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Can $A \sin (kx - \omega t)$ be used for longitudinal wave?
Question: One particle in the medium has its equilibrium position at $x = 1.00 \ m $. Show that the speed of this particle at $t=0.882 \ s$ is $4.9 \ m/s$.
I attempted at using the formula
$$y=A \sin(kx- \omega t)$$
and then I tried to use
$$\frac{dy}{dt}=\frac{dy}{dx} \times \frac{dx}{dt}$$
to evaluate how $y$ changes with respect to time (which is what the question asks).
But when I plug in $x=1.00$ and $t=0.882$, then I get a negative $\frac{dy}{dx}$, but that contradicts with the fact that in the graph the tangent has a positive slope.
So, the question: Is it wrong to use $y = A \sin (kx - \omega t)$ in this question where the wave is longitudinal?
| You can use $y(x,t)=A\sin(kx-\omega t+\phi)$ for longitudinal waves. But understand its meaning: $y$ denotes the longitudinal displacement of the particle whose equilibrium position is $x$, at time $t$.
The problem wants you to evaluate the speed of the particle whose equilibrium position is $x=1$ (let's label the particle as "$1$" since there is only one particle whose equilibrium position is $x=1$). Therefore, $f_1(t)\equiv y(1,t)=A\sin(k-\omega t+\phi)$ gives the displacement of "$1$" and $v_1(t)\equiv \frac{df}{dt}(t)=-A\omega \cos(k-\omega t + \phi)$ gives the velocity of "$1$" at time $t$.
The values of $A$, $k$, $\omega$, and $\phi$ can be determined using the given graphs.
Response to comments are made here
You directly differentiated with respect to time (maybe partial derivative?) ...
$\frac{\partial y}{\partial t}(x,t)$ gives the velocity of the particle whose equilibrium position is $x$, at time $t$. I defined a function called $f_x(t)=y(x,t)$ to specifically denote the displacement of the particle whose equilibrium position is $x$. But the essence is the same: $\frac{\partial y}{\partial t}(x,t)=\frac{df_{x}}{dt}(t)$.
What is the reason why the use of chain rule here is not valid?
Chain rule makes sense in cases such as the one shown below. Here, $y$ is a function of $h$ which in turn is a function of $x$ and one would like to know how much $y$ changes when one varies $x$.
$$y=y(h(x)) \Rightarrow \frac{dy}{dx}=\frac{dy}{dh} \times \frac{dh}{dx}$$
To the contrary, in our wave equation, $x$ and $t$ are independent variables. The $x$ in $y(x,t)$ identifies the particle whose displacement we would like to determine. In other words, $x$ identifies the particle of interest. If you specify $x=1$, the function $y$ thinks "Oh, he's referring to the particle whose equilibrium position is $x=1$. Cool. At what time would you like to know its displacement?". Time is an additional piece of information that is not determined by $x$.
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Could speed of light be variable and time be absolute? I get my "demonstration" of time dilation from the textbook thought experiment.
A laser is mounted on a cart with a reflective ceiling. At $t=0$ the cart starts moving and the laser is fired. When the laser is reflected back at the starting point the (thought) experiment stops.
Now, two different observers, one sharing the frame of the cart and another standing on the ground perpendicular to the cart will observe two different things. For the first one, the laser bounces back and then down in a straight line. For the second one, the light travels in a triangular pattern which is longer than the path observed by the first guy.
Given that the speed of light is constant, the time has to dilate/contract.
Why is the speed of light held constant here? Could we work out a physics where time is absolute but the maximum speed of light variable?
| There is a physics where time is constant and the speed of light is not: Newtonian with an ether.
Galilean transformations in Newtonian physics have time as a fixed parameter that ticks away uniformly for all points and states of motion.
For the speed of light to be frame dependent, one needs the ether in which it propagates at $c$. The ether then defines a preferred reference frame against which all motion is absolute.
So in the train experiment, the station (which is solidal with the ether ;-) sees light take a long path at $c$. Meanwhile, the observer on the train sees light propagating vertically more slowly:
$$ c' = \frac c {\sqrt{1+(\frac v c)^2}}$$
Now if you want to find a theory that is fully Lorentz invariant such that it correctly matches the train-light-clock experiment and the other standard S.R. scenarios, then: No. Relativity is the Lorentz invariant solution.
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Is Griffiths simply wrong here? (Electrostatic Boundary Conditions)
In the above illustration, shouldn't $E_{above}$ and $E_{below}$ be in opposite directions? If not, how did Griffiths end up the following equation? From the above directions, shouldn't the flux add up?
| Griffiths in correct. The flux through the top portion of the box is not just $E_{\small{top}}^\top$ but actually $\vec E\cdot (A\vec n) $ where $\hat n$ is perpendicular to the surface and points out: in your diagram, $\hat n$ would be along $+{\hat z}$ for the top portion of the flux calculation.
For the bottom of the box the $\hat n$ vector points along $-\hat z$ so the flux through through those sides of the box work out to
$$
(E_{\small{top}}^{\top} - E_{\small{bottom}}^\top)A \tag{1}
$$
when $\vec E_{\small{bottom}}$ points along $+\hat z$: the minus sign in (1) comes from the $-\hat z$ for the direction of the surface element $d\vec A$ at the bottom of the box.
Thus, ignoring the thin sides of the box because the flux through those is arbitrarily small, you have
$$
(E_{\small{top}}^{\top} - E_{\small{bottom}}^\top)A=\frac{1}{\epsilon_0} \sigma A
$$
and the area cancels out.
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What causes a black hole ringdown and why can it prove the no-hair theorem? When two black holes merge, they will produce the so-called ringdown before forming a new black hole, even the newly created black hole produces some sort of tones due to the force of the impact, but it is barely audible for LIGO and other instruments to pick up. Anyway, I want to know how is the loud but short-lived overtone produced and why can it be used to validate that black holes do not have other properties beside mass, angular momentum and charge.
| The ringdown is simply the black hole formed in the merger settling down to a final stationary form. According to the no-hair theorem, this stationary form has to be described by the Kerr family of metrics described by the mass and angular momentum of the final black hole (since the intitial system is assumed to be electrically neutral).
The characteristic spectrum of this ringdown can be calculated in general relativity and is determined only by the mass and angular momentum of the final Kerr black hole. Consequently, if one measures the characteristic frequency and decay time of the dominant mode, one can uniquely determine the mass and angular momentum of the final black hole, leading to a prediction for the frequencies of all other modes present in the ringdown.
If one would observe other modes in the ringdown that do not belong to this spectrum, then this falsifies this prediction, and means that the final black hole and its ringdown are not described by the Kerr metric and general relativity. This is often presented as a would be failure of the no hair theorem, which tells us that the final black hole has to be Kerr. (However, it could equally be interpretted as a failure of GR itself.)
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Conservation of Angular Momentum for Rigid Bodies I have question about Conservation of Angular Momentum of Rigid Bodies. I've been doing some examples from Hibbeler's book, and noticed that in this chapter about Conservation of Angular Momentum of Rigid Bodies, there are some examples where we sum all the angular momentums about some fixed point O, but when they are writing equation, they write $I_G$ (inertion in point G, center of mass). Why $I_G$ ? Why they didn't write $I_O$ (inertion in point O, the point about which we are writing this equation of conservation). So this confuses me. I mean, there are also examples where they do it like I expect (Inertion about point O).
Look at this example.
As you can see, we are writing conservation of angular momentum about point A. So:
$\sum{H_{A1}}$ = $\sum{H_{A2}}$.
And as you can see they calculated $I_G$. Why? Why they didn't wrote about point A, and wrote $I_A$ = $I_G + md^2$ (where $d$ is distance of center of mass of body from point A).
So this confuses me.
Whoever helps, thanks in advance!
| As long as point A isn't moving, you can do it either way. In vector form, angular momentum about A is
$$ \boldsymbol{H}_A = \mathbf{I}_G \boldsymbol{\omega} + \boldsymbol{c} \times \boldsymbol{p} = \mathbf{I}_G \boldsymbol{\omega} + \boldsymbol{c} \times ( m \,\boldsymbol{v}_G ) = \mathbf{I}_G \boldsymbol{\omega} - m \boldsymbol{c} \times (\boldsymbol{c} \times \boldsymbol{\omega}) = \mathbf{I}_A \boldsymbol{\omega}$$
This works because $\require{cancel} \boldsymbol{v}_G = \cancel{ \boldsymbol{v}_A }+ \boldsymbol{\omega} \times \boldsymbol{c} $, where $\boldsymbol{c}$ is the vector to the center of mass from A.
But since the bullet isn't rotating, it doesn't make sense to assign a rotation and calculate $\mathbf{I}_A \boldsymbol{\omega}$ for the bullet. It makes more sense to just use $$\boldsymbol{H}_A = \cancel{ \mathbf{I}_G \boldsymbol{\omega}} + \boldsymbol{c} \times m \boldsymbol{v} $$
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String Landscape, De Sitter vacua and Broken Supersymmetry If we assume that the swampland conjectures, etc. regarding de sitter vacuas existence in the string / F-theory landscape turn out to be incorrect (and therefore we can assume the problem is well-posed), would all such solutions have broken supersymmetry?
There are a number of obvious “difficulties” with formulating (at least for the case of generic qft on curved spacetime) particle theories on de Sitter like spaces (ex. non-trivial hamiltonian from vanishing of globally timelike killing vectors, etc.). One feature of de Sitter space, that is probably obvious but likely not trivial (at least from first glance) is the non-existence of positive conserved energy and its requirement for broken supersymmetry.
Have I made an error, or is this a “just so” theorem regarding de Sitter vacua and the necessity for broken supersymmetry?
| You're right. Realistic string compactifications require four macroscopic dimensions. For supersymmetry to exist in a given background a non zero globally defined Killing spinor is required. In the d=4 case such a spinor should be (locally) the generator of the $Spin(4,1)$ group, the problem is that $Spin(4,1)$ has no Majorana representations (condition required to realize supersymmetry in a unitary way). Nevertheless, non unitary realizations of supersymmetry over de-Sitter space are possible.
There are papers in the literature claiming to surpass many of the mathematical subtleties. But to my poor knowledge no one is really popular among the experts. See https://arxiv.org/abs/1403.5038
Polyakov arguments about fundamental instabilities in de-Sitter cosmologies are also very nice and strongly physical. See for example https://arxiv.org/abs/1209.4135, https://arxiv.org/abs/0709.2899 or https://www.youtube.com/watch?v=Gn87Fu8QHYI.
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Postulate of constancy of speed of light in vacuum I'm not of course questioning the constancy of the speed of light, just the way the postulate about it is worded. It is often stated that the speed of light is independent of the motion of the source. Einstein himself said "and also introduce another postulate, which is only apparently irreconcilable with the former, namely, that light is always propagated in empty space with a definite speed $c$ which is independent of the state of motion of the emitting body."
But even in classical physics, isn't the speed of a wave independent of the motion of the source? Once the wave is emitted, its speed just depends on the medium. If I have a loudspeaker moving toward me, the frequency of the sound will be higher but not its speed. If I am moving toward the loudspeaker, both the frequency and the speed of the sound will be higher.
This postulate of relativity is sometimes stated in terms of the state of motion of the observer, but it is frequently stated just in terms of the source, which is what I am questioning.
| There is no medium. The equations stand alone. The possibility that empty space may itself be a stationary medium for somebody has been disproved by experiment.
Hence, any electromagnetic propagation in empty space always seems to any inertial observer to proceed at the universal constant speed irrespective of where it has come from or where else it might be going.
But for the observer measuring it, his observations depend on the source, not on other observers. The postulate says how it depends on the source.
And of course, counter-intuitively, says it doesn't.
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Do colliding black holes violate time reversal symmetry? Two black holes can collide and merge into one bigger black hole, but not split into two. Does this mean colliding black holes violate time reversal symmetry?
Related: Do black holes violate T-symmetry? Based on the answer to that question, time-reversing a black hole yields a white hole. However, that seems to imply that white holes are very unstable because they can spontaneously split into two, which would then split into four, ad infinitum, and the universe would be covered with tiny white holes all over.
| Yes, to the same extent as falling egg violates time reversal symmetry (have you seen a broken egg spontaneously jump from the floor to your cup?).
This is called "second law of thermodynamics" and it is not time-symmetric.
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Connection between Poisson bracket and Anti-commutator? Canonical quantization promotes Poisson brackets in classical mechanics to commutators in quantum mechanics. Is there any classical counterpart similar to the Poisson bracket for the anticommutator?
| FWIW, in classical Hamiltonian formalism, the phase space is a supermanifold with a super-Poisson bracket $\{\cdot,\cdot\}_{SPB}$. The super-Poisson bracket corresponds to a super-commutator $\frac{1}{i\hbar}[\cdot,\cdot]_{SC}$ upon quantization. The super-commutator is a commutator (an anticommutator) in the Grassmann-even (Grassmann-odd) sector, respectively.
See also my Phys.SE answer here.
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How gravity time dilation affect radio wave communication? I'm not in any way an expert so sorry if my question is silly.
So i was wondering about a hypothetical situation. I'm orbiting a black hole and I'm trying to contact via radio with someone outside the gravitational effect of that black hole. As far I understand the radio wave similarly to light can be bent but cannot be slowed down by gravity. So is the communication possible? How would time dilation due to black hole gravity affect my attempt to communicate?
Regards
Sebastian
| First of all let's make the situation more concrete. The effect of gravity extends to infinity, although it drops off as a factor of $(\sim \frac{1}{r^2})$ according to Newtonian mechanics(which suffices in terms of building our intuition here). When you are just outside the event horizon of the black hole, all the signals you send can eventually reach any observer who is not him/herself inside a black hole. A radio wave travels at the speed of light in vacuum so in your example it would be the same as sending a beam of light.
Let's think about the signal just outside the event horizon and somewhere far away from there. Because of the gravitational well, the clock of the source of the beam will run slower than the clock at the receiving end of the signal. This creates a shift in the signal as it passes through space. This is what creates a redshift in the signal.
For a Schwarzschild black hole, the frequency difference is given by the formula $$
\frac{\lambda_{\infty}}{\lambda_{e}}=\left(1-\frac{r_S}{R}\right)^{-\frac{1}{2}}
$$
Where $\lambda_{\infty}$ the wavelength far away from the black hole,$\lambda_e$ the wavelength of the emitted wave, $r_S$ the radius of the black hole and $R$ the distance from the centre of the black hole.
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Are the muon/tau neutrinos produced in the Sun? If not, then where? I was reading about Solar Neutrinos, and apparently they are all Electron Neutrinos.
However, there are two other types of neutrinos, the Muon and Tau Neutrinos. Does the Sun produce them? If not, are they produced naturally anywhere in the universe?
| There can also be blackbody (thermal) antineutrino and neutrinos of all flavors, which are emitted during type-II supernovae core collapse.
Here is a figure from H.-Thomas Janka, Neutrinos from type-II supernovae and the neutrino-driven supernova mechanism (reprint from: Conference Proceedings Vol. 40: "Frontier Objects in Astrophysics and Particle Physics").
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What happens to matter when it is converted into energy? According to Einstein’s equation
$$E=mc^2$$
Matter can be converted into Energy. An example of this is a nuclear reaction. What happens to the matter in the process? Do the atoms/subatomic particles just vanish? Any insights into this process are appreciated.
| To start with the $m$ in $E=mc^2$ is the relativistic mass, and in particle physics this is out of use it causes confusion. One uses four vectors , where the "length" of the four vector is the invariant mass, uniquely identified with elementary particles, and with systems of elementary particles.
Four vectors are good in keeping track of energy, as one of the components of the four vector is the energy.
$$\overrightarrow{P} =
\left[\begin{matrix}E\\ p_xc\\ p_yc\\ p_zc\end{matrix}\right] =
\left[\begin{matrix}E\\ \overrightarrow{p}c\end{matrix}\right]$$
It is vector algebra and allows rigorous calculations.
The elementary particles in the table of of particle physics have fixed masses seen in the table and their four vector always has the length of that mass:
$$\sqrt{P\cdot {P}} = \sqrt{E^2 - (pc)^2} = m_0c^2$$
What happens to the matter in the process? Do the atoms/subatomic particles just vanish
All matter is composed out of elementary particles, in bound systems where there are a large number of four vectors to add to get to the invariant mass of the composite system.
Look at how complicated a proton is
That is why it has a large mass of almost 1000 Mev whereas the constituent quarks and antiquarks have masses of mev and the gluons zero mass. The added four vectors give the mass of the proton ( in a complicated to calculate way)
There are quantum number conservation rules that matter composed out of elementary particles has to obey. Charge, baryon number lepton number etc have to be conserved. So one way of turning mass to energy is by annihilation, particle hits antiparticle , quantum numbers add up to zero (by antiparticle definition)and then other pairs of particles and radiation can appear ,taking away kinetic energy. Proton antiproton annihilating even if they have small momenta, create a large number of pions, of course following quantum number conservation.
In nuclear physics because of the binding energy curve of the periodic table of elements ,
It is possible to get energy by fission or by fusion of particular elements, which has led to the atomic bomb and the hydrogen bomb. The four vectors representing a nucleus can break up into four vectors of other nuclei and release energy in the form of radiation or kinetic energy of the new nuclei.
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Maximum speed in a spring-mass system I am studying energy right now and I can use only gravitational potential energy, elastic energy and kinetic energy to solve some problems. My doubt is how can I prove that the maximum speed in a mass hanging on a spring is reached at the middle of the elongation of the spring.
The exercise shows the following situation: A block with mass $m$ is attached to an ideal spring (no mass) of an elastic constant $k$. The block is released when the spring is on his natural state. Consider no friction and that the system is conservative.
The question ask for the maximum speed of the block using energy only. I solved it already (with $m=0.1$ $[kg]$ and $k=10\left[\frac{N}{m}\right ]$ and assuming that the point $0[m]$ is when the spring reaches it maximum elongation and the block is released from a height of $h\;[m])$. Then, I got that the maximum speed is $1[\frac{m}{s}]$, but I assumed that the maximum speed is reached at the middle, so I want to know why this is true in these ideal conditions.
| Maximum speed occurs at the mean position of spring mass system as Diffrentiation of velocity, acceleration is 0 there.
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Time in Lorentz transformation While doing Lorentz transformation for position
$$x'=\gamma(x-ut)$$
Here I understand that instead of just length contraction we also use the term
$$ut$$
To include the distances between the frames of reference.
But while doing Lorentz transformation for time that is
$$t'=\gamma\left(t-\frac{ux}{c^2}\right)$$
What is the physical interpretation of the term. $$\frac{ux}{c^2}$$ like that of $$ut$$ in above
I know the derivation but it doesn't give a clue about its actual meaning as if just pops out of nowhere.
| To show how the term in question accounts for synchronization, consider Einstein's famous train thought experiment. The station observer sees the light that strikes the front of the frame moving toward the observer in the middle of the train car with relative speed (c+u) and for the light from the back of the train it's (c-u). Both light waves travel the same distance L, where L is half the length of the train car as determined by the station observer. So to the station observer, the time difference for reaching the middle of the car, which the car observer interprets as the lack of synchronization of the two light flashes, is Δt = L/(c-u) - L/(c+u) = (2Lu/c2) ɣ2. (I'm considering here the station frame of reference to have the unprimed coordinates and the train frame of reference to have the primed coordinates). Now 2L is the length of the car, D, as determined by the station observer, so we have Δt = (Du/c2) ɣ2. It's starting to look familiar. The ɣ2 factor goes away if we take account of length contraction, D=D'/ɣ and Δt=ɣΔt' (the train observer's time difference is a proper time interval as he is in the same location for each time measurement). So finally we get Δt'=D'u/c2, the amount by which the signals are not synchronized, expressed in terms of the train observer's coordinates.
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Eigenvalue corresponding to the stationary state If $M$ denotes the transition matrix of a Markov chain, then the vector $x$ that satisfies $Mx=x$ is the stationary distribution or stationary state. However, this paper seems to use the term stationary state in a different sense, wherein the stationary state satisfies $Mx=0$ (see Equation 4). What is the difference in the physical interpretation between the stationary state referring to the eigenvector corresponding to the eigenvalue 1 and one which corresponds to the eigenvalue 0? Is there a possible explanation for the use of the same terminology to mean different things? Terminology for something as basic as a Markov chain would have been uniform at this point.
| I am going to answer my own question since I was able to find the answer very shortly, after bit of searching on Google.
If $x$ is the stationary state, then the usual Markov matrix or transition matrix satisfies $Mx=x$. In other words, $M_{ij}$ is just the transition probability from state $i$ to $j$.
There is another matrix associated with the same Markov chain called the generator matrix which satisfies $Nx=0$ where $N_{ij}=M_{ij}$ if $i\neq j$ and $N_{ii}=- \sum_j M_{ij}$. Unfortunately, the generator matrix is also sometimes called the Markov matrix or transition matrix.
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Does work expend a permanent magnet? My understanding of a permanent magnet is that it has the potential to do work. This would seem to me to imply it has a kind of magnetic charge similar to how a battery is charged with energy. Does this imply that that using the magnet to do work will expend this charge causing the magnet to become less magnetized with each use? If not how can this be explained given conservation of energy?
| The magnet may perform work while attracting another magnet or a piece of iron, but all that work is reversed when you take the two magnets and pry them loose from one another and separate them again. Energy is conserved in this process.
To make a chunk of iron into a permanent magnet requires you to perform work upon it, so as to align most of the magnetic domains in the iron in the same direction. Using the resulting magnet to attract other pieces of iron does not directly destroy the alignment of the magnetic domains within the magnetized chunk of iron (which process also requires work), but there are other processes that can do this- like subjecting the magnetized iron to a rapidly alternating magnetic field, or heating it up above a certain critical temperature whereupon the domain alignments get thermally jiggled into randomness.
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What is the Eigenvalue of $T^2$ ($SU(3)$ Casimir)? For example, in $SU(2)$, $\hat{S}^2|s,m_s>=\bar{h}^2 s(s+1)|s,m_s>$.
What about in $SU(3)$, $\hat{T}^2|T,m_3,m_8>=?|T,m_3,m_8>$ where $\hat{T}^2=\sum_i^8 T_iT_i $, $T_i = \frac{\lambda_i}{2} $, $\lambda_i$ is $SU(3)$ generator.
| Like for $SU(2)$ the Casimirs of $SU(3)$ provide extra labels to the irreducible representations that can be useful in distinguishing irreps. that have the same dimension. Note I talk in the plural because for $SU(3)$ one can construct two Casimirs:
$$\hat{C}_{1} = \sum_{k= 1}^{8}T^{k}T^{k} \qquad \textrm{ and } \qquad \hat{C}_{2} = \sum_{klm = 1}^{8}d_{klm} T^{k}T^{l}T^{m}$$
where the $d_{klm}$ are the symmetric structure constants.
To answer your question the Casimir $\hat{C}_{1}$ acts on $SU(3)$ states labelled by weights $(n, m)$ to return the eigenvalue $\left(n^{2} + m^{2} + 3n + 3m + nm\right)/3$:
$$\hat{C}_{1} |(n, m)\rangle = \frac{1}{3}\left(n^{2} + m^{2} + 3n + 3m + nm\right)|(n, m)\rangle.$$
In particular for the states in the triplet, with labels $(1, 0)$ we get an eigenvalue $C_{1} = \frac{4}{3}$ (and it turns out that $C_{2} = \frac{10}{9}$).
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Origin of terms in the Nernst-Planck equation We know the Nernst Planck equation is
$$ \frac{\partial c}{\partial t} = - \nabla \cdot J \quad | \quad J = -\left[ D \nabla c - u c + \frac{Dze}{k_\mathrm{B} T}c\left(\nabla \phi+\frac{\partial \mathbf A}{\partial t}\right) \right] $$ $$\iff\frac{\partial c}{\partial t} = \nabla \cdot \left[ D \nabla c - u c + \frac{Dze}{k_\mathrm{B} T}c\left(\nabla \phi+\frac{\partial \mathbf A}{\partial t}\right) \right] $$
We know this is basically a continuity equation where a partial of a conserved density quantity is equal to the divergence of its flux. The first term is derived from Ficks Law, the second term is simply a convection term flux=velocity*density and the final term is due to Einstein's Relation stating
$$ D = \mu \, k_\text{B} T$$
Where
μ is the "mobility", or the ratio of the particle's terminal drift velocity to an applied force, μ = vd/F and we know F=EQ and V=ED so the third term is derived from these equations.
I am having trouble with the modified Nernst Planck equation for porous media where we have
$$ \nabla (\kappa_{eff}\:\ln C)=\mbox{Current Density}$$
where $$ \kappa_{eff}=\frac{2RT\kappa}{F}\left(t_+-1\right)\left(1+\frac{d \ln f}{d \ln C}\right)=D\left(1+\frac{d \ln f}{d \ln C}\right)$$
Where does this term come from? Logarithm of the activity would imply a chemical potential but how is this related to diffusion?
| I see now that for Chemical Systems of non-ideal solutions/mixtures Ficks First Law is:
$$ J_i = - \frac{D c_i}{RT} \frac{\partial \mu_i}{\partial x}$$
This is a restatement of the First Law as $$ J_i = - \frac{D c_i}{RT} \frac{\partial \mu_i}{\partial x}= - \frac{D c_i}{RT} \frac{RT\partial \ln C}{\partial x} \implies d \ln C =\frac{dC}{C}\implies J_i = - D \frac{\partial C}{\partial x}.$$
| {
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Confused about what a wave is When a wave of something, let's say light or some electromagnetic wave is given, I am confused because I do not understand if shape of a wave represents projectile of it or some value that possess at certain positions. I researched a lot but I have no idea what a wave really is.
My question is: what is a wave?; what defines a wave?
| You are asking whether the shape of the EM wave represents a projectile (projection) of it, or some value that the EM wave posesses at certain points.
You are confused because you see these images of EM waves.
In reality, it is the value of the EM field at any given point in space that these diagrams represent for the electric and magnetic field.
In light propagation, oscillation does not mean any movement in space. It is the value of the electromagnetic field, at one given point in space, that oscillates. For electromagnetic waves, there is no matter or photons that go up and down. Instead, you have to imagine that there is a little arrow associated to each point in space: this little arrow is the electric field direction. Another arrow, at the same point, is the magnetic field. These two arrows change size and direction with time, and in fact they oscillate.
How does light oscillate?
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Does $Δs^2 = - Δ\tau^2$ imply two distinct metrics for spacelike and timelike displacements? Reading Misner-Thorne-Wheeler concerning the metric for spacelike and timelike displacements it seems to me that two different metrics must be distinguished, one metric for spacelike and one for timelike displacements - is this a wrong interpretation, or are there two metrics?
My question concerns the combination of three citations:
On page 20, I read in the section IIA "Coordinate-free language":
…The proper distance $s_{AB}$ (spacelike separation) or proper time
$\tau_{AB}$ (timelike separation) is given by …
On page 21, section IIB "Language of coordinates", shorter:
From any event A to any other nearby event B there is a proper
distance $s_{AB}$ or proper time $\tau_{AB}$ given in suitable (local
Lorentz) coordinates by $$ s_{AB}^2 = -\tau_{AB}^2 = -[x^0 (B) - x^0
(A)]^2 + [x^1 (B) - x^1 (A)]^2 + [x^2 (B) - x^2 (A)]^2 + [x^3 (B) -
x^3 (A)]^2. $$
And finally, on page 305 (at equation 13.1):
In the language of coordinates, "metric" is a set of ten functions of
position $g_{\mu\nu} (x^a) $, such that the expression
$$ Δs^2 = -Δ\tau^2 = g_{\mu\nu} (x^\alpha)Δx^\mu Δx^\nu$$
gives the interval between any event $x^\alpha$ and any nearby event
$x^\alpha + Δx^\alpha$.
The first citation explains (correctly) that proper distance corresponds to spacelike separation and proper time corresponds to timelike separation.
In the second citation it is said that for any event there is one of two possibilities, either a proper distance $s$ or a proper time $\tau$, and the square of $s$ equals the square of $\tau$ with opposite sign. This seems logic because for timelike displacements, $s^2$ would give a negative square, and $s$ itself would be imaginary, and vice versa.
But, with respect to the third citation this would mean that
$$ g_{\mu\nu} (x^\alpha) Δx^\mu Δx^\nu$$
is only referring to spacelike displacements, as the squared proper time is $Δ\tau^2$ and not $-Δ\tau^2$, and that it is not applying to timelike displacements where the metric has the opposite sign:
$$- g_{\mu\nu} (x^\alpha) Δx^\mu Δx^\nu$$
So my question is: Must we distinguish between one spacelike metric and one timelike metric?
| Yes, you can talk about there being two metrics, which are coincident for a proper relativistic spacetime - but whose distinction becomes important when considering the classical limit. This is similar to the case for quantum mechanics, where the classical limit requires us to bifurcate the wave function to separate positional and momental wave functions, so that in $\hbar \rightarrow 0$ it reduces to the case of a classical probability scenario with classical Bayesian agent primed with incomplete information about the system (hence leading naturally to the interpretation of $\hbar$ as being an information resolution limit for the universe, just as $c$ is information speed limit).
In particular, we have indeed that
$$d\tau^2 = dt^2 - \frac{1}{c^2} (dx^2 + dy^2 + dz^2)$$
as the time-like metric, and
$$ds^2 = -c^2 d\tau^2 = (dx^2 + dy^2 + dz^2) - c^2 dt^2$$
as the space-like metric. When $c \rightarrow \infty$, we get that $d\tau^2 = dt^2$ and $ds^2 = dx^2 + dy^2 + dz^2$, i.e. two separate spatial and temporal metrics, corresponding to the distinct space and distinct time of Newtonian mechanics. (Note that for the spacelike case, we must ensure $dt \rightarrow 0$ so the interval remains spacelike, otherwise it blows up because you in effect now have a category error.)
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Help Understanding this Free Body Diagram (Eotvos Experiment) I am reading through Hartle's "Introduction to Einstein's General Relativity" and it discusses the Eotvos Experiment in Chapter 6. The free-body diagram (shown below) has me a little puzzled.
This experiment was designed to see if there's any difference in inertial mass mI (think Newton's 2nd law) and gravitational mass mG (think Newton's law of gravity). The setup is this: Imagine two masses of equal weight, connected by a rigid rod. This rod is then suspended by a fiber and hangs freely. At first, you would (incorrectly) think there are only two forces acting on the masses; the force of gravity mGg pulling down straight down to the floor, and tension T pointing straight up along the fiber.
BUT, we are on Earth, and Earth is rotating; thus, there is also centripetal force mIa acting the masses.
So in the free body diagram below, we have three forces acting on the rod/masses: tension, gravity, and centripetal force.
The book/diagram claims that the fiber hangs at a small angle such that a small component of the gravitational force can balance the centripetal acceleration. From the figure though, I don't see how any component of gravity can cancel the centripetal acceleration. If you broke gravity up into X and Y components, one component is pointing WITH the centripetal acceleration, and another component is perpendicular to it - there's no component of gravity that can possibly cancel the centripetal force, is there? What am I missing here? How can gravity cancel the centripetal force?
| This isn’t a proper free body diagram, because the $ma$ line doesn’t represent a force. Only $T$ and $mg$ are forces. Only forces should be in a free body diagram.
Their resulting net force has to equal that $ma$ line because the Earth is rotating.
Now that you see that, note that $T$, the hanging direction, isn’t aligned to $mg$. That (small) difference is what’s being discussed.
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Are matter particles really excitations of fields? So it is said that particles are excitations of fields. We are given examples like photons for electromagnetic field, gravitons for gravity, etc. We are also told that normal matter is an excitation. However the former are all virtual particles that can never be detected. So how are particles like protons that can be detected on the same footing?
| You are confused because of the difference between static EM fields and EM radiation.
An electromagnetic field (also EM field) is a classical (i.e. non-quantum) field produced by moving electric charges.[1] It is the field described by classical electrodynamics and is the classical counterpart to the quantized electromagnetic field tensor in quantum electrodynamics.
https://en.wikipedia.org/wiki/Electromagnetic_field
Now in your example, the magnets' static EM fields are what are represented by these virtual photons. These virtual photons are not real, they are just a mathematical description of a phenomenon called the static EM field. In reality we do not really know how these static fields affect each other (or other particles), we just see that they do in experiments. We describe this with virtual particles.
In physics, electromagnetic radiation (EM radiation or EMR) refers to the waves (or their quanta, photons) of the electromagnetic field, propagating (radiating) through space, carrying electromagnetic radiant energy.[1] It includes radio waves, microwaves, infrared, (visible) light, ultraviolet, X-rays, and gamma rays.[2]
https://en.wikipedia.org/wiki/Electromagnetic_radiation
Now EM waves are what are made up (QFT) of real photons, the very quanta of light. These photons can be detected in experiments.
| {
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Do silicon solar cells act like an LED when you flip the voltage, or are we just seeing black body radiation? In science communicator Steve Mould's Youtube video Why all solar panels are secretly LEDs, he supplies a voltage across a solar cell. Using an IR camera, we see IR emission coming from the solar cell, and he claims that is LED behavior.
Is it not more likely that the IR we're seeing is blackbody radiation? My understanding of LEDs is that indirect semiconductors (i.e. silicon) make for terrible light-emitters, and the energy will be emitted non-radiatively by a phonon.
| Indeed, silicon does not act as an infrared LED. That is because the band gap is indirect, the maximum of the valence band and the minimum of the conduction band are not at the same point in $k$-space. Momentum must be conserved in recombination, and this is impossible because the infrared photon has negligible momentum.
This is why one uses GaAs for IR LEDs.
But he is using a lot of current, so maybe enough to get some near-IR photons anyway.
| {
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What makes a wheel spin? I don't fully grasp what makes a wheel much easier to move than to push a solid block.
The pressure at the point of contact between a wheel and the ground must be pretty enormous compared to the pressure created by a block of same material and mass as the wheel.
Friction is defined as the product of normal force exerted on the object and the coefficient of friction between the object and ground. So I assume that for two identical objects of infinite masses this parameter does not make any difference.
Given these circumstances, I don't understand the physics behind it. Am I missing some other attributes of a wheel that makes it easier to move?
| Let's see if we can answer your question by drawing a free body diagram for these two objects.
From the picture we can see, that in order for body to move, external force must be greater than friction force. If objects have the same mass and are in identical surroundings, they will be equally hard to move. (iff torque would be zero!)
We must however note that forces are not collinear, thus torque is produced.
Now let's focus on our second observation. Torque on a round object causes it to rotate. The same thing happens to object 2.
If you imagine the rectangular object rotating around it's center of mass, you can see, that it's lower left corner is in the way of the ground. Therefore that part of the body acts on the ground and by Newton's 3rd law, the ground acts on the corner of the body.
Since frictional force is dependent on normal force, which increases, it makes the block even harder to move.
(Note that if that normal force exceeds the weight of the object, it's COM will move in positive y direction by Newton's second law)
What happens with COM of a round object, when it rotates? It stays at exactly the same height, so any torque produced does not lift up our object, requiring work, nor does it 'add' to frictional force.
In short:
Objects having circular cross-section are easier to displace as torque resulting from non-collinear forces does not increase normal force on the object, which would yield to higher frictional force.
| {
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If a ball is thrown to a person standing on a frictionless surface, is the impulse of the thrower equal to that of the catcher? If a person throws a ball, exerting a given impulse does the person that catches the ball receive the same impulse assuming that the catcher moves. Is the impulse that the catcher receives less than the impulse that the thrower receives because the ball continues to move with the catcher or does the catcher receive all of the impulse, to begin with and then return momentum to the ball as they pull it along in their hand?
| Momentum is conserved in any inertial frame. You choose an inertial frame to write the equations for momentum conservation. In the catcher's frame obviously the ball's momentum is less than that in thrower's frame due to relatively slow motion.
| {
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Is the Four-gradient of a Scalar Field a Four-Vector? Consider a scalar field $\phi$ as a function of spacetime coordinates $x^\mu$. The four-gradient of $\phi$ is given by
\begin{equation}
\frac{\partial \phi}{\partial x^\mu} = \left( \frac{\partial \phi}{\partial t}, \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}\right)
\end{equation}
I'm studying a little bit of classical field theory and in the topic of real scalar fields, it is common to introduce four-gradients of these fields.
My question then is: The components of this four-gradient are not the components of a four-vector. For example, take the first component, and considering $t = t(x',t')$
\begin{align}
\frac{\partial \phi}{\partial t} &= \frac{\partial \phi}{\partial t'}\frac{\partial t'}{\partial t} + \frac{\partial \phi}{\partial x'}\frac{\partial x'}{\partial t}\\ &= \gamma \left( \frac{\partial \phi}{\partial t'} - v \frac{\partial \phi}{\partial x'} \right)
\end{align}
EDIT: In some books I've read, the four-gradient $\partial _\mu$ is defined as a four-vector, and since the D'alambertian operator $\Box = \partial _\mu \partial ^\mu$ is a lorentz invariant, the four-gradients must be four-vectors. But my computations give me this unsatisfactory result. Am I doing computational mistake or what? Please help me.
| The four gradient is a four vector but it transforms covariantly, rather than contravariantly. This makes it a "covector".
It also has a contravariant form, obtained by multiplying it by the metric, which transforms like 4 four position or four momentum.
| {
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What does the arbitrary constant in cosine equation of displacement in S.H.M say? The phase and phase constant in a displacement time equation show from where the particle has started.
In my school textbook, first the displacement equation was given as :-
$$x= A\sin(\omega t+\phi)$$
where $\phi$ is the phase constant.
But then it said if the particle is at extreme position then we add $\pi/2$ because obviously displacement is maximum at $\pi/2$
So now the equation at extreme should be :-
$$x=A\sin\left(\omega t+\frac{\pi}{2}\right)$$
$$x=A\cos(\omega t)$$
But in my textbook the equation is :-
$$x=A\cos(\omega t + \phi')$$
It says that $\phi '$ is another arbitrary constant. But technically $\phi$ is $\sin ^{-1} (x/A)$, here $x$ will be $A$ and we get $\pi/2$ so no constant remains. But what is this $\phi '$ constant and on which thing it depends?
| It depends on the initial conditions. If its at maximum displacement at $t=0$ then the equation is $x=A\cos{\omega t}$. If it is at equilibrium position and maximum velocity at $t=0$ then $x=A\sin{\omega t}$. In general, the solution is $x=a\cos{\omega t} + b\sin{\omega t}$ which simplifies to either $A\cos{(\omega t + \phi ')}$ or $A\sin{(\omega t +\phi)}$ using the harmonic addition formula.
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On the "spectrum" of an operator in quantum mechanics Very simple question, I'm new to this. I'm reading Griffiths book on QM and have a question about the "spectrum" of an observable operator. Does the spectrum of an operator require specification of a particular system? Or is the spectrum of an operator just every possible eigenvalue that can be obtained by every possible eigenfunction of an operator?
| In general, the spectrum of an operator depends on the Hilbert space (= space of possible wavefunctions) on which it is defined. For example, the Hamiltonian for a free particle in one dimension $H = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}$ has eigenvalues that depend on the boundary conditions for the wavefunction $\psi$. In a box (= infinite potential well) of length $a$, the eigenvalues are $E_n = \frac{\hbar^2}{2m}\left(\frac{n\pi}{a}\right)^2$ which depend on $a$, even though the expression for the operator does not. For the same operator without any boundary conditions, any real number is an eigenvalue.
The Hamiltonian is a special operator because it defines the system. A "system" is nothing more than a choice of Hilbert space and a Hamiltonian defined on it. The spectrum of other operators, such as position $x$ and momentum $p = \frac{\hbar}{i} \frac{\partial}{\partial x}$, depends on the Hilbert space but not on the Hamiltonian.
| {
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Is the coefficient of drag the same on two similar objects? if I had two objects, scaled perfectly to each other so that one is $5$ times the size of another while keeping the shape the same, would their coefficients of drag, $C_d$, be the same in the formula $D = \frac12 \rho C_d S V^2$?
| Not necessarily.
If we take the fluid dynamical system under consideration to be only a function of fluid density $\rho$, some characteristic speed $V$, fluid viscosity $\mu$, and some characteristic length $R$, then it has to be the case from dimensional analysis that the drag coefficient is only a function of the Reynolds number:
$$C_d = f(Re)$$
Since altering the characteristic length of the object changes the Reynolds number, you'd expect the drag coefficient to change as the size of the object changes.
That being said, at high enough Reynolds numbers, the drag coefficient "paradoxically" becomes independent of the Reynolds number (and by extension of everything), so size doesn't affect the drag coefficient in that scenario. You should certainly expect a dependence at low Reynolds numbers, where the drag coefficient is asymptotically inversely proportional to the Reynolds number.
For a more detailed discussion on the mathematical derivations of why this is so, feel free to read my notes on lift/drag here.
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$V$-$I$ characteristic of a solar cell please explain the VI characteristics of a solar cell. The characteristics is given in my book without any explanation. How can the Voltage decrease on increasing current shouldn't it be opposite.
Solar Cell I-V characteristics
(Image from Electrical 4 U - Characteristics of a Solar Cell and Parameters of a Solar Cell)
| VI characteristic graph of diode with and without illumination (breakdown not shown):
When light falls on the depletion region of the diode, electron-hole pairs are generated that try to reach their parent nuclei. The electrons move towards the n-side and the holes to the p-side. This leads to a photovoltage that tries to make current flow in reverse direction.
Let $V_p$ be the photovoltage and $V$ be the external applied voltage. Then under illuminated conditions:
• If $V=0$, there's a negative current (i.e., a reverse current) in the circuit due to the photovoltage.
• In forward-biased, if $V_p>V$ then the current is negative and if $V>V_p$ then the current is positive.
• In reverse-bias ($V<0$), the photovoltage is largely responsible for the negative current; the current does not increase much on increasing the applied voltage (until reaching breakdown).
OP's graph just seems concerned with the magnitude of current in case of solar cell (quadrant 4 in my diagram).
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How does the current remain the same in a circuit? I understand when we say current, we mean charge (protons/electrons) passing past a point per second. And the charges have energy due to the e.m.f. of the power supply.
Now tell me, if a lamp has resistance and you hook it in the circuit, how will the current stay the same? The charges obviously lose energy in the lamp and so become SLOWER, which should mean current decreases, right?
[Edit]
All answers explained a bit of everything, so it was hard to choose one. If YOU are looking for an answer, please check the others too, in case the accepted one doesn't answer your question.
| The question
The charges obviously lose energy in the lamp and so become SLOWER, which should mean current decreases, right?
shows that you have a misconception about the motion of the conduction electrons.
If you were correct then to maintain the same current around a circuit by a miracle more conduction electrons would need to contribute to the conduction process as you went around the circuit.
If this did not happen then the conduction electrons would move slower and slower and . . . . . eventually stop?
In fact what happens is that the electrons gain kinetic energy (and lose electric potential energy) between collision with the lattice (bound) ions from the electric field in the wire and then lose that extra kinetic energy to the lattice ions upon collision with them.
The net effect is that the temperature of the material increases as the internal kinetic energy of the material has increased (the lattice ions vibrate more) and the conduction electrons move along the conductor with a constant average speed.
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2-D water jet on oblique plane We have a 2-D water jet on oblique plane. We are ignoring gravity and want to calculate the force on the plane. The explanation I have been given goes as follows. We assume inviscid flow. Image is attached.
Using Bernoulli's theorem on the free surface streamline:
$$\frac{1}{2}V^2+\frac{p_{atm}}{\rho}=\frac{1}{2}U^2+\frac{p_{atm}}{\rho}$$
and so far enough along, we must have the flow velocity to be constant. We also have conservation of mass implying that $$Va=Va_1+Va_2$$ These two parts I understand. We then find the momentum flux parallel and perpendicular to the plate:
parallel component of: $$\int_S \rho \underline{u}(\underline{u}.\underline{n})+p\underline{u} dS=0$$ where the other terms in the momentum equation disappear since we have steady flow and are neglecting gravity. We then say this is equal to $$\rho aV^2\cos(\beta)=\rho a_2V^2+\rho a_1V^2$$ We do a similar argument to find the perpendicular component of the momentum equation.
I am not sure about this part. Are we not neglecting the term from $$\int_S p\underline{u} dS$$ when finding the parallel component, if so why?
Here is the full derivation for reference:
| The pressure doesn't figure in the parallel component because it is acting perpendicular to the wall (and, moreover, at the inlet and outlet, the pressure is atmospheric).
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To what extent is standard quantum mechanics actually non-relativistic? I often hear that while QFT is a relativistic theory, standard quantum mechanics is not. But fields aren't inherently relativistic, you can easily construct non-relativistic QFTs, a relativistic QFT is one with a Lorentz invariant Lagrangian.
In the same way, it doesn't seem to me that there is anything inherently non-relativistic about the standard description of quantum mechanics, and that the actual non-relativistic part is the Schrödinger equation.
As a matter of fact if we substitute the Schrödinger equation axiom with dynamics described purely by quantum channels, with the only requirement that an evolved quantum state is still a quantum state and that the theory is linear, we do recover some statements that resemble locality in the Lorentz sense, like the no communication theorem.
Are fields necessary to obtain a relativistic theory? To what extent is standard QM non-relativistic? If QM is inherently non-relativistic even without the Galilean dynamics given by the Schrödinger equation, why should one care about non-locality in Bell inequalities and "spooky action at a distance"? Those would be expected in a non-relativistic theory, just like the gravitation of the Sun on the Earth is spooky action at a distance in Newtonian gravity.
|
the actual non-relativistic part is the Schrödinger equation
Indeed. Hence, people tried to come up with a Lorentz-invariant evolution equation for the wave function such as the Klein-Gordan and Dirac equations (as mentioned by anna v). However, interpretation of solutions of these equations as probability densities are problematic (in case of the Klein-Gordon equation, the predicted probabilities won't be positive definite - though the Dirac equation 'fixes' this particular problem).
We've come to realize that these equations are not to be understood as evolution equations for the quantum state, but in the sense of quantum field theory, where (in the Schrödinger picture), their (classical) solutions correspond to the configurations of the system (ie the equivalent of positions in single-particle QM), with states being wave-functionals on that solution space.
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Normalization of the action in Special Relativity The action for a massive point particle in Special Relativity is given as
$$A =-mc^2\int d\tau,$$
Where $\tau$ represents the proper time, and $m$ represents the (rest) mass. From what I could understand, the Action must not change with respect to the reference frame, and hence it can be written as $$something\int d\tau$$ but why must the something need to be proportional to mass?
| I prefer to think like this. Inertial observer ($X$) is at rest in his/hers reference frame. The world-line of $X$ is the longest possible route between any two events. This follows since, in its frame, $X$ is moving fully along the temporal axis, therefore $ds=cd\tau$ (displacement in time; $c$ is the speed of light), and $d\mathbf{r}=\mathbf{0}$, and one should bear in mind that, in general, the 4d distance between two events, within the causality cone, is $ds=\sqrt{c^2 d\tau^2 - dr^2}$.
We can therefore describe the path taken by the observer $\bar{x}^\mu=\bar{x}^\mu\left(\tau\right)$ as the (unique) path that is longest possible path between events $A$ and $B$ (that lie on the world-line). Since length of a curve (in 4d sense) is an invariant quantity, the universal description of the path of the observer is, a world-line that maximizes/minimizes:
$S\left[\bar{x}\right]\propto\int^B_A cd\tau$
Where $cd\tau$ gives a length of a small segment of the world-line. We can then fix the sign of the scaling constant in order to demand, that this quantity, which we shall call `action', needs to be minimized to arrive at the correct world-line:
$S\left[\bar{x}\right]=-\alpha\int^B_A cd\tau,\quad \alpha>0$
The above prescription must be valid for all observers, including those that observe $X$ as moving at small velocity $\mathbf{v}\: \left(\left|\mathbf{v}\right|\ll c\right)$. Call any such observer as being in the lab-rame. In lab-frame we can describe $X$ with classical action, thus, in the limit $\left|\mathbf{v}\right|/c\to 0$:
$S=-\alpha\int^B_A cd\tau \to \int^B_A \frac{m\left|\mathbf{v}\right|^2}{2} dt$
Next, the world-line of $X$ in the lab-frame is, by construction $\bar{x}^\mu=\left(ct,\,\bar{\mathbf{r}}\left(t\right)\right)^\mu,\:d\bar{\mathbf{r}}/dt=\mathbf{v}$. The 4d distance between any two (close) events on $\bar{x}$ is:
$cd\tau=\sqrt{c^2-\left|\mathbf{v}\right|^2}dt$
Thus, in the limit $\left|\mathbf{v}\right|/c\to 0$
$-\alpha\int^B_A \sqrt{c^2-\left|\mathbf{v}\right|^2}dt \to \int^B_A \frac{m\left|\mathbf{v}\right|^2}{2} dt$
This fixes $\alpha$, though, of course, the specific value of $\alpha$ is more of a convention.
| {
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If magnetic field lines don't exist, what are these iron filings doing around a magnet? Obviously the iron filings can be seen aligning themselves along the virtual magnetic field lines produced by the permanent magnet, the virtual magnetic field line is made of electromagnetic field due to the alignment of electrons in the magnet but why the patterns, why lines? Do these lines have thickness? Are they due to interference pattern?
| The iron filings themselves repel each other and create a pattern that can be observed visually as lines.
Liquid metals create mountains that end with peaks as the active forces dissipate, so you can see the concentration is highest nearest the origin.
| {
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Heisenberg's principle in classical world I have a ball with me and I have been observing it's position which remains constant at a place only, hence it must be largely localised, further as it remains on the same place from so long, it's velocity must be largely localised to 0 only.
Now with both momentum and position know to any degree, Where is the uncertainty??
With the understanding of quantum mechanics, we must expect the "fixed" ball to have any arbitrary momentum, but that is not the case.
Where am I going wrong??
| First, while the amount of uncertainty required by the Principle is significant at quantum mechanical scales, it's tiny on macroscopic scales. Are you certain where the ball is, down to an atom's radius?
Second, the uncertainty is regarding the object's momentum, not its velocity. If you have an object that's 10,000 times the mass of an electron, then it can have 1/10,000 the uncertainty in its velocity, while having the same uncertainty in its momentum. A ball is larger than an electron by a factor larger than Avogadro's Number, so even a tiny amount of uncertainty in its velocity results in an uncertainty in it momentum that is massive by quantum mechanical standards.
| {
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Explicit calculation of spin connection through Cartan's first structure equation Given the metric
$$
ds^2 = F(r)^2dr^2 + r^2d\theta^2 + r^2 \sin^2(\theta)\, d\phi^2,
$$
I'm trying to find the corresponding spin connections $\omega^a_{\ b}$ using the first structure equation:
$$
de + \omega e = 0.
$$
I found the vielbeins $e$ and their exterior derivatives $de$:
$$
de^1 = 0, \quad de^2 = drd\theta, \quad de^3 = sin(\theta)drd\phi + r\cos(\theta)d\theta d\phi,
$$
but I am stuck on actually working out the $\omega$.
I read through Zee's 'GR in a nutshell', and he does the same calculation but just says: "In general, write $\omega^a_{\,b} = \omega^a_{\,bc}e^c = \omega^a_{\,b\mu}dx^\mu$. Plug this into the first structure equation and match terms. How do I actually go about calculating $\omega^1_{\, 2}$, $\omega^1_{\, 3}$, and $\omega^2_{\, 3}$ at this point?
| As you have, the first step is to identify $e^r = F(r)\mathrm{d}r, e^\theta = r\mathrm{d}\theta$ and $e^\phi = r\sin \theta \mathrm{d}\phi$. The trick is to then take the derivatives but re-express them in terms of $e$ again. Thus,
$$\mathrm{d}e^r = 0, \quad \mathrm{d}e^\theta = -\mathrm{d\theta} \wedge \mathrm{d}r = -\frac{1}{rF(r)}e^\theta \wedge e^r$$
and,
$$\mathrm{de^\phi} = -\sin\theta \mathrm{d\phi} \wedge \mathrm{d}r - r\cos\theta \mathrm{d}\phi \wedge \mathrm{d}\theta = -\frac{1}{rF(r)} e^\phi \wedge e^r - \frac{\cot \theta}{r^2} e^\phi \wedge e^\theta.$$
Now let's take an example of using Cartan's first equation. We have $\mathrm{d}e^a + \omega^a_b \wedge e^b = 0$ and if we choose $a=\theta$ the equations read,
$$\frac{1}{rF(r)}e^\theta \wedge e^r = \omega^\theta_r \wedge e^r + \omega^\theta_\theta \wedge e^\theta + \omega^\theta_\phi \wedge e^\phi.$$
We have $\omega^\theta_\theta = 0$ by anti-symmetry. We can identify now $\omega^\theta_r = -\omega^r_\theta = \frac{1}{rF(r)}e^\theta$. Notice the last term we could choose $\omega^\theta_\phi = 0$ however Cartan's equations are a system of equations, so we are not free to make this choice yet without considering the other equations. We can at best say $\omega^\theta_\phi$ is proportional to $\mathrm{d}\phi$ to ensure $\omega^\theta_\phi \wedge e^\phi = 0$. As it turns out, we don't have $\omega^\theta_\phi = 0$ because of the $a = \phi$ equation, which will give you $\omega^\theta_\phi = -r^{-2}\cot\theta\, e^\phi$.
I hope this elucidates how to use Cartan's structure equation. Computing the Ricci tensor is then much simpler, as rather than solving for components you're just plugging in and computing.
| {
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Gravitational Wave - What is waving? Two kinds of wave transmission are:
*
*Light waves, where a substance (photon) travels as a wave.
*An attached rope, like at the gym, that is "waved" up and down. Here, no substance travels to a new spot, but adjacent parts transmit the energy to others.
QUESTION: Which method do gravitational waves propagate by?
| A gravitational wave is a wave in a tensor field called the “metric of spacetime”. This metric determines the geometry of spacetime by specifying how far apart spacetime events are. The wave just consists of changes at each point in the values of the components of this field.
It is very similar to a classical electromagnetic wave, which is just changes in the values of components of a different field, one that has nothing to do with spacetime geometry.
| {
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How to find expectation value from probability density? How to find expectation value from a given probability density?
E.g. given to find expectation value of $x^2$ from probability density $\rho(x)= 1/(2(hx)^{1/2}) $.
This is not a homework question, I just want to understand how to solve questions like this, you may use your own example.
| According to expected value basic properties:
$$ {\displaystyle \operatorname {E} [g(X)]=\int _{\mathbb {R} }g(x)f(x)\,dx.} $$
it should be :
$$
E[x^2] = \int x^2\,\frac{1}{{{2}{\sqrt{{h}{x}}}}}\,dx
$$
| {
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Wave function of $s$-band with odd parity How does a wave function of a $s$-band state with odd parity looks like (in real space)? To keep it simple, the restriction to a linear chain and a state at the $\vec{k}=0$-point might be useful.
My understanding is somehow like described in https://www.ensc-rennes.fr/wp-content/uploads/2015/04/JFH-Intro-Solide.pdf. When we want to construct the crystals wave function in a tight binding fashion, we start with all electrons in their atomic s-orbital. With next-neighbour interactions, we get the cases of neighbouring orbitals being in the same or oposite phase, what gives us (with two atoms in a unit cell) one bounding and one antibounding band (see p.14 in the reference).
As I have learned from topological considerations of semiconductors, each band in a crystal which has inversion symmetry (thus also our linear chain) should have a well defined parity (at least at $\vec{k}=0$). If we put the origin of this inversion into one atom, both states (bound and antibound) have positive parity. Just if we would put the origin of the inversion exactly between two atoms, the parity of the antibound state would be odd. However, as far as I know, the origin of the unit cells in crystals and thereby the origin of the symmetry operators of the point group are defined within an atom. So how do we get a s-band with an asymmetric wave function at $\vec{k}=0$?In my opinion that is just possible by a p-like atomic orbital.
| While my picture of band creation and the spatial extend of the corresponding wave functions was right, the assumption that the origin/center for inversion symmetry is always on an atom CAN be wrong. While considering a linear chain, the choice of the inversion center can be on an atom or halfway between two atoms. For a diamond lattice the inversion-center must be halfway between two atoms, while all rotations etc. are defined with respect to an origin at an atom. This shift of the coordinate system regarding different symmetry-operators is named "nonsymmorphic" space group and is rather irrelevant for further group-theory ... thus this detail is rarely mentioned.
| {
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If sound passes through material, vibration is produced. So are electromagnetic waves produced too? Sound means vibration of molecules and vibration produces electromagnetic waves. So, this means that sound produces electromagnetic waves directly.
Is this possible?
| A sound wave passing through a medium (e.g. air) indeed displaces molecules by a distance of a few nanometers. It seems reasonable that it should also displace the atoms, and thus electrons and protons in the process, which are charged particles and should radiate by Larmor's equation when undergoing acceleration.
Let us assume the sound frequency is on the order of kHz (which is in our audible range). Then molecules are accelerated by
$$a \approx \left(10^3\, \mathrm{Hz}\right)^2\times \left(10^{-9}\, \mathrm{m}\right) \approx 10^{-3}\, \mathrm{m/s}^2 $$
Then the predicted power of the radiation produced by Larmor's equation is ridiculously small
$$P = \frac{2}{3}\frac{q^2 a^3}{c^3} \sim 10^{-73}\, \mathrm{W} $$
Even if one multiplies this by the number of molecules of air in a $\mathrm{m}^3$, $N\approx 10^{25}$, this would never be detectable. Therefore, this effect may well exist, but it is absolutely negligible in all respects.
N.b. My answer focused on direct effects of the acceleration of air molecules due to a sound wave. As other answers mention correctly, there are interesting secondary effects of (especially large-amplitude) sound waves involving EM radiation. Among these are sonoluminescence and heating of the air by sound dissipation leading to increased thermal radiation.
| {
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Observing from a true stationary point in space is there a "True time" we can observe? Aren't all points on planets, in solar systems, in galaxies moving through space subject to different amounts of time dilation due to gravity and velocity when observed from a stationary point in space? Do we generally have a mathematical transformation to "true universal time" versus our dilated time with regards to how we view the universe? Does it in effect "lense" our view of the universe?
| As trula's answer reveals there is no place in the universe and never has been that allows a static observer. Think of it like we're all under water floating around and nobody is excluded.
We can't have an objective "point" from which to measure all others. All measurements are relative.
20 meters floating in strong current is different from 20 meters in placid water.
When we get in a strong current we might not notice it unless we have something to compare it to. Normally we could see the riverbank passing by more quickly but if there is no riverbank we have to judge by the bodies in the water. It's hard to know what currents are moving them though unless we get in their position.
If you pass another human by under water they may seem to be going fast but maybe it's you that is moving quickly and they think they are slow.
If we go to their position to find out what's going on we give up our own position and can't compare anymore.
Each person measures things from their own perspective and there is no fixed time.
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Why is it that lampblack repels water but graphite doesn't? I have been playing around with lampblack recently, and one property that I particularly like in them is their Hydrophobicity. It's interesting to watch water drops just bouncing off the surface.
I reasoned that this was because lampblack was carbon, which is hydrophobic. I wondered if Graphite could also repel water since it was also made up of carbon. I took a pencil and made a thick patch of graphite on a paper and placed a water drop on it. But I couldn't see any hydrophobicity.
Why is it that the lampblack repels water whereas graphite doesn't; even though both are made up of carbon?
| the lampblack contains partially-burned hydrocarbons which are strongly hydrophobic.
| {
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Torque and force Does the force due to torque on the edge of a wheel depend on the mass of the the wheel or is it always $\tau / R$?
| A car engine puts effort into producing a torque about the wheel's axle. This torque makes the wheel turn.
If the wheel is heavy (large mass, thus large moment of inertia), then it is harder for the engine to produce this torque.
But if the engine succeeds in producing the torque, then that torque causes a force at the road (at the contact point between road and wheel). And the relation is:
$$\tau=rF_\perp$$
This expression always applies. And it does not involve the wheel mass. That mass influenced the effort required to produce the torque, but not the force caused by that torque.
The answer to your question is no.
Note that in other situations you might think og the torque being produced by the force. Such as when turning a bolt with a wrench. Then the situation is reversed. Then it takes effort to produce the force (effort by your hand for example). And the heavier the wheel, the larger the necessary effort.
Still, in this situation as well, the torque that this force causes does not depend on the mass, but again follows the expression above.
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The motion of an electron accelerating in a uniform magnetic field As far as I know, when electrons travel perpendicular to a uniform magnetic field, the Lorentz force makes the electron undergo circular motion. As this electron undergoes circular motion, it emits EM radiation, so it goes through a spiral trajectory with a continuously decreasing orbital radius.
*
*As this electron emits EM radiation, its energy should decrease. However, as far as I researched, the velocity of the electron should remain constant as the magnetic field does no work on the electron. Does this mean that the loss of the kinetic energy of the electron comes from its loss of mass? If it does, what is the formula for the mass of an electron (or any charged particles) that is accelerating in a uniform magnetic field and thus emitting EM radiation?
*Where can I find some videos that actually show the spiral trajectory of an electron?
*How fast should an electron be moving or strong the magnetic field be for the electron to undergo a spiral path like the one shown below?
| The constant speed of electron is by assuming electron does not emit EM radiation.You just mixed up with it.
| {
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Why does the same proportion of a radioactive substance decay per time period? (half life) Just wondering, if decay is random, why does the activity half every half life, as in, why does it have to reduce by the same proportion in the same time period?
| The word random in this context does not mean totally without order.
What it means is that one cannot predict exactly when a particular unstable nucleus will decay although there is an underlying probability of decay of an unstable nucleus in a specified interval of time.
An interval of time which is often used for a particular species of unstable nucleus is the half-life.
The probability that an unstable nucleus will decay in a time interval equal to one half life is $\frac 12$.
If an unstable nucleus does not decay in that time interval then the probability that it will decay during the next time interval of the same length is still $\frac 12$ . . . etc.
You will note that this is similar to the toss of a coin with two outcomes heads and tails each with a probability of $\frac 12$.
However an insight to the randomness of coin tossing might be shown in that although the probability of tossing a head is $\frac 12$, then if one tossed a coin $100$ times there is a fairly low probability, $0.07959$, of the outcome being exactly $50$ heads and $50$ tails.
So what you have is a number of possible outcomes, number of heads + number of tails $=100$, for which you can predict the probability of them happening but you cannot say for certain, probability $=1$, which of those outcomes will actually occur.
In the context of radioactive decay on average half a sample of unstable nuclei will decay in one half life and then on average half the remaining unstable nuclei will decay during the next interval of one half life etc.
With samples of billions and billions of unstable nuclei the statistical fluctuations about “one half will decay during a time interval of one half life” will be small.
As the number of unstable nuclei becomes smaller the statistical fluctuation about one half will get larger.
Just think about what you would predict about the decay of $3$ unstable nuclei in an interval of one half life. They could all not decay for ten half lives although the probability of the happening is small.
| {
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Would connecting two pressurized gas tanks together exchange gases? While the answer to the title is inevitably "Yes", allow me to clarify. I have no background in physics, so please bare with me and I'll be happy to add any clarifications requested in the comments.
If I have 2 tanks:
*
*rated for 100,000 kPa, currently at 100,000 kPa, contains Gas#1
*rated for 2,000 kPa, currently at 1,000 kPa, contains Gas#2
Let's say hypothetically their valves are connected together with an infinitely strong pipe and some form of pressure regulator that prevents Tank #2 from being pressurized beyond 2,000 kPa. Let's also say the gases inside are of equal density/volume/weight/etc.
The question is:
If I opened the valves to these 2 tanks, and Tank #1 pressurizes Tank #2, will the gases exchange between the tanks? In other words, which one of these scenarios would be correct:
Scenario 1:
Tank #1 is now at 90,000 kPa, contains Gas#1
Tank #2 is now at 2,000 kPa, contains 50/50 Gas#1/Gas#2
-
Scenario 2:
Tank #1 is now at 90,000 kPa, contains Gas#1/Gas#2 (ratio unimportant)
Tank #2 is now at 2,000 kPa, contains Gas#1/Gas#2 (ratio unimportant)
Thank you in advance!
| Note that two gas in cylinder will diffuse with each other and come to same pressure and temperature. Also will occupy whole volume of two container.
Mathematically finding pressure
$$P(V_1+V_2)=(n_1+n_2)RT$$(for final combination).
Now for final temperature $$n_1C_1(T-T_1)=n_2C_1(T_2-T)$$
Finally solving these
$$P=(R(n_1T_1+n_2T_2))/(V_1+V_2)$$
Hope this helps you.
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Double slit for electrons (two beams or one)? I understand the double slit for waves but for electrons do we have a beam for each slit so each beam is responsible le to shoot electrons through its own slit.
Or do we have just one beam? Which slit do we place the beam to? If it’s in the middle, wouldn’t all the electrons hit the wall between the two slits?
Sorry Ian very confused as how to carry out the experiment for electrons
| There is only one source or one beam of electrons. They are directed toward the slits, with some hitting the middle and some making contact with the edges of the openings.
| {
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Is it possible to reduce the speed and frequency of a light wave to zero in a liquid medium? Assume that two mirrors are located in a huge liquid medium – say, deep down on an ocean floor – with a refractive index of $n'$ as measured by an observer $A$ standing on the beach platform at rest WRT the liquid. One of the mirrors is at rest WRT $A$, and the other mirror recedes from $A$ at $v$. Remember that the mirrors are parallel to each other both perpendicular to the velocity vector.
Now, suppose that a tiny laser diode attached to the lowest part of the first mirror sends a photon towards the moving one at an angle nearly perpendicular to the surfaces of the mirrors. The photon, after being bounced between the mirrors for $i$ times, is received by a detector set on an upper side of the same mirror. [See the Figure.] Each time the photon hits the moving mirror, its speed reduces as measured by observer $A$.
Can we say that, from the perspective of $A$, the velocity of the photon, complying with the relativistic velocity addition, reduces each time it hits the moving mirror till the speed, possibly along with the light frequency, approaches zero if the number of bounces is great enough? Is it correct to use the relativistic velocity addition formula in this example?
According to the figure, $c'_1$ is calculated to be:
$$c'_1=\frac{c}{n'} \tag{1}$$
Using the relativistic velocity addition formula, for $c'_2$ we can write:
$$c'_2=\frac{c'_1-v}{1-\frac{vc'_1}{c^2}}=\frac{\frac{c}{n'}-v}{1-\frac{v}{n'c}} \tag{2}$$
Similarly, for $c'_3$, $c'_4$, and $c'_i$, we have:
$$c'_3=\frac{c'_2-v}{1-\frac{vc'_2}{c^2}} \tag{3}$$
$$c'_4=\frac{c'_3-v}{1-\frac{vc'_3}{c^2}} \tag{4}$$
$$c'_i=\frac{c'_{i-1}-v}{1-\frac{vc'_{i-1}}{c^2}} \tag{5}$$
As we see, we can arbitrarily reduce the speed of light if we repeat the calculation enough. Does this mean that the speed of light can be as slow as a turtle in a medium by this means? Do the accurate calculations become more complicated than those done here by considering the fact that the refractive index varies as a function of frequency as well as velocity $v$?
| In other words, does speed of light (after reflection) depend on speed of mirrors as well as speed of water or not? Clearly the answer is NO.
Let's redo this experiment, but this time in space not water. Simply enough we can see that speed of light will be c no matter what. Wavelength is the only thing which is affected by velocity of mirrors and that's because of Doppler effect.
Water is no exception(why would it be?). Light just doesn't care about what it sees at boundaries. Its speed from A perspective will be $c/n$ before and after reflections. From perspective of moving mirror however, light's speed will oscillate between $\frac{c/n-v}{1-\frac{cv}{nc^2}}$ (before reflection) and $\frac{c/n+v}{1+\frac{cv}{nc^2}}$ (after reflection) where $v$ is speed of water.
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Fringing of electric field I have read that in a capacitor with charged parallel plates the electric field lines are parallel in the middle, but they tend to bend outwards (causing a "fringe") towards the ends of the parallel plates. Can someone explain why this really happens? Does it happen due to the lack of symmetry, which is usually present in an infinitely long charged plate? It is to some extent obvious that the electric field isn't uniform at the ends, but why should they bend outwards only, can't they bend inwards?
| How is the field produced? By charges on the surface. If you go to the quantum frame, it is excess electrons on one plate and excess positive charge (holes) on the other plate. Think of the electric field generated by an electron. It goes radially out. In an infinite plate capacitor the addition of the fields, because of symmetry becomes vertical. Given dimensions, the electrons at the edge will be having lines radially out, the positive charges on the other plate will go to meet them again radially out, because that is the geometry of the point charges. On the side that is in the air, there are no fields to add towards the vertical and the shape is like the field shape in two dimensions of a pair of +- , in the line perpendicular to the edge.
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Age of Universe from Friedmann Equation - How to actually solve the integral? The Friedmann equation for a flat universe can be written as
$$
H(t)=\frac{\dot{a}}{a}=H_0\sqrt{\Omega_{m,0}\cdot a^{-3}+\Omega_{\Lambda,0}}=H(a)
$$
To calculate the age of the universe, many books jump directly to the result. But there should be some sort of integral in between. I assume one can do the following:
$$
t=\int_0^{t_0}\!\mathrm{d}t=\int_0^{1}\!\mathrm{d}a\frac{\mathrm{d}t}{\mathrm{d}a}=\int_0^{1}\!\frac{\mathrm{d}a}{\dot{a}}
$$
with $\dot{a}$ from above expressin for $H$.
But how is this integral solved? Mathematica did something for hours but did not came up with a result. Most books and wikipedia pages skip directly to the result
$$
t_0=\frac{1}{3H_0\sqrt{\Omega_\Lambda}}\log{\frac{1+\sqrt{\Omega_\Lambda}}{1-\sqrt{\Omega_\Lambda}}}
$$
which leads to the well known result of ~13 billion years (depending on the DM density).
Again: But how is the integral solved?
| In a flat universe where $\Omega_m+\Omega_{\Lambda}=1$ you can first simplify to
$$H=\frac{\dot{a}}{a}=H_0\sqrt{(1-\Omega_{\Lambda}) a^{-3}+\Omega_{\Lambda}}$$
then you solve for $a$ and get ⁽¹⁾
$$a=\sqrt[3]{\frac{1}{\Omega \Lambda }-1} \ \sinh ^{\frac{2}{3}}\left(\frac{3}{2} H_0 \ t \ \sqrt{\Omega \Lambda }\right)$$
where you solve for $t$ and get
$$t=\frac{2 \sinh ^{-1}\left(\frac{a \sqrt{\frac{a}{\sqrt[3]{\frac{1}{\Omega \Lambda }-1}}}}{\sqrt[3]{\frac{1}{\Omega \Lambda }-1}}\right)}{3 H_0 \sqrt{\Omega \Lambda }}$$
which, when you set $a=1$ for the present scale factor, can be further simplified to
$$t_0=\frac{2 \sinh ^{-1}\left(\frac{1}{\sqrt{\frac{1}{\Omega \Lambda }-1}}\right)}{3 H_0 \sqrt{\Omega \Lambda }}$$
which is equivalent to the expression your source gave.
| {
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Why does the galvanometer read 0 at a certain point?
We are studying potentiometers in class, and our teacher wasn't able to explain why the galvanometer would read 0 at a certain point along the wire. I think I'm not quite understanding how the current is flowing through the circuit because of the orientation of the cells. Somehow they're supposed to 'cancel each other out' but I'm not sure how this works.
If anyone could explain this I would be very grateful!
| The driver cell in the OP's figure is connected to the potentiometer wire AB which is acting as a voltage divider with the output at the jockey. It may help to rotate the figure. (It may also help to set the bottom potential to zero, ground.) Then that part of the figure looks like this:
Now, if the output voltage $V_{\rm out}$ is equal to the unknown emf of cell X, then there won't be a current in the galvanometer.
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Ionizing radiation in thermal radiation According to the black-body radiation equation, the spectrum extends to infinitely high frequency (although its intensity gets small quickly towards high frequency).
(1) How do you roughly estimate the ionizing radiation power in common high power thermal sources like a 2KW heater without fan, then make sure it is safe considering it is used for years instead of 0.1 second/year in the case of DR X-ray scan? (First the portion of power getting radiated instead of convected needs to be estimated, second this seems not to be black-body and will it have a similar radiation curve that extends to infinite frequency?)
(2) In the photoelectric effect, there are things like cut-off frequency; so, why doesn't thermal radiation - which is quantum mechanical in microscopic level - have a cut-off frequency, i.e. it doesn't radiate X-ray and gamma-ray at all?
| for 1) In this link there is the black body formula of Planck's law that gives the
the power per unit solid angle and per unit of area normal to the propagation
So for a given temperature of the body one can calculate the power, in your case for high enough frequencies, take a $Δ(ν)$ .
Once you have the power you will have the energy per second. Divide the power with the average $hν$ of your $Δ(ν)$ and you can estimate how many photons of that approximate frequency radiate per second.
You will find that one has to wait for a loooo..ng time for an X-ray photon to come our of your radiator. Better worry about the Xrays created by the muons passing your body at the rate of 1 every $cm^2$ every minute, much worse when you fly in an airplane.
for 2) Quantum mechanics gives probabilities for limits. When the probability gets very very small, that is a limit.To get enough energy for an x-ray the material would have to synchronize so that random vibrations would quantum mechanically allow for a quantum level to exist to have enough energy for an x-ray photon to come out.How improbable that is is reflected in the Planck formula.
Also a limit is given by energy conservation laws , also included in the formula.
The basic line is that it is very improbable, as the other answers state.
| {
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How to properly calibrate phase axis to represent an optical interferometric stability curve? Assume we want to measure how stable an interferometer is.
We send two ultrashort pulses (left figure), delayed by the fixed time $\Delta t=T$ relative to each other, into a spectrometer to observe an interference pattern (middle figure). The distance between the chosen two peaks in this interferometric pattern is $2\pi/T$. Now we want to measure how stable this pattern is over a long-term period (many hours), i.e. how much it moves around (left or right) in the frequency domain over time. In other words, we want to obtain a graph similar to what is schematically shown in the right figure. How should this properly done?
I assume that one should choose a peak in the spectrum (e.g. red dot in the middle figure), and plot its shifts over time. I also reckon though that the shifts will have different sensitivity depending on the time delay between pulses (due to the reciprocal dependence indicted in the middle figure). Then, how should it be properly taken into account? Any other nuances that should be taken into account?
| This is really an engineering problem, not a physics problem. Nevertheless there are probably half a dozen if not more ways to skin the cat.
The way I would do it is capture the initial spectrum as a baseline. Then at regular intervals, seconds, minutes hours (you should decide based on the expected spectrum of your interferometer disturbance source) capture another spectrum and calculate the cross correlation. This would provide you with an overall measure of spectral stability relative to your baseline.
| {
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Stability of plum pudding model My textbook says that the plum pudding model(Thomson's model) should be electrostatically unstable. Why is that so?
| No system can be electrostatically stable. As Wikipedia explains,
Earnshaw’s theorem states that a collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges.
It can be understood intuitively as a consequence that electrostatic fields are divergenceless.
For a particle to be in a stable equilibrium, small perturbations ("pushes") on the particle in any direction should not break the equilibrium; the particle should "fall back" to its previous position. This means that the force field lines around the particle's equilibrium position should all point inwards, towards that position. If all of the surrounding field lines point towards the equilibrium point, then the divergence of the field at that point must be negative (i.e. that point acts as a sink). However, Gauss's law says that the divergence of any possible electric force field is zero in free space.
| {
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What determines how transparent a material is? Every time light hits a material, both reflection and refraction occurs. How does a material determine how much light gets reflected/ refracted e.g. glass vs silver? So far what I could find is light gets scattered by atoms in the material in all directions, but they all cancel out except in these 2 directions. But that does not explain the different degrees of reflection and refraction.
| Classical light is the super position of zillions of photons with the energy of $hν$, mathematically this means their wave functions are added to create light and the images carried by its variations.
Lets take glass:
In order for the material to be transparent, it means that the image information carried by the superposition goes through carrying the same phases and energy photons, so that the colors and shapes pass through unchanged. This means that the photons should scatter elastically with the whole solid state crystal lattice that composes the glass. i.e. individual wave function solutions are photon +lattice elastic scattering.
There are variations on this, from color material, which absorbs some frequency photons and leaves others to scatter thus changing the balance, to opaque which do not carry images through but light goes through in a combination of absorption and reemission scaters. Completely opaque materials absorb all the photons, or reflect them, the energy turning into lattice vibrations and in the end macroscopically heat.
The same goes for reflection, for glass the elastic scattering in the back direction generates reflection. For a silver backed mirror the reflection is elastic scattering of photons at the first atomic level of the silver lattice, but the forward part is absorbed into lattice vibrations.
| {
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Why in quantum hall effect longitudinal resistance and conductivity can be simultaneously zero? Why in quantum hall effect longitudinal resistance and conductivity can be simultaneously zero? I am puzzled about it. What's the physics meaning of it?
| The resistivity and conductivity are tensors
\begin{align}
\rho &= \begin{pmatrix}\rho_{xx} & \rho_{xy}\\\rho_{yx} & \rho_{yy}\end{pmatrix}\\
\sigma &= \begin{pmatrix}\sigma_{xx} & \sigma_{xy}\\\sigma_{yx} & \sigma_{yy}\end{pmatrix}
\end{align}
linearly relating the electric field and current density:
\begin{align}
E_a &= \rho_{ab}J_b,\\
J_a &= \sigma_{ab}E_b.
\end{align}
As matrices, $\rho$ and $\sigma$ are inverses,
\begin{align}
\sigma = \rho^{-1}
\end{align}
so that
\begin{align}
\sigma = \frac{1}{\det \rho}\begin{pmatrix}\rho_{yy} & -\rho_{xy}\\-\rho_{yx} & \rho_{xx}\end{pmatrix}.
\end{align}
If the longitudinal components $\rho_{xx}$, $\rho_{yy}$ of the resistivity are zero, and if the transverse $\rho_{xy}$ and $\rho_{yx}$ are non-zero so that $\det \rho \neq 0$, then the longitudinal components of $\sigma$ are also zero.
Since the longitudinal conductivity is zero, an electric field applied along the $x$ (or $y$) direction does not lead to a current in this direction. Instead, a transverse current is produced. Since the longitudinal resistivity is zero, if we pass a current along the $x$ (or $y$) direction, we don't observe a potential difference along this direction. Instead, we would measure a transverse (Hall) voltage.
Quoting Steve Girvin:
Notice that, paradoxically, the system looks insulating since $\sigma_{xx} = 0$ and yet it
looks like a perfect conductor since $\rho_{xx} = 0$.
| {
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Clarification of the concept "less resistance means less heating" in a wire So my textbook says that the reason cables that are suppose to carry high currents, are thicker that those that are meant to carry lesser current, is that "less resistance (of the wire) means less heating..."? Is this even true?
Isn't CURRENT the reason wires heat up? If we decrease resistance, more current flows, and that should produce more heating!
| The rate of heating of a wire is the power dissipated in the wire which is
$$P=I^{2}R$$
The resistance of wiring is much less than the resistance of the loads that the wiring supplies current to. Therefore the size of the wire has little effect on the current supplied to the loads (within reason). In other words, we can consider the current to be constant for different wire sizes within a limited range of sizes.
For a fixed current load ($I$ = constant), the greater the wire (conductor) resistance the greater the heating of the wire, the less the resistance the less the heating of the wire. So a larger conductor for a fixed load current produces less heat because its resistance is lower.
Finally, overheating of undersized conductors is a concern because it can cause the failure of the insulation on the conductors due to melting or long term thermal degradation if temperatures exceed the insulation temperature rating. Since insulation may be relied upon to reduce the risk of fire and electric shock, overheating of the insulation increases the risk of fire and electric shock.
Hope this helps.
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Do black holes move through space? I know it was already asked here:
Does a black hole move through space? What happens to other things around it?
And it might be a very stupid question, but here it is:
From a relativistic perspective, do black holes move through space, or is it the space around them that is curved in such a way that for us they seem to move?
I know there is no absolute frame of reference in relativity, but let's say the standpoint of one blackhole, I would think time is frozen, so without time how can things move?
| Yes. Here is a geometric perspective.
Take e.g. the Schwarzschild metric in coordinates $x^\mu=(t,r,\phi,\theta)$:
$$
g_{\mu\nu}dx^\mu dx^\nu=-(1-r_S/r) dt^2 + (1-r_S/r)^{-1}dr^2 + r^2 (d\theta^2+ \sin^2\theta d\phi^2) \qquad \text{(Schwarzschild)}
$$
where $r_S$ is the Schwarzschild radius. The geometry as $r\to +\infty$ will look like Minkowski space in spherical coordinates (for the spacelike part):
$$
g_{\mu\nu}dx^\mu dx^\nu=- dt^2 +dr^2 + r^2 (d\theta^2+ \sin^2\theta d\phi^2) \qquad \text{(Minkowski)}\,.
$$
Consider changing to Cartesian coordinates $(r,\phi,\theta)\to(x,y,z)$, doing a boost (say along the $x$ direction), and changing back to spherical coordinates. The Minkowski geometry will look exactly the same.
However, doing the same for the Schwarzschild geometry will give you a different geometry! (Which I will not write down...) The new geometry corresponds to a boosted Schwarzschild black hole, one which moves at constant velocity relative to the distant observer at $r\to \infty$.
In fact, the same argument tells you that any asymptotically flat black hole can move relative to an observer far away from the black hole. (Of course they don't all have to move at constant speeds; it's just that one can construct constantly moving black hole geometries from the immobile ones by the above argument without actually calculating anything.)
| {
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Dumping hot liquid water in sink When I make pasta or potatoes, I dispose of the hot water (around 80-100°C) in a metal sink.
Am I justified to first pour regular hot tap water on the sink to reduce the amount of steam generated, or is this pure superstition ?
If it's not, what is the physical explanation ?
| Steam is water in the gaseous phase and is invisible. It occurs when water boils which, at one atmosphere, is a temperature of 100 C.
What you are seeing coming up from your sink is not steam, but water vapor that is condensing into liquid water droplets. You can see the same thing above a hot, but not boiling, cup of coffee. It is an accelerated form of evaporation of water at the surface of the liquid due to its high temperature. Evaporation occurs at the surface of the liquid even for liquids at room temperature, though the rate is slow. It is due to the escape of the higher kinetic energy water molecules that exist at the surface. The higher the temperature of the water, the greater the kinetic energy of the molecules at the surface and the greater the rate of evaporation. Then when the water vapor mixes with the cooler air above the surface it condenses into water droplets.
Hope this helps.
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How should I treat background radiation in an experiment? I am performing an experiment to verify the inverse-square law for the intensity of a beam of $\gamma$ rays emitted by a sample of $^{60}\text{Co}$. The set-up is as follows:
*
*A Geiger-Müller tube connected to an electronic counter is set up.
*A background reading is first taken, by performing 10 measurements of $100\text{ s}$ each with no source nearby.
*The average count is taken: in my experiment, I have a value of $c_{BG} = 28.5$, with a standard deviation, $\sigma_{BG}=3.837$, represented as $28.5 \pm 3.837$.
*Now, the radioactive source is placed at a fixed position, and the GM tube is placed at a distance $d\text{ cm}$ away from the source.
*10 readings lasting $100\text{ s}$ each are taken, and the average count for 100 seconds, $\langle c_d \rangle$ is calculated at this distance $d$.
*Steps 4 and 5 are repeated for other values of $d$.
Now, for each distance $d_i$, I have an average count over 100 seconds, $\langle c_i\rangle$, and an associated standard deviation of the 10 readings, $\sigma_i$.
I then have a set of $\left\{c_i \pm \sigma_i\right\}$.
My question is therefore this: can I simply subtract $c_{BG} \pm \sigma_{BG} $ from each $c_i \pm \sigma_i$, and then propagate the errors to find the average count rate, $\left\langle N\left(c_{{d}_{i}}\right)\right\rangle$ at each $d_i$? If not, how else should I go about dealing with the background radiation?
My end goal is to plot the logarithm of the average count rate, $\ln(\left\langle N\left(c_{{d}_{i}}\right)\right\rangle)$, on $\ln d$ and determine the gradient—ideally $\thicksim -2$.
| If you think of the problem in terms of random variables, it becomes easy. You have two random variables: The first one $X_1$ for the background noise, and the second one $X_2$ for the radioactive decay. Both follow a Poisson distribution.
\begin{align}
X_1 &\sim Pois(\lambda_1) \\
X_2 &\sim Pois(\lambda_2)
\end{align}
For the Poisson distribution we know that the decay rate is equal to the expectation value as well as to the variance, $\lambda = E[X_1] = Var[X_1]$. Thus, the expectation value is a "good" estimator for the rate parameter. Hence, we use $\lambda_1 \approx \bar{N}_{BG}$.
Now, we can show that the sum of two independent Poissonian random variables is again Poissonian,
$$
X_1 + X_2 \sim Pois(\lambda_1 + \lambda_2)
$$
(Note that here we assume that the decay product is not radioactive)
Thus, the estimate of $\lambda_2$ is given
$$
E[X_1 + X_2] = E[X_1] + E[X_2] \Rightarrow \lambda_2 = E[X_2] = \ldots$$
For the standard deviation of $\lambda_2$ we use $Var[X_1 + X_2] = Var[X_1] + Var[X_2]$. Solving for $Var[X_2]$ and taking the square root, we obtain
$$
\sigma_2 = \sqrt{Var[X_1 + X_2] - Var[X_1]}
= \sqrt{E[X_1 + X_2] - E[X_1]}
$$
| {
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Do eigenfunctions of the position and momentum operators vary from one problem to another? Now the eigenfunctions of the Hamiltonian clearly differ from one problem to another since the Hamiltonian depends on the potential and hence for a different potential we get a different eigenvalue equation for the Hamiltonian hence the eigenfunctions are different each time.
However, the eigenvalue equation for the position and momentum operators don't change (since they don't depends on the potential), so is it always the case that the eigenfunctions of momentum and position operators (which are continuous) are always the same? My gut tells me that no since the eigenfunctions must also satisfy boundary conditions which differ from one problem to another, but Griffiths book only solved this one case for the eigenfunctions of position and momentum operators and later used them as a standard eigenbasis of position/momentum for all types of problems. Even in the end of chapter problems he never asks us to solve the eigenvalue equation for the position and momentum operators for, say, the infinite well. Which left me with the aformentioned question: Are they always the same?
| Yes they do vary. Say you're doing quantum mechanics on a circle. $\hat p\equiv -i\hbar \partial_x$ where $x\sim x+2\pi R$ is the coordinate along the circle, will then have discrete eigenvalues $k$ and complex exponential eigenfunctions: we have
$$
\hat p e^{ikx/\hbar}=k e^{ikx/\hbar}\,,
$$
but $e^{ikx/\hbar}$ will be a function on the circle iff it is periodic. Therefore
$$
k\frac{R}{\hbar}\in\mathbb Z\,.
$$
In contrast if you are considering quantum mechanics on say $\mathbb R^3$ then $\hat p$ will have continuous spectrum, so its "eigenvalues" $k\in\mathbb R^{3}$.
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The Enigma of Universal Gravitation Forces This is taken from a book called "Physical Paradoxes and Sophisms" by V. N. Lange.
1.22. The Enigma of Universal Gravitation Forces
The law of gravitation can be written $F=\gamma\frac{m_1m_2}{R^2}$.
By analyzing this relationship we can easily arrive at some interesting conclusions: as the distance between the bodies tends to zero the force of their mutual attraction must rise without limit to infinity.Why then can we lift up, without much effort, one body from the surface of another body (e.g., a stone from the Earth) or stand up after sitting on a chair?
| Consider the gravitational acceleration $g$ at a surface point $p$ of a massive compact body $M$ whose density $\rho$ is assumed to be bounded from above. Then $g$ is finite rather than infinite.
*
*If $M$ is a spherically symmetric ball, this is true due to Newton's shell theorem. The effective distance $r>0$ in Newton's formula then becomes the radius of $M$.
*Even without spherically symmetric this is true. At the surface point $p$, one can sum up hemispheres with center $p$ and radius $r$. The $1/r^2$-singularity in Newton's formula gets cancelled by a $r^2$-area-dependence of the hemisphere with thickness $dr$, so that $g$ is not infinite but grows (at most) linearly with the size of $M$. A similar estimate can be found from Gauss' law for Newtonian gravity.
If you like OP's question, you may also enjoy reading this Phys.SE post.
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What does Heisenberg's uncertainty principle tell about nature? I agree with the fact that the principle points out to the inaccuracy in the measurement of the two quantities of the particles (momentum and position).
But measurements apart, does it explain anything about how nature works, in general? As in, I think the particle would have some exact value of momentum at that point in space (if not, please explain why).
So why not just tell that 'okay it does possess some momentum at that position, but I can't tell what that exact value is'?
Edit: I understood that the principle points out at nature as a whole, in general, and does not just point out at measurements
| The HUP is what tells us that the world is fundamentally QM in nature. All the elementary particles that you see defined in the SM are behaving in a way that will obey this simple rule, and it is not just about momentum and position (though that is a pair of observables or physical quantities that is commonly used in examples), but is true for any pair of observables.
In quantum mechanics, the uncertainty principle (also known as Heisenberg's uncertainty principle) is any of a variety of mathematical inequalities[1] asserting a fundamental limit to the precision with which the values for certain pairs of physical quantities of a particle, known as complementary variables or canonically conjugate variables such as position x and momentum p, can be predicted from initial conditions, or, depending on interpretation, to what extent such conjugate properties maintain their approximate meaning, as the mathematical framework of quantum physics does not support the notion of simultaneously well-defined conjugate properties expressed by a single value. The uncertainty principle implies that it is in general not possible to predict the value of a quantity with arbitrary certainty, even if all initial conditions are specified.
https://en.wikipedia.org/wiki/Uncertainty_principle
The HUP is important because it tells us that the error is not in our measuring devices, but that nature is weird (when looking at it with a classical view).
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Time period of piston performing SHM Suppose I have a closed piston-cylinder system which contains an ideal gas,and it is being compressed adiabatically by the piston which has a vertical orientation,and in doing so,the piston is actually undergoing SHM. Now if I use a similar piston in a horizontal orientation (keeping in mind the dimensions of the cylinder), would the time period of the SHM be affected? Would anything change in this orientation? The equation $pV^\gamma=\text{constant}$ would still be valid, but would anything change at all?
| The equilibrium position of the piston would change due to the weight of piston. However it would not interfere with time period or frequency as it is acting throughout the process. In the second position equilibrium position would be different. For more accuracy you could also take into account the change in external pressure in the two cases . Yet it will also have no effect on time period but only on equilibrium position.
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In Rotational Dynamics do we have to consider the torque of pseudo force in Instantaneous Axis of Rotation frame? In the attached image have I done everything right? If yes then why there is no torque of Pseudo Force?
| There will be a pseudo force acting on the COM of that body. How ever this will not provide any torque because, the instantaneous acceleration of the Point of contact will be $\frac{v^2}{R}$ acting upwards. So when the pseudo force is applied on the centre of mass, it will be pointed downwards passing through the IAOR. Hence no torque will be generated.
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Raising and Lowering Indices of Basis Tensors? Can you raise and lower the indices of basis vectors and dual basis vectors? I know I can write
$$V^\mu = g^{\mu\nu} V_\nu $$
but can I also write
$$\vec{V} = V^\alpha \hat{e}_\alpha = V_\beta \hat{e}^\beta?$$
| The equation $$\vec{V} = V^\alpha \hat{e}_\alpha = V_\beta \hat{e}^\beta$$ does not make sense, you are equating two different objects, a vector and a covector. If you take a finite dimensional vector space, $V$, the dual space $V^*$ is the set of all linear functions that map elements of $V$ to the field over which it is defined.
The metric tensor is a special object that, when defined, provides a bijective map between the elements of $V$ and $V^*$. That is, each element in $V$ has a unique covector in $V^*$ and the metric is a map that relates them, this is why you are allowed to mnemonically "raise" and "lower" indices with the metric (but not with any old rank-2 tensor).
If you pick a basis in $V$ you can also define a special basis in $V$* such that
$$\hat e^i(\tilde e_j)=\delta^i_j$$
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If we ignore energy and talk about force, Why do objects bounce lower when they hit a softer surface? If an object hits a soft surface it will bounce lower compared to the object hitting a hard surface, isn't the impulse in the first case equal to the impulse in the second case, so why does the object bounce lower when it hits a softer surface?
| No, the final speed is lower with a softer surface, so the impulse is lower.
If you throw a ball at a hard surface, the ball will not stop instantly. It will compress as it slows down. The energy of that compression will then be released, pushing the ball back in the direction opposite of the original motion. In the theoretical case of perfect elasticity, all of the energy will be retained by the ball, and the final speed will be the same as the original.
When a ball hits a soft surface, some of the energy is absorbed by the surface, rather than the ball, deforming. If the surface is elastic, then that energy will be restored to the ball. But if the surface is inelastic, that energy will be dissipated into heat. The ball will leave with less energy, and thus less speed.
| {
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Is there an analogy for Wilson loops/lines in statistical mechanics? When reformulated in Euclidean space, quantum field theory bears some strong resemblance to statistical mechanics: for example a scalar field $\phi$ can be seen as a spin $s$ in Landau theory, and the source $J$ in path integral formalism can be seen as an external magnetic field. What about Wilson loops/lines?
In Peskin & Schroeder, the Wilson line is defined as:
$$U(z,y) := \exp \left[ - ie \int_P dx^\mu A_\mu (x) \right] \tag{1}$$
with the integral taken along any path running from $y$ to $z$.
| Gauge theories often appear in condensed matter. For introduction I recommend read brilliant introduction article by professor Wen Topological order: from long-range entangled quantum matter to an unification of light and electrons.
In gauge theory, observables are gauge-invariant operators (because gauge symmetry is not symmetry! It is redundancy in our description.). In gauge theory without matter such objects are Wilson and t'Hooft loops. So in any theory with gauge symmetry this observables are natural and very important.
For concrete examples consult with Field Theories of Condensed Matter Physics. See 8.4 about disordered spin states, 9.6 the Ising gauge theory, 9.9 Boundary conditions and topology,....
Ising gauge theory
Here I follow Z 2 gauge theory by Subir Sachdev.
System is defined through Hamiltonian:
There are two gapped phases in the theory, which are necessarily separated by a phase transition. Remarkably, unlike all previously known cases, this phase transition was not required by the presence of a broken symmetry in one of the phases: there was no local order parameter characterizing the phase transition. Instead, Wegner argued for the presence of a phase transition using the behavior of the Wegner-Wilson loop operator $W_C$, which is the product of $\sigma_z$ on the links of any closed contour C on the direct square lattice:
In this example, Wilson loop is nonlocal order parameter for phase transition.
| {
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Minimal coupling to electric dipole form When deriving the electric dipole form of the semiclassical hamiltonian from the minimal coupling form, we define a new state vector as:
$$\psi_{old}=e^{i\textbf{A}\cdot\textbf{x}/\hbar}\psi_{new}=\textit{u}(\textbf{x},t)\psi_{new}$$
we now put this into the SE:
$$i\hbar\frac{\partial}{\partial t}\psi_{old}=\left[\frac{\textbf{p}^2}{2m}-\frac{e}{m}\textbf{A}\cdot\textbf{p}+\frac{e^2}{2m}\textbf{A}^2\right]\psi_{old} $$
and obtain the following:
$$i\hbar \textit{u}(\textbf{x},t)\left[\frac{ie}{\hbar}\dot{\textbf{A}}\cdot\textbf{x}+\frac{\partial}{\partial t}\right]\psi_{new}=\left[\frac{\textbf{p}^2}{2m}-\frac{e}{m}\textbf{A}\cdot\textbf{p}+\frac{e^2}{2m}\textbf{A}^2\right]\textit{u}(\textbf{x},t)\psi_{new}$$
My question is relatied to the left side of the previous equation. Namely, after inserting the expression for the $\psi_{old}$ into the second equation, why wasn't $\frac{\partial}{\partial t}$ applied to the coordinate operator $\textbf{x}$, to give an additional term - $\frac{ie}{\hbar}\textbf{A}\cdot\dot{\textbf{x}}$?
So, I guess that the gist of the question is what does the time derivative operator $\frac{\partial}{\partial t}$ do when applied to the coordinate operator $\textbf{x}$?
| Here you work in the Schrödinger representation, where the time dependence is encoded in the wave function. (This is opposed to the Heisenberg representation, where the time evolution is shifted to the operators.) Thus the derivative of any operator in respect to time is zero, unless it contains explicit time dependence, as the vector-potential in this case.
| {
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What is a simple argument to prove that the stars in the sky are further away from the Earth than the Moon? How do we know, without using modern equipment, that the stars are further away than the moon in the night sky? Further, is there a simple and actionable argument to prove that this is indeed the case? Additionally, I would like to know how people in earlier times knew this fact without having access to modern equipment, including telescopes.
| If you take it as a given that the Earth is revolving around its axis once per day than you could make the argument that anything that does not move (after subtracting the apparent movement due to the Earth's rotation) must be far away, or it would fall down.
Satellites have a short orbital period, the Moon a longer one, everything else must be "very very far away". Of course things are complicated by Earth's orbit around the Sun; if that's not a given then one is stuck with Ptolemy's curly trajectories for the planets, but the general idea still holds.
| {
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Confusion about cells in parallel I cant understand why cells in parallel will last longer. Could someone please explain?
| A battery is like a store in which you shop for electrons. If two are in parallel you have more stuff you can buy simultaneously; that is you can have two guys shopping at the same time, then get on the road and go home. If they are connected in series first you have to visit one store and then you may go to the next after which you can go home, only you do the shopping, so it takes longer to empty it.
| {
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If work is a scalar measurement, why do we sometimes represent it as the product of force (a vector) and distance (scalar)? Consider an object being pushed 3/4 of the distance around a circular track. The work done on the object would be the distance of 3/4 the track’s circumference times the force applied to the object (given that it was pushed at a constant force). Since we are multiplying a vector by a scalar, why is work a scalar measurement? Or would the work done on the object actually just be force times displacement? Thanks.
| The reason for this is that your understanding of the definition of work contains an error:
"product of ... distance (scalar)"
Work is not defined using a product of force and a "distance", but a displacement. A displacement is the difference of two positions, which are points, and displacements are vectors. So it is actually the dot product of two vectors, which, in turn, is a scalar.
That said, I think what you may be asking about here is that there is sometimes seen a simple scalar-only formula that looks like
$$W = Fd$$
where we involve only the (scalar) magnitudes of both force and displacement. This formula only works in one dimension(*), or else that the force and displacement are acting along the same line. Otherwise, it looks like
$$W = Fd \cos \theta$$
where $\theta$ is now the angle between their lines of action, and this now is exactly the dot product of vectors.
There is no "vector times scalar" formula for work. Such a thing would, indeed, as you suggest, be a vector, and work is not a vector.
(*) Technically, in one dimension it is also a dot product of two vectors, but there is little functional difference between a vector of dimension 1 and a scalar, mathematically. That said, I still think it may be useful to keep this distinction in mind as a matter of conceptual clarity. A vector of dimension 1 has a solitary component:
$$\mathbf{v} = \langle v_x \rangle$$
but it belongs to a different "data type", so to speak, than scalars (real numbers) do and this has some important algebraic consequences such as that you cannot "sensibly" add a one-dimensional vectorial quantity and a scalar real number.
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How does amplitude affect photon's intensity as a particle? Considering the scenario were a photon acts as a particle, how does amplitude affect the photon?
Does it increase its intensity? How do you visualise this?
| For a higher amplitude, just visualize more photons per unit volume (i.e., a higher number density and a higher energy density). But each individual photon has an energy that depends only on the frequency, not on the amplitude. Having lots of low-energy photons around doesn’t tend to ionize atoms if none of them have enough energy to eject an electron.
One exception to this general picture would be multi-photon ionization, but this is suppressed by factors of the fine-structure constant so it is comparatively rare.
| {
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What does "up to a total derivative" really mean and how should I know when to use it? I am a mathematician who is taking a quantum field theory course without much prior pyhsics. We have had the term "up to a total derivative" a few times, yet every time I asked what it meant I didn't really grasp it.
As an example, for our last tutorial we were given the Lagrangian
$$ \mathcal{L} = i\psi^*\partial_0\psi - \frac{1}{2m}\nabla\psi^*\cdot\nabla\psi,\tag{1}
$$
but then immediately in the tutorial it was given that this is equivalent (up to a total derivative) to
$$ \mathcal{L} = \frac{i}{2}(\psi^*\partial_0\psi - (\partial_0\psi^*)\psi) - \frac{1}{2m}\nabla\psi^*\cdot\nabla\psi.\tag{2}
$$
The things I really don't understand are:
*
*how exactly are these things the same? (/what does "up to total derivative" mean)
*how do I know when I should try to convert something to another thing through a total derivative?
| *
*The Euler-Lagrange (EL) equations are not affected by total derivative terms, cf. e.g. this Phys.SE post.
*In OP's concrete example the Lagrangian density (2) is preferred as it is manifestly real. See also this related Phys.SE post.
| {
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Conservation of energy of 2 identical Rolling Disks with and without friction My physics book claims that if two identical disks moving at the same velocity travel up nearly identical hills, with the second hill not having friction, then the disk rolling up the first hill will travel to a greater height. Given that the disks started with the same kinetic energy while rolling at the base of the hill, shouldn't they reach the same height (i.e. same potential energy) as a result?
| Your second disk cannot convert its rotational energy to gravitational potential energy.
If I understand you correctly, both disks start out with some forward motion, and some rotation.
Disk 1 will slow down as it travels uphill, converting both kinetic energy from its forward motion and kinetic energy from its rotation into gravitational potential energy (height). When it reaches the top, it will have stopped moving, which means also no rotation. Its total energy remains constant in the whole situation, except this energy is now in the form of gravitational potential energy.
Disk 2 will slow down as it travels uphill, but only its forward motion will slow down. As it moves without friction, its rotation is unaffected. It can't convert its rotational energy (a form of kinetic energy) into gravitational potential energy/height.
When it reaches its top, it will have stopped moving forward, but will still be merrily rotating. At that point, its total energy is still equal of that of disk 1, only its total energy is distributed, between gravitational potential energy, and (kinetic) rotation-energy.
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Confusion regarding properties of Poisson Brackets I have just started learning about Poisson Brackets, and came across the following property
$$\{q_i,q_j\}=0$$
And
$$\{p_i,p_j\}=0.$$
Where $p$ and $q$ are respectively the momentum and position coordinates i.e. phase space coordinates.
Now Poisson Brackets are defined as
$$\{F,G\}=\frac{\partial F}{\partial q_i}\frac{\partial G}{\partial p_i}-\frac{\partial G}{\partial q_i}\frac{\partial F}{\partial p_i}$$
$i$ and $j$ here stand for the $i$'th and $j$'th spatial coordinates.
$$\{q_i,q_j\}=0$$
$$\Rightarrow \{q_i,q_j\}=\frac{\partial q_i}{\partial q_i}\frac{\partial q_j}{\partial p_i}-\frac{\partial q_j}{\partial q_i}\frac{\partial q_i}{\partial p_i} =0$$
But I am having a hard time proving it.
I know that the second term $(\frac{\partial q_j}{\partial q_i})$ is zero because the i'th and j'th spatial coordinates are orthogonal and hence, there is no change in $q_i$ on changing $q_j$. However I don't know how to prove the first term to be zero, and that is where I need help.
To summarise, my question is prove that
$$\frac{\partial q_i}{\partial q_i}\frac{\partial q_j}{\partial p_i}=0$$
Any help will be deeply appreciated.
| As $p$ and $q$ do not depend functionally on one another
$$ \frac{\partial q_i}{\partial p_j} = 0$$ and also $$ \frac{\partial p_i}{\partial q_j} = 0$$ for all $i,j$
| {
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Why can vector components not be resolved by Laws of Vector Addition? A vector at any angle can be thought of as resultant of two vector components (namely sin and cos).
But a vector can also be thought of resultant or sum of two vectors following Triangle Law of Addition or Parallelogram Law of Addition, as a vector in reality could be the sum of two vectors which are NOT 90°.The only difference here will be that it is not necessary that components will be at right angle.
In other words why do we take components as perpendicular to each other and not any other angle (using Triangle Law and Parallelogram Law).
| Indeed, any vector can be resolved in terms of two components (in $n$-dimensional space in terms of $n$ components). For this being possible the components should be linearly independent, i.e. in your case they should not be parallel. The advantage of using two orthogonal/perpendicular components is that their scalar product is zero, which simplifies the math when calculating the coefficients:
\begin{array}
\mathbf{F} = F_x\mathbf{e}_x + F_y\mathbf{e}_y \Longrightarrow F_x = \mathbf{e}_x\cdot \mathbf{F}, F_y = \mathbf{e}_y\cdot \mathbf{F}.
\end{array}
Indeed,
\begin{equation}
\mathbf{e}_x\cdot \mathbf{F} = \mathbf{e}_x (F_x\mathbf{e}_x + F_y\mathbf{e}_y) = F_x\mathbf{e}_x \cdot\mathbf{e}_x + F_y\mathbf{e}_x \cdot\mathbf{e}_y = F_x,
\end{equation}
and similarly for $F_y$, since
\begin{equation}
\mathbf{e}_x \cdot\mathbf{e}_x = 1, \mathbf{e}_y \cdot\mathbf{e}_y = 1, \mathbf{e}_x \cdot\mathbf{e}_y = 0.
\end{equation}
For non-orthogal vectors $\mathbf{e}_x \cdot\mathbf{e}_y \neq 0$, the math becomes a bit more complicated and the interpretation of the projections as coordinates is less intuitive.
There are however some cases where using non-orthogonal components is beneficial, notably when dealing with non-orthogonal crystal lattices, such that of graphene (a hexagonal lattice.)
| {
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Difference between projection and component of a vector in product? This is a very basic question about dot products or scalar products:-
If I want to move a block and I apply a force parallel to displacement, the block will move and some work will be done. So in the formula will be $W= F\cdot S$, here we won't calculate the force mg of the block but the force we applied (the parallel force).
Now let's say that the force is not parallel and is at some angle from the horizontal
So in this case the work done will be the projection of the force $F_1$ on the $x$ axis, because that is how dot products are defined (As Projections).
But can't we say that the block moved due to its horizontal component $F_1\cos(\theta)$ and the answer would be same. And obviously we won't count $F_1\sin(\theta)$ as the work is not done by it.
So why do we say projection and not component in dot or vector product ?
| Projection and component mean the same thing in this context.
Let me however note that a force is a part of the physical reality, existing in the world (at least for the purposes of mechanics), whereas a projection is a mathematical object/construct. From this point of view the work is done by the force (or by its component), but not by the projection.
| {
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Experiment for jammers I'm going to grade 11 and I have to write an essay in physics. One requirement is to do an experiment which - in the best case - also includes measurements.
I would like to write about jammers, which cut connections by interference signals. Since it is not legal, I will not build or use a real jammer.
Instead I am searching for an experiment that demonstrates how jammers work.
Can anyone help me with finding a good but practical experiment? ;)
Best regards
Julius
| Do you have access to a laboratory with a water wave channel?
If yes, you could reproduce this soliton experiment . Solitons are wavepackets, and one could use them sequentially to carry information, a Morse code for example, dependent on the distance between them.
Jamming could be a water sprout , disturbing the flow of information.
If you do not have access to a lab, you could reproduce an effect of sending a coded message by water disturbance by using a trough, and creating the wave packet by letting marbles fall at the beginning of the water line, with time distance between falls creating the Morse code..
| {
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What is the meaning of a "permutation -invariant " MHV amplitude? When reading my notes, I read that the difference between 4 points colour-ordered MHV amplitude and gravitational MHV amplitude is that the gravitational MHV amplitude is permutation-invariant, unlike the latter.
I don't understand what this means. Is it that when calculating the MHV amplitude I am not meant to take the permutations into consideration?
Does this mean that, using a 3-point diagram as an example to measure the MHV amplitude we would start by only considering:
$$A= \sqrt2 \epsilon_1 \epsilon_2 \epsilon_3 p_1 $$
instead of the "correct" $A$?:
$$A= \sqrt 2 \left( \epsilon_1 \epsilon_2 \epsilon_3 p_1 +\epsilon_2 \epsilon_3 \epsilon_1 p_2+\epsilon_3 \epsilon_1 \epsilon_2 p_3\right)$$
| By little group scaling, the three-particle MHV amplitude for three massless spin-1 particles is given by
$$ \mathcal{M}^{--+} =\frac{\langle 12 \rangle^3}{\langle 13 \rangle\langle 32 \rangle}~.$$
This is not permutation invariant in the sense that if I exchange legs one and two I pick up a minus sign. The equivalent graviton amplitude doesn't pick up a minus sign, since it is squared, and hence is permutation invariant.
An interesting side note, the amplitude above violates bose symmetry, which can only be cured if the coupling constant also picks up a minus sign under the exchange: the amplitude above must be for colour charged spin-1 particles (gluons) and must vanish for photons.
| {
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What is *physical meaning*? What do we mean when we talk about physical meaning of a quantity, an equation, theory, etc.? Should the physical meaning touch on the relation between the math and the real world? Or does it have more to do with how the equation/theory is used by physicists?
Background
For the immediate background that prompted me to ask this question see the discussion that followed answers to this question.
This forum contains nearly 3000 questions of the type What is the physical meaning of X... but do we know what we are asking?
Opinion
I think the question is important, because it defines the special place of physics among other disciplines. When we ask about a physical meaning of something we really ask how this something is related to the real world, as opposed to purely mathematical reasoning. Mathematicians and biologists do not question mathematical or biological meaning of their objects of study, since it is obvious. Yet, physicists must justify their calculations by basing them on the experimental data and making experimental predictions (as opposed to mathematicians). In the same time physicists cannot do experiments without developing complex mathematical models (unlike biologists or chemists - even though these are often more knowledgeable about complex statistical methods than an average physicist.)
| Let me first ask you a question; what do you think I mean by
$$\mathbf{F} = m \mathbf a$$
?
From a mathematical point of view, the equation expresses the relationship between two vectors.
However, a physicist, when using mathematics to understand nature, makes mapping between physical concepts and mathematical objects. For example for the above case, there is a measurable quantity & a physical concept called force and we are denoting it by a mathematical object, namely a vector, so the mapping is
$$\text{Force (measurable quantity)} \to \vec{F} (\text{mathematical object}).$$
Now, coming back to your question, what do they mean by "physical meaning of a mathematical expression" is the inverse mapping of the above relations, i.e
$$ \vec{F} (\text{mathematical object}) \to \text{Force (measurable quantity)}$$
| {
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Why the imaginary constant on the spin-connection? I have been studying the spin-connection for the Dirac equation. The covariant derivative is defined as:
$$D_\mu \psi = (\partial_\mu - {i \over 4} {\omega_\mu}^{ab} \sigma_{ab}) \psi$$
where $\sigma_{ab} = i[\gamma^a,\gamma^b]$ and ${\omega_\mu}^{ab}$ is the spin-connection.
When working out the consequence I find that this $i$ factor before the spin-connection leads to pseudo-vector terms which treat left and right differently. (Perhaps this is good as maybe left and right handed fermions move differently under gravity but this seems to go against the idea that all things moves the same under gravity)
So my very technical question is, is this $i$ really necessary. Or would the theory still be self consistent if we removed the $i$? What is the justification for this $i$? It would be a different theory but would it be self-consistent? Could we even add a factor of $\gamma^5$ before the spin-connection and keep it self-consistent? (Since this would be the same except having a factor of -1 for right-handed fermions?) Basically I'm thinking of any ways that this equation can be changed and remain consistent with general covariance.
| The $i$ is necessary.
In order to derive the Dirac equations in curved spacetime, the approach is the following: We know how Lorentz vectors can be transported in a generic curved spacetime using covariant derivatives. We use the fact that bilinears of $\psi$(a spinor) should transform like a vector. This then gives rise to the above formula.
You may wanna look at this paper for derivation.
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The electric field should be in circular coil. But why do current flows in whole circuit? We know changing magnetic flux induces electric field which makes current to flow .here in the below picture the flux is changing through only circular coil bout not through the rectangular part.so what induces electric field in rectangular part for current to complete the circuit ..Note that the electric field is required inside awire to make the current flow.
| By moving a magnet, voltage will be induced in the circular part of the coil (which is in the range of magnetic field). This voltage will drive the current in rest of the part of wire.
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Energy conservation in reflection of light from a perfect mirror I came across a question where a light source is shined on a mirror attached to a spring which is attached to a rigid support. The question goes:
A perfectly reflecting mirror of mass $M$ mounted on a spring constitutes a spring-mass system of angular frequency $\Omega$ such that $\frac{4\pi M\Omega}h=10^{24}\ \mathrm{m^{-2}}$ with $h$ as Planck's constant. $N$ photons of wavelength $\lambda=8\pi\times10^{-6}\ \mathrm m$ strike the mirror simultaneously at normal incidence such that the mirror gets displaced by $1\ \mathrm{\mu m}$. If the value of $N$ is $x\times10^{12}$, then the value of $x$ is $\_\_\_$. [Consider the spring as massless]
Now, in the solution the question has been solved by equating the change in momentum of the photons to the change in momentum of the mirror. However, what escapes me is how does the mirror move at all if all the photons are being reflected perfectly? If the mirror is indeed perfectly reflecting then the net energy incident must be equal to the net energy reflected. So, how can the mirror move if it does not take any energy from the light?
However if I do assume each photon gives up a bit of its energy, thus changing its wavelength, then the incoming momentum will not be equal to the outgoing momentum. But this would lead to a contradiction, as we assumed the mirror was perfectly reflecting.
I am puzzled. I think the only plausible answer to this is that 'there cannot be a perfectly reflecting mirror', but if that is the case, what would happen if we imagined one? In the same way that a perfectly black body does not exist but we can always imagine one.
| I'm not totally sure, but this sounds like a problem concerning the frame of reference. In the center of mass frame there should be no energy transfer for an elastic collision. But as the mass of the mirror is not given, the center of mass frame may not even be approximately the lab frame where the mirror is at rest initially.
Differently phrased: One shouldn't interpret the term "perfectly reflecting" as preserving the energy of the light, as this is dependent on the frame. It just means that there is no absorption going on.
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Is De Broglie's formula $\textbf{p}=\hbar \textbf{k}$ applicable to a discrete wave number system? I don't know if my question has sense at all but while doing my homework there appeared in my mind this question.
Say, for a particle in a box, the confinement makes that the wave number k is discrete, depending on integer numbers $\textbf{n}=(n_{x}, n_{y},n_{z})$.
If De Broglie's formula holds, it doesn't mean that momentum $\textbf{p}=\hbar \textbf{k}$ is also discrete?
| No. You can't use the De Broglie formula here because De Broglie's formula is true only for a free particle $(V(x)=0$ everywhere$)$.
For the infinite square well, we can write down the energy eigenvalue equation as:
$$\frac{-\hbar^2}{2m}\frac{\partial^{2}\psi(x)}{\partial x^2}+V(x)\psi(x)=E\psi(x)$$ such that $V(x)=0 $ inside the box and $V(x)\rightarrow\infty$ outside. Now, clearly we can see that the Energy operator $\hat{E}$ doesn't commute with $\hat p$ (although they do commute for a free particle potential.) If you try to use the definition of momentum operator $ \hat p_x=(\hbar/i)\partial_x $ on the stationary solutions of infinite well, you'll see that none of them is an eigenfunction of $\hat p$.
The probable reason for your confusion here is naming the constant $2mE/\hbar^2$ as $\textbf k$. Here, $\textbf k$ is not the wavenumber as we had for a free particle. Just name it as something else, and logically try to argue if it still makes sense to call it the same wavenumber. The definition $\textbf p =\hbar\textbf k$ doesn't extend here.
Also, if you are thinking about three dimensions, you can extend the argument very similarly to higher dimensions by symmetry.
| {
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Why are there no even harmonics in a closed pipe? I have seen a diagram on sites such as hyperphysics.com that show that there is a missing bit every time so that it makes every harmonic odd. I was hoping I could get a more intuitive explanation. We recently got introduced to harmonics and standing waves in class and I am a bit lost :)
| Musical notes are made by strings and by pipes. Standard discussion of harmonics and standing waves focusses on strings, but pipes are just a little more subtle.
For a wave on a string you plot the displacement and get sine waves and that's obvious as the plot looks like the excited string.
For a wave in a pipe the displacement is still described by sine waves but it is longitudinal instead of transverse so the graph is no longer a picture. You could also graph the pressure in the wave - it's basically the same as the density - and if you do that you get the equivalent cosines: in the peaks of displacement the nearby molecules have all moved by the same amount so the density is unchanged, whereas around the region of zero displacement the gradient means that some molecules are displaced further than others leading to variations in density. Displacement nodes are pressure antinodes, and vice versa.
The end of a string is simple. It's clamped. the displacement is zero.
The end of a pipe is ambiguous. If it is closed then the longitudinal displacement is zero: a displacement node. But if it is open then it must always be at atmospheric pressure: a pressure node and thus a displacement antinode.
So instruments involving pipes fall into two types: similar at both ends (like a flute, which is open-open) for which the harmonics are similar to a string, or closed at one end and open at the other (like a trumpet: the lips and mouthpiece form a closed end) which have a displacement node at one end and a pressure node (and thus displacement antinode) at the other. This requires $1 \over 4$ of a wavelengt to fit along the length, or $3 \over 4$, or $5 \over 4$... the odd harmonics.
| {
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How do we know that a non-rechargeable battery obeys the law of conservation of energy? For a rechargeable battery, I can show that energy is conserved by
*
*Discharging the battery.
*Measuring the energy required to charge the battery.
*Measuring the energy I get out from a second discharge.
For a non-rechargeable battery, how do I experimentally determine the potential energy before I discharge the battery?
I would prefer an answer that works for any battery chemistry.
| This will provide an upper bound:
*
*Measure the mass of the battery.
*Count all the atoms in the battery (good luck!).
*Compute the mass of all the atoms under the assumption of the atoms are non-interacting.
*Since E=mc^2, the difference between the actual mass and the computed mass is an upper bound on the potential energy that could be discharged from the battery.
If discharging the battery produces less energy than this upper bound, that is evidence the battery is consistent with the law of conservation of energy.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How can a red light photon be different from a blue light photon? How can photons have different energies if they have the same rest mass (zero) and same speed (speed of light)?
| Let me add a few things.
*
*A photon is an elementary particle, and as long as it propagates, it is in a superposition of states, meaning it is in a superposition of frequencies, and does not have a well defined frequency. You cannot know its frequency until you interact with it or absorb it.
As a quantum mechanical entity photons can be in superposition
Does a single white photon exist?
*A photon, as long as it propagates, could be viewed from different reference frames, and since there is no universal reference frame, the red wavelength photon could be viewed as blue from another reference frame. You cannot know its frequency until you interact with it or absorb it.
Why does the motion of the emitter (doppler shift) impact the energy of the photons
*Let's say you emit a blue wavelength photon, and it travels in expanding space, and undergoes cosmological redshift. The absorber will see it as a red wavelength photon. Who is right, would you call that a blue or a red wavelength photon?
https://en.wikipedia.org/wiki/Redshift
| {
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"timestamp": "2023-03-29T00:00:00",
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Moment of inertia tensor and symmetry of the object What information does the moment of inertia tensor give on the structure of an item.
I was told that its eigenvectors give the principal axes of the object.
Do you know more about this?
| Using the inertia tensor $I_{jk}$ of an object, you can construct an ellipsoid satisfying the equation $$1=x_{j}I_{jk}x_{k} .$$ As you said, the eigenvectors of $I_{jk}$ are the principal axis $\vec{r}_{j}$ of your object with the respective moment of inertia $\theta_{j}$ as eigenvalue. Thus, changing coordinates to the principal axis of the object diagonalises the inertia tensor and the equation for the ellipsoid is then $$1=\theta_{1}x_{1}^{2}+\theta_{2}x_{2}^{2}+\theta_{3}x_{3}^{2} .$$ The semi-axles of this ellipsoid are parallel to the principal axles of your object and their length is the inverse square root of the moment of inertia in that direction $a_{j}=\frac{1}{\sqrt{\theta_{j}}}$.
| {
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Why does frequency remain the same when waves travel from one medium to another? I was reading about reflection and refraction on BBC Bitesize and I can't understand why frequency is a constant in the wave speed equation. I can't visualise the idea of it. I know that wave speed and wavelength are proportional to each other but how can I tell the speed of a wave by looking at a random oscillation?
Here's where I got confused: https://www.bbc.co.uk/bitesize/guides/zw42ng8/revision/2 (the bottom of the page about the water)
| Imagine a rope attached to the bottom of a swimming pool. You, above the surface, grab it and shake it back and forth to create waves along the rope. The part of the rope just above the surface has to be moving back and forth at the same frequency as the part of the rope just below the surface. Otherwise the rope would have a break at the surface.
| {
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Why does the dielectric field not cancel out the capacitor's field? When a conductor is in a region with electric field, free charges will move until they balance out the external electric field. However in dielectrics this does not happen. I know that charges are bounded to the atoms, and there is only a small portion that will be near the surface of the the capacitor, but should we not also consider the small electric fields inside the polarized atoms? They may add up and cancel out the external field.
|
why does not the dielectric field cancel out the capacitor's field?
The polarization of the dielectric in the capacitor does reduce the effective electric field of the capacitor, but doesn't completely cancel it out. The reason is the molecules of the dielectric material are not perfectly polarized by the capacitor's electric field.
See the diagrams below taken from the following link on the Hyperphysics website: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dielec.html
Note that the polarization of the dielectric by the capacitor's electric field does not result in perfect alignment of the dipoles with the electric field. The greater the dielectric constant $k$ of the dielectric, the better the alignment and cancelling effect. The effective electric field is then
$$E_{effective}=E-E_{polarization}=\frac{σ}{kε_0}$$
Where $σ$= charged density = $\frac{Q}{A}$
$ε_0$ = permittivity of free space.
Hope this helps.
| {
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Why atmospheric pressure on a liquid and not on a solid? An open cylindrical container (radius = 0.5 m) has 100 kg of water.
a) Find the water pressure at the bottom of the container.
b) Find the pressure exerted by the container on the floor (ignore the mass of the container)
What confuses me, in a) We simply use p = p0 + ρgh
then in b) p = F/A = ρgh then we already had a larger pressure in a) and now, in b) the pressure is much smaller ( but the water inside the container, at the bottom yields a much larger pressure??)
Can someone enligthen me please? I feel I'm missing a key concept here. Thanks in advance!
| In my opinion, it should have been explicitly mentioned in part (b) that the value of excess pressure is being asked. By excess pressure, I mean the difference between the pressure on the ground below the container and the pressure on the ground elsewhere. This way, the answer $\rho g h$ is justified.
| {
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Why a ball doesn't stop when it collides with a wall? If a ball collides with a ball of same mass the first ball stops and the second ball gets the velocity of first ball.The first ball stops due to the reaction force acting on it. But when a ball collides with a wall why doesn't it stop due to the reaction force?
| Because the reaction force is larger.
The force needed to make an object reach a certain speed is the same as the force needed to slow it down from that speed to zero over the same time. This is Newton's 2nd law.
So, when two equal and movable objects collide, the (action) force that is required to make one object speed up to a certain speed is exactly the same as the (reaction) force required to slow the impacting object down to precisely zero.
When two non-equal objects collide, this is not necessarily so, and thus you will not see any of the objects stopping. Instead you might see that when
*
*a lighter object hits a heavier object, the action/reaction forces between them are larger than they would be between equal objects. And thus, the reaction force that stops the lighter object is larger than what is needed to precisely slow it down to zero. Thus, the lighter object slows down to zero and then start moving backwards, as if it bounces off.
*When an object hits a wall, you can consider this a collision between a light object and a very heavy object, and so, this bouncing-back effect is much stronger (the reaction force is much bigger). It might actually be so strong that the impacting light object bounces backwards with exactly the speed it came with, just directed oppositely.
| {
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If an observer changes their velocity, is it equivalent to a momentum shift? Suppose we have a world with one massive free particle and one observer. If the observer changes their velocity with respect to the particle, will this have the same effect on $\lvert\Psi(r,t)\rvert^2$ as if $\langle p\rangle$ was changed locally (for example, using compton scattering if we added another particle)?
Part of me says, yes: the "wavelength of $\Psi$" is related to momentum by the de Broglie relation, and momentum is relative (assuming we have just a Galilean transform here).
Part of me says, no: due to the uncertainty principle, the probability of finding a free particle anywhere would seem to be uniform and negligible over all space. From this argument, the wave function shouldn't change, because for a free particle, it wouldn't have a well-defined wavelength to begin with.
I know I'm missing some conceptual connection, because these can't both be true.
I am currently studying an intro QM course, so please take my undergraduate level of understanding into account!
| The relation between the two cases is given by a unitary transformation. This means that $|\psi(x,t)|^2$ is necessarily preserved.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/541376",
"timestamp": "2023-03-29T00:00:00",
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Newton 3rd law of motion If I took both hands and hold as I am greeting someonei.e.namaste. Now, I push both hands with same force say 10N , there is no change in position.But Newton 3rd law has said,equal and opposite forces are on different bodies and they never cancel each other (action reaction). So, why they don't move? (Here , they- hands)
| When you push on your right hand with your left you feel a force back on your right hand.
This is the 'equal and opposite' force that comes from Newton's third law. However, at the same time you are pushing with your right hand you are also pushing with your left and feel a force back on your left hand, again due to Newton's third law.
While it is true that the equal and opposite forces act on different bodies and as such can't cancel there are actually two pairs of Newton's third law forces with a force and its reaction force (equal and opposite force) from each hand.
These can cancel.
If you were to stop pushing actively with one hand, say the left, then both hands would move left as the only forces now are the force to the left from the right hand and the reaction force from the left hand on the right hand.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/541557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove the Jackson-Feenberg identity in quantum mechanics? The Jackson-Feenberg identity for evaluating the expectation of kinetic energy is frequently used in the theory of quantum liquids and is given by:
$$<\hat{T}>=-\frac{\hbar^2}{8m}\int d^3r[(\nabla^2\psi^*)\psi+\psi^*(\nabla^2\psi)-2(\nabla\psi)\cdot\nabla(\psi^*)],$$ where $\psi$ is a normalized eigenstate.
How to prove this?
I tried to operate $\hat{T}=-\frac{\hbar^2}{2m}\nabla^2$on $|\psi\rangle $ and integrate after operating from left by $\langle\psi^*$|, and noticing $(fg)''=fg''+gf''+2g'f'$, and the integrand I obtained is like $fg''$. I got $\nabla^2({\psi}{\psi}^*)-\psi\nabla^2\psi^*-2(\nabla\psi\cdot\nabla\psi^*)$ as my integrand, apart from the multiplicative constant -$\frac{\hbar^2}{2m}$ outside the integral. This leads to nowhere and I think I am not on the right track. All papers I came across just applied it. How can we derive this?
| Just integrate by parts a few times to get it in the desired form. For the first term in the identity,
$$\int d\mathbf{r} \, (\nabla^2 \psi^*) \psi = - \int d\mathbf{r} \, (\nabla \psi^*) \cdot (\nabla \psi) = \int d\mathbf{r} \, \psi^* \nabla^2 \psi.$$
For the third term in the identity,
$$- 2 \int d \mathbf{r} (\nabla \psi^*) \cdot (\nabla \psi) = 2 \int d \mathbf{r} \, \psi^* \nabla^2 \psi.$$
Therefore, putting all the terms together, we get
$$- \frac{\hbar^2}{8m} \int d \mathbf{r} \, (1 + 1 + 2) \psi^* \nabla^2 \psi = - \frac{\hbar^2}{2m} \int d \mathbf{r} \, \psi^* \nabla^2 \psi = \langle \hat{T} \rangle$$
as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why entropy in ideal BEC is independent of density I am confused about one statement from the book Bose-Einstein Condensation in Dilute Gases.
In the chapter discussing superfluid, the auther claims:
The pressure and the entropy density $\frac{S}{V}$ of ideal Bose-Einstein condensed gas depend on temperature but not on density.
Here, ideal means no interaction, homogeneous external field or no external field, and infinite volume.
This statement is perhaps easily obtained from the quantum partition function.
For classical particles, assuming the number of micro states for one particle is $a$, then for $n$ non-interacting particles, the number of microstates would be $a^n$. Therefore the entropy is dependent on the total particle number. Given that the volume is fixed, the entropy is dependent on density.
For the BEC case, it's different in the fact that symmetry and indistinguishability must be considered. And these two effects will effectively reduce the number of allowed microstates.
However, I cannot obtain the above statement directly without writing down the partition function and hence no clearer picture or intuition. Could anyone help to better understand this?
| For an homogeneous ideal gas in 3D below $T_c$ we get
\begin{align}
p &= \zeta(5/2) \left(
\frac{m}{2\pi \hbar^2}\right)^{3/2} \;
(k_B T)^{5/2}\propto T^{5/2}
\\
\frac{S}{N} &= k_B \frac{5}{2}
\frac{\zeta(5/2)}{\zeta(3/2)}
\left(\frac{T}{T_c}\right)^{3/2}
\propto \left(\frac{T}{T_c}\right)^{3/2}
\propto \left(\frac{T}{n^{2/3}}\right)^{3/2}
= \frac{T^{3/2} }{n} = \frac{T^{3/2} }{N} V
\end{align}
where $\zeta(\alpha) = \sum_{j=1}^{\infty} j^{-\alpha}$ is the Riemann zeta function. Hence, below $T_c$ the entropy does not depend on the total number of particles. The reason for this is that only particles in the excited states contribute to $S$. Consequently, the number of particles in the macroscopically occupied state is irrelevant.
Why do the particles in the ground state do not contribute to the entropy?
The energy of the ground state is arbitrary, and we usually choose $E_0=0$ -- the energy in the ground state is zero. Now, the specific heat is given by $C = \frac{\partial E}{\partial T}$ which yields
$$
C
\propto \frac{E}{T}
\propto T^\alpha
$$
where we used $E \propto T^{\alpha +1}$. Since we also know $C = T \frac{\partial S}{\partial T}$ we obtain
$$
S = \int \frac{C}{T} dT
\propto T^\alpha
\propto \frac{E}{T}
$$
Therefore, a particle in the ground state does not contribute to the entropy.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is energy density of two $P$ W lamps the same for one $2P$ W lamp? Do two lamps of power $P$ produce the same energy output as single lamp of power $2P$, assuming ideal environment? E.g. lamps do not occlude each other, distance between lamps is negligible, no monochromatic interference, same input energy to output energy ratio.
In my basic understanding it should be the same, and adding more lamps (or increasing their power) should scale energy output linearly, but something makes me unsure about that (I'm possibly wrongly thinking about perceived brightness, which is nonlinear).
| Radiant flux does add linearly - two lamps with radiant flux P will sum to a total flux of 2P and match the single lamp with 2P.
With lamps, we should all be careful to separate electrical power from radiant power. Different lamps will have a different conversion rate of electrical power to radiant power (especially in the visible wavelengths - many lamps produce heat). Two lamps of the same electrical power may not produce the same radiant power output as one lamp of twice the electrical power.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/542002",
"timestamp": "2023-03-29T00:00:00",
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How can we calculate the change in entropy of a block and its surroundings, when the block is placed in a heath bath? Let's say we have a block of lead (of heat capacity $C$) at temperature $T$, and a heat bath of temperature $T_0$. The block is placed in the bath and it cools to $T_0$ (a clearly irreversible process). How can we go about working out the entropy change of both the block itself and of the surroundings (i.e. the heat bath)?
First of all, work doesn't make sense in this context and so it's clear that the block is only in thermal contact with the heat bath (so only heat can be exchanged). The first law of thermodynamics tells us that:
$$dU=dQ+dW=TdS-pdV$$
$$dU=dQ=TdS$$
From this we have a differential expression for $dS$, and armed with the heat capacity of the block, we can perform the integration to find $\Delta S$ for the block, and we can follow a similar method for the bath.
There are two issues I have with this:
*
*How can $dU=dQ$ when the former is a total differential (because $U$ is a function of state) and the latter isn't ($Q$ is not a function of state and depends on path)?
*This leads us to $dS=\frac{dQ}{T}$. However, we know from Clausius' theorem that $dS>\frac{dQ}{T}$, unless $dQ=dQ_{rev}$, i.e. the process is reversible, which in this case it clearly isn't.
| Just calculate $\Delta S=mc \ln(T_2/T_1)$ for the block, assuming it's an incompressible solid (we'll derive this below). Entropy is a state function so the process of how you aquired this change doesn't matter. Now to answer your questions:
*
*This is true because the path/process determines $Q$, which changes the state of the system, hence changing $U$.
*You are correct, this is irreversible if you just drop a hot rock in cold water. But you could drop the rock in many infinitesimally cooler baths, thus cooling it reversibly, and $dS=dQ/T$ would apply for each of these increments. Now integrate over all these infinitesimally small coolings, with $dQ=mc\ dT$, and get $\Delta S=mc \ln(T_2/T_1)$. Entropy is a state function, so we don't care about the path we took to get from initial to final state.
| {
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