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Interacting CFT fixed point of an RG flow Suppose we have a gauge theory defined in the UV and it flows to an interacting CFT in the IR, i.e. the beta function vanishes for some finite value of the coupling. I am confused about the meaning of this. Isn't a CFT by definition scale independent? But we have the CFT only at a particular energy scale where the beta function vanishes. So what do we mean when we say that at some energy scale the theory does not depend on the energy scale?
|
Your confusion arises from the following fact:
In the situation that you describe, the beta function vanishes at some finite value of the coupling. But it does not vanish at a finite value of the energy!
When you go down in energy from the UV to the IR, the coupling increases from zero to its fixed point value. But it never exactly reaches the fixed-point value. The closer you get to the fixed point, the slower the coupling runs with energy (that's the definition of a beta function approaching zero). You only get to the IR interacting CFT in the strict limit of zero energy.
As you correctly pointed out, you can never have a CFT at a given energy scale. A CFT is by definition scale-free.
|
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Fastest numerical method to solve Lindblad Master Equation? The Lindblad Master Equation is a generalization of the Schrodinger Equation for open quantum systems, given by
$$
\frac{\mathrm{d} \rho}{\mathrm{d}t} = -i \left[ H, \rho\right] + \sum_k \gamma_k \left( L_k \rho L_k^\dagger - \frac{1}{2} \left\{ L_k^\dagger L_k, \rho \right\}\right) = \mathcal{L}(\rho)
$$
where $\rho$ is the density matrix describing the state of the system, $H$ is some Hamiltonian, and $L_i$ are the jump operators. All of these objects are of size $N \times N$ where $N$ is the Hilbert size of the system.
Question: Assuming that everything is time-independent (constant Hamiltonian and jump operators), what would be the fastest numerical method to solve this equation? More specifically, starting from a known initial density matrix $\rho_0$, how can I get the density matrix at time $t_f$, $\rho(t_f)$?
Starting point: I already know of two possible methods.
(1) Computing the complete Lindbladian propagator, i.e. $\Lambda(t_f) = \exp(t_f \mathcal{L})$ and then doing the simple superoperator-matrix product $\rho_f = \Lambda(t_f) \rho_0$. However, $\mathcal{L}$ is a superoperator of size $N \times N \times N\times N$, so computing its exponential does not seem numerically very fast.
(2) Making some Kraus operator propagation, i.e.
$$
\rho(t+\mathrm{d}t) = \sum_\nu M_\nu \rho(t) M_\nu^\dagger
$$
where $M_\nu$ are the Kraus operators, given by
$$
M_0 = I - \mathrm{d}t \left(i H + \frac{1}{2} \sum_k L_k^\dagger L_k\right) \quad \text{and} \quad M_\nu = \sqrt{\mathrm{d}t} L_\nu
$$
which is a trace preserving method up to second order in $\mathrm{d}t$. The good thing about this is that it only requires matrix-matrix products. However, we do need to take sufficiently many time steps, and thus perform quite a lot of matrix-matrix products.
Is there any known fast method for such problems?
|
You can use a Python package called QuTiP which will allow you to solve the LME. It also has different numerical methods to solve your problem, you can take a look at them and compare. Here is some documentation (http://qutip.org/docs/3.1.0/guide/dynamics/dynamics-master.html) There are also plenty of examples to aid you in using the package.
|
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How to solve this problem only using the kinematics of rotational motion? This is a problem from my introductory physics textbook:
A wheel of moment of inertia $I$ and radius $r$ is free to rotate about its centre. A string is wrapped over its rim and a block of mass m is attached to the free end of the string. The system is released from rest. Find the speed of the block as it descends through a height $h$.
The answer in my book utilizes energy considerations, reasoning that "the gravitational potential energy lost by the block must be equal to the kinetic energy gained by the block and the wheel." Thus,
$$mgh=\frac{mv^2}{2}+\frac{Iv^2}{2r^2}$$
$$\rightarrow v=\sqrt{\frac{2mgh}{m+I/r^2}}$$
My question is, is it possible to solve this question using only the equations of kinematics of rotational motion, viz. $$\omega = \omega_0+\alpha t$$ $$\Delta \theta=\omega_0t+1/2\alpha t^2$$ $$\omega^2=\omega_0^2+2\alpha \Delta \theta$$
As an analogy, consider the case when the wheel was massless. Then, the equation from energy considerations would have been: $$mgh=\frac{mv^2}{2}$$
Solving the above, we get, $$v=\sqrt{2gh}$$
When we use the equation $v^2=u^2+2gh$, setting $u=0$ for the system starting from rest, we again get, $v=\sqrt{2gh}$.
|
To write a more complete answer for other readers: I would use Newton's second law and its rotational equivalent. Use $F= ma$ for the block of mass $m$ (since there is only translation, and no rotation), and then use $\tau = I \alpha$ for the wheel (since there is only rotation about its center, and no translation).
The tension force exerted on the string as pointed out by Bernhard will be $T$, and this force acts on the mass (such that $\Sigma F_{\rm mass} = mg -T$) and also provides a torque on the wheel about its center, such that $\Sigma \tau = RT$. Since there is no slipping, use the fact that $a_{\rm mass} = R \alpha_{\rm wheel}$. System of equations, solve for $\alpha$.
Then, relate the distance travelled for the bloc, $h$, and the angular displacement by knowing that $h=R\theta$. After that you can use the $\theta = \tfrac{1}{2}\alpha t^2$ equation.
|
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What does it mean to treat space and time on equal footing? I often read from textbooks that in relativity, space and time are treated on an equal
footing. What do authors mean when they say this?
Are there any examples that show space and time are treated on an equal footing? Conversely, what examples show that space and time are not treated on an equal footing?
|
After some thought, this is what I understand:
In Newtonian physics, a particle's path can be specified by $x^i(t)$ where the time $t$ can be seen as an independent parameter. The space coordinates $x^i(t)$ are dependent variables that depend on $t$. We thus say that space and time are not treated on an equal footing.
In relativity, a particle's worldline is specified by $x^\mu(\lambda)$ where $\lambda$ is an independent parameter (often taken as the particle's proper time). Both space and time coordinates $x^\mu(t)$ are dependent variables that depend on $\lambda$. We thus say space and time are treated on an equal footing.
|
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Is the torque on a gyroscope a conventional virtual force perpendicular to rotation? So I am having some difficulty understanding gyroscopic precession. I understand that mathematically by convention torque is perpendicular to the force and so is angular momentum but surely that force is a true force acting outwards as this is what occurs in gyroscopic precession. My question is is this torque a conventional virtual force perpendicular to rotation?
|
This is a good question. Gyroscopic precession is also what has baffled me the most of all classical mechanics I've encountered.
The force comes from the inertia of the spinning mass. Gravity tries to make the gyroscope (the top) tilt and fall straight down. But while falling down it also spins. The particles at the lower part of the periphery thus experience falling sideways. As they all have this tendency, they collectively turn and the gyroscope as a whole starts turning in a horizontal plane.
In the next moment this exact same thing happens. And just like with circular motion, the turning takes place as infinitesimal changes while the system simultaneously adjust, so that there is no magnitude change but only a direction change. In that same way the gyroscope doesn't fall down (no angular displacement change) but only turns.
Source
So no, no involved torques are virtual here. All are real forces and torques. But they appear in an unintuitive manner, just like how unintuitive classical circular motion is where a centripetal force pulls inwards but still the object never comes nearer the centre.
Vsauce has a quite good explanation here (from 6:25): https://youtu.be/XHGKIzCcVa0?t=385.
|
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Two Cylinders on Ramp Suppose I have two cylinders: a light one and a heavy one. Now, I let the cylinders roll down a ramp without slipping. My question is, which one will get to the bottom of the ramp first, and why?
|
Assuming that the cylenders are identical in appearence and just made of material of different densities. Then the torque about the COM will be unequal but the acceleration would be equal.
Instead ofequation just percieve it as
$$\tau(torque) \propto Mass$$ as gravitational torque and other parameters are equal
$$ \tau(torque)=I\alpha$$
$$I \propto Mass$$
So you get $\alpha$ independant of mass.
Now as these are equal for both cylenders.
So any kinematical or rotational calculation you would do should be equal for both.
If the cylenders are not identical then please elaborate your question.
|
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How could I see the distance light traveled from an airplane? It was nighttime. I was flying on an airplane. As we were landing we passed over a highway. I saw cars below with their headlights on. I could see that the light from their headlights only lit up a certain distance in front of them. Any object that fell past that distance would have probably looked dark to the driver.
I'm confused as to how I was able to see this thousands of feet up in an airplane. The light from the cars' headlights only traveled a few feet in front of the cars but also traveled thousands of feet to the airplane I was on?
How is this possible?
Also, how could I see the beam-light structure of the light, if the light is able to be seen by an observer in any direction? I feel like none of this makes any sense and we were never taught anything to clarify this in E&M. Maybe I just missed something truly important.
|
we were never taught anything to clarify this in E&M.
The basic is that our eyes see objects from light reflected from the objects.
Light coming from a source, the headlights of a car are a source, after a certain distance opens up spherically , which means the energy of the beam is dispersed in a wider and wider volume . We see the lighted objects by the reflection of light from them. At a distance where the beam energy per cubic centimeter is very low, there is not enough light reflected by objects, to reach our eyes and be registered.
What you draw is the extent of the beam in reflecting from objects on the ground, as far as the energy density is large enough to reflect from objects (in this case the road and other cars). The beam itself that you observe is the reflected light from the air and dust in front of the car and mainly the tarmac, on which part of the light shines.
|
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Why does particle measurement cause quantum wavefunctions to collapse When we attempt to measure a certain property of a particle, how and why does its wave function collapse? I've tried to find answers on my own, but they've been far too complicated for me to comprehend. Would appreciate any answer with limited complex jargon, and more simplistic explanation, if possible.
|
A simple way to look at it is to remember that the wavefunction describing a particle or a system of particles is a mathematical function, a solution of a specific quantum mechanical wave equation, with specific boundary conditions . When a measurement is made, an interaction with the system happens that changes the boundary conditions. This means that a new mathematical function will describe the particle or the system of particles after the measurement, and this is what is called "collapse", a misleading term. A wavefunction is not a balloon.
|
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Picking a direction for currents when doing nodal analysis involving capacitors Let's assume that I was given the following circuit:
Where the capacitor is fully charged at 6V, and a switch (not shown) will be closed at time t=0. From this, let's say that I pick the following directions of my currents in order to do nodal analysis (assume ground is on the middle bottom node):
My simplified KCL equations would end up being:
$I_{capacitor}(t) = -\frac{1}{4k} V_c(t)$
Since the capacitor is discharging, the equation becomes:
$ -C \frac{dV_c(t)}{dt} = -\frac{1}{4k} V_c(t) $
Solving this differential equation yields:
$ V_c(t) = 6e^{2.5t} $
This is clearly wrong, as this solution diverges.
However, if I reverse the directions of the resistor currents to this:
Then my simplified DE equation becomes:
$ -C \frac{dV_c(t)}{dt} = \frac{1}{4k} V_c(t) $
Which results in the solution:
$ V_c(t) = 6e^{-2.5t} $
Which is the correct solution.
So how do I know which direction to make my currents before I even solve the differential equation?
|
You don’t change the sign because the capacitor is discharging. The fundamental equation presupposes the current flowing into the capacitor (as you have it in your drawing). The negative sign will take care of itself in the solution (discharging).
$$ C \frac{dV_c(t)}{dt} = -\frac{1}{4k} V_c(t) $$
The fundamental capacitor equation being,
$$i(t)=C\frac{dv}{dt}$$
If you want to define your current as leaving the capacitor then you need to include a negative sign in your dv/dt term.
|
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Kinetic energy and curvature In quantum mechanics, the kinetic energy of a particle described by the wave function $\psi$, is related to the curvature of the $\psi$. This is easily seen, but I have confused my self with the negative sign. That is: $\hat{T} = -\frac{\hbar^2}{2m}\nabla^2$, is the kinetic energy operator. So what I gather is, that the greater the curvature of $\psi$, the lower the kinetic energy, due to the minus sign. I know this can not be right.
|
The "curvature" is a local property of the wave function, but there is no concept in standard QM as "the local value of the kinetic energy" (see e.g. this).
The kinetic energy is one of the eigenvalues of the $T=p\cdot p$ operator (neglecting the factor $1/2$ and setting $\hbar=m=1$).
To understand the minus sign in $T=-\nabla^2$, you can use the set of eigenstates of the momentum operator $p$ (i.e. the plane waves, which are also eigenstates of $T$). By following the reasoning of AccidentalTaylorExpansion, you discover that the minus is needed to ensure the positivity of the kinetic energy eigenvalues:
$$
T e^{i k\cdot x} = -\nabla^2 e^{i k\cdot x} = |k|^2 e^{i k\cdot x} \, ,
$$
so you see that the generic eigenvalue of $T$, namely $|k|^2$, is positive.
Consider now the more general case in which $\psi$ is not a plane wane, i.e. it is not an eigenstate of $T$. In this case, the only thing you can do is to find the average kinetic energy $\langle T \rangle$ on such a state $\psi$ via
$$
\langle T \rangle = \int d^3x \, \psi^*(x) T \psi(x) = -\int d^3x \, \psi^*(x) \nabla^2 \psi(x)
$$
You can do an integration "by parts", assuming that $\psi\rightarrow0$ at spatial infinity, and check that $\langle T \rangle $ is always positive:
$$
\langle T \rangle = -\int d^3x \, \psi^*(x) \nabla^2 \psi(x)
=\int d^3x \, \nabla\psi^*(x)\cdot \nabla \psi(x)
=\int d^3x \, |\nabla \psi(x)|^2 >0
$$
Moreover: just notice that the kinetic energy is $T = p\cdot p$ and that $p =- i \nabla$, so it is clear (at least formally) that $T$ should have a minus sign.
|
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Why is the tension on both sides of an Atwood machine identical?
The field forces $F_{g1}$ and $F_{g2}$ push down on Block 1 and Block 2, respectivley, where
$$F_{g1}=m_1g$$$$F_{g2}=m_2g$$
Since the pully system reverses the direction of each force, wouldn't the following be true?
$$T_1 = F_{g2} = m_2g$$$$T_2 = F_{g1} = m_1g$$
And since $m_1 \neq m_2$, wouldn't $T_1 \neq T_2$?
My textbook states that tension is the same throughout the whole string, but I can't wrap my head around why this is so. If $m_1 \neq m_2$, how could the same force $T$ accelerate both of them an equal amount? Wouldn't Block 2 require a greater force?
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Its because the pulley in an atwood's machine is an Ideal pulley. It has no mass and its frictionless. This means that the rope is only going to slip over the pulley freely without rotating it at all. In that case the rope is completely isolated from the pulley and tension should be uniform throughout.
Note that frictionless means that no friction between rope and pulley
Had it been a rolling pulley (one with friction along rim so that rope does not slip), the tension on both ends would be different due to friction. and it would be that difference that help it turn.
Everywhere in mechanics a frictionless pulley implies that its a "slipping" pulley and not a rolling pulley (as we see in our daily life). These frictionless pulleys are kept only to change the direction of pull keeping the tension in the string same.
|
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Can coldness be converted to heat energy? We know that the heat can be converted into heat energy with the help of thermoelectric generators, but why can't we generate energy from coldness?
Like the temperature of the universe in 1 K, can this be used in the near future to be used as an energy resource for probes or satellites?
Here is the link to the article that made me think about this. Somewhere in the middle it is written that scientists can harness the cold energy using some active input method.
I think this article is poorly written.
|
There's a more recent scientific study about this topic, which was then covered in articles like this and this. I also asked a similar question a few months ago.
Basically, this is called "thermoradiative photovoltaics" and involves generating energy by emitting heat (as infrared rays) to a heat sink. The proposed technology would use the Earth as a heat source and the night sky as the heat sink. In a sense it's the opposite of traditional solar panels, which is why some have referred to it as "anti-solar panels".
|
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Spin part of the angular momentum from the Lagrangian For fermions of spin $1/2$ the angular momentum has following form:
$$
\mathcal{J}_z = \int d^{3}x \ \psi^{\dagger} (x) \left[i(- x \partial_y + y \partial_x) + i\sigma^{xy} \right] \psi(x)
$$
Here the first term is orbital part and the latter one is the spin part of angular momentum.
However, in the general prescription for derivation of the angular momentum:
$$
\mathcal{J}^{ij} = \int d^{3} x (x^i T^{0 j} - x^j T^{0 i}).
$$
I cannot see, where the spin part can actually arise.
For example for the photon field, where:
$$
T^{\mu \nu} = F^{\mu}_{\alpha} F^{\mu \alpha} - \frac{1}{4} g^{\mu \nu} F_{\alpha \beta} F^{\alpha \beta} .
$$
This procedure seems to provide only the orbital part of the angular momentum, and no the spin. Or it is implicitly included in this expression?
For the spin $1/2$ field the term $\sigma^{xy}$ emerges, when one considers corrected energy tensor https://en.wikipedia.org/wiki/Belinfante%E2%80%93Rosenfeld_stress%E2%80%93energy_tensor. (formula 6 in https://arxiv.org/abs/1508.06349).
However, for the spin-1 theory this expression incorporates everything and is the most symmetric one.
I would strongly appreciate any help and comments!
|
Your observation is correct. The total angular momentum, integrated over volume, contains both contributions as pointed out in the other answer. However it takes the overall for of an orbital, position dependent, angular momentum while clearly the electromagnetic potential has internal AM as well.
My answer is written down in a peer reviewed, published paper, https://arxiv.org/abs/physics/0106078. Based on the so called Fermi lagrangian I derive a theory in which spin and angular momentum are separate observables.
|
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Why does ponytail-style hair oscillate horizontally, but not vertically when jogging? Many people with long hair tie their hair to ponytail-style:
Closely observing the movement of their hair when they are running, I have noticed that the ponytail oscillates only horizontally, that is, in "left-right direction". Never I have seen movement in vertical "up-down" direction or the third direction (away-and-back from the jogger's back). Why is the horizontal direction the only oscillation?
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This phenomenon happens with ear plugs too. I like to think of it by considering what each step of the run does to the motion of the hair. Hair in a ponytail has a fulcrum where it is attached to the head, so it can freely move like a pendulum in its plane.
When someone takes a step forward with their right foot and pushes off, it pushes the body to the left which then provides an equal an opposite force to the right to the hair higher up the body. I am unsure if this exact force exchange process described is correct.
Small impulses build up to oscillate the hair at its natural frequency (dependent on the material properties of the human hair and head pendulum). If the human drives this pendulum at different driving forces, it will oscillate at different amplitudes and frequencies, where there will be a maximum amplitude at its natural frequency for a particular driving force of the human's run or walk.
|
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Can spacetime be curved even in absence of any source? Einstein's equation in absense of any source (i.e., $T_{ab}=0$) $$R_{ab}-\frac{1}{2}g_{ab}R=0$$ has the solution $$R_{ab}=0.$$
But I think $R_{ab}=0$ does not imply that all components of the Riemann-Christoffel curvature tensor $R^c_{dab}$ be zero (or does it?). From this can I conclude that spacetime can be curved even in absence of any source?
|
This is a simple answer:
I would view this in the same light as the following question:
Does
$$ {\bf \nabla \cdot E} = \frac{\rho}{\epsilon_0} $$
imply zero electric field in region with no charge density?
To which the answer is clearly, "No".
And as an example: The astronauts on the moon. They were there in a pretty good vacuum dropping feathers and hammers, which then took off on like geodesics.
|
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Are we using a hybrid coordinate system to define four velocity? Ordinarily in Newtonian physics, velocity is defined as $${\vec v}=\frac{d{\vec x}(t)}{dt}$$ where we use the coordinates of an observer and the universal time $t$.
When we dive into special relativity, we define the spatial components of the four-velocity as $$v^i=\frac{dx^i(\tau)}{d\tau}$$ instead of $$v^i=\frac{dx^i}{dt}.$$
*
*But in the usual definition (second equation), are we not using a hybrid system of spacetime coordinates? I mean, while $x^i(\tau)$ are the coordinates measured by the stationary observer, $\tau$ is the proper time of the moving observer.
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In Newtonian mechanics, $t$ is the time for the object that is moving, as it is in SR. The difference is that any inertial frame believes that it is also their own time. It is an approximation that is corrected by SR.
Only after that correction some important quantities are conserved. For example, momentum before and after a collision, as observed by different inertial frames. If defined as $m\frac{d\mathbf x}{dt}$, it is not conserved (we think it is because discrepancy for usual velocities is completely neglible).
|
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Physical meaning of $\frac{π}{8}$ in Poiseuille's equation Recently I read about the Poiseuille's equation which relates the flow rate of a viscous fluid to coefficient of viscosity ($\nu$), pressure per unit length($\frac{P}{l}$) and radius of the tube ($r$) in which the fluid is flowing. The equation is
$$\frac{V}{t}=\frac{πPr^4}{8\nu l},$$
where $V$ denotes volume of the fluid.
One thing which I don't understand in this equation is that why do we have the constant term as $\frac{π}{8}$ . Is there some physical significance of this specific number here or is it just a mathematical convention?
|
The appearance of the irrational $\pi$ comes from the assumption, that the pipe is perfectly circular.
This in itself is of course a nonphysical assumption, so it is not surprising to get a irrational result.
|
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Relation between uncertainty in measurement of mean and standard error on the mean I have a series of individual (time) measurements with a certain uncertainty each per measurement, which is the same for all the measurements (±one frame). I have understood that the uncertainty on the arithmetic mean of these measurements will be equal to that same uncertainty of the measurements (±one frame in this case). I am completely lost on how this relates to the standard error from the mean (I suspect the standard error is an additional error on top on the uncertainty that comes from the underlying parent distribution and true value). How would one combine the uncertainty in the mean from measurement with the standard error, to give a "final" error (±) for the mean?
|
In order to understand what is happening it is helpful to know the true distribution. Thus, I simulated $N=100$ data points from a normal distribution with mean value $\mu=21$ and standard deviation $\sigma=3$. The data looks like this
The red line is the average value, and the blue line is a Gaussian (least square) fit to the data.
Formula: y ~ k * exp(-1/2 * (x - mu)^2/sigma^2)
Parameters:
Estimate Std. Error t value Pr(>|t|)
mu 20.93766 0.35034 59.764 3.18e-16 ***
sigma 3.01581 0.35599 8.472 2.08e-06 ***
k 0.12922 0.01302 9.922 3.90e-07 ***
Note that the center of the Gaussian fit is not equal to the average value $\bar x$, because the fit does not weight the data points linearly.
Next, I repeat the calculation $nLoop = 1000$ times. In each iteration I draw $N=100$ samples from a normal distribution and calculate the average value. Hence, after the loop I obtained $\{\bar x_1, \bar x_2, \ldots, \bar x_{1000}\}$.
Now the key point for your question: The central limit theorem tells us that the average value $\bar{\bar x} = \frac{1}{1000} \sum_{i=1}^{1000} \bar x_i$ is expected to be Gaussian, $N(\mu, \sigma/10)$, which is the red curve
Note that I use the center of the fit as estimator for the mean value and not the average value. Usually, the fit provides a more accurate estimate -- although this it is not true for this particular sample (my first $N=100$ data points). If I include the 1000 sample averages in the plot I get
It is obvious that the standard deviation of $\bar x$ (red) is much smaller than the standard deviation of the original data (blue).
|
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How to transform velocity 4-vectors to Zero Momentum Frame I have a particle $p$ with speed $u$ in lab frame approaching a stationary particle $q$.
The $p^{\mu}$ and $q^{\mu}$ velocity 4-vectors are:
$$p_{LAB}^{\mu}=\gamma_u(c, u, 0, 0)$$
$$q_{LAB}^{\mu}=(c, 0, 0, 0)$$
To get to ZMF, I need a standard lorentz boost with speed $v=u/2$:
$$p_{ZMF}^{\mu}=
\begin{pmatrix}
\gamma_{v} & -\gamma_v \beta_v & 0 & 0\\
-\gamma_{v} \beta_v & \gamma_v & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
\end{pmatrix}\begin{pmatrix}
\gamma_{u}c\\
\gamma_{u}u\\
0\\
0\\
\end{pmatrix}=\gamma_u \gamma_{\frac{u}{2}}
\begin{pmatrix}
c-\frac{u^2}{2c}\\
\frac{u}{2}\\
0\\
0\\
\end{pmatrix}
$$
and
$$q_{ZMF}^{\mu}=
\begin{pmatrix}
\gamma_{v} & -\gamma_v \beta_v & 0 & 0\\
-\gamma_{v} \beta_v & \gamma_v & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
\end{pmatrix}\begin{pmatrix}
c\\
0\\
0\\
0\\
\end{pmatrix}=\gamma_{\frac{u}{2}}
\begin{pmatrix}
c\\
-\frac{u}{2}\\
0\\
0\\
\end{pmatrix}
$$
The magnitude of the first spatial component of $p_{ZMF}^{\mu}$ is a $gamma_u$ times more thanthe first spatial component of $q_{ZMF}^{\mu}$. I would expect that in the ZMF, they are opposite sign but otherwise equal. Is this expectation wrong, and if not, what am I doing wrong?
|
Your intuition is correct, in the zero momentum frame you'd assume both objects to be moving towards you at the same speed. (I'm assuming, of course, that the two objects are identical.) So the first question to ask yourself is "What should this speed be?" The assumption you made ($v=u/2$) is incorrect, as while $q$ will move towards you with a speed $u/2$, $p$ will not be moving with a speed $u/2$, because of the relativistic velocity addition law.
An easy way to figure it out is to actually equate the speeds of $p$ and $q$ in this new frame: if the frame is moving rightwards (say that's the direction of $p$'s motion) with a speed $v$, then it will see:
$$p\text{ moving with velocity }= \frac{u-v}{1-uv/c^2} \quad \text{and $q$ moving with velocity}= -v.$$
Equating the magnitudes it's very easy to see that you get a quadratic equation in $v$ $$\frac{u}{c^2} v^2 - 2v + u = 0,$$ which you can solve for $v$. Interestingly, when the speeds are much smaller than the speed of light (i.e. in the "non-relativistic case) the first term vanishes since $uv/c^2 \to 0$, and you get back the classical result $v = u/2$!
You can solve the above quadratic equation to get two solutions for $v$, I invite you to find out why only one of them is physically acceptable. This is the boost velocity needed to get to the zero-momentum frame.
|
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Applications of representation theory of finite group in physics? Well, I have just finished my study on basic representation theory of finite group from a pure math course. After tortured a lot by abstract constructions, I would like to know the real application of this theory, however, it seems to me not many topics in physics are related to representations of finite group. Anyone could provide some examples on the application?
|
In the study of equilibrium configurations and vibrational properties of molecules, clusters and solids, finite groups representations play a key role to extract all consequences of point symmetries of the average atomic positions.
Also electronic states in molecules, clusters and solids provide a different example of applied representation group theory.
In both cases, the classification of vibrational and electronic states in terms of irreducible representation of the relevant symmetry group allows a complete characterization of such states and it is pivotal for the analysis of experimental data.
|
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|
Finding the metric tensor from a 2D line element A 2D space has coordinates $x^1$ and $x^2$ with line element
$$\mathrm{d} l^{2}=\left(\mathrm{d} x^{1}\right)^{2}+\left(x^{1}\right)^{2}\left(\mathrm{~d} x^{2}\right)^{2}.$$
I'm looking to find the metric tensor and its dual metric. I understand the basics of summing over $g_{ij}$ for $n=2$ and using the Kronecker delta but the second non $dx^1$ term is tripping me up. I don't know how to use it in summing as its not a derivative.
|
Your question is a bit unclear but as I understand it you want to derive the metric tensor of this line element and its inverse. You have to remember that $dl^2 = g_{i j}dx^i dx^j$. So your metric is $\text{diag}(1,(x^1)^2)$. And so its inverse is $\text{diag}(1,(x^1)^{-2})$.
Edit for clarity :
\begin{equation}
dl^2=g_{i j}dx^i dx^j=g_{11}(dx^1)^2+2g_{12}dx^1 dx^2+g_{22}(dx^2)^2
\end{equation}
So one has $g_{11}=1$, $g_{12}=0$ and $g_{22}=(x^1)^2$. Since the metric is diagonal finding its inverse is straightforward : $g^{11}=1$, $g^{12}=g^{21}=0$ and $g^{22}=(x^1)^{-2}$
|
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|
Unruh particles When we accelerate, an event horizon forms behind us resulting in Unruh radiation. In this kind of scenario, the existence of the radiation particles themselves is observer dependent.
My question is: If the existence of the particles is observer dependent, then what do the particles, themselves as observers, observe?
|
A photon is not an observer, and can't be because its proper time is always zero and it has no frame of reference in which it is at rest. However, Unruh radiation is black body radiation, so it does include particles with charge and mass. To get an observer from this, you can get low-probability interactions among these particles that form a Boltzmann brain. This is known as the formation of a Boltzmann brain by nucleation, and it's the scenario commonly envisioned when Boltzmann brains are discussed in the context of the distant future of our universe, which is de Sitter.
What an accelerated observer sees as a detection of a quantum of Unruh radiation, an inertial observer sees as the emission of a quantum by the detector. I don't see why this analysis would change if the inertial observer is a Boltzmann brain. The Boltzmann brain can also say, "I think, therefore I am," but according to an inertial observer the brain never existed and never had that thought.
|
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How come the formula $W=Fd$ doesn't apply for energy stored in springs? I always thought that work is like the energy transferred and it is given by $W=Fd$, but this concept gets problematic for springs.
If the force $F$ is applied to a spring which compresses it by a length $d$, then apparently the energy stored in the spring is $$E_{p}=\frac{1}{2}kd^2=\frac{Fd}{2}$$
Why is the energy transferred to the spring not $Fd$?
|
Because $W = Fd$ only holds for a very special case. The general definition of work
is given via
$$W = \int_\gamma \vec F (\vec r) \cdot \text{d}\vec r$$
where $\gamma$ represents a trajectory in $\mathbb{R}^3$ and $\vec F (\vec r )$ represents a vector field.
The case where $W=Fd$ holds is when $\vec F (\vec r)$ is constant over all space and the trajectory is parallel to the force. For example when pulling a stone of mass $m$ to height $d$ in a straight line along the $z$-axis we get
$$
W = \int_\gamma \vec F (\vec r) \cdot \text{d}\vec r = \int_0^d mg\ \text d z = mgd\ \hat{=}\ Fd.
$$
But the force when pushing or pulling a spring is proportional to how much you pulled it - it depends on the position. So we see when elongating a spring by a length $l$ along the $x$-axis we get
$$W = \int_\gamma \vec F (\vec r) \cdot \text{d}\vec r = \int_0 ^l kx\ \text d x = \frac 1 2 k l^2\ \hat{=}\ \frac{F(l)\,l} 2
$$
Note: The signs depends on the system you look at, so for certain systems you would have a minus sign with the force.
|
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Angular velocity of a particle moving in 3D I have a particle trajectory where particle position is available at discrete time steps with respect of (0,0,0) in 3D. Time step is 0.05 sec. For reference, positions are shown in following image.
Particle velocity can be computed easily. My goal is to calculate angular velocity for each time step. Here are the different ways I have tried:
*
*Velocity vector
For each consecutive time step we can calculate change in direction of velocity vector. let's suppose velocity at point 1 is $\bf{v_{1}}$ and at point 2 $\bf{v_{2}}$. We can calculate change in angle by
$$
\theta = acos(\frac{\bf{v_{1}}.\bf{v_{2}}}{v_{1}v_{2}})
$$
I am using unit velocity vectors to compute this angle. Finally we can divide this by time step to acquire angular velocity.
*Position vector
We can repeat the same procedure but we can use position vector.
*Wikipedia Page Formula
I have also tried using this formula as listed at Wikipedia page:
$$
\omega = \frac{\bf{r} \times \bf{v}}{r^{2}}
$$
My confusion with this formula is the units. Is it per time step and if so do I need to multiply with 20 or not.
My issue is not with the calculation itself because I can manipulate rotation matrix and quaternion. My issue is that I do not know which vectors are useful here to calculate the angular velocity and how do I use them. Any comment or suggestion would be help full.
|
You have a time series $\mathbf r (n\Delta t)$ where $\Delta t = 0.05$ seconds. To apply the formula
$$\mathbf \omega = \frac {\mathbf r \times \mathbf v}{r^2}$$
at time $t=n\Delta t$ start with
$$\mathbf v(t) \approx \frac{\mathbf r(t+\Delta t) - \mathbf r(t)}{\Delta t} = 20(\mathbf r(t+\Delta t) - \mathbf r(t))$$
so
$$\mathbf \omega (t) \approx 20 \frac {\mathbf r(t) \times \mathbf r(t+\Delta t)}{r^2}$$
since $\mathbf r(t) \times \mathbf r(t)=0$. The units here are radians per second. If you want radians per time step omit the factor of $20$.
Note that this gives you the instantaneous angular velocity about the origin - the angular velocity about any other point will be different.
|
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|
Relativistic energy of harmonic oscillator What is the relativistic energy of an harmonic oscillator:
$$\frac{m_0 c^2}{\sqrt{(1-\frac{v^2}{c^2})}}+\frac{1}{2}kx^2$$
Or
$$\frac{{m_0 c^2}+\frac{1}{2}kx^2 }{\sqrt{(1-\frac{v^2}{c^2})}}$$
I think the first one is true but I need an exact logic or derivation.
|
The energy in special relativity (SR) is defined to be $\gamma m_0c^2$ via the scalar product between 4-momenta. So... I think that $kx^2/2$ would then be encompassed in $\gamma m_0c^2$. You should be able to answer this with an application of math by using Hamilton's action principle and the Euler-Lagrange equations.
|
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|
Mathematical Definition of Power I am a high school student who was playing around with some equations, and I derived a formula for which cannot physically imagine.
\begin{align}
W & = \vec F \cdot \vec r
\\
\frac{dW}{dt} & = \frac{d}{dt}[\vec F \cdot \vec r] = \frac{d\vec F}{dt} \cdot \vec r + \vec F \cdot \frac{d\vec r}{dt}
\\
\implies & \boxed{P = \frac{d\vec F}{dt} \cdot \vec r + \vec F \cdot \frac{d\vec r}{dt}}
\end{align}
I differentiated Work using its vector form formula $\vec F \cdot \vec r$
So I got this formula by applying the product rule. If in this formula $\frac{d\vec F}{dt}=0$ (Force is constant), than formula just becomes $P = \vec F \cdot \frac{d\vec r}{dt}$ which makes total sense, but this formula also suggests that if $\frac{d\vec r}{dt}=0$ then the formula for power becomes $P =\frac{d\vec F}{dt} \cdot \vec r$, which implies that if the velocity is zero that doesn't necessarily mean that Power of the object will also be zero!
But I don't find this in my high school textbook and I can't think of an example on that top of my head where this situation is true.
From what I have heard and read, if the velocity of the object is zero then power is also zero.
Can someone please clear my supposed misconception or give me an example of the situation where this happens?
|
Work is defined as $W = \int_{}^{} \vec F \cdot d \vec r = \int_{}^{} \vec F \cdot \vec v \enspace dt$. Power, P, is dW/dt = $\vec F \cdot \vec v$.
Your relationship for work is incorrect, so your relationship for power (boxed-in relationship in your question) is not correct.
|
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Is any particle allowed within the laws of physics? In the same way wormholes are possible but not guaranteed to occur naturally, is a theoretical particle possible such as one with a mass of a tonne and a charge of -3, or are there upper and lower limits of a particle's properties?
Related: Is there a theoretical upper bound on the mass any new particles can have?
|
I'm not sure I understand what you mean by lower bounds, since there are massless, and also chargeless particles, like the photon. Let's stick to upper bounds.
When it comes to charge, hadronic resonances have some pretty high charges like 2, and nuclei go on and on... Don't get started on elementary versus composite.
A droplet of strange matter could be pretty big. But, no, you want elementary?
A very heavy particle would be so unstable to decay it would be hardly detectable, unless it didn't much couple to our world, in which case it would be a cosmological entity...
In any case, if you are talking about 1000 kg, so, then, 11 orders of magnitude higher than the Planck mass (2.176434(24)×$10^{−8}$ kg), its Compton wavelength would be $10^{-21}$ smaller than its Schwarzschild radius, so it would qualify as a mini-black-hole, rather than a quantum particle. Can you compute this?
Most people appreciate small/pointlike objects heavier than the Planck mass are easier to think of as black holes than as particles.
|
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Why do low-energy waves produce heat, but high-energy waves do not? Radio waves, microwaves and infrared are known to produce heat and even cause burns, while visible light and ultraviolet are not. This seems counterintuitive to me, as the latter contain the highest amount of energy.
Why is this? Does it have to do with the quantity of waves rather than the energy?
|
The amount of heat generated by an EM radiation source depends on many factors, including how much of it is absorbed by the material. Of course in the microwave and infrared, absorption is high in liquids and solids and those sources rapidly produce heating by rotational and vibrational excitations being rapidly converted to kinetic energy. High frequencies can produce heating also, of course, but at the visible and above, the absorption of photons starts to have chemical consequences instead of just rotational/vibrational consequences. For instances, instead of ultra-violet light producing an excited vibrational state in a molecule, it will cause an electron to be ejected or for a chemical bond to break. This may be less likely to be translated into kinetic energy in the short term.
Very high frequency radiation can sometimes pass through material, or it will have chemical consequences instead of thermal consequences. But some high frequency radiation will be converted to thermal energy quickly. Highly radioactive substances emit gamma radiation and, and if handled (which they never should, of course) are known to cause a heating around them.
|
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Gravitational Binding energy of a sphere of 2 uniform densities So I know that the gravitational binding energy of a sphere of uniform density can be given by:
$$U=-\frac{16}{3}G\pi^2\rho^2\int_0^Rr^4dr$$
Which if integrated gives:
$$U=-\frac{3GM^2}{5R}$$
As desired. But say I had a density function given by:
$$\rho(r)=\begin{cases}\rho_a&\text{ for } r\leq r_a\\\\\rho_b&\text { for } r_a<r\leq R\end{cases}$$
Could I then write that:
$$U=-\frac{16}{3}G\pi^2\int_0^R\rho(r)^2r^4dr$$
$$U=-\frac{16}{3}G\pi^2\left[\int_0^{r_a}\rho_a^2r^4dr+\int_{r_a}^R\rho_b^2r^4dr\right]$$
Or am I missing some nuance? I feel like I am because I don't think I'm taking into account the mass of the first density in the second integral, but I am honestly not sure. Any clarification would be much appreciated.
|
Could I then write that...
No, it's more complicated than that. You didn't show how you got your first integral, but one way to do it is as in Wikipedia:
Imagine that it is pulled apart by successively moving spherical shells to infinity, the outermost first, and find the total energy needed for that.
The mass $dm$ of a shell between $r$ and $r+dr$ is
$$dm_\text{shell}=\begin{cases}
4\pi r^2 \rho_a\,dr, & 0<r<r_a \\
4\pi r^2 \rho_b\,dr, & r_a<r<R
\end{cases}$$
and the mass inside this shell is
$$m_\text{interior}=\begin{cases}
\frac43 \pi r^3 \rho_a, & 0<r<r_a \\
\frac43 \pi ra^3 \rho_a + \frac43 \pi (r^3-r_a^3)\rho_b, & r_a<r<R.
\end{cases}$$
The potential energy between these is
$$dU=-G\frac{m_\text{interior}dm_\text{shell}}{r}.$$
Integrate this over the entire sphere in two parts, $0<r<r_a$ and $r_a<r<R$.
|
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Does PIXE only measure external composition? When performing X-ray spectroscopy, say with PIXE, do atoms inside the metal have vacancies induced or only the outer layer?
For example, a quarter has a shell of 75% copper and 25% nickel, but a bulk composition of 8.33% nickel and 91.67% copper. If I perform X-ray spectroscopy, would I be measuring the shell composition or the bulk composition?
|
Typical Particle-Induced X-ray Emission (PIXE) uses an incident proton or alpha particle of a few MeV. Electronic stopping of this particle transfers energy to electrons in the sample, exciting them up from valence states. Electrons relaxing back down into those empty states release the characteristic x-rays.
So, the first question is how deep in does a ~1MeV proton or alpha go? Roughly speaking this is on the order of 1 to a few (maybe 10) microns, depending of course on just what the substrate is. So that is an indication of the maximum depth that you might be able to produce an x-ray.
Then, will the x-ray actually get out? Generally speaking, from a few microns the answer is yes. So, it is the range of the incident particle that determines the depth of the probe. And, it is the near-surface region.
|
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Law of refraction In my textbook it is given that "The refractive index of a substance does not depend on the angle of incidence"
But
Refreactive index =sine of angle of incidence/sine of angle of refraction
It is clear from above relation that refractive index depends on angle of incidence.Where am I wrong?
|
The accepted answer is great. Just wanted to distill it down to the most basic statements that could be made.
Refractive index is a property of a material.
Angle of incident is a property of a light ray.
|
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Difference in the equlibrium of the buckets Let there be a bucket of mass on earth resting on floor with water reaching the brim. And now consider another identical bucket with water till brim in outer space with no gravitational influence, and also a 3rd identical bucket with water till brim falling freely near Earth's surface. Here, as we know according to newton's laws, 1'st and 2'nd bucket are in equilibrium as they have no net force acting on them. And the 3rd bucket is accelerating at $g$ in $m/s^2$.
But we tend to say 2'nd and 3'rd buckets are equvalently in identical conditions (saying this because pressure difference pattern in 2'nd and 3'rd buckets is identical).
What is this supposed to be?
|
Well, the second bucket in outer space is in static equilibrium and the third one (that's freely falling) is in dynamic equilibrium. And according to my knowledge, the first one, on the ground, is also in static equilibrium. It would be helpful if you could explain what you meant by "pressure difference pattern in the first and the third buckets is identical".
|
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Does the Earth's magnetic field reversal and the corresponding magnetostrictive stress and strain make any contribution to the Earth's oblateness? If the changing magnetic field in a power grid's transformer can induce stress and strain on the transformer core, then can the changing magnetic field of the Earth, similarly, place stresses and strains on the body of the Earth?
If it can, is it possible to estimate the size of these stresses and strains; and therefore discount the possibility that when the field collapses (during a reversal) that the oblateness of the Earth doesn't alter to such an extent that it becomes thinner at the equator than the poles and hence unstable?
|
As far as I know, magnetostrictive materials are unusual. I doubt there are enough to be noticeable.
Also, Earth's magnetic field is weak, and shape changes even in strong fields are generally small.
Even so, it might not make any difference. Magnetic fields change over long times. Rocks flow over long times, especially deep underground where temperatures are high. So if rocks did change shape and raise a portion of the earth, it would flow back to normal. But that may not really be right if the time scale for rock flow is longer than magnetic field change times.
|
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Dark standard model linked though Higgs is it possible to have a dark std model (ie image of the std model) linked to the std model only through the Higgs? That would allow dark matter condensation, dark galaxies, stars, planets and biology.
Is there any astronomical evidence for (or search for) large condensed dark matter structures though gravitational lensing? "distant Light bending around nothing?"
all fanciful, but fun to think about.
|
Yes. Actually, it is also possible to have such a Dark Standard Model that is linked to the Standard Model only through gravity.
Original theories for the nature of dark matter assumed all of the dark matter to be made of a single particle. WIMPs and Axions are the two prototype examples. Since these have not yet been found, it is perhaps no surprise that also more complex "Dark Sectors" are being looked into. A simple extension beyond a single dark matter particle for example includes a fifth force that lives in that dark sector. "Dark Photons" are one particular and currently rather popular model of that class. The problem with all these more complex models is though that the experimental constraints are minimal or non-existent, and one enters philosophical thin ice when developing such complex models that have no foundation in any experimental data.
Entirely dark galaxies are unlikely. The science of structure formation in combination with extensive gravitational lensing studies and analysis of the cosmic microwave background rule out the hypothesis that most of dark matter would be in clumps exceeding a few solar masses.
Concerning lower-mass clumps, there are also significant constraints, again from gravitational lensing. The corresponding observations typically are referred to "MACHO searches", for MAssive Compact Halo Objects. It is thus not yet ruled out completely that there could be dark stars or dark planets, but indeed it is rather unlikely, given the strong constraints from these searches.
To summarize, while it is allowed by current data that there could be dark aliens wandering around their planets, existing constraints make this quite unlikely.
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Formula for centripetal acceleration: simple proof that does not use calculus? I teach physics to 16-year-old students who do not know calculus and the derivates. When I explain the formula for centripetal acceleration in circular uniform motion, I use this picture:
Here,
$$\vec{a}_{\text{av}}=\frac{\Delta \vec{v}}{\Delta t}=\frac{\vec{v}_2-\vec{v}_1}{\Delta t}$$
and
$$\vec{v}_1=(v\cos\phi){\bf \hat x}+(v\sin\phi){\bf \hat y}, \quad \vec{v}_2=(v\cos\phi){\bf \hat x}+(-v\sin\phi){\bf \hat y}.$$
Combining these equations gives
$$\vec{a}_{\text{av}}=\frac{\Delta \vec{v}}{\Delta t}=\frac{-2v\sin\phi}{\Delta t}{\bf \hat y}, \tag 1$$
which shows that the average acceleration is towards the center of the circle.
Using $\Delta t=d/v=2r\phi/v$, where $d$ is the distance along the curve between points $1$ and $2$, gives
$$\vec{a}_{\text{av}}=-\frac{v^2}{r}\left(\frac{\sin \phi}{\phi}\right){\bf \hat y}.$$
As $\phi\to 0$, $\sin \phi/\phi\to 1$, so
$$\vec{a}_{\text{cp}}=-\frac{v^2}{r}{\bf \hat y}, \tag 2$$
which shows that the centripetal acceleration is towards the center of the circle.
Does there exist another simple proof of Equation $(2)$, in particular, that the centripetal acceleration is towards the center of the circle?
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A very simple derivation that requires no calculus is the following. We use the fact that there is a perfect mathematical analogy in which the velocity vector is to the radius vector as the acceleration is to the velocity. Writing $r$, $v$, and $a$ for the magnitudes, the analogy allows us to infer that if $v=2\pi r/T$, then $a=2\pi v/T$. Eliminating $T$ then gives $a=v^2/r$.
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Lorentz transformation of continuity equation In a particular reference frame with coordinates $x^\mu$, we can define a current density 4-vector $J^\mu=(c\rho,\vec{J})$ where $\rho$ is the charge density and $\vec{J}$ is the current density.
The continuity equation in this frame is then $$\frac{\partial J^\mu}{\partial x^\mu}=0.$$
How can it be shown that this equation holds in a Lorentz boosted frame with coordinates $x'^\mu=\Lambda^\mu_\nu x ^\nu$,
i.e. $$\frac{\partial J'^\mu}{\partial x'^\mu}=0$$ is true?
|
From $J'^\mu=\Lambda^\mu_\nu J^\nu$ take derivative on both sides
$$\frac{\partial J'^\mu}{\partial x'^\mu}=\Lambda^\mu_\nu \frac{\partial J^\nu}{\partial x'^\mu}$$
Note that $\frac{\partial}{\partial x'^\mu}=\Lambda^\sigma_\mu\frac{\partial}{\partial x^\sigma}$ Then we have
$$\frac{\partial J'^\mu}{\partial x'^\mu}=\Lambda^\mu_\nu \frac{\partial J^\nu}{\partial x'^\mu}=\Lambda^\mu_\nu\Lambda^\sigma_\mu\frac{\partial J^\nu}{\partial x^\sigma}=\delta^\sigma_\nu\frac{\partial J^\nu}{\partial x^\sigma}=\frac{\partial J^\nu}{\partial x^\nu}=0$$
|
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Why is it easier to handle a cup upside down on the finger tip? If I try to handle a tumbler or cup on my fingertip (as shown in fig), it is quite hard to do so (and the cup falls most often).
And when I did the same experiment but this time the cup is upside down (as shown in fig), it was quite stable and I could handle it easily.
In both the cases, the normal force as well as the weight of that cup is the same but in first case it falls down and in the other it is stable.
I guess that it is falling because of some torque but why is there no torque when it is upside down.
What is the reason behind this?
|
Dr jh has explained by taking torque about the center of mass but it could be explained better if we take it about the point of contact of your finger and the cup.
When the cup is upright gravity provides a torque by a force acting on its center of mass which is above your finger. Small perturbations will cause rotation about the point of contact and it rotates away from its original position. It's an unstable equilibrium and it's kind of like balancing a pencil on your fingertips.
When it's upside down it's a stable equilibrium and any perturbations would be restored back to its original position with the help of gravity. Kind of like a pendulum.
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Please explain the meaning of below statement Newtons second law is a local law.
(In the book,it says that it means that it applies to a particle at a particular instant without taking into consideration any history of the particle or its motion.)
Um, I couldn't understand what do they mean by " taking into consideration any history of the particle or its motion ".
If possible ,please explain it with an example.
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Yeah, it's quite confusing to read that. Even me, myself couldn't understand it first. Newton's second law is the relation between force and acceleration. It says that " the acceleration is directly proportional to the applied force, with the mass of the body being constant ".
And I guess this is what your NCERT Textbook says-
The second law of motion is a local relation
which means that force F at a point in space
(location of the particle) at a certain instant
of time is related to a at that point at that
instant. Acceleration here and now is
determined by the force here and now, not by
any history of the motion of the particle.
This means that the acceleration of the body does not depend on the previous states or positions of the body. It only depends on the Force applied at that exact point of time that we are talking about. In short we can say that it tells us to ignore what the body was doing in the past and concentrate on the present.
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Why is acceleration directed inward when an object rotates in a circle? Somebody (in a video about physics) said that acceleration goes in if you would rotate a ball on a rope around yourself.
The other man (ex Navy SEAL, on YouTube too) said that obviously it goes out, because if you release the ball, it's going to fly in outward direction.
Then somebody said that the second man doesn't know physics; acceleration goes in.
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A moving object continues in a straight line unless a force is applied to it.
If a ball is whirled in a circle at the end of a string, it is caused to move in a circle by the pull of the string. If the string breaks the ball proceeds in a straight line unless gravity pulls it downward. The ball's straight line is a tangent to the circle. See the previous drawings showing that.
If there was a centrifugal force the released ball would move from its position directly away from the center of the circle like the symbol for Mars. It does not do that.
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Superconducting energy storage and voltage I've been reading up on superconducting magnetic energy storage, and while I know how the energy is stored ($E=\frac12LI^2$) I have no idea how the discharging works. Specifically, as per this paper 1 a voltage is applied across the superconducting coil, and the power output is given by $VI$, where $I$ is the current circulating in the superconductor. But I thought the notion of applying a voltage across a superconductor didn't make sense, as per this physics SE question.
Furthermore, I'm a bit uncertain as to how you just apply a voltage across something. For example, you could presumably put a 12V battery across the coil, but wouldn't that just result in your battery being charged? What if you wanted to deliver electrical power elsewhere?
1 P. Tixador, "Superconducting Magnetic Energy Storage: Status and
Perspective", IEEE/CSC & ESAS EUROPEAN SUPERCONDUCTIVITY NEWS FORUM, No. 3, January 2008.
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Normally a superconducting magnet (a coil) is closed. If a resistor is connected at two points on the coil and the coil is opened between those points, current will flow through the resistor. The back EMF caused by the current through the resistance will cause a voltage between the ends of the coil. This in turn causes the current in the coil to drop. When current in the coil changes, self-inductance of the coil induces a back EMF that opposes the change. So, the rate of energy transfer to the resistor is controlled by the coil geometry, the amount of current in the coil, and the value of the resistor.
In practice, current from the superconducting coil can be switched rapidly to drive an AC current in an external coil, allowing energy to be transferred out inductively instead of resistively. The portion of the system that deals with the switching, then converting the power to a useful form, is called a power conditioning unit. There is a lot of information online about power conditioning units for superconducting systems, such as this.
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Work done when you bring an bringing an object down from a height I am going to explain this question through an example.
Suppose I lift an object I apply a force $mg $ then I apply additional force, that would be $ma $ so total force would be $m(g+a)$. My doubt is that the work done by a person in lifting a box by applying more force than the weight that should be $m(g+a) *h $ right?
I would love to know if my answer is right or if someone could kindly correct me
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That is correct. Also, since the gravitational PE increases by $mgh$ it is clear that the kinetic energy increases by $mah$
|
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Is there an accepted Lagrangian for the transport equation? Perhaps because it is so simple, I have not seen a lagrangian form of the transport equation
$$(\partial_t + a \partial_x)q = 0.$$
This equation is first order, which makes obtaining it from the Euler-Lagrange equation a bit tricky.
It would appear that the lagrangian $\frac{m}{2}q_{t}^2 + \frac{T}{2}q_{xt}$ yields $q_{tt} + \frac{T}{m}q_{xt}=0$, which in turn yields the transport equation with constant forcing
$$q_t + \frac{T}{m}q_x + c =0$$
after integrating with respect to $t$ (the constant c comes from here).
Alternatively we might consider the lagrangian $q^2(q_{tt} + q_{xx}) + q(q_t^2 + q_x^2)$, which via the Euler-Lagrange equations yields
$$-q_t^2 - q_x^2 = 0 $$
which factors into the two transport equations describing motion in opposite directions.
Is there some classic approach or equation that I am missing?
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I may be wrong, but it seems to me, that lagrangian, satisfying this equation is not possible in principle, at least classically.
Consider the Euler-Lagrange equation:
$$
\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{x}}\right) -\frac{\partial L}{\partial x} = 0
$$
Note, that it is invariant under the time reversal $t \rightarrow -t$.
The advection equation is by itself not invariant under the time reversal. The Lagrangian, producing the wave equation incorporates right and left moving wave as well.
Higher order Lagrangians of form $L(x, \dot{x}, \ddot{x}, \ldots)$, will be invariant under the time reversal, because the signs in the $\frac{d}{dt}$ and $\ddot{x}$ will cancel each other.
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Would there be any gravity inside a hollow planet made of a spherical shell of basalt crust? I am writing a science-fiction story, which contains a 'crazy' planet. I would like some input into the consequences of this physics according to standard physics.
The planet in question is Earth-sized and it orbits a Sun-like star. Within the story, this planet has been taken over by a malignant 'force' and has been hollowed out, so that there is one mile of basalt crust forming a spherical shell, and the inside is a void filled with nitrogen. This force keeps the planet intact, but does not affect the planet in any other way (i.e. any standard-physics problems regarding the integrity of this spherical shell can be ignored, as the story explains them in other ways). The force has no mass.
Given this setup, what would the gravity be like at the center of the planet? In the story, the protagonist is able to fly around in a large bubble of zero-gravity at the middle. Would this be possible? (Assume that the protagonist can survive the atmosphere.) Could he, assuming he has a suitable raft, get around by blowing?
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If you neglect the gravity of the nitrogen gas, there is no net gravitational force anywhere inside. This is a result of the famous “Shell Theorem” for inverse-square forces. At any point, you are attracted by all the atoms in the shell, but the vector sum of all these forces — in different directions, and having different magnitudes because they are caused by atoms at various distances from you — turns out to be zero. This assumes a perfectly spherical shell of uniform mass per unit area.
If you want to take the nitrogen gas into account, at radius $r$ from the center you are attracted by the mass of all the nitrogen in a sphere of radius $r$ around the center, as if this mass were concentrated in a point at the center. The nitrogen at larger radii doesn’t attract you for the same reason that the shell doesn’t.
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Why don't opaque objects reflect light? My sister was doing a quiz and I tried to point her in the right direction by giving her scenarios to imagine. One of the questions in the quiz was:
Which of the following objects do not reflect light:
*
*Polished metal
*Mirror
*Undisturbed water
*Book
She suggested that the answer was "undisturbed water" and that made sense to me too.
But the answer given was "book", which didn't make sense to me. How can you even see the book if it didn't reflect light in the first place?
Is this terrible framing by her teacher or am I having a conceptual misunderstanding?
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The question is asking "which of the following objects will you not see a reflection?".
A distinction (albeit poorly) is being made between specular reflection and diffuse reflection.
The objects in options 1-3 will exhibit specular reflection, while option 4 "a book" will exhibit diffuse reflection. So the correct option will be "4 Book" since this object will not exhibit specular reflection, whereas "1. Polished metal, 2. Mirror" and "3. Undisturbed water" all exhibit specular reflection.
You are correct and the question should probably have been worded similar to this:
"Which of the following objects would exhibit diffuse reflection, as oppose to specular reflection?"
Now with the understanding that the question posed by the teacher was probably at an elementary school level, it should be noted that a more technical answer (and more accurate answer) should explain these two forms of reflection in detail, so see more in the links below. But to briefly summarize:
Diffuse reflection:
Diffuse reflection is the reflection of light or other waves or particles from a surface such that a ray incident on the surface is scattered at many angles rather than at just one angle as in the case of specular reflection.
Specular reflection is described as:
Specular reflection, or regular reflection, is the mirror-like reflection of waves, such as light, from a surface.
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Example of an infinite volume Lindblad system What is an explicit example of a Lindbladian
\begin{align*}
L(\rho) = - i \lbrack H_A, \rho \rbrack + G \sum_{j} V_j \rho V_j^* - \frac{1}{2}(V_j^* V_j \rho + \rho V_j^* V_j)
\end{align*}
acting on the space of trace class operators on some Hilbert space $\mathcal{H}$ such that there is a trace-class operator $\rho_\infty$ with $L(\rho_\infty ) = 0$.
EDIT: As pointed out in an answer I did not specify what I mean by infinite volume. I mean that the Hilbert space is for example $l^2(\mathbb{Z})$ with the standard orthonormal basis and that the state $\rho_\infty$ should be an element of the trace-class operators on $l^2(\mathbb{Z})$. I.e. that the system is not just a single mode, but physically extended.
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In the following paper https://arxiv.org/abs/2206.09879 (work I did in connection to asking the question) several examples of infinite volume Lindblad systems are discussed.
In the single particle sector, consider the Hilbert space $l^2(\mathbb{Z})$ and some space of operators on that Hilbert space, e.g. trace-class, Hilbert-Schimdt, compact or bounded operators. Then consider the Lindbladian $\mathcal{L}$ defined on one of these space, for example with a nearest neighbor hopping Hamiltonian
$ H = \sum_{k \in \mathbb{Z}} \mid k \rangle \langle k+1\mid + \mid k+1\rangle \langle k \mid $ and Lindblad operators $V_k = \mid k \rangle \langle k\mid$ for each $k \in \mathbb{Z}$.
Since the Lindbladian is a super operator a lot of mathematical complications arise. I.e. one needs to think very carefully about which space of operators the Lindbladian acts on and many properties that are true in finite dimensions are not true in infinite dimensions.
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Analytic solution to Kepler's Problem, exegesis From 'Solving Kepler's Problem' by Colwell, the first analytic solution to Kepler's Problem used a theorem of Lagrange, and later Burmann, to invert Kepler's equation. When you look on the internet for a proof you find these lines that begin the section on Burmann's theorem (copied straight from Whitaker: A Course in Modern Analysis) Burmann's Theorem
Given $f(z)$ analytic on a region, $\phi(a)=b$, and $\phi'(a) \neq 0$, then Taylor's theorem gives:
$$\phi(z)-b = \phi'(a)(z-a)+\frac{\phi''(a)}{2!}(z-a)^2+...\tag{1}$$
If it is legitimate to revert this series the result is:
$$z-a=\frac{\phi(z)-b}{\phi'(a)}-\frac{1}{2}\frac{\phi''(a)}{\phi'(a)^3}[\phi(z)-b]^2+...\tag{2}$$
How do you get from the first equation to the second?
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Basically you suppose that $z\in B_\epsilon(a)\subseteq S$
($S$ in Wolfram), for small $\epsilon$. In other words you suppose that the distance (The metric for this case) $|z-a|$ is small enough (which you would have to anyways if you neglect the higher order Terms of the Taylor-Series). It follows that
$$
\phi'(a)={\rm lim_{z\longrightarrow a}} \frac{\phi(z)-\phi(a)}{z-a} \approx \frac{\phi(z)-\phi(a)}{z-a} := \frac{\phi(z)-b}{z-a}.
$$
Plugging this in your Taylor-Series
$$
\phi(z)-b=\sum_{n=1}^\infty \frac{\phi^{(n)}(a)}{n!}(z-a)^n,
$$
gets you
$$
\phi'(a)(z-a)=\sum_{n=1}^\infty \frac{1}{n!}\frac{\phi^{(n)}(a)}{[\phi'(a)]^n}(\phi(z)-b)^n,\\
\Leftrightarrow z-a=\sum_{n=1}^\infty \frac{1}{n!}\frac{\phi^{(n)}(a)}{[\phi'(a)]^{(n+1)}}(\phi(z)-b)^n.
$$
I'm not that sure about the $(-1)^{n+1}$, which according to Wolfram should be in the sum. Maybe somehow
$$
\phi'(a) \approx \frac{\phi(z)-b}{a-z}.
$$
is true?? Because plugging that in you get:
$$
\phi'(a)(a-z)=\sum_{n=1}^\infty \frac{(-1)^n}{n!}\frac{\phi^{(n)}(a)}{[\phi'(a)]^n}(\phi(z)-b)^n,\\
\Leftrightarrow z-a=\sum_{n=1}^\infty \frac{(-1)^{(n+1)}}{n!}\frac{\phi^{(n)}(a)}{[\phi'(a)]^{(n+1)}}(\phi(z)-b)^n.
$$
Which is exactly your second equation.
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Which graph shows how the power $P$ dissipated in the load resistor varies with the resistance of the load resistor?
^ question
Which graph shows how the power P dissipated in the load resistor varies with the resistance of the load resistor?
The answer is A but what I need help in understanding is why. From what I understand the answer can't be A because since R increases, the power dissipated increases as $P = I^2R$ and the current $I$ is the same for both resistances because in series the current is equal for both components whenever it changes, so the line has an increasing gradient - but what exactly is the reason that causes the gradient on the graph to decrease again? If the resistance keeps increasing and so does the power dissipated across it, just why does it randomly start decreasing? I feel like $E=V+Ir$ has to do something with it, but using it only lets me infer that $Ir = E-V$ thus with increasing terminal potential difference there's a smaller value for the lost volts. How that's supposed to help me exactly I don't understand.
I'll appreciate any response.
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Seems like I've got an answer (if the working is incorrect that would invalidate this answer, in that case please leave a comment, thanks).
Power dissipated:
$P = I^2R$
$I = V/R_{T} = 12/(2+R)$
$P = 12/(2+R) * R$
choosing values 0.5 through 4 for $R$ we can infer by the results that the gradient should increase and that the maximum of the graph should be at $R = 2$ and after that (for values of $R$ where $R > 2$) the magnitude of the total power dissipated decreases hence the gradient on a graph would then decrease.
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How can I make sense of the work-energy theorem in the situation of a force applied on a disk? Pretend that we are in space and that gravitational attraction is negligible. Now, imagine that there are two disks that are exactly the same. However, on the first disk, a force is applied tangentially and on the second disk, the same force is applied to its center of mass. The force is exerted over the same distance for both.
Now here is where I'm having problems making sense of the situation. Since the force is the same for both disks the horizontal acceleration should be the same. After travelling a distance $d$, both disks should have the same final horizontal velocity. The work done should also be the same.
However, since the force is tangential on the first disks, there must be a torque and the first disk would have gained an angular velocity. But $W = Δk_{rotation} + Δk_{translation}$. The work and the kinetic energy of translation can't be the same if disk 1 has rotational kinetic energy.
One of the assumptions I have made must therefore be wrong. Either the work done on the disks isn't the same, or the horizontal acceleration isn't the same, or disk 1 has no angular velocity.
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However, on the first disk, a force is applied tangentially and on the second disk, the same force is applied to its center of mass. The force is exerted over the same distance for both.
It is not possible for the force to be applied for both the same amount of time and over the same distance. The tangentially applied force will cover a larger distance over the same time, or conversely will cover the same distance in a shorter time.
If you match time then their ending momentum will be the same, but the tangential disk will have higher energy corresponding to the rotational energy and the additional distance over which the force was applied.
If you match the distance that the material at the point of contact with the force moves (this is not the same as the distance that the CoM moves for the tangential disk) then the tangential force will be applied only briefly. The tangential disk will have less momentum but the same energy. That energy will be split between center of mass KE and rotational energy.
|
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Why does air pressure decrease with altitude? I am looking to find the reason: why air pressure decreases with altitude? Has it to do with the fact that gravitational force is less at higher altitude due to the greater distance between the masses? Does earth’s spin cause a centrifugal force? Are the molecules at higher altitude pushing onto the molecules of air at lower altitudes thus increasing their pressure? Is the earths air pressure higher at the poles than at the equator?
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I edited this question on the first day, in response to a few comments that pointed out a misunderstanding, but it didn't register. I sincerely apologize for that.
As pointed out by other answers, the pressure due to any fluid, compressible or not, increases with depth. This is due to the greater mass and thus weight of the fluid above.
What's interesting is that the pressure of water increases linearly with depth, while that of air does not.
The gravitational field strength drops down to only 88% even at the height of the ISS. The drop in pressure has more to do with the fact that unlike water, air is a compressible fluid. As you move further down the atmosphere, there is a greater weight of air above pushing down on the air below, so the density, and thus the air pressure, increases. Basically, the density $\rho$ is a function of $h$. so you have to integrate density over the height instead of simply multiplying.
$$P=g\int\rho\mathrm{d}h$$
or $$P=\int g\rho\mathrm{d}h$$ if you want to account for the change in gravitational field, however small
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Magnetic dipole moment of volume current derivation The multipole expansion of the magnetic potential yields the dipole term :
$$\mathbf A_{dipole}(\mathbf{r}) =\frac{\mu_0}{4\pi r^{2}}\int (\hat {\mathbf r} \cdot {\mathbf r}') \mathbf J {dV}'$$
How do I get from this expression to the final expression :
$$\mathbf A_{dipole}(\mathbf{r}) =\frac{\mu_0}{4\pi r^{2}}\left (\frac{1}{2}\int ( {\mathbf r}'\times \mathbf J) \ {dV}'\right) \times \hat{\mathbf r} $$
Thanks in advance.
|
There is in identity for the triple product:
$$
\mathbf{r}^{'} \times (\mathbf{J} \times \hat{\mathbf{r}}) =
\mathbf{J} (\mathbf{r}^{'} \cdot \hat{\mathbf{r}})
-\hat{\mathbf{r}} (\mathbf{r}^{'} \cdot \mathbf{J})
$$
Substitutring it in the exprerssion, one is left with the integral:
$$
\int dV^{'} \hat{\mathbf{r}} (\mathbf{r}^{'} \cdot \mathbf{J})
$$
Which vanishes, as averaging over all directions.
(there would in any dimensionality something like $\int d \theta \cos \theta$, where $\theta$ is angle between $J$ and $\mathbf{r}^{'}$).
However, I do not see, from where does the factor $1/2$ arise.
|
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Effect of gravity in space According to general relativity, I know that time flows at a different rate in the presence of gravity which is nothing but the curvature of spacetime. And is called time dilation.
My doubt is, are there similar effects in space as well? I mean, when there is a curve in space-time, it's not just in time but also in space. So are there any effects like length contraction or 'Space dilation' under the effect of gravity? If one measures the distance between two points as 1m in a strong gravitational field, will it be different from the 1m at a place in weaker gravity?
|
If you are talking about gravitational time dilation and length contraction yes there are analogous formulations like in special relativity.
But in this case, they are rather more complicated since you are no longer working on flat spacetime. You need to define the metric. For simplicity if you choose a non-rotating black hole
$$ds^2 = -\Big(1-\frac{2GM}{rc^2}\Big)c^2 dt^2 + \Big(1-\frac{2GM}{rc^2}\Big)^{-1}dr^2 + r^2 d\Omega^2$$
where $d\Omega^2 = d\theta^2 + \sin^2 \theta d\phi^2$
you can calculate length contraction by taking $dt = 0$, $d\phi = 0$ and $d\theta = 0$ so
$$ds^2 = \Big(1-\frac{2GM}{rc^2}\Big)^{-1}dr^2$$
and this gives
$$ds = \frac{1}{\sqrt{\Big(1-\frac{2GM}{rc^2}\Big)}}dr$$
Similar relation for time dilation will hold, you can calculate it now pretty straightforwardly.
|
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Why is the net force acting on a massless body zero? I know that massless bodies can accelerate (in theory) even with the net force equaling to zero. But, why cannot there be a net force on a massless object? Why does it always have to be zero as a resultant in the end?
I'm talking about object whose mass is assumed to be zero, i.e. $m\to0$
|
If we start from Newton's Second Law
$$F=ma,$$
then we can see that if the mass is zero, then the total force must be as well:
$$F = ma = (0)a = 0.$$
Then again, this leaves acceleration completely undefined since $a = F/m = 0/0.$ This is why every massless object in a physics problem--whether rope, spring, or pulley--is attached to something with mass. The combined object has a non-zero total mass and so can have a sensible acceleration. The massless object simply follows the massive object it is attached to. Newtonian physics doesn't work with massless objects. You can try taking limits as mass goes to zero, as was attempted when the deflection of light by gravity was predicted according to Newtonian gravity theory, but this resulted in an incorrect prediction.
|
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How does a capacitor get charged instantly in AC whereas it takes infinite time in DC? I have seen when a capacitor is connected to a dc source it takes infinte time to charge, but when connected to ac it takes the potential of the source instantly,
probably the approach in the books is not adequate, please clarify,
Here is a link that mentions the time constant for DC https://www.electronics-tutorials.ws/accircuits/ac-capacitance.html
And here is one that describes the AC https://physicscatalyst.com/elecmagnetism/growth-and-delay-charge-R-C-circuit.php
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Capacitors always take time to charge. In practice, when a capacitors is ~99% charged , we can call it fully charged. The exponential which is used to describe the charging of a capacitors does not make sense when time is very large because charge can never be less than charge of an electron while in the exponential equation, for a large enough time you can get charge less than charge of an electron which is meaningless.
Having that said, the exponential is a very good approximation for short time. In AC (just like DC) the capacitors need some time to charge. It is this extra time that causes the voltage across them to lag behind.
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Can magnetic loops with no source current knot or link? The answer to this question is obviously no. I would like to pose a variation of that question. Suppose a simply connected domain of a 3-d vacuum space has no source current. Does there exist a case where two closed loops of magnetic field residing in that domain knot into a link? (Either static or dynamic answer would be great.)
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As @benrg mentioned in his comment, there are papers on exactly this topic, e.g.,
*
*Carlos Hoyos, Nilanjan Sircar and Jacob Sonnenschein, New knotted solutions of Maxwell’s equations.
*Hridesh Kedia, Iwo Bialynicki-Birula, Daniel Peralta-Salas,
and William T.M. Irvine, Tying knots in light fields.
There, the electromagnetic field is constructed using the Batesman's constuct, where the fields $({\bf E},{\bf B})$ is encoded by the Riemann-Silberstein vector
$${\bf F = E}+ i{\bf B}=\nabla\alpha\times\nabla\beta$$
where $\alpha$ and $\beta$ are complex functions of $(t,x,y,z)$ and $i$ is the imaginary unit. This is then substituted in to the Maxwell's equations to obtain an equation constraining $(\alpha,\beta)$. The solutions of linked field lines are then constructed through some transformations such as conformal mapping.
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Is inertia and gravity determined by relativistic mass or invariant mass? As far as I know, mass fundamentally determines inertia and the gravitational force. But since there are two types of mass, which mass determines which? From what I have read so far, and correct me if I'm wrong, the relativistic mass determines the inertia, but not the gravitational force. Then why does one determine inertia and another determine gravity? Also since relativistic mass represents the total mass-energy of an object taking into account the kinetic energy, does that imply that the gravity is not determined by the total energy content of an object, but only by its invariant mass which doesn't take into account its kinetic energy?
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As stated is not advised to call the relativistic energy $m\gamma$ relativistic mass. Mass these days strictly refers to total energy in the rest frame divided by $c^2$.
$m\gamma$ is indeed the source of gravity and it determines inertia, as the relativistic momentum is $m\gamma \vec v$. Newton's second law is replaced by $$\dot {\vec p} = m \gamma {\vec a} + \gamma^3 \left( {\vec v} \cdot {\vec a} \right) {\vec v} ~.$$
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Why different methods to solve this question gives different values of time While solving the following question.Why do we get different values of time by different approaches.
An elevator whose floor to ceiling distance is $2.7m$ starts ascending with a constant acceleration of $1.2m/s^{2}$.Two sec after it starts, a bolt begins to fall from the ceiling of the elevator.Find the bolt's free fall time.
Approach 1 :- In lift's frame of reference.
Applying $S=ut+\frac12 at^2$ on bolt we get,
$2.7=0+\frac12 (9.8)t^2$
$t=0.742 sec$
Approach 2:- In ground frame of reference.
Let bolt takes ‘t’ time to fall
$\vec{s_{b,l}}=-2.7\vec j$
$\vec{s_{l,g}}=\frac {1.2}{2}((t+2)^2-4) \vec j$
As $\vec{s_{b,l}}+\vec{s_{l,g}}=\vec{s_{b,g}}$
So, $\vec{s_{b,g}}=(\frac {1.2}{2}((t+2)^2-4)-2.7)\vec j$
Velocity of bolt after $2$ sec $=2.4m/s$
Applying $S=ut+\frac12 at^2$ on bolt we get,
$\frac {1.2}{2}((t+2)^2-4)-2.7=2.4t-\frac12 (9.8)t^2$
On solving $t=0.7006$ sec
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In approach 1 which is from the lift frame, you are doing the calculation as if the frame is inertial. Meaning, you are using only the gravitational acceleration $g$ and are ignoring the centrifugal part, namely the presence of the lift acceleration.
$$\text{Inertial frame:}\qquad a=g$$
This is how the bolt will fall if dropped here on Earth. This is how it would fall if the lift wasn't accelerating. But the lift is accelerating up also. The bolt will reach the floor quicker since the floor is accelerating upwards towards the bolt. The lift acceleratino must be added so that you'll use an adjusted frame-specific acceleration in the formula:
$$\text{Non-inertial lift frame:} \qquad a=g+a_\text{lift}$$
In general you must always add (or remove) the centrifugal acceleration part when using a non-inertial frame. This is the part that distinguishes inertial from non-inertial in our calculations. Then the calculation fits:
$$S=ut+\frac12 \underbrace{(g+a_\text{lift})}_a t^2\quad \Leftrightarrow \quad t=0.7006\,\mathrm s$$
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Why is the $i\epsilon$-prescription necessary in the Klein-Gordon propagator? When evaluating the Klein-Gordon propagator, in the book by P&S, p. 31, I see that, it is customary to shift the poles and add $i\epsilon$ in the denominator. I don't understand, why this is necessary. Why can't we just use complex analysis? What is wrong in the following steps?
\begin{align}
\int \frac{e^{ibz}}{z^2-a^2}\, dz &= (2\pi i) \left[\lim_{z\rightarrow a} (z-a) \frac{e^{ibz}}{z^2-a^2} + \lim_{z\rightarrow -a} (z+a) \frac{e^{ibz}}{z^2-a^2}\right] [\mathrm{Residue~theorem}]\nonumber\\
%
&= (2\pi i) \left[\lim_{z\rightarrow a} \frac{e^{ibz}}{z+a} + \lim_{z\rightarrow -a} \frac{e^{ibz}}{z-a}\right]\nonumber\\
%
&= (2\pi i) \left[ \frac{e^{iba}}{2\,a} - \frac{e^{-iba}}{2\,a}\right]\nonumber\\
%
&= \frac{i\pi}{a} \left[ e^{iba} - e^{-iba}\right]\nonumber\\
%
&= - \frac{2\, \pi\, \sin{ba}}{a}
\end{align}
What goes wrong in proceeding this way? Can't we just do the integration $p^0$ as is done for the $z$-variable? Obviously, $a$ will be function of $\vec{p}$ and $m$.
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Note that the original integral you are trying to compute is over the real line, not over a closed contour, so the Cauchy theorem does not apply until you find a suitable way to close the contour. Due to the presence of the exponential factor $e^{ibz}$, as you have written it, one can close the contour in the upper half plane if $\mathrm{Re}\, b>0$. Let's assume that's the case. Now your two poles are actually on the real line, so we also need to specify which way to pass around them. Since you are closing the contour above, and you are picking up both of the residues, you are implying that you are passing below these two poles. If you passed above them, they would be outside your contour and would not contribute. Since you are passing below your two poles, we could equivalently describe what you did by saying that the two poles are shifted upwards on the complex plane by an infinitesimal amount $+i\epsilon$. This would guarantee that you pass below them as you integrate along the real axis. So you see that you also actually have included some $\epsilon$s in your calculation too, although you didn't acknowledge it.
For calculations in QFT, there is a correct physical prescription for which way to go around the poles, which is called the Feynman prescription, and differs from what you did above. This is covered well in P&S.
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What is the non-relativistic limit of the quantised electromagnetic field? I’m not a physicist so this question may be naive ... For a real scalar field, quantisation yields the Klein-Gordon equation and the non-relativistic limit of this gives the Schrödinger field. What is the equivalent equation or field starting from the electromagnetic field? I.e. quantising the EM field (E- and B-fields) and dispensing with Lorentz covariance gives what kind of equation or field? I.e. what’s the equivalent of the Schrödinger equation/field for an EM field (rather than a scalar field).
I have an application that observes the structure of the EM field but not Lorentz covariance.
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As electromagnetic fields in vacuum always move at speed $c$ there is no nonrelativistic approximation to electromagnetism. In other words photons have zero rest mass and therefore no rest frame.
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Number of variables in the Hamilton-Jacobi equation In Goldstein's Classical Mechanics, while introducing the Hamilton-Jacobi equation, he argues that the equation $$H(q_1, ... , q_n; \frac{\partial S}{\partial q_1}, ..., \frac{\partial S}{\partial q_n}; t) + \frac{\partial S}{\partial t} = 0$$ is a partial differential equation in $(n + 1)$ variables $q_1, ... , q_n; t$.
He then proceeds to say that the solution (if it exists) will be of the form $$S(q_1, ... , q_n; \alpha_1, ... , \alpha_{n+1}; t)$$ where the quantities $\alpha_1, ... , \alpha_{n+1}$ are the $(n + 1)$ constants of integration.
How is time a variable? Isn't it the parameter we're integrating over?
Perhaps this warrants some context. He introduces the Hamilton-Jacobi equation with the motivation to find a canonical transformation that relates the canonical coordinates at a time $t$ -- $(q(t), p(t))$ -- and the initial coordinates $(q_o, p_o)$ at $t = 0$. I hence get that time must be a variable here. However, it is still the parameter we integrate the Hamilton-Jacobi over in order to get $S$, right? Where does the $(n+1)^{th}$ constant of integration come from?
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the action function in the HJ formalism is dependent on time through the upper bound.
$$S(q_i,t,\alpha_i,t_0)=\int_{t_0}^tL(q_i,\partial_tq_i,t')dt'$$
in cases of Lagrangian that is independent (explicitly) of time then the constant of integration conjugate to $t$ let's call it $\alpha_t$ is simply the Energy.
read more here:
https://en.wikipedia.org/wiki/Hamilton%E2%80%93Jacobi_equation#Hamilton's_principal_function
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Will this spaceship collide with the star? (time dilation) I thought of the above thought experiment and arrive on 2 conflicting conclusions. I can't seem to identify the flaw in my reasoning.
Suppose there is a star 4 light years from earth that has will explode and turn into a white dwarf in 3 years (as measured in the earth frame). A spaceship travels to the star at 86% the speed of light.
According to earth's frame of reference, the journey to the star will take 4.5 years so the star will have turned into a white dwarf.
according to the ship's frame of reference, however, the journey will only take 2.25 years. Moreover, since the star is travelling relative to the ship in its own frame, the event of the star exploding will actually take 6 years. So the spaceship will collide with a white dwarf instead of a star.
I thought this had something to do with simultaneity, but I know that the events must be same in all frames of reference. My collusions imply that collisions occur between different bodies depending on the frame of reference, which can't possibly be true, can it? Where am I going wrong?
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$t=0$, $x=0$
B: $t=3$, $x=4$
C: $t=4/v\approx 4.62$, $x=4$
*Lorentz transform to the ship frame:
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Why don't radio waves sent by electronic devices intefere with each other? You know how phones, computers and other electronic devices that use wireless communication use radio waves to communicate?...well since almost everyone has gotten a smart phone wouldn't radio waves from different smart phones interfere with each other...well...they don't (other wise we wouldn't be able to call people)... But I want to know why radio waves from different electronic devices or from two different sources don't interfere with each other... if there are two explanations then pls tell me both of them (specially if one of them is a quantum mechanical theory)..help will be appreciated...thanx in advance
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For a significant and stable interference pattern, two such waves would have to be of the same frequency and of comparable amplitude. There is an area in North Miami where I get poor reception on my car radio. I think the several radio and TV transmitters in the area are over-driving the circuits in my radio.
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Is the brick-concrete connection the 'weakest link' in brick made buildings? When an earthquake strikes,do buildings made of bricks and concrete break mostly on concrete-brick connection spots?If Yes does the brick on the picture prevents this from occuring?
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As far as I know, yes, brick-concrete connection is the weakest link. The connection would break if torque about the interface surpasses the the limit to which the interface is designed/expected to bear the torque.
The above design provied a counter torque about the interface due to normal force from the inclines at the interface(upto the limit that the inclines don't break). This prevents the interface from being a weak-link.
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Why is there a specific negative sign in front of the $m_{12}$ term of the 2HDM Higgs potential? Why is there a specific negative sign in front of the $m_{12}$ term of the 2HDM Higgs potential?
(but not for the $m_{11}$ and $m_{22}$)
See for example: https://arxiv.org/abs/1106.0034
Eq. (2) Page 6:
$$
V = m_{11}^2\Phi_1^\dagger\Phi_1 + m_{22}^2\Phi_2^\dagger\Phi_2 -m_{12}^2(\Phi_1^\dagger\Phi_2+\Phi_2^\dagger\Phi_1) + \frac{\lambda_1}2(\Phi_1^\dagger\Phi_1)^2+\frac{\lambda_2}2(\Phi_2^\dagger\Phi_2)^2+\lambda_3\Phi_1^\dagger\Phi_1\Phi_2^\dagger\Phi_2+\lambda_4\Phi_1^\dagger\Phi_2\Phi_2^\dagger\Phi_1+\frac{\lambda_5}2\left[(\Phi_1^\dagger\Phi_2)^2+(\Phi_2^\dagger\Phi_1)^2\right].\tag{2}
$$
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The authors are writing down the most general potential consistent with the symmetries of the problem. The constants $m_{ij}^2$ are free parameters, and the sign is purely conventional. For example, if I were to write the most general linear function of $x$, I could write $f(x)=\alpha+\beta x$ for some parameters, or $f(x)=-\alpha+\beta x$, or $f(x)=\alpha-\beta x$, etc. All these parametrizations are equivalent, as $\alpha,\beta$ are free parameters and thus I am free to define their sign however I want.
In the case at hand this freedom in choosing the parametrization becomes even more clear due to the fact that we can change the sign in front of $m_{12}^2$ by the field redefinition $\Phi_1\mapsto-\Phi_1$ (or same with $\Phi_2$), which flips the sign of $(\Phi^\dagger_1\Phi_2+\Phi^\dagger_2\Phi_1)$, but leaves the rest of the Lagrangian invariant. The sign of $m^2_{12}$ is irrelevant, as it depends on our conventions. (Note that the signs of the eigenvalues of the mass matrix don't care about the sign of $m_{12}^2$, so this sign has nothing to do with symmetry breaking!) The authors are just choosing a particular sign that they found convenient for some reason. But there is no physics behind it.
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How does universe expand when cosmological constant is zero? From what I learned, Einstein believed in a static universe but from his general relativity equations universe must collapse under gravity. Hence Einstein adjusted this gravity with cosmological contact which is a kind of anti-gravitational effect. But later it was discovered that the universe is expanding. Einstein was ashamed by his cosmological constant and threw it out from his equations.
My doubt is how the removal of cosmological constant agrees with an expanding universe ?. When there's nothing to counteract gravity, shouldn't the universe be contracting?.
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The universe can expand just without a cosmological constant.
Assuming the universe to be spatially homogeneous and isotropic, and combining this with the Einstein field equations produces the two Friedmann equations$$\frac{\dot{a}(t)}{a(t)} = \frac{8\pi G}{3}\rho - \frac{k}{a^2(t)}+\frac{\Lambda}{3}$$ and $$\frac{\ddot{a}(t)}{a(t)}=-\frac{4\pi G}{3}(\rho+3p)+\frac{\Lambda}{3}$$ where $k=+1,0,-1$ depending on curvature. $\Lambda$ is the cosmological constant.
if we want $\ddot{a}(t)=\dot{a}(t)=0$ (no expansion) and $\Lambda=0$, then the first equation implies $\frac{8\pi G}{3}\rho a^2(t) = k$. This will not work if $k=0, -1$ since the left side is nonzero and positive. The second equation leads to $\rho+3p=0$: for any positive density there has to be negative pressure even if we are just thinking of the contents of the universe as pressure-free dust. So it looks like $\dot{a}(t) \neq 0$... unless one adds a suitable nonzero value of $\Lambda$ to make things stand still.
Universe is expanding.
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How to compute gauge potential $A$ from the field strength $F$? Let $F=dA+A \wedge A$ be the field strength that solves vaccum Yang-Mills equation.
The question is: how to recover the gauge potential $A$? Is there any standard way? or any theorem stating the solvability? Suppose the metric is $g=g_{\mu \nu}dx^{\mu}dx^{\nu}=\eta_{ab}e^ae^b$, $e^a$ is tetrad basis.
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You can't recover the gauge potential uniquely without specifying more information, because of gauge invariance. $A$ is not uniquely defined by $F$. For any gauge transformation $g: \Sigma \to G$, the transformed connection $A^g = g^{-1}A g + g^{-1}dg$ has curvature $F_{A^g} = g^{-1}F_A g$.
It's physically a bit weird to specify $F_A$ in non-Abelian gauge theories. Gauge transforms don't leave $F_A$ invariant, so $F_A$ isn't an observable. But if one insists that $F_{A^g} = F_A$, then all connections $A^g$ with $g$ in the centralizer of $F_A$ give the same curvature.
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Can an electron have a spin opposite to the hole? Consider the simplest case - a gapped system where the electron in the valence band is excited to the conduction band. In this process, is the spin conserved? Or to word it differently, can the excited electron have a different orientation of spin compared to the hole?
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Yes, the Spin of the electron will be the same (as it is essentially the same electron of the same shell).
[
Moreover, the spin of the electron will not matter at all, as the other electron of different atom 'lives' in splited subshell (different).
So, Auf Bau'S principle will not be violated
]
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How can $1/V$ be equal to $0$ in Boyle's Law? In relation to ideal gases, Boyle's Law states that pressure is inversely proportional to volume under constant temperature. In other words,
$$P \propto 1/V$$
Below is a graph that plots pressure, $P$, against inverse volume, $1/V$.
How can $1/V$ ever equal zero? How is this possible?
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You've got to start with the ideal gas law
$$pV=mRT$$
Where, in this version, $m$ is the mass of the gas, $R$ is the specific gas constant and $T$ is the temperature. If $m$ and $T$ are constant, then $mRT$ = constant = $C$.
$$p=\frac{C}{V}$$
It is important to note that Boyle's law only applies to a closed system, i.e., a system where $m$ is constant. So even if the volume increases there are always the same number of gas molecules within that volume. Although the density of the gas (molecules per unit volume) keeps decreasing with increasing volume, it never becomes zero. And as long as there are gas molecules, there will be collisions between the molecules and any surfaces within or bounding that volume resulting in pressure proportional to the collision rate.
So if the volume were infinite, it would simply mean that an infinite amount of time would be required for a collision to occur. The rate of collisions approaches zero, but can never actually become zero as long as there are gas molecules in the volume.
Hope this helps.
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Period behavior near separatrix in Hamiltonian system Given the periodic potential Hamiltonian $H=\frac{p^2}{2} - \omega_0^2 \cos(q)$ I would like to show that near the separatrix the period has this behavior: $T(E)\sim |\log(\delta E)|$ with $\delta E=|E-\omega_0^2|$.
More generally given an Hamiltonian system of the form $H=\frac{p^2}{2} + V(q)$ with $V''(q^*)\ne 0$ for a non stable fixed point, I would like to show that near the separatrix we get the same kind of law.
I could prove that $p$ is a solution on the separatrix and found an infinite period. Then I tried doing different development of $E$ to first order and second order but didn't get any result. Do you have any idea on how to do that for the first case and then maybe the general case?
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The period is calculated by the integral
$$
\sqrt{2}\int_0^{2\pi} \frac{d q}{ \sqrt{\omega_0 ^2 \cos (q)+E}}
$$
which can be represented by special functions. After applying a replacement $E\to \delta E+\omega^2_0$, you need to expand this integral around separatrix $\delta E=0$, the leading term is $-2\omega_0^{-1}\ln(\delta E)$. Thus the leading term of period around $\delta E=0$ will be $T\sim-2\omega_0^{-1}\ln(\delta E)$.
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Speed, acceleration, accelerating acceleration, etc. How do we know where to stop? I am not a physicist.
Suppose a body A is falling towards body B in a vacuum. We know that A's speed will increase. However, as A draws near to B, the force of gravity will increase so the rate at which A accelerates will increase. Also we can presume that B's motion is affected by A. So now we have multiple levels of acceleration.
Question
I understand that, for practical purposes we can usually neglect smaller effects. My question is: In Nature do we ever get to the end of this apparently bottomless pit of derivatives?
Considerations
A and B are of comparable but non-identical mass. They will therefore accelerate towards one another at differing rates.
When they get close enough, they can no longer be assumed to be a dimensionless point wrt gravity.
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In your question, you are tracking both ${\bf r}_A$ and ${\bf r}_B$ (the position vectors in a fixed coordinate system), with something like:
$$ m_a\ddot {\bf r}_A = km_am_b\frac{{\bf r}_A - {\bf r}_B}{||{\bf r}_A - {\bf r}_B||^3}$$
$$ m_b\ddot {\bf r}_B = km_am_b\frac{{\bf r}_B - {\bf r}_A}{||{\bf r}_A - {\bf r}_B||^3}$$
which is not an endless pit of derivates, but rather a endless cycle of second derivatives.
However, there is a coordinate transformation that helps. If you rewrite the equations in terms of:
$$ {\bf R} = \frac{m_a{\bf r}_A + m_b{\bf r}_B}{m_a+m_b}$$
$$ {\bf r} = {\bf r}_A - {\bf r}_B $$
you should find:
$$ \ddot {\bf R} = 0 $$
and
$$ \frac{m_am_b}{m_a + m_b}\ddot{\bf r} = km_am_b\frac{{\bf r}}{||r||^3}$$
which breaks the cycle.
The first coordinate is the evolution of the center-of-mass: since there are no external forces, it moves at constant velocity.
The other coordinate is just the separation, which works when you use the reduced mass ($\mu$), defined via:
$$ \frac 1 {\mu} = \frac 1 {m_a} + \frac 1 {m_b} $$
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Net force on links in chain? Suppose there is a chain with 5 links in it where each link has a mass of $m=0.1kg$, and the chain is being accelerated upward at $2.5 m/s^2$. I want to find the net force on each link in the chain.
I would have thought that the net force is the sum of the force from gravity (i.e., its weight) with the upward acceleration. But the book answer has 0.25N, indicating that only the upward acceleration is relevant.
After thinking about it, I realized that the downward force on each link is being offset since each link is supported by the link above it (with the top one supported by the rope accelerating it upward). Thus for each $W=mg$ force vector downward, there is a normal force going upward to offset it because of the normal force from the supporting link above it, leaving only the upward acceleration of the chain as a whole as the net force, which gives $F=(0.1kg)(2.5m/s^2)=0.25 N. $ Is my reasoning correct?
Thanks for any help.
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The reason each link will have the same net force is because:
*
*Each link has the same acceleration, and;
*Each link has the same mass.
Since the net force is $\Sigma F=ma$ and $m$ and $a$ are the same for each case, the net force on 1 link is the same as the net force on each other link.
So yes, each link will have a net force of $\Sigma F_\mathrm{link}=T_i-T_i'-mg=ma$ where $m$ is the link mass, and $T_i$ is the force pulling up that specific link (each link will have its own magnitude for $T_i$), and $T_i'$ the force pulling up the link below (Newton's third law action-reaction pair).
I would have thought that the net force is the sum of the force from gravity (i.e., its weight) with the upward acceleration.
The net force on the system is $\Sigma F_\mathrm{system} = T-Mg$ where $T$ is the upwards force pulling everything up and $M$ is the mass of the five chain links, $M=5m$.
The net force of a link is $\Sigma F_\mathrm{link}=T_i-T_i'-mg$ where $m$ is the mass of 1 chain link and $T_i$ is a force pulling up that specific chain link (different for each chain), and $T_i'$ is the force pulling up the link below.
The following FBD should clear it up:
You can also calculate the net force on the system, namely:
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Peskin and Schroeder, where is the mass in the denominator of the simple harmonic oscillator Hamiltonian? This relates to page 20 of Peskin and Schroeder.
They state that the Fourier transform of the Klein-Gordon field satisfies the following:
$$\left[\frac{\partial^2}{\partial t^2}+(|\vec p|^2+m^2)\right]\phi(\vec p,t)=0 \tag{2.21},$$
which is the equation of motion of a simple harmonic oscillator with frequency:
$$\omega_\vec p=\sqrt{|\vec p|^2+m^2} \tag{2.22}.$$
This is fine, however their next equation is the Hamiltonian for the simple harmonic oscillator:
$$H_{SHO}=\frac{1}{2}p^2+\frac{1}{2}\omega^2\phi^2,$$
which, confusingly to me, does not have a mass $m$ in the denominator of the kinetic term. I have searched around a bit online and not found any reference to this, have I missed something?
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Neither does it have a mass in the numerator for the $\phi^2$ term! Peskin & Schroeder just do not bother with a constant $m$ is this context. As you can see, this part introduces you to the ladder operators, in order to apply the formalism to the Klein-Gordon hamiltonian. No need to worry about $m$'s, which are irrelevant to the commutation relations anyway, set it to 1 and work your way through the SHO properties.
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Where does the law of conservation of momentum apply? Take the scenario of a snowball hitting a tree and stopping. Initially, the snowball had momentum but now neither the snowball nor tree have momentum, so momentum is lost (thus the law of conservation of momentum is violated?). Or since the tree has such a large mass, is the velocity of the tree is so small that it's hardly noticeable?
If the explanation is the latter, this wouldn't hold for a fixed object of smaller mass. So in that case, how would the law of conservation of momentum hold?
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Let give a bigger picture. Assume the snowball appears (due to atmospheric processes) at some height "h". From this moment, gravity accelerates it against earth (discard wind forces). In the same way, earth is accelerated towards the snowball. Gravitational forces are the same for snowball and earth in modulus but opposite sense, same for momentum, accelerations are not. Total momentum is always zero.
After some time, the snowball and earth collides. Their momentum cancel, final velocity after collision vanishes and the final object snowball+earth is again at rest (regarding this subject), returning to its initial stage.
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Can thermodynamics be considered logical? One of the laws says that heat won't flow from cool to warm and at the same time this same theory claims that there is a finite (albeit tiny) chance that it will, because there is always such a microstate.
We can also have a situation where all air molecules in the room can be found in the left side of the room and none in the right side, because it is one of the microstates therefore it can happen and the entropy will drop. So how can we say that the entropy always increases when it can decrease too sometimes?
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I think it is a matter of probabilities. The macrostate that is seen when the system is in equilibrium has enormously large number of microstates than the macrostate where all gases are on one side of the system.
'Enormously large' still does not feel large enough, its actually mind bogglingly large. Thus, the probability of the system getting divided into two compartments spontaneously would be impossible statistically.
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What makes a standing wave a wave? Well, this is my physics professor's question and it really made me think a lot about standing waves, realising I don't actually understand it. What makes a standing wave a wave? How could I explain it? What actually makes wave a wave?
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As John Rennie says, a wave is defined in physics to be a function which satisfies a wave equation. I just want to add that what makes this a wave in common-speak is that the solutions are cyclic (i.e. repeating) functions. This is of course already clear to anyone familiar with the maths in John Rennie's answer which specifies the simplest solutions as sin waves, and that all solutions are sums of sin wave solutions (given by Fourier analysis). This does of course include standing waves, just as he describes.
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What experiment confirms $\mathbf{J}^2 = \hbar^2 j(j+1)$? I learned that if we measure the spin angular momentum of an electron in
one direction $J_z$, we get $\pm \frac{1}{2} \hbar$. But if we measure
the magnitude of the angular momentum $\mathbf{J}^2$, we should get
$\frac{3}{4} \hbar^2$. What experiment gives the latter result?
As @user1585635 notes, measuring $J_x$, $J_y$, and $J_z$ separately and
summing their squares gives $\frac{3}{4} \hbar^2$. This is not what I'm
looking for. First, if I measure $J_z$, measure angular momenta in
three directions separately, and measure $J_z$ again, the two
measurements of $J_z$ aren't guaranteed to be the same, since $J_z$,
$J_x$, and $J_y$ don't commute. But $\mathbf{J}^2$ commutes with $J_z$.
Second, when the spin is not $\frac{1}{2}$, say it's $j$, summing the
squares of components of $\mathbf{J}$ gives 3ℏ²², where ² should
be ℏ²(+1) doesn't always give $\hbar^2 j(j+1)$. (Thanks to
@MichaelSeifert for pointing that out)
This question is not a duplicate of
Why is orbital angular momentum quantized according to $I= \hbar \sqrt{\ell(\ell+1)}$?.
That question is about how $\mathbf{J}^2$ is derived mathematically.
Mine is about how it is confirmed experimentally.
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The Zeeman effect, discovered in the 19th century, was understood in quantum theory by Lande and Sommerfeld early on. Essentially, in a magnetic field, the energy levels of atoms are split by differences of magnetic moments, which are proportional to $J_z$, or the corresponding S and L components comprising J. This then allows to count the degenerate states so split of a system with fixed J (S, L, etc...) in a mixed population, graphically and dramatically. So at the most primitive level, the effect is a handle on degeneracy, $2j+1$.
This degeneracy, $2j+1$, is an easy counting observable, emerging as an eigenvalue of the formal operator
$$
\sqrt{4\mathbf{J}^2/\hbar^2+1} ,
$$
the ready hands-on manifestation of the quantum number j, which bounds the eigenvalues' magnitude $|j_z|$ from above. Even if you had no idea about quantization, the discreteness of these eigenvalues would cue you in to the quantum nature of the lines' spacings, and the integer nature of all these quantities. In the 1920s, Wigner and his cohorts changed the face of physics fully understanding these regularities in terms of the theory of the rotation group (Their injection of theory was termed "Gruppenpest" --Group pestilence--by the experimental community at the time).
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$P=\epsilon_o \chi E$ or $\epsilon_o \chi E_o$ Suppose the polarisation inside a dielectric is given by $P$, then is it related to the electric field as $\vec{P}=\epsilon_o \chi \vec{E}$ where $E$ is the field inside the dielectric or is is $E$ the original field that would have been present in that region in absence of the dielectric?
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$E$ is the field inside the dielectric. "Original field" is called "electric displacement" and usually denoted by $\mathbf D$, not $\mathbf E$.
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What causes pain when experiencing a static shock? What exactly causes the pain in my fingertip when I get a static shock from touching something? Is it the current flowing through my nerves or is it a localized thermal burn from the spark itself?
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It is the flow of tiny amounts of current through the nerve endings that creates the pain of a shock. The brevity of the (tiny) heat pulse is not detectable by your nerves unless the current is really big (which does not happen with shocks caused by static electric discharges)- which current might also be big enough to render you unconscious, in which case you probably will not experience (or remember) the pain or the heat flash.
There is a youtube video of a brown bear that climbed up a high-voltage power pole and before he could be rescued, shorted himself across two of the wires. At the instant this occurs, a flash of steam and smoke is blown out of his pelt as he falls unconscious to the ground. In this case, the heat pulse was sufficient to boil the water out of his hide, but not enough to kill him. By some miracle, he survived both the shock and the fall.
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Plotting the Bethe-Bloch Formula We had in our scriptum the following formula for the Bethe-Bloch formula:
$$-\frac{dE}{dx} = K\frac{\rho Z}{A} \frac{z^2}{\beta^2} \left[ \ln\left( \frac{2m\gamma^2\beta^2}{I} - \beta^2 \right) \right], $$
where I set $c = 1$ for convenience, $K$ is a constant, $I$ is the average ionisation potential, and $\rho$ denotes the density of the material.
My question is: I have often seen plots which show on the $x$-axis $\beta\gamma$, but I am not sure how one would plot $-\frac{dE}{dx}$ as a function of $\beta\gamma$, given that in front of the $\ln$, we have the factor $\frac{1}{\beta^2}$ (and not $\frac{1}{\beta^2\gamma^2}$), and also in the $\ln$, we have $-\beta^2$ and not $-\left( \beta\gamma\right)^2$.
I would love hear some opinions on this!
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If you know $\beta$ you can compute $\gamma = (1 - \beta^2)^{-1/2}$, so it is just a matter of calculating $\beta \gamma$ for each value of $\beta$ you are interested in. And if you know $\beta\gamma$, you can extract $\beta$ from it, e. g. by plotting $\beta\gamma$ versus $\beta$ (again, $c = 1$):
When you know $\beta$, you can put it back into the Bethe-Bloch Formula.
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Mutual inductance - induced magnetic flux in the primary Let there be two coils, L1 (with self inductance L1), and L2, with self inductance L2.
The first coil is connected to a sinusoidal supply, and the second one is connected to a resistor load, as shown in the image:
As we know There is a changing current in coil 1, therefore a changing magnetic flux (which goes through the second coil).
Because of the changing magnetic flux, we have an emf induced in the second coil, and because it is a close circuit with a resistor we have also got a changing current in the second coil.
My questions -
As the formula shows the induced emf in the second coil is:
$\varepsilon_2 =-L_2\frac{\mathrm{d}I_2 }{\mathrm{d} t} - L_{21}\frac{\mathrm{d} I_1}{\mathrm{d} t}$
but because of the changing current in the second coil, we have a changing magnetic flux which induces an emf on the first coil, therefore we have a changing current in the first coil, a magnetic flux which induces a changing current in the second coil and so on.... the first coil induces an emf on the second and the second on the first...
So how come we don't take into account these infinity number of emf on each other into this formula?
Is my assumption of induced emf which creates an induced emf on the other coil which then again creates an induced emf on the first coil even true? is this process goes on and on?
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The infinite process you suggest exists but only for non-zero intervals of time. At a single time instant, there are only two induced effects on coil 1: that due to the coil 1, and that due to coil 2. That is what the equation
$\varepsilon_2 =-L_2\frac{\mathrm{d}I_2 }{\mathrm{d} t} - L_{21}\frac{\mathrm{d} I_1}{\mathrm{d} t}$
correctly takes into account.
When we solve the equations numerically or using some formal method, the functions $I_1(t), I_2(t)$ can be determined, in which the infinite number of effects of the past events can be seen. But this is not necessary for writing the equations down or solving them.
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Does Schwarzschild metric in Kruskal-Szekeres coordinates admit asymptotic ($r \to +\infty$) timelike observers? I thank in advance whoever will answer my question.
Schwarzschild metric in Schwarzschild coordinates in $\mathbb{R}^{1,3}$ is [1]:
$$ds^2=-\bigg(1-\displaystyle\frac{2M}{r}\bigg)dt^2+\bigg(1-\frac{2M}{r}\bigg)^{-1}dr^2+r^2d\Omega^2.$$
When $r\to+\infty$ then $g_{\mu\nu} \to \eta_{\mu\nu}$, with $\eta_{\mu\nu}$ Minkowski metric (which obviously admits timelike observers e.g. $(1,0,0,0)$). By operating the change of coordinate of Kruskal-Szekeres we have [1]:
$$ ds^2=\frac{32M^3}{r}e^{-r/2M}(-dT^2+dX^2)+r^2d\Omega^2$$
with $r=2M(1+W_0[(X^2-T^2)/e])$ and $W_0$ Lambert function. In this case, when $r \to +\infty$ we get $ds^2 \to r^2d\Omega^2>0$ i.e. any vector becomes spacelike there. How can this discrepancy be explained?
Thanks a lot.
References
[1] http://bascom.brynmawr.edu/physics/courses/325/Problem_Sets/Review_PS_3.pdf
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I don't think there are any $r\to\infty$ asymptotic timelike observers. There seems to be a limit of that sort in some coordinates, like Eddington-Finkelstein $(u,r)$, but it's an illusion. Conformal spatial infinity is a single point with no time dimension, and its causal past only contains past infinity.
The $r\to\infty$ limit of stationary worldlines on the Penrose diagram covers past and future null infinity, with a crossover at spatial infinity. This is the same thing that happens at $r=2M$, and the meaning is the same: the observer can't be timelike in the limit, and has to be null instead.
In Minkowski space, you could put the (future) observer at $(t{-}r)(s)=s$, with $t{+}r$ constant and tending to $+\infty$. The "redshift" seen by this observer is the same as that seen by an observer at finite $r$, if you take $s$ to be its "proper time". In the K-S case, a stationary worldline at finite $r$ is given by
$$(T{\pm}X)(t) = \pm R \exp(\pm t/4M)$$
where $t$ is the Schwarzschild time and $R=\sqrt{X^2-T^2}$, and the observer at infinity can be at
$$(T{-}X)(s) = -R_\infty \exp(-s/4M)$$
where $R_\infty$ is any positive value (adjustable by shifting the origin of $s$). The "redshift" this observer sees is the limit of redshifts seen by stationary timelike observers.
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Partial derivative in Newtons Second law Newton's second law states Force is the time derivative of momentum. But is it a total derivative or partial derivative? What is the reason behind it?
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Newton second law
$$\frac{d}{dt}(m\,\vec v)=m\,\frac{d\vec v}{dt}=\vec{F}\tag 1$$
the velocity $~\vec v$ is in general depending on the generalized coordinates $~\vec q~$ the velocity of the generalized coordinates $~\vec{\dot{q}}~$ and the time $~t$
$\Rightarrow$
the total derivative is
$$d\vec v=\frac{\partial \vec v}{\partial \vec q}\,d\vec q+
\frac{\partial \vec v}{\partial \vec{\dot{q}}}\,d\vec{\dot{q}}+\frac{\partial \vec v}{\partial t}\,dt$$
Eq. (1)
$$m\,\left(\frac{\partial \vec v}{\partial \vec q}\,\frac{d\vec q}{dt}+
\frac{\partial \vec v}{\partial \vec{\dot{q}}}\,\frac{d\vec{\dot{q}}}{dt}+\frac{\partial \vec v}{\partial t}\right)=\vec{F}$$
but if you obtain only the partial derivative , you get:
$$m\,\frac{\partial \vec v}{\partial t}=\vec{F}$$
this is not correct
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Polarity in a magnetized Möbius strip When a flat iron or Alnico washer is magnetized one of the faces develops a north polarity and the other, south. The geometric shape here is simple.
However, when a standard Möbius strip (or one of given thickness, radii of curvature and torsion of edges) is magnetized, which regions develop a north polarity and which regions south and according to which geometrical or other mass distribution criterion/law?
I am curious to know because such a Möbius strip (of rectangular section) has only one surface and only one edge.
Is magnetic polarity and strength distribution after magnetization influenced by changed geometry ( by homeomorphism ) ?
It may be easy to make a flattened thin Möbius strip looking like a recycling symbol to apply a magnetizing current. Thanks in advance for references.
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For a 'uniformly' magnetized material, the longitudinal (down the length) component can be continuous around the loop, but both the transverse (across the width) and perpendicular (through the thickness) components must encounter a 180-degree change somewhere around the loop.
If the magnetic anisotropy is high, the magnetization may make just a single sharp transition (domain wall). The type of domain wall will be a Neel wall for transverse magnetization or a Bloch wall for perpendicular magnetization.
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If a jet engine is bolted to the equator, does the Earth speed up? If a jet engine is bolted to the equator near ground level and run with the exhaust pointing west, does the earth speed up, albeit imperceptibly? Or does the Earth's atmosphere absorb the energy of the exhaust, and transfer it back to the ground, canceling any effect?
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As other answers have stated there would be a small change in the Earth's rotation rate and a change in the opposite direction to the atmosphere's rotation rate. The size of the change would be limited by the increased friction between the earth and the atmosphere, and so the length of a day would settle to a new value if the jet engine were left running long enough. After the jet engine is turned off the increased friction would return the Earth to its usual rotation rate and the length of the day would revert to its natural value.
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If the sea surface were absolutely calm should the Sun reflection be the area of a circle instead a ribbon? Although waves produced on the sea can cause different points of the sea surface to reflect sunlight towards the same observer, how is that kind of ribbon image produced? Why isn't the reflection stretched also perpendicularly to the line of sight by wavyness of the sea?
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First off, some terminology: the 'ribbon' is known as sun glitter (and, when observed from space, as sunglint). A good explanation is at this NOAA page, but the basic dynamics is that the light reflects specularly off of the water, but the variable angles of the surface deflects it in different directions.
If you were to see the sun glitter at normal incidence (i.e. the Sun at the zenith, observed from above the water, either from a plane or from space) then it would look completely round. As the elevation of the Sun decreases and it moves from the zenith to the horizon, the glitter becomes more and more elliptical, until it becomes a ribbon as in your image.
On the other hand, if the water were completely still, with a smooth and horizontal surface, then the sun glitter ribbon would be gone, and you would just see a direct mirror image of the Sun.
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Lorentz indices to label rotation irreducible representations Consider the $A_\mu\in\left(\frac{1}{2},\frac{1}{2}\right)$, the vector representation of the restricted Lorentz group. One can decompose this vector under spatial rotations as $A_\mu\in 0\oplus 1$ where $A_0$ transforms as a 3D scalar and $A_i$ transforms as a 3D vector. If say we had $\psi_a\in\left(\frac{1}{2},0\right)\oplus\left(0,\frac{1}{2}\right)$ a Dirac spinor, under rotations it transforms as $\psi_a=\frac{1}{2}_L\oplus\frac{1}{2}_R$ where $a=1,2$ will be the left part and $a=3,4$ will be the right part.
The question I have is when we consider higher representations, say a $\frac{3}{2}$ particle, $\psi_{\mu a}\in\left(\frac{1}{2},\frac{1}{2}\right)\otimes\left[\left(\frac{1}{2},0\right)\oplus\left(0,\frac{1}{2}\right)\right]$. It is prety clear that under rotations it transforms as $\psi_{\mu a}\in\left(\frac{1}{2}\oplus\frac{1}{2}\oplus\frac{3}{2}\right)_L\oplus\left(\frac{1}{2}\oplus\frac{1}{2}\oplus\frac{3}{2}\right)_R$. However I don't see how I can distribute the indices $\mu$ and $a$ under rotations.
I am also interested in the general concept.
$\textbf{Edit: (to give more details for what my question is)}$
If we take the scalar part of the $\left(\frac{1}{2},\frac{1}{2}\right)=0\oplus1$, and multiply it by the the Dirac spinor we get $\frac{1}{2}_L\oplus\frac{1}{2}_R$. So I would say the part $\psi_{0a}$ transforms as $\frac{1}{2}_L\oplus\frac{1}{2}_R$.
What's left is $\psi_{ia}$ which transforms as $\left(\frac{1}{2}\oplus\frac{3}{2}\right)_L\oplus\left(\frac{1}{2}\oplus\frac{3}{2}\right)_R$. My question is: What part of the indices $i,a$ transforms as $\left(\frac{3}{2}\right)_L\oplus\left(\frac{3}{2}\right)_R$?
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This is a close duplicate.
$\psi_{\mu~\alpha}$, in $$\left(\frac{1}{2},\frac{1}{2}\right)\otimes\left[\left(\frac{1}{2},0\right)\oplus\left(0,\frac{1}{2}\right)\right],$$ has 16 components; so you project out the spinor piece, $\gamma\cdot \psi_\alpha =0$, $ \left[\left(\frac{1}{2},0\right)\oplus\left(0,\frac{1}{2}\right)\right]$,which removes 4 components,
leaving you with 12:
$$
\left[\left({1},\frac{1}{2}\right)\oplus\left(\frac{1}{2},{1} \right)\right].
$$
However, 4 of these remaining components are gauge (local susy) degrees of freedom, $\partial_\mu \epsilon_\alpha$, that do not enter the R-S action, so they are projected out as in vector gauge theories, leaving you with just 8,
$$ \left[\left(\frac{3}{2},0\right)\oplus\left(0,\frac{3}{2}\right)\right],$$
a left-handed and a right-handed quartet.
NB On-shell, one may go on: the absence of $\partial_0 \psi_{0~\alpha}$ from the action allows fixing $\psi_0=0$, reducing the 8 components to 4, and the Majorana condition reduces them to only 2 d.o.f., the extreme polarization states, analogous to photons, explained in Freedman & van Proeyen.
|
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Normalization Constant in Time Evolution of Density Matrix Given the Hamiltonian:
$$%H = \omega \left(|0\rangle \langle1| + |1\rangle \langle0| \right) = \begin{bmatrix}
0 & \omega \\
\omega & 0
\end{bmatrix}$$, I want to find the final state $\rho(t_f)$of the given density operator:
$$\rho(0) =|0\rangle \langle0| = \begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix} $$
To do so I started by stating:
\begin{equation}
\rho(t_f) = U\rho(0)U^\dagger\\
U=e^{-i\frac{t_f}{\hbar}H} \approx 1-i\frac{t_f}{\hbar}H\; \; \Rightarrow \;\;U^\dagger \approx 1+i\frac{t_f}{\hbar}H
\end{equation}
Although once I compute $\rho(t_f)$ using the above formula I obtain a non normalized state:
$$Tr(\rho(t_f))\neq 1 \; \; \forall \omega \neq 0$$
Of course this problem could be solved if out of nowhere I multiplied my $\rho(t_f)$ with a normalization constant N:
$$N = \frac{1}{Tr(\rho(t_f))}$$
My question is: is there something wrong with my thought process or calculations? Or do I really just have to introduce a new normalization constant? I would not mind an explanation in the option that the latter was the case(even if just as a reference).
I worked with it for a bit, and this is what I got:
P.S.
As suggested, I fully expand the U operator:
$$\%mathbf{U}=e^{-i\frac{t_f}{\hbar}\mathbf{H}} = \sum_n^\infty \left(\frac{c^n}{n!}\mathbf{H}^n \right)$$
Where for simplification I defined $c =i\frac{t_f}{\hbar}$.
By introducing a new operator denoted as $\mathbf{H}'$ ($\mathbf{H'} = \frac{1}{\omega}\mathbf{H}$), I notice the property:
$$\mathbf{H}^n=\left\{\begin{matrix}\omega^{n} \mathbf{I},& if \;\; n = even \\
\omega^n \mathbf{H}',& \; \; \; \; if \;\; n = odd
\end{matrix}\right.$$
Hence, the problem to solve becomes:
$$\mathbf{\rho}(t_f) = -\left(\sum_n^\infty \frac{c^n}{n!}\mathbf{H}^n \right)\rho(0) \left(\sum_n^\infty \frac{c^n}{n!}\mathbf{H}^n \right)$$
$$=%-\left( \sum_n^\infty\frac{c^{2n}}{2n!}\omega^{2n}\mathbf{I} + \frac{c^{2n+1}}{(2n+1)!}\omega^{2n+1}\mathbf{H'}\right)\rho(0)\left( \sum_n^\infty\frac{c^{2n}}{2n!}\omega^{2n}\mathbf{I} + \frac{c^{2n+1}}{(2n+1)!}\omega^{2n+1}\mathbf{H'}\right)$$
$$=-\left[ \sum_n^\infty \omega^{2n} \left( \frac{c^{2n}}{2n!}\mathbf{I} + \frac{c^{2n+1}}{(2n+1)!}\omega\mathbf{H'}\right )\right]\rho(0)\left[ \sum_n^\infty \omega^{2n} \left( \frac{c^{2n}}{2n!}\mathbf{I} + \frac{c^{2n+1}}{(2n+1)!}\omega\mathbf{H'}\right )\right]$$
$$=-\left( \sum_n^\infty \omega^{2n} \begin{bmatrix}
\frac{c^{2n}}{2n!} & \omega\frac{c^{2n+1}}{(2n+1)!}\\
\omega\frac{c^{2n+1}}{(2n+1)!}&\frac{c^{2n}}{2n!}
\end{bmatrix} \right)\left( \sum_n^\infty \omega^{2n} \begin{bmatrix}
\frac{c^{2n}}{2n!} & \omega\frac{c^{2n+1}}{(2n+1)!}\\
0&0
\end{bmatrix} \right)$$
$$%= \sum_n^\infty \omega^{4n}\begin{pmatrix}\frac{c^{4n}}{\left(2n\right)!\left(2n\right)!}&\omega\frac{c^{4n+1}}{\left(2n\right)!\left(2n+1\right)!}\\ \omega \frac{c^{4n+1}}{\left(2n\right)!\left(2n+1\right)!}& \omega^2\frac{c^{4n+2}}{\left(2n+1\right)!\left(2n+1\right)!}\end{pmatrix} = \mathbf{\rho}(t_f)$$
|
Additionally to Norbert Schuch's answer and comments, where he points out your problem of normalization, I want to add a brief note about the exact calculations that can be performed in this example.
First, for convenience, I want to take the factor $\omega$ out of the definition of the Hamiltonian. Now we have to notice (you also stated it in a comment) that
$$ H^{n}=\begin{cases} |0\rangle\langle 0| + |1\rangle\langle 1| &\quad \text{for } n \text{ even} \\ H &\quad \text{for } n \text{ odd} \end{cases} \quad .$$
To proceed, the exponential of an operator is defined by
$$e^{c\,H} \equiv \sum\limits_n \frac{c^n}{n!}\, H^n \quad,$$
for $c\in\mathbb{C}$.
You can use this relation for your operators $U$ and $U^\dagger$. To make us of the elaborated properties of the Hamiltonian, you have to split the series into even and odd terms. You'll find a very simple expression for these operators. From this, you can calculate $U\, \rho(0) \,U^\dagger$ and thus find a form of $\rho(t)$.
Edit: Of course, if you calculate the exact $\rho(t)$, then $\mathrm{Tr}\rho(t) = 1$.
Edit 2: I think it would be easier to first consider only the expansion of $U$, which reads ($\hbar=1$):
$$U(t) = \underbrace{\sum\limits_{n=0}^{\infty} \frac{(-i\,\omega\,t)^{2n}}{(2n)!} H^{2n}}_{\text{even}} + \underbrace{\sum\limits_{n=0}^{\infty} \frac{(-i\,\omega\,t)^{2n+1}}{(2n+1)!} H^{2n+1}}_{\text{odd}} \quad.$$
If you now use the properties of $H$ for the even and odd series, you will find
$$ U(t) = \sum\limits_{n=0}^{\infty} \frac{(-i\,\omega\,t)^{2n}}{(2n)!}\, \left(|0\rangle\langle 0| +|1\rangle\langle 1| \right)
+ \sum\limits_{n=0}^{\infty} \frac{(-i\,\omega\,t)^{2n+1}}{(2n+1)!}\, H \quad.$$
Now you can simplify this expression with the help of the sine and cosine function, i.e. their respective series expansions. From there it is easy to obtain $U^{\dagger}(t)$ and also straightforward to calculate $\rho(t)$. Still, if you have questions, let me know.
|
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Identifying inertial system without already having an inertial system? Learning some basic physics principles via Spivak's Physics for Mathematicians, and he rephrases Newton's first law as
Newton's First Law: There is at least one coordinate system — an inertial system — in which any object not acted upon by any forces has constant velocity.
In practice, do we have a way to measure forces that doesn't rely on choice of coordinate system?
The only way I can think of to measure forces relies on choosing a coordinate system (an inertial frame). For instance, I see a lead ball placed atop of a spring compress the spring, and the deceleration here is informing me about the force. However, I am only able to observe deceleration because I have put myself at the centre of a coordinate system, not the ball.
So my current understanding is that to pick an inertial system we first need to know the forces in the system, but to identify forces in a system we first need to pick an inertial system. To break this circular dependency, shouldn't there be a way to define forces independent of the inertial system chosen?
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In practice, do we have a way to measure forces that doesn't rely on choice of coordinate system?
Yes, we use accelerometers (more properly inertial measurement units which measure acceleration on three axes and rotation about three axes).
Regardless of your coordinate system, if your accelerometer reads 0 then the object is moving inertially. An inertial frame is then a system of coordinates where all such objects have straight worldlines.
Note, this implies that inertial frames are free falling frames, which is different from how Newton originally consider things, but in modern physics it has been found to be better.
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Why do we not observe a greater Casimir force than we do? I am very new to quantum field theory, so forgive me if this question is a bit silly. The Casimir force is usually explained by the zero point energy of the field. You assume that the frequencies of the field are quantized between the two plates, perform some regularization, and out pops, for the electromagnetic field, $$F=-\frac{\pi^2\hbar c}{240a^4}A$$
where $a$ is the separation and $A$ is the area of the plates. However, what if we have multiple fields? For an ordinary scalar field, I believe the Casimir force for that only differs by a factor of $2$ (due to the polarizations of light), so we have $$F=-\frac{\pi^2\hbar c}{480a^4}A$$
In a world with both of these fields, I'd assume the total Casimir force would be their sum of each contribution. In the real world, we have a bunch more fields than just the electromagnetic one (including a scalar Higgs field)! I would assume that each of these would produce a Casimir force in the same manner as the scalar field and the electromagnetic field, and that the total Casimir force is their sum. However, we only measure the Casimir force due to the electromagnetic field. Why is this? Is there a flaw in my reasoning?
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Metal plates impose a boundary condition on the electromagnetic field, because metal is made of charged particles which interact with an electromagnetic field. But those metal plates do not impose a boundary condition on the Higgs field, which extends through conductors.
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|
In QFT why does the degree of the interaction terms in Lagrangian start from 3? I'm new to QFT so it's not obvious to me why there is no quadratic interaction terms in Lagrangians.
For example, the Lagrangian for a real scalar field is
$$L=\frac{1}{2}\partial_\mu \phi \partial^\mu \phi-\frac{1}{2}m^2\phi^2-\sum_{n\geq 3}\frac{\lambda_n}{n!}\phi^n.$$
What's the reason that we can't add terms like $g\phi^2$ to the free field Lagrangian?
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Given that an interaction of the $n^{th}$ order implies the existence of an $n$ pointed vertex.
ex. $L_1 = g \phi^3$ in a real scalar field theory gives rise to a $3$ pointed vertex.
If the interaction term had form $$L_1 = g\phi^2$$ this would just be a $2$ pointed vertex. If we allow conservation laws in this theory then a particle going into the vertex with energy and momentum $(p_i, E_i)$ will have to be equivalent to the outgoing one, making $(p_f, E_f) = (p_i, E_i)$. a.k.a, it's the same particle and nothing changed.
It is interesting to see that given a free Lagrangian
$$L_0 = \frac{1}{2}\partial^\mu\phi\partial_\mu\phi + \frac{1}{2}m^2\phi^2,$$ by applying the previous interpretation here, $m$ could be seen as the coupling strength between a free non-interacting particle and, nothing.
|
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Einstein's light speed postulate I have seen two statements of Einstein's 1905 light speed postulate; for instance, in Andrew Steane's Relativity Made Relatively Easy:
*
*There is a finite maximum speed for signals.
*There is an inertial reference frame in which the speed of light in vacuum is independent of the motion of the source.
Does anyone have a proof that these statements of the postulate are equivalent? Can their equivalence be shown without resorting to the relativity postulate?
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I like this, so I'm going to try. I'm going to add in "light transmits signals", since it seems like that equivalency is not what the OP is getting at. Also, I'm going to say "there is a maximum speed for everything". I think those two things come from reasonable definitions of signals, light, and what a "source" is:
1->2: Go to a frame in which the speed of light is maximized. If the speed of light is maximized, it is constant, which implies the speed of the frame must be constant as well (relative to some other frame).
2->1: Say we are in this frame, and the speed of light is c, regardless of the motion of the source. Now consider that the source is some kind of laser, moving in the direction of the beam at a speed v. If v > c, we can't in any reasonable way measure the speed c - it is either equal to v, violating the assumption, or the light doesn't get emitted, implicitly violating the assumption. So (1).
I'm not incredibly happy with that - for example, I would rather have used Newton's first last to get 1->2. But I think it's reasonably good.
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Have anyone seen an unbound electron absorb/emit a photon via Bremsstrahlung process? I know free electron cannot absorb or emit photon as it cannot satisfy both the conservation of energy and momentum at once but how about in Bremsstrahlung process? The presence of a positively charged particle is used to deflect the trajectory of an unbounded electron causing it to decelerate/accelerate depending if it's inwards or outwards, then it should emit photon no?
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Bremsstrahlun photons were used in order to identify electrons when studyig elementary particle interactions in cloud chambers and bubble chambers.
Strong magnetic fields were imposed so that charged particles would bend and their momentum measured. The photon from the bending electrons pair-produces in the chamber and thus identify the photon
In this paper one such example is given in page 19,
This unassuming picture is one of the bubble chamber ‘greats’: a tiny electromagnetic shower demonstrates the existence of a very simple, but rare, collision between an anti-muon neutrino ̄νμbeam particle and an atomic electrone−in the bubble chamber liquid
The first e− is the one that was knocked out of an atom by the ̄νμ. The two little e+e− pairs are produced by bremsstrahlung photons emitted from this first electron, thus identifying it.
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Question regarding Lorentz Transformation and Space Contraction- Contradiction I stumbled upon this question regarding Special relativity- and have seemed to reach a contradiction.
I am trying to find the distance that the ball travels
I am obviously not looking for the numerical answer, but I'm trying to understand what should be my intuition when looking at these types of problems.
The train is moving at a velocity of $c/2$ relative to earth, while the ball is moving at a velocity of $c/3$ relative to the train.
The train has a proper length of $L_0$.
Now, when trying to find the distance that the ball travels regarding earth, I see two approaches:
*
*If we set $t=t'=0$ as the time where the ball is at the back if the train ($x=x'=0$), we find that the time he travels in the train's frame of reference is $3L_0/c$ and the distance is $L_0$. Using Lorentz transformation we find the distance traveled is $5L_0/\sqrt{3}$ in the earth frame of reference.
*In the train's frame, the ball is moving at a speed of $c/3$ a distance of $L_0$. Using the proper length we are able to calculate the train's length in the earth's frame as $\sqrt{3}L_0/2$. Seeing as the ball starts at the back of the train (in both frames) and reaches the front of the train, this is the balls distance traveled.
What am I missing here? Which approach should I use and where does this contradiction come from?
Thanks a lot in advance!
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When dealing with problems in Special Relativity its best to deal with individual events. So let's consider two events here: the ball leaves the "back end" of the train, and the ball arrives at the "front end" of the train.
Now let's see what we know: The ball leaves the back end ($x=0=x'$) at $t=0=t'$, and the front end of the train is at $x'=L_0$, the proper length. Using just this information, we can figure out everything else. For starters, in the train's rest frame, the ball covers a distance of $L_0$ with a speed of $c/3$ so, as you point out, the ball will hit the front end at $$t' = \frac{3L_0}{c}.$$ To make this clearer, let's make a table:
\begin{array} {|c|c|}\hline \textbf{Event} & \text{Train Frame} & \text{Earth Frame} \\ \hline \text{Ball leaves the back} &\,\,\quad t'=0,\quad\quad\,\, x'=0 \quad\quad & t=0, \quad x=0 \\
\hline \text{Ball arrives at the front} & t'=3L_0/c, \quad x'= L_0 & t = {\color{red}?}, \quad x = {\color{red}?} \\ \hline \end{array}
We can now use the information we have to find both $t$ and $x$, the coordinates of the second event ("Ball arrives at the front of the train") in the Earth Frame. Let's write the Lorentz Transformations in terms of the difference of the events: \begin{aligned}\Delta x &= \gamma (\Delta x' + v \Delta t')\\ ~\\\Delta t &= \gamma \left( \Delta t' + \frac{v}{c^2}\Delta x'\right)\end{aligned}
From the table, you should be able to see that $\Delta t' = 3L_0/c$ and $\Delta x' = L_0$. Given that $v=c/2$, you can show that \begin{aligned}\Delta x &= \frac{5}{\sqrt{3}}L_0\\ \Delta t &= \frac{7}{\sqrt{3}}\frac{L_0}{c} \end{aligned}
So your first approach is correct, the distance between the events in the Earth frame is indeed $5L_0/\sqrt{3}$. However, we can now see why the second approach is wrong: the ball doesn't just cover the length of the train. The front wall of the train is also moving! Therefore, the distance covered by the ball in the Earth frame is given by $$\text{Length of the train (in Earth frame)} + \text{Distance covered by the front wall (in Earth frame)}$$
Clearly (as you pointed out) $$\text{Length of the train (in Earth frame)} = \frac{L_0}{\gamma} = \frac{\sqrt{3}}{2}L_0,$$ and the distance the front wall covers in the time $\Delta t$ is $$\text{Distance covered by the front wall (in Earth frame)} = v\Delta t = \frac{7}{2\sqrt{3}}L_0.$$
Adding them up, you can see that $$\text{Distance covered by ball in Earth frame} = \frac{L_0}{2}\left(\sqrt{3} + \frac{7}{\sqrt{3}}\right) = \frac{5}{\sqrt{3}}L_0,$$ as you'd imagine, since it's exactly what we calculated above.
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Is there a correlation between the Earth's spin, the moon, and temperature? Based on NASA's arcticle, changes to land, ice sheet, ocean, and mantle flow affect Earth's spin.
Does the moon's elliptical orbit around Earth also affect Earth's spin? What effect does distance play?
The moon has also been drifting away continuously from the Earth. I've read that's due to the Earth's slowing axial rotation.
Would the moon's exodus from the Earth eventually reverse with Earth's recent trend of faster spining?
Would a faster spinning Earth increase the friction between the ocean and land, increasing the global temperature?
Does global warming perpetuate a cycle of faster rotation and and warmer oceans?
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This is a change that's happening over bullions of years. So the changes exist, but are so small they have little relevance on a timescale of years or centuries, probably not millennia either.
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|
Calculation of time for electronic transitions How do we determine the time for electronic transitions in atoms or in semiconductor devices, from one energy level to another?
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For an absorption process, the transition time (in order of $femto second$) is usually estimated using uncertainty principle:
$$
\Delta t = \frac{\hbar}{\Delta E} = \frac{\hbar}{E_2 - E_1};
$$
Where $E_1$ and $E_2$ are the energies of the two levels.
For emision process, the life time of electron is much longer (in the range of pico seconds ~ nano seconds). The mechanisms cause the emission are very complicated often coupled to phonon, and therefore depends on temperature.
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|
Amount of matter in two infinite universes I don't know exactly where to find any other information on this question so I thought I would ask here. If there were two infinitely large universes one where 20% of the space in the universe was taken by matter and nothing else and one with 30% of its space taken up by matter and nothing else would you be able tell the difference of matter percentage between the universes or would they seem to have the same amount.
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Because percentages are dimensionless numbers (each representing only a portion of the particular object of which it is a percentage), they are not what the mathematician Cantor (extremely important in astronomy and cosmology) called "countably infinite": As a result, the information you may be seeking, which I'd imagine might be a comparison of their sizes, cannot be elaborated from the information that you've provided about the universes concerned. (The comparison, from the specific information that you've provided so far, would simply be that both universes would be infinite.)
You might edit your question to make it more specific.
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Why does a body in SHM have more probability (of being observed) at the extreme positions and not at other positions? There is a question on the number of times a body comes to a place in a simple harmonic motion. Have a look:
I thought that the answer was B because in each oscillation a body is at the extremes only once, while for other positions, it comes two times. But the correct answer is C which I don't know why.
How come this is true?
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At the ends, the oscillator slows down. So the probability of finding it at ends is high.
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Why is the tension equal to the spring force here?
Here the block is oscillating and to solve this question I took the tention in the string to be equal to the spring force
But if that's the case a particle in the junction of the spring and the rope will expirence a net force of zero yet it still it goes up and down every oscillation
So the tention should be higher than the spring force at some intervals
But most questions like these take them to be the same value
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As you seem to be having trouble with a general concept rather than asking for a homework solution, I will attempt an explanation. In an ideal situation they are equal. It takes more force, as the mass's weight carries it downwards, to stretch the spring further therefore more tension on the string. So as spring force increases so does string tension, in fact they must be equal per Newton's third law of motion.
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What information is sufficient for describing a thermodynamic system? For a single-component system, why are the energy, volume, and number of particles sufficient for describing the thermodynamics of the system? Why just three variables and those three variables in particular?
In the book that I am using (Callen, $\textit{Thermodynamics and an Introduction to Thermostatistics}$) he postulates that the macroscopic equilibrium state is characterized by the energy, volume, and particle numbers of its components, but what is the reason for this?
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A flippant, but not entirely inaccurate, answer is that those three variables were chosen because it was convenient.
From the Wikipedia page for Thermodynamic State:
A thermodynamic system is a macroscopic object, the microscopic
details of which are not explicitly considered in its thermodynamic
description. The number of state variables required to specify the
thermodynamic state depends on the system, and is not always known in
advance of experiment; it is usually found from experimental evidence.
The number is always two or more; usually it is not more than some
dozen. Though the number of state variables is fixed by experiment,
there remains choice of which of them to use for a particular
convenient description; a given thermodynamic system may be
alternatively identified by several different choices of the set of
state variables.
|
{
"language": "en",
"url": "https://physics.stackexchange.com/questions/611534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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|
Validity of Ampère's law in terms of $H$ We know that the auxiliary magnetic field $\bf{H}$ is
$$\mathbf{H}=\frac{1}{\mu_{0}} \mathbf{B}-\mathbf{M}$$
and
$$\mathbf{\nabla} \times \mathbf{H}=\mathbf{J}_{f}$$
but this differential equation is generally not valid at the boundary of a magnetized body due to an abrupt change in the magnetisation $\mathbf{M}$ at the boundary.
From this differential equation follows
$$\oint \mathbf{H} \cdot d \mathbf{l}=I_{f_{\mathrm{enc}}}$$
Is this integral equation valid if applied across a boundary of a magnetized material,(that is one end of the integral is inside the magnetised body and the other is out of it)? Or does it inherit any problems from its differential form?
EDIT: I think my question isn't clear so I'll add:
We derive the loop integral from the curl using stokes theorem. However the curl is not defined on the boundary so is the loop integral well defined? Another way of saying is that: can I use stokes theorem when the curl of function isn't defined on some region? –
|
Just use Ampère's law for $\mathbf B$ to get the total current. Then subtract the bound current from the result to get the free current, if needed.
|
{
"language": "en",
"url": "https://physics.stackexchange.com/questions/611662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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|
Converting energy from units of joules into units of wavenumber If we have an expression for energy $E$ in Joules why is it that if we wish to convert the energy into wavenumber we divide the expression by $hc$?
I know that $$E = \frac{hc}{\lambda}$$
So surely this implies that $$\frac{E}{hc} = \frac{1}{\lambda}$$
and since $$ k = \frac{2\pi}{\lambda}$$
Would it not make sense that in order to convert energy in joules to wavenumber we multiply $E$ by $\frac{2\pi}{hc} = \frac{1}{\hbar c}$.
So my question is why in converting energy from the units of joules to the units of wavenumber do we divide $E$ by $ hc $ rather than by $\hbar c$
|
From the Planck relation
$$E=h\nu = \frac{hc}{\lambda} = hc\tilde{\nu} $$
where in spectroscopy $\tilde{\nu}$ is the wavenumber and it has the following relation with angular wavenumber $k$
$$\tilde{\nu}=\frac{k}{2\pi}$$
|
{
"language": "en",
"url": "https://physics.stackexchange.com/questions/611777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
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