
It is not simply proportional to Youngs modulus. In stretching a wire we do not only make it longer, but we also allow its thickness $W$ to change. Young's modulus therefore needs to know Poisson's ratio, which is defined by
$$
\frac{dW}{W}= -\sigma_{\rm Poisson} \frac{dL}{L}
$$
and is given by
$$
\sigma_{\rm Poisson}= \frac 12 \frac{\lambda}{\lambda+\mu}.
$$
The bulk modulus would be relevent to stretching, only if we prevent the wire from getting thinner.
|
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