Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
Massless string vs massless spring in a mass-spring system
Two masses connected by a massless spring, on a frictionless surface , and a force of $60$N is applied to the 15kg mass such that it accelerates at 2 $\frac{m}{s^2}$. What is the acceleration of the $10kg$ mass?
I came across this question. I first thought that that the $10$kg was constrained to move at the same acceleration. But when I work it out, I get $a_2$ = 3 $\frac{m}{s^2}$. And it is the correct answer according to the book.
What I am unable to understand is, isn’t the $10$kg mass constrained to move at the same acceleration as the $15$kg mass? I thought we could replace the massless spring by (or treat it as) a massless string and results would be the same. Am I making a fundamental mistake?
| The problem is poorly stated. If a 60 Nt force is applied to a 15 kg mass, the acceleration will be 4 m/s/s. The 10 kg mass will start slowly and accelerate as the spring is stretched. The two masses will then oscillate relative to each other. At some later instant when the force from the spring is 30 Nt, the 15 kg will be accelerating at 2 m/s/s and the 10 kg at 3 m/s/s.
| {
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Cars' wear resistance, quietness and water drainage dependence on thread pattern on tires? Why can tires with fine thread patterns increase cars' wear resistance, quietness and water drainage? What is the logic behind?
The following is the tire ad I saw:
Nowadays, under the premise of pursuing comfort, most of Sedan / Saloon's tires have begun to adopt the design of fine tread patterns, that is, a very small gap is drawn between the tire blocks by laser or other manufacturing processes. The central Z-shaped tread pattern of Yokohama L752R adopts the design of the fine tread pattern that has been popular among sedan / saloon tires, which can increase its wear resistance and also increase the quietness. The side is also designed with a fine tread pattern to increase the car's drainage and wear resistance.
| Here is how this works. When a rubber tire rotates and comes into contact with the pavement, the round rubber tire gets deformed (pressed flat against the flat pavement).
Since the tire wants to remain round, this causes the tire surface to "scrub" against the pavement which grinds away at the tire surface, wearing off the rubber and turning it into dust. And when the tire's rolling action takes the tire surface out of contact with the pavement, the deformed part of the tire snaps back into its round shape and this "snap" radiates sound into the tire, the wheel, the suspension and into the body of the car where you hear it.
By putting lots of thin open spaces between little patches of tire surface, all those individual parts of the tire can more easily conform to the pavement without deformation: the "scrubbing" action is lessened, and wear is reduced. And the tire surface can "unscrub" without snapping back, reducing the noise output.
| {
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What is the ideal damping ratio for a building? I am a high school student investigating how the mass ratio of a pendulum tuned mass damper inside a building (ratio of the mass of the pendulum tuned mass damper to the mass of the structure) affects the damping ratio when the building vibrates. How do I select the ideal mass ratio for a structure based on the damping ratios I've gotten? In other words, what value of the damping ratio is ideal with regards to the safety/comfort for the citizens inside the structure?
| The optimum damping ratio is that which produces a critically damped system. Critical damping dissipates the system energy in the shortest possible time and does not support oscillation. For a critically damped system, the damping ratio is equal to 1.
| {
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The fluctuation-dissipation theorem In Giuliani & Vignale's Quantum Theory of the Electron Liquid, in page 126, they point out that the absorption and emission spectra are related by $$S_{AA^\dagger}(-\omega)=e^{-\beta\hbar\omega}S_{A^\dagger A}(\omega)$$ and that to obtain such relation, you should just start off from the definition $$S_{AA^\dagger}(\omega)=\sum_{nm}P_m|A_{mn}|^2\delta(\omega-\omega_{mn})$$
and take $\omega\rightarrow-\omega$ and interchange $n$ and $m$. I know that $\omega_{mn}=-\omega_{nm}$ and $\hat A$ is an hermitian operator.
But so far I haven't managed to understand how taking such considerations can lead one to the first relation. Any help would be welcome, thank you!
| The averaging here is a thermal trace:
$$S_{AA^\dagger}(t)= Tr[e^{-\beta H}A(t)A^\dagger(0)].$$
You need to permute it in order to change the order of the operators.
| {
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How can a photon collide with an electron? Whenever I study the photoelectric effect and the Compton effect, I have always had a question about how a photon can possibly collide with an electron given their unmeasureably small size. Every textbook I've read says that the photo-electrons are emitted because the photons collided with them. But since the photons and electrons virtually have no size, how can they even collide? I have searched for the answer on the internet but I couldn't find any satisfying one.
| Let's start with a classical picture, where an electron possesses a negatively electric charges, $q=-e$, and light is an electro-magnetic field. Hence, in the classical description we expect that there exists an interaction between these two objects, because
*
*there exists an interaction between the electric field of the light and the charge, $\vec F = q\cdot \vec E$,
*there exists an interaction between the magnetic field of the light and the charge, $\vec F = q \vec v \times \vec B$,
Now, if we use a quantum picture and think of light as being composed of photons, we have to account for this interaction. This is done by using an interaction cross-section or interaction strength/amplitude. The math becomes involved -- its called quantum electro dynamics (QED).
| {
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Cross-sectional area of a Gaussian beam of particles On page 47 of the seventh edition of Particles and Nuclei, the authors write the following:
For a Gaussian distribution of the beam particles around the beam centre (with horizontal and vertical standard deviations $\sigma_x$ and $\sigma_y$ respectively), $A$ [the cross-sectional area] is given by \begin{equation}
A=4\pi\sigma_x\sigma_y.
\end{equation}
This formula isn't proven or even motivated, however. Could anyone help me to understand how this result comes to be?
| I read your question again.
I can't say why a particle beam would have the same Gaussian beam shape as a laser. But in lasers, this is the most common and desirable beam. If nothing else, it has the most collimated beam because other modes have larger diffraction effects. Diffraction is often important for lasers.
The intensity profile across the center of a Gaussian beam has is a bell shaped curve. Because of this, there is no obvious beam diameter. The beam fades away as you get farther from the center, without any definite edge. The beam diameter is a matter of convention.
For a bell shaped curve, the standard deviation, $\sigma$, is an obvious way to speak of beam radius. And this is one of the usual conventions for beam radius. At this radius, the E field is $1/e$ of the central value, and the intensity has dropped to $1/e^2$. Another convention for beam diameter is Full Width at Half Max (FWHM).
There is plenty of intensity outside these diameters. Typically lenses used to focus the beam must have an aperture 1.5 times the beam $\sigma$ based diameter. At that distance from the center, the intensity is about 1% of the central intensity. This is low enough that diffraction effects from truncating the beam are usually negligible.
Not all Gaussian beams are circular. Some are elliptical. These have two different beam radii, $\sigma_x$ and $\sigma_y$.
Given this, a convention for the area would reasonably be
$$A = \pi \sigma_x \sigma_y$$
It appears this author is using $2\sigma$ as his convention. So he gets
$$A = 4\pi \sigma_x \sigma_y$$
| {
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What's the meaning of a continuity equation with $\nabla^2 \rho$ on the right-hand side? I stumbled upon a continuity equation with a $\nabla^2$ term on the right-hand side:
$$ \partial_t \rho + \nabla (\vec b \rho) = D \nabla^2 \rho , $$
where $b$ denotes the forward velocity and $D$ is a constant.
What's the meaning of such a diffusion equation?
Some background:
Since, we have particle number conservation, we have
$$ \partial_t \rho + \nabla (\vec v \rho) = 0 , $$
where $v$ denotes the ordinary flux velocity. Moreover, if there are sources, we have
$$ \partial_t \rho + \nabla (\vec v \rho) = \sigma . $$
| It is the so called convection–diffusion equation (but it is also known under other names). You may find more information in the above linked wikipedia page, but, in brief, it is an equation which combines a convective cause of time variation at one point (the $\nabla (\vec b \rho)$ term), with a diffusive process, controlled by the Laplacian term.
The convective term in the continuity equation captures the information about the flux of the quantity $\rho$ which locally moves with velocity $\vec b$. When integrated over a closed surface, the net time variation of $\int \rho$ over the volume is due to the surface integral of the current $\rho \vec b $.
The diffusive term with the Laplacian, provides a different mechanism for the net time variation of $\int \rho$ over the volume: the presence of a diffusive flux, not accompanied by a macroscopic field of velocity, proportional to the local gradient of $\rho$, according to a Fick's-like law ( current = $-D \nabla \rho$).
| {
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Explicit form of a qubit state and relative phase In quantum computing, we can write our wave function as:
$$
|\psi\rangle=\alpha|0\rangle+\beta|1\rangle
$$
which can be rewritten as
$$
|\psi\rangle=\cos(\theta/2)|0\rangle+e^{i\phi}\sin(\theta/2)|1\rangle
$$
ignoring the global phase. This can be represented in a Bloch sphere where theta and phi are angles on the sphere. That being said, my question is, how is this rewritten equation found and why is there a $e^{i\phi}$ term multiplied by the $\sin(\theta/2)$? And what does this additional phase represent? And finally, why is it on the sine rather than the cosine? Thank you! To preface, I am currently an undergraduate physics junior who is doing research on QC/QI.
| This is covered in very many textbooks. One example is the book of John Townsend. Anyways the point is if you write
$$
\vert\psi\rangle = \alpha \vert 0\rangle +\beta \vert 1\rangle
$$
you need $\vert\alpha\vert^2+\vert\beta\vert^2=1$ for normalization. Thus
$$
\alpha=e^{i a}\cos\frac{1}{2}\theta\, ,\qquad
\beta=e^{i b}\sin\frac{1}{2}\theta
$$
will do the trick so one writes
\begin{align}
\vert\psi\rangle =e^{i a}\left(\cos(\textstyle\frac{1}{2}\theta)\vert 0\rangle + e^{i\phi} \sin(\textstyle\frac{1}{2}\theta)\vert 1\rangle\right)
\tag{1}
\end{align}
with $\phi=b-a$ and choose $a=0$ since the factor $e^{i a}$ is an overall phase. It is conventional to choose the coefficient of $\vert 0\rangle$ to be real.
The half angle is related to the rotation properties of spin states. Alternatively, opposite spin states in 3D space are orthogonal in spin-1/2 space so set $\theta=0$ (North pole state) to get $\vert 0\rangle$ and $\theta=\pi$ (South pole state) to get $\vert 1\rangle$, which are indeed orthogonal.
This is the most general form. The relative phase $e^{i\phi}$ controls the interference between the two states. You can see its effects by computing and comparing the average values of $\hat \sigma_x$ and $\hat \sigma_y$ on (1): these average values are explicitly $\phi$-dependent.
Moreover the eigenstates of $\hat \sigma_x$ and $\hat \sigma_y$ only differ by a relative phase factor.
| {
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What is the meaning of a number changing process? On my lecturer's notes on Dark Matter I am told:
"It is customary to define densities normalised by the time dependent volume $V (t) = a(t)^3$. The reason for this is that, in the absence of number changing processes, the comoving number density remains constant with time evolution."
I don't understand what is meant by a number changing process.
The image is given as an example of the preservation of the comoving number density.
Initially I thought a number changing process was one in which the number of the particles changed as the process evolved but I believe the image is trying to emphasise that the number of particles is kept constant.
| You are right. The number of particles does not change, but the volume increases as $V(t) = V_0 a^3(t)$, because of expansion.
Naively one can compute the number density as
$$
n(t) = \frac{N(t)}{V(t)}=\frac{N_0}{V_0 a^3(t)},
$$
where I used the fact that $N(t)=N_0$ in the absence of process that change the number of particles.
However, if we spice up things by normalizing the volume by the expansion factor $a^3(t)$ in the previous expression, we have
$$
\hat{n}(t) = \frac{N_0}{V_0 a^3(t)/a^3(t)}= \frac{N_0}{V_0},
$$
which is indeed the definition of comoving number density and it is manifestly time-independent.
| {
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Calculating work of an object moving up a slope If an object is being pushed across a horizontal surface, the equation for the work done is $W = F s$, where $s$ is the horizontal displacement.
If an object is being lifted to a height of $h$, the equation for the work done is $W = F h$, where $h$ is the vertical displacement.
If an object is being pushed up a slope, or if a human is moving up a flight of stairs, however, only the vertical displacement is concerned in calculating work, while the horizontal displacement is omitted. Why is this so?
(With hindsight, I think this is a conceptual misunderstanding that only arises because it is quoted out of context!)
| It really depends on what forces exactly you are looking at. If it's friction & gravity, then total work done against them is :
$$ \begin{align}
W &= F_g h + F_{fr}~\ell \\
&= F_g h + F_{fr}~ \frac {h}{\sin \theta} \\
&= h \left( F_g + F_{fr}~ {\sin \theta}^{-1} \right)
\end{align}
$$
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Does wave function always mean position in time? Does wave function always mean position in time? If we take an entangled state wave function $\frac1{\sqrt2} (|0>|0>+|1>|1>)$, we see nothing about position and time.
| A wavefunction $\psi(x,t)$ is the coefficient of a normalised vector in Hilbert space $\big(|\psi(t)\rangle\big)$ when expressed in the position basis $\big(|x\rangle\big)$. In other words,
$$|\psi(t)\rangle=\int dx|x\rangle\langle x|\psi(t)\rangle=\int dx|x \rangle\psi(x,t)$$
In your case, what you have is a vector in Hilbert space. If you project it onto space you’ll get the wavefunction.
| {
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How to write singlets and triplets in second quantization for fermions? It has been a long time I haven't done this and I am having a hard time writing things down in second quantization notation. Let us have a $n$-body system where the spin part and orbital parts are decoupled.
If want to write a two particle state, let's say a triplet state $\alpha$, with parallel spins, an electron with energy $\epsilon_1$ and another with $\epsilon_2$, I would write something like
$$|\alpha\rangle=c^\dagger_{1\uparrow}c^\dagger_{2\uparrow}|\emptyset\rangle$$
where $|\emptyset\rangle$ is the void state, and $\uparrow,\downarrow$ are the spin states.
Same for a singlet $\beta$ with two electrons in the same energy
$$|\beta\rangle=c^\dagger_{1\uparrow}c^\dagger_{1\downarrow}|\emptyset\rangle$$
But how do I distinguish a state $\gamma$ with $m_z=0$ (different energies, opposite spins)? What does this do?
$$|\gamma\rangle=c^\dagger_{1\uparrow}c^\dagger_{2\downarrow}|\emptyset\rangle$$
This could be either a triplet (with opposite spins) or a singlet, what am I missing?
Edit: I now realize that there has to be a difference in the Fock space, between writing $|n_{1\uparrow},n_{1\downarrow},n_{2\uparrow},n_{2\downarrow}\rangle=|1001\rangle$ and $|0,1,1,0\rangle$ but I don't know how to interpret these states in terms of the singlet and the unparalleled triplet
| The point is that the wavefunction is symmetric in spin for the triplet and antisymmetric for the singlet. So in your notation, the triplet is
$$\frac1{\sqrt2}(c^\dagger_{1\uparrow}c^\dagger_{2\downarrow} + c^\dagger_{1\downarrow}c^\dagger_{2\uparrow})|\emptyset\rangle$$
and the singlet is
$$\frac1{\sqrt2}(c^\dagger_{1\uparrow}c^\dagger_{2\downarrow} - c^\dagger_{1\downarrow}c^\dagger_{2\uparrow})|\emptyset\rangle.$$
| {
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How to compute expectation value $\langle e^{iH}\rangle$ for quadratic Hamiltonians? I have a rather basic, but actually non-trivial question:
We consider a bosonic system with creation operators $\hat{a}_i^\dagger$ and annihilation operators $\hat{a}_j$ and vacuum state $|0\rangle$ with $\hat{a}_i|0\rangle=0$. We consider a quadratic Hamiltonian
\begin{align}
\hat{H}=\sum_{ij}(A_{ij}\hat{a}_i\hat{a}_j+A^*_{ij}\hat{a}^\dagger_i\hat{a}^\dagger_j+B_{ij}\hat{a}_i^\dagger\hat{a}_j)\,.
\end{align}
Hermiticity requires $B_{ij}=B^*_{ji}$. How can I compute the expectation value:
\begin{align*}
\langle 0|e^{i\hat{H}}|0\rangle
\end{align*}
In many situations, it is useful to express the problem in terms of Hermitian quadrature operators $\hat{q}_i=\frac{1}{\sqrt{2}}(\hat{a}^\dagger_i+\hat{a}_j)$ and $\hat{p}_i=\frac{1}{\sqrt{2}}(\hat{a}^\dagger_i-\hat{a}_j)$ to define $\hat{\xi}^a=(\hat{q}_1,\dots,\hat{q}_N,\hat{p}_1,\dots,\hat{p}_N,)$, such that $$\hat{H}=\frac{1}{2}h_{ab}\hat{\xi}^a\hat{\xi}^b$$ with implied sums over $a$ and $b$. It is also useful to define the symplectic form $\Omega^{ab}$, such that $[\hat{\xi}^a,\hat{\xi}^b]=i\Omega^{ab}$, which gives rise to the symplectic generator $K^a{}_b=\Omega^{ac}h_{cb}$. We only require that $h_{ab}$ is symmetric and real.
| Start diagonalizing H. You can follow this:
https://arxiv.org/pdf/0908.0787.pdf
Then use the power series of an exponential. You get the result in the diagonal basis. Make the inverse transformation and it is done.
| {
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Fraunhofer line width What sets the width of Fraunhofer lines on the solar spectrum ?
I first thought of Doppler broadening, but numerical applications result in much too high temperatures. For instance, using these data, I find a $\Delta \lambda =$ 0.01nm line width on the 630.25nm line of iron, corresponding to a temperature of
$$
T = \frac{mc^2}{k_B} \frac{\Delta \lambda ^2}{\lambda ^2} \simeq 100\, 000 \, {\rm K}
$$
which is way above the Sun's photosphere temperature.
Is there something wrong with the above calculation, or is the line width coming from something else ?
| Your calculation sounds legit… Line profile looks similar in other high resolution data. I don't have any complete answer ; my guess would be unresolved Zeeman effect. It could be unresolved due to variation in B field over the field of view, even if the spectrometer itself is of high enough resolution.
Other reasons I could imaging for line broadening :
*
*Collisionnal broadening (limited time in an energy level between collisions $\Rightarrow$ uncertainty in energy)
*Stark effect (I highly doubt there are such electric fields there).
None of them looks convincing to me.
| {
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Is there any operator in quantum mechanics that measure an observable with non-zero uncertainty? What does a measurement do?
The answer is: If the detector is designed to measure some observable O, it will leave the measured object, at least for an instant, in a zero-uncertainty state.
I want to know, in the context of quantum mechanics and base on Hilbert space, can define an operator to measure an observable with non-zero uncertainty?
| The uncertainty of an observable depends on the state $|\psi\rangle$ of the system that is being measured. The expectation value of the observable $A$ is given by
$$\langle A\rangle=\langle\psi|A|\psi\rangle$$
and the uncertainty is given by
$$(\Delta A)^2=\langle (A-\langle A\rangle)^2\rangle=\langle A^2\rangle-\langle A\rangle^2$$
Therefore, you can have a state with non-zero uncertainty in $A$ when $\langle \psi|(A-\langle A\rangle)^2|\psi\rangle\neq0$. There are many examples of systems like these. For example, a Gaussian wave packet has non-zero uncertainties in both position and momentum.
In terms of measurement, of course if you measure an observable the system is in an eigenstate of the observable and hence has $0$ uncertainly for that observable. But that does not make the new state a state of "non-zero uncertainty" in general, because the state could have uncertainty with respect to other observables. More explicitly, if we measure $A$ for our state so that now $|\psi\rangle=|a\rangle$, we now have $\Delta A=0$, but there could be (definitely is?) another observable $B$ such that $\Delta B\neq0$ for this new state $|a\rangle$.
If you want to make an observable where state after measuring $A$ has a non-zero uncertainty in $A$, then that is impossible. This is because the state after measuring $A$ has to be an eigenstate of $A$, so then $\Delta A=0$ after measurement always.
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Why does the cosmological constant problem use the expectation value of the QFT vacuum energy? As I understand it, the famous 120 orders of magnitude discrepancy between the observed cosmological constant and the calculated QFT vacuum energy density relies on the vacuum Einstein field equations
$$G_{\mu\nu} + \Lambda g_{\mu\nu} = \left< 0 |T_{\mu\nu}| 0 \right>.$$
The QFT vacuum energy expectation value $\left<\rho_{vac}\right>$ is formally infinite, but some natural cutoff is taken.
The quantum expectation value is the average value over many measurements. So why can we say that the vacuum is being measured? And if we can't say that there is wavefunction collapse, why should GR, a classical theory, care?
Of course this brings up the measurement problem. It is true that virtual particles can be realized, like for example in Hawking radiation, but this relies on some sort of wavefunction collapse due to interaction with the black hole. Since most of the universe is nearly empty, can we really say that vacuum fluctuations have an effect away from regions of appreciable matter and gravity?
For reference, I am coming at this from a mathematics background and the physics level of, say, a first graduate course in GR and QFT.
| It is expected that a future theory of everything will solve this discrepancy which depends, as you observe , in mathematically comparing classical with quantum values .
At the moment the only theories that can quantize gravity and at the same time embed the standard model of particle physics, which is expressed in quantum field theory terms, are string theories.
The question Can/has string theory solved cosmological constant problem? addresses this. A good answer exist by dr Motl, a known string theorist.
The gist as far as your question goes is that, yes, to mix classical with quantum and to consider it a big problem when there are discrepancies is not correct. There are possible solutions once gravity is quantized.
| {
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Why do small animals appear to move faster than larger ones? I am keen to understand why smaller creatures move relatively faster than larger ones. Not only do they move faster, but their metabolism runs at a faster rate, they seem to process information faster (try swatting a fly!) - their entire lives seem to burn more rapidly living shorter lifespans. I’ve read about Kleibers scaling law, but this explanation alone does not seem to offer a fully satisfactory answer.
| There are essentially two reasons. First, as they are smaller we are likely to look at them from nearer, creating the illusion of greater speed which results from greater angular speed subtended at the eye (just as when travelling by train or car, near objects cross our field of vision more quickly).
Second, they are lighter, body mass going with the cube of linear size. Muscular force goes with the cross section area, that is the square of linear size. Consequently they do in fact accelerate much more rapidly than larger animals, and greater acceleration can also produce an illusion of greater speed.
| {
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Why does frictional force depend on normal reaction and not the weight of a body? One possible explanation I came across for this was that if you have an external force and you press on the body, the body's frictional resistance to motion increases and hence it should depend on normal reaction and not weight.
| The two are same if the path is horizontal.
If the path is inclined/curved then it becomes necessary to distinguish between the two in the play out of friction in situations where we draw the force equilibrium diagram.
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Graphical explanation for length contraction I can understand the mathematical explanation for the reason why there should be a length contraction, but I fail to understand it intuitively. That is why I tried to explain it using spacetime diagrams, but for some reason, I was unable to do so.
Let us use the following procedure to measure the length of the rod from S's reference frame: S moves with $\vec{v} = v \hat x$ and sets it's watch to $t = 0$ when it is in one end of the rod, and looks at it's watch when it is in the other end of the rod and sets $t = t_2$. Let us also put one end of the rod, the one which $S$ visits first, to the origin of $S'$.
We have two events
$$e_1: \quad (t_1', x_1') = (t_1', 0) \quad and \quad (t_1, x_1) = (0,0),$$
$$e_2. \quad (t_2', x_2') = (t_2', L_0)) \quad and \quad (t_2, x_2 = (t_2, 0))$$
Note that, the rod is stationary wrt S' and both events happen at the origin of S wrt S.
Since we are relying on $t_2$ to calculate the length of the rod, graphically (see the above figure),
$$t_2 = \sqrt{L_0^2 + (t_2')^2}.$$
If we just cheat a bit (to see whether we are on the right track) and use Lorentz transformations, we can see that $t_2' = \frac{t_2}{\sqrt{1-v^2}}$, which means that the above equation implies
$$t_2 = \sqrt{v^2 + 1}t_2' \quad \Rightarrow \quad x_2 = \sqrt{v^2 + 1}L_o,$$
which is clearly wrong.
Question:
What am I doing wrong?
| For an intuitive explanation you need a better diagram. The spacecraft measures length L at equal time in the spacecraft frame.
The spacecraft’s clock is in the bow. The spacecraft and Earth set their clocks to zero when the bow passes the Earth clock. Earth uses radar to measure the distance, $l$, from bow to stern, by sending a signal at time $-l$, which returns at time on the Earth clock. The same signal is used to determine the proper length, $L$, as measured on the spaceship. Using the Doppler shift, the outgoing signal passes the bow at time on the spacecraft’s clock. The returning signal reaches the bow at time $-l/k$. You should already have
$$k^2 = \frac{1+v}{1-v}$$
$$\gamma = \frac{1}{\sqrt{1-v^2}}$$
So the spacecraft’s proper length is
$$L=\frac{kl+l/k}{2} = \frac{k^2+1}{2k}l=\frac{(1+v) + (1-v)}{2(1-v)}\sqrt{\frac{1-v}{1+v}l} = \gamma l$$
| {
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When Is It Appropriate To Use The Ladder Operator Method in Quantum Mechanics? I'm trying to understand when it is intuitively obvious that the ladder method would be best used to tackle a problem in quantum mechanics.
| Ladder operators might be one method of a solution if
*
*the system has a discrete set of eigenvalues of an observable operator (hamiltonian, angular momentum, etc)
*you can establish a non-zero commutator relationship between the ladder operators (usually an operator and its adjoint)
*the observable operator can be constructed from the ladder operators, but generally don't commute with the observable.
*look for symmetry in the system
This is probably not an exhaustive list, but the best teacher here is experience. After you see several examples, you begin to get a sense that ladder operators might work.
It's not really "intuitively obvious" without the experience. "Obvious" to an expert = "quite elegant" to the first year grad student = "WHAT?" to the sophomore.
| {
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Where did the work done by smaller force go? Suppose I have a spring of spring constant 150 N/m.One person pulls with it a force of 15 N. The extension produced is 0.1 m. Now another person comes and pulls with a force of 30 N (The first person is still there). The final extension is 0.3 m.The initial and final potential energies of the spring are 0.75 J and 6.75 J respectively. The work done by the bigger force is 6 J which is exactly equal to the change in P.E of the spring. So where did the work done by the smaller force (when they were pulling it together) go? Why didn't it help to further increase the potential energy of the spring?
|
The work done by the bigger force is 6J which is exactly equal to the change in P.E of the spring.
Actually, this is incorrect. If we don't want to be adding kinetic energy and worrying about that, then the total force from the people must at all times equal the force from the spring. This means that the force from the second person is only 30 N at full extension and is less than 30 N before then. Here is a plot of the forces vs displacement. The blue line is the total force from both people, the yellow line is the force from the first person, and the green line is the force from the second person.
In this plot the work is the area under the curve. It is immediately obvious that the area under the blue curve is the sum of the areas under the curve for the yellow and for the green person.
What is not so obvious is that the work done by the first person is actually more than the work done by the second. The area under the green curve is 3.0 J, not 6.0 J. The area under the yellow curve is 3.75 J, with 0.75 J under the initial triangular portion from 0 to 0.1 m and the remaining 3.0 J in the flat section.
So, indeed, the work done by the smaller force is essential for calculating the total energy that went into the spring. The calculation of 6.0 J from the second person is simply incorrect. It is incorrect by a factor of 2 which basically is the difference between incorrectly assuming that the 30 N force was applied throughout the movement instead of correctly recognizing that it is only 30 N at full extension. Of course, alternatively you could consider the change in KE that would result from a constant 30 N force, as described by @Aaron Stevens in his answer.
| {
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Why are domains formed as separate units? I was reading about exchange coupling and domain formation in ferromagnetic materials. As far as I can understand, some of the dipoles in the ferromagnetic material align themselves in a group called a domain, and several such domains are formed with enough randomness to prevent magnetisation of the material.
But I'm slightly confused in the part that the book assumes domains formed as pockets of dipoles in one direction, separated by a boundary and all the dipoles in a domain point in the same direction. I am unable to imagine intuitively why the phenomenon suddenly stops at the boundary between two domains. Shouldn't dipoles of one domain affect those of other domains?
| The interaction of atoms in a ferromagnet is typically modeled as an exchange interaction. The exchange interaction is related to the Pauli exclusion principle, which prohibits electron clouds of neighboring atoms from overlapping when the unpaired electron spins are aligned. This prohibition decreases the electrostatic potential energy of the spin-aligned configuration relative to the spin-antialigned configuration (in which the electron clouds do overlap), making the spin-aligned configuration more energetically favorable.
As you can see, in the exchange interaction, each atom is only affected by its nearest neighbors. So it doesn't know, and can't affect, the spin orientation of the atoms in other domains. There are only a few atoms in the material that do experience any kind of misalignment with their neighbors. These are the atoms near the magnetic domain walls. As the domain wall is crossed, the orientation of the local spin rotates smoothly from the alignment of one domain to the alignment of the other. So the phenomenon doesn't suddenly stop at the boundaries; it's a very gradual transition on the microscopic scale. See the following illustration, where A and C are domains and B is a domain wall:
| {
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Why are these quantization forms different? I'm confused about the form of the quantized electromagnetic field.
Sometimes, one writes the quantized electric field as
$$\hat E = \vec E_0(\hat a + \hat a^†)$$
, which can been seen the website below in the section, "Mathematical formulation 1".
https://en.wikipedia.org/wiki/Jaynes%E2%80%93Cummings_model
But as I know, the quantized electric field as
$$\hat E(r) = i\vec E_0(\hat a(k)e^{ikr} - \hat a^†(k)e^{-ikr})$$
for a given wavevector $k$ and position $r$.
This can be found in the below website.
https://en.wikipedia.org/wiki/Quantization_of_the_electromagnetic_field
I wonder why the two forms are different.
Please help me out.
| The first expression, used in the Jaynes-Cummings model, is a description of the electric field at just one single point in space. This is useful when the region of interest (the size of an interacting atom or molecule, for instance) is small compared with the wavelength of the field; $E$ is nearly constant over this region. However, this first expression is really just an approximation of the second expression; the second expression includes the additional spatial dependence that is relevant on longer scales. (The other differences just correspond to phase factor differences in the definitions of the raising and lowering operators, and so have no physical import.)
| {
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Which force makes the contact point of an object in rolling motion rise? When an object is rolling without slipping, the point at the bottom has no tangential velocity. When the point at the bottom rotates and rises, what force lets the point rise with tangential velocity from having no tangential velocity from before? I read that static friction is zero for rolling objects on a flat surface, so which other force could do this?
| In rolling motion it is not friction that is zero. Work done by friction is zero because all the points are just momentarily in contact with the rough surface.
Technically, as you said, the points are lifted up the next moment they come in contact with the floor. So, no displacement occurs in the direction of frictional force.
I think your confusion is about the tangential acceleration of the rolling body. The frictional force acting along the tangent at the point of contact increases the tangential velocity when it leaves the surface
Frictional force acts, and is necessary for pure rolling. If you try to let a ball roll in a smooth surface, it will slide rather than roll. A certain amount of frictional force is a prerequisite for rolling motion.
| {
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Equation of state for ideal gas from Helmholtz free-energy Starting from the definition of Helmholtz free energy: $$F:=U-TS$$ (where $U$ is the internal energy , $T$ temperature and $S$ entropy) we derive in few steps the following relation: $$F=-T\int \frac{U}{T^2}\mathrm d T+ \text{constant} \tag{1}$$
Now, we know also that Maxwell relations holds so at $T=\text{constant}$ we have: $$P=-\frac{\partial F}{\partial V} \tag{2}$$
In ideal gas the internal energy have the following form: $$U=\frac{3}{2} NT \tag{3}$$
If i substitute $(3)$ in $(1)$ and put the result in $(2)$ i should find the classical equation of state for ideal gas: $$ PV=NT \tag{4}$$ ...but from calculation i don't find this. Where is the error in my steps? It could be in the value of the constant ?
Yes, wrong word. Anyway we can say something about this function a posteriori :
$$
p = - \frac{\partial F}{\partial V} = -C'(V) T.
$$
Using (4) in this equation we obtain $$ C(V) = N * ln(V)+ constant$$
| In 1) there is additive "constant" of integration. The integration is only over $T$, the terms may depend also on volume $V$ which can be arbitrary. Therefore the "constant" in that integration over $T$ can be actually a function of $V$:
$$
F(T,V) = -T\int \frac{U}{T^2}dT + C(V)T.
$$
Since the first term, for an ideal gas, does not depend on volume, the only part relevant for calculating pressure from $F$ is the second term:
$$
p = - \frac{\partial F}{\partial V} = -C'(V) T.
$$
The conclusion is, we cannot infer the familiar equation of state of ideal gas $p = nc_V T / V$ just from knowing $U = nc_V T$. The above result suggests large class of functions of $C(V)$ is consistent with $U=nc_VT$. But we did find at least that pressure must be proportional to temperature $T$.
In Callen there is a rationale for this - the equation $U=nc_VT$ is not the fundamental form, that is, $U$ is not expressed as function of its natural parameters $S,V$. If it was, we should be able to derive the equation of state from it.
| {
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In a vacuum, can you see light which is not travelling towards you?
In air, when there is light propagating in a direction, we can still see it even when it is not primarily travelling in our direction, because a small part of the light hits the air molecules, and changes its direction; it travels towards us.
Does this mean that, in a vacuum, you would not be able to see light which is not travelling towards you?
| In a universe that is shaped like the three-dimensional surface of a basketball whose space would be expanding at a rate initially almost exponential but eventually only quasi-inertial, like each of the local universes in at least one "bouncing" and inflationary cosmology (Nikodem J. Poplawski's "cosmology with torsion", described in numerous papers written between 2010 and 2020 that are available free on Arxiv), photons would (since they each have an infinitesimal relativistic mass) be orbiting through the curvature of the aforementioned surface's volume, so that you would eventually be able to see even light that had been emitted from the curved region to your rear, if you'd somehow wait (and survive) long enough after having somehow made an appearance on the scene.
However, regarding a question (at https://astronomy.stackexchange.com/questions/19013/statistical-techniques-for-estimating-distribution-of-mass) on the Astronomy Stack Exchange, Pela's answer describing the most standard procedure for estimating mass distributions shows a disproportionately large distribution of mass in the brightest and most red-shifted clusters of galaxies, which suggests that the probability that any photon you might see would not have been involved in one or more refractions might be very low.
| {
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QFT: 2d massless fermion propagator in coordinate space. Where am I wrong? I am trying to compute the fermion massless propagator in two dimension:
$i \int \frac{d^2k}{(2 \pi)^2} e^{ik \cdot x} \frac{k^0 \gamma^0 + k_1 \gamma^1}{k^2 + i \eta} = \frac{i}{(2 \pi^2)} \int dk^1 e^{ik^1x^1} \int dk^0 e^{-ik^0x^0} \frac{k^0 \gamma^0 - k^1 \gamma^1}{(k^0)^2 - (k^1)^2 + i \eta} = \frac{i}{(2 \pi^2)} \int dk^1 e^{ik^1x^1} (2 \pi i) e^{-i(k^1-i \eta)x^0} \frac{(k^1-i\eta)\gamma^0 - k^1\gamma^1}{2k^1-i \eta} = - \frac{1}{4\pi} (\gamma^0- \gamma^1) \int dk^1 e^{k^1(x^1-x^0) - \eta x^0} = -\frac{1}{4 \pi}(\gamma^0 - \gamma^1) \delta(x^0 - x^1).$
Where I put $\eta = 0$ where it is possible and I supposed that $x^0 > 0$.
The correct result, reported by Peskin-Schoreder is:
$- \frac{i}{2 \pi} \frac{\gamma^\mu x_{\mu}}{x^2}.$
Where am I wrong?
| From the $+i\eta$ in the denominator, I'm assuming you're using the $(1,-1)$ signature but then I'm not sure why you wrote $ik\cdot x=-ik^0x^0+ik^1x^1$. More importantly, your second mistake is in the claim of a pole at $k_1-i\eta$. The poles for the $k^0$ integration are at $\pm(|k^1|-i\eta)$, i.e. they involve the magnitude $|k^1|$, not the variable $k^1$ directly.
Besides, it is rather messy to work with several gamma matrices in the integral. A cleaner approach is to push them to the side by writing:
$$
i\int \frac{d^2k}{(2\pi)^2}\frac{\gamma^\mu k_\mu}{k^2+i\eta}e^{ikx}=-i\gamma^\mu\partial_\mu \int \frac{d^2k}{(2\pi)^2}\frac{i}{k^2+i\eta}e^{ikx}
$$
Now you can leave the $-i\gamma^\mu\partial_\mu$ factor alone and work with the integral:
$$
G(x)= \int \frac{d^2k}{(2\pi)^2}\frac{i}{k^2+i\eta}e^{ikx}
$$
You may recognize this as the propagator of a massless scalar. There are already several answers on this site concerning the evaluation of this integral. See, for example, here.
The idea is still the same, depending on the sign of $x^0$ we can choose to close the contour either in the upper or lower plane. But it may be easier to do a Wick rotation and do the integration in Euclidean space. The result is, with the $i\eta$ prescription for $\ln(r)$ omitted :
$$
G(r)=\frac{1}{2\pi}\ln(r)
$$
Here $r=\sqrt{x^2}=\sqrt{x_0^2-x_1^2}$. The result for your original integral is:
\begin{align}
-i\gamma^\mu\partial_\mu \frac{1}{2\pi}\ln(r)&=-\frac{i}{2\pi} \gamma^\mu \frac{1}{r}\partial_\mu \sqrt{x_\mu x^\mu} \\
&= -\frac{i}{2\pi} \frac{\gamma^\mu x_\mu}{x^2}
\end{align}
| {
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For a semiconductor, why does a band gap larger than the incident photon energy mean material transparency? If a photon with less energy than a band gap hits the band gap, why will the material be transparent? Although the photon cannot raise an electron in the conduction band to the valence band, why isn't the photon absorbed and turned into heat?
| The answer is contained in the question: in order to be absorbed the photon needs to raise an electron from the valence band to the conduction band.
Absorption here means transmitting the photon energy and momentum to the semiconductor. Being an electromagnetic wave, the photon is coupled to the charge in the semiconductor, and the only possible change of the charge state is moving an electron to a different energy state. This is not a problem in metals, where the conduction band is only partially filled and electrons can be excited to any energy within this band, which is why metals are reflecting in the visual spectrum.
| {
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Why don't we use rapidity instead of velocity? In school we learn that we can add velocities together, and then later on we learn that it's not correct and that there is a speed limit. Why create all this confusion when we could just use rapidity to begin with?
Rapidity is defined as $w = \mathrm{arctanh}(v / c)$, where $v$ is velocity and $c$ is the speed of light in a vacuum. Rapidities can be summed and have no upper bound. At non-relativistic speeds it acts proportional to velocity.
In fact, at non-relativistic speeds, we could substitute $v$ for $wc$ (rapidity times speed of light), and one could hardly tell the difference. The ISS moves rather fast at a velocity of 7660 m/s (27,576 km/h), and has a $wc$ of about 7660.0000016667 m/s. Why can't we just substitute velocity for rapidity in real-world and classroom use, and end the confusion about why there is a speed limit once and for all?
| As well as issues of practicality, it doesn't answer the question of why there's a speed limit. (It can't as it's just a mathematical transformation.). The question becomes 'in the formula for $w$, why do you take $c=3 \times 10^8$ m/s'?
| {
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If I apply a constant force to an object until I've reversed its starting velocity, does its final position remain unchanged?
A body of mass $m$ has initial velocity $v_0$ in the positive $x$-direction. It is acted on by a constant force $F$ for time $t$ until the velocity becomes zero; the force continues to act on the body until its velocity becomes $−v_0$ in the same amount of time. Write an expression for the total distance the body travels in terms of the variables indicated.
I let $t'$ denote the time at which the velocity goes to zero so I can use $t$ as a variable. Then I found
$$a = \frac{\Delta v}{\Delta t} = \frac{-2v_0}{2t'} = -\frac{v_0}{t'}$$
Substituting this expression for $a$ into $s(t) = \frac{1}{2} a t^2 + v_0 t$ yields
$$\begin{align}
s(t) & = \left(-\frac{1}{2} \frac{v_0}{t'}\right) t^2 + v_0 t \\
s(2t') & = \left(-\frac{1}{2}\frac{v_0}{t'}\right)(2t')^2 + v_0 (2t')^2 = 0
\end{align}$$
This makes intuitive sense to me. If the positive $x$ direction is up and the force is gravity, then if we shoot a ball upward with velocity $v_0$, it will decelerate to velocity $-v_0$ exactly when it returns to the launch position.
To double check, I tried taking (from $a = -\frac{v_0}{t'}$ and $F = ma$) $a = \frac{F}{m}$ and $v_0 = -\frac{F}{m}t'$. I get the same answer when substituting into $s(t)$:
$$\begin{align}
s(t) & = \frac{1}{2} a t^2 + v_0 t \\
& = \left(\frac{1}{2}\frac{F}{m}\right) t^2 -\left(\frac{F}{m}t'\right) t \\
\implies s(2t') &= \left(\frac{1}{2}\frac{F}{m}\right) (2t')^2 -\left(\frac{F}{m}t'\right) (2t') =0
\end{align}$$
My textbook, however, says without comment that the answer is $\frac{F}{m}(t)^2$, or in my notation, $\frac{F}{m}(t')^2$. Am I wrong, or is the textbook wrong, or could this be an ambiguity in wording—e.g. is there a way to construe "total distance the body travels" as "the furthest position the body reaches from its starting point"? Or could it be that the question is asking not about the object's location at time $t = 2t'$ but about its location as a function of time $s(t)$? In that case my answer is found above and would still be wrong.
| The total distance is a sum of two distances: $S_{1}$ and $S_{2}$, where the first is till the velocity goes to zero, the second: from zero to $-V_{0}$. $S_{1}=V_{0}t+\frac{1}{2}at^{2}$. Here $a=\frac{-V_{0}}{t}$. So $S_{1}=V_{0}t-\frac{V_{0}t}{2}=\frac{V_{0}t}{2}$.
The same for $S_{2}$ but with zero initial speed and positive acceleration: $S_{2}=\frac{1}{2}at^{2}=\frac{1}{2}\frac{V_{0}}{t}t^{2}=\frac{V_{0}t}{2}$
Total $S=S_{1}+S_{2}=V_{0}t=\frac{F}{m}t*t=\frac{F}{m}t^{2}$.
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Is the potential energy stored in a spring proportional to the displacement or the square of it? Suppose a mass of $M$ kg is hanging from a spring in earth. The mass will stretch the spring about $x$ m. So the change in the gravitational potential energy is $mgx$ J (supposing $x$ to be very small compared to the radius of earth).
And this amount of energy will be stored in the spring as potential energy. So,
Change of gavitational energy = $mgx$ = potential energy stored in the spring
And it seems that the potential energy stored in a spring is proportional to displacement $x$. But the potential energy in a spring is $U=\frac{1}{2}kx^{2}$ and so it's proportional to $x^2$, the square of displacement. So surely I am wrong somewhere. But where am I wrong?
| Note: The potential energy stored in a spring is proportional to the square of the displacement from equilibrium.
When you attach a mass to an unstretched spring, there will be a new equilibrium position for that mass on that now-stretched spring. About this new equilibrium position, you will have simple harmonic motion.
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How does cutting a spring increase spring constant? I know that on cutting a spring into n equal pieces, spring constant becomes n times.
But I have no idea why this happens.
Please clarify the reasons
| For a given deformation, the distance change between two adjacent particles (molecules/atoms) is more if you decrease the length of the spring. Thus, if you keep the displacement small enough so as the intermolecular force is linearly proportional to the intermolecular distance, the force required to produce the same deformation in a short spring is more compared to a longer spring. If you cut a spring into $n$ pieces, the distance change between two particles would have to be $n$ times more to keep the total deformation the same, and linearity tells us that the force required will be $n$ times greater.
Note that the above model may fail as in a real metal spring, there are grain boundaries, dislocations, etc. But it is presents a good intutive picture.
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Would a perfectly collimated laser beam have a flat intensity distribution? I'm trying to simulate a collimated laser beam in some FORTRAN and am wondering about the intensity distribution of a perfectly collimated beam: would it be a flat distribution (equal intensity across the beam) or it would be more of a gaussian? I know that a perfectly collimated beam wouldn't disperse, but would this translate to equal intensity across the beam?
| "Top hat" optics can provide a nearly flat intensity distribution across a beam. A Gaussian beam is nearly flat near its center, so simply passing it through a circular aperture yields a nearly flat beam. A beam that is slightly nonuniform can be corrected by passing it through a customized gray-level filter to reduce intensity in the high-intensity regions. In practice, you can create a beam with intensity pattern and/or phase pattern you want, using a combination of phase and intensity spatial light modulators.
| {
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Does the centre of charge behave as a point charge? We can find the centre of charge of a system of charges in much the same way we find the centre of mass of a system of masses
Suppose we have two charges of the same polarity and equal in magnitude that are placed at -x and +x along the x axis. The centre of charge would be at the origin.
Would this centre of charge, having magnitude twice as either of the individual charges, behave as the two other charges would combinedly? Like how gravity can be assumed to apply on the centre of mass?
If so, then can we say that for objects with a constant charge density that their charge can be assumed to be concentrated at their geometric centres?
If not, then what is the significance in defining a quantity as centre of charge?
| No. For a simple intuitive example take a long rod containing a charge (uniformly distributed throughout its length). If you place a test charge a small one that does not adversely affect the charge configuration on rod, it will obviously suffer more attraction/repulsion from the elemental charge closer to it than from one away from it.
So, as you can observe, the behaviour of elemental charges charges about the centre of charge, is not symmetric, and where there is no symmetry, such generalisations are not possible.
And as Mr. Harish Chandra Rajpoot said, the centre of charge can explain behaviour of an extended body in an electric field. Besides that, there is not much significance for the quantity
| {
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What is the range of Pauli's exclusion principle？ In many introductions to the pauli's exclusion principle, it only said that two identical fermions cannot be in the same quantum state, but it seems that there is no explanation of the range of those two fermions. What is the scope of application of the principle of exclusion? Can it be all electrons in an atom, or can it be electrons in a whole conductor, or can it be a larger range?
| In principle it covers all Fermions in the Universe. Not two Fermions share the same quantum numbers. In a material with many moles of electrons each one of them has different values of energy level, etc. Of course, you have to consider, for example, that two electrons with the same n, l, m and spin numbers orbit two identical nuclei. They have, however, different quantum numbers since given a reference frame and the description of the system by some rather complicated quantum state vector, they would differ in their quantum numbers. The same applies for more complicated systems. So, final example, fermions in a collapsing star resist the collapse due to Pauli's exclusion principle even though they are in a huge system with not a very nicely defined quantum state vector.
| {
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Approximation from discrete Kronecker Delta to continuum Dirac Delta I am working on second quantization of the Dirac field with discrete momentum I was asked to compute the creation/annihilation anticommutator by imposing the anticommutators on $\psi$ i.e.
$$ \{\psi_a(\vec{x}),\psi^{\dagger}_b(\vec{y})\} = \delta^{(3)}(\vec{x}-\vec{y})\delta_{ab}$$
I start with:
$$\psi(\vec{x}) = \sum_{r,\vec{k}} \sqrt{\dfrac{m}{VE_{\vec{k}}}}\bigg[ c_k(\vec{k})\,u_r(\vec{k}) \,e^{-i x \cdot p} + d^{\dagger}_r(\vec{k})\,v_r(\vec{k})\, e^{ix \cdot p} \bigg] $$
Since I am trying to isolate $c$ and $d$ I try to multiply for the expoencial $e^{ix\cdot p'}$ and integrate over the volume, and I will apear an integral like this
$$\iiint_V e^{-i \,x \cdot (p'-p)} \,d^3x$$
Is there any way approximate this? Like doing:
$$\iiint_V e^{-i \,x \cdot (p'-p)} \,d^3x \approx V\delta_{p,\,p'}$$
|
Cited from Mahan's book. It may help you.
| {
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How to model a quantum circuit Let's say we have a system of 2 qubits, which are entangled in an unknown Bell basis configuration. Since the qubits are in a Bell configuration, each state is orthogonal to every other state, and thus must be distinguishable from each other.
My understanding is that there are 4 measurement operators (see below) that can be simultaneously applied to a single qubit (because they commute). And this is why we can unambiguously identify which Bell state the qubits are in, even if Bob and Alice measure their respective qubits with the 4 measurement operators, individually and concurrently with each other.
Question
How does one come up with a quantum circuit for something like this? How do we go about mapping an arbitrary measurement to one of the well known quantum gates?
On a related note, let's say I have an arbitrary unitary matrix. How does one map that to a quantum gate?
Bell configuration:
*
*$\vert{T_1}\rangle = \frac{1}{\sqrt{2}} (\vert{10}\rangle - \vert{01}\rangle)$
*$\vert{T_2}\rangle = \frac{1}{\sqrt{2}} (\vert{10}\rangle + \vert{01}\rangle)$
*$\vert{T_3}\rangle = \frac{1}{\sqrt{2}} (\vert{00}\rangle + \vert{11}\rangle)$
*$\vert{T_4}\rangle = \frac{1}{\sqrt{2}} (\vert{00}\rangle - \vert{11}\rangle)$
Distinguishability:
*
*$\langle T_i \vert T_j \rangle = \delta_{i,j}$
Measurement operators:
*
*$M_{T_i} = \vert{T_i}\rangle\langle{T_i}\vert$
*Commutation: $[M_{T_i}, M_{T_j}] = 0$
| Measurement is usually just defined as a gate on it's own. The "well known gate" that measurement maps to is simply the measurement gate. Its one of the few places where some difficult, hard to model interactions with an environment are allowed in a quantum circuit, so we just separate that part out.
For example, in this circuit
You can clearly see two measurement gates. The output of said gates is shown to be classical information (wires with two lines), and used later to determine whether or not a gate will be applied.
With that said, you can still simplify the measurement process. For your problem, if you apply a CNOT gate and a Hadamard gate, a Bell state will map onto a state in the 2-qubit computational basis, which turns an arbitrary measurement into a fairly standard one.
| {
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Is 1 Joule the work done to lift ~100g through a distance of 1m? I am seeing many videos saying that 1 Joule is the work done to lift ~100g through a distance of 1m (like this one https://youtu.be/BYpZSdSEk4A?t=348).
The idea is that 100g has a gravitational force downward of 1 Newton. So lifting it means applying 1 Newton over 1m, and therefore is 1 Joule.
But if I apply 1 Newton upward, and gravitational force is 1 Newton downward, the object shouldn't move.
So I believe you need to apply a force greater that 1 Newton in order to lift it.
Or am I missing something?
Would it be more accurate to say that 1 Joule is the work done by gravity to make fall 100g by 1 meter?
| In your example of lifting an object, if the upward external force is exactly the same as the weight in magnitude, then the object is still in perfect equilibrium. And since the initial velocity of it was zero, its velocity would still remain zero because equilibrium means no acceleration. So, there would be no movement. So, in order to actually lift the object, you do need to provide an upward force which is at least slightly greater than the weight of the object.
| {
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Ideal gas temperature definition I have been doing some statistical mechanics and at the beginning of the course, I had seen this statement:
$$T^\circ (K) = 273.16\frac{\lim_{V\rightarrow \infty} (PV)_{\text{system}}}{\lim_{V\rightarrow \infty} (PV)_{\text{triple point of water}}}$$
Where, $T^{0}$ is the ideal gas temperature in Kelvin, $P$ is the pressure, and $V$ is the molar volume.
I don't understand this definition of the temperature of a gas in the ideal regime of $V\to \infty$, and why this is considered a rigorous definition of temperature of an ideal gas.
My question is, what is the inspiration behind such a definition of ideal temperature? Why does the triple point of water come into the picture here? Is this trying to tell us that the term $$\frac{\lim_{V\to \infty} (PV)_{\text{triple point of water}}}{273.16} = R$$ Where $R$ is the ideal gas constant? If yes, then how?
| The limit $V \to +\infty$ comes from the fact that for a given number of moles/gas particles, effects of interactions go down as the volume increases, because molecules are less and less likely to run into each other. This means that $\lim_{V \to \infty} (PV)$ can be deduced from the $PV$ of the equivalent ideal gas. But for an ideal gas, $PV \propto T$. So all things put together, $\lim_{V \to \infty} (PV)_{\mathrm{system}} \propto T_{\mathrm{system}}$.
The other terms just correspond to the definition of temperature in Kelvins. By (historical) convention, we have decided that the so-called triple point of water (which correspond to the unique values of $P$ and $T$ such that all three phases of water exist in equilibrium at the same time) was $T = 273.16\,\mathrm{K}$ (the pressure is not relevant here). So by convention/definition, if the system sits at the same temperature as the triple point of water, its temperature is $T_{\mathrm{system}} = 273.16\,\mathrm{K}$. You can check indeed in your formula that because the numerator and the denominator cancel when the system is at the triple point of water, the formula yields $T = 273.16\,\mathrm{K}$.
| {
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electrostatics- work done related question A point charge $q$ is at the center of an uncharged spherical conducting shell of inner radius $a$ and outer radius $b$. How much work would it take to move the charge out to infinity (through a tiny hole drilled in the shell)? I could not understand the question. Please help me solve this out.
| You just need to find out the potential at the center of the shell. Work required to take the charge to infinity is simply negative of that potential assuming zero potential at infinity.
The shell is made of conductive material which means all the charge will accumulate at surface. If you imagine a spherical gaussian surface of radius $R$ such that $$a<R<b$$
Flux through this surface is zero because electric field inside a conductor is zero. This implies that net charge inside the region enclosed by the gaussian surface is zero. So, charge accumulated at the inner surface of the conductor is $-q$ and by conservation of charge for conductive sphere, charge accumulated on outer surface is $+q$
I think I have simplified the problem enough for you to be able to solve it. You just have to calculate potential due to the two surfaces, you can assume these surfaces to be thin shells with uniform charge. If you don't know about gauss's law or conductor's properties, then I suggest that you go through the theory before attempting problems.
| {
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Why does the ideal gas law exactly match the van't Hoff law for osmotic pressure? The van't Hoff law for osmotic pressure $\Pi$ is
$$\Pi V=nRT$$
which looks similar to the ideal gas law
$$PV = nRT.$$
Why is this? Also, in biology textbooks, the van't Hoff law is usually instead written as
$$\Pi=CRT =\frac{NC_m RT}M$$
where $C_m$ is the mass concentration, $N$ the number of ions, and $R$ the ideal gas constant. Why?
| I'm answering your 2nd question:
It's a really easy proof :
You have $C_m=\frac{m}V$ and $C=\frac{n}V$ where $n=\frac{m}M$, thus $C=\frac{C_m}M$
Therefore :
$$\Pi=\frac{C_m R T}{M}$$
For $N$ the ions number it's related to biology, the ions are interacting and they cause this pressure, For example if it's $\mathrm{NaCl}$ the ions are $N=2$
Thus Van't hoff law is written in Biology and Biophysics textbooks :
$$\Pi =\frac{NC_mRT}M$$
| {
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Independent Elements of Elastic Stiffness and Compliance Tensor for ALL Space Groups In short: Does anybody know if there exists a compendium, a document, a book or a stone tablet listing the independent elements of the elastic stiffness and compliance tensors ( that is, naming the elements, e.g., $C_{1111},C_{2222},\ldots$ that are independent and giving the values of the dependent elements as functions of the independent ones ) for all space groups? Thanks very much!!
In long: Recently I was doing some calculations that involved handling the elastic stiffness tensor from elasticity theory, namely $C_{ijkl}$, for materials with quite specific symmetries. As you probably know, $C_{ijkl}$ has, in general, 21 independent elements. But, depending on the symmetries of the crystal you are regarding, this can change dramatically, down to 2 independent elements for isotropic materials.
I thought to my-self: Well, there is only a finite number of space groups out there, so somebody should have one day calculated the independent elements of $C_{ijkl}$ for all space groups that exist. And, while we are at it, also for its inverse, the compliance tensor $S_{ijkl}$. But I was not succesfull in finding such a compendium anywhere and had to do the calculation on my own. Poor me. Perhaps we can help future generations with that information?
| You might want to have a look at the figure at the bottom of:
https://serc.carleton.edu/NAGTWorkshops/mineralogy/mineral_physics/tensors.html
this also suggests that tetragonal cases can be split into two subgroups with slightly different symmetries.
| {
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Does a fan rotating with a uniform angular velocity consume electrical energy? Work done on a rotating body is equal to the change in its kinetic energy. When an electric fan rotates with a constant angular velocity, then its kinetic energy doesn't change. Does it mean that it doesn't consume electrical energy?
| As stated in the other answers, it is true that a fan rotating with a uniform angular velocity consumes electric energy due to the presence of energy dissipation. But it's not only due to the energy transferred to the air molecules (as others state as "air drag"), but also due to other factors like - friction in the bearing, Joule heating and electromagnetic damping in the motor's coil. Electromagnetic damping also has some useful applications (see Eddy current brake).
It must be noted that when we turn off the fan, electromagnetic damping is present only when the AC motor has a permanent magnet. If the magnetic field for the rotation of the shaft is produced by loops of wires instead of a permanent magnet, it would also become zero. In this case the fan is brought to rest solely by friction and loss of energy to air molecules.
| {
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Can the total momentum in the centre of mass frame be non-zero? I have a question regarding the usage of the centre-of-mass definition (which I thought required the total momentum, $p_T=0$) in the solution to the question given below:
A particular centre-of-mass energy is needed to create a new particle. We will do the calculation in a so-called fixed-target configuration.
A particle of mass $m_1$ and total energy $E_1$ in the lab frame hits a stationary particle of mass $m_2$. Show that the required particle energy for a given $s$ is:
$$E_1=\frac{s-m_1^2c^4-m_2^2c^4}{2m_2c^2}$$
where $s$ is the square of the centre-of-mass energy. This is often called a ‘fixed target’ configuration as experiments were historically often done by colliding a beam of particles with a stationary target material.
The solution says:
In the fixed target experiment, the total energy $E_T=E_1+m_2c^2$ and the total momentum magnitude is $p_Tc=\sqrt{E_1^2-m_1^2c^4}$. The square of the centre-of-mass energy
is, therefore,
$$s=m_T^2c^4=E_T^2-p_T^2c^2=E_1^2+2E_1m_2c^2+m_2^2c^4-E_1^2+m_1^2c^4$$
$$=2E_1m_2c^2+m_1^2c^4+m_2^2c^4$$
Rearranging, gives the required particle energy:
$$E_1=\frac{s-m_1^2c^4-m_2^2c^4}{2m_2c^2}$$
But, in the centre-of-mass frame $p_T=0$, as shown in this image* below:
So therefore,
$$p_T=\sqrt{E_1^2-m_1^2c^4}=0$$
and as a result,
$$E_1=m_1c^2$$
My question is, why is the author using the centre-of-mass definition (which I thought required zero total momentum) when the total momentum is actually non-zero (only one particle is stationary)?
*Image in body is from ICL dept. of Physics
| The author explicitly states
We will do the calculation in a so-called fixed-target configuration.
So the calculation is done in what is usually called the laboratory frame, and calls it fixed-target in order to emphasize that the target is at rest in this frame.
Because of Lorenz transofrmations, any inertial frame can be used as long as energy and momentum are conserved in the given frame.
The center of mass is used only in order to define s
where s is the square of the centre-of-mass energy.
The calculation is in the lab. s is the "length" of the sum of the four vectors describing the particles involved, for any inertial frame, this is invariant for all inertial frames.
| {
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Confusion about an example of inertia related to bus When the bus is stationary the passengers are also stationary. When the bus starts moving the part of the body (lower part) in contact with bus starts moving, but due to inertia the upper part remains stationary and thus he gets reclined back. If he is standing he will fall backward.
I cannot understand that how lower part starts moving while upper part remains stationary? My understanding is that our complete body is at rest - rather an unbalanced force acts on us in backward direction when bus starts?
Can anyone explain the scenario?
| The bus floor makes an horizontal force on the feet of the people. As the point of application of the force is at some distance from the centre of mass, there is a torque.
To an inertial frame outside the bus, people inside are not falling backwards, they are tilting, what is the consequence of a torque.
| {
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Are Dirac and Wu-Yang monopoles the same thing? When reading about the Wu-Yang monopole, it seems to me that the method that Wu & Young used (using patches and gauge transformations to cover $S^2$) is more systematic and general than Dirac's method for the Dirac monopole analysis (and for reaching the Dirac quantisation condition), but it overall seems to me that the two methods describe the same thing: the Dirac monopole and the same quantisation condition.
Is there a difference in the physical results? Also, in what way is the Wu-Young method more general? For example, I know that the gauge group involved here is $U(1)$ (because of the electromagnetic field), so how does the method generalize to other groups? I have seen the derivation of the 't Hooft-Polyakov monopole, which has to do with $SU(2)$, but the method used there has nothing to do with the Wu-Yang method and the end result is a non-singular monopole.
| *
*On one hand, Dirac-type and Wu-Yang-type magnetic monopoles usually refer to different mathematical descriptions of the same$^1$ underlying class of physical phenomenon in various space dimensions and with various gauge group:
*
*A Dirac-type monopole uses singular Dirac-strings/delta-distributions and globally defined singular gauge fields on $\mathbb{R}^d$.
*The Wu-Yang bundle construction avoids singularities by considering a punctured space $\mathbb{R}^d\backslash \{0\}$ and locally defined gauge potentials.
*On the other hand, a 't Hooft-Polyakov-type monopole is a globally defined, regular, classical field configuration on $\mathbb{R}^d$, cf. my Phys.SE answer here.
--
$^1$ However, some authors call the $U(1)$-monopole in 3D for a Dirac monopole and the $SU(2)$-monopole in 5D for a Yang monopole, etc.
| {
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Why is electric flux through a cube the same as electric flux through a spherical shell? If a point charge $q$ is placed inside a cube (at the center), the electric flux comes out to be $q/\varepsilon_0$, which is same as that if the charge $q$ was placed at the center of a spherical shell.
The area vector for each infinitesimal area of the shell is parallel to the electric field vector, arising from the point charge, which makes the cosine of the dot product unity, which is understandable. But for the cube, the electric field vector is parallel to the area vector (of one face) at one point only, i.e., as we move away from centre of the face, the angle between area vector and electric field vector changes, i.e., they are no more parallel, still the flux remains the same?
To be precise, I guess, I am having some doubt about the angles between the electric field vector and the area vector for the cube.
| From Gauss's law
$$\int\vec{E}.d\vec{s}=\frac{q_{in}}{\epsilon_{0}}$$
So the flux through both of the surfaces would be same as the charge inside both of the surfaces is same.
If we approach the problem through integral, you mislooked the angle between area vector and electric field in the case of cube.
| {
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Can we have pressure with zero net force on a 2d plane? From Wikipedia:
Pressure is the force applied perpendicular to the surface of an object per unit area over which that force is distributed.
Suppose we apply 2 equal and opposite forces on a 2d plane with area 'A' perpendicular to it ($F_a$ & $F_b = F$)
Will we say that pressure is $0$ or $\frac{2F}{A}$
Suppose we have static fluid in a container, then force applied by water on top of a cross-section (because of its weight) is equal the force being applied from the downside as it is a static fluid
Now as in the example of the plane if the answer is that pressure is 0 then the fluid at some depth also has equal and opposite forces then pressure even at some depth should be zero
And if it is $\frac{2F}{A}$ then the pressure should be multiplied by 2 which is not done in the derivation of "variation of pressure vertically with depth"
| id you excert a force in one direction , you always have the opposite force it is called actio= reaction, otherwise , with only one force you accelerate your plane in the direction of the force.
as picture: you standing on the floor, exert a pressure of your weight divided by the area of your feet or shoes, but the floor presses against your feet.
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Selections rules for spin what do we mean by the selection rule $\Delta S=0$?
Can you give me some example for hydrogen atom?
For example if I want to go from $1s$ to $2p$ how can I calculate $S$ for $1s$ or for $2p$?
| This question has already been asked in different forms. [Refer : 1) Selection rule ΔS=0: Why does a photon not interact with an electrons spin? 2) Does a photon interact with the spin of an electron?].
I will try to give you a summary of the very good points brought up. $\Delta S = 0$ actually is not a binding rule. Basically, in the semi-classical model of light where you consider oscillating electric and magnetic dipoles, only the magnetic dipoles can change the spin of the electron. This process is called 'spin-flip'. It is the same mechanism for the famous Hydrogen 21 cm astronomical limit. This is a well known case of spin flip and where the selection rule actually breaks down. In fact, spin-flip is very possible and observed in PL spectra of many materials - although, you would need a high intensity pulse (of a very short duration) because magnetic dipole interactions are very weak compared to electric dipole interactions (because in the former, the velocity is restricted by the speed of light - Read 'Fine structure constant in Hydrogen').
Therefore, in optical processes, you typically consider only electric dipole interactions which are stronger in strength, and in this case, as you can see, the spin selection rule $\Delta S = 0$ is very much true.
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Why is there an electric potential drop in electric circuits? I know a battery creates a potential difference, making an electric field that exerts a force on the electrons, who start moving. But why is there a potential drop after a resistor for example? How does it go in hand with electric potential being a scalar assigned to a point in space? How can a resistor change the potential of all the points in the conductor succeeding him? Or am I looking at it in a wrong way? I think I'm considering electric potential from an electrostatics point of view and it gets me nowhere.
| The relations between currents, electric and magnetic fields in a circuit have to follow Maxwell equations. There is nothing there about resistances.
So it is possible to have a current in a closed circuit without any eletric field, (so no potential drop) since the relation between $\mathbf B$ and $\mathbf E$ are fulfilled. It is the case of superconducting.
So it is the other way around. If there is an electric potential drop (so an electric field) between two point of a circuit where a current is present, there is some resistance there. It is a definition of resistance. $$R = \frac {\Delta V}{I}$$
| {
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Does This System Of Electric Charges Violate Conservation Of Energy? Imagine a system of 3 point charges and some kind of circular "rail".
All charges have the same magnitude - 2 are positive and one negative.
One positive charge is placed at some distance from the center of the circular rail.
The two other, are "glued" together so they must move together, and they are placed on the rail which
force them to move only on the rail path.
Now imagine i give them a slight push.
Calling the glued charges the "system", We can say that the force between the system and the positive charge outside the rail is both attractive and repulsive (depending on what part of the rail the system is) but we can identify that the Parallel component of the force (with respect to the velocity vector) will always be in the direction of motion.
Thinking it through, i think you will agree that the charges on the rail will accelerate every 1 period -
therefore "generating" energy !
Can you spot what's "wrong" here ?
In this figure i used rectangular charges just to make it more clear, The "Paradox" is of course about point charges.
| In the bottom position, if there is some space between the two charges the positive charge at the bottom will push the positive charge away and attract the negative charge. This tends to push the two charges clockwise.
At the top the negative charge will be pulled towards fixed positive charge, while the positive charge will be repelled, so the force is anticlockwise. The force at the sides is weak because there is a force on both charges, but one there will be a greater force on the charge closer to the fixed charge. Again the force is anticlockwise.
So the moving charges will be speeding up at the top and sides, but decelerating at the bottom. I expect that they would all balance out over a full cycle.
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Is the average force calculated from $F(x)$ the same as that calculated from $F(t)$? Say a force is doing work on an object in one dimension. I could calculate the average force over the distance with
$$\frac{1}{\Delta{x}}\int_{x_1}^{x_2} F(x) \text dx$$
If I also formulated force as a function of $t$, I could calculate the average force over the total time period with
$$ \frac{1}{\Delta{t}} \left| \int_{t_1}^{t_2} F(t) \text dt \right|$$
Obviously, the integrals themselves are not equal. One represents work, and the other represents impulse; $ \Delta{p} \neq \Delta{K} $.
However, if $x_1$ corresponds to $t_1$ and $x_2$ corresponds to $t_2$, could I set the above expressions equal to one another? Does averaging over the interval make them equal? Put another way, I've never seen texts differentiate between the $F_{ave}$ in $W = F_{ave} \cdot \Delta{x}$ and the $F_{ave}$ in $\Delta{p} = F_{ave} \Delta{t}$. Is there a difference? If not, I'm thinking they can relate work to $\Delta{t}$.
| The answer is no, they are not the same things.
There is a tacit assumption when texts say things like "the average force," about what the average is performed over. Averaging over position is not the same as averaging over time.
You can always construct simple examples in which it just so happens that they are the same, but that is of limited utility.
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How does focal length affect magnification? My answer would be the longer the focal length, higher the magnification will be, resulting in a larger image. But in a ray diagram, how does it look? I am searching for a comparison of ray diagram between short focal length and long focal length but didn't manage to get anything. My high school textbook didn't explain much about focal length also. I wanted to know more about how does different focal length affect the image in a telescope as well. But Google shows only the simple ray diagrams. Thanks in advance.
| If you are using the lens as a magnifying glass the standard formula $\frac{1}{u}-\frac{1}{v}=\frac{1}{f}$, where $u$ and $v$ are the distances from the lens to the object and image respectively and $f$ is the focal length, together with the formula $m=\frac{v}{u}$, show that in theory you can get any desired magnification from any converging lens. In practice, to achieve a high magnification the object needs to be close to the focal point of the lens. This makes the image larger, but also further away. For high magnification with the image being fairly close you need a short focal length.
For a telescope the magnification is (objective focal length) / (eyepiece focal length). But for a good quality image neither can be too short.
Standard ray tracing diagrams are a better approximation for longer focal lengths, because they assume a lens/mirror is flat, but still works like a curved mirror/lens.
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Acceleration in photoelectric effect In all the books I read, only the initial speed with which an electron is removed from a plate is mentioned, what happens to the acceleration?
Consider a plate (charge neutral and not connected to anything) and light of appropriate wavelength hits the plate
Suppose that an electron is removed from the plate, it leaves a hole (a positive charge) in the plate, therefore the removed electron will feel a force of attraction with the plate and therefore there is an acceleration
I am right?
If n electrons are removed, then there are n holes, therefore the acceleration remains constant for n electrons?
| After being ejected from the material, the electron is treated as a free particle, therefore acceleration is zero.
The metallic plate is macroscopic, namely it has $10^{23}$ and more electrons. Then, for all purposes, it remains neutral. If you want, you can think that it is connected to ground. Then, the hole which is left by the ejected electron is immediately filled.
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De Broglie relationship What wavelength De Broglie relationship represents, if particle's motion is given by its group velocity (which is superposition of waves of many wavelengths)
| Assuming the electron is being modelled as a wavepacket, then there is no single momentum given by the de Broglie relation. We know this from the Heisenberg uncertainty principle:
$$
\Delta x \Delta p\geq\frac{\hbar}{2}
$$
Which tells that there is uncertainty in the position and the momentum of the wavepacket as can be seen in the diagram below:
Where the width of the wavepacket is directly related to the uncertainty of the position (on the left) and to the uncertainty of the momentum (on the right).
The momentum given by the group velocity ($p=mv_g$) is the momentum of the entire wavepacket and does not correspond to any meaningful wavelength.
| {
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What are some areas of physics, where the concept of "natural integral" may arise? Natural integral (as we will define it) is a distinguished antiderivative of a function that can be understood as interpolation of the sequence of consecutive derivatives to the $-1$. It has a naturally defined integration constant. While it is possible to define natural integral in various ways, it all boils down to the following property:
$$f^{(-1)}(x)=\int_0^x f(t) \, dt+\frac{1}{2} \left(\int_{-\infty }^0 f(t) \, dt-\int_0^{\infty } f(t) \, dt\right),$$
where the integrals in the brackets should be understood in the sense of regularization, if they diverge. The $0$ is not important here and can be replaced by any point, this will not affect the answer.
That said, I wonder, if such natural antiderivative ever appears in physical applications?
UPDATE
As @Qmechanic pointed out, it appears in many areas of physics. I prefer examples from classical mechanics or elementary quantum theory.
| The natural antiderivative $$f^{(-1)}(x)~:=~\frac{1}{2}\int_{\mathbb{R}}\!\mathrm{d}x^{\prime} ~{\rm sgn}(x\!-\!x^{\prime})~f(x^{\prime})$$ is the most symmetric choice of integration constant. The kernel ${\rm sgn}(x\!-\!x^{\prime})$ appears all over physics. It is the Fourier transform of ${\rm PV}\frac{1}{k}$ up to a multiplicative constant. Examples:
*
*The symplectic potential $\frac{1}{2}z^I\omega_{IJ}\dot{z}^J$ in the Lagrangian leads to the Feynman propagator/Greens function $$G^{IJ}_F(t\!-\!t^{\prime})~=~\frac{1}{2}\omega^{IJ} {\rm sgn}(t\!-\!t^{\prime}).$$
See also my related Phys.SE answer here.
*The Poisson commutation relations $\{\phi(x),\phi(y)\}=\frac{1}{2}{\rm sgn}(x\!-\!x^{\prime})$ for a self-dual boson field becomes $$\{\Phi[f],\Phi[g]\}=\frac{1}{2}\int_{\mathbb{R}}\!\mathrm{d}x \int_{\mathbb{R}}\!\mathrm{d}x^{\prime}~{\rm sgn}(x\!-\!x^{\prime})f(x) g(x^{\prime})$$
in terms of test functions $f,g$.
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Since computer screens can't display many colors, where can I go to see the full color gamut? I found out recently that computer screens are only able to display a subset of all the colors visible to the human eye. Naturally one of my first questions was what do the other colors look like, but this is one of the few questions that I cannot Google. I have been able to find a few objects around my house that I'm pretty sure have colors that my screen can't display, but I'd like to see the full gamut. I will have to find it IRL, so my question is where can I find it? Is my local library likely to have a picture? I live in Boston so I was thinking I could try the Science Museum or the MIT Museum, but neither seems to have an exhibit on color.
| The colors you can see form (the positive orthant of) a three-dimensional real vector space. (A color is defined by three real numbers, describing the intensity of stimulation of the three types of cones.) A computer can display a finite number of colors. So you'll never get close to displaying the "entire gamut".
On the other hand, you could certainly ask for a computer that can display a set of colors that are "dense" in the sense that for any given color X, there exists a displayable color Y with the property that you can't subjectively notice the difference betweeen X and Y. How big that set of colors would have to be --- and whether it is greater or less than the number that your computer already displays --- seems like more of a biology question than a physics question.
| {
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Evaluating $\sigma^{\mu\nu}F_{\mu\nu}=i\alpha \cdot E+\Sigma\cdot B$ matrix, spin dependent term in quadratic Dirac equation I derive the quadratic form of Dirac equation as follows
$$\lbrace[i\not \partial-e\not A]^2-m^2\rbrace\psi=\lbrace\left( i\partial-e A\right)^2 + \frac{1}{2i} \sigma^{\mu\nu}F_{\mu \nu}-m^2\rbrace\psi=0$$
And I need to find the form of the spin dependent term to get the final expression
$$g \frac{e}{2} \frac{\sigma^{\mu\nu}}{2}F_{\mu \nu}=-g\frac{e}{2}\left(i\vec{\alpha}\cdot\mathbf{E}+\vec{\Sigma}\cdot\mathbf{B}\right)$$
But I don't get this expression.
I'm using the Dirac representation with these quantities
$$\vec{\alpha}=\begin{pmatrix}
0 & \vec{\sigma}\\
\vec{\sigma} & 0
\end{pmatrix} \ \ \ \ \ \vec{\Sigma}=\begin{pmatrix}
\vec{\sigma}& 0\\
0&\vec{\sigma}
\end{pmatrix}$$
Where $\vec{\sigma}=(\sigma_x,\sigma_y,\sigma_z)$ is the Pauli matrix vector.
I constructed the electromagnetic tensor term by term, using the definition $F_{\mu\nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}$ with the metric tensor $g^{\mu\nu}=\textrm{diag}(+1,-1,-1,-1)$ and I get
$$F_{\mu\nu}=\begin{pmatrix}
0 & E_x&E_y&E_z\\
-E_x&0&B_z & -B_y\\
-E_y&-B_z&0&B_x\\
-E_z&B_y&-B_x&0
\end{pmatrix}$$
I evaluate the $\sigma^{\mu\nu}$ matrix starting from its definition in terms of gamma matrices $\sigma^{\mu\nu}=\frac{i}{2}\left[\gamma^\mu,\gamma^\nu\right]$
$$\sigma^{00}=\frac{i}{2}[\gamma^0,\gamma^0]=0$$
$$\sigma^{0i}=\frac{i}{2}[\gamma^0,\gamma^i]=\frac{i}{2}[\gamma^0,\gamma^0\alpha_i]=\frac{i}{2}[\alpha_i-\gamma^0\alpha_i\gamma^0]=\frac{i}{2}2\alpha_i=i\alpha_i$$
$$\sigma^{ij}=\frac{i}{2}[\gamma^i,\gamma^j]=[\gamma^0\alpha_i,\gamma^0\alpha_j]=\frac{i}{2}\gamma^0(\alpha_i\gamma^0\alpha_j-\alpha_j\gamma^0\alpha_i)=\frac{i}{2}
\begin{pmatrix}
-[\sigma_i,\sigma_j] &0\\
0&-[\sigma_i,\sigma_j]
\end{pmatrix}=\epsilon_{ijk}\begin{pmatrix}
\sigma_k &0\\
0&\sigma_k
\end{pmatrix}=\epsilon_{ijk}\Sigma_k$$
And the remaining terms follow by the antisymmetry property $\sigma^{\mu\nu}=-\sigma^{\nu\mu}$
$$\sigma^{\mu\nu}=\begin{pmatrix}
0 & 2\alpha_x & 2\alpha_y & 2\alpha_z\\
-2\alpha_x&0&\Sigma_z & -\Sigma_y\\
-2\alpha_x&-\Sigma_z&0&\Sigma_x\\
-2\alpha_x&\Sigma_y&-\Sigma_x&0
\end{pmatrix}$$
Now, my questions are:
"Why these calculations do not yield the correct result?"
"What I should do to obtain the correct result? What I'm missing?"
$$\frac{\sigma^{\mu\nu}}{2}F_{\mu \nu}=-\left(i\vec{\alpha}\cdot\mathbf{E}+\vec{\sigma}\cdot\mathbf{B}\right)$$
| You did not quite explain how you failed to obtain the target result.
I would not like to spoil the fun of catching your factors and signs involved, so I will strictly deal with significant proportionalities.
$$ \sigma^{\mu\nu} F_{\mu \nu}= \sigma^{0i} F_{0 i}+\sigma^{i0} F_{i 0}+ \sigma^{ij} F_{ij}=2\sigma^{0i} F_{0 i} + \sigma^{ij} F_{ij} . $$
Now,
$$
\sigma^{0i} F_{0 i} \propto \alpha_i E_i,
$$
and
$$
\sigma^{ij} F_{ij} \propto \epsilon_{ijk}\Sigma_k ~~ \epsilon_{ijm} B_m =2 \Sigma_k B_k,
$$
by virtue of the 2-index Levi-Civita contraction identity.
Proceed to fix numerical normalizations, if needs be by assuming sparse special constant EM fields.
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Intrinsic carrier concentration and bandgap My understanding is that the intrinsic carrier concentration of a wide bandgap material tends to be lower than that of a narrow bandgap material.
$$ n_i = \left(N_cN_v\right)^{1/2}e^{\left(\frac{-E_g}{2kT}\right)} $$
and the open circuit voltage of a solar cell is
$$ V_{OC} = \frac{kT}{e}ln\left(\frac{J_{SC}}{J_o}\right) $$
where
$$ J_o = e\left(\frac{D}{L}\right)\left(\frac{n_i^2}{N}\right) $$ multiplied by recombination loss factors.
This makes sense to me since greater bandgap yields greater intrinsic carrier concentration, and greater intrinsic carrier concentration yields greater $ J_o $ and greater open circuit voltage.
However, I recently learned that a wide bandgap material, AlGaInP (Aluminium gallium indium phosphide), has a pretty high intrinsic carrier concentration due to its aluminum content.
How can I understand this situation where you have a wide bandgap AND high intrinsic carrier concentration??
| The other terms in the equation for $n_i$ also matter. $N_c$ and $N_v$ depend on the effective mass. In particular, they are proportional to the effective mass to the three halves power. All things being equal, a higher band gap means a higher effective mass:**
This is something you can derive from k.p theory with a little hand waving. See for example this and this. (Sorry that I don't have a better reference handy.)
So, the equation for $n_i$ has two forces that act in different directions as $E_g$ increases: the exponential gets smaller, but $N_c N_v$ gets larger. Normally, the exponential wins. But evidently AlGaInP bucks the trend; the effective mass must be high enough that it overpowers the exponential.
Offhand, I don't know why the aluminum would have this effect, but maybe that's obvious to someone with more semiconductor knowledge than myself. Still, the takeaway is that the band gap acts in multiple ways, so it's not always the case that a larger bandgap means a smaller $n_i$.
** Of course, there are exceptions to this, but it is useful as a general rule.
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How do you measure the parity of a particle? When one looks up elementary particles on Wikipedia, as one of their properties their parity is stated. For example the Proton has parity $+1$, while a Pion has parity $-1$. I understand that you have to define the parity of some particles to have a reference to measure the parity of all particles (since one could just redefine $+1\mapsto -1$).
How is the parity of a particle measured in experiment?
| If you look at the particle data group tables, mesons for example you will see for each resonance listed its parity, in the same line as the name and the mass of the resonance. Like the mass, it is an observation from measuring the resonance in experiments, laboriously in experiments over the years .
for example, the parity of the pi- was measured in a specific experiment:
The reaction π−+d→2n has been observed by detecting the two neutrons in coincidence with slow negative mesons incident on a liquid deuterium target. The observed angular correlation of the two neutrons confirms the identification of the process. The process is therefore not forbidden, and this fact may be used to establish the odd relative parity of the pion and the nucleon.
Received 8 June 1954
So it is angular correlations in combination with known parities that establish the unknown ones.Once there is a list of known parity particles and resonances, the parity of new states can be measured , as is proposed to measure the parity of the Higgs in future experiments.
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Do gravitational waves cause time dilation or not? There are a lot of questions on this site about gravitational waves and time dilation, and some of the answers are contradictory.
I have read this question:
Do gravitational waves cause time dilatation?
where Tom Andersen says:
In other words, if there was a beam of gravity waves, and one person was in the waves, the other not, the person who experienced the waves would have a small difference in their watch as compared to the person who was not in the wave zone.
Can a gravitational wave produce oscillating time dilation?
where peterh - Reinstate Monica says:
As you can see, it changes only the space coordinates. And only the transversal ones. If there is a change also in the time coordinate, it is not a gravitational wave any more. So, the short answer in the literal sense is a clear no.
Do gravitational waves affect the flow rate of time?
where G. Smith says:
I am reasonably sure that they do cause time to slow down and speed up in an oscillatory way for nearby observers.
So for the sake of argument, let's say there is a non-planar GW and there are two photon-clocks, one of them is in the way of the GW, the other is not affected by the GW.
As the GW passes through one of the clocks, the mirrors will come closer and farther in an oscillatory way, because of the GWs effect of stretching and squeezing spacetime itself. Thus, the clock that is affected by the GW, will seem to relatively (compared to the other clock) tick slower and faster.
Question:
*
*Do gravitational waves cause time dilation or not?
| I agree with G.Smith. A gravitational wave passing by causes distortions of space-time which means distortions of space and time. This distortion can be imagined as a gravitational potential well which inevitably involves time dilation. - Remember the Shapiro time delay. The only difference regarding curvature is that here we have Ricci curvature whereas gravitational waves are due to Weyl curvature.
| {
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Is the zero-point energy of helium stronger than other liquids to disfavour freezing? Under normal atmospheric pressures, liquid helium does not freeze even when cooled very close to absolute zero. This is attributed to the uncertainty principle or due to zero-point energy.
But the quantum uncertainty or zero-point energy is not an exclusive feature of liquid helium only. Then, why should it stop the freezing of helium but not that of other liquids? If it is strong in helium, then why?
| For the sake of simplicity, I will answer the question for the bosonic species He(4). Although there are some subtleties for the Fermionic species He(3), due to the presence of total spin-$\frac{1}{2}$, the main message is the same.
The key points are summarized here as follows:
*
*The energy contribution from the zero-point motion is seven times larger than the depth of the attractive potential between two He(4) atoms. Therefore, the zero-point motion is enough to destroy any crystalline structure of He(4).
*Helium is special because of the combination of its small mass and the value of binding energy.
*The zero-point energy for the other gases is either comparable or far smaller than the depth of the attractive potential that holds the atoms.
Now we can be quantitative using the harmonic oscillator model. The potential between two atoms is short-ranged repulsive and it becomes attractive for the long-range. Near the potential minimum, the attractive potential can be modeled via the Lennard-Jones potential $-$
$$V(r) = \epsilon_0\left(\frac{d^{12}}{r^{12}}-2\frac{d^6}{r^6}\right),$$
where the parameters $\epsilon_0$ is the trap-depth, i.e., the minimum potential and $d$ is the interatomic separation at the minimum potential.
Since the question involves comparison with other gases, below I put the parameters of He(4) and the closest noble gas neon
$$\begin{array}{|l|c|c|}
\hline
\text{Gas Name} & \text{$\epsilon_0$ [meV]} & \text{$d$ [nm]} \\ \hline
\text{He(4)} & 1.03 & 0.265 \\ \hline
\text{Neon} & 3.94 & 0.296 \\ \hline
\end{array}$$
Now, using the parameters from the above table, we can estimate the zero-point energy in three-dimensions $E_0 = \frac{3}{2}\hbar \omega_0$, assuming an fcc crystal lattice. The oscillation frequency can be estimated as
$$\omega_0 = \sqrt{\frac{4k}{m}},$$
where
$$k = \frac{1}{2}\frac{d^2}{dr^2}V(r) = \frac{36\epsilon_0}{d^2}.$$
This expression leads to a $E_0 \approx 7 $ meV for He(4), while the binding energy for atoms is $\approx 1.03$ meV. Therefore the zero-point energy is enough to destroy any crystalline structure of He(4). And this is the reason why He(4) is not found in crystal form, at normal pressure. However, if we compare the binding energy 3.94 meV and the zero-point energy $\approx 4$ meV of neon, we see that the gas can be put into crystal form at relatively small pressure.
To understand the effect of pressure, we look at the following phase diagram of He(4), where we see that the liquid/gas forms continue down to ~0 K, if the pressure remains below 25 atm. The figure distinguishes the two phases He-I and He-II separated by the black line. The superfluid fraction is shown to increase dramatically as the temperature drops.
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Simultaneity and special relativity Suppose, in inertial reference frame $F_1$, observers A and B are at rest, each having torch, and are separated by some distance and we have put machine M at middle of A and B.
Machine M has light bulbs on both sides ,right and left, so that if it catches light from A which is at left ,then machine M glows left light bulb, similar with right bulb.Also, if it senses both reaching at same instant of time then it start to make noise.
Now consider another inertial reference frame $F_2$ which is moving at constant speed $v$ with respect to $F_1$ to the right.
Now ,in frame $F_1$ , both A and B turn on their torches at same instant of time, say $t=0$ and both rays reach at M at $t=t_1$, and machine M makes noise indicating that those events were "simultaneous" in $F_1$.
Now we know these events are not simultaneous in $F_2$, in other words ,person seating in $F_2$ will say ,"I should not hear sound from machine M."
But somehow machine makes noise.(or it doesn't make noise?)
So does this mean according to $F_2$ ,machine is malfunctioning?
| A machine $M$ is not equidistant between $A$ and $B$ in $F_2$, so receiving their signals at the same time means they didn't send them at the same time, and they didn't. Totally self consistent.
| {
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Lorentz Transformation Proof - Special Relativity This is from A.P. French Special Relativity book, Chapter 3 (page 78)
Setup of the proof: $S$ and $S'$ be inertial reference frame. $S'$ move to the right with respect to $S$ at velocity $v$.
Let co-ordinates in $S$ be $(x,t)$ and co-ordinates in $S'$ be $(x',t')$
Equation (3-8) in the book, he writes that transformation will be of the form:
$x = ax' + bt'$
and by symmetry of the reference frames as implied by relativity principle,
$x' = ax - bt$
My question:
How does symmetry of reference frame argument lead to the above conclusion? For e.g. why can't the second equation above be of the form
$x' = -ax - bt$ or maybe $x' = -ax + bt$. These equations look as symmetric to me (mathematically!) as the one author uses.
(I know I'm wrong but want to understand more clearly why am I wrong)
Thanks
| Without loss of generality, let's assume $v$ is positive. From the viewpoint of the S' frame, the S frame is moving to the left with velocity $v$. (Equivalently, S moves to the right with velocity $-v$). Now, if we make a video of this, and play the video backwards, it will look like S is moving to the right with velocity $v$.
Playing the video backwards is equivalent to replacing $t$ with $-t$. So the transformation from S coordinates to S' coordinates is the same as the transformation from S' coordinates to S coordinates, but with time reversed.
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How is it physically possible that the electric field of some charge distributions does not attenuate with the distance? Let's consider for instance an infinite plane sheet of charge: you know that its E-field is vertical and its Absolute value is $\sigma / 2 \epsilon _0$, which is not dependent on the observer position.
How is this physically possible? An observer may put himself at an infinite distance from all charges and he will receive the same E-field. It seems strange.
| The answer is simple. It is not physically possible. No charge distribution can extend without limit. Every charge system is bounded. If you are farther away from a flat charge distribution than it's size, the electric field will attenuate with distance.
| {
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What causes neutrinos to be weakly coupled? When reading about "active-sterile mixing", which requires some Dirac mass ($m_D$) and some Majorana masses ($M_R$) to be very small but not zero, the seesaw limit model is discussed ($M_R \gg m_D$).
In this paper (Light Sterile Neutrinos: A White Paper), it is mentioned that the active-sterile mixing matrix, when squared can be given by:
$$\Theta^2 \sim \frac{m_{1,2,3}}{m_{4,5,...}} \sim \frac{m_D ^2}{M^2 _R} \tag{1}$$
where $m_{1,2,3}$ and $m_{4,5,...}$ are the neutrino masses.
My doubt lies in the following statement that is preceded by eq. $(1)$:
Given $m_{1,2,3} \lt 10^{-1}$ eV, the mostly sterile states are very weakly coupled unless $m_{4,5,...} \lt 10$ eV and $\Theta^2 \gt 10^{-2}$.
Why were these the values chosen for the mass of the neutrinos? Why are those the conditions for the states to be weakly coupled? What causes them to be weakly coupled, just the value of the masses?
| There are two types of neutrinos there: active ones (part of the $SU(2)_L$ doublet, carrying the weak charge) and sterile ones (singlets of the SM).
The neutrinos mass matrix via the Dirac coupling mixes the two so that the active ones become a little sterile and the sterile ones become a little bit active (inherit some interactions with SM charged leptons). The strength of the mixing (hence the new interactions of the sterile neutrinos) is quantified by the angle you have shown.
The quoted limit on the mass of the heavy (quasi-sterile) states, or equivalently on the angle, reflect the current experimental sensitivity. Note that weakly in "the mostly sterile states are very weakly coupled unless" has to be understood as feebly; it's not the actual weak charge.
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"timestamp": "2023-03-29T00:00:00",
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Why does adding a photon to the crystal of a semiconductor, gives a vertical transition in the reduced zone scheme? Why does adding a photon to the system, gives a vertical transition in the reduced zone scheme?
Considering me, it's due to the fact that a photon does not change de $k$-vector, is that correct? And why is that?
| A band structure is really a plot of energy versus momentum. Optical transitions are vertical in such a plot because the photon momentum can be neglected. The massless photon at the same kinetic energy has much less momentum than the massive electron, except in the ultra relativistic case.
| {
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Where is leverage calculated from in jointed system (car wishbone/spring rate vs wheel rate) Please let me know if this is on the right SE site.
In a car suspension, the spring typically acts partway along the wishbone:
As such, there's formulae for working out the leverage ratio that exists in the suspension. The method I've been using involves the ratio between the length of the lower wishbone (d1+d2 in this picture) and the length between the inner pivot and the spring mount (d2).
However, in my research I've found a number of different ways of measuring the leverage ratio. The point of contention seems to be where the outer end of the lever is. The second method I've seen includes the additional length between the end of the wishbone and the centrepoint of the tyre. They measure the lever arm as b+c in this picture:
In terms of the physics, is the lever measured as 'b' or 'b+c' in the above mechanism?
Sorry if I haven't explained that very well, feels like a messy description. I've asked the question on a number of automotive forums, but they seem to rely on 'I've been told it works like this', without explaining any of the underlying principles as to why.
Edit: Or is it more accurate to view it as a compound series of levers?
Lever 1: the ratio between c and d
Lever 2: the ratio between a and b
| Leverage ratio
The leverage ratio $i_L$ is the relation between the spring deflection $d_S$ and
wheel deflection $d_W$ .
$$i_L=\frac{d_S}{d_W}$$
The leverage ratio is depending which wheel suspension you have and of the geometry of the wheel suspension.
Example: Mac Pherson strut
you start with the static equilibrium, thus we just look for the dynamic deflections.
Point B is hinge joint thus the suspension can rotate about the x axis .
Point w Wheel will move on a circle to w' and also point c to c'
The components of the vector $\vec{R_{w'}}$ are:
$$\vec{R_{w'}}=\left[ \begin {array}{c} 0\\ -\cos \left( \varphi
\right) \left( d_{{1}}+d_{{2}} \right) \\ \sin
\left( \varphi \right) \left( d_{{1}}+d_{{2}} \right) \end {array}
\right]
$$
thus the deflection is:
$$d_W=\parallel\vec{R_{w'}}\parallel=d_1+d_2$$
The components of the vector $\vec{R_{c'}}$ are:
$$\vec{R_{c'}}=\left[ \begin {array}{c} 0\\ -\cos \left( \varphi
\right) d_{{2}}\\ \sin \left( \varphi \right) d_{{
2}}\end {array} \right]
$$
the spring deflection is $$d_S=\parallel\vec{R}_{CA}\parallel-\parallel\vec{R}_{c'}\parallel$$
but because our calculation is from the static equilibrium, the spring deflection is:
$$d_S\mapsto -\parallel\vec{R}_{c'}\parallel=-d_2$$
The leverage ratio for this suspension is:
$$i_L=\frac{d_S}{d_W}=-\frac{d_2}{d_1+d_2}$$
and the spring force is:
$$\vec{F}_S\approx-\frac{d_2}{d_1+d_2}\,d_w\,\begin{bmatrix}
\cos(\theta) \\
\sin(\theta)\\
\end{bmatrix}$$
| {
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How do you combine two spatial modes of light into one spatial mode in optics? Is there a lossless way to combine light coming from two different single photon sources into one spatial mode? Either free space of fiber would be fine.
Let's assume the wavelength and polarization are the same in both input spatial modes.
The catch is that the intensity at the single output should be roughly the same as the sum of the intensities at the two inputs. That is why I believe a 1x2 fiber coupler cannot be used because a 1x2 coupler is just a 2x2 coupler with one output chopped off - therefore, the half of light that would normally go into this output is lost.
| No, two single photons coming from two different sources cannot be combined to form a superposition that represents one photon with one spatial mode. In other words,
$$ \text{two photon state} \neq |a\rangle + |b\rangle . $$
Even though the two photons come from different sources, they still give you a two-photon state. Therefore, what you have is
$$ \text{two photon state} = |a\rangle|b\rangle . $$
As a result, there is no way to form the superposition of their spatial modes.
You can send then along the same path by changing the polarization of one and combine them with a polarizing beam splitter, but they will each carry their own degrees of freedom, without any superposition.
| {
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Measurement of current and measurement of momentum of the electrons Is a measurement of the current flowing through some material a measurement of the momentum of the electrons? Does their wavefunction collapse to an (approximate?) wavefunction of momentum?
| Current is measure of average momentum(or velocity $v_d$) gained by the electron due to applied Electric field in a very short period of time($\tau$) between the collision with other electrons or ions in the lattice.
Since at room temperature the electrons are very frequently undergoing collisions, the electrons can described only as statistical ensemble due to decoherence.
Yes the electrons occupy momentum eigen states with some probabilities which are given by the distribution function which depends on temperature.
| {
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How nerve signals preserve quantum coherence? Roger Penrose in this interview says he was trying to find out "... how it is that nerve signals could possibly preserve quantum coherence".
What does he mean by that?
| Penrose believes that the apparently non-deterministic nature of human thought and free will can be reconciled with the laws of physics if quantum uncertainty is somehow involved.
A colleague of his has developed the idea that the brain holds its long-term memory inside microtubule structures within the neurons (these microtubules occur in pretty much all eukaryotic cells including plants and animals and act as both a scaffolding and a transport highway). There is some (highly controversial) evidence that mircotubules exhibit properties which can only be explained via quantum mechanisms such as coherence (it is worth noting that other macromolecules such as chlorophyll are subject to related quantum-weird phenomena).
The problem Penrose faces is how a nerve impulse can transfer the information, encoded via quantum coherence, to and from a given microtubule.
One has to say that, ingenious though the theory is, it is a very long shot and is not taken seriously by mainstream science. Even if the philosophical insight is valid, the microtubule memory mechanism remains wholly speculative. Its QM arguments have also been subject to heavy criticism and accusations of fundamental flaws, but then, that is true of a great many findings related to quantum weirdness.
Still, if anybody can prove the mainstream overly pessimistic, Penrose is far the best qualified to do so; the conformal geometry of spacetime and twistor theory make for a pretty good CV. (Recall too that Shing-Tung Yau first rose to fame for disproving the Calabi conjecture, before winning the Fields medal for tuning his result on its head and proving it true after all, thus giving string theorists their Calabi-Yau manifolds. Maths and physics at this level is full of surprises.)
| {
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How does gravity smooth a planet? Over time, gravity (on a macro scale) and erosion (on a micro scale) will work together to smooth out a planet. How does the gravity work? Is it like erosion on a massive scale (ie pulling down the outside of a mountain) or does it un-deform the planet like unsqueezing a water ballon (ie by pulling the inside of a mountain and forcing the sea bed up to make room)? I have included some crude sketches to illustrate what I mean.
Super Erosion:
Un-deforming:
| For things the size of planets, gravity smooths them out through erosion and, to make things really smooth, wave action, underwater sedimentation, and tides.
For very dense objects like white dwarf and neutron stars and black holes, gravity is so overwhelmingly powerful that it smooths them out directly, without having to rely on erosion or anything else.
| {
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What will happen if I bring a positively charged body near the negative terminal of battery? Will the electrons flow from negative terminal of the battery to the charged body so as to neutralized a charged body..thus causing a decrease in pd of battery
| (a) I'll assume that the positive charged body is a conductor and that it is touched on to the negative terminal of the battery.
(b) Electrons will then flow from the negative battery terminal on to the body. The same amount of negative charge will be pumped through the battery from its positive terminal (and/or positive charge in the other direction).
(c) The flow of charge will stop (after a very short time) when the pd between the battery terminals is restored to its previous value. This value is the emf of the cell, in other words the work the battery can do, per coulomb of charge flowing through it, because of the chemical reactions inside it.
(d) The battery as a whole will have lost some electrons to the body, and will be at a positive potential with respect to infinity, but the pd between its terminals will not change (except for a very short-lived dip owing to its internal resistance).
| {
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On particle diffraction and its relation to the statistical interpretation of the wave function Particles can be diffracted due to their quantum nature and that is understood by their wave-like behavior. Clearly seen in e.g. plane wave solutions of the Schrodinger equation or a superposition of states which can be seen as a wave packet.
Statistical interpretation tells that those wave functions are merely to describe probability amplitudes, so one can argue they are not ontic/ physical waves.
On the basis of these two statements, why/how can de Broglie wavelength be used as in the classical wave approach to determine if a quantum object will be diffracted by a slit, lattice, nucleus etc. ?
| In a section of Heisenberg's The Physical Principles of Quantum Theory; he emphasizes Duane's perspective on corpuscular picture of diffraction/reflection and points out for example for a grating, if it is known that the particle will hit a certain $$\Delta x $$ of the grating then the momentum will become uncertain by an amount proportional to $$\frac{h}{\Delta x}$$
The reflection direction will therefore be uncertain according to the relation above. The value of this uncertainty can be calculated from the resolving power of a grating of
$$
\frac{\Delta x}{d}
$$
lines; d being a plane ruled grating's constant.
When,
$$
\Delta x < < d
$$
the interference maxima are no longer visible and the trajectory can be compared to that of a classical particle.
| {
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Why isnt the water from bottle dropping in the tub(volume of bottle/tub doesnt matter) until the level equalises? I can't seem to understand why water in the bottle wont drop down until the water level equalizes so pressure is same in any horizontal plane. As you can see just at tip of bottle,the pressure inside bottle would be higher due to the water level above it
And yes the bottle is closed from the top, so I assume the answer has to do something with the pressure created due to that?
Edit:- ok so pressure inside the little pocket is $P_{outside}$ (I'll just call it $P_o$)
from the time the bottle was opened
so, the pressure at the tip should be $P_o + D \cdot g \cdot h$ where $h$ is height of water level from tip to the level in the bottle and $D$ is density
but from the bigger container's perspective, the pressure there is $P_o$
| The pressure in the bottle (at the top), Pi, is lower than the pressure outside: Po = Pi + Dgh.
| {
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Can a fan cause a liquid to cool below room temperature? I enjoy the occasional hot drink, but place it below a small fan in order to cool it to a drinkable temperature. Unfortunately, as expected, I commonly forget about my drink, and it ends up very cold.
In fact, it ends up so cold that it feels much colder than I would expect given my relatively warm room ($\sim \mathrm{25^\circ C}$). However, this could be an illusion caused by room temperature still being less than body temperature, or the mug being fairly cold.
I’m wondering if the fan is able to cool the liquid below room temperature. I’m aware that evaporation works on the warmest molecules, and leaves the remaining liquid cooler, but I’m not sure that it can ever become cooler than the surrounding air temperature?
| As it's been asked about "fan", the fan can't cool anything by itself (imagine using a fan inside vacuum XD), it brings the fluid outside the container to motion hence helping to transfer the heat from container to the moving fluid(air in usual cases).
If we use much more cooler fluid we must be able bring down the temperature significantly.
| {
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"url": "https://physics.stackexchange.com/questions/556234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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D'Alembert derivation of Lagrange Equation - why can it use both virtual and normal differentials? In "Classical Mechanics" by Goldstein and "A Students Guide to Lagrangians and Hamiltonians" by Hamill I noticed that both the virtual displacement derivatives and the normal displacement derivatives are used at different points of the proof, as shown below. My question is why can this mixing of real & virtual derivatives be done?
To simplify the equations it is assumed there is only one mass and one associated generalised variable with $x=x(q,t)$, with $\dot{x}$ meaning differential with respect to time.
The virtual displacement $\delta x$ is used to set up the virtual work equation via:
$$\delta x = \frac {\partial x}{\partial q} \delta q, \qquad \delta t=0, \tag{1}$$
being substituted into ($F$ is force, $a$ is acceleration):
$$(F/m) \delta x = a \delta x = a \frac {\partial x}{\partial q} \delta q.\tag{2}$$
The following equations (3) and (4) are used to transform the acceleration $a$ in the rhs of (2) into a form based on the kinetic energy $T$, using the usual velocity differential equation with possible explicit $t$-dependence:
$$v=\dot{x} = \frac {\partial x}{\partial q} \dot{q} + \frac {\partial x}{\partial t} \tag{3}$$
to derive:
$$ \frac {\partial v}{\partial \dot{q}} = \frac {\partial x}{\partial q}. \tag{4}$$
So it looks like virtual displacements are used in (1) & (2) and real displacements
$$\delta x = \frac {\partial x}{\partial q} \delta q + \frac {\partial x}{\partial t} \delta t \tag{5}$$
are used in (3) & (4) parts of d'Alembert derivation of Lagrange equations.
| *
*On one hand, holonomic constraints and the Lagrangian itself are certainly allowed to have explicit time dependence, cf. e.g. the last term in OP's eq. (3).
*On the other hand, it's a well-established fact that the relevant (infinitesimal) displacements in the d'Alembert's principle and the principle of stationary action -- the so-called (infinitesimal) virtual displacements -- are frozen in time $\delta t=0$. See e.g. this, this & this related Phys.SE posts.
| {
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Evaluating sum of torques for different choices of origin when solving equilibrium problem My textbook says,
When applying equilibrium conditions for a rigid body, we are free to choose any point as the origin of the reference frame.
(source)
I am trying to understand this by looking at the following picture (from an exercise problem), in which the ball is in static equilibrium because of the applied horizontal force $\vec F$ and the friction between the ball and the surface.
(source)
If we pick the point where the surface and the ball are in contact as the origin, then I can see how the torques cancel out. We have a negative torque $F \cos \theta$ acting on the center of the ball at a distance of $R$ and a positive torque from gravity $ mg \sin \theta$ acting again on the center of the ball at distance $R$. So
$$\sum \tau = 0 = - F R \cos \theta + mg R \sin \theta$$
and we can begin solving for the external force.
Alternatively, let's try defining the center of the ball as the origin. Since the applied force and gravity both act on the center of the ball, they provide zero torque. Likewise, the normal force from the surface is pointed directly at the ball's center and therefore provides zero torque.
The only nonzero torque I can see is provided by the frictional force, whose magnitude is $mg \cos \theta$, and acts at a distance of $R$ in the positive (CCW) direction.
$$\sum \tau' = 0 = \mu_s mg R \cos \theta$$
Since the friction force is nonzero, what force causes the torque to balance when choosing the center of the ball as the origin?
| The force of static friction is given by $$|\vec F_f| \le \mu_s |\vec N|$$ Since this is an inequality, you cannot use this to find the magnitude of friction, only whether $\mu_s$ is large enough to provide the necessary friction. Instead, the way to find static friction is based on the fact that it is static, i.e. not moving, so you can just use Newton's Second Law to solve for friction by solving $$\sum \vec F = 0$$
If you solve for the external force using your first equation, you will find that the friction is zero, which is consistent with your second equation.
| {
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Does it actually take infinite (observer) time for someone to fall into a black hole? If you were to watch your friend approach a black hole, I understand that you'd see their clock slow until they appear frozen and redshift within a few seconds. But if you were to detect the increasingly long wavelengths coming off of your friend, you would still see them frozen above the event horizon for infinite time. I want to know if this is due to actual time dilation and the friend is actually above the event horizon for infinite time or if this is because of some Doppler effect and the friend actually fell in but you're only observing the past events of the friend. Put another way, given any finite time, could you theoretically go fast enough to save your friend before they fall in? Or would you only realize that your friend fell in a long time ago?
| It is not because of the doppler shift, but because of the gravitational time dilation. In the frame of an external stationary observer the infalling observer indeed never reaches the horizon.
If both observers carry entangled quantum bits, and the external observer makes a measurement on his first, he also determines what the infalling observer will measure on his; should the infalling observer make his measurements at or after he fell through the horizon, the external observer has an infinite amount of time to make his measurement and thereby define what the infalling observer will measure on his.
| {
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Do virtual electron-positron pairs have mass? When a photon produces an electron-positron pair, do both these particles have mass? Why or why not?
|
Do virtual electron-positron pairs have mass?
Virtual particles are within an integral depicted by a Feynman diagram
Only lines entering or leaving the diagram represent observable particles. Here two electrons enter, exchange a photon, and then exit. The time and space axes are usually not indicated. The vertical direction indicates the progress of time upward, but the horizontal spacing does not give the distance between the particles.
The virtual lines are described by a fourvector, and the "length" of a four vector is the mass for a free particle.In an integral this mass is variable within the limits of integration.
you ask:
When a photon produces an electron-positron pair, do both these particles have mass? Why or why not?
A photon has zero mass, and conservation of energy and momentum do not allow the decay into massive particles.
| {
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How do you hang a bottle off a toothpick? I came across this video: https://www.youtube.com/watch?v=mHsVxNMFWwA
How is this possible? I'm guessing the vertical toothpick is exerting an upward force on the table toothpick to balance the torque. But how? I am confus
| An easier way to understand why the arrangement is static is to think about energy. The arrangement will be static if it is in a minimum of energy. Addition of the two toothpicks to the rope/bottle arrangement introduces a geometric constraint that makes this arrangement a minimum of energy.
I don't know if you are familiar with the concept of virtual work, but basically, to analyze if a situation is at an energy minimum, we presume we move the assembly as allowed by its geometry by a very small distance, and see if energy is increased or decreased. If energy increases for all possible movements, then we are at an energy minimum and have static equilibrium.
First, consider the initial configuration, with only 1 toothpick, the rope and the bottle. If the bottle goes down. energy is reduced. Since the toothpick is completely free to rotate about the table, the bottle can go down, and we are not at equilibrium.
In the second situation, the two additional toothpicks add a constraint. If the "original" toothpick is rotated a bit, the bottle will go up, as the new "vertical" toothpick would force the new horizontal toothpick, in the rope, to move towards the table. Since we are pivoting about the table, this horizontal movement has to be along an arc and the horizontal movement has to be accompanied by a vertical movement in the up direction, which will raise the bottle a bit. This movement is therefore not allowed to occur spontaneously.
Note that this requires enough friction with the table. If there is no friction, a combination sliding/rotating movement is allowed and the arrangement is not stable. This is related in forces term to the fact that there must be an horizontal friction between the vertical toothpick and the original table-supported toothpick, as the "vertical" toothpick is not completely vertical.
| {
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How does RF communication between electronic circuit elements affect the circuit? The other day, a friend and I were looking at a simple resistor-capacitor-inductor circuit. As we were looking at the circuit, he mentioned that the capacitors and inductors "talked" to each other through the air. From this, I assume he meant that the capacitors and inductors were exchanging EMF waves, and that it was an essential part of how these circuits operated.
I was surprised by this because I didn't remember learning about this in my circuits class, but it made some sense because I figured every electronic device is emitting some sort of EMF wave. I just always assumed these were of minor importance to the circuit.
My question is how do these circuit elements communicate with each other in simple RCL circuits? Is it primarily through the voltage/current local to the wires, or are there important EMF considerations.
To clarify, I understand that at some level, the voltage/current in the wires is being controlled through EMF waves since that is how the charge of an electron is communicated to near by electrons. I am referring to EMF waves here to be over macroscopic distances (between circuit elements).
| One form of this interaction which is real and has consequences is called radiated electromagnetic interference. This happens when the circuits are carrying high frequency signals, under which conditions a simple piece of wire in the circuit or a trace on a PC board becomes an antenna, broadcasting EM waves. When those waves strike other circuit elements, they can induce unwanted flows of electricity in them, with the potential to completely upset the operation of the circuit.
| {
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Does energy conservation imply time invariance? This is similar to this question: Is the converse of Noether's first theorem true: Every conservation law has a symmetry?. However, the answer given there is very technical and general. I am only interested in the specific case of energy conservation (mostly because dark energy seems to break energy conservation / time invariance).
| You may possibly want this explained with regards to Noether's theorem or something similar, but the answer is yes if we're dealing with quantum mechanics. In quantum mechanics the time-translation operator is given as
\begin{equation}
\hat{T}(t) = \exp{(-i\hat{H}t/\hbar)}
\end{equation}
such that $T(t_{0})|\psi(t)\rangle = |\psi(t + t_{0})\rangle$.
In the Heisenberg picture we know
\begin{equation}
\frac{\text{d}\hat{A}(t)}{\text{d}t} = \frac{i}{\hbar}[\hat{H}, \hat{A}]
\end{equation}
Energy conservation imples $\frac{\text{d}\hat{H}}{\text{d}t} = 0$ which is easily shown by the above equation:
\begin{equation}
\frac{\text{d}\hat{H}}{\text{d}t} = [\hat{H}, \hat{H}] = 0
\end{equation}
We can rephrase this in terms of the time-translation operator as
\begin{equation}
\begin{split}
[e^{i\hat{H}t/\hbar}, \hat{H}] &= \\
[\hat{T}(t), \hat{H}] &= 0
\end{split}
\end{equation}
We can start with this final equation and work through the above derivation backwards to show that time-translational symmetry does indeed imply energy conservation.
| {
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How do I get data of nuclear mass of each nuclide? I have been searching for a table of nuclear mass for a long time, but what I got are mostly data of atomic mass. The mass of electrons and the atomic binding energy might bring error to the calculation involving nuclear mass.
| You need a chart of the nuclides which you can access on line. There used to be posters of this but i don't know if its still sold. LiveChart has an interactive display and Wikipedia has a chart on their web site but it doesn't have your data included. The chart on Wikipedia is not complete but gives you an idea of how big the Chart of the Nuclides is. There is a very useful chart here that appears to be complete. You can scroll in and out looking for a nucleus. Once you find one you want you can click on the square and it lists a large amount of data for each nucleus. There are literally thousands of nuclides.
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Are there any more physically representative analytical expression for a slit or edge than a step-function? What about ${\rm erf}(x)$ for example? Historically slits have been invaluable in teaching, research, and theory validation in both electromagnetic and quantum mechanics but conceptually they differ from what we can actually build based on the physical properties of matter because the canonical slit perfectly absorbs an incident wave in an infinitely thin layer without reflection or induced phase shifts.
This simple implementation also results in discontinuities in the amplitude at the slit edge. We know this is wrong but we still get pretty good results when matching up the resulting calculated pattern produced from a hard-edged infinitely thin theoretical binary slit that just multiplies the incoming wave by either unity inside the slit opening or zero outside.
Question: Have more sophisticated analytical models of slits been suggested that will function similarly but take steps to be more physically realistic in terms of finite thickness and reduced discontinuity?
Just as an illustrative example $\frac{1}{2} \text{erf}\left(\frac{x+1}{\sigma}\right) - \frac{1}{2}\text{erf}\left(\frac{x-1}{\sigma}\right)$ looks a little "softer" than a pair of step functions, but I don't know if it is any better or worse in terms of the wave mechanics.
Notes:
*
*As pointed out in comments I'm really asking about modeling single edges; this could apply to the edges of rectangular or circular apertures, or even diffraction from a single straight edge.
*Answers that addresses either an electromagnetic wave or a matter wave (e.g. atoms) are welcome.
*I've asked about analytical models for slits rather than constructs used in finite element analysis, but there may be something to be learned from those impedance-matched constructs.
*Answers to In the double slit experiment what, exactly, is a slit? don't go far enough to answer this question.
| A reasonable mathematical model for a gradual transition from one state to another is the $\tanh$ function.
Thus
$$ T(x) =\frac{1}{2}\left( 1 + \tanh(x/t)\right)$$
smoothly transitions from $T=0$ when $x<0$ to $T=1$ for $x>0$, with a characteristic "width" for the transition of $\pm t$ about $T=0.5$ at $x=0$. A sharp edge is recovered by allowing $t \rightarrow 0$. The plot below shows $T(x)$ for $t=1$.
An alternative is the logistic sigmoid.
$$T(x) = \frac{1}{1 + \exp(-2x/t)}$$
where $t$ has a similar definition. The plot below shows this function for $t=1$.
| {
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Why doesn't water boil in the oven? I put a pot of water in the oven at $\mathrm{500^\circ F}$ ($\mathrm{260^\circ C}$ , $\mathrm{533 K}$). Over time most of the water evaporated away but it never boiled. Why doesn't it boil?
| Short answer is that it boils, but it boils differently because it's either evaporation from a liquid surface in low temperature or "bulk evaporation" aka. boiling, due to temperature gradient. Now because oven heats more or less uniformly all sides of pot is heated the same, thus eliminating clear temperature gradient. Without temperature gradient "strong visual" bulk evaporation is impossible. Besides in oven air is heated to high degree, thus producing higher pressure to escaping water vapor molecules, so water becomes a little-bit super-heated, which may raise water's boiling point a bit. This is second reason why you don't see standard boiling effects as with boiling kettle.
| {
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Expansion Postulate Quantum Mechanics How does the expansion postulate allow predictions to be made about measurement outcomes?
I understand the postulate as:
$$
ψ =\sum_{n} a_n φ_n
$$
with coefficients calculated by:
$$
a_n =\int φ_n^*ψdτ.
$$
I think that:
$$
|a_n|^2
$$
is the probability of the system being in state φ, but I do not think that is the answer to the question.
| The postulates of quantum mechanics say the following about measurements. Consider a physical quantity represented by an operator $\hat{O}$, whose eigenvalue equation is $\hat{O}\phi_n=\lambda_n\phi_n$, for eigenvalues $\lambda_n$ and eigenstates $\phi_n$. The outcome of a measurement of this physical quantity is then one of the eigenvalues of the associated operator.
If you are measuring this property for an arbitrary state $\Psi$, then expanding this state in the basis of eigenstates of the operator $\hat{O}$ gives:
$$
\Psi=\sum_n a_n\phi_n,
$$
where the expansion coefficients are as you wrote. Then quantum mechanics says that probability of obtaining $\lambda_n$ when measuring the property associated with the operator $\hat{O}$ is:
$$
P(\lambda_n)=|a_n|^2.
$$
For degenerate eigenvalues, you need to sum over the degenerate subspace to obtain the total probability.
| {
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Why exactly do we feel a shock when we place our hand into a conducting solution? I have a very naive question.
Suppose you have pure water in a flask, and you place two ends of a copper wire (which are connected to a battery) into the water.
If you were to place your hand into the water, you would not feel any shock, as pure water does not conduct electricity.
However, if you add an electrolyte like common salt to the same water, you would probably feel a shock.
Adding salt makes the solution conducting. When the two wires are placed in the solution, the ions are attracted to the end of the wire which has an opposite charge.
However, what does the movement of those ions have to do with whether or not your hand feels a shock? Shouldn't whether you feel a shock just depend on what resistance your hand offers?
| Whether you feel a shock depends on how much current flows through your hand. How much current flows through your hand depends on the resistance along the whole path of the circuit, not just in your hand.
For example, if you put a battery on a table and then put a resistor a few inches away from the battery, there would be no current flowing through the resistor, because air is a very bad conductor. But if you now connect that same resistor, with the same resistance, to the battery using wires, now there is current flowing through the resistor, because the resistance along the rest of the path has decreased tremendously. This is despite the fact that the resistance of the resistor didn't change.
| {
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Does it make sense to say that something is almost infinite? If yes, then why? I remember hearing someone say "almost infinite" in this YouTube video. At 1:23, he says that "almost infinite" pieces of vertical lines are placed along $X$ length.
As someone who hasn't studied very much math, "almost infinite" sounds like nonsense. Either something ends or it doesn't, there really isn't a spectrum of unending-ness.
Why not infinite?
| In layman's terms, something is “almost infinite” if it is so large that
it would make no difference if it was any larger. This can be formalized
with the mathematical notion of limit, as shown in previous answers.
Here, I would just like to add a simple illustration. Here is a picture
of my 35 mm lens:
See the infinity marking I highlighted on the focusing distance scale?
This indicates the correct focus for photographing a subject that is
infinitely far. Whether it is a mountain range a few kilometers away or
a star field a few parsecs away makes no difference. As far as the lens
is concerned, anything further than 50 m or so may be considered
“at infinity”.
This can be understood by looking at the lens equation: a subject at
infinity would produce an image at the lens’ image-side focal point (in
the sense of a mathematical limit). If the distance to the subject is
much larger than the focal length, then the position of the image is
also, to a very good approximation, at that focal point.
How far is infinity obviously depends on the context. A greater film or
sensor resolution, a better lens quality, a longer focal lens, or a
larger aperture, all push “infinity” further away. It could be argued
that the hyperfocal distance is the shortest distance that
could be considered at infinity.
| {
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Continuity equation in QM I found this question in a quantum mechanics exam:
What is the physical interpretation of the continuity equation $\frac{\partial\rho}{\partial t}+\frac{\partial j}{\partial x}=0$? Here $\rho(x,t)$ is the probability density and $j(x,t)$ is the probability current.
I assume they want a one liner like "probability is conserved". But to be honest I do not understand this. Can any one help me here? What's the one liner they are looking for and why? Many thanks!
| The continuity equation in 3-dimensions is
$$\frac{\partial \rho}{\partial t} + \vec{\nabla}·\vec{j}=0$$
where the second term is the divergence of $\vec{j}$. By integrating this equation within a fixed volume $V$ whose boundary is $\partial V$, and applying the divergence theorem, we get the integral form of the continuity equation:
$$\frac{d}{dt}\iiint\limits_{V}{\rho dV} + \iint\limits_{\partial V}{\vec{j}·\vec{dS}} =0$$
where the surface integral is over the closed surface $\partial V$ with $\vec{dS}$ defined as pointing normally outward. This equation states that the time rate of change of the probability within volume V is equal to the probability flux entering volume V across the boundary $\partial V$. This is a statement of conservation of probability.
| {
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Symmetry factor of certain 1-loop diagrams in $\phi^4$-theory I have to derive a formula for the symmetry factor of the diagrams of the form
in $\phi^4$-theory, where $\phi$ is a real scalar field. By symmetry factor I mean only the number of possible contractions, which lead to the same diagram (without the factor $1/n!$ for $n$th order of pertubation theory and without the factor $1/4!$ for each vertex from the Lagrangian).
So let $n$ be the number of external legs. For each diagram, we have a factor $(n/2)!$ from the interchangeability of the internal points. Furthermore, we get $(4!/2)^{n/2}$ to connect each pair of external lines to one of the vertices. What is left is the number of ways to connect the left internal lines, in order to get the circle...
In the first diagram, this gives a factor of $1$. In the 2nd diagram, we have a factor of $2$ and for the 3rd diagram, we have a factor of $2\cdot 2\cdot 2=4\cdot 2$. In a diagram with 4 pairs of external legs, we can simply see that we would get a factor of $6\cdot 4\cdot 2$. Therefore, we get a factor of $(n-2)!!$ for each diagram, for completing the circle.
In total, I find
$$S=(n/2)!\bigg (\frac{4!}{2}\bigg )^{n/2}(n-2)!!$$
However, I should have found
$$S=\bigg (\frac{4!}{2}\bigg )^{n/2}(n-1)!$$
according to the solution, which is clearly different from my expression. So, where is my error?
| OP's exercise seems to conflate the number $n$ of external legs and the number $m=\frac{n}{2}$ of 4-vertices. The symmetry of an $m$-gon is $S(m\text{-gon})=2m=n$. The symmetry factor of the relevant 1-loop Feynman diagram is then $S=2^m S(m\text{-gon})=2^mn.$ For each vertex, there is $\begin{pmatrix}4\cr 2 \end{pmatrix}=6$ ways to choose a pair of halflines that participates in the loop. Since the order matters, this makes $2\times 6=12$ ordered pairs. We can order the vertices along the loop in $\frac{m!}{S(m\text{-gon})}=\frac{m!}{n}$ ways. Hence the sought-for number of contractions are $\#=12^m \frac{m!}{n}$. One may check that the resulting factor in the Feynman diagram becomes $\frac{\#}{m!(4!)^m}=\frac{1}{S}$ as it should.
| {
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Understanding quantum cross sections as areas In scattering cross sections we deal with $d\sigma/d\Omega$, incident area per scattered solid angle. When a particle scatters into a small finite $\Delta\Omega$, the incident particle was in a small finite area $\Delta\sigma$. However, in QM the incident state is a plane wave / asymptotic momentum eigenstate, so it's totally delocalized in position space. Isn't the probability for the incident particle to be found in the area $\Delta\sigma$ therefore zero (a small area out of an infinite plane)? If we integrate $d\Omega$ we'd find the total probability to be zero, which is absurd. Where did this reasoning go wrong?
It seems to me it would make more sense to define $dP/d\Omega$ instead of $d\sigma/d\Omega$. In a scattering there is some probability the final momentum angle is in some $d\Omega$. This would then integrate to 1. But obviously this is not done and the cross-sectional area $\sigma$ is somehow necessary.
| Here's my take on the thing, based on what is expressed in chapter 11 of N. Zettili's Quantum mechanics: Concepts and applications. The scattering cross section is defined as the number of particles $d\sigma$ scattered into an element of solid angle $d\Omega$ defined by the angles $(\theta, \varphi)$. This is related with the incident flux of particles $J_{inc}$ as
$$\frac{d\sigma(\theta, \varphi)}{d\Omega}=\frac{1}{J_{inc}}\frac{dN(\theta, \phi)}{d\Omega}$$
where $dN$ is the number of particles scattered into an element of solid angle. The incident flux can be calculated as
$$J_{inc}=\vert A\vert^2\frac{\hbar k_0}{\mu}$$
where $\mu$ is the reduced mass of the system, $A$ is a normalization constant and $k_0$ is the wave number for the incident wave.
Now, what you're saying isn't exactly wrong, you're just forgetting that position and probability are actually bound in a quantum system. So, this is information is stored in the wave number somehow, but specially in the probability amplitude, and surely in the number of particles $dN$.
| {
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Intuition of sound wave impedance and admitance The impedance of a soundwave is given by both:
$$\frac{\mathcal{P}}{V}=\rho_0c$$
Where $\mathcal{P}$ is the pressure wave, $V$ is the velocity vector of particles, $\rho_0$ is the static value of the gas density and $c$ is the speed of sound\wave.
It is very easy for me to understand an impedance in an electrical circuit. The impedance of a capacitor is a measure of how will a capacitor influence the current (flow of electrons through a conductor). In particular, impedance is the negative influence on the current while admittance is a positive influence.
Is there an intuitive explanation with respect to a sound wave? Looking at the $\rho_0c$ term it seems to me that this is a medium-wise constant value. Is that correct? Is there another\better intuition?
| Impedances in circuits are extrinsic while in your questions you are showing the intrinsic acoustic impedance. As you pointed out, it is a material property and intuitively you can say that it shows you how easy it is for a given pressure to obtain a particular particle velocity. I suggest that you check my previous answer on the topic.
| {
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Dispersion Relation of Quasiparticle in Superfluid I am now reading a book "Superconductivity, Superfluids, and Condensates" by Jame F. Annett. I am stuck at the part of writing dispersion relation of the quasiparticle in superfluid here. The author does not seem to give any explanation about this. He somehow mentioned the Fermi's golden rule previously, but I could not see how we can use it to prove this dispersion relation.
Here I attached the picture of that page. The equation that I could not follow is (2.83)
|
According to the dispersion relation above, which was obtained by neutron scattering, we can see firstly see that the quasiparticle has the linear dispersion relation at low energy.
Imagine now we drap an object of mass $M$ in the fluid (that object can be a defect at the tube wall), the momentum of the object is initially $P$. It can emit the phonon (quasiparticle - zero sound mode) with the energy $\hbar \omega_k$ and momentum $\hbar\vec{k}$.
Write down conservation of energy, we obtain
$$
\frac{\vec{P}^2}{2M}=\frac{(\vec{P}-\hbar\vec{k})^2}{2M} + \hbar \omega_k
$$
Hence,
$$
\omega_k = \vec{V}\cdot\vec{k}-\frac{\hbar k^2}{2M}
$$
In the large $M$ limit, we finally arrived at
$$
\omega_k = \vec{V}\cdot\vec{k}
$$
| {
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Why is a projectile trajectory parabolic while that of a satellite elliptic? I understand that the parabolic trajectory is an approximation of a more elliptical trajectory, since acceleration due to gravity is taken to be a constant for a projectile. However I'm intrigued to know that what changes in kinetic and potential energy contribute to the question? What I mean is I read that for a trajectory under a central force, the reciprocal of radial distance is $$\frac{1}{r}=-\frac{mK}{L^2}+A \cos\theta$$ where $$A^2 = \frac{m^2K^2}{L^4}+\frac{2mE}{L^2}.$$
For an ellipse clearly the first term should be greater than $A$ but somehow that doesn't seem to make a lot of sense to me. Further for a parabola of I equate the first term with $A$, which again, seems a little weird. Where am I going wrong?
Please feel free to criticise me but it's a request not to vote close the question unnecessarily. My question regarding fusors were vote closed simply because a few didn't know about the existence of such a thing.
| We consider the mechanical energy of the object in orbit due to a central force to be
$$E=\frac{1}{2}mv^2 + U(r)$$
where $m$ is the reduced mass of the object, $v$ is the instantaneous speed, and $U(r)$ is the instantaneous potential energy of the system due to the central force.
There are three general values for the mechanical energy:
E>0
This energy results in a hyperbolic orbit, which is an open path because no matter how far away you go there is always some kinetic energy to keep moving away.
E=0
This energy results in a parabolic orbit, again, an open path. There is always kinetic energy to keep moving until $r\to \infty.$
Note that this is a different energy/orbit calculation than for the constant force situation (F=mg) because the potential energy zero is arbitrary for F=mg.
E<0
This is the elliptical orbit, which is a closed path. There are turning points because there is a minimum value of kinetic energy related to the (constant) angular momentum, and the kinetic energy is always less than the magnitude of the potential energy.
If the kinetic energy, $K$, is constant, the orbit will be circular and the total mechanical energy will be $E= - K$.
| {
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Why does the energy of an electron inrease with its shell number inside an atom? According to this:
$ E = \frac{-13.6 Z^2}{n^2} $
the energy of an electron is, well, higher the farer it is away from the core. I found this confusing as I need to put less energy to release an electron when it is on the 5th shell than when it is on the 1st shell. Am I right?
Concluding, one would say "The more energy an electron has the less energy I need to release it".
This wouldn't be the case if "they" didn't introduce the minus sign. It probably has a reason and I'm probably neither the first nor the last who asks about that. But, sorry, I still have to ask: Why did they do that? Why is a minus better than a plus?
|
Why did they do that? Why is a minus better than a plus?
Firstly, understand that when the electron is 'infinitely' far from the nucleus its energy is $0$ (assuming it is also stationary).
But empirically we also know that when an electron 'falls' from a higher orbital ($n=n_1$) to a lower one ($n=n_2$ where $n_2<n_1$) energy is released in the form of a photon emission, so $\Delta E <0$. And indeed:
$$\Delta E=E_2-E_1=-13.6 Z^2\Big(\frac{1}{n_2^2}-\frac{1}{n_1^2}\Big)<0$$
So the choice of sign is correct.
| {
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Ideal wire and resistance I am having a bit of trouble understanding what an ideal wire is.
Let's assume there is a positive charge on the positive terminal of the battery and a negative on the other side which will give us the same results. Now electric potential is given as integral of E.dr. Therefore the potential will decrease with the distance .How is it possible that there is same potential across a wire. Electrons would not move if this was the case.
Also it is said that potential is dropped across a resistor and all the energy lost is made in heat energy,and that the battery provides this energy,now when an electron moves closer to a proton because of Coulomb attraction potential energy is also lost but no one has to provide it in this case?
Also when it moves closer energy is lost and similar happens in a resistance although resistance provides an obstruction the energy lost in that should be different how can we equate these two and make them heat dissipation?Can someone answer these questions
| If you short an ideal voltage source (V) with an ideal wire (R=0) then an infinite current I will flow such that V=IR . In practice a fire or explosion will result to remind you that idealisations are approximations to reality.
| {
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Is it possible for two observers to observe different wavefunctions for one electron? Suppose there are 2 scientists who have decided to measure the location of an electron at a same fixed time. Is possible that while one observes the wavepacket localized at (position=x) while the other observes the wavepacket localized at (position=y). The condition however is position x is not equal to y.Please dont confuse about the degree of localization which can be quite varying depending upon which measurement-momentum or position is given priority
:( I have little to no experience with quantum superposition and gaussian wavepackets.......kindly manage with my rough knowledge
| In order to observe an electron one must interact with it in some way. For example one could shine light at it so that it scattered the light, or one could arrange for it to hit something like a multi-channel array (a charge detector with many small elements). The various observers will study some sort of large-scale signal such as a current from the array or else the light hitting a camera. They will all agree on what large-scale signal was seen, and they will agree on the chain of inference which determines what information it gives about the electron. So, in summary, the answer to your question is that they all get the same answer.
| {
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Why isn't the work done by gravity positive in this situation? I want to find the work done by the force of gravity to move an object of mass $m$ from infinity to a point $P$ at distance $r_p$ from a body of mass $M$ (that I assume fixed). The formula should be
\begin{equation*}
W = \int_{\infty}^{P} \vec{F}(r) \cdot d\vec{r}
\end{equation*}
The force is conservative, so I can assume a straight path, and since the force on $m$ is directed toward the mass $M$, it is parallel to the displacement and the dot product is positive, equal to magnitude of the force.
\begin{equation*}
\vec{F}(r) \cdot d\vec{r} = G\frac{Mm}{r^2}dr
\end{equation*}
But doing so result in a negative work
\begin{equation*}
\int_{\infty}^{P} G\frac{Mm}{r^2} dr = -G\frac{mM}{r_p}
\end{equation*}
and this is wrong since it should be positive. What accounts for the discrepancy?
Note that Halliday and Resnick in their book "Physics" (chapter 13), do almost the same thing, because they compute the work from the point $P$ to infinity. In their case the math checks out because the force is antiparallel to the displacement and the dot product is negative, resulting in a negative work (which is ok because in that case the movement is against the force). Here it should be positive, but the problem is that not only the integral bounds are switched, but also the sign of the integrand, and so the overall sign is not changed.
| $\begin{equation*}
W = \int_{\infty}^{P} \vec{F}(r) \cdot d\vec{r}
\end{equation*}$
$\vec r = r \,\hat r$ is the displacement from the centre of the mass $M$ and $d\vec r=dr\,\hat r$ is incremental displacement.
The gravitational force is attractive and so in the opposite direction to $\hat r$ thus $\vec F(r) = -G \dfrac {Mm}{r^2} \hat r$
The integral now becomes $\displaystyle \int _\infty^r \left (-G \dfrac {Mm}{r^2} \hat r\right)\cdot (dr\,\hat r)\int _\infty^r -G \dfrac {Mm}{r^2}\,dr$ which will give you your required positive value.
| {
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in photoelectric effect, where do electrons go when they are ejected from metal? When light shines on a metal, electrons are ejected from the surface of the metal provided the energy of the photons are greater than the work function of the metal.
My question is:
*
*Where do the ejected electrons go? Are they running freely within the metal (ie ejected from the atom they were bound to) or are they physically ejected from the metal altogether? Is it a combination of the two? On a standard diagram (shown below), they seem to be leaving the metal altogether.
*How does this affect the charge of the metal? If electrons are indeed ejected from the metal, I would assume the metal would be positively charged.
| the electrons are actually ejected from the surface of the metal. If the metal is surrounded by a vacuum, the electrons form a cloud close by called a space charge. If there is a positively-charged piece of metal close by, the loose electrons are attracted to it and their flow is called a photoelectric current.
| {
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