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Does work done on a spring = elastic potential energy? So the UK exam board specifications (AQA GCSE) clearly state
"...the work done on the spring and the elastic potential energy stored are equal"
Here's my problem,
So work done = Force x displacement
Force = Spring constant x extension
Elastic potential energy = 0.5 x spring constant x extension squared
Extension = displacement for a strecthed spring
However if I take some sample values and calculate the work done on the spring, and then the elastic potential energy stored. The Elastic potential is always exactly half the work done. This contradicts the statement in the specifications.
I have looked all over the place but can't find a satisfactory answer to this question. What am I missing? I get that elastic potential energy is equal to area under the Fx graph but why does that not equal the work done?
| The work is $Fx$ if the force is constant during the displacement from 0 to $x$.
In the case of an elastic spring, the force is a function of the displacement: $F = kx$
So the work for a small displacement when the spring is streched in a given $x$ position: $\Delta W = kx\Delta x$
Integrating from 0 to x: $W = \frac{1}{2}kx^2$, which is the stored elastic energy.
| {
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Can Schrödinger's cat be filmed? Before opening the box, the observer does not know if the cat is alive or dead, however, a camera placed internally "knows" all the time, which is really happening.
Does this camera cancel the result of the experiment, even if it is not consulted?
EDITING:
I made this edition just to clarify my doubts.
The camera could be replaced by a piezoelectric attached to the wall of the box, that contains the experiment. Both, the (flat) crystal and the wall, record the cat's heartbeat. When the observer opens the box, both, its walls and the piezoelectric already know the history of the cat's life in advance, regardless of being consulted.
This invalidates the observer's opinion in the moment he opens the box?
Why the inner walls of the box, which are also "observers", do not invalidate the experiment as it was designed?
| Yes, it is possible to put a movie camera in the box with the cat. But when you open the box and look at the movie, you will see the cat die or not die, but not both. And that will agree with what you see when you then look at the cat: it will be either dead or alive. The movie film will be entangled with the cat. By opening the box and looking, you randomly select one state or the other, and that's all you see is the selected state. Some call this "collapse of the wavefunction", and some call it "branching of the world". For most practical purposes it's the same thing.
| {
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Null geodetic on different metric relation between affine parameter If i have a null geodetic on a $g_{\mu \nu}$ metric, it's also null in a generic $\hat{g}=\exp{2\omega(x)}g_{\mu \nu}$ for any $\exp{2\omega(x)}\geq 0$?
What's the relation between affine parameter?
| Consider a null geodesic on $g$ with affine parameter $\tau$. We have the following equations
$$
\frac{d^2x^\mu}{d\tau^2} + \Gamma^\mu_{\nu\rho} \frac{d x^\nu}{d\tau} \frac{d x^\rho}{ d \tau} = 0 , \qquad g_{\nu\rho} \frac{d x^\nu}{d\tau} \frac{d x^\rho}{ d \tau} = 0 .
$$
For the metric ${\tilde g} = e^{2\omega} g$, we have
$$
{\tilde \Gamma}^\mu_{\nu\rho} = \Gamma^\mu_{\nu\rho} + \delta^\mu_\nu \partial_\rho \omega + \delta^\mu_\rho \partial_\nu \omega - g_{\nu\rho} \nabla^\mu \omega .
$$
Thus, the geodesic equation on the conformal metric takes the form
\begin{align}
\frac{d^2x^\mu}{d\tau^2} + {\tilde \Gamma}^\mu_{\nu\rho} \frac{d x^\nu}{d\tau} \frac{d x^\rho}{ d \tau} &= \frac{d x^\mu}{d\tau} \left( 2 \partial_\rho \omega \frac{d x^\rho}{ d \tau} \right) + \left( \frac{d^2x^\mu}{d\tau^2} + \Gamma^\mu_{\nu\rho} \frac{d x^\nu}{d\tau} \frac{d x^\rho}{ d \tau} \right) - \nabla^\mu \omega \left( g_{\nu\rho} \frac{d x^\nu}{d\tau} \frac{d x^\rho}{ d \tau} \right) \\
&= \frac{d x^\mu}{d\tau} \left( 2 \frac{d \omega}{d\tau} \right) .
\end{align}
Thus,
$$
\frac{d^2x^\mu}{d\tau^2} + \tilde \Gamma^\mu_{\nu\rho} \frac{d x^\nu}{d\tau} \frac{d x^\rho}{ d \tau} = \frac{d x^\mu}{d\tau} \left( 2 \frac{d \omega}{d\tau} \right).
$$
Thus, we find that same curve $x^\mu(\tau)$ is a geodesic (also null - reader should verify!) in the new metric as well! However, $\tau$ is not an affine parameter on the new metric. Suppose that the affine parameter is ${\tilde \tau} \equiv {\tilde \tau}(\tau)$. Then, writing out the equation in terms of ${\tilde \tau}$
\begin{align}
\frac{d^2x^\mu}{d \tilde \tau^2} + \tilde \Gamma^\mu_{\nu\rho} \frac{d x^\nu}{d\tilde \tau} \frac{d x^\rho}{d\tilde \tau} &= \frac{d^2\tau}{d\tilde \tau^2} \frac{dx^\mu}{d \tau} + \left( \frac{d\tau}{d\tilde \tau} \right)^2 \left( \frac{d^2x^\mu}{d\tau^2} + \tilde \Gamma^\mu_{\nu\rho} \frac{d x^\nu}{d \tau} \frac{d x^\rho}{d \tau} \right) \\
&= \left[ \frac{d^2\tau}{d\tilde \tau^2} + 2 \frac{d\tau}{d\tilde \tau} \frac{d \omega}{d\tilde \tau} \right] \frac{dx^\mu}{d \tau}.
\end{align}
Then, ${\tilde \tau}$ is an affine parameter iff.
$$
\frac{d^2\tau}{d\tilde \tau^2} + 2 \frac{d\tau}{d\tilde \tau} \frac{d \omega}{d\tilde \tau} = 0 \quad \implies \quad \frac{d^2 {\tilde \tau} }{ d \tau^2 } - 2 \frac{ d {\tilde \tau } }{ d \tau } \frac{d \omega}{d\tau} = 0 .
$$
This solves to
$$
{\tilde \tau}(\tau) = a + b \int_0^\tau e^{2 \omega(x(\tau')) } d\tau'.
$$
The constants $a$ and $b$ represent the freedom in the definition of the affine parameter of the form $\tau \to a \tau + b$. These can be fixed by picking an origin and a scale for the affine parameter along the geodesic. For instance, pick two points $x_0$ and $x_1$ on the geodesic and we choose our parameterization so that $x(\tau=0) = x_0$ and $x(\tau=1) = x_1$. Suppose that in the new affine parameterization, we wish to have the same choices, i.e. $x(\tilde \tau=0) = x_0$ and $x(\tilde \tau=1) = x_1$. In other words, we need ${\tilde \tau}(0)=0$ and ${\tilde \tau}(1) = 1$. It follows that $a=0$ and $b= \frac{1}{ \int_0^1 e^{2 \omega(x(\tau')) } d\tau'}$ so that
$$
{\tilde \tau}(\tau) = \frac{\int_0^\tau e^{2 \omega(x(\tau')) } d\tau'
}{ \int_0^1 e^{2 \omega(x(\tau')) } d\tau'}.
$$
| {
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Is there such a thing as a white laser pointer? I can't seem to find a white laser pointer for sale anywhere. Is it just me or is it somehow not possible to make it? I see only color laser pointers like red, green, blue, etc.
| It's not possible to make.
white is a mixture of all the visible spectrum whereas a laser can only be of a single wavelength (ie a single color in the rainbow)
See Wikipedia white light article for more details on white light
| {
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Are humans special in that they collapse wave functions? First of all, I don't really believe that humans are special. So I know the answer must be that they are not.
But the way quantum mechanics is described is that all particles exist as clouds when humans are not looking but exist as particles when humans look at them. How is this consistent with the fact that humans are not special (I mean..humans are made of the same particles, so it's not like they can do anything special)?
| What is really meant by an observer in quantum mechanics is not an actual conscious human, but a macroscopic object, contact with which causes collapse of the wave function. That is, the collapse is supposed to occur, even if there are no human observing it - in fact, this happens most of the time, since even in a lab, the collapse of a wave function is caused by the equipment, sandwiched between the human and the microscopic object being studied. This leaves open the question of whether a falling tree makes a sound, if there are no humans to hear it... but this is already philosophy rather than physics.
To summarized: from the point of view of physics, humans are not special.
| {
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Inverse-square laws and point particles It's my understanding that many inverse-square laws can be explained as a central point emitting "interaction rays" in all directions equally. And that when another object with some area is "impacted" by those rays, it will then feel an effect proportional to the amount of rays that hit it. This amount can be found geometrically to be the inverse-square of the distance between the objects.
These laws are often used to predict the behavior of tiny particles like electrons, protons, etc. Some of these objects are sometimes conjectured to be point particles.
But then they would have no area. Which means that no matter how close or far they were to the central emitter, they would only ever be hit with a single ray. And the inverse-square law would not be observed. Interaction would be the same at all distances.
Does this mean that particles that follow inverse-square laws cannot be points? That they must have some non-zero area, no matter how 'elementary' they may be?
| The language of "interaction rays" is wrong and conceptually misleading, but there is just enough validity in the question to be worth an answer, at least in so far as quantum electrodynamics is concerned (gravity is understood as a geometric effect, and the idea has no application there). The electromagnetic force is transmitted by photons (conceived as massless "point" particles, not rays). We can represent the transmission of a photon from a charge to a spherical surface centred centred on the charge in this diagram.
Since the photon may land randomly anywhere on the surface, the force transmitted by many photons to each part of the sphere is inversely proportional to the area of the sphere.
This is contrasted in qed, in which the force is transmitted by gluons and the gluon-gluon interaction is also taken into account.
The result is that the strong force does not decrease with distance, resulting in quark confinement because energies would rise to the point of particle creation when quarks are separated (e.g. by collisions in high energy scattering).
| {
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How does the view of an observer near an event horizon change as a function of the observer's position, velocity, and acceleration? I see that this topic has been well covered from the perspective of a) a stationary observer dangling on the end of a rope and and b) an observer free-falling from infinity. What I'm still unsure about are the roles of velocity vs. acceleration. Is it all about velocity? Does an observer who is stationary near the horizon and only just starting to freefall see the same as a)? Also, for an observer who is descending at close to escape velocity but then fires his/her retrorockets to cancel the acceleration and achieve constant velocity, does he/she still see the same as b)?
| What you see is what lies on your past light cone. The past light cone only depends on spacetime position, not velocity or higher derivatives.
What you see also depends on your velocity in the sense that it is "distorted" in different ways by Doppler shift and aberration. But given a (infinite-precision, omnidirectional) photograph of what someone at a particular location with a particular velocity sees, you can derive what someone at the same location with any other velocity will see just from that picture, without needing to know anything about the 4D spacetime it was originally derived from.
Acceleration doesn't affect what an idealized camera/eye sees.
Does an observer who is stationary near the horizon and only just starting to freefall see the same as a)? Also, for an observer who is descending at close to escape velocity but then fires his/her retrorockets to cancel the acceleration and achieve constant velocity, does he/she still see the same as b)?
Yes to both questions, and also, if you ignore Doppler shift and aberration, all four of them see the same thing when they're at the same spacetime location even if they don't match their velocities.
| {
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Imaginary time & predictions Is the imaginary time just a different convention to express the time evolution to make the calculations easier? Hawking also said that
"It turns out that a mathematical model involving imaginary time predicts not only effects we have already observed but also effects we have not been able to measure yet nevertheless believe in for other reasons." (From "The Universe In a Nutshell", UK, Bantam, 2001 edition, page 59, last paragraph.)
What are the predictions that we haven't been able to measure yet?
Thanks!!
| The main things that imaginary times gives us are "instantons". These are solutions of the classical equations that exist only in imaginary time. They describe tunneling processes that are classically forbidden, but quantum mechanically allowed.
For simple systems we can derive these tunneling processes by other means such as the WKB approximation, and the instanton predictions agree with both experiment and these alternative derivations. In field theory the alternative methods are harder to do, and for many effects have not been done.
There is at least one imaginary time prediction that we do not want to verify, and this is the decay of a "false vacuum state". Such a decay in our own universe would be fatal to all existing matter, and the life that depends on it.
| {
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What’s the motivation for using Ricci scalar in action term for spacetime? Why do we use Ricci scalar in the action equation for the spacetime? Why don’t we use other functions? Is it just intuition? What forces us to use that?
| $R$ is the lowest-order (and thus simplest) scalar that can be formed from the Riemann curvature tensor. That plus a constant term (the cosmological constant) seem sufficient to explain what we observe. It is possible that the Lagrangian density is more complicated, but physicists choose the simplest explanation that works.
Some physicists think that when we have an accepted theory of quantum gravity, these two terms will turn out to be just the beginning of an expansion of the action in an infinite series of powers of the curvature. These higher-order terms will be significant only at the Planck scale.
| {
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Is equation of continuity valid, while dealing with vertical pipes? Let try an experiment.
If water enter through an end $A$ with some velocity say $v_1$,and leaving end $B$ with speed $v_2$ in a UNIFORM cylindrical tube $AB$ (which is completely filled with water).
If we consider 3 cases.
*
*tube is horizontal
*tube is vertical with $A$ upward
*tube is vertical with $B$ upward
And we go through the experiment and we got our results, which is $v_1~=~v_2$
So for which case this is valid to?
Edit
Due to lack of logical answers
If we consider Bernoulli's equation
$$P+\rho gh+\frac{1}{2}\rho v^2= \text{Constant}$$
So in the case of vertical pipe, we consider two points $A$ and $B$ and now applying Bernoulli's equation here as follows.
Let's assume $A$ to be upward, and take $B$ as reference level and applying Bernoulli's equation.
$$P_a+\rho gh+\frac{1}{2}\rho v_1^2 = P_b+\frac{1}{2}\rho v_2^2$$
If both are exposed to the atmosphere, then $P_a = P_b = $P_{\text{atm}}$
Then we get
$$\rho gh = \frac{1}{2}\rho(v_2^2 - v_1^2)$$ which implies that $2gh +v_1^2 = v_2^2$
So, finally this will prove that $v_1$ never equal to $v_2$ in vertical pipe, but if we consider equation of continuity then mass going in is same as mass coming out so according to equation of continuity
$$\Delta m = \rho A_1 v_1\Delta t = \rho A_2 v_2 \Delta t$$ which implies $A_1 v_1 = A_2 v_2$, and in our case $A_1 = A_2$ then according to equation of continuity $v_1 = v_2$.
Thus, according to continuity $v_1 = v_2$ in vertical pipe case and according to Bernoulli's equation $v_1$ never equal to $v_2$.
How can this be possible? Please guys help me out. Please go through the question and then answer it?
| If the tube is vertical, the fluid can't be filling the tube, and the area of the stream must be changing, thus allowing a velocity change to be consistent with mass conservation. In the case where A is at the bottom, if the fluid fills the tube, the pressure can't be atmospheric at the bottom. Therefore, you can't assume both the same pressure at the ends of the tube as well as a constant area of the fluid stream.
| {
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Lagrange equation for the pendulum This is Wilberforce Pendulum that has spring & weight:
https://faraday.physics.utoronto.ca/IYearLab/WilberforceRefBerg2of8.pdf
Wilberforce pendulum is a system of a spring hanging vertically, and a weight
with moment of inertia is hanging.
The system keeps transferring vibration between the spring's vibration and the tortion.
The paper solves with the Lagrange Equation.
However, the Lagrange equation in the paper does not contain the term $mgz$.
Why didn't this take the "gravitational potential energy" into account?
| The Lagrangian in question contains the potential energy,
$$U(\theta, z) = \frac12 k~ z^2 + \frac12 \delta~ \theta^2 + \frac12 \epsilon~z~\theta.$$
If we simultaneously add offsets $z = \bar z + \alpha, \theta = \bar \theta + \beta$ and absorb any constant terms into the arbitrary choice of zero for the energy, this becomes
$$\begin{align}
U(\theta, z) &= \frac12 k~ \bar z^2 + k \alpha \bar z \\
&~+ \frac12 \delta~ \bar \theta^2 + \delta\beta~\bar\theta \\
&~+\frac12 \epsilon~\bar z~\bar \theta + \frac12 \epsilon \beta ~\bar z + \frac12 \epsilon \alpha ~\bar \theta.
\end{align}
$$In particular choosing to solve the linear system of equations $$
\frac12 \epsilon\beta + k\alpha = mg,\\
\frac12 \epsilon\alpha + \delta\beta = 0,
$$
yields a potential energy which looks entirely like the above Lagrangian but adding a term $m g \bar z.$ Therefore I would just say that probably they are letting the pendulum come perfectly to equilibrium and measuring $z,\theta$ as displacements from that equilibrium; the term $m g z$ was “folded into” the quadratic potentials and did not have to be specially treated.
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Anomalous global symmetry in non-gauge theories I’m a bit confused on the effects of anomalous global symmetries. So take for instance the following theory
$$\mathscr{L}=\partial_\mu\phi\partial^\mu\phi^*+i\bar{\psi}\gamma_\mu\partial^\mu\psi-y \phi\bar{\psi}\psi+\text{h.c}-V(\phi)$$
with $V(\phi)=m^2|\phi|^2+\lambda |\phi|^4$
It has two global symmetries $U_V(1)$ with $\psi\to e^{i\theta}\psi$ and $U_A(1)$ with $\psi\to e^{I\gamma_5\theta}\psi$ and $\phi\to e^{-2 i\theta}\phi$.
These symmetries have significant physical consequences; of course $U_A(1)$ forbids a mass for $\psi$,also the interplay of $U_V(1)$ and $U_A(1)$ forbid $\phi$ from decaying since decaying into two fermions is forbidden by helicity consideration, and other decays are forbidden by either $U_V(1)$ or $U_A(1)$.
However we would usually consider the $U_A(1)$ to be anomalous; certainly it can’t be gauged. But it's unclear to me what physical affect this anomaly actually has. If $U_V(1)$ was gauged, then we would have $$\partial_\mu J^A_\mu=-\frac{g^2}{16\pi^2}F_{\mu\nu}\tilde{F}^{\mu\nu},$$
which would allow violations of $U_A(1)$.
However, when $U_V(1)$ is just a global symmetry, it seems like there is no physical consequence of the “anomaly”.
So my question is: are $U_A(1)$ and $U_V(1)$ good symmetries of the theory I described? If not, what observable consequences does this have? I understand that anomalies come from regularisation ambiguities so perhaps a different way to phase my question is: Is there a regularisation scheme that respects both $U_A(1)$ and $U_V(1)$ and if not what observables are ambiguous?
| There is no anomaly problem with this system --- except that as written it does not have a continuous $U_A(1)$ symmetry. You need to include a term $i\bar\psi \gamma^5 \psi$ term in addition to the $\bar\psi\psi$ term. With that included it is a simple model that can be be used for illustrating chiral symmetry breaking.
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Why can we not take gravity to act at the center of mass in this case? I found this problem when I was trying to prove that the gravitational attraction on an object is the gravitational attraction on the center of mass of that object (I had doubts on it). And in doing so I ran into this problem which shows that the effect on the center of mass isn't equal to the effect on the whole thing
So why don't the two values equal...If the image I posted isn't clear heres everything non mathematical that was on the image. 'A' means the area of the circular face of the cylinder.' L' is the distance from the center of the circular face of the cylinder to the surface of Object A. Also the surface plane of the circular face of the cylinder is parallel to any tangential surface of Object A
| I didn't look over your work (check-my-work questions are off topic), but the conceptual error is still obvious. The gravitational field isn't uniform, so the force acts on the center of gravity, not the center of mass. For uniform fields these are the same thing, but in general they are not.
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What's exactly the new definition of kilogram, second and meter? Could one explain this?
Technically a kilogram (kg) is now defined:
[…] by taking the fixed numerical value of the Planck constant h to be 6.626 070 15 × 10–34 when expressed in the unit J s, which is equal to kg m2 s–1, where the metre and the second are defined in terms of c and ΔνCs.
Does that mean that 1 kg = 1 Planck constant?
And what exactly is the new definition of the second and the meter?
| The SI system is now defined entirely by physical constants. There are no more “prototype” artifacts. How it works is thus:
https://www.bipm.org/utils/common/pdf/si-brochure/SI-Brochure-9-concise-EN.pdf
The SI is the system of units in which:
• the unperturbed ground state hyperfine transition frequency of the caesium 133 atom $\Delta \nu_{Cs}$ is $9 \ 192 \ 631 \ 770 \text{ Hz}$,
• the speed of light in vacuum $c$ is $299 \ 792 \ 458 \text{ m/s}$,
• the Planck constant $h$ is $6.626 \ 070 \ 15 × 10^{−34} \text{ J s}$,
So we make a caesium atomic clock and when that clock ticks 9192631770 times, that is $1\text{ s}$. This is our SI unit of time.
Now, we want to get a unit of distance. We could use a physical prototype but physical prototypes can be damaged or distorted and are by definition not possible to distribute.
So instead we can use a universal physical constant, in this case the speed of light. We simply define the meter to be the length such that the speed of light in a vacuum is exactly $299792458\text{ m/s}$. Since we already have a definition for a second, fixing the speed of light defines the meter. Any experiment that we previously would have used to measure the speed of light now becomes a measurement of the length of a meter. This is good because we can measure the speed of light very accurately and it allows a meter that cannot be damaged or distorted and which anyone can replicate.
Then we take the same approach for mass. If we define Planck’s constant to be exactly $6.62607015 × 10^{−34} {\rm\ kg\ m^2\ s^{-1}}$ then, since we already have fixed the second and the meter this defines the kilogram. Any experiment that we previously would have used to measure Planck’s constant now becomes a measurement of the mass of a kilogram. It has all of advantages of the meter, except currently Planck constant experiments are not as precise as speed of light experiments.
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How does property of material effect it's interaction in Faraday's Law? We all know that faraday law mathematically states:
$$ V = - \frac{\partial \phi_B }{\partial t}$$
But I think this law is a bit weird because it says that the voltage developed across any conductor due to changing flux is same.. which is a bit counter intuitive for me. I mean, is there no dependence of how the material is but I'm pretty sure different materials have different responses to coming in contact to magnetic field.
So, what exactly am I missing here? Taking this to the extreme we could even say take some sort of closed gaussian surface and talk about the flux change in it and hence the voltage generated on it's boundary is equal to magnetic flux through it... so is the voltage generated independent of material existing at all?
I bring up the previous point, because the explanation behind the law is that in when we have changing magnetic flux, the force pushes charged particles to different ends of a conductor and hence generates a potential difference which leads to a current.
| The Faraday law is no more than a different formulation of the Maxwell-Faraday equation: $$\oint_{\partial\Sigma}\mathbf{E}\cdot d\mathbf{l} = -\int_\Sigma\frac{\partial\mathbf{B}}{\partial t}\cdot d\mathbf{A}$$
where $\Sigma$ is any closed surface, $d\mathbf{A}$ a vector element of this surface; $\partial\Sigma$ the contour of this surface and $d\mathbf{l}$ an infinitesimal vector element of this contour. It leads to Faraday law if we define the electromotive force $V$ $$V = \oint_{\partial\Sigma}\mathbf{E}\cdot d\mathbf{l}$$
and the flux : $$\phi_B = \int_\Sigma\mathbf{B}\cdot d\mathbf{A}$$
The important point of all this is that $\Sigma$ can be any closed surface, so the electromotive force generated is indeed independent of material existing at all. But if material exists and if it is a conductor, this force will push its electrons and it generates a current.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How does the mere movement of gluons keep quarks together? You know the strong force (the one that keeps quarks together). Well it works by exchanging gluons right?
So how does that force keep the quarks together? I mean you can imagine that process as three people passing balls between them right? Well as far as I know that throwing of the ball wouldn't force those 3 persons to stay within a range. I had this one idea that when a gluon is emitted it results in a force that pushes the quark in the opposite direction but that would be towards the outside of the quark right? Pls explain this to me. Any help would be helpful and greatly appreciated.
| You have stumbled into one of the most interesting questions of QED and QCD, that is, how can we model the attractive and repulsive forces by the exchange of the massless mediators (photon and gluon respectively)? The answer is mathematically very complicated and when we look for an explanation in our everyday classical view, there is a very nice analogy:
These are very nice classical analogical explanations of how momentum conservation laws can be obeyed by the exchange of the mediator particles (in your case gluons). For repulsive forces, it is easier to understand by throwing balls at each other, but attractive forces are a little bit harder to understand classically, these boomerangs can give a nice analogy.
How can photons cause charges to attract?
All internal lines in a Feynman diagram are force carriers, i.e. transfer dp/dt by construction,not only the gauge bosons. See the diagram for compton scattering for example.
Lattice QCD goes for direct solutions on the lattice, and therefore the concept of virtual particles is not necessary. It is a different calculational approach , although the article involves quark propagators in the calculations.
Are force carrying particles always virtual particles?
It is very important to understand that usually these are mediator exchanges are described using a mathematical model that uses virtual particles (like virtual photons), although in the case of lattice QCD virtual particles are not necessary.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/582710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is string theory the boundary theory of M-Theory? Looking at various AdS/CFT correspondences, we find that some (n-1) dimensional field theories on the boundary of $AdS_n$ with $N=\frac{8}{n-3}$ supersymmetries are equivalent to M-Theory in $AdS_n \times S_{11-n}$.
(e.g. for $n=7$ we get 6D $N=(2,0)$ superconformal CFT.)
Setting $n=11$, seems to suggest that there is a 10 dimensional N=1 supersymmetric theory on the boundary of $AdS_{11}$ that is equivalent to M-Theory in $AdS_{11}$.
The natural theory in 10 dimensions with $N=1$ supersymmetry is superstring theory (or one of them).
Therefor could 11D M-Theory and 10D superstring theory be related by the AdS/CFT correspondence? If not what 10D theory living on the boudary of $AdS_{11}$ would be equivalent to M-Theory on $AdS_{11}$?
Therefor is superstring theory not a theory of the bulk, but actually a theory of what happens on the surface of a black hole? i.e. string theory is just the theory of the hologram?
| Heres a simple way to see that the answer is no. The theory on the boundary has to be superconformal, but the classification of superconformal symmetry says that SCFTs terminate in dimension 6. For M-theory there can only be 3d and 6d SCFTs duals because these theories are the conformal fixed points of the worldvolume theories of M2 and M5 branes which are the only branes in the theory.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the analogue for symplectic structure in case of spin variables? According to some (e.g. Haroche and Raimond in Exploring the quantum: atoms, cavities and photons), the quantum world consists (mainly) of spins and harmonic oscillators.
For harmonic oscillators (i.e. bosons), it is well known that they can be appropriately described in $(x,p)$ phase space, which satisfies a 'symplectic' structure (see e.g. Gaussian states in continuous variable
quantum information). A system with coupled bosonic modes can be symplectically diagonalized into the eigenmodes.
My question is: is there a similar structure for spin states, living in $(\sigma^x,\sigma^y,\sigma^z)$ space (without resorting to a Holstein transformation or similar)? For simplicity, I'm mainly interested in the spin-1/2 case.
| The phase space for spin is the two-sphere $S^2$ with the symplectic form being the area 2-form
$$
\omega= J \sin\theta d\theta\wedge d\phi.
$$
Here $\theta$ and $\phi$ are the polar angles. Then, with
$$
S_x= J \sin\theta \cos\phi,\\
S_y= J \sin\theta \sin\phi,\\
S_z= J \cos\theta,
$$
we have $\{S_x,S_y\}= S_z$ etc.
| {
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"source": "stackexchange",
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Find the potential difference given the function of electric field I have the following problem:
Suppose that, as a function of x, an electric field has an x component
$E_x=6x^2y$
where the electric field is measured in volts per meter and the distances are measured in meters. Find the potential difference between the origin and the point x=3 on the x axis.
I have done the following:
$x=0\;\;\; E_{x0}=0\;\;\; V_0=0\\
x=3\;\;\; E_{x3}=54y\\~\\
V=-\int_{0}^{3}E\ dl\cos0=-\int_{0}^{3}E_{x3}dx=-\int_{0}^{3}54ydx=-54y[x]_0^3=-54y*3=-162y$
However, the answer in the book says that the potential difference is $0V$. Should I assume that there is no change in electric field across y-axis, hence y=0?
| If you are moving along the x-axis then $y=0$ and so $E_{\rm x}=0$.
This means that the potential along the x-axis is a constant and so the potential difference between any two points on the x-axis is zero.
| {
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What if Jupiter or the Sun was made of rock, like Earth and Mars, rather than gas? Jupiter is a "gas giant". If it was (significantly) bigger the pressure from gravitation would ignite a fusion process and it would become a star, which is basically what happened to the sun.
However, what if a body the size of Jupiter or the Sun was made of rocks like Earth and Mars are - what would happen then? Somewhere around iron (lead?) fusion can no longer take place and there is plenty of heavier-than-iron material on Earth.
Or is there something that would prevent such a large body of rocks to form?
| I hope it is ok to link to other stackexchange communities, as there is an excellent answer to be found in the worldbuilding community: Is there a theoretical maximum size for rocky planets. The consensus seems to be that the maximum size for Earth-like planets is at around twice the radius of Earth.
| {
"language": "en",
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"source": "stackexchange",
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When excess electrons are in an insulator they don't move. Why is this? Let's say you have an insulator that is electrically neutral(has no net charge). Let's say you are able to add additional electrons into the same insulator resulting in the insulator having a net negative charge. These electrons that were added to the insulator will stay where they are and not move. How is this possible when same charges repel each other? Shouldn't those excess electrons added to the insulator repel away from each other?
| Consider the following diagram of the quantum mechanical densities of electron states in different types of material:
The Fermi energy $E_F$ is the energy level new electrons are added at. Loosely said, the states below this energy are filled, the states above are empty, with a smooth transition around $E_F$. The conducting electrons are those in this smooth transition.
On the far left is a metal. There are plenty of states available around $E_F$, so adding an electron simply puts it among the already conducting electrons.
On the far right is an insulator. The Fermi energy is in a gap between the bands with no available electron states. The lower band is filled and cannot conduct, the upper band is empty and cannot conduct. If you add an electron to this material, you can only add it in the upper band, since the lower band is already full. When you do this, you effectively lift $E_F$ up to the upper band, and you no longer have a perfect insulator. The electron you added can "move" just fine, and conduct electricity. This is similar to the n-type situation in the diagram.
See also the Wikipedia article on the band gap.
| {
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Question from Relative Error Q -A satellite orbits Earth at a height of 560 km. In many computations, the Earth-Sun distance of 1.5 X 10⁸ km is used to approximate the distance of satellite from the Sun. What is the maximum relative error of this approximation?
Please explain the main concept in detail.
| If the distance between the earth and the sun is given as $1.5 \times 10^8$ km then you can assume this figure has a precision of $\pm 0.05 \times 10^8$ km, which is $\pm 5$ million km. If the satellite height is stated to be $560$ km then you can assume this figure has a precision of $\pm 5$ km. I'll let you take it from there.
| {
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Does light have a mass? Could you please argue with work and gravity? Or however you like. I just don't get it. Thanks.
A physics professor talked about if light had a mass, it should do work ($W=N\cdot m = \frac{kg\cdot m}{s^2}\cdot m$) to get out of the gravitational field. But because we somehow can see it doesn't, light doesn't have a mass.
| Mass is defined in relativity by $m^2=E^2-p^2$, where $E$ is the mass-energy and $c=1$. A ray of light or an EM plane wave has zero mass. However, mass is not additive, and a collection of light rays, or a more complicated wave pattern can have nonzero mass.
It's not really correct to argue about whether light has mass based on gravitational effects as it emerges from a gravitational field. General relativity doesn't describe gravity as an interaction involving mass or mass-energy, it describes it as an interaction involving the stress-energy tensor. For example, parallel pencil beams of light happen to have zero gravitational interaction with each other according to GR, but we don't conclude from that that a pencil beam of light has no mass or no mass-energy. Such a beam will interact gravitationally with other things.
The answer by anna v is conceptually confused. None of this has anything to do with quantum mechanics or wave-particle duality.
Even in the context of newtonian gravity, it doesn't really make sense to talk about a projectile doing work to escape from a gravity well. Work is done on the projectile. The projectile's force on the planet does zero work in the rest frame of the planet.
| {
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Sand leaking out of bag and skater "A person skating on a frictionless icy surface is holding a sandbag. The sandbag has a small hole at the bottom, from which the sand starts to leak. As the sand leaks from the sandbag, the speed of the skater..."
The answer was that the speed of the skater stays the same. I thought that the speed of the skater must increase because now the mass of the sandbag is less than the mass before it leaked. What is wrong with my intuition here? The system I'm considering is the skater and sandbag together.
$p_{system}(t_i) = (m_{skater}+m_{sandbag})v_i \\
p_{system}(t_f) = (m_{skater}+m_{sandbag})v_f$
Since the mass of the sandbag is smaller, the $v_f$ must increase, no?
| A given force will propel a lighter body to a greater velocity.
Once moving, if we could just vanish part of the body's mass, it might speed up. But we can't. The sand, although jettisoned, still exists and takes its kinetic energy along with it.
So no speed up.
This ignores air resistance of course. If the leak changes the shape of the bag, this might affect the slowing effect of air resistance.
This assumes still air of course. If the body is being propelled by the wind...
| {
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Why is there a special connection between entropy and heat? As I understand it:
$dS = \frac{1}{T}dU + \frac{p}{T}dV$
(for a thermodynamic system where $dN=0$) and since for an ideal gas $pV=Nk_BT$ and $U=C_VT$ we can say
$dS = \frac{C_V}{U}dU + \frac{Nk_B}{V}dV$
so ultimately entropy changes are caused by changes in proportional changes internal energy $U$ (proportional to the total internal energy already) and changes in volume (proportional to the total volume already).
This makes sense to me, especially the volume part since there will be more available microstates for a larger volume (and the proportionality bits are there to make entropy an extensive quantity).
What I would like to know is whether $dS=\frac{dQ_{rev}}{T}$ is just essentially a 'backward engineered formula' which accounts for both ways entropy can change since $dQ_{rev}=dU+pdV$, or whether it suggests there is some 'special' link between entropy and reversible heat transfer.
At the moment, I don't think there is a special connection since for a Joule expansion there is no heat transfer but there is an entropy increase due to the volume increase, and even though we can model this expansion as an equivalent reversible isothermal expansion with a certain %$Q_{rev}$, in 'reality' that isn't what is happening.
Am I correct to think of thermodynamic entropy as fundamentally to do with proportional changes $dU$ and $dV$ or should I be thinking more fundamentally in terms of heat instead?
| Let the work done on the system be $\delta W$ while its internal energy change be $dU$, assume that the system may also exchange energy with a reservoir that is at temperature $T_r$. Then for an arbitrary process the entropy change $dS$ of the system satisfies $dS \ge \frac{dU-\delta W}{T_r}$. The equality sign holds for a reversible process.
When the process is reversible then the system's internal temperature is equal to that of its surroundings, here $T=T_r$.
By the 1st law the heat exchanged with the surroundings is, of course, $\delta Q = dU-\delta W$, so you have Clausius's formula that $dS \ge \frac{\delta Q}{T_r}$ and also $dS = \frac{\delta Q_{rev}}{T_r}=\frac{\delta Q_{rev}}{T}$
(Note that "$T_r$" or "$T$" is in the denominator and not "$dT$".)
Of course one can always say that $\delta S_r = \frac{\delta Q}{T_r}=\frac{dU-\delta W}{T_r}$ is the entropy transferred from the reservoir to the system but unless the process is reversible it is not true that $\delta S_r$ is all the entropy change $dS=\frac{dU}{T}-\frac{\delta W}{T}$ in the system; in fact $dS = \delta S_r +\sigma$ where $\sigma \ge 0$ is the internally generated entropy due to dissipation and is never negative.
| {
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Calculating with creation and annihilation operators I'm currently taking a quantum mechanics lecture and am having trouble with the mathematical formalism.
I have to calculate the following:
$$\langle n+2|b^\dagger b^\dagger |n\rangle$$
and
$$\langle n+1| \left(b^\dagger bb^\dagger + \frac {\hat{1}}{2}\right) |n \rangle$$
for the first one I can use the notation $ b^\dagger|n \rangle = \sqrt{n+1} |n+1 \rangle$ right? So I would get
$$\langle n+2|b^\dagger b^\dagger |n\rangle = \langle n+2|n+1|n\rangle $$
if I'm correct. Would I be done there? And how do I solve the second one?
Many thanks in advance!
Edit: Thanks for the help everyone. The answer to the first one would be: $$\langle n+2|b^\dagger b^\dagger|n \rangle = \langle n+2|b^\dagger \sqrt{n+1} | n+1 \rangle$$
$$ =\sqrt{n+1} \langle n+2|b^\dagger|n+1 \rangle = \sqrt{n+1} \langle n+2|\sqrt{n+2}|n+2 \rangle $$
$$ =\sqrt{(n+1)(n+2)} \langle n+2|n+2 \rangle = \sqrt{(n+1)(n+2)} $$
And for the second:
$$\langle n+1| \left( b^\dagger bb^\dagger + \frac {\hat{1}}{2} \right) |n \rangle = \langle n+1|b^\dagger bb^\dagger |n \rangle + \langle n+1| \frac 12 |n \rangle $$
$$ = \langle n+1|b^\dagger b \sqrt{n+1} |n+1 \rangle + \frac 12 \langle n+1|n \rangle $$
$$ = \sqrt{n+1} \cdot \langle n+1|b^\dagger \sqrt{n+1} |n \rangle + \frac 12 \langle n+1|n \rangle $$
$$ = (n+1) \cdot \langle n+1|\sqrt{n+1} |n+1 \rangle + \frac 12 \langle n+1|n \rangle $$
$$ = (n+1) \sqrt{n+1} + \frac 12 \langle n+1|n \rangle $$
$$ = (n+1)^\frac 32 $$
Can I simplify the last expression? Provided it's correct.
Edit: Made a mistake in the original question.
Edit2: In an answer below the second term can be simplified via the Kronecker-Delta notation, see Andreas Mastronikolis answer.
| Just continually act the operator on a bra or ket. The first one goes as follows:
$$\begin{align}
\langle n+2 |b^\dagger b^\dagger |n\rangle &= \sqrt{n+1}\,\langle n+2|b^\dagger|n+1\rangle \\
&=\sqrt{(n+1)(n+2)}\, \langle n+2|n+2\rangle \\
&= \sqrt{(n+1)(n+2)}
\end{align}$$
because, in this case, $\delta_{mn}=\langle m | n\rangle$. Can you attempt the next one?
| {
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Proof that the electric field in a dielectric decreases by a factor We have a linear homogeneous dielectric material half filling a parallel plate capacitor. It's said that the field inside is reduced by a factor ,so we have
$\mathbf
{E}=\frac{1}{\epsilon_{r}}
\mathbf{E}_{\mathrm{vac}}$.
What's its proof?
( If anyone tries to prove it using the similaritiy of $E$ and $D$ ,I think that is incorrect due to non zero curl of $D$ in this case) .
| Gauss Law suggests that $ \nabla . D = \rho $. When there is no charge in a defined boundary. Ingoing and outgoing $D$ are equal. In your case, there are no charges outside of electrodes, thus, there will be no change of $D$ along the paths between the electrodes. Since $D = \epsilon_r . \epsilon_0 . E$ . As a result, the electric field will depend on the $\epsilon_r$ to keep $D$ constant.
| {
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Is there any physical interpretation of the constant which is seen in the constraint curve of an adiabatic process? What is the $C$ in $ PV^{\gamma} = C$? I always saw it as a result out of the mathematical calculations that we do but I recently saw this video which made me think that the constant may have more meaning that meets the eye.
See this video at 4:11
He writes $ S = PV^{\gamma}$ .. but where exactly is this equation from? I don't think I've seen it anywhere else.
| Start your process with the gas having volume and pressure $V_0, p_0$. If the process is adiabatic and reversible then $pV^\gamma = p_0V_0^\gamma$, in other words $C=p_0V_0^\gamma$.
| {
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If objects in motion experience time differently, how does my body stay synced when I move my legs or arms? If I move my swing my arm really fast, the matter in my arm should experience time slower than the matter in my body.
So how does my body still sync with each other?
And a more general question that derives from this: A lot of matter move at different speeds inside our body, how does anything ever stay synced?
| Your arm is traveling at a tiny percentage of the speed of light. The speed of light 300 million meters per second, so even if your arm were moving at a ridiculous speed of 300 m/s, it would still be only one millionth of the speed of light. So even if there were a linear effect, it would be tiny. But the effect isn't linear. The exact amount is given by the Lorentz factor, but for small speeds, it can be approximated as $\frac {v^2}2$. This shows up in the formula for kinetic energy: in Newtonian physics, it's $\frac {v^2}2m$, which is an approximation of the relativistic amount. Using this approximation, your arm would be off by one part in 2 (million)^2, or one part in 2 trillion. Over the course of sixty years, your arm would experience one fewer millisecond. If you only reach 3 m/s, it would be one part in 60 quadrillion.
| {
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The motion of an object moving up and down through an infinite plane Consider an infinitely large plane which exerts gravity onto a point mass with mass $m$ and located at height $h_0$. The point mass moves perpendicularly through the plane without undergoing any friction.
It can be easily derived that the force exerted by the plane is a constant and thus independent of the height at which the point mass is located. I am going to take for granted the result of the calculation, which is $|\vec{F}| = 2Gm\pi\sigma$.
Then it is an obvious conclusion that the $a-h$ graph will look as follows:
Now to specify the problem I will provide the initial conditions: at $t = 0$ $h=-h_0 \ (h_0 > 0)$ and $v = 0$.
I want to find the relationship between $h$ and $t$, i.e. draw the graph of $h(t)$.
What I thought was that initially, while the object is at $-h_0 \leq h < 0$, $$\frac{1}{2}gt^2=0-(-h_0) \Rightarrow t = \sqrt{\frac{2h_0}{g}}$$
and when the object becomes located at $0 \leq h < h_0$ for the first time, the following equations will be valid:
(calling the magnitude of velocity at $h=0$ as $v_0$) $$\frac{1}{2}mv_0^2 = mgh \Rightarrow v_0 = \sqrt{2gh}$$
and as the motion is of uniform acceleration,
$$v(t) = v_0 - gt \ (g = 2G\pi\sigma)$$
$$\Rightarrow -v_0 = v_0 - gt \Rightarrow t = \frac{2v_0}{g}$$
Substituting the expression for $v_0$ into the above, $$t = 2\frac{\sqrt{2gh}}{g} = 2\sqrt{\frac{2h_0}{g}},$$ which is exactly the twice of the time taken for the first interval ($-h_0 \leq h < 0$). So we can deduce the following graph.
Let's then perform simple integrations to find $v(t)$ and thus $h(t)$. The main problems here would be to find appropriate integration constants so that at the end both $v(t)$ and $h(t)$ are a continuous function. With a bit of physical intuition about the initial conditions, we can conclude that the graphs will look as follows:
Thus the generalisation would be
$$h(t) = -h_0 + \frac{1}{2}gt^2, \ if \ 0 \leq t \leq \Delta t$$
$$h(t) = (-1)^{n+1} v_0 (t-(2n-1)\Delta t) + \frac{(-1)^n g}{2}(t-(2n-1)\Delta t)^2 \ for \ n \in N, \ if \ t \geq \Delta t$$
This is what I came up by myself, and it wasn't even a proper problem in textbooks. I would like you to check if there is any error in my analysis. Also please feel free to tell me if there is any better or simpler (perhaps less algebraic and more physical?) way to reach the same conclusion.
| You are obtaining non-harmonic oscillations in potential of type $V(x) = \alpha |x|$. This is correct, and the problem makes perfect sense. I do not want to go through the details of your calculations though - I think it is against the rules of this site.
| {
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Antiproton synthesis For a pion minus hitting a stationary proton, what are the other particles if an antiproton is to be created among them? A positive pion is possible but the total rest mass energy of the final state is comparable to the initial particles.
| Begin with the incomplete reaction
$$p + \pi^- \to \bar{p}\ +\ ?$$
Then write the hadrons as their quark-composites.
$$uud + \bar{u}d \to \bar{u}\bar{u}\bar{d}\ +\ ?$$
By counting the quarks and anti-quarks you find,
3 $u$ and 3 $d$ quarks need to be added on the right side.
There are many ways to do this.
Just play around with baryons (3 quarks) and mesons (quark and anti-quark).
Some examples are:
$$\begin{align}
uud + \bar{u}d &\to \bar{u}\bar{u}\bar{d} + uud + udd \\
uud + \bar{u}d &\to \bar{u}\bar{u}\bar{d} + uuu + ddd \\
uud + \bar{u}d &\to \bar{u}\bar{u}\bar{d} + uuu + udd + \bar{u}d \\
uud + \bar{u}d &\to \bar{u}\bar{u}\bar{d} + uud + ddd + u\bar{d} \\
uud + \bar{u}d &\to \bar{u}\bar{u}\bar{d} + uus + udd + d\bar{s} \\
uud + \bar{u}d &\to \bar{u}\bar{u}\bar{d} + uud + uds + d\bar{s}
\end{align}$$
Translating the quark-composites back to baryons and mesons you get:
$$\begin{align}
p + \pi^- &\to \bar{p} + p + n \\
p + \pi^- &\to \bar{p} + \Delta^{++} + \Delta^- \\
p + \pi^- &\to \bar{p} + \Delta^{++} + n + \pi^- \\
p + \pi^- &\to \bar{p} + p + \Delta^- + \pi^+ \\
p + \pi^- &\to \bar{p} + \Sigma^+ + n + K^0 \\
p + \pi^- &\to \bar{p} + p + \Lambda^0 + K^0
\end{align}$$
Which of of these reactions actually happen,
depends on how much energy the incident $\pi^-$ brings in.
| {
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What happens to $A$'s magnitude change in uniform circular motion?
Looking at this picture from Kleppner's mechanics text, Δa is added to $A$ and it is suggested that:
"This causes a change of direction but leaves the magnitude practically unaltered if $ΔA$ is small."
What is practically happening in this sense? When looking at it from a calculus perspective, this makes total sense and the math lines up...but looking at it from a physical perspective my intuition tells me there is some sort of 'stretching' that goes on in the arm/bar/thread that is being rotated via perpendicular force (taking into account friction, a constant force must be needed to maintain uniform velocity).
In a real scenario, what is this 'stretching'? At some point does the arm break?
What if we are dealing with the rotation of a particle via an applied field? What is the 'stretching' that goes on here?
| Uniform circular motion dilemma
I think this link answers my question and provides some good examples of effects of force as well.
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What does the orbit of Pluto look like from the perspective of Neptune? Pluto is locked in 3:2 resonance with Neptune, so that means that it should be possible to generate a simple looping animation that shows what movements Pluto would make given 2 orbits of Neptune.
I can find some animations of Pluto's orbit, but none that show Neptune locked in Position
Wikipedia has a similar animation for Haumea's orbit with Neptune locked in position:
https://en.wikipedia.org/wiki/Orbital_resonance#/media/File:Haumea.GIF
What does Pluto's orbit look like with Neptune locked?
I found this: orbitsimulator.com/gravity/articles/pluto.html But it doesn't explain the time scale, and implies that it isn't in medium term 3:2 resonance, which contradicts a lot of what I've read elsewhere.
| I do not know if this is what you are looking for, but I thought it might be helpful.
https://upload.wikimedia.org/wikipedia/commons/thumb/a/af/Animation_of_Pluto_orbit.gif/440px-Animation_of_Pluto_orbit.gif
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What does it mean for particles to "be" the irreducible unitary representations of the Poincare group? I am studying QFT. My question is as the title says. I have read Weinberg and Schwartz about this topic and I am still confused. I do understand the meanings of the words "Poincaré group", "representation", "unitary", and "irreducible", individually. But I am confused about what it means for it to "be" a particle. I'm sorry I'm not sure how to make this question less open-ended, because I don't even know where my lack of understanding lies.
| I apologize for not reading all of the answers. In "classical" QFT (e.g., QED and the current standard model that extends QED to electroweak theory -- except that real neutrinos observed by experiment flavor mix and thus effectively are not massless -- as well as QCD), the Poincaire group describes the requirement of Einstein special relativity (as contrasted with Galileo relativity of Newtonian physics). The field operators that create and destroy particle states in QFT must in terms of all observable quantities in the theory obey special relativity and thus the simplest requirement is to pick states that are irreducible for spatial coordinate system transformations (e.g., the parameters of a field space-time point -- x y z t as one representation of a space-time point) and boosts as required by special relativity -- which classically is the Poincaire group of transformations. Note that non-invariant elements may appear in the theory provided these do not make non-invariant observables (i.e., nominally what could be measured in an experiment).
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Transition between 2 energy levels - wave function picture Suppose we have a system that has discrete energy levels (e.g. hydrogen atom, potential well) and the stationary solutions for the wave function are $\psi_n$. I would assume that there should be a way in which one can model the transition $2\rightarrow 1$ by using as initial condition the stationary state $\psi_2$ for the Schrodinger equation (temporal version). However I did not manage to find materials that cover this approach.
I assume that the Schrodinger eq. should not be changed (for a spontaneous transition there is no need of a photon to trigger it). From this, it should follow that a stationary wave function should evolve in another one of lower energy, or, if the transition is possible on multiple lower energy levels, I assume that the solution of the Schrodinger eq. is a superposition of the possible lower energy states.
So, in order to conclude the question, are there any materials on this approach that I did not manage to find? If so, I would appreciate suggestions on the topic.
| On place to look for information is by reading about Rabi oscillations, which arise when a two-level system is driven by a near-resonant electromagnetic field. For some situations one could find solutions in terms of wave functions that are exact or nearly exact.
More generally, the way to approach this problem is by considering a joint wave function of the two-level system (or atom) and the electromagnetic field. The electromagnetic field is the missing component in the reasoning, and it is also the presence of an infinite number of modes of this field with the energies close to the level splitting that is responsible for the spontaneous emission
$$
|1\rangle_{atom}\prod_k|0\rangle_{k} \rightarrow |0\rangle_{atom}\sum_k\left[|1\rangle_k\prod_{q, q\neq k}|0\rangle_q\right]
$$
The infinite number of the field modes here means that one needs to take thermodynamic limit, since otherwise one would expect the wave function revival, when the atom returns to its excited state.
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Magnetization for Paramagnetic Materials If we consider the magnetization for paramagnetic materials, then we obtain $$M = -n\frac{\partial F}{\partial B} \propto B_J(x),$$ where $$x \equiv g\left( JLS\right) \cdot J \cdot \frac{\mu_B\cdot B}{k_B T}$$ is an auxiliary variable and $B_J( \ . )$ the Brillouin function. Now, what exactly is $J$ in this context? I thought that $\hat J = \hat L \ \oplus \ \hat S $, s.t. $J$ would be its quantum number. But then, how can $J$ be an integer, as is shown in the plot in our lecture:
| Well, I can't see why it bothers you. Mathematically we have all these combinations, with different degeneracies, of S (1/2, 1,...)
Physically, integer S can be achieved either by decimating pairs of spin-1/2 particles (where a m=0 sector is accessible) or by considering a bosonic paramagnet, obtained by analyzing the unit cell of some crystal lattice, that sometimes has bosonic character; if this guy has spin-1 and its ground-state is singlet-like, there is a physical realization.
I'd like people to add concrete examples of materials that behave like this in nature.
| {
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Direct and in-direct measurments I have seen and heard this many times that a certain (physical) measurement is "direct" or "indirect". What is the characteristic or definition that sharply separates these two notions?
| There is no sharp distinction. The usage of these terms simply is one of context to indicate that one measurement requires less calculation to infer the desired quantity from the experimental data when compared to the other method.
The reason for that is a deeply philosophical one: Every scientific measurement is always an indirect one. For example, if you want to measure the mass of a stone, you put it on a balance, sure, but really that is not a direct measurement either, if you have a digital balance, your actual measurement is of the electronics creating an electrical signal on the display, which in turn you don't see directly either but you see photons from the display hitting your eye, which really is a measurement of the response of your retina to it's current illumination, which really is a measurement of your synapses propagating electrical signals ... Etc.
The one exception I know is in contemporary dark matter detection, where "indirect" became a standing term that refers to the search for astrophysical signals, and "direct" a term to refer to searches for signals generated in laboratory experiments. But that's just unfortunate and inaccurate jargon.
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Current induced in a superconducting coil Does the magnitude of current induced in a shorted superconducting coil depend on the external flux's rate of change $\frac{d\Phi_{EXT}}{dt}$ ?
Assume that initially the flux through the coil is zero and the initial current circulating in that coil is zero, too, but at some later time, an external flux source attempts to change the flux threading that shorted ideal coil to $\Phi_{EXT}$.
| Since the total flux through the superconducting loop remains constant at $0$ (the reason for this is that any change in flux requires a nonzero emf around the loop, which requires in infinite current, so magnetic flux through the loop cannot change), the flux from the self-inductance $L$ of the loop must be equal and opposite to the external flux. We conclude that $I=\frac{\Phi}{L}$.
To answer the question, no: current only depends on the flux, not the rate of change.
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Negative critical exponent $\alpha$ for Superfluid helium at lambda point Due to the positive critical exponent of the transition in liquid helium I would expect there to be no peak at the transition $t=0$. Since the $t$ dependent part of the specific heat should go to 0 as $C \approx A \, t^{0.0127} + B$. What is going on here? Is this understood? It seems that the critical exponent $\alpha$ (defined by $C\sim t^{-\alpha}$) as fitted would be positive. Yet RG/Bootstrap predicts a negative value of $\alpha$.
Can the negative value of $\alpha$ be found by the experiment in any way? Or does its value just come from theoretical predictions?
| The coefficient $A$ satisfies $A<0$. Also note that the background contribution $B$ is different for $T<T_c$ and $T>T_c$.
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Does the time-independent Schrodinger equation in 1D have an exact and general solution? The (time-independent) Schrödinger equation is for sure the most important equation in quantum mechanics:
$$-\frac{\hbar^2}{2m}\nabla^{2}\psi(\vec{r}\,)+V(\vec{r}\,)\psi(\vec{r}\,)=E\,\psi(\vec{r}\,).$$
Let’s consider the one-dimensional equation,
$$\frac{d^2\psi(x)}{dx^{2}}-\frac{2m}{\hbar^2}\left[V(x)-E\,\right]\psi(x)=0.$$
We can also rewrite the equation as
$$\frac{d^{2}psi(x)}{dx^2}+ S(x)\psi(x)=0,$$
where
$$S(x)=-\frac{2m}{\hbar^{2}}\left[V(x)-E\,\right].$$
Is there is an exact general eigenvalue-eigenfunction solution for such equation? And if it’s not possible to get the exact eigenvalue-eigenfunction solution, can the equation be solved in a pure mathematical sense? I’m not talking here about the WKB approximation method; I’m talking about an exact and general solution. So does this equation have analytical solutions?
| There is no general solution. But this is not only due to different possible potential functions, but also due to the boundary conditions. No differential equation can be solved without boundary conditions, and those can vary according to the problem.
Moreover the case distinction $E<0$ and $E>0$ leads to very different solutions. The first leads to descriptions of bound states, whereas the second one leads to scattering solutions, both very different.
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The derivative of the unit velocity vector
The set up:
An intertial frame Y-X used to describes trajectory of an insect on some rigid body using some relative vectors. Symbols: $ \vec{r_a}$ is is the vector connecting the origin to some point on the rigid body, $ \vec{r_b} $ is the vector connecting origin to the insect and $ \vec{r } $ is the vector connecting the reference to the insect. The relation between vectors:
$$ \vec{r_b} = \vec{r_a} + \vec{r } $$
In a video lecture about corollis force by professor M.S.Sivakumar, I don't get a formula at 19:12 which is used:
$$ |v_{rel} | \frac{ d \hat v_{rel} }{dt} = \vec{\omega} \times \vec{v_{rel}} $$
With,
$$ v_{rel} = \frac{ d|r| }{dt} \hat{r}$$
Where $ \hat{r} $ is a unit vector connecting the reference to the insect $|r|$ is the length of the whole vector connecting the reference to insect.
In a previous post, I had it explained to me the relation about the time rate change of basis is related to the angular velocity by the equation $ \frac{d}{dt} \hat{u} = \omega \times \hat{u}$. However, I do not understand how that idea extends to this case as we are talking about the basis of velocity since $ \omega$ which was used initially was regarding the angular change of the position vectors.
References:
Previous stack post
Lecture Series on Mechanics of Solids by Prof.M.S.Sivakumar, Department of Applied Mechanics, I.I.T.Madras.
| Since $\vec v_{rel}$ is a scalar multiple of $\vec r$ we have $\hat v_{rel} = \hat r$, so
$\displaystyle \frac {d \hat v_{rel}}{dt} = \frac {d \hat r}{dt} = \vec \omega \times \hat r
\\ \displaystyle \Rightarrow |\vec v_{rel}| \frac {d \hat v_{rel}}{dt} = |\vec v_{rel}| ( \vec \omega \times \hat r ) = \vec \omega \times (|\vec v_{rel}| \hat r) = \vec \omega \times \vec v_{rel}$
| {
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Why is the potential not infinite? One way to calculate potential (using infinity as our reference point) is to sum all the contributions of charges that are around.
Let's say I want to calculate the potential at some point on charged surface.
At that point, there is some charge (can be infinitesimal) and that charge should contribute something divided by zero (since the distance is zero) to the potential at that point. Using that logic, every point on charge distribution should have infinite potential.
What is flawed with this argument?
| Point charges are just an abstraction, a limit model, like surface charge distribution in a infinitely thin layer.
You can think at a point charge as the limit for $r \rightarrow 0$ of a continuous distribution in a volume. As you look inside the small charges, the charge enclosed in the volume decreases and goes to zero when you're looking to a point.
The actual field depends on the distribution of the charges inside the small volume. If we assume a uniform charge distribution inside the spherical volume, we end up with an electric field
$\mathbf{e}(\mathbf{r}) = \left\{ \begin{matrix} \frac{q}{4\pi\varepsilon}\frac{\mathbf{r}}{R^3}\quad r \le R \\ \frac{q}{4\pi\varepsilon} \frac{\mathbf{r}}{r^3} \quad r \ge R \end{matrix}\right.$
and potential $\phi (\mathbf{r}) $ s.t. $\nabla \phi(\mathbf{r}) = \mathbf{e}(\mathbf{r})$
$\phi(\mathbf{r}) = \left\{ \begin{matrix} \frac{q}{8\pi\varepsilon}\frac{r^2}{R^3}-\frac{3}{8}\frac{q}{\pi\varepsilon}\frac{1}{R}\quad r \le R \\ -\frac{q}{4\pi\varepsilon} \frac{1}{r} \qquad \quad \quad r \ge R \end{matrix}\right.$
where the constant $\frac{3}{8}\frac{q}{\pi\varepsilon}\frac{1}{R}$ (if I made no mistake) appears to get a continuous potential field going to zero at infinity (this condition at infinity sets the value of the irrelevant additive constant for our choice of the potential).
You can use other distributions, with the given total charge. The results change only locally inside the charge itself. But if you use a model of point charges your main interest is not the distribution inside the small volume, and thus every distribution inside the small mass with the same total charge should be equivalent for your model and your goal.
The same applies to surface distributions, and other singular charge distributions. You can be rigourous mollifying the differential equations to introduce point charges to regularize delta-functions.
If you wish to start with the Maxwell's equations as the fundamental equations of electromagnetism, a point charge of intensity $q$, located in $\mathbf{r}_0$ is represented by a charge density
$\rho(\mathbf{r}) = q \delta( \mathbf{r} - \mathbf{r}_0)$
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Is the Von Neumann entropy of two initially non-interacting systems always increasing? Let $A$ and $B$ be two systems that does not interact initially ($t=0$), i.e., the density matrix of the initial total system is given by $\rho(0) = \rho_A (0) \otimes \rho_B (0)$. Suppose that interaction between the two systems is turned on after $t=0$. Then the density matrices of each system can be obtained by partial tracing: $\rho_A(t) = \text{Tr}_B \rho(t)$ and $\rho_B(t) = \text{Tr}_A \rho(t)$. I want to show that the von Neumann entropy $S(\rho_A(t)) + S(\rho_B(t))$ increases with $t$ (though I'm not sure if this is true). I tried to calculate the time derivative of $S(\rho_A(t)) + S(\rho_B(t))$ directly by using $i\hbar\dot{\rho(t)} = [H, \rho(t)]$, but it resulted in a seemingly useless messy equation. Can anyone tell me whether the statement is correct? If it is, how can I approach to prove it? I appreciate any help.
| This is not true. The interaction $H$ (unless it has irrational eigenvalues) will have a recurrence time $T$ at which $e^{-iHT/\hbar}=\mathrm{Id}$.
At that time, the entropy will be equal to the one at $t=0$. Thus, it cannot increase all the time - unless it stays constant (which it will generally not do).
As an example, consider an initially unentangled state $\lvert\uparrow\rangle\otimes\lvert\downarrow\rangle$ and the Heisenberg interaction $\vec S_1\cdot\vec S_2$.
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How would quarks behave in the event of 'The Big Rip'? I have always heard that you can't get a quark by itself because "the energy required to split them apart is enough to create another." But, in the case of The Big Rip, the idea is that phantom dark energy would tear molecules and atoms apart because it's a much stronger variant of dark energy. So in this case, would protons and neutrons stay together because as phantom dark energy tried to pull them apart the energy outputted would just turn back into more quarks to complete them? Or would the universe just devolve into a quark soup?
| In that scenario, the universe wold probably devolve into a quark-gluon plasma.
I was just reading a paper by Caldwell, Kamionkowski, and Weinberg which explores this very question. The paper is on arxiv here.
Part of the abstract reads
"Here, we explore the consequences that follow if the
dark energy is phantom energy, in which the sum of the pressure and energy density is negative.
The positive phantom-energy density becomes infinite in finite time, overcoming all other forms of
matter, such that the gravitational repulsion rapidly brings our brief epoch of cosmic structure to
a close. The phantom energy rips apart the Milky Way, solar system, Earth, and ultimately the
molecules, atoms, nuclei, and nucleons of which we are composed, before the death of the Universe
in a “Big Rip”.
| {
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Discharging of a capacitor Consider a charged capacitor with its positive plate holding charge Q. Now I join the capacitor to an circuit with resistance R . So the capacitor starts to discharge. Small charge $q$ flows out of positive plate in a small time $dt$ . My textbook says that the instantaneous current that flows is equal to $d$(Q-$q$)/$dt$ $=$ $-dq/dt$ . But I feel that as $q$ amount of charge has flown out the instantaneous current should be only $dq/dt$
| Firstly, you show some confusion between $Q$ an $q$. They are effectively the same thing, the charge stored by the capacitor. Since it is changing, it is customary to use lower case.
$dq/dt$ is the rate of change of $q$. As the capacitor discharges, $q$ falls; thus its rate of change $dq/dt$ is negative.
However the current $i$ flowing out is a positive current. In the equation $i = dq/dt$, we therefore also require the $dq/dt$ side to be positive. But it is negative, so we add an extra minus sign to yield a minus times a minus, written as $i = -dq/dt$.
This may still seem difficult or confusing to justify, so let's look at it a different way.
Consider the opposite situation, with a charging current $i$ where $q$ is increasing and $dq/dt$ is positive. Now we can happily write $i = dq/dt$ with no complications.
But that charging current is flowing the opposite way to the discharge current, so for the discharge situation we would write the current flowing in as $-i$. And, since $di/dt$ is then negative too, we can happily write $-i = dq/dt$.
But we do not want to express it as a negatively flowing charge current, we want a positive flowing discharge current going the other way. So we move the minus sign across and write $i = -dq/dt$.
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How is it that for a closed subshell configuration $L=S=0$? For a closed subshell configuration of a many-electron atom, $M_L=\sum_i m_{\ell_i}=0$ and $M_S=\sum_i m_{s_i}=0$. But I do not understand why does it necessarily mean $L=S=0$. The values $M_L=M_S=0$ are compatible with nonzero values of L and S. Then how does $M_L=M_S=0$ enforce $L=S=0$? I looked at all my books (Bransden, Liboff etc), all of them did a poor job explaining this, in my opinion.
| As you say, for a closed subshell $M_L=M_S=0$
And this is true whatever direction you happen to have chosen for the $z$ axis.
If a vector has a $z$ component for any $z$ direction this can only be because it has length zero, as opposed to happening to be in a particular orientation where it lies entirely in the $xy$ plane, which is how the $L>0, M=0$ possibilities arise.
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Understanding Born's rule for non-hermitian Hamiltonians Say I have a non-Hermitian hamiltonian, such as one might have in an incomplete description of a system where the states are allowed to decay. Then probabilities are not conserved since magnitudes will decrease with time. In particular, say I have an eigenfunction $\mid \psi \rangle$ with $\langle \psi \mid \psi \rangle \sim e^{-kt}$, so that $\psi$ is normalized at $t=0$ (before any decay), but the probability of encountering $\psi$ decreases with increasing $t$.
My question is about the probability of finding $\psi$. If I pick a generic state $\mid \alpha \rangle$, I can calculate the probability that measuring $\psi$ gives $\mid \alpha \rangle$ by $\mid \langle \alpha \mid \psi \rangle \mid^2$. Now if I merely want to find my probability of finding $\mid \psi \rangle$ at all, then I would do the same as above but with $\alpha = \psi$, hence I compute $\mid \langle \psi \mid \psi \rangle \mid^2$. this feel very weird since now I am looking at $\mid \psi \mid^4$ as opposed to $\mid \psi \mid^2$, which is usually what we interpret at the probability.
What is going on here? (Note that I bring up a system with decay so that it genuinely matters what the exponent is. In a usual system, $\mid \psi \mid^2 = 1$ so it doesn't matter what power is chosen.)
| Suppose you have $\langle \psi(t) | \psi(0) \rangle = A e^{-k t}$, for some normalization constant $A$.
Then the probability for the system to be in state $|\psi \rangle$ at time $t$ is
\begin{equation}
P_\psi(t) = |\langle \psi(t) | \psi(0) \rangle |^2 = |A|^2 e^{-2kt}.
\end{equation}
I feel this answer is a little glib but I am not sure where you are stuck actually.
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Contradiction in phase of reflected longitudinal waves While studying the fundamentals of sound waves in organ pipe, I noted that the fact about phase of reflected waves is contradicting while referring multiple sources
This book of mine describes the reflection from a rigid surface/closed end to be in phase
Whereas this one describes the reflection from a closed end to be 180° out of phase
I found the same issue while referring some online portals on this topic. Why are they contradicting each other?
| Its a bit weird to compare the phase of a wave and its reflection, since their directions are different. Both the sources you have put up are saying the same thing: Compressions are reflected as compressions, and rarefactions are reflected as rarefactions.
Now because their directions are different, the phase difference is continuously changing, so I dont really know what the books mean when they say there is no phase difference, or there is $180^ \circ$ phase difference. The important thing to understand is the bold statement above.
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What is the physical meaning of the complex field amplitude? I see that the harmonic field is sometimes written in exponential form. But sometimes the complex amplitude of this form is just a constant and in others (like when talking about modes) is dependent on the position. Why sometimes is dependent and what does it represent in both cases?
First case $E(\vec{r},t)=he^{-i\omega t}$
Second case $E(\vec{r},t)=h(\vec{r})e^{-i\omega t}$
Thanks
| *
*As explained in detail in What is the physical significance of the imaginary part when plane waves are represented as $e^{i(kx-\omega t)}$?, when complex field amplitudes like $E(\vec r,t) = h(\vec r)e^{-i\omega t}$ are presented, there is a broad convention that the physical field is obtained as its real part, $E_\mathrm{phys}(\vec r,t) = \mathrm{Re}\left[h(\vec r)e^{-i\omega t}\right]$.
*The amplitude will sometimes be space-dependent and sometimes be spatially uniform because sometimes we are interested in configurations where the field is uniform, and sometimes we do care about the spatial dependence.
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Where is the energy involved in osmosis? Osmosis creates pressure on the side of the membrane with higher concentration. But where does the energy for this come from?
| Consider a U shaped tube with a membrane in the middle, permeable to water but not to salt. It is partially filled with water. The height of the water surface on both sides of the tube will be the same, as they both feel the same atmospheric pressure.
Now we add salt to the left side of the tube, which fully dissolves (and it can't go through the membrane so it all stays there). We observe that the water level on the left rises and on the right side it falls. So there was a net movement of water molecules from right to left.
This movement went against gravity, as it lifted some of the solution, which is now weighting down on the rest of the liquid, on top of the atmospheric pressure already there. Yet the solution stays risen.
Such net movement, implies an attraction between water and salt. Salt dissolves mostly because of entropic effects (rather than, say, forming more stable bonds). Once dissolved, the water molecules are attracted to the charges of the ions, which are stronger than the mere dipole moment other water molecules have.
It is this attractive force that pushes water to the left, and causes the solution to rise. In other words, this attraction supplies the necessary energy to make the solution rise.
| {
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Operator norm and Action We define the norm of the operator as $\left\lVert A \right\rVert = \sup \frac{\left\lVert A\psi \right\rVert}{ \left\lVert A \right\rVert} = \sup \left\lVert A\psi \right\rVert$ for $A ∈ L(H)$. It is said that $||A||$ measures the magnitude of the action of $A$.
What is meant by the action of $A$ and what is sup in this equation?
Also, how can we check if $A$ is bounded or unbounded with this statement?
| The sup in the equation is a supremum over all states out of the Hilbert space $\mathcal{H}$. In other words, you pick the state $|\psi\rangle$ out of $\mathcal{H}$ for which the number $\frac{\mid\mid A|\psi\rangle\mid\mid}{\mid\mid |\psi\rangle\mid\mid}$ becomes the largest. It can be shown that the operator norm of $A$ corresponds to the largest eigenvalue of $\sqrt{A^{\dagger}A}$. As an operator acting on $\mathcal{H}$, $A$ maps one state out of it to another, which may be not normalized since you have no guarantee on $A$ being unitary. So you can think of the action of $A$ as a measure for “how much” non-normalized $A|\psi\rangle$ is, given that $|\psi\rangle$ is, i.e. $\langle\psi | \psi \rangle=1$.
| {
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Why is the path of light not visible in vacuum? When light is directed towards our eyes we detect the source and locate it. In Huygens' principle it is said that every point on wavefront acts as a source of secondary waves which again travel in all possible directions.
If light is directed not towards our eyes, like path of eyesight being perpendicular to path of light, light from the source of secondary waves can reach our eyes and we can locate that source (that source being in path of light). Then why can't we see the path of light in vacuum?
| The following may be useful to consider.
It is true that Huygen's principle states that every point on a wavefront acts as a source of secondary waves which travel in all possible directions. However, those individual waves are not seen as separate, but only as a superposition.
In other words, say you are looking at a source that is very far away, it will arrive at your eyes essentially as plane waves. If one was to construct Huygen's wavelets at the halfway point between the source and your eyes, those waves would again combine to become plane waves when they reach your eyes. So, if you are looking at the perpendicular direction, the wavefront will not enter your eyes.
However, if you imagine placing a barrier with a small hole half way between the source and your eyes, it might be possible to see the source if you are not looking directly at the it due to diffraction since only part of the plane wavefront is allowed to pass through the hole.
The below figure shows the two cases.
I hope this helps.
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Confusion on wave packet and creation operator in Mark Srednicki's book In Mark Srednicki's QFT book, section $5$, he mentions following things:
$a^{\dagger}({\bf k})$ creates a particle with momentum $k$ and is given by
\begin{equation}
a^{\dagger}(k)=-i\int d^3x [e^{ikx}\partial_{0}\phi(x)-\phi(x)\partial_0(e^{ikx})].\tag{5.2}
\end{equation}
In the next, he defines another operator $a_1^{\dagger}$ (see equation 5.6) near momentum $k_1$ by
\begin{equation}
a_1^{\dagger}\equiv\int d^3k f_1({\bf k})a^{\dagger}({\bf k}),\tag{5.6}
\end{equation}
where
\begin{equation}
f_1({\bf k})\propto \exp{[-({\bf k}-{\bf k}_1)^2/4\sigma]}\tag{5.7}
\end{equation}
is an appropriate wave packet.
My confusion is: what is the physical meaning of $a_1^{\dagger}$? And what does the "wave packet" mean here? I guess $a_1^{\dagger}$ is some operator that creates one-particle state of momentum "near" the given $k_1$, but why is the integral defined in whole momentum space?
| $a^{\dagger}_k$ creates a particle with a definite momentum $k$; definite in the sense of Dirac delta function.
While $a^{\dagger}_1$ creates a state in which the momentum is not definite but almost smeared over $3\sigma$ range about $k_1$. Wave packet is proper term to refer such kind of state since this is how we create wave packet in QM.
Also such wave packet are used while approximating initial and final state in LSZ formula since in a scattering setup we can't produce particle having only single momentum value because of experimental limitations.
| {
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If gravitational field is not real, then am I gaining energy? I don’t know much about General Theory Of Relativity but I have heard that it does not consider gravitational fields like Newtonian Mechanics.
If an object were to be free falling, then according to general relativity that object would be in an inertial frame and everything around it will be accelerating at g rate.
So, if everything around is accelerating at g,then they must be gaining velocity, meaning they must be gaining energy or mass?
This is obviously not true, so I am confused as to how this works out.
| In the frame of reference of a freely falling body, you are gaining speed or energy(assuming you are standing on the ground), because the normal force is accelerating you and doing work on you(in that frame there will be no gravitational force). If that normal force was not there then you to would be in free fall.
Now in your own frame, you are not accelerating, despite there is an upward normal force because there is some force that is similar to a pseudo force acting on you. That is gravity.
Additional Note: One important thing is that Gravitational effects are not exactly equal to the effects caused due to acceleration inside an accelerating frame in space. The equivalence principle works only locally. In a variable gravitational field, we can do nonlocal experiments and find whether it is due to gravity or not. In this example since you are small compared to the radius of the earth, we can assume uniform gravity.
| {
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How does the whole universe agree on the laws of physics? How is it possible that the every particle in the universe agrees on the laws of physics?
What enforces those laws? Might the laws change slightly across the universe in the same way the cosmic microwave background radiation (CMBR) does?
| You seem to attribute sentience and intention to lifeless particles - which they do not possess (or need not posses to describe the universe in a meaningful way, c.f. Occam's razor).
A "law" in physics is kind of the opposite of a "law" in the legal sense. There isn't any "physics police" enforcing obedience of the Universe, but rather it's the Universe behaving the way it does, and physicists (and everyone else) trying to make sense of that behaviour and describe it as uncomplicated as sensible with what is then called a law.
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Can a black hole be characterized by an observer who is inside the black hole? According to the no-hair theorem a black hole can be completely characterized by three parameters: mass, electric charge, and angular momentum. Can any of these parameters be determined by an observer who is inside the event horizon of the black hole?
|
Can any of these parameters be determined by an observer who is inside the event horizon of the black hole?
Yes. As long as they can make non-local observations (i.e. they are not confined to a small windowless spaceship) then an observer inside the event horizon can in principle gather just as much information about the black hole as an external observer. If anything, they might have more information than an external observer because they may have a less distorted/red-shifted view of other objects that have passed the event horizon.
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Why is the electrostatic force felt in straight lines? When two positive charges are kept close, they get repelled in the direction of a line joining both the charges. Why is it so?
Also, why is the repulsion in a straight path?
In both the cases, the potential energy of the charge which gets repelled decreases. What makes it repel in a straight line such that the line passes through both charges?
| Following R.W. Bird answer:
Consider an isolated system of two particles:
Since system is isolated, angular momentum (and linear) is conserved.
I.e. $\vec\tau_{net}=\Sigma\space \vec r\times\vec F=0$
But clearly from figure $\vec\tau_{net}=\vec r\space\times\space\vec F_y\ne 0$
Thus as R.W. Bird noted, the system violates the conservation of angular momentum.
| {
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Dirac delta equalities in physics Earlier I asked this question on the Math Exchange but I'm looking for a physics point of view. How do you interpret an equation like $$x^n \delta(x) = 0, \qquad n\in \mathbb{N},$$ around $x=0$? Why does it suffice to show the integral of this expression is zero around the singularity to show the equality is valid? There are many definitions to the Dirac delta "function" so I get extremely confused with this notion. In other words, do you treat equality of a Dirac delta function and some other expression as an "integral equality" because just inputting the value doesn't make sense?
| There may be many different definitions of the Dirac delta, but they all share the common feature that they only make sense inside an integral, and that
$$\int_{-\infty}^\infty f(x)\delta(x-a)dx = f(a)\qquad (*)$$
My preferred perspective is that the collection of symbols on the left-hand side of $(*)$ is defined to mean $f(a)$, in the sense that the integrand is not a true function so the integral is not meant to be taken literally.
Another point of view is that the Dirac delta is the limit of a sequence of increasingly tall and narrow Gaussians,
$$\delta(x) = \lim_{\epsilon\rightarrow 0^+} \frac{e^{-x^2/\epsilon}}{\sqrt{\epsilon\pi}} \qquad (**)$$
For $x\neq 0$, the above limit evaluates to zero. For $x=0$, the limit does not exist (or is $\infty$, if you prefer). However,
$$\lim_{\epsilon\rightarrow 0^+}\int \frac{e^{-x^2/\epsilon}}{\sqrt{\epsilon\pi}} dx = \lim_{\epsilon\rightarrow 0^+}1 = 1$$
With a bit of work, one can show that
$$\lim_{\epsilon\rightarrow 0^+} \int f(x) \frac{e^{-(x-a)^2/\epsilon}}{\sqrt{\epsilon\pi}} dx = f(a)$$
Therefore, we are to understand the expression $(**)$ as a limit which is to be taken only after an integral sign has been applied.
In both cases, the Dirac delta is an object which is only well-defined inside an integral. As a result, $x^n\delta(x)$ is defined in the same way.
| {
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Why does the $\mathbf{B}$-field of a cylindrical magnet have no $\phi$-component I have a cylindrical permanent magnet with uniform magnetization $\mathbf{M}=\mathbf{a_z}M$, length $L$ and Diameter $D$.
I'm wondering why the $\mathbf{B}$-field created by this uniform magnetization has no $\phi$-component, that is, the field lines don't "circulate" in the magnet.
The field only has a $r$-component and a $z$-component, so the field lines only point inward towards the center, outward away from the center and up or down along the $z$-axis. I believe it has something to do with
$$\displaystyle\oint \mathbf{B} \cdot d\mathbf{l}=\mu_0I$$
but it would be really helpful if someone could explain why this is the case.
| A possible way to see it is the following. A standard result from electromagnetic theory is that a magnetization is equivalent to a current distribution, generating the same field, given by a volume current density
$$\boldsymbol{J} = \boldsymbol{\nabla}\times \boldsymbol{M}$$
and a surface current density
$$\boldsymbol{K} = \boldsymbol{M}\times \boldsymbol{\hat{n}}$$
where $\boldsymbol{\hat{n}}$ is a unit vector orthogonal to the surface.
For a uniform magnetization in a cylinder, the curl of $\boldsymbol{M}$ is zero and there exist only an equivalent surface current similar to the current distribution in a solenoid with its axis parallel to the cylinder axis. From the Biot-Savart law, the field of such a current distribution has zero azimuthal component.
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Normal ordering of hamiltonian
I came across this in the lecture notes of quantum field theory by David Tong. Inside time ordering interactions aren’t taken to be normal ordered. Interaction hamiltonian should be normal ordered otherwise it is not well defined (due to ordering ambiguity and related singularities). Most standard QFT textbooks don’t address this issue. Am i missing something here or normal ordering was assumed?
|
Most standard QFT textbooks don’t address this issue. Am i missing something here or normal ordering was assumed?
What? No. What could you possibly mean by "most standard QFT textbooks..."?
Check out the section titled "Wick's Theorem" in Peskin and Schroeder's textbook titled: "An Introduction to Quantum Field Theory" (first edition, copyright 1995).
In the first edition of the above-mentioned textbook, Wick's Theorem is discussed in section 4.3 on page 88.
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Galileos statement about object thrown in horizontal direction arriving on the floor at the same time as if free fall Is this statement correct? I intuitively believe it is wrong. Simplest example would be shooting a bullet.
I just read this from my little brothers textbook. Is the textbook garbage or did i miss out on something.
Thank you for your help!
| The statement is correct as long as the ground is horizontal and aerodynamic forces such as drag or lift can be ignored (so it doesn't apply when throwing a frisbee, for example). The effect of gravity on a bullet's path is known as bullet drop.
| {
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The depicted shape of subatomic particles I physics books and such, I understand why they show atoms as spheres because they have the electron cloud. But why are protons, neutrons, and electrons spheres? I guess people say that because of calculations in nuclear physics, they turn out to be a sphere, but why naturally. Also, are quarks and neutrinos also spheres it are they string-like things?
| By machine-gunning protons with electron bullets, you can look at the angles at which the electrons bounce off them and determine something called a structure function for the proton. If there are no preferred directions for the electrons to bounce off in, then you are justified in modelling a proton as an object with a finite diameter, which is radially symmetric- in other words, sphere-like. This is why protons and neutrons are commonly represented as little three-dimensional spheres.
But if your bullets (electrons) carry enough punch, then they can penetrate those little spheres (protons or neutrons) and reveal whether or not there are smaller things (quarks) running around inside them. You then get a structure function that no longer represents a featureless sphere because the bullets are now blasting their way inside and bouncing off the internal parts of the sphere which are not uniformly distributed inside it.
So for low-energy electron bullets, a proton is a tiny sphere but for high-energy electrons it is a complicated thing with internal constituents.
When you shoot electron bullets at other electron bullets, to the limits of the precision of the experiment the electrons reveal no internal structure and behave as if they were not tiny spheres with finite (but tiny) diameters but instead infinitesimal points of essentially zero size. Likewise for neutrinos.
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Does sugar have a quasi-liquid film on its surface at room temperature? Background
According to an article from Physics Today, ice is slippery because there is a “liquid or liquid-like layer” on its surface.
There are 3 mechanisms that can cause this layer to exist, each playing a role that depends on the temperature, and other factors. These mechanisms are:
*
*Pressure melting (the least dominant)
*Frictional heating
*Premelting
Pressure melting and frictional heating can cause a liquid water film to form on top of the ice, while premelting can cause a liquid-like layer to form.
Premelting
Premelting refers to the formation of quasi-liquid layers (QLLs) on the surface of a solid crystal even below its melting point. This quasi-liquid layer forms because molecules on the surface are bonded to fewer other molecules when compared to molecules which are below the surface. Therefore, molecules on the surface are less firmly held in place and can move around more.
Question
If premelting occurs when molecules on the surface of a crystal move around because they are bonded to relatively fewer other molecules, does this happen with other crystals such as sugar at room temperature?
Just like water molecules in ice, sucrose molecules are held in place in a sugar crystal via hydrogen bonds. So premelting at room temperature might be possible, perhaps with the quasi-liquid film having a lower thickness in the case of sugar.
On the other hand, according to this paper, the melting point of sucrose is 185 °C. Perhaps this is too high for premelting to occur at all at room temperature?
| If so, I don't think it's a significant effect. Otherwise, sugar would not flow as freely as it does. However, sugar is somewhat hygroscopic, and sugar that has absorbed water vapour from the atmosphere can stick together in lumps. In warm humid climates, that happens fairly quickly to sugar that's kept in an unsealed bowl. In cool dry climates, it takes much longer for lumps to form.
Raw & brown sugar is more clumpy than pure sucrose, due to the presence of molasses, which contains some water.
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Would an astronaut in this spacecraft feel weightless?
A spacecraft is placed in orbit around Saturn so that it is
Saturn-stationary (the Saturn equivalent of geostationary – the
spacecraft is always over the same point on Saturn’s surface on the
equator).
Information the question provided:
mass of saturn = $5.68\times 10^{26} kg$
mass of spacecraft = $2.0 \times 10^{3}kg$
period of rotation of Saturn = $10$ hours $15$ minutes
Information I calculated:
radius of orbit = $1.1 × 10^8m$
Now part d)
Would an astronaut in this spacecraft feel weightless? Explain your answer.
I am unsure how to answer this question. I guess I first need to define "weightlessness"? From what I know the sensation of weightlessness is the absence of normal force? Or its the sensation that you feel that you weight less than your normal weight? I also calculated $g = 3.13m/s^{2}$ if thats any useful?
| Feeling weightless is like saying that you feel not a normal force on yourself preventing free-falling.
I think that this is the starting point for formulating Einstein Weak Equivalence Principle.
| {
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Can we prove the Earth rotation with a disk mounted in its center on a frictionless axis? Can somebody prove the rotation of Earth if it places an uniform disk with a hole in its the center on an axis and orient it paralelly to Sun ecliptical disk?Just assure himself that the disk on the axis is in equilibrium and the friction between the disk and axis at the center hole of the disk is negligable. The disk should preserve its absolute orientation with time while the Earth would change its orientation due rotation and this should be visible after several minutes or hours by looking a labeled part of the disk regarding the floor?
| If the disk is spinning then it will act as a gyroscope and will try to keep its axis oriented in a constant direction in space. This is called a Foucault gyroscope and it can be used to demonstrate the rotation of the earth, although great care must be taken to minimise friction and allow the axis of the gyroscope to rotate freely. See this Wikipedia article for more details.
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How to show connection between measured data (sound in $\rm dB$) and the size of speakers? How can we relate the measured data of speakers with its SIZE (speaker)? I have the data on white noise and pink noise (FFT and 1/3 octave), I'm a bit confused about how to show the relationship (connection) between them.
Here I have 2 speakers, one is a car speaker and the other is a large one.
| In general you cannot say anything because there is no direct relashionship between the size of the speaker and the output level. Think for example of a speaker box containing three speakers, the bass, the middle range and high frequency one. If you apply a high frequency signal (in the audio range) the big bass speaker will barely move whereas the most output comes from the thinny high frequency speaker. This is just one example. There are other things that makes the relationship complicated.
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Why is the $y$-component of acceleration twice the $x$-component?
We were given this question. the answer said that when $m$ released, the $y$ component of acceleration of $m$ should be $2$ times the $x$ component of acceleration of $m$. I can get that the $x$ components of acceleration of $M$ and $m$ are equal but I can't understand why $y$ and $x$ components aren't equal. if $m$ moves $x$ meters $M$ has to move $x$ meters as well at the same time. and thus shouldn’t the $x$ and $y$ components of acceleration be equal?
| The pulleys make it trickey. You can see how this works by looking at the lengths of the individual rope pieces. When the cart moves 1 meter to the right the bottom part gets 1 meter shorter. The top part also gets 1 meter shorter. Since the rope can't change length the left part (that is attached to $m$) gets longer by 2 meters. So for every meter that the cart moves to the right the mass drops by 2 meter.
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Importance of complex functions in quantum mechanics In quantum mechanics, we work with the space $\mathcal{H} = L^2(\mathbb{R})$ of functions with complex value square integrable. Thus Hermitian operators will play a central role since they have a real mean, to which we can give a physical interpretation:
$$\overline{\langle A\rangle} = \overline{\langle x|Ax\rangle} = \langle Ax|x\rangle = \langle x|A^\dagger x\rangle = \langle x|A x\rangle = \langle A\rangle$$
But I don't understand why we work with functions with complex values, it seems much more natural to me to work directly with real functions. Is it possible to study quantum mechanics with real-valued functions? What are the problems that will arise?
| Since the solutions to the time-dependent SE are of the form $\psi(x)e^{-iEt/\hbar}$ it is difficult to avoid complex wave functions.
Moreover a coherent state
$$
\vert \alpha\rangle=\sum_{n=0}^\infty e^{-\vert\alpha\vert^2/2}\frac{\alpha^n}{\sqrt{n!}}\vert n\rangle
$$
for instance is a complex combination of eigenfunctions of the harmonic oscillator whenever $\alpha\in \mathbb{C}$. Even if you start with real $\alpha$, the state will evolve to a complex combination.
The probability densities of $(\psi_a(x)+\psi_b(x))/\sqrt{2}$ and $ (\psi_a(x)+i\psi_b(x))/\sqrt{2}$ are completely different (assuming the eigenfunctions are real).
Eigenstates of $\sigma_y$, describing a system with spin align along $\pm \hat y$ are complex combinations of the usual eigenstates of $\sigma_z$ describing systems with spin quantized along $\hat z$.
The list is long. One cannot just do with reals without arbitrarily complicated and convoluted gymnastics.
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Might the Kaluza-Klein scalar provide a solution to the dark puzzles? Kaluza-Klein theories of a five-dimensional spacetime yield not only the equations of general relativity and electromagnetism, but also a scalar field. This scalar field, sometimes quantised as the radion or dilaton, is thought not to exist.
Given today's twin puzzles of Universal expansion, dubbed dark energy, and gravitational anomalies on the galactic scale, dubbed dark matter, (how) can we be sure that the Kaluza-Klein scalar is not involved in either of them?
| Just wanted to add to the other answer, that the current outlook for a Kaluza-Klein-type theory being the answer for dark matter, and dark energy also, isn't compelling. There are of course people who've considered such models because they're phenomenologically interesting, but they're either under stringent constraints or have theoretical issues, especially for dark energy in both late-time cosmology and inflationary cosmology (see Section 5 of Modified Gravity and Cosmology for a quick discussion on the current outlook).
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Why is this paper stack not flat? So we have lots of paper for photocopy. They come in a bundle. What I have noticed is that There are sine curves or maybe cosine when I saw them from the side. That's Look like:
Is there any explanation for these curves?
| Let's assume that they were originally perfectly stacked. If you apply some nonuniform external force to the whole stack, some sheets might slightly slide around (which is true in this case, as visible in the photo; note sides of the stack). This can be viewed as a deformation of the whole stack. Due to being stacked, upper sheets will press down on the lower layers, thus creating friction, which ultimately prevents the whole stack to reverse the deformation completely.
To compensate these inner stresses, paper can either compress or decompress in its own plane or bend in the third dimension. Apparently paper prefers to bend more than to stretch or compress.
(By the way, sine and cosine is basically the same, just shifted by $\pi /2$ )
| {
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Any good references on the analytic structure of scattering amplitudes? In papers they often say things about the analytic structure of S matrices - things like resonances are poles on the unphysical sheet, particle channels cause a square root branch cut etc.
I've seen this demonstrated in a couple of simple cases but I was wondering if there was any book/notes where this is talked about in general?
| In general analyticity is consequence of causality and unitarity.
A classic text is Eden, Landshoff and Polkinghorne The analytic S-Matrix. It focusses on $S$-matrix elements, but discusses the analytic properties of individual Feynman diagrams.
| {
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Does a rest charge on the surface of the earth emit radiation Suppose we have an observer $O$ in free fall and electric charge $e$ on the surface of the earth. As $O$ is in inertial frame, Maxwell's equations are valid for $O$. However, according to general relativity the electric charge $e$ is accelerated. Does the observer $O$ see radiation emitted by the charge $e$? Has any experience been made?
| https://en.wikipedia.org/wiki/Paradox_of_radiation_of_charged_particles_in_a_gravitational_field#:~:text=The%20paradox%20of%20a%20charge,the%20context%20of%20general%20relativity.&text=Maxwell's%20equations%20say%20that%20an,stationary%20particles%20in%20gravitational%20fields.
This will answer fully, but quick answer, for a falling object it will observe radiation being emitted from the particle on the ground, but for an observer on the ground it will not observe radiation being emitted from the charged particle on the ground. An observer on the ground will also NOT observe a falling object due to gravity to emit radiation as it is not really accelerating
| {
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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?
| In Special Relativity, there is the invariant interval defined as $$\Delta s^2=c^2\Delta t^2-\Delta x^2$$ (for relative motion in the x-direction only). Here $\Delta t$ and $\Delta x$ are the difference in t and x for two events in some frame of reference. It has the same value in any other inertial reference frame using that frame's coordinates t' and x' to describe the same two events.
As t and x both appear in the equation in a similar fashion, one might say that time and space are being treated on an equal footing.
| {
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A problem containing constrained motion of strings and a block Here is the question :-
Two methods came to my mind while trying to solve it, which are:
*
*I assumed the velocity of M as v(upwards). Then, as the strings are inextensible the cosine component of v should be equal to u, if that doesn't happen then the strings will stretch or slack, which we don't want to happen.
Hence,
$$vcos\theta=u$$
Which gives us $v=u/cos\theta$ as our answer.
*The point on M which is attached to both strings will have 2 velocities which will look like,
Its net velocity can be given by $2ucos\theta$ along the dotted normal. And as that point is on the block, the block will also move with the same velocity. This gives us $v=2ucos\theta$ as our answer.
The 1st method gives the correct answer but the 2nd method does not and I am looking for the reason behind it.
| Generally, when it comes to constraint relation of inextensible strings the components of velocity ,acceleration etc,of a particle attached to it are taken along the direction of the string as its length cannot change.It is wrong to take the component of the veleocity of the string in the direction of motion of the object.Furthermore, your equation would imply that at $\theta$=0 the particle is moving with 2v but the string is moving at v, meaning the string is not taut which is impossible.
| {
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Center Point In Circular Motion Suppose a point initially located at (x,y) moves to (x',y') in a circular motion with angular velocity $w$. Then, the center of this circular motion (x*,y*) can be found by the following:
where $\theta = w \Delta t$.
I really do not understand how this relation holds. Do you have any ideas?
| This isn't a complete answer, but it may help show the formula is reasonable.
First there is a line segment connecting $(x,y)$ and $(x^{'},y^{'})$. The center of the circle is somewhere on the line that bisects that segment and is perpendicular to it.
The center of the segment is $$(x_0,y_0) = \left(\frac{x+x^{'}}{2},\frac{y+y^{'}}{2}\right)$$
The vector from $(x,y)$ to $(x^{'},y^{'})$ is
$$\left(x^{'}-x, y^{'} - y\right)$$
You can do a dot product to show that this vector is perpendicular to that one.
$$(X,Y) = (y^{'} - y, x-x^{'})$$
So the line containing the center of the circle is the set of vectors
$$(x_0,y_0) + a (X,Y)$$
where a is a real number. So you need to find $a_0$, the value of a that matches the center of the circle.
$$(x^{*},y^{*}) = \left(\frac{x+x^{'}}{2},\frac{y+y^{'}}{2}\right) + a_0 * (y^{'} - y, x-x^{'})$$
To do that, you might think about lines from the center of the circle that pass through $(x,y)$ and $(x^{'},y^{'})$.
| {
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How exactly do quantum numbers increase in relation to energy levels in more than one dimension? When you have increasing energy levels in 2 or 3 dimensions how does the values of quantum numbers $n$ increase for each dimension?
For example if you have ground state $E_1$ then you have for $(n_x,n_y,n_z)$ is $(1,1,1)$ in 3 dimensions.
So does this mean for energy $E_2$ you have $(2,2,2)$ ?
I've only learnt it in 1 dimension so i don't understand how it works in higher dimensions?
|
So does this mean for energy $E_2$ you have $(2,2,2)$?
This is wrong! The 3-Dimensional box (and 2-Dimensional) have degeneracy is energy state that there 2 or more level for which the energy is same. Like for $E_2$, We have $(2,1,1),(1,2,1),(1,1,2)$.
Note that for 3-D box, The energy given by
$$E_{n_1,n_2,n_3}=(n^2_1+n^2_2+n^2_3)\frac{\pi^2\hbar^2}{2mL^2}$$
| {
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How to solve this problem without using energy considerations? This is a problem from my introductory physics textbook:
A cylinder is released from rest from the top of an incline of inclination $\theta$ and length $l$. If the cylinder rolls without slipping, what will be its speed when it reaches the bottom.
This is of course easy to solve using the principle of conservation of mechanical energy. The change in potential energy, $mgl\sin \theta$, must be equal to the kinetic energy at the bottom, i.e., $$\begin{align} mgl\sin \theta &=\frac{1}{2}I\omega ^2 +\frac{1}{2}mv^2 \\
&= \frac{1}{4}mv^2+\frac{1}{2}mv^2 = \frac{3}{4}mv^2 \end{align}$$
Solving for v,
$$v=\sqrt{4/3gl\sin \theta}$$
My question is, how do we solve this problem without using energy considerations, i.e., while taking a force-based approach?
| You use force and torque relationships to do the evaluation. You can find this approach discussed in the Halliday and Resnick Physics textbooks.
It is important to recognize that for rolling without slipping $v = r \omega$ and that allows for the relatively simple evaluation using force and torque provided by Ali; this relationship is not true if the object slips.
For a force to do work the force must act through a distance. For rolling without slipping the force of friction does no work because there is no relative motion of the instantaneous point of contact and the surface. Rolling friction provides a force and a torque but does no work. That is why rolling friction has no effect on the energy balance you provided. As another example, for a fluid moving in a pipe assuming the no-slip condition at the pipe walls, the force of friction from the pipe does no work on the fluid.
If the object slips (slides), the energy approach must account for the work done by friction. For a rigid body, all the friction goes into affecting the kinetic energy since there can be no change in the internal energy of a rigid body. You can find the evaluation for the rigid body sliding case- using energy and force/torque- in some physics mechanics textbooks such as Analytical Mechanics by Fowles.
In reality, the object is not rigid and "heating" effects should be considered.
| {
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Time dependent Schrödinger equation with time independent potential and separation of variables suppose we have a potential that's independent of time $V(x,t) = V(x)$
so in Schrödinger equation we get:
$$i\hbar \frac{\partial \Psi (x,t)}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi (x,t)}{\partial x^2}+V(x)\Psi(x,t)$$
since the LHS involves a variation of $\Psi$ with $x$ and the LHS involves a variation of $\Psi$ with $t$ can we say that a basis for the solutions will be with the form $\Psi(x,t)=\psi(x)T(t)$?
| Yes, see for example this small set of notes that outlines how the Schrodinger equation comes apart into two separate equations: $$-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x)=E\psi(x) \tag{1},$$ and $$i\hbar\frac{d\phi(t)}{dt}=E\phi(t) \tag{2}.$$
| {
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Problem with direction convention with angular velocity and angular displacement Example 16.17, on page 380 of the dynamics part of the 14th edition of R.C. Hibbeler's Engineering Mechanics: Statics & Dynamics, states:
The crankshaft $AB$ turns with a clockwise angular acceleration of $20\;\text{rad/s}^2$. Determine the acceleration of the piston at the instant $AB$ is in the position shown. At this instant $\omega_{AB} = 10\;\text{rad/s}$ and $\omega_{BC} = 2.43\;\text{rad/s}$.
The book gives the answer $a_C = -13.5\;\text{ft/s}^2$ using the relative acceleration equation method.
If I do it like this:
Let $AC = x, \angle BAC = \theta$,
$$x = \sqrt{0.25^2+0.75^2+2\times0.25\times0.75\times \cos(\theta + \sin^{-1}(\sin(\theta)/3))}\tag{1}$$
$$\frac {dx}{dt} = \frac {dx}{d\theta} \frac {d\theta}{dt}$$
$$\frac {d^2x}{dt^2} = \frac {d^2x}{d\theta^2}\left(\frac {d\theta}{dt}\right)^2 + \frac {dx}{d\theta} \frac {d^2\theta}{dt^2} = \frac {d^2x}{d\theta^2}\omega^2 + \frac {dx}{d\theta} \alpha$$
Then $a_C = \frac {d^2x}{dt^2}$ evaluated at $\theta=\frac {\pi}{4},\;\omega=10\;\text{rad/s},\;\alpha=20\;\text{rad/s}^2$ which gives $a_C = -22.3\;\text{ft/s}^2$.
What should be the signs of $\omega$ and $\alpha$ be and why, given the convention implicitly specified in Eq (1) on $\theta$?
| After some closer inspection, I finally noticed the mistake! There is an inconsistency with direction. Assume counter clockwise is positive. When your $\theta$ INCREASES in magnitude, then your rod ${AB}$ is moving COUNTER CLOCKWISE, so $\omega$ is POSITIVE.
Now if your rod is to move clockwise, then $\omega$ must be NEGATIVE! Same thing with $\alpha$ -- 'stick' a negative in front of it too.
Negate $\alpha$ and you shall arrive at the correct answer. Cheers.
If you choose clockwise to be positive, it gets a bit more complicated, because of the way your equation is set up.
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Does spin really have no classical analogue? It is often stated that the property of spin is purely quantum mechanical and that there is no classical analog. To my mind, I would assume that this means that the classical $\hbar\rightarrow 0$ limit vanishes for any spin-observable.
However, I have been learning about spin coherent states recently (quantum states with minimum uncertainty), which do have a classical limit for the spin. Schematically, you can write down an $SU(2)$ coherent state, use it to take the expectation value of some spin-operator $\mathcal{O}$ to find
$$
\langle \mathcal{\hat{O}}\rangle = s\hbar*\mathcal{O},
$$
which has a well defined classical limit provided you take $s\rightarrow \infty$ as you take $\hbar\rightarrow 0$, keeping $s\hbar$ fixed. This has many physical applications, the result usually being some classical angular momentum value. For example, one can consider a black hole as a particle with quantum spin $s$ whose classical limit is a Kerr black hole with angular momentum $s\hbar*\mathcal{O}$.
Why then do people say that spin has no classical analog?
| An essential difference is that there is no representation of spin in ordinary $3D$ space$^\dagger$. Unlike the spherical harmonics $r^\ell Y_{\ell m}(\theta,\varphi)$ which can be expressed in terms of spherical (and eventually Cartesian) coordinates, such a representation in terms of "physical" coordinates is not possible for spin-$1/2$ (or half-integered spin in general).
$^\dagger$ see Gatland, I.R., 2006. Integer versus half-integer angular momentum. American journal of physics, 74(3), pp.191-192.
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Cylinder’s kinetic energy and the arc-length A uniform cylinder of mass m and radius r rolls without slipping on a parabolic surface which $y=kx^2$
then how to write cylinder’s kinetic energy by arc-length $l$. since it's not a circular surface with constant $R$， I don't know how to work out kinetic energy in this case only by arc-length.
Should arc-length be ${\theta\times}kx^2$ here?
| To get you started.
Start with $dl^2=dx^2+dy^2$, take the square root and factor out $dx$. You can calculate
$\frac{dy}{dx}$.
Integrate to get $l$ in terms of $x$ and use the $y(x)$ equation to get $l$ in terms of $y$. Use the potential and kinetic energy equations from there.
| {
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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.
|
Edit- here is the link to article that made me think about this.
somewhere in middle it is written that scientists can harness the cold
energy using some active input method.
The following statement from the article is poorly worded:
“Essentially, a sky-facing surface passes its heat to the atmosphere as thermal radiation, losing some of its heat to space and reaching a cooler temperature than the surrounding air”.
You don’t “lose” heat to space. Heat is not something you lose or store. Heat is defined as energy transfer due solely to temperature difference.
What appears to be happening is that there is a decrease in the internal microscopic kinetic energy of the atoms/molecules of the material of the device due to heat transfer by thermal radiation to the night sky. As a result the temperature of the device drops below the temperature of the surrounding air. Then there is heat transfer from the surrounding air to the device and that is what generates electricity.
Bottom line: It is the thermal energy of the surrounding air that is harnessed to generate electricity. The reduction in the temperature of the device below the temperature of the surrounding air is what enables that harnessing of energy. It is not the thermal energy of space that is harnessed.
Hope this helps.
| {
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Error propagation for cube and then square root I have a variable $z$ and I know its error value $\Delta z$.
So $z = 4.480$ and $\Delta z = 0.168$. I need to find $y + \Delta y$ such that
$$y + \Delta y = (z+\Delta z)^{3/2}$$
So in this case, what is $y$ and $\Delta y$?
I am finding that
$$y = z^{3/2} \tag 1$$ and $$\Delta y = \frac{3}{2} z^{1/2} \Delta z \tag 2$$
Are these equations correct?
| I do not understand why you want to evaluate $y + \Delta y$ as $(z + \Delta z)^{3/2}$? You can evaluate the uncertainty in y(z) where y(z) is a function of the random variable z; for your case $y = z^{3/2}$. If this is the case, the following applies.
I assume 4.480 is the mean for z and $\Delta z$ 0f 0.168 is the standard deviation for z?
You can find discussions of how to evaluate the uncertainty for a function of a random variable in many statistics texts, and use that information to evaluate the uncertainty in y for your function as its standard deviation $\Delta y$. For more complicated functions, you can to do a Taylor series expansion. For example see Dougherty's text on Probability and Statistics or Meyer's text Data Analysis for Scientists and Engineers.
For your function, we have $\Delta y = y_{mean} (m^2 {\Delta z^2/z_m}^2)^{1/2}$ where $m = 3/2$, and $y_{mean}$ = $z_{mean}^{3/2}$. With this I calculate $y_{mean}$ of 9.48 and $\Delta y$ of 0.53; same result as Penguino provides in his answer.
Then you can express your answer for y with uncertainty as $y_{mean} \pm \Delta y$; 9.48 $\pm$ 0.53.
| {
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Is Light our limit? Suppose something existed faster than light will we be able to perceive it?
And even if we encounter it wouldn't seem to travel with speed of light?
| There exist hypothetical particles named Tachyons that would travel faster than light. They are merely hypothetical since Einstein's relativity does not forbid their existence but there is no experimental support for their existence at all.
If such particles existed they would to us also appear to travel faster than light. We would first observe them when they are inside our detector and only later would we see the light that reflected of the particle as it aproached our detector.
The following image shows the spacetime diagram of a tachyon and how it would be observed, I copied this image from this webpage
Notice that the order in which the particles are observed by the observer do not follow chronological order. Event 0, which happened long before event 6 is only observed after event 6.
Note: This image claims for the light of the particle to be blueshifted as it aproaches the observer and redshifted as it moves away. I personally am not so sure whether that is truly what happens since the equation for relativistic redshift is
$\lambda_{recieved} = \lambda_{emitted} \sqrt{\frac{1+\beta}{1-\beta}}$ which for $\beta > 1$ gives imaginary $\lambda_{recieved}$ which I would not know how to interpret.
| {
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Advanced/Retarded Eddington-Finkelstein coordinates & Black/White Holes The Schwarzschild spacetime is described by $$ds^2=-(1-\frac{r^*}{r})c^2dt^2+(1-\frac{r^*}{r})^{-1}dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2,$$ where $r^*$ is the Schwarzschild radius.
The advanced Eddington-Finkelstein (EF) coordinate system is a transformation of the Schwarschild coordinates $(t,r,\theta,\phi)$ where the time coordinate is transformed using
$$c\bar{t}=ct+r^*\ln|r-r^*|.$$
While the retarded EF coordinate system is transformed from Schwarzschild coordinates using $$c\bar{t}=ct-r^*\ln|r-r^*|.$$
I then read that advanced EF coordinate system describe black holes and retarded EF coordinate system describe white holes. Why is that so?
My understanding is that these two coordinate systems are just two convenient ways to describe Schwarzschild spacetime where the gravitational source has a radius smaller than the Schwarzschild radius. What is the connection to black holes and white holes?
| The different coordinate systems cover different parts of the full solution, which are relevant to different sorts of objects.
The full Schwarzschild vacuum solution has four 4D regions, and 3D boundaries (event horizons) between them. They're conventionally labeled with roman numerals:
Region II is the black hole interior, IV is the white hole interior, and I and III are exteriors. Only one of these exterior regions is present in black holes that form from collapsing matter, and it's arbitrarily taken to be I. A white hole that was simply a time reversal of a black hole would also have only one exterior region. An "eternal grey hole" is the only sort of object in which all four of these regions would actually be physically relevant at once.
The $r>r^*$ part of the Schwarzschild coordinates covers region I or III (normally taken to be I) and the $0<r<r^*$ part covers region II or IV (usually taken to be II, if this part of the coordinate map is used, which it typically isn't). The Schwarzschild coordinates don't cover the event horizons at all; $r=r^*$ is a coordinate singularity. The lowercase $t$ and $r$ coordinates shown in this image are Schwarzschild coordinates in units where $r^*=1$. ($T$ and $X$ are Kruskal-Szekeres coordinates.)
Ingoing coordinates (Eddington-Finkelstein and otherwise) cover regions I and II (or III and II) and the boundary between them. Outgoing coordinates cover regions I and IV (or III and IV) and the boundary between them.
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Bragg-Williams theory of phase transition
Hi I am currently studying Bragg wiliam theory and I dont understand how they derive the equations f(T,m). Actually i dont understand how they got S/N given by equation 8. Can someone explain?
| Well, as usual the entropy is the logarithm of the number of microstates. If the total number of spins is $N$ and there are $N_{\rm up}$ "$+$ spins", then this corresponds to a total number of possible configurations given by $\frac{N!}{N_{\rm up}!(N-N_{\rm up})!} = {}^NC_{N_{\rm up}}$ (the number of ways of choosing which of the $N$ spins are "$+$ spins"). This is exactly the first identity in (8).
The second identity in (8) follows from the fact that the magnetization density is
$$
m= \frac{N_{\rm up} - N_{\rm down}}{N} = \frac{2 N_{\rm up} - N}{N} = 2 \frac{N_{\rm up}}{N} - 1,
$$
so that
$$
N_{\rm up} = \frac{(1+m)N}{2}.
$$
Of course, this way of computing the entropy totally ignores the fact that the different configurations do not have the same energy, which is why the Bragg-Williams theory only provides an approximation.
Then, they go on getting another approximation, this time for the energy, writing
$$
E = -J\sum_{\langle i,j \rangle} \sigma_i\sigma_j \approx -J\sum_{\langle i,j \rangle} m^2 = -\frac12 JNz m^2,
$$
since there are $\frac12 N z$ pairs of nearest neighbors.
Once they have the (approximate) entropy and energy, they compute the free energy by the usual thermodynamic relation.
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Time dependence of operators In Griffiths's Introduction to Quantum Mechanics, while studying the time evolution of the expectation value of position, the author wrote:
$$\langle x\rangle=\int_{-\infty}^{+\infty}x|\Psi(x,t)|^2\,dx.$$
So
$$\frac{d\langle x\rangle}{dt}=\int x\frac{\partial}{\partial t}|\Psi(x,t)|^2\,dx.$$
Did he just assume that $x$ has no time dependence? And why?
| The are two formulations of quantum mechanics :
*
*Schrödinger representation. The time evolution is encoded in the state vector, wavefunction - $\Psi(x,t)$, and the observables(operators) are constant in time
*Heisenberg representation. Now the operators evolve in time, and the state vectors are time-independent, kept fixed.
In the case of interacting theories there is a hybrid Interaction representation. Here the operators evolve with the non-interacting Hamiltonian $H_0$, and the states evolve via the interaction part $H_I$.
So in your case the author uses the Schrödinger representation.
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Maxwell equations: Lagrangian or Eulerian description I am just wondering the 4 Maxwell equations (i.e Fadaray Law, Maxwell-Ampere) are Lagrangian or Eulerian description? Does it really matter?
| The distinction between the two descriptions you consider assumes that there is a field of velocities describing the motion of the particles of the continuous body.
Adopting one or the other description respectively means (a) to use the initial position of the particles to label the integral curves of the field (Lagrangian description) or (b) to refer to the istantaneous positions of particles in a given rest space (Eulerian description).
The electromagnetic field does not provide this field of velocities so that the choice does not make much sense. Or maybe, to some extent we can say that only the Eulerian description can be adopted.
The EM field is not a continuous body or a fluid, even if it shares some features with these physical systems.
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Are the potentials of the electrolytes equal in Galvanic cell? My book defines Electrode Potential for a Galvanic Cell as follows:
A potential difference that develops between the electrode and the electrolyte is called Electrode Potential
$$E= V_{electrode} - V_{electrolyte}\tag1$$
It further defines cell potential as:
The potential difference between the two electrodes of a galvanic cell is called Cell Potential.
Mathematically, I interpret it as
$$E_{Cell} = V_{electrode,anode} - V_{electrode,cathode}\tag2$$
However my book further mentions that
The cell potential is the difference between the electrode potential of the anode and cathode.
I interpret this as:
$$E_{Cell} = E_{anode} - E_{cathode}$$
From (1)
$$E_{cell}=(V_{electrode,anode} - V_{electrolyte,anode} )- (V_{electrode,cathode} - V_{electrolyte,cathode})$$
To keep the above consistent with (2),
$$V_{electrolyte,anode} = V_{electrolyte,cathode}\tag4$$
I find (4) illogical as there is no basis for the two electrolytes to be at same potential.
What is going on here?
| Yes, the potentials of electrolytes are equal. If they are of different type and connected by a salt-bridge then they are at same potential, because we assume that the salt bridge offers no resistance to electrons, it is essentially short-circuiting the two electrolytes. If the electrolytes are the same then no salt bridge is needed, and they are obviously at same potential.
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Is information conserved in quantum mechanics (after wave function collapse)? I have heard in popular science that there is a law of "conservation of information." Some times this is described as: for any event that happens, there is enough information to reconstruct the original state. So, for example, if you knew the exact positions of the atoms that flew off a burning piece of paper (and everything else near by that's interacting with these atoms), that you could reconstruct the information on the paper.
Is this true when quantum measurement is taken into account? Can we really reconstruct the past completely even though much of it has collapsed to a particular configuration due to QM?
EDIT:
Just to clarify, of course it is clear that the wavefunction itself (without it collapsing) conserves information. The question is if information is conserved after collapse.
| Short answer: the collapse of a wavefunction destroys information.
As you correctly said, as long as the quantum state evolves according to the Schrodinger equation, information is conserved.
If we adopt an interpretation of quantum mechanics in which collapse happens upon measurement (the Copenhagen interpretation), then even in the simplest case we can see that information would be lost upon collapse.
For example, suppose your system is in a superposition of spin up and spin down states. If you measure it to be spin up, there is no way for you to find out whether it was in a pure spin up state, or in a superposition. Hence, information is lost.
Clarification: in the above scenario, it's even "worse" than just you not being able to find out the initial state. The state of the whole universe (you, the system, the measuring device, etc.) will be the same whether or not the initial state was a pure spin up state or a superposition.
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Averge of the mean field theory hamiltoninan So I am trying to solve a problem where I have to calculate E and I am given the hamiltonian so I can compute the averge of H which is then E.
H is defined as: $ H = -J*d*m\sum{\sigma_i} + \frac{JdNm^2}{2}$
Here J is a constant d is also a constant, N is number of particles so also a constant. m is magnetisation and sigma is in which direction the spin is pointing. If it points up it is +1 and if it points down it is -1. How should I do to calculate this?
| Taking average on both sides of the equation we get:
$<H> = -Jdm<\sum \sigma_i> + \frac{JdNm^2}{2}$
By definition $m = \frac{<\sum \sigma_i>}{N}$. Substituting it in the above equation we get:
$<H> = -\frac{JdNm^2}{2}$
This is the average energy.
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Can two photons excite an electron consecutively? I know that a photons energy is quantized and that it can excite a bounded electron from one energy state to the other whic depends upon the energy the photon carries my question is that can two photons consequtively excite an electron from an initial state (say E1) to E2 and then E3 without the electron falling from E2 to E1 in between.
| To add to the answer by @CrazyGoblin. Every state can be characterized by a lifetime, $\tau_i$, which characterizes the rate of relaxation $\Gamma_{i\rightarrow 0}=1/\tau_i$ to the ground state. Inducing an absorption to a higher energy state requires that the absorption happens faster than the relaxation. In other words, the rate of absorption, given by the Fermi Golden rule, $\Gamma_{i\rightarrow j}$, should be higher than the rate of relaxation to the ground state (preferably much higher, for the effect to be clearly observable):
$$
\Gamma_{i\rightarrow j} \gg \Gamma_{i\rightarrow 0}.
$$
Since the absorption rate is proportional to the square of the matrix element, i.e., to the square of the field inducing the absorption, this usually requires very strong optical fields, and as such belongs to the domain of nonlinear optics. The absorption from the ground state obviously has the advantage that the ground state has infinite lifetime, and the absorption can be easily observed even for relatively low fields.
Finally, let me note that given strong fields and appropriate non-dipolar couplings, one can observed two- and multi-photon absorption even with no intermediate levels present.
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The Dirac comb with one $\delta$-function removed Has anyone ever encountered the Dirac comb/Shah function with one removed $\delta$-function,
$$
V(x)=\frac{\hbar^2\kappa}{m}\sum_{n\neq0}\delta(x-an),
$$
in any literature? I want to find the solution of the Schrödinger equation with such potential but currently, I'm experiencing some difficulties. I believe that this potential describes the lattice defect so I tried to find any appearance of it in the corresponding literature so I could get some ideas, but I failed.
| This is an interesting, and likely solvable, generalization of scattering from a delta -potential. I would suggest formulating it differently:
*
*Taking unmodified Dirac comb as the potential for calculating the Basis functions (i.e., the Bloch waves):
$$
V(x) = \lambda\sum_{n=-\infty}^{+\infty}\delta(x-an)
$$
*and adding an impurity potential for $n=0$:
$$
V_{imp} = \lambda_1\delta(x)
$$
This way the problem is more general, since the impurity can have arbitrary potential strength, whereas the case suggested in the question corresponds $\lambda_1=-\lambda$, which is a vacancy rather than impurity. Moreover, in the limit $\lambda \rightarrow 0$ one should recover the solution for plane waves scattered by a delta-potential.
A very similar, although superficially looking very different, is a problem of an impurity in the tight-binding Hamiltonian, which actually corresponds to the Dirac comb with negative $\lambda$ and negative energies:
$$
H_0=\sum_{i}\left(tc^\dagger_i c_{i+1} + h.c.\right),\\
H_{imp} = \epsilon_0 c^\dagger_ic_i.
$$
This one is solved in many books.
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What is angular velocity in 3-dimensional space? (Revised) If I'm not mistaken, there are analogies between the translational dynamics of a rigid body and the rotational dynamics of that body. For example, the position of a rigid body is analogous to the orientation of that body. What property of a rigid body, from a rotational aspect, is analogous to the rigid body's velocity? (I would ask the similar question with respect to acceleration, but then this question would get disqualified by the moderators.)
In 2-dimensional space, I'm guessing that angular-velocity is analogous to velocity. But in 3-dimensional space, the way an object can be rotating can be much more complex. For example, the body may be rotating about an axis which itself is also rotating about some other axis. I imagine these axes all stemming from the center of mass of the rigid body.
Note that a single axis/angle pair is enough to describe a body's orientation at any given time, but it is not always sufficient to describe how that body is rotating in 3-dimensional space.
| If I understood, you are making a kind of analogy table, where position in translational dynamics is equivalent to angular orientation in rotational dynamics.
Following this line, what is the analogy pair of linear velocity?
There is a problem because velocity is conserved for translational movement without applied forces, but angular velocity is not. The rotating body can wobble.
But a good pair analogy is linear momentum, that is also conserved, and angular momentum. Even a wobbling rigid body has a constant angular momentum without applied torques.
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Why is it that we ignore height difference when applying Bernoulli in an airfoil I learn physics myself and sorry if this is a very simple question
*
*Why is it that we can apply Bernoulli on above and below the plane even if the are not in the same streamline?
*Why do we ignore height difference when doing so?
Any help whatsoever is highly appreciated.
| Your first question finds an answer in this Wikipedia article :
if the fluid flow is irrotational, the total pressure on every
streamline is the same and Bernoulli's principle can be summarized as
"total pressure is constant everywhere in the fluid flow". It is
reasonable to assume that irrotational flow exists in any situation
where a large body of fluid is flowing past a solid body. Examples are
aircraft in flight, and ships moving in open bodies of water
And for your second question, I would say that the effect of the height difference is negligible compared to the velocity and pressure terms.
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Justification for "reducing integrals" in the virial expansion for gases I have been following this document (https://sites.chem.utoronto.ca/chemistry/jmschofi/chm427/gases.pdf) regarding the virial expansion of gases and on finding the virial coefficients.
On Page 7, they do the following:
$$Z_3 = \int _V d\mathbf{r}_1 d\mathbf{r}_2 d\mathbf{r}_3 f_{12}f_{23}$$
where $f_{ij} = f(|\mathbf{r}_i-\mathbf{r}_j|)$, the Meyer-f function. We are integrating over the entire phase space, with $\mathbf{r}$ being the 3D position vector.
They say that the above integral is reducible, and reduce it as,
$$Z_3 = \int _V d\mathbf{r}_1 d\mathbf{r}_2 d\mathbf{r}_3 f_{12}f_{23} = \int _V d\mathbf{r}_{12} d\mathbf{r}_{2} d\mathbf{r}_{23} f_{12}f_{23}\\=V\left(\int _V d\mathbf{r}_{12}f_{12}\right)\left( \int _Vd\mathbf{r}_{23}f_{23}\right)$$
My question is, why is this last step valid?
I understand substitution (like $u$-substitution, or trig substitution when we are learning basic integrals) of variables within an integral sign, but this seems to be a bit more than that, right?
I made the substitution, $$\mathbf{r}_1 \rightarrow \mathbf{r}_{12}\\ \mathbf{r}_{3} \rightarrow \mathbf{r}_{23}$$
But, if $r_{12} = r_1 - r_2$, then shouldn't $d\mathbf{r}_{12} = d\mathbf{r}_1 - d\mathbf{r}_2$? Furthermore, if $f_{12}$ and $f_{23}$ have a dependence on $\mathbf{r}_2$, why are we ignoring that dependence and just integrating it out?
I am really struggling to understand why that step is allowed.
I would appreciate any advice you have for me.
| This is not a single variable integral and you should recall what you learn in multiple variable calculus --- the change of variable should follow by Jacobian. Obviously here the Jacobian is 1. Then after change of variable $f_{12}(r)=f_{12}(r_1-r_2)=f_{12}(r_{12})$ would depend on $r_{12}$ only and has nothing to do with $r_2$ and you could integrate $r_2$ out.
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Why is are induced electric field's non-conservative while static electric fields are conservative? I have learned that the $E$-field induced by changing magnetic flux, such as in 'motional emf', is non-conservative in nature.
I am also aware that static $E$-fields are conservative in nature.
What is the reason for this difference in the nature of the $E$-fields?
| The reason static field is conservative is this: static field means, by definition, total Coulomb field of any static configuration of charges. We know Coulomb field is conservative (because of Coulomb's law).
If tommorow there was a non-conservative field discovered that is due to and associated with static charges, that would be a big change, as we could extract energy from it without moving any of the static charges. But nobody has discovered yet such non-conservative field of static charges, and we believe it doesn't exist. It would be a so-called "free lunch" - free energy extraction with no change in configuration of the charged system. In reality, extraction of energy leaves systems changed.
If charges move with acceleration, non-conservative field (circulation integral is not zero) may be present and this can be verified, for example, if magnetic field inside a solenoid changes. Then the induced electric field is observed to be present and acting on current in the coil, it is non-conservative, it goes in circles. Now we can extract energy from that as well, but there is no free lunch this time - the charges are moving, hence work can be done on/extracted from them. Extracting energy from non-conservative electric field is associated with forces on moving charges and thus configuration of the charges in space is affected adequately to the amount of energy extracted. This prevents the free lunch because eventually, the source of the energy is drained.
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Is $(L^2, L_z)$ a complete set of commuting observables? According to the main definition we define a (C.S.C.O.) complete set of commuting observables $(A,B,C, \dots)$ if:
*
*Every commutator between the operators of the list is $0$
*If we fix the eigenvalues of the operators there exists a unique eigenvector with these eigenvalues.
(Anyway, there is a reference for the exact formal definition of this concept ? In every textbook I have this concept is introduced with just a brief discussion on the subject.)
If I follow blindly this definition I conclude that ($L^2$, $L_z$) is a CSCO, because if a fix a value of $l$ and a value of $m$ there exists a unique eigenvector (namely a unique spherical harmonic for every fixed value of $l$ and $m$).
But if this set is complete, why in the study of the Hydrogen atom I can add to the set the Hamiltonian $H$ ?
For myself the set must not be complete, because if I fix just one value of $l$ or either $m$, I can clearly notice the degeneracy.
I even think that the latter reasoning may serve as a method to find that the set of observables is not complete, but I haven't found any reference in the literature.
So, what parts of my reasoning are wrong ?
| Regarding the Hydrogen atom, being in an eigenspace of both $L^2$ and $L_z$ means knowing the type of orbital the electron is in ($s$, $p$, $d$, etc.) - this gives the $l$ label - and also which specific orbital it is in ($p_x$, $p_y$, $d_{x^2-y^2}$, etc.) - this gives the $m$ label - see here.
However, every shell (labelled by $n$) has an $s$-orbital, every shell with $n\geq 2$ has a $p_x$, $p_y$ and $p_z$ orbital, etc. In other words, knowing that the electron "is in a $p_x$ orbital" doesn't give complete information, the remaining information is given by specifying which eigenspace of the Hamiltonian we are in (this decides the $n$ label).
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Ball rotating in a circle without rope/string - why doesn't the ball fall after the peak? I am currently learning about the problem of a ball rotating in a vertical circle - where we are interested in determining the minimal height $h$ required in order for a ball sliding from rest down a frictionless track (which consists of of a slope followed by a vertical circle) not to fall before completing a full round on the circular part. As far as I understand, the way to do it is to determine the minimal velocity required for the ball to reach the peak of the circle (marked by $v=?$ in the figure below) without falling off.
My question is -
What guarantees that once the ball reaches the peak it will not fall in any other point on the circle that is more distant than the peak?
Looking forward to your help. Thanks!
| Suppose the car reached the top point with a velocity $v$. The centripetal acceleration must be $g \implies g = \frac{v^2}{r}$
If the track was removed, the problem was to find the trajectory of a particle with an horizontal initial velocity $v$, having a constant acceleration $g$ downwards. We know that the solution is a parabola in this case.
Considering the origin the center of the circle, the equation of the parabola comes from:
$x = vt$ and
$y = r - \frac{1}{2}gt^2$
$$y = r - \frac{1}{2}g\frac{x^2}{v^2}$$
Substituting the value of $g$
$$y_{par} = r - \frac{1}{2}\frac{x^2}{r} = r\left(1 - \frac{1}{2}\frac{x^2}{r^2}\right)$$
The equation of the circle is:
$y_{cir} = \sqrt{r^2 - x^2} = r\sqrt{1 - \frac{x^2}{r^2}}$
It is necessary to proof that the "natural path" of the parabola is always higher than the circle (they are equal to $r$ for $x=0$).
Expanding the square root in a Taylor series at x = 0:
$$y_{circ} = r\left(1 - \frac{1}{2}\frac{x^2}{r^2} - ...\right)$$
The first 2 terms are the same as the parabola, but follows an infinity of negative terms that makes $y_{cir} < y_{par}$
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Show that the initial speed of gun and the shell are in the inverse ratio of their masses A shell of mass $m$ is fired horizontally by a gun of mass $M$ which is free to recoil
and which is on horizontal ground. Show that the initial speed of gun and the
shell are in the inverse ratio of their masses.
Here is my attempt is there anything wrong please tell me?
$v$: velocity of shell relative to the gun
$u$: velocity of the gun
them by conservation of momentum
$m(v-u)-Mu=0$
since $z=v-u$ is initial speed of shell, $\frac{m}{M}= \frac{u}{z}$
is it correct?
| Both velocities (shell and gun) have to be with respect to an observer stationary on the ground (an inertial reference frame). So your use of (v - u) for the shell for the conservation of momentum is correct, since this is the velocity of the shell in the inertial frame.
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