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Why is the South Pole Telescope located exactly at the South Pole? I read that there is less atmospheric interference for the telescope at the South Pole because the atmosphere is thin and there is less water vapor in the air. However this seems to be true for many locations on Antarctica? Are there any other reasons that this telescope is located at exactly the South Pole?
| There is no scientific reason that the South Pole Telescope is located so close (within a few 100 meters, really) of the geographic south pole. It might as well be located a few hundred miles to either direction, no difference.
The reason is purely practical: On the South Pole, there exists very significant infrastructure in the form of the permanently staffed Amundsen-Scott South Pole Station, including power generators, an airfield, telecommunication channels to get the data off to warmer offices ;) and of course permanent scientific and technical staff.
Alternative permanent research areas are mostly near the coast of Antarctica, and there indeed the seeing would be much worse. The two exceptions are the Russian Vostok station and the French/Italian Concordia station, but the South Pole Telescope is mostly US-funded. What is more, there are dedicated pots of funding available for research in this location, as opposed to e.g. some high-altitude site in Chile.
Without a real scientific reason, it is clear that the question of why the Amundsen-Scott station is exactly at the pole is not a scientific one, but is geo-political in the context of the emerging cold war in the 1950s.
That said, the extremely clean and dry air and the high altitude (almost 3km) at the south pole are crucial for a millimeter-telescope. But this location also means that the telescope has always the same sky above. That would be a show-stopper for a survey telescope, but is an advantage if you want to collect detailed ("deep") information about whatever area in front of you. Also it is nice when building a telescope since a simple azimuthal mount at the pole is in fact an equatorial mount there.
Finally, there is a long history of telescopes and instruments at the south pole for exactly these reasons. That in turn means that any new telescope, such as the South Pole Telescope, can now capitalize on a well-studied site, and can observe the same patches of sky to build on previous observations.
| {
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"timestamp": "2023-03-29T00:00:00",
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Green's function approach in Tight-Binding I am studying single-particle Green's functions using Economou's textbook, and am interested in using them to calculate surface states in tight-binding models. What I don't really understand is what benefit the Green's function method provides Green, as opposed to just using the Hamiltonian itself. Ham
| One example quantity that one would like to know from a given Hamiltonian
is the density-of-states (DOS), which is related directly to measurable quantities, e.g. conductance. Instead of diagonalizing the Hamiltonian
to know the DOS, the trace of the imaginary part of the Green function then
gives the DOS.
By the way, the Green function is defined through Hamiltonian. So the question is a bit confusing.
| {
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When the angle of incidence is less than the critical angle, why is some light still reflected? When the angle of incidence is less than the critical angle, why is some light still reflected? When the angle of incidence is greater than the critical all light is reflected, why is not all light transmitted when it's less?
| Light is an electromagnetic wave, and as such is governed by the Maxwell equations. These equations (considering only the linear mediums 1 and 2) give certain boundary conditions:
$$
(\text{i})\ \epsilon_1E^{\perp}_1=\epsilon_2E^{\perp}_2\ \ ;\ \ (\text{ii})\ \bf{E}^{\parallel}_1=\bf{E}^{\parallel}_2
$$
$$
(\text{iii})\ B^{\perp}_1=B^{\perp}_2\ \ ;\ \ (\text{iv})\ \frac{1}{\mu_1}\bf{B}^{\parallel}_1=\frac{1}{\mu_2}\bf{B}^{\parallel}_2
$$
These conditions determine how the light behaves at the surface, and note that they imply the total E-parallel component in medium 1 is equal to the E-parallel component in medium 2. Mathematically,
$$
\bf{E}^{\parallel}_1=\bf{E}^{\parallel}_{\text{incident}}+\bf{E}^{\parallel}_{reflected}=\bf{E}^{\parallel}_{transmitted}
$$
This happens independently of the angle of incidence, hence there must be a reflected wave in order to have a refraction. You can calculate the amplitude for each of the components and check that they obey it. I suggest reading chapter 9 of Griffiths' "Introduction to Electrodynamics" for a better understanding.
| {
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What is the difference between $\psi$ and $|\psi\rangle$? My understanding is that $\psi(\vec{r}, t)$ and $|\psi(\vec r,t)\rangle$ are the same thing yet one expressed as a wave function and the other expressed as a vector in the Hilbert space. Is this true? Or is there a deeper difference between the two notations?
| Velut Luna gives the main answer. One can see this because we have the probability expectation $1~=~\langle\psi(t)|\psi(t)\rangle$ and with the completion sum $\mathbb I~=~\int d^3r|\vec r\rangle\langle \vec r|$ we then have
$$
1~=~\langle\psi(t)|\psi(t)\rangle~=~\langle\psi(t)|\left(\int d^3r|\vec r\rangle\langle\vec r|\right)|\psi(t)\rangle~=~\int d^3r\langle\psi(t)| \vec r\rangle\langle\vec r|\psi(t)\rangle.
$$
In the wave function form we have unity of probability as
$$
\int d^3r\psi^*(\vec r,t)\psi(\vec r,t).
$$
the identification is obvious.
| {
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What determines invariant mass? The only answer I have been able to find is that energy determines mass but photons have energy yet are still massless. Furthermore this then leads to the question of what determines invariant energy, which I would think to be mass. So in total what are the circumstances that determine fundamental attributes of particles such as mass and energy?
| Very succinctly: a system's total energy, as measured from its rest frame, is the invariant mass. The rest frame is determined by the frame wherein a system's total momentum is nought.
You are probably a little confused by the photon and other so-called massless objects. Such objects are always measured to have a speed of $c$ relative to any observer, so there is no rest frame for such systems.
This succinct definition naturally leads to the formulas for rest mass determination in Yly's answer. For massless particles, the total energy is all associated with the particle's momentum.
I suspect you may be reading too much into the word "mass". See my answer here for some more thoughts along these lines.
| {
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Are there superconformal field theories in 10D? I've heard that there is a belief that interacting conformal field theories do not exist in dimensions greater than 6, and in 6D the only known nontrivial CFTs are superconformal field theories. What is the argument that these theories can't exist in higher dimensions? I have a sense that 10D is rather special, and I wonder if there could exist a superconformal field theory in 10 spacetime dimensions.
An explanation of the general argument for/against higher dimensional (super)conformal field theories, and/or links to references discussing this would be very helpful.
| As said in the comments, there are no superconformal field theories in $D>6$ dimensions. The references for this result are
*
*Werner Nahm, Supersymmetries and their Representations, Nucl.Phys. B135 (1978) 149.
*Shiraz Minwalla, Restrictions imposed by superconformal invariance on quantum field theories, Adv. Theor. Math. Phys. 2, 781 (1998) (arXiv:hep-th/9712074).
| {
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Partial Derivatives Relation on Thermodynamics and minus sign
My problem is to show that
$$ \left( \frac{\partial C_{V}}{\partial V} \right)_{T} = -T \left[ \frac{\partial (\alpha/\kappa_{T})}{\partial T} \right]_{V}$$
where
$C_{V}$ is molar specific heat capacity of constant volume and $$\alpha := \frac{1}{V} \left(\frac{\partial V}{\partial T} \right)_{P}$$ which is Thermal Expansion coefficient and $$\kappa_{T}:=-\frac{1}{V} \left(\frac{\partial V}{\partial P} \right)_{T}$$ is isothermal compressibility.
My attempt: First, $\displaystyle{C_{V}=T\left( \frac{\partial S}{\partial T} \right)_{V}}$. So by chain rule, $$\left( \frac{\partial C_{V}}{\partial V} \right)_{T}=T\left( \frac{\partial}{\partial V} \left( \frac{\partial S}{\partial T} \right)_{V} \right)_{T}+\left( \frac{\partial T}{\partial V} \right)_{T} \left( \frac{\partial S}{\partial T} \right)_{V}=T\left( \frac{\partial}{\partial V} \left( \frac{\partial S}{\partial T} \right)_{V} \right)_{T}$$
We can interchange the order of differentiation so $\displaystyle{\left( \frac{\partial C_{V}}{\partial V} \right)_{T}=T\left( \frac{\partial}{\partial T} \left( \frac{\partial S}{\partial V} \right)_{T} \right)_{V}}$. Since $\displaystyle{\left( \frac{\partial S}{\partial V} \right)_{T}=\left( \frac{\partial P}{\partial T} \right)_{V}}$ by Maxwell's Relation and using the fact that $\displaystyle{\alpha/\kappa_{T}=-\left(\frac{\partial V}{\partial T} \right)_{P} \left(\frac{\partial P}{\partial V} \right)_{T}=\left( \frac{\partial P}{\partial T} \right)_{V}}$, I get the result without the minus sign. Where does minus sign come from??
| I started with $$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$ So, $$\frac{\partial ^2U}{\partial T \partial V}=\left(\frac{\partial C_v}{\partial V}\right)_T=-\left(
\frac{\partial \left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]}{\partial T}\right)_V$$
The rest is easy.
| {
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Could a planet with a strong magnetic field exert a diamagnetic force on an orbiting moon? Here is a question from the world building stack exchange.
https://worldbuilding.stackexchange.com/questions/79003/making-a-slow-orbit-around-a-large-gas-giant
Requested: a means to have a moon of Jupiter orbit closely but slowly. Gravity precludes this.
If there were some force that opposed gravity, making the attractive force weaker, then closer slower orbits become possible. If gravity were completely opposed the object could hover.
Which is like diamagnetic levitation. The question:
*
*Is the magnetic field (magnetic gradient?) of Jupiter adequate to exert a repulsive force on a hypothetical diamagnetic satellite?
*A superconducting satellite?
*Could this force be on the order of gravity such that a slower orbit for this satellite is possible at the same distance from Jupiter?
| May be if the satellite have some electric charge the Lorentz's force applies, and act in the radial direction.
I think that to have a diamagnetic satellite has any effect in radial acceleration, may be in $\hat{\phi}$ direction (in spherical coordinate system with the origin in Jupiter's center).
| {
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Why energy corresponding to most probable speed is not equal to most probable energy? According to Maxwell speed distribution of molecules of gas at temperature $T$, most probable speed is given by
$$v=\sqrt{\frac{2kT}{m}} \, .$$
The corresponding energy is
$$E= kT \, . $$
According to the Maxwell Boltzmann energy distribution, the most probable energy is
$$E_p = \frac{kT}{2} \, .$$
Why are these different?
| Lets think about exactly what the Maxwell-Boltzmann distribution tells us. If we are being lazy we might say that it gives us the probability that a particle has a speed $v$, however since $v$ can take a continuous range of values, the probability that it takes any given value is $0$. Instead we say that it gives us the probability that a particle has a speed in a narrow range form $v$ to $v+\mathrm{d}v$. Now it is the probabilities, and not the probability densities, that physically matter and so should be independent of how we choose to represent them. Therefore
$$
f(v)\mathrm{d}v = f(E)\mathrm{d}E
$$
and not, as you might have naively though $f(v)=f(E)$. This is important because, since the relationship between $v$ and $E$ is non-linear, its sizes gets distorted as we make the transformation and $\mathrm{d}v\ne\mathrm{d}E$. This means that, even at corresponding energies and velocities, $f(v) \ne f(E)$. In fact, as @gautampk points out, since $v\propto E^{\frac{1}{2}}$, $\mathrm{d}v\propto \frac{1}{2}\mathrm{d}E$, leading to exactly the discrepancy you noticed.
This is related to a concept called the density of states, a function which counts the "number of ways" to have an energy in the range $E$ to $E+\mathrm{d}E$, or alternatively how much the length $\mathrm{d}v$ has been squashed and stretched. It is the $\sqrt{E}$ bit in the Boltzmann distribution for energy (along with some of the constants out the front). This is actually one of the rare moments where quantum mechanics makes life simpler. Our gas of particles is really quantum mechanical if we look closely enough and a gas of quantum particles in a finite volume has a discrete set of energy states. The density of states then really is counting how many of these discrete states have an energy in a certain range.
| {
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Crash simulation on Mythbusters I remember an episode of mythbusters where they were busting myths to do with a head on collision between two cars.
They said that instead of crashing two cars into each other at 50mph they would crash a car into a stationary object at 100mph because the energy involved in the crash would be the same.
Later on they corrected themselves to say that the energy is not the same. But I can't figure out why this would be the case?
Can someone explain if these two scenarios are the same or not. And why?
| The Mythbusters hosts realized that the important issue in a head-on collision is what is happening inside your vehicle. There is indeed twice the kinetic energy involved with one car colliding with an immovable object at 80 mph vs. two cars colliding head-on at 40 mph. However, Newton's third law requires the immovable barrier to provide an equal and opposite force when it stops a car that collides with it. This means that from the standpoint of an individual driver, there is no difference between two cars colliding head-on at 40 mph, and a single car colliding with an immovable barrier at 40 mph. In principle, you and your car experience the same stopping force in both instances.
| {
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Monopole Spherical Harmonics I was following Yakov Shnir his book about magnetic monopoles, there they derive the monopole spherical harmonics. I will sketch the derivation briefly,
The starting equation is the eigenvalue problem of the total angular momentum,
\begin{align}
\hat{\textbf{J}}^2 Y_{\mu lm}(\theta, \phi) &= \hbar^2 \Big\{\frac{-1}{\sin^2 \theta} \Big[ \sin \theta \frac{\partial}{\partial \theta} \big( \sin \theta \frac{\partial}{\partial \theta} \big) + \big( \frac{\partial}{\partial \varphi} - i\mu (1-\cos \theta )\big)^2 \Big]+ \mu^2\Big\} Y_{\mu lm}(\theta, \phi)\\&= \hbar^2\lambda Y_{\mu lm}(\theta, \phi).
\end{align}
The $\theta$-parameter can be replaced by the relation $x=\cos\theta$ and the $\phi$ dependence can be eliminated by $Y_{\mu lm}(\theta, \phi)= \exp(i(\mu+m)\phi) P(\theta)$,
\begin{equation}
\label{eq: diff eq angular part1}
\Big\{ -(1-x^2)\frac{d}{dx^2} + 2x \frac{d}{dx} + \frac{(m+\mu x)^2}{1-x^2} +\mu^2 \Big\} P(x) = \lambda P(x).
\end{equation}
To solve the problem we want to rewrite it as the Euler's hypergeometric function. To obtain that form $P(x)$ has to be rewritten as,
\begin{equation}
P(x) = (1-x)^{-\frac{\mu +m}{2}} (1+x)^{-\frac{\mu -m}{2}} F(x),
\end{equation}
equation (\ref{eq: diff eq angular part1}) will then become,
\begin{equation}
(1-x^2) \frac{d^2F}{dx^2} + 2 ( m+(\mu-1)x) \frac{dF}{dx} + (\mu - \mu^2 + \lambda)F = 0.
\end{equation}
If now the transformation $z= (1+x)/2$ is applied, then we have rewritten the differential equation as the Euler's hypergeometric differential equation,
\begin{align}
z(1-z) F'' &+ \big\{ m-\mu + 1 + 2z(\mu-1) \big\} F' + (\mu -\mu^2 + \lambda) F \\
& = z(1-z) F'' + \big\{ c-(a+b+1)z \big\}F' -abF = 0,
\end{align}
where $c=m-\mu+1$,$~ab = \mu^2 -\mu -\lambda$ and $a+b+1 = 2(1- \mu)$. The solution of this differential equations is the hypergeometric function $_2F_1(a;b;c;z) $ which is defined as,
\begin{equation}
_2F_1(a;b;c;z) = \sum_{n=0}^{\infty} \frac{(a)_n(b)_n}{(c)_n} \frac{z^n}{n!} \equiv \sum_{n=0}^{\infty} \alpha_n z^n.
\end{equation}
Now my question is the following, they consider that $a$ is negative integer such that the sum terminates, my question now is why?
Because if the sum not terminates then will the convergence radius be 1, but since $\theta$ is between $0$ and $\pi$ and by definiton of $z$ you still have a bijection from $z$ to the range of $\theta$.
This indeed doesn't say anything about being square integrable but how could one see this from the hypergeometric function/series? In fact I am searching for an analytical argument why the quantum numbers only differ in integers?
Thanks in advance.
| This is not a direct answer to the question asked, but rather a different approach to the problem. It's much easier to derive the monopole harmonics from a geometric consideration of the Hopf map ${\rm Hopf}: S^3\to S^2$, which is the same thing as the Hopf bundle $\pi: {\rm SU}(2) \to {\rm SU}(2)/{\rm U}(1)$. Then one immediately sees monopole spherical harmonics coincide with the ${\rm SU}(2)$ represenation matrices. Recall that the ordinary spherical harmonics are given in terms of these matrices as
$$
Y^L_m(\theta, \phi)= \sqrt{\frac{2L+1}{4\pi}}[D^L_{m,0}(\theta, \phi,\psi)]^*
$$
where, on the LHS, $\theta$ $\phi$ are the spherical polar angles on $S^2$. On the RHS $\theta$, $\phi$, $\psi$ are the Euler-angle co-ordinates on $S^3$. The Hopf map simply takes $(\theta, \phi, \psi)$ to $(\theta, \phi)$. Because
$$
D^L_{m,n}(\theta, \phi,\psi)= <L,m|\exp(-i\phi J_z)\exp(-i\theta J)\exp(-i \psi J_z)|L,n>
$$
the RHS does not depend on $\psi$ when $n=0$.
For a monople of strength $\int B d(Area)= 4\pi \Lambda $, where $4\pi \Lambda $ must be an integer multiple of $2\pi$, the monopole harmonics are
$$
{\mathcal Y}^J_{m, \Lambda} (\theta, \phi,\psi)= \sqrt{\frac{2J+1}{4\pi}}[D^J_{m\Lambda }(\theta,\phi,\psi)]^*
$$
Now both sides depend on $\psi$, so one must choose a $\psi$ for each pair $(\theta, \phi)$. This is just a choice of gauge.
For a derivation and explanantion of all this see page 278 in my online lecture notes https://courses.physics.illinois.edu/phys509/sp2017/bmaster.pdf.
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Is Frobscottle from the movie 'The BFG' less dense than air? For those who have either read the book, or watched the movie "The BFG", you would know Frobscottle as a green drink the giant uses, and has bubbles fizzing "in the wrong way", which is downwards. Assuming the bubbles to be filled with air, and that gravitational force on the bubble is greater than buoyant force, does this imply Frobscottle is less dense than air? Furthermore, is a liquid possible that is less dense than air?
| It implies that the author made it up without worrying about physics. If Frobscottle was less dense that air, it wouldn't stay in a cup. It would float like a helium balloon.
There is a liquid less dense than air, but it is nothing you could put air bubbles in. It is 3 monolayers of $He^3$ adsorbed on graphite at temperatures below 80 milliKelvin. First, the liquid is only 3 atoms deep. Second, air freezes solid at that temperature.
See http://www.u-tokyo.ac.jp/en/utokyo-research/research-news/lowest-density-liquid-in-nature/ and https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.109.235306.
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Bell Inequalities - Expectation Values I'm currently reading Loopholes in Bell Inequality Tests of Local Realism by Jan-Ake Larsson
https://arxiv.org/abs/1407.0363
On Page 6, Equation 7, he has a short proof, where I am having a hard time seeing through the math. I'll re-state here for convenience:
$ \lvert E(A_{2}B_{1}) - E(A_{2}B_{2}) \rvert = \lvert E(A_{2}B_{1} + A_{2}B_{1}A_{1}B_{2}) \rvert \leq E(\lvert A_{2}B_{1}(1 + A_{1}B_{2}) \rvert ) = 1 + E(A_{1}B_{2}) $
where
$ A_{1}B_{1} = -1 $
and
$E(A) = \int A( \lambda ) \rho ( \lambda) d \lambda $
I'm not seeing how he goes from the 2nd expression on the LHS of the less-than-or-equal sign to the third expression on the RHS of the less-than-or-equal sign. I suspect I am missing something on the properties of expectation values. Mainly, I think the absolute signs (|) moving from outside of the E()'s to the inside of the E()'s have me the most confused. Further, I don't see how he goes from the 3rd expression to the 4th, either.
Can anyone offer any clarification here?
| From 2nd to 3rd expression:
It is well known that:
$$\left|\int f(x)dx\right| \leq \int \left|f(x)\right| dx.$$
Here you can find the reference for this fact.
Now, you have that:
$$\lvert E(A_{2}B_{1} + A_{2}B_{1}A_{1}B_{2}) \rvert = \left|\int (A_{2}B_{1} + A_{2}B_{1}A_{1}B_{2})\rho(\lambda)d\lambda \right| \leq \\
\leq \int \left|(A_{2}B_{1} + A_{2}B_{1}A_{1}B_{2})\rho(\lambda)\right|d\lambda = \int \left|A_{2}B_{1}(1 + A_{1}B_{2})\rho(\lambda)\right|d\lambda.$$
Since $\rho(\lambda) \geq 0$ by definition (it is a distribution), then $\rho(\lambda) = |\rho(\lambda)|$, and hence:
$$\int \left|A_{2}B_{1}(1 + A_{1}B_{2})\rho(\lambda)\right|d\lambda = \\
\int \left|A_{2}B_{1}(1 + A_{1}B_{2})\right|\rho(\lambda)d\lambda = E(|A_{2}B_{1}(1 + A_{1}B_{2})|).$$
From 3rd to 4th expression:
Suppose that $B_1=1$. Then:
$$E(|A_{2}B_{1}(1 + A_{1}B_{2})|) = E(|A_{2}(1 + A_{1}B_{2})|) = E(|A_{2}||1 + A_{1}B_{2}|).$$
Since $A_2 \in \{-1, +1\}$, then $|A_2| = 1$, and hence:
$$E(|A_{2}B_{1}(1 + A_{1}B_{2})|) = E(|1 + A_{1}B_{2}|).$$
Since $B_1 = 1$, then $A_1 = -1$, and the term $1 + A_{1}B_{2}$ can be equal to $0$ or to $2$. In each case, it is positive, so we can drop the modulus:
$$E(|A_{2}B_{1}(1 + A_{1}B_{2})|) = E(1 + A_{1}B_{2}).$$
Last steps are easy...
$$E(|A_{2}B_{1}(1 + A_{1}B_{2})|) = E(1 + A_{1}B_{2}) = E(1) + E(A_1 B_2) = 1 +E(A_1 B_2).$$
Now suppose that $B_1 = -1$. Then:
$$E(|-A_{2}(1 + A_{1}B_{2})|) = E(|A_{2}(1 + A_{1}B_{2})|).$$
Using the same reasonings of above, we get that:
$$E(|-A_{2}(1 + A_{1}B_{2})|) = E(|1 + A_{1}B_{2}|).$$
$A_1$ is $1$, and $1 + A_{1}B_{2}$ can be $0$ or $2$. We can drop the modulus... etc... etc... Even in this case, we get that $E(|A_{2}B_{1}(1 + A_{1}B_{2})|) = 1 + E(A_{1}B_{2})$.
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Friction on cars It is known that friction is given as :
$F_{friction}=\mu F_n$ , where $F_n$ is the normal force, and $\mu$ is coefficient of friction.
For a car travelling down a hill with constant velocity, the component of the gravitational force which is parallel to the cars velocity must be equal and opposite to the frictional force, whereby the frictional force opposes the motion of the car.
However, when the car is going up the hill, for a constant velocity to be obtained, the frictional force must be going up the hill, in the same direction as the motion of the car, and equal and opposite to the gravitational force which is antiparallel to the cars velocity.
I thought friction always opposes motion?
How can a car accelerate with the same force (i.e. friction) which also causes it to slow down. If there is no friction, a car cannot accelerate?
| When your wheel is powered by the engine the wheel pushes backward and reactionary friction force opposing this pushes your car forward.Same as the case when we walk.The feet push backwards and the reactionary frictional force pushes us forward.The friction is in a direction opposite to the direction we are pushing in and in the direction we want to move.
When the car is moving without powering the wheel the wheel is not pushing back but is simply rolling forward and the frictional force develops accordingly to oppose this.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does an increase in entropy always result in an increase in heat, or can there be increased entropy without an increase in heat? Most situations I can think of where entropy increases also results in an increase in heat, but just wondering if that is a rule. Are there any cases where heat does not increase with entropy?
| Your question has one unclear definition: heat. I guess that is temperature.
Here is one simple example. A box is filled with gas and the box wall is well insulated. If, somehow, the box volume can be doubled, the entropy increases and the gas temperature decreases.
| {
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Metamaterial : Snell's law and Fermat principle How do we deduce Snell's law using Fermat's principal in case of metamaterials? Metamaterials have negative refractive index. This makes the refracted ray of light bend on same side of normal as incident ray. But according to Fermat principle, the light could have taken a shorter path. Without the light 'bending backward', it would have a path which took lesser time, like one of the possible path I have shown below in red.
| What is the "travel time" of a beam of light in a medium with negative refractive index? Do you save time by traveling a greater distance? Time = distance/speed, and speed = c/n. When $n<0$, the part of the trajectory in the meta-material will contribute "negative time" to the over all travel time of the beam.
Remember, Fermat's "shortest" means least time, not least distance. And it's just a mathematical construct.
| {
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Does energy conservation implies the mass conservation? According to Noether's theorem, every symmetry implies and conserved quantity. And, from Einstein's equation, every mass have an amount of energy associated.
Can it say that the mass conservation is a consequence of energy conservation associated to homogeneity of time?
PD.: is not duplicated question
| In classical mechanics mass is conserved, conservation of mass is not valid in relativistic physics unless you consider the so-called "invariant mass" of a closed system. The sum of the masses of particles can change due to interactions. In classical mechanics this sum of masses is conserved and this follows from conservation of energy.
If we have an initial state of $N$ free particles and a final state of $M$ free particles then the total energy is conserved. We then have:
$$\sum_{j=1}^{N}m_j v_j^2 = \sum_{j=1}^{M}m'_j w_j^{2}$$
where the $v_j$ and $m_j $are the initial velocities and masses, while the $w_j$ and $m'_j$ are the final velocities and masses. The particles being free particles only have a kinetic energy, but this still a completely general statement because in the intermediary stage, arbitrary interactions can have taken place. We're then only neglecting radiation escaping from the system (note that radiation is an extremely relativistic phenomenon).
Then suppose that the exact same process is observed in another frame that moves with velocity $\vec{u}$ relative to the original frame. According to Galilean invariance, we also have conservation of energy in this new frame, therefore:
$$\sum_{j=1}^{N}m_j (\vec{v}_j-\vec{u})^2 = \sum_{j=1}^{M}m'_j (\vec{w_j}-\vec{u})^{2}$$
Expanding out the square of each term:
$$ (\vec{v}_j-\vec{u})^2 = v_j^2 - 2\vec{v}_j\cdot\vec{u} + u^2$$
on both sides, and using that $\vec{u}$ is arbitrary, yields conservation of momentum and conservation of mass.
| {
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How are the constants for the NFW profile determined? I'm trying to understand the NFW profile and how it causes a flat rotation curve.
$$ \rho(r) = \frac{\rho_0}{\frac{r}{r_s}\left(1+\frac{r}{r_s}\right)^2} $$
I think I've got why it causes a flat rotation curve. The density seems to be inversely proportional to $r^3$ when $r >> r_s$ leading to a linearly increasing mass in proportion to radius since volume is proportional to $r^3$.
However I realized I can't really prove my statement above as I don't really understand what $r_s$ and $\rho_0$ are. Thus I was hoping to see a numerical example to get my head around the relationships.
While looking for one I came across this answer by Kyle Oman applying the NFW profile to calculate the dark matter density in the Solar System and was wondering where he got his $r_s$ and $\rho_0$ values or how one calculates them themselves.
| Why does it matter what their values are? $r_s$ is some characteristic scale length of the system where it switches from a $r^{-1}$ dependence to a $r^{-3}$ dependence and $\rho_0$ is a normalisation constant that ensures that the total mass within some radius is correct.
The values of these constants are obtained by fitting the function to an observed density profile or modelling the rotation curve of a galaxy.
If
$$
\rho(r) = \frac{\rho_0}{\frac{r}{r_s}(1+\frac{r}{r_s})^2}
$$
Then the mass of a shell of thickness $\Delta r$ at radius $r$ is
$$\Delta M = \frac{\rho_0}{\frac{r}{r_s}(1+\frac{r}{r_s})^2}\ 4\pi r^2 \Delta r\ .$$
When $r \ll r_s$
$$\Delta M \sim 4\pi r_s r \rho_0\ \Delta r$$
and by integrating shells, the total mass inside radius $r$ increases as $r^2$.
When $r\gg r_s$
$$\Delta M \sim 4\pi \frac{r_s^{3}}{r} \rho_0\ \Delta r$$
and the total mass within $r$ increases as $\ln r$.
Typical values found for the Milky Way dark halo are $r_h \simeq 12$ kpc and $\rho_0 \simeq 10^{-2}$ $M_{\odot}/{\rm pc}^{3}$. See for example Sofue (2012) who does exactly the calculation that Kyle Oman did in the Physics SE question that you referred to.
| {
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What is the energy of a single charge system? I will try to limit the question in the case of the electric fields, but is something that applies also to the magnetic ones.
There are two ways to express the energy in a capacitor:
*
*By Voltage : $U = 1/2 CV^2 $
*And by Field : $U = 1/2 \varepsilon E^2Ad$, With Energy Density: $u = 1/2 \varepsilon E^2 $
Unless i understood everything wrong and these two are NOT the same quantity, i have the following question. When we have two charges placed at points A and B, then in order to calculate the energy of the system, we will take the first charge, place it at point A WITHOUT doing any work, and then we will calculate the work needed to place the second charge at point B.
The weird thing to me here, is that while we have placed the first charge, without generating any work the system will still have the energy held in the field of the charge!
There is obviously something that i miss, but what?
| Besides the things mentioned in the other answers, here is one more thing to consider:
The energy of the single charge at some point in space can be thought of as arrived at by starting with an infinitesimal charge, and adding (from infinity) a little bit of charge. That will take a little bit of work. The next bit of charge will take more work, and so it continues. In other words - a single (finite) charge (assumed distributed over a small region of space - you can't have a finite charge in an infinitesimal volume) must have some energy associated with it.
| {
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Size of mercury barometer and effect on its reading I was thinking that since atmospheric pressure is 760mmHg what would happen if I shrink a mercury barometer until it's shorter that 760mm in height. What would happen? Would the barometer retain it's ratio of mercury height? Will the mercury fully fill the barometer? Is 760mmHg a constant and not affected by the size of the barometer (or diameter)? Thanks in advance.
| A vacuum must exist above the meniscus of the column so that the mercury
can rise above 760mm (or 30 in.) which is what a mercury barometer reads
at sea level. In other words, the mercury column needs headroom to rise
with increasing ambient atmospheric pressure.
A barometer capillary tube with a ball pediment cistern must have an overall
length of 33 inches to allow for the column to rise to 31 inches (which is
rare here in the Midwest, but may read higher if your location is below sea level). Search "barometer ball pediment tube" for clarification.
| {
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Why young modulus value doesn't change even if we change the parameters? The Young modulus of steel is determined using a length of steel wire and is found to have the value EE. Another experiment is carried out using a wire of the same steel, but of half the length and half the diameter.
What value is obtained for the Young modulus in the 2nd experiment?
Why doesn't yhe values change even if we are changing the parameters
| Young's Modulus is measuring the ratio of stress to strain.
Stress is a measure of the forces in a material. It has units of force per unit area. This is the reason why changing the diameter does not affect the experiment. If the wire is thinner you may have to apply a different force to get the same strain (elongation/stretching).
The stress on the other hand, is the variable we are concerned with, and will depend on both the applied load and the geometry. We have determined through experiments that many materials have a region where stress is linearly proportional to strain. The slope of that linear relationship is known as Young's Modulus of elasticity.
Also consider how tension acts when considering what a change in length will do. In a (ideal) wire, tension is the same through the whole thing. Because of that; shortening the rope will not change the applied load (but it will change the total length; but not the % elongation).
| {
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Field due to internal Induced charge on a conductor to an external point? A charge q is located at a distance r from the center of a conducting sphere with inner radius 2r. The charge induces charges on the inner surface of the sphere according to Gauss' law .
The electric field at point p is to be approximated.
Inside the material of the conducting sphere, the electric field due to induced charge will cancel out the electric field due to the charge inside the sphere. Accordingly the electric field lines will begin at induced charge and terminate at the inner charge.
Therefore the field due to internal induced charge on the point p must be zero , (note it may be nonzero due to external induced charge but the problem specifies internal)
The solution however says it to be $kq/17r^2$ and not zero
Isn't the electrostatic system shielded from the conductor?
| The electric field at the point P is solely due to the charges on the outer surface of the sphere [Suppose this was not true, for the sake of contradiction.The only way this can happen is if the magnitude of induced charge on the inner surface of the sphere is not equal to the q itself.If this happens, then by Gauss' law, we have a non-zero electric field in the meat of the conductor, which is impossible. This proves that the field at the point P is solely due to the charges on the outer surface of the sphere].
Thus, the the field due to the charge has to be cancelled by the charges induced on the inner surface. The induced charges have to be opposite to that of the charge q. The electric field due to the charge q at the point P from Coulomb's law is $kq/17r^2$. So, the field due to the induced charges on the inner surface has to be $kq/17r^2$ in magnitude but opposite in direction to that of the field direction due ti charge q alone.
| {
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Gauss Law Doubt I have a doubt regarding whether or not I can take a spherical surface as a Gaussian surface when the taken Gaussian surface completely overlaps a symmetrical spherical charge distribution of same radius.
Won't it violate the fact that a Gaussian surface can not have charges on it.
Or would be okay for a continuous distribution.
Finally when we say continuous do we we mean continuous on the surface or do we mean continuous into the surface and out of the surface.
| Gaussian surface is an imaginary plane which we draw so that we would know how much of electric field lines are passing through that imaginary area we took(flux).Hence it will give perfectly correct answer. This can sbe proved by using Coulombs law and Gaussian law for the same surface
And the charge "+q" is already distributed on the surface in the figure
1)Applying coulombs law $$\sigma=q\left(4\pi r^2\right)$$=The charge density by area.
By coulombs law ,$$E=-\frac{dV}{dr}$$Hence total potential at the center
is $$\frac{K\sigma\left(4\pi r^2\right)}{r}$$
$$E=-\frac{d\left(\frac{K\sigma\left(4\pi r^2\right)}{r}\right)}{dr}$$
$$E=\frac{\sigma}{\epsilon_{0}}$$
2) by Gauss law
$$\oint E dS=\phi$$
Now according to gauss law$\phi$ for any closed symmetric surface is $$\frac{q}{\epsilon_{0}}$$
So And distribution of the charge is on the surface so
$$E4\pi r^2=\frac{q}{\epsilon_{0}}$$
$$E=\frac{\sigma}{\epsilon_{0}}$$
Hence proved Gauss law is correct for this case also
| {
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Why do colours of object change due to incident light? A leaf is green, a pen is blue and so on because those objects absorb all colours and reflect only one colour. However when red light is incident on these objects, their colour becomes reddish. Why is that the case?
| In an ideal situation, i.e. if
*
*really all colors except one are absorbed
*your red light is perfect (one wavelength)
the objects should actually appear black, i.e. no light is reflected.
In the real world a blue pen will reflect a small amount of red (and other colors). That's why it appears reddish if you shine a red light on it. Likely it would appear darker than if you shine blue or white light at the same intensity on it.
| {
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Does this make sense from mass-energy equivalence standpoint? Over the weekend I was (in over my head) discussing mass/energy and I proposed this thought experiment.
Given two separate but equal quantities of matter - one being wood, the other being gasoline - do they both have the same total energy? I am being assured that the gasoline has more every than the wood but if the mass is equal shouldn't they have the same theoretical energy? What is the name for that type of energy? Is it simply a matter of the gasoline being in a form that is more readily extractable (explosion vs burning)?
| What is storing energy?
There is a mass-energy equivalence, which since they have the same mass is equal. We have no idea how to get anything but trivial fractions of this energy through the below mechanisms.
There are nuclear bonds, which since they are both mostly hydrocarbons are probably pretty similar, though having different fractions of the different elements will mean there is a difference. We can make nuclear reactions, but wood and gas are not used in any known fusion or fission process.
There are Chemical bonds, and here we get interesting differences. We can break those bonds and then the energy released from the bonds is more than it costs to break the bonds, and a chain reaction happens. Gasoline has more energy in it's bonds than wood (why it burns better).
The chemical bonds are trivial to the nuclear bonds and the nuclear bonds are pretty trivial to the energy-mass equivalence, so if you made a not quite strict accounting the energy could be said to be equal because all the chemical and even nuclear energy could be rounded off. But if you mean usable energy only the chemical energy is at all important so the gasoline has more.
| {
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Would a handspinner spin indefinitly in space? I'm having a argument with a colleague, I don't know how to explain to him that if you spin a handspinner in space it will spin indefinitly (if you don't hold it). I agree that if you hold it, it will slow down because of the friction with the center part.
Would it theoreticaly spin forever?
| Space is not a perfect vacuum, there are gases, electromagnetic fields and so on. The rotating object would be slowing down its rotation and in time, its motion will get erratic, driven by fluctuations of the environment.
| {
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How to evaluate the critical exponents of modified van der Waals equation? The given modified van der Waals equation is $$(P+(a/v)^{n})(v-b)=RT$$ where $(n>1)$. What is the physical significance of the power $n$ in the above equation. How could one evaluate the critical constants and exponents for this modified equation ? Suggest some reference to read this
| The modified Van Der Walls Equation is
$(P+\frac{an^2}{V^2})(V-nb)=nRT$......... $(1)$
where $a$ is a Van Der Walls constant whose value depends upon intramolecular forces of attraction within the gas. $b$ is the correction term for Volume of gas.
Putting $n=1$ in $(1)$,
$(P+\frac{a}{V^2})(V-b)=RT$......... $(1)$ or
$V^3-(\frac{RT}{P}+b)V^2+\frac{a}{P}V-\frac{ab}{P}=0$.......... $(2)$
This equation has three real roots. Also, at critical condition
All three roots are equal
So, $V_1=V_2=V_3=V_c$ or
$(V-V_c)^3=0$ or
$V^3+3VV_{c}^2-3V^2V_{c}-V_{c}^3=0$.........$(3)$
Now, on comparing the coefficients of $(2)$ and $(3)$,you get
$V_{c}^3=\frac{ab}{P_c}$,
$\frac{RT_c}{P_c}+b=3V_c$ and
$\frac{a}{P_c}=3V_{c}^2$
Solving,
$V_c=3b$
$P_c=\frac{a}{27b^2}$
$T_c=\frac{8a}{27Rb}$
where $V_c, P_c$ and $T_c$ are critical Volume, Pressure and Temperature respectively.
Hope this helps, brother.
| {
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Master Equation under a classical fluctuating noise I have a system as a qubit with Hamiltonian
$H_S = \frac{\Delta}{2}\sigma_z$
The interaction Hamiltonian is
$H_I = \frac{V(t)}{2}\sigma_z$ where $V(t)$ is a stochastic fluctuating variable. One can for example assume it as a random telegraph noise(RTN). In this case, what is the general prescription to write down the master equation for the qubit?
| The book by Klyatskin discusses general functional methods for dealing with gaussian and telegraph noises.
However, let me make a few general remarks:
*
*Telegraph noise is not delta-correlated (white), which means that one would get an integral equation, rather than usual master equation.
*Including noise as an external force, without taking into account its respose to the qubit dynamics, breakes the fluctuation-dissipation theorem. Thus, the resulting equation will not contain dissipative/relaxation terms (again, it is not a master equation, properly speaking).
*Writing a master equation for a two-level system is a bit of overkill, since Bloch equations are enough for a two-level system.
| {
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Correct definition of an 'acoustic mode'? I am reading 'The Oxford Solid State Basics' by S.H.Simon in which on page 92 defines an acoustic mode as:
... any mode that has linear dispersion as $k\rightarrow 0$.
Whilst on page 94 he defines it as:
... one mode will be acoustic (goes to zero energy at $k=0$).
Unless all modes that tend to zero do so linearly and vice versa then these two definitions don't overlap. Thus my question is as follows: does one of these conditions imply the other and if not what is the correct definition for an acoustic mode?
| No, one does not imply the other, and I disagree with the first definition.
For example, the dispersion relation of the ZA mode in graphene goes to zero like $x^2$, so energy goes to zero as $k \to 0$ but does not do so linearly.
The 'A' in 'ZA' stands for acoustic, so that's an example of a nonlinear acoustic mode.
(That said, the first definition has some merit. The slope of a linear dispersion relation as $k \to 0$ is the speed of sound, which is a constant -- at least in isotropic materials. "Acoustic" modes get their name because they behave like sound at long wavelengths, and non-linear dispersion relations don't have a speed of sound. So there is logic in saying that non-linear dispersion relations are not acoustic. However, I don't think that's the common definition.)
| {
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How to explain the relationship between wave's amplitude and intensity? I have the following statement which I don't know how to explain:
Suppose I have 2 identical monochromatic waves (same intensity and phase) shooting into the same receiver. If each wave's intensity is I, based on energy conservation I would expect the 2 waves together will bring a total intensity of 2I. For example if each wave carries 100mW power, I'm expecting a 200mW total power on the receiver side.
However the summation using phasor gives a different result: if we consider that intensity is proportional to the square of wave's amplitude, the square root of 100 gives 10 for the amplitude of 1 wave (for simplicity I use 10 as amplitude which included the constants), adding the other wave of the same phase which gives 20 as amplitude of the new beam, then square it to get intensity which is 400mW instead of 200 in the above example.
If we keep going by summing 4 50mW waves with the same convention, we get (4 x sqrt(50))^2 = 800mW ... which the logic is obviously not correct.
My ultimate goal is to sum up power of beams with different intensities and phases. Phasor addition works great if I have amplitude and phase, but when I try to use intensity to get wave's amplitude I got the above dilemma. Could someone point out where my logic go wrong, and please explain the way to do power summation with intensity and phase known for each wave? Thanks!
| Your initial assumption that you can simply sum intensity (power) is incorrect.
The law of superposition of Maxwell's equations says that you can sum electric fields and magnetic fields -- or voltage and current in a transmission line or circuit context. So if you consider total voltage $V(t)=V_1(t)+V_2(t)$ and total current $I(t)=I_1(t)+I_2(t)$, then total power is $P(t)=VI=(V_1+V_2)(I_1+I_2)=V_1I_1+V_2I_2+V_1I_2+V_2I_1$. Notice those cross terms.
| {
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How to calculate altitude from current temperature and pressure? In a certain project, I need to calculate the altitude of the current location given the current location temperature and current location pressure. Temperature, pressure and altitude of a 'reference-level' could be provided if necessary (using a fixed sea-level pressure constant is also acceptable). This project is done between altitudes of-100 meters to 2000 meters above sea level.
This website uses the 'hypsometric formula':
$$h=\frac{((\frac{P_0}{P})^\frac{1}{5.257}-1)\times(T+273.15)}{0.0065}$$
given current location pressure, $P$, pressure at sea level, $P_0$, and current location temperature in Celsius, $T$.
However, I was also told by my friend that finding the altitude could also be calculated by the 'barometric formula':
$$h=44330\times\left(1-\left(\frac{P}{P_0}\right)^\frac{1}{5.255}\right)$$
which is obviously not equivalent to the first equation. Furthermore, this formula doesn't allow changes in temperature.
The 'barometric formula' given in Wikipedia is also different;
$$P=P_b\times\left[ \frac{T_b}{T_b+L_b\times(h-h_b)} \right] ^ \frac{g_0M}{R^*L_b}$$
This formula uses more constant values including the universal gas constant, $R^*$, the gravitational acceleration, $g_0$ and the molar mass of Earth's air, $M$. However, this formula isn't what I was looking for because it appears that the temperature at current location isn't taken into account.
My question is what equation is used to calculate current location altitude given current location temperature and current location pressure (or, if no such equation exists, best suited in range from sea level to 2 km altitude).
| All these formulas are actually equivalent when $P_b = P_0$, $T_b = 273.15 + T$, $T = 15$, $h_b = 0$ (sea-level), $L_b = -0.0065$ the standard lapse rate, and hence $\frac{g_0M}{R^*L_b} = -5.25579$.
| {
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Where does the force actual act?
Why are the two mass of $m_1$ and $m_2$ not multiplied by minus one?
I know that two minus multiplied gives you plus by but I mean the two masses are attracting so they should have a sign like so $$F_g=\frac{G(-m_1) \times (-m_2)}{r^2}$$
I ask these because equation would actual give the right explanation as to what the equation is actual doing, meaning that the two mass are attracting.
|
but I mean the two masses are attracting so they should have a sign like so $F_g=\frac{G(-m_1) \times (-m_2)}{r^2}$
I am not following the logic. Attraction implies reduction in distance, and not reduction in mass. Therefore the distance related quantities should contain the negative and not the mass ones. Maybe if you group the equation as
$$ \vec{F} = \left( \frac{ G m_1 m_2}{\| r \|^2} \right) \left(- \frac{ \vec{r} }{ \| \vec{r} \| }\right) $$
it would make more sense to you. The first part is the magnitude and the second part the direction.
| {
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Experimental proof pions are pseudoscalar particles In nuclear physics lessons I've been taught that pions are pseudoscalar particles and thus their intrinsic parity is odd. The professor said that an experimental proof of this can be derived observing the process:
$$
\pi^- + D \quad \longrightarrow \quad n + n
$$
where the pion is considered as an s-wave ($\ell = 0$ ) and the deuteron is in triplet state ($s=1$).
I can't understand why this decay can prove pions have a negative intrinsic parity.
| The answer is based on a comment by Cosmas Zachos.
The pion has no spin, and so can't flip the spin of either nucleon in the deuteron. The nucleons in the deuteron have their spins aligned ($S=1$), so the two neutrons in the final state must be in a spin triplet as well. In order for the final two neutrons to be antisymmetric under exchange, their wavefunction must have $L=\text{odd}$, which means the final state has odd parity.
If the final state has odd parity, and the $\pi^-\rm D$ are initially in an $s$-wave state, then the parities of the $\pi^-$ and the $\rm D$ must be opposite. The negative parity of the $\pi$ follows from the positive parity of the deuteron.
This seems to have been shown originally by Steinberger.
| {
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Easily approximate center of mass of a person when sitting What is the simplest way to get the center of mass of a human body when sitting? I'm especially interested in getting this when sitting in a chair, so this center of mass would include the chair. I'm trying to make an exercise device for myself that attaches to a lift that I installed on my ceiling. I use a wheelchair and don't get much movement other than pushing my wheelchair, so I'm hoping this will improve my health.
My intention is to have a bar stabilizing the chair, but I don't want a lot of torque/stress (axle) on the chair. You can think of the axle as the rod that might connect the inner gimbal of a gyroscope. But this will only rotate on one axis. I will simply be able to change my pitch with this chair hanging.
| This link should help:
https://www.faa.gov/data_research/research/med_humanfacs/oamtechreports/1960s/media/AM62-14.pdf
For a natural sitting position with hands in the lap, the center of gravity is about (from what I can make out from the text - it is not clear) 8 3/8" and 9 1/8" from the horizontal and vertical reference points, which I believe are taken to be the horizontal and vertical surfaces of the seat. This would put the centre of mass somewhere around the position of your navel.
Including a wheelchair in the centre of mass would shift it downward and rearward probably by a small amount - maybe somewhere closer to your lap, but it all would depend on your mass and the mass of your wheelchair, etc.
| {
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Time-dependent harmonic oscillator I am dealing with the problem to solve the following Schroedinger equation:
$$i\hslash\partial_{t} \Psi = ( -\nabla^2 +w^2(t) )\Psi$$
where the frequency of the oscillator depends on time. I tried to resolve it by using the Kruskal approach, as described in the following paper
"An Exact Quantum Theory of the TimeDependent Harmonic Oscillator and
of a Charged Particle in a TimeDependent Electromagnetic Field",
H. R. Lewis and W. B. Riesenfeld,
but without any results.
In particular I am having problems to find the $\rho$ function.
In my case the frequency $w^2(t)$ is:
$$e^{4(t+Ce^{t})}$$ where $C$ is a generic constant. Can you help me?
| (No mass term and no factors of 1/2? It is weird to me that you included $\hbar$ and not these other factors)
It seems like you may be able to use separation of variables. Just assume (I am going to just use 1 dimension right now as it should be the same process for 3)
$$\Psi = \psi(t)\phi(x)$$
This gives:
$$i \hbar \;\partial_t \psi (t) \phi(x) = -\hbar^2\psi(t)\partial_x^2\phi(x)+\omega^2(t)\psi(t)\phi(x)$$
Now divide by $\psi(t)\phi(x)$
$$ i \hbar \;\frac{\partial_t \psi (t)}{\psi(t)} = -\hbar^2\frac{\partial_x^2\phi(x)}{\phi(x)}+\omega^2(t) $$
Now assume $\frac{\phi_x^2 \psi (t)}{\phi(x)}$ is equal to a constant $K$ and you get two differential equations:
$$-i\hbar \frac{\partial_t \psi (t)}{\psi(t)} = \hbar^2 K + \omega^2(t)$$
$$\partial_x^2\phi(x) = K \phi(x)$$
The second one (the spatial one) is easily solved to be a sum of exponentials and the first one (I believe) can also be solved, perhaps using an integrating factor since it is only first order. You probably won't be able to put it in closed form, but it will be an integral solution that can be numerically calculated or put into an expansion form.
I haven't looked at the details of this but I think what I say will work....
| {
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How does the addition of two wavefunctions develop in time? Two time dependent wavefunctions:
$\Psi _1(t)= \psi_1*exp(\frac{-i * E_1}{\hbar}*t)$
$\Psi _2(t)= \psi_2*exp(\frac{-i * E_2}{\hbar}*t)$
Both a solution to the timeindependent (note "in") Schrödinger eq. with the same H. We know they are solutions. Furthermore $E_1$ and $E_2$ are different.
$\mid \psi_1exp(\frac{-i E_1}{\hbar}t) + \psi_2exp(\frac{-i E_2}{\hbar}t)\mid ^2 $
$= \mid \psi_1exp(\frac{-i E_1}{\hbar}*t)\mid^2 + \mid\psi_2exp(\frac{-i E_2}{\hbar}t)\mid^2 + 2 \mid\psi_1exp(\frac{-i E_1}{\hbar}t) \psi_2exp(\frac{-i E_2}{\hbar}t) \mid$
$= \mid \psi_1exp(\frac{-i E_1}{\hbar}*t)\mid^2 + \mid\psi_2exp(\frac{-i E_2}{\hbar}t)\mid^2 + 2 \mid\psi_1 \psi_2exp(\frac{-i (E_2-E_1)}{\hbar}t) \mid$
Is the following correct?:
$\mid \psi_1exp(\frac{-i E_1}{\hbar}*t)\mid^2 = \mid\psi_1^{(*)}\psi_1\mid = \mid\psi_1\mid^2* exp(\frac{-i E_1}{\hbar}t) * exp(\frac{i E_1}{\hbar}t) = \mid\psi_1\mid^2$
leading to:
$\mid \psi_1exp(\frac{-i E_1}{\hbar}t) + \psi_2exp(\frac{-i E_2}{\hbar}t)\mid ^2 = \mid\psi_1\mid^2 + \mid\psi_2\mid^2 + 2 \mid\psi_1 \psi_2exp(\frac{-i (E_2-E_1)}{\hbar}t) \mid$
meaning that:
$\mid\Psi _1(t)+\Psi _2(t)\mid $ ocilliates with $\frac{\hbar}{(E_2-E_2)}$ ?
| You obviously mean that
$$\begin{aligned}
H\psi_1 &= E_1\psi_1,\\
H\psi_2 &= E_2\psi_2.
\end{aligned}$$
Then, the solution of the Schrödinger equation (I will use units such that $\hbar=1$ throughout: old habit of a former theoretical physicist!!)
$$H\Psi_k = i\dfrac{\partial \Psi_k}{\partial t}$$
with the initial condition
$$\Psi_k(t=0) = \psi_k$$
is indeed your
$$\Psi_k = \psi_k \exp(-iE_k t).$$
Then you consider a superposition of these two states, $\Psi = \Psi_1 + \Psi_2$, which is also solution of the Schrödinger equation for the initial condition
$$\Psi(t=0) = \psi_1 + \psi_2,$$
i.e. a superposition of a state of energy $E_1$ and a state of energy $E_2$.
Your calculation of $|\Psi_1 + \Psi_2|^2$ is a wee bit incorrect:
$$|\Psi_1 + \Psi_2|^2 = |\Psi_1|^2 + |\Psi_2|^2 + \underbrace{\Psi_1 {\Psi_2}^{\!*} + {\Psi_1}^{\!*}\Psi_2}_{2\Re\Psi_1{\Psi_2}^{\!*}}$$
where $\Re$ denotes the real part. That is to say
$$|\Psi_1 + \Psi_2|^2 = |\psi_1|^2 + |\psi_2|^2+2\Re \psi_1{\psi_2}^{\!*}\exp[ i(E_2-E_1)t]$$
This does not change your conclusion though: oscillations with a pulsation $E_2 - E_1$.
Right, now that this is all clear and correct, could you elaborate on your question? I can't guess what you want to know!
| {
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Are all atomic collisions elastic? If this is the case, why? In kinetic theory of gases it is considered all atomic collisions to be elastic. But if collisions are non-elastic the molecules must lose energy.
|
Are all atomic collisions elastic? If this is the case, why?
No, this is one of the key approximations for an Ideal Gas.
I will quote you the very first line on that page word for word; it says "An ideal gas is a theoretical gas composed of many randomly moving point particles whose only interaction is perfectly elastic collision."
Why do we need to make this approximation you might well ask?
The answer is basically so that we can apply the Equation of State to real gases which enables us to calculate their thermodynamic quantities such as the Temperature, Pressure of the gas or its Internal Energy (but for an ideal gas the internal energy is purely kinetic energy for reasons I will explain below).
I won't go into the details too much but depending how far you want to take this you can approximate real gases even better by Van der Waals equation.
As mentioned in the comment real gas molecules (when we don't approximate the gas as 'ideal') lose or gain kinetic energy (a non elastic collision) when they collide.
Not part of your question but real gas molecules also have potential energy due to intermolecular forces between the molecules (which are assumed to be zero for an ideal gas).
| {
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How to prove the constant speed of light using Lorentz transform? I read the light-clock example in my book which proved the time dilation formula by assuming that the speed of light is constant for all observers. But I've trouble in understanding it the other way around. Lorentz transformation is just a correction to Newtonian mechanics to account for the constant speed of light for all observers, right? I have trouble understanding how does applying this correction preserve the speed of light for all observers.
Can we start by assuming that the Lorentz transformation formulas are true and then prove that two observers $A$ and $B$ will see a light pulse moving at the same speed $c$ regardless of their relative velocity with respect to each other?
| In short yes. You can try to solve it yourself. Take 2 observes A and B, moving at velocity v wrt each other. A sees a light pulse, traveling as x = ct (which means light's velocity as seen by A is dx/dt = c). Now use Lorentz transformation to find out coordinates of the pulse as seen by B. you'd see it comes to be c again.
| {
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Is there any Electric Current on (Solar) Coronal Loops? I'm studying Plasma Astrophysics and now I have came across the following problem, which I have been trying to solve on my own, and came to a "non-logical" conclusion?
Here is the Problem:
Let the Coronal Loop have the following Magnetic Field:
$$ B_{x} = B_o e^{-kz}cos(kx)$$ and $$ B_z = -B_o e^{-kz}sin(kx) $$
for z>0 , and |x| < pi/2k
After finding the Magnetic field lines, it says find the Current and Density distribution on z dependence.
From $ J = \bigtriangledown x B$
One can find that $J=0$ ??
From Equilibrium Equation the Density has no dependence on z , except the gravitational field, which is weak?
Any help?
|
One can find that J=0 ??
Your math is okay. What you are being asked to examine is a special case of a coronal loop constructed from what is called a potential field. You often see the converse, namely a non-potential field, in discussions of dynamo theory.
It is perfectly fine, mathematically, to have $\nabla \times \mathbf{B} = 0$. Whether that describes reality is another issue but it is okay mathematically. One of the consequences is that this geometry describes a force-free field, i.e., one that satisfies $\mathbf{j} \times \mathbf{B} = 0$.
A general force-free field is defined by assuming:
$$
\nabla \times \mathbf{B} = \alpha \left( \mathbf{r} \right) \ \mathbf{B}
$$
where $\alpha \left( \mathbf{r} \right)$ is a scalar function of position/altitude. Under these assumptions, a potential field is the limit where $\alpha \left( \mathbf{r} \right) \rightarrow 0$ while the non-potential field corresponds to $\alpha \left( \mathbf{r} \right) \neq 0$. Using vector calculus and $\nabla \cdot \mathbf{B} = 0$, one can show that:
$$
\mathbf{B} \cdot \nabla \ \alpha \left( \mathbf{r} \right) = 0
$$
From Equilibrium Equation the Density has no dependence on z , except the gravitational field, which is weak? Any help?
Under these conditions, one can describe the number density and pressure using hydrodynamics, similar to how one examines neutral atmospheres, i.e.:
$$
n\left( r \right) \propto e^{-r/h}
$$
where $h$ is the characteristic scale height and $r$ is the altitude.
Side Note
While I have not seen your specific example used, I have seen something similar given by:
$$
\begin{align}
B_{x} & = B_{o} \ e^{- l \ z} \ \sin{\left( k x \right)} \\
B_{y} & = B_{o} \ e^{- l \ z} \ \sin{\left( k x \right)} \\
B_{z} & = B_{o} \ e^{- l \ z} \ \cos{\left( k x \right)} \\
\end{align}
$$
which has $\nabla \times \mathbf{B} \neq 0$. As you probably already noticed, if you change the sign of your z-component from your example you would also have $\nabla \times \mathbf{B} \neq 0$.
| {
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Gauge covariant derivative I have seen distinct definitions of gauge covariant derivative (in Yang-Mills theory)
$$ D_\mu \phi = (\partial_\mu + igA_\mu) \phi $$
vs
$$ D_\mu \phi = \partial_\mu \phi + ig[A_\mu,\phi] .$$
I guess the first is the common definition, which is the same as the covariant derivative in QED. What is about the second?
| As AccidentalFourierTransform points out, the second expression is the non-abelian generalization of the former. The first one is only valid for the abelian case (QED), while in general Yang Mills the fields are matrices transforming in some representation of the gauge group and the correct form is the latter.
| {
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Heisenberg Uncertainty Principle Applied to an infinite square well I appreciate the statement of Heisenberg's Uncertainty Principle. However, I am a bit confused as to how exactly it applies to the quantum mechanical situation of an infinite square well.
I understand how to apply Schrödinger's equation and appreciate that energy Eigenvalues can be deduced to be
$$E_n=\frac{n^2\hbar^2\pi^2}{2mL^2}.$$
However, I have read somewhere that the reason that the quantum particle cannot have $n = 0$—in other words, $E = 0$—is because by having zero energy we also have a definite momentum with no uncertainty, and by the Heisenberg uncertainty principle this should lead to an infinite uncertainty in the position of the particle. However, this cannot the case be in an infinite well, as we know the particle should be somewhere in the box by definition. Therefore $n$ can only be greater than or equal to one.
Surely when $n = 1$ we have the energy as
$$E_1 = \frac{\hbar^2 \pi^2}{2mL^2},$$
which is also a known energy, and so why does this (as well as the other integer values of $n$) does not violate the uncertainty principle?
| Provided that we have an infinite square well with potential
$$
V(x)=\begin{cases}
0, \text{ if }0<x<L,\\
+\infty, \text{ otherwise}
\end{cases}
$$
we obtain the wave functions as
$$
\psi_n(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{\pi n x}{L}\right),
$$
where $n$ is a positive integer.
it is now straightforward to calculate the necessary averages (using the trigonometric formulae for double angle and the integration by parts):
$$
\langle n |\hat{p}|n\rangle=\int_0^L\psi_n(x)\hat{p}\psi_n(x)dx=0,\\
\langle n |\hat{p}^2|n\rangle=\int_0^L\psi_n(x)\hat{p}^2\psi_n(x)dx=\frac{\hbar^2\pi^2n^2}{L^2},\\
\langle n |\hat{x}|n\rangle=\int_0^Lx\psi_n^2(x)dx=\frac{L}{2},\\
\langle n |\hat{x}^2|n\rangle=\int_0^Lx^2\psi_n^2(x)dx=L^2\left(\frac{1}{12}-\frac{1}{2\pi^2n^2}\right).
$$
We can now evaluate the standard deviations of momentum and position as
$$
\sigma_p=\sqrt{\langle n |\hat{p}^2|n\rangle - \langle n |\hat{p}|n\rangle^2}=\frac{\hbar \pi n}{L},\\
\sigma_x=\sqrt{\langle n |\hat{x}^2|n\rangle - \langle n |\hat{x}|n\rangle^2}=L\sqrt{\frac{1}{12}-\frac{1}{2\pi^2n^2}}
$$
Thus, we have
$$
\sigma_x\sigma_p=\hbar\pi n\sqrt{\frac{1}{12}-\frac{1}{2\pi^2n^2}}
=\frac{\hbar}{2}\sqrt{\frac{6\pi^2n^2-1}{3}}
$$
Since $\pi^2\approx 10$, for the ground state ($n=1$) we have
$$
\sigma_x\sigma_p\approx\frac{\hbar}{2}\sqrt{\frac{59}{3}}>\frac{\hbar}{2}.
$$
For higher $n$ the inequality is even stronger. However, if we took $n=0$, we would have a nonsense imaginary answer (of course, the wave function is identically zero for $n=0$).
| {
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Will a Cylinder placed on a frictionless inclined plane keep on slipping at its place or skid and slide down the plane? I've been wondering after learning about rolling without slipping and how it needs static friction for an object to start rolling but my question is that if theoretically the surface is frictionless then due to the torque of the weight will the Cylinder keep rolling at a fixed place or slide down etc.
Would appreciate a lot if I got to know what really happened as it's I don't seem to find much on the web.
| The cylinder won't experience any torque, because gravity can be though of as acting at the cylinder's center and the normal force from the ramp points directly toward the center. So the cylinder's angular velocity won't change. If it is released without any rotation, then it will simply slide down the ramp without ever rotating.
| {
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Evaluating Potential Energy Integral in Quantum Chemical Calculations My question is what are the steps for taking an integral of the following form?
$$\int e^{-\alpha|\mathbf r- \mathbf R_a|^2} {1\over|\mathbf r- \mathbf R_b|} e^{-\beta|\mathbf r- \mathbf R_b|^2} dV$$
This integral is commonly seen when attempting to do Quantum Chemistry calculations with a Gaussian type basis set. I have tried to use wolfram alpha to solve this problem but it fails to give a solution.
| Let's call this integral $C = C({\bf R}_a,{\bf R}_b,\alpha,\beta)$, since the arguments are the free variables in the integral. Let's also assume that we're working in $\mathbb R^3$.
Our first attack consists of performing the substitution ${\bf r}\rightarrow{\bf r}+{\bf R}_b$. This won't change the volume element, so the result is now:
$$
C = \int d^3{\bf r}\ \ e^{-\alpha|\mathbf r- \Delta \mathbf R|^2} {1\over r} e^{-\beta r^2}
$$
where $r = |{\bf r}|$ and we have defined $\Delta{\bf R}\equiv {\bf R}_a-{\bf R}_b$. That looks a lot simpler, doesn't it? We see that the integral only depends on the separation between a and b. $C = C(\Delta{\bf R},\alpha,\beta)$ But there's that pesky magnitude in the first exponential...
We can handle this if we work in spherical coordinates, so that $d^3{\bf r}=r^2\sin\theta \ dr\ d\theta\ d\phi$. Don't forget the Jacobian factor in the volume element or you will get 7 years of bad luck. Now, recall that
$$
\begin{align}
|{\bf r}-\Delta{\bf R}|^2&=({\bf r}-\Delta{\bf R})\cdot({\bf r}-\Delta{\bf R})\\
&=r^2+\Delta R^2-2r\Delta R\cos\theta
\end{align}
$$
If we plug this into the integral, along with the volume element, use the substitution $u=\cos\theta$, and integrate over $\phi$, then we get:
$$
C = 2\pi e^{-\alpha\Delta R^2}\int_0^\infty dr \ re^{-(\alpha+\beta)r^2}\int_{-1}^1du \ e^{2\alpha r\Delta R u}
$$
The $u$ integral is pretty simple, leaving us with
$$
C = \frac{\pi}{a}\frac{e^{-\alpha\Delta R^2}}{\Delta R}\int_0^\infty dr\ \left(e^{-(\alpha+\beta)r^2+2\alpha r\Delta R}-e^{-(\alpha+\beta)r^2-2\alpha r\Delta R}\right)
$$
Now we have two incomplete Gaussian integrals, so the results will introduce error functions. Note that in both integrals we can complete the square and write them as:
$$
\begin{align}
e^{\frac{\alpha^2}{\alpha+\beta}\Delta R^2}\int_0^\infty dr e^{-(\alpha+\beta)(r\pm \frac{\alpha\Delta r}{\alpha+\beta})^2}&=e^{\frac{\alpha^2}{\alpha+\beta}\Delta R^2}\int_{\pm \frac{\alpha\Delta r}{\alpha+\beta}}^\infty dr \ e^{-(\alpha+\beta)r^2}\\
&=\frac{1}{2}\sqrt\frac{\pi}{\alpha+\beta}\text{erfc}(\pm\frac{\alpha\Delta R}{\sqrt{\alpha+\beta}})
\end{align}
$$
where we have introduced the complimentary error function. Recalling the identity $\text{erfc}(-x)-\text{erfc}(x)=2\text{erf}(x)$, we can plug everything in to find:
$$
C(\Delta R,\alpha,\beta)=\frac{\pi^{3/2}e^{-\frac{\alpha\beta}{\alpha+\beta}\Delta R^2}}{\alpha\sqrt{\alpha+\beta}\Delta R}\text{erf}(\frac{\alpha\Delta R}{\sqrt{\alpha+\beta}})
$$
One can quickly check that the units for both answers are $Length^2$, as they should be. The argument of the error function must be dimensionless. All is well in the world.
Hope that helps!
| {
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Scintillator for adjusting sputter gun lenses What do you use to adjust the beam of you sputter cannon?
We have a test plate with a scintillating coating, that is now almost completely worn out, and I would like to replace it, but I am unable to find out what the luminescent coating was, or any supplier of similar plates. Google was strangely devoid of results.
Does anyone have any pointers to either how to make a suitable UHV-compartible coating or where to procure a new test plate?
| One simple way is to use a tarnished (oxidized) copper plate, you will see the spot where the ions clean the surface by observing a color change.
other crystals that show luminescense from dopant atoms are commonly called "phosphors". You can find a few names here
http://www.proxivision.de/products/phosphor-screen.html
A common one would be YAG:Ce.
| {
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Why do different letters sound different? If one sings the letter "A" and "M" at the same volume and pitch, the two letters are still differentiable. If both pitch and volume are the same however, shouldn't the sound be the exact same?
| The basic frequency is determined by the vocal cords. They make the air flow pulsate with a frequency of 100 Hz to 200 Hz. The pulses are short, so there are overtones upp to several kHz.
The mouth and tongue make the vocal tract resonant at different frequency ranges. Those are called formants. Have a look at the formant map here: https://www2.warwick.ac.uk/fac/sci/physics/staff/academic/bell/sonify/ttm/sound_files/
| {
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Definition of conjugate momentum in QFT My lecture notes define the conjugate momentum of a scalar field via:
$$\pi = \dot{\psi}$$
Where
$$\psi = \int \frac{d^3p}{(2\pi)^{3}}\frac{1}{\sqrt{2E_p}}\left(a_p e^{i\vec{p}\cdot \vec x} + a_p^\dagger e^{-i\vec p \cdot \vec x}\right) $$
and claim that this gives
$$\pi = \int \frac{d^3p}{(2\pi)^{3}}\sqrt{\frac{E_p}{2}}\left(a_p e^{i\vec p \cdot \vec x} + a_p^\dagger e^{-i\vec p \cdot \vec x}\right)$$
whilst working in Schodinger picture. But clearly $\psi$ doesn't even depend on time. Am I right thinking that what is stated in my lecture notes is wrong and the definition
$$\pi = \dot{\psi}$$
is valid in the Heisenberg picture only? And in order to obtain the above expressions, which are in Schrodinger picture, one needs to take the Heisenberg picture expressions:
$$\psi = \int \frac{d^3p}{(2\pi)^{3}}\frac{1}{\sqrt{2E_p}}\left(a_p e^{-ip\cdot x} + a_p^\dagger e^{ip\cdot x}\right) $$
$$\pi = \int \frac{d^3p}{(2\pi)^{3}}\sqrt{\frac{E_p}{2}}\left(a_p e^{-ip\cdot x} + a_p^\dagger e^{ip\cdot x}\right)$$
(where I now used the 4-vector notation) and then turn them into Schrodinger picture?
| I think I know what your problem is. You are forgetting that the time dependency can be implicit and doesn’t have to only be explicit. For instance, $\psi$ might depend on time because $x$ and/or $p$ depend on time. In this case the derivative will not be zero.
Also, the definition must be valid in both pictures since deriving a function with respect to time is the same in the matrix representation as deriving the operator with respect to time.
| {
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Can the 7-10 rule of thumb for radiation be understood theoretically? Is there a way to understand where the 7-10 Rule of Thumb for nuclear radiation comes from? A seven fold increase in time after explosion results in a 10 fold reduction in exposure rate.
From a FEMA page on responding to "nuclear threats":
From the exposure rate determined by a survey instrument, future exposure rates may be predicted from a basic rule known as the "7:10 Rule of Thumb."
The 7:10 Rule of Thumb states that for every 7-fold increase in time after detonation, there is a 10-fold decrease in the exposure rate. In other words, when the amount of time is multiplied by 7, the exposure rate is divided by 10. For example, let's say that 2 hours after detonation the exposure rate is 400 R/hr. After 14 hours, the exposure rate will be 1/10 as much, or 40 R/hr.
The exposure rate must be expressed in the same unit as the time increase. For example, if the time increase is expressed in hours, the exposure rate must be expressed as the radiation exposure per hour.
| Theory behind the rule? Local fallout from a groundburst is a million parts irradiated dirt (light elements) and one part plutonium nastiness. Each component has its own half-life, so the only way to estimate decay is empirical observation. My bet is that someone in the late 1950s plotted radiation versus time on log-log graph paper and derived the following:
$$\text{radiation} = 10^{(\log_{10}(\text{radiation}_1) - 1.183 \times \log_{10}(\text{time}))}$$
The text in Kearny's book has all the hallmarks of a poorly-constructed word problem. The graph doesn't match the text. Other posters misinterpreted it too.
The result of this formula, in case you want to argue, is as follows:
*
*At 1 hour since blast, radiation = 1000 {for example}
*At 7 hours 100
*At 49 hours 10
*At 343 hours 1
*At 2401 hours 0.1
"With each seven-fold increase in time since detonation, there is a ten-fold reduction in radiation."
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Schwarzschild coordinates beyond the event horizon We can write down the metric of the Schwarzschild black hole in Schwarzschild coordinates.
On page 6 of the notes by Leonard Susskind of a course given at the Perimeter Institute titled 'Black Holes and Holography.' we find the following:
However, note that the Schwarzschild coordinates are only formally valid for $r > R_{s}$, and must be analytically continued within the event horizon.
Which aspect of the metric in Schwarzschild coordinates indicates that the coordinates are only valid outside the event horizon?
| Coordinates are not sacred objects in GR. Any coordinate system is just as good as any other coordinate system. So to ask whether the Schwarzschild coordinates are valid or not is a meaningless question ${}^1$.
However it is reasonable to ask if coordinates have an intuitive meaning for some specified observer. So for example if we take an observer far from the massive object then the Schwarzschild time coordinate is the time as measured by that observer's clock, and the Schwarzschild radial coordinate is the circumference of a circle centred on the object divided by $2\pi$. Both these are intuitively meaningful measurements for our observer.
The problem with the interior of the black hole is that for our external observer anything falling into the black hole takes an infinite time even to reach the event horizon, let alone pass through it, so that makes us sceptical about any physical meaning for the Schwarzschild time inside the event horizon. And indeed if we write the Schwarzschild metric:
$$ ds^2 = -\left(1-\frac{r_s}{r}\right)dt^2 + \frac{dr^2}{1-\frac{r_s}{r}}+r^2d\theta^2 + r^2\sin^2\theta d\phi^2 $$
We find that the sign of the $dt^2$ and $dr^2$ terms changes as we move through the horizon. Since the sign tells us whether a term in the metric is spacelike or timelike this means inside the horizon $t$ behaves like a spatial coordinate and $r$ like a time coordinate.
This doesn't mean anything freaky, like time turning into space and vice versa as the more lurid popular science articles would have you believe, it just means the coordinates don't have the intuitive meaning that we associate with them outside the event horizon. However as I said right at the outset, they remain perfectly good coordinates and we just have to be careful about interpreting them.
${}^1$ the coordinates are singular at the horizon, i.e. at $r=r_s$, so they are not useful exactly at the horizon.
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Electron concentration in compensated semiconductor - different from intrinsic? I am having trouble understanding the formula for total electron concentration in the conduction band (under the Boltzmann approximation):
$$n_{0}=\frac{N_{d}-N_{a}}{2}+\sqrt{\left(\frac{N_{d}-N_{a}}{2}\right)^{2}+n_{i}^{2}} $$
While I understand the steps in the derivation, I am having trouble understanding the case where
$$N_d =N_a$$
in which the result is:
$$n_0 =n_i $$
I will refer to the figure below to explain myself.
From my understanding, if the semiconductor was intrinsic, the electrons in the conduction band would consist only of "thermal electrons", (see figure). If I dope the semiconductor with donors, I will also have the donor electrons regardless of the presence of acceptors (as far as I understand...).
The reason I say this is because the acceptors create holes, but they don't "suck up" the electrons from the conduction band (only from the valence band).
Therefore, I cannot find a reason why Na would cancel out Nd. I would expect the following result instead:
$$n_0 = n_i + N_d$$
| As it turns out, according to this explanation (section 2.6.4.1 'Dopants and impurities'), in room temp the donor electrons more readily "migrate" (energetically) to the acceptor states than to the valence band. Therefore, contrary to the OP, donors and acceptros do 'cancel' out. This is indicated by the red arrows in the figure below.
Moreover, if:
$$N_d /N_a \gg n_i $$
and the temperature is about room temp (therefore there is full ionization)
then the carrier concentrations are give by:
For n-type:
$$n_0 =N_d-N_a$$
For p-type:
$$p_0 =N_a-N_d$$
which perfectly demonstrates the 'cancelling out', or compensation.
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Ohm's law and current due to magnetisation I'm reading Landau's Electrodynamic of continuous media, specifically the following paragraph of section §29 (The magnetic field of a constant current):
If a conductor carries a non-zero total current, the mean current density in it can be written as $\rho{\bf v}= c\ {\bf curl}\ {\bf M}+{\bf j}$. The first term, resulting from the magnetisation of the medium, makes no contribution to the total current, so that the net charge through a cross-section of the body is given by the integral $\int {\bf j}\cdot d{\bf f}$ of the second term. The quantity ${\bf j}$ is called the conduction current density.$\dagger$ The statements made in §20 apply to this current; in particular, the energy dissipated per unit time and volume is $\bf{E}\cdot\bf{j}$.
In section §20 appears the Ohm's law. My question is: why magnetisation current ${\bf j_M}= c\ {\bf curl}\ {\bf M}$ doesn't contribute to Ohm's law?
In other words: let ${\bf j_t}={\bf j_f}+{\bf j_b}$ be the total current density, with ${\bf j_b}=c\ {\bf curl}\ {\bf M}$ the bounded (magnetisation) current density. Is ${\bf j_f}=\sigma {\bf E}$ the Ohm's law?
| Ohms law applies to free charges. The magnetization current applies to bound charges. The total current is the sum of the two.
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What am I failing to understand about a light in a material? In a material, a photon's velocity becomes slower, photon's wavelength becomes shorter, but photon's frequency doesn't change.
If there is a material that makes low frequency photons have very short wavelength photons, and if I stick my finger in the material, do I get a radiation burn? What kind of effects can I have from the short wavelength photons in a material except for the refraction?
|
In a material, a photon's velocity becomes slower, photon's wavelength becomes shorter, but photon's frequency doesn't change.
The photon always travels with velocity c and its frequency and wavelength are fixed unless there are inelastic interactions.
The classical wave emerges from a confluence of photons, and it is the classical electromagnetic wave that goes with smaller than c velocity in transparent materials. Because of the way the classical em wave emerges from the constituent photons one can qualitatively say that in a transparent material the photons scatter elastically and travel a longer path than the optical ray defining the light they build up.
This makes no sense "If there is a material that makes low frequency photons have very short wavelength photons".
Low frequency has large wavelengths, period, by construction.
A light beam can give up all its energy, photons included, to some solid, laser light can melt metals, and yes , your finger will be burned , but that has nothing to do with the original photons' frequency, except the delivery of so much energy that the metal can melt.
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Simultaneity in Newtonian mechanics How would Newtonian mechanics answer the train and moving light question?
The setup is:
A train is moving in the positive x_axis with speed c/2. A person stands in the middle of the train. There are two light bulbs at both ends of the train. The light goes off at the same time (absolute time in Newtonian physics). The person standing in the middle of the train would perceive both lights independently.
Outside the train there is a stationary observer. Let's assume the train is already to the "right" of the observer (in x_axis) when the lights go off. Would the stationary observer observe the rear light before the front light?
The reason why I am asking this is that the relativity of simultaneity is often attributed ONLY to special relativity. Here, would Newtonian mechanics also predict that the stationary observer observes different simultaneity than the moving observer in the train?
| Rather than think in terms of what someone will observe, think in terms of what would be measured in their frame of reference assuming the frame to be equipped with suitably distributed clocks.
As an example, even in the Newtonian realm, suppose an someone is at the origin of his frame of reference and there are lamps at x = 10 million and x = 20 million miles that turn on at the same time. The light from the closer lamp will reach the observer before the light from the further lamp. He would observe them not to be simultaneous but they really are simultaneous in his frame of reference.
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Changing magnetic flux In the figure, there is a metallic ring inside which there is a dotted region through which constant magnetic field is passing. A wire with sliding contact is moved from position (1) to (2). Will the galvanometer show any deflection?
In this problem, when we take wire from (1) to (2) there is a change in magnetic flux through the loop since initially there was no magnetic field inside and afterwards there is. So, shouldn't there be an induced current in the circuit?
| Imagine a wire configuration between 1 and 2, where the sliding contacts are not touching one another. Since you want to evaluate the emf along the conductor, you'd have to choose one closed path encompassing the galvanometer, the wires and one of the two sections of the ring, in order to connect the two sliding contacts. If you choose to include the left part of the ring on the path, while varying from configuration 1 to 2 the area enveloped by the loop never includes the magnetic field. Thus the flux is identically zero during the process, resulting in zero emf. Now let's see what happens if you choose a path which includes the right part of the ring; imagine sliding the contacts back to 1. While doing that, the portion of the ring included in the path increases, until it includes the whole ring in configuration 1. Therefore the surface enveloped by the loop contains the same non-zero magnetic flux during all the process; since the flux is constant, $d\phi/dt=0$ and the emf is 0.
| {
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How to tell if a thermodynamic cycle is reversible without calculating entropy change? Consider the Carnot cycle, consisting of two reversible, isothermal processes and two isentropic processes. It is reversible, pretty much by definition.
Now consider the Lenoir cycle, consisting of an isochoric compression (heat addition), followed by an isentropic expansion, followed by an isobaric compression (heat loss). I calculated the entropy created by this cycle and found it to be strictly positive.
However it's not clear intuitively why this cycle should be irreversible. Is heat change at constant volume or at constant pressure necessarily irreversible?
| If the isochoric and isobaric transformation are performed reversibly, i.e. quasistatically and without heat dissipation caused by friction or other effects, then your cycle will be reversible.
This is true for every thermodynamic cycle you can draw in the $PV$ plane: if every step is performed reversibly, then the cycle is reversible.
The peculiarity of the Carnot cycle is that it is the only reversible engine that operates between two heat sources only.
You can easily see how many different heat sources you are using if you draw the cycle into the $TS$ diagram (picture from Wikipedia):
In this case, it is easy to verify that the change in entropy of the surroundings is
$$\Delta S_{surr} = -\frac{Q_H}{T_H}+\frac{Q_C}{T_C} =0$$
So that the engine is indeed reversible.
But now let's take your Lenoir cycle in the $TS$ diagram (picture from Wikipedia):
As you can see, during $1 \rightarrow 2 $ and $3 \rightarrow 1$ you are cutting infinitely many isotherms. The formula you have to use is in this case
$$\Delta S_{surr} = -\int_1^2 \frac{\delta Q}{T} + \int_3^1 \frac{\delta Q}{T}$$
But this time you cannot take out $T$ from the integral like you would do with a Carnot cycle, because it is not a constant.
What you can do is to assume that step $1 \rightarrow 2 $ and $3 \rightarrow 1$ are performed reversibly: in this case, $\Delta S_{surr}=0$ by definition.
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Is the classification of particles into matter and anti-matter arbitrary? It is well known that every fundamental particle has a corresponding antiparticle, and that – except for particles which are their own antiparticle – for practically every pair of particles and antiparticles, one type of particle predominates. We call this particle matter, and its antiparticle is considered to be antimatter.
However, is there any intrinsic difference between matter and antimatter, or are these terms defined based on nothing more than the balance that we so happened to end up with? For example, is there any reason that we should place an up quark and a charm quark in the same category, and not an up quark and a charm antiquark? Is there any reason we should group together up quarks and electrons rather than up quarks and positrons?
| You are confusing two things at least.
This is the particle table:
There exists a mirror antiparticle table such that when the quantum numbers of particle and antiparticle are added each sum is identically zero, and particle and antiparticle have the same mass.
When a particle meets an antiparticle , and both are at rest, they annihilate into energy, that comes out with new particle pairs with different particle quantum numbers. For example : an e+ scattering on an e- with enough center of mass energy can give a proton antiproton pair.
The table is such that the standard model of particle physics makes sense, and then one has protons, neutrons made up of quarks so that the charges make sense with the data. Baryon number is carried by quarks and is +/1/3 so you cannot add an antiquark and expect to get a baryon, etc.etc.
The standard model encapsulates the data and is very successful most of the ime in fitting new experimental data.
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Classifications of quasiparticles Different particles can be represented as different irreducible representations of Poincare group. Can we classify quasiparticles using irreducible representations of some group? Also, quasiparticles are low energy excitations of the system, and it is closely related to the ground states of the system, so I am wondering if there is any relations we can use to help us classify quasiparticles from ground state symmetry. Also, is the irreducible representation of some group argument can apply to those excitations of topological origin?
| The group representation argument can not be applied to those excitations of topological origin. Group theory is not the most general classification criterion of quasiparticles. Group theory is useful when symmetry plays an important role in the discussion. For example, Goldstone modes in the symmetry breaking phases can be classified by group representation theory, boundary modes in the symmetry protected topological phases can be classified by group cohomology theory. But if the system has no symmetry (or the symmetry is not relevant to the discussion), then group theory may not be very useful. For example, different anyon excitations in topologically ordered phases are not classified by their group representations: they may all belong to the trivial representation for example, but still, they are distinct topological excitations and can not transit from one to another. In this case, we use category theory to classify different topological excitations. In fact, category theory is a more general classification criterion of quasiparticles. Because the representations of a group also form a category, so the group representation classification is part of the categorical classification. Moreover, the categorical classification also captures the topological excitations nicely, which demonstrate its advantage over the group theory classification.
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Why is light bent but not accelerated? Light is bent near a mass (for example when passing close to the sun as demonstrated in the famous sun eclipse of 1919). I interpret this as an effect of gravity on the light.
However, it seems (to me, at least) that light is not accelerated when it travels directly toward the (bary-)center of the sun. The same gravitational force applies yet the speed of light remains constant (viz. $c$).
What am I missing?
| Currently, there is no evidence that photons have mass, and it is generally accepted that they are massless particles.
Nonetheless, gravitation does affect the path of photons, because the bending of space-time causes all particles to travel on curved paths, including massless ones. But that does not mean than light will be accelerated. The speed of light (299,792,458 m/s) is an absolute maximum, and it may not decrease from that nor may it increase.
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Does Violating Cosmic Censorship Really Mean Violating Causality? As I understand it, the basic motivation behind ruling out a naked singularity is that we don't know what is happening at a singularity and thus, we won't be able to predict anything in the universe if there is no horizon around such an unknown region. But the reason we don't understand what is happening at the singularity is that we don't have a theory of quantum gravity. But when we have a theory of quantum gravity, this limitation should go away. And thus, causality should be preserved even with naked singularities.
It is a very cultural fact that we don't know how to deal with singularities without horizons at this stage. Thus, it seems quite naive to assume that causality would actually be violated if horizons don't cover the singularity. Though, I believe under some restricted energy conditions the censorship conjecture has been proven and thus, the censorship might be correct due to some other than causality reasons but causality doesn't seem to force the censorship.
| There are closed timelike curves in the interior of the Kerr horizon. The obvious way to see this is if you go through the center of the ring singularity (thus, not intersecting the ring singularity), the Boyer-Lindquist $r$ goes negative, and the Boyer-Lindquist $\phi$ becomes timelike. Since, by construction, the orbits of $\phi$ are closed, this means that they are closed timelike curves.
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How does one come up with the model of torque? In school, we learn that $T = Fd$, where $T$ is the torque, $F$ is the force perpendicular to the moment arm, and $d$ is the length of the moment arm. If I was the first physicist to come up with this model for torque, what might my train of thought be? In other words, how does one come up with this model?
| The way you worded the question sounds to me like you're asking more for an intuitive understanding of why someone thought Fd would be a useful enough quantity to give it its own name, why it's taught to kids as a sort of fundamental equation rather than just another step in a derivation, and also where it fits into the big picture.
If that's the case, however, I can't tell you what the actual line of thinking really was, but I can at least offer you one view of why we consider it to be an important/fundamental quantity.
You can think of torque as being the rotational analog of linear force. We observe some simple relationships such as:
$$\omega_{\text{avg}}=\frac{v_{\text{tangential}}}{r}$$
$$\alpha_{\text{avg}}=\frac{a_{\text{tangential}}}{r}$$
and one might wonder, if we simply multiply the $a_{\text{tangential}}$ by m, will the result give us something akin to rotational force?
Well it turns out if you do you get:
$m\alpha_{\text{avg}}=\frac{ma}{r}$
Of course the ma quantity is our familiar linear force, however if you multiply both sides by $r^{2}$ you will find "moment of inertia" on the left:
$(mr^{2})\alpha_{\text{avg}}=\frac{F}{r}(r^{2})=Fr$
therefore it is reasonable to think of $(mr^{2})\alpha$ as some sort of "angular force" which we've named "torque" usually given the symbol $\tau$, and to think of $mr^{2}$ almost as a sort of "angular mass" which we have given the name "Moment of Inertia" with the symbol I.
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Equation of motion for a falling slender bar
I have a few question about equations of the uniform slender bar motion.
The bar is released from rest on a horizontal surface as shown in the figure and falls to the left. The surface is rough so friction forces are applied at bottom end of the bar. The coefficient of friction is μ and the mass of the bar is 'm'
First, I want to know the angle β when the end of the bottom end of the bar begins to slip. I know that if the x-axis direction force gets bigger than the static friction force it will start to slip. But I can't figure out the equation about the x-axis direction force as function of θ.
Second, I want to know the angle α when the bottom end of the bar starts to lift off the ground. This situation will happen when the normal force is zero. But I can't figure out the equation of the normal force as a function of θ.
I want to know the equation to calculate the alpha and beta.
The problems are not related, it's a different problems.
Any hints or opinion would be appreciated.
| Given that the bar is narrow, it would not have to tip very far for the center of mass to be vertically outside the base and therefore topple.
Hint for calculating alpha: consider how fast the top of the bar would have to accelerate in order for the bottom to lift off of the ground, and that it should rotate around the center of mass if this happens.
Hint for mass: the force of friction is equal to μ times R, and f=ma, the acceleration does not depend on mass, only the absolute value for friction will change.
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Why are we able to break a vector along it's components or in other words why is it that a vector exists along $x$, $y$ and $z$ axis? Does a 3 d vector exist in three dimensions at once? It seems to me that a vector always changes axis along which it is to fit into the scenario. For example: Electric field in $x-y$ plane when passes through $y-z$ plane only uses it's $x$ component.
| Yes, a vector exists in three dimensions at once. Imagine we apply force to a body diagonally then it will also move diagonally, which means it is moving in both x and y directions. If you block that object from moving in only y direction then you will see it is moving in only x direction while we are still applying the force diagonally.
If we know the amount of force we are applying in this case then by resolving it, we can find that how much of the total force is acting in x axis and in Y axis because some part of the total force is acting in X direction and some part is acting in Y direction at once.
Same goes for three dimensions.
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A question related to time and motion I have a theory, I think that we cannot travel at speeds faster than light because, as we know,when you travel at speeds near light time passes slowly and that if we go further than light time may pause perhaps and that if time pauses its impossible to have motion because in 0 seconds ( I mean no time has passed) you cannot travel any distance.
I have also another theory, I think that if we go a little high than light speed except pausing time may reverse, and if time reverses an object will never exists in space but continuously go back in time and reach big bang. The object will never exist in space but actually in a TIME dimension ( I know dimension word is wrong but I couldn't think of a word except this).
Can anyone one of these 'theories' be true, even very tiny bit true? Please point out my mistakes. I am just a kid though of 9th grade. But, I really wonder could these be true or not.
| PBS Spacetime's YouTube channel has several videos related to your theories. There are specific methods physicists use to examine Light and Time. One of them is the Penrose diagram:
Here are some of the videos where they talk about Time, Black Holes, and Superluminal Travel... all of which examine how the geometry of spacetime is utilized to draw conclusions on fringe topics:
Superluminal Time Travel + Time Warp Challenge Answer
How Time Becomes Space Inside a Black Hole
The Geometry of Causality
What Happens at the Event Horizon?
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Why does the annihilation operator acting on the ground state in Quantum Field Theory gives a zero? One of the main motivations for Quantum Field Theory after Dirac Equation is that the Dirac equation predicts negative energy states which leads to the ground state being unbounded which ultimately leads to the sea of negative energy electrons. It is said that the field theoretic approach cures it of this problem.
But every QFT book I have seen often defines the ground state $|0\rangle$ as:
$$ a|0\rangle=0 $$
but I can't see the basis for this definition ?
Of course, once I define the ground state as this it will automatically never lead to negative energy states? Why isn't this arbitrary? Why could I not do the same with Dirac Equation and define its ground state as such and hence remove all such negative energy states?
| For the Klein Gordon equation, when we interpret it as being the equation for a wavefunction of a particle, we have the negative energy states as a solution.
This problem vanishes when we move to field theory and think of KG eqn as the differential equation for a classical field. Write the Hamiltonian for it and it is positive definite.
Write the Hamiltonian in terms of ladder operators and require the condition that norm of the states must be positive. You will see that the eignevalues of the number operator must be positive.
Hence, the problem for -ve energy states is solved for KG by interpreting it as the equation for a field.
The Dirac equation, in a sense is already a quantum field and the negative energy states thus, cannot be eliminated. Well, actually you can.
By the way,
Majorana modified Dirac's equation to include only positive energy states. There is an article by Frank Wilczek of MIT on this topic.
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Non-constant tension in rope Could somebody explain to me intuitively how tension is not the same in a rope with mass? My physics teacher (when regarding a massless string) told me that the tension is always equal because if you pull at one side more then the other side has to pull just as much to keep it in equilibrium resulting in the tensions being equal. This fits really well intuitively with me, but if the string has a non-negligible mass how can I adapt this idea? Or replace it if necessary?
| Imagine a rope having a finite mass and is placed in gravity free space and having a constant velocity.Now take the rope as your system and exert an external force on one end of the rope then the rope will bear some tension at the point of application of force.As rope is the system tension is the internal force.The whole rope will now be having same acceleration.At the point of application of force imagine an infinitesimal section of rope,by Newton's third law we know that the section will exert an equal and opposite force on the agent applying the force.As the whole rope is constrained to move with that infinitesimal section, then that section would apply a force on the adjacent infinitesimal section of the rope but this force would be less than the external force itself as the section of point of application of external force is itself accelerated and the acceleration of whole rope is same.In the same manner force applied by these infinitesimal sections would decrease in a defined order which can calculated using basics of calculus.Hence,tension in a rope having finite mass is not the same everywhere.
| {
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Brownian motion in a box It is often said that the Brownian motion for a particle in a box, thus a finite domain, is described by a uniform probability distribution in the longtime limit. One may easily imagine this maybe intuitively, but is there actually an easy way of showing this? How is it that even though there are boundaries, no bias is introduced in the longtime limit of where the particle may be in the box?
If one assumes the discrete case, so the box is populated with only a finite number of points the Brownian particle can occupy, does the above statement still hold?
Please feel free to give references that you see fit for such questions, I imagine these are all solved problems and that's why most people often state them using the adverb trivially.
| I think the distribution remains flat for diffusion in a box, unless the boundary conditions are really strange or unphysical. Let me try with this not-so-accurate argument. Free diffusion gives a probability distribution which starts at $t=0$, for a single particle, as a delta function centered at the initial position, and evolves as a Gaussian $G_{free}$ with a variance that grows linearly in time.
Now, if you have a box around the particle, I'd say that the probability distribution $G_{box}$ at a time $t$ will be some sort of "folded" version of the Gaussian $G_{free}$ (with the tails of $G_{free}$ i)folded back for reflective boundary conditions, or ii) copied and pasted on the on the side for periodic boundary conditions, or iii) just removed for sticky boundaries iv)... ). This can create a strange $G_{box}$ at finite times, but eventually in the long run everything will be smoothed out and $G_{box}$ will become flat.
(On the other hand, if your box is not a rigid box but a harmonic potential well (the particle feels a force linear with displacement from the center), then $G_{box}$ will remain always the same Gaussian.)
| {
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Tipping point on fence panels? I deal in temporary fence panels - and my concern is the tipping point of our product out in the field.
Panels are 6' tall x 12' long (63#)
Stands are 23" long x 6" wide
We use sand bags (30#-40#) on the bases, but for whatever reason, they don't always stay on (or people take them).
Is a 23" base an optimum base for this height - or rather the "minimum effective dose" to achieve a stable panel? I understand that increasing it to 30" (15" on each side) would make it stable, but is it necessary for that height and still remain stable?
We are reviewing this as we are looking at a taller product (8') and I don't believe the same stands can be used on both heights, as the added height will greatly increase the chances of a tip over.
So, maybe my actual question how do I determine my center of gravity on the upright fence - so that I can accurately define the necessary base width?
| If the height of the fence is increased by a factor $k$, this will increase the area by the same factor, and also the wind force on it. (I assume for simplicity that windspeed does not vary with height, which is probably not true.) The average height at which the wind force is applied is also $k$ times as large. So overall the toppling moment (force x lever arm) is $k^2$ times as large.
To counteract this, you would have to make the righting moment $k^2$ times as large. Presumably increasing the height will increase the weight of the fence by the same factor $k$. (You may have to use more material to increase strength, in which case this increase is more than $k$.) If you also increase the weight of the sandbags by $k$, then the total weight is $k$ times as much. As Lewis states, you would also have to increase the size of the base by $k$ in order to make the righting moment $k^2$ times bigger also, giving the same stability.
If an increase by factor $k$ in the size of the base or the total weight of fence plus sandbags is impractical, any combination giving a product of $k^2$ will do just as well.
The joint keeping the base perpendicular to the panels will be under $k^2$ times as much stress, so this may need to be strengthened.
| {
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How to integrate out the $W$-boson fields? What does it mathematically mean to 'integrate out' the $W$-boson fields to obtain the Fermi Lagrangian from the electroweak theory? How does one achieve this mathematically? It will be helpful if someone can explain this both in the path integral formalism and the operator formalism of quantum field theory.
In particular, I know that the $W$-boson propagator, $\frac{1}{p^2-m_w^2}$ in the limit $p^2\ll m_w^2$ becomes $-\frac{1}{m_w^2}$. But why is such an approximation called 'integrating out' the fields? What are we really doing to the EOM when we are making this approximation?
| Heuristically, there are two ways.
*
*In the electroweak Lagrangian, you substitute for W by its classical solution. That is, you find the Euler-Lagrange equation for W, solve for it, and plug the solution in Lagrangian. In doing this, the W fields are "integrated out".
*When an internal line of the W boson is present in a Feynman diagram, you erase it. For instance, the tree level diagram of electron scattering with neutrino involves a W propagator. Erasing it, you get a four-point vertex of fermions. In terms of Lagrangian, you have an operator of four spinors, which is a dimension 6 operator. The numerical prefactor of it can be fixed by matching the scattering amplitude.
| {
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Meanings of the word "phase" I have been confused at points due to multiple uses of the word "phase".
*
*Mainly, when I think of a phase diagram, I think of a graph relating temperature to pressure, and segments the possible combinations of these values into regions in which a particular substance is "solid", "liquid", etc.
*This is something completely different from phase space, in the dynamical systems sense, where each point in the space represents the state of a dynamical system.
*There are also notions of phase and phase velocity.
Am I correct in assuming that the words "phase" in these contexts have nothing to do with each other, and are there other meanings that I shouldn't conflate?
| Right. Other than the generic meaning of phase being the state of a system, these two specific meanings are independent. As a third example, the phase associated with wave motion would be another separate definition.
| {
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Why large excavated holes can create air currents? My friend shared this post on my wall, but I got doubt: How can a hole like this create air currents which suck helicopterss also? I didn't really understand.
| In the link that reproduces the photo it is said that temperatures in Siberia on the ground can be as low as -40C.
The chimney effect is used for generating power , so this might act as a solar updraft tower :
The solar updraft tower (SUT) is a renewable-energy power plant for generating electricity from low temperature solar heat. Sunshine heats the air beneath a very wide greenhouse-like roofed collector structure surrounding the central base of a very tall chimney tower. The resulting convection causes a hot air updraft in the tower by the chimney effect.
This demonstrates that winds with a lot of power can be generated by temperature differentials.
As the open mine is closed but there exists tunnel mining, air will be renewed from where the tunnels are being used and a strong updraft due to maybe 60C or 70C temperature differential from top to bottom could create unexpected and dangerous for helicopter air turbulence for a helicopter.
| {
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In summer, should I close curtains during the day? In summer, that is when it's warmer outside the house than inside, one want to prevent the air in the house to warm up too much. Let's consider that all the windows are kept closed, that they are double panes and made of "Low-E" glass. During the day, does closing the curtains on the windows reduce the heating of the house? If so, does the color of the curtains have a significant impact?
Intuitively, I would think that once the sunlight goes through the window, it's too late to prevent the greenhouse effect. However, the curtains could create a layer of warm air that would improve the isolation, but maybe the convection would prevent this layer to form.
| Windows are transparent to visible light, but opaque to infrared. If your curtains are dark, they will absorb the visible and emit infrared. This will cause your house to warm as the light doesn't escape so easily. If they are lighter, they will be more reflective to visible light and will bounce it back through the window. This is a rather gross simplification of the process, but this is the basic idea.
| {
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Help in an integration step in QFT by Lewis H. Ryder There is an integration step I can not figure out and is frustrating.
We start from the equality
$$ \dfrac{\partial^2\phi}{\partial x^2} = \dfrac{\partial V}{\partial \phi} $$
and by integration process we are supposed to get (eq. 10.8 in the textbook):
$$ \dfrac{1}{2}\left(\dfrac{\partial\phi}{\partial x}\right) ^2 = V(\phi)
$$
Maybe I'm just overcomplicating it but I can not understand how this is done.
$\phi =\phi(x,t)$ but for this case $\dfrac{\partial\phi}{\partial t}=0$ and $\phi$ approaches zeroes of $V(\phi)$ when $x\rightarrow\pm\infty$.
Then my idea was to integrate by $d\phi$ both sides to get the RHS of eq. 10.8 and for the LHS I tried integrating by parts using $$d\phi=\dfrac{\partial\phi}{\partial x}dx$$
but got no success yet, and also the fact is that I don't even think what I'm doing is correct since is $V(\phi)$ and not $\phi$ what tends to zero when $x\rightarrow \pm \infty$.
Any help would be appreciated
| Another possible way which might be more clear for you is to do the following:
Begin with
$$ \frac{\partial^2 \phi}{\partial x^2} = \frac{\partial V}{\partial \phi}.$$
Multiplying the above by $\partial \phi / \partial x$ and moving $\partial V/\partial \phi$ to the other side yields
$$ \label{eq:1} \tag{**} \frac{\partial \phi}{\partial x} \frac{\partial^2 \phi}{\partial x^2} - \frac{\partial \phi}{\partial x} \frac{\partial V}{\partial \phi} =0.$$
Note:
The first expression on the LHS of the above equation can be equivalently expressed as $$ \frac{\partial \phi}{\partial x} \frac{\partial^2 \phi}{\partial x^2} \equiv \frac{1}{2}\frac{\partial}{\partial x} \left( \frac{\partial \phi}{\partial x} \right) ^2.$$
Combining the pieces, we can express Eq. (\ref{eq:1}) as
$$ \frac{\partial }{\partial x} \left( \frac{1}{2}\left( \frac{\partial \phi}{\partial x}\right)^2 - V(\phi) \right) = 0. $$
Hopefully you can take it from here. Note The integration constant is zero according to pg. 393 Lewis Ryder; QFT.
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Why does a massless, frictionless piston move from high pressure to low pressure? Consider an ideal gas kept in a rigid cylinder with a movable massless, frictionless piston at the top. Let the pressure inside the cylinder be $P$ at pressure exerted by the surrounding on the cylinder be $p$. Let $$ P>p $$
Now since the piston is massless net force on it must be 0 (since $m=0$). This implies that the piston can move with any acceleration. But it moves in only one direction that is outwards. Why doesn't it moves inwards?
| Piston being massless never makes sense because it is the same as talking about thin air.
Massless in the context that any force due to the elastic collisions between the particles and the piston can move the piston, i.e. the piston is maneuverable.
| {
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Why do we say Sun curves space and the Earth moves following the geodesic? Why do we say sun curves space even if the Ricci Scalar for a Schwarzchild metric, the solution of Einsteins Equations for the Sun, is equal to zero.
The Ricci scalar for the constant time slice is also zero.
Can we define curvature using a single quantity? If yes then what is it?
| According to the mathematical definition of the curvature, it represents the commutator of covariant derivative. More precisely, $$[\nabla_\mu,\nabla_\nu]A^\rho=R^\rho_{\kappa\mu\nu}A^\kappa$$
Where $R^\rho_{\kappa\mu\nu}$ is the Riemann curvature tensor. It is the single quantity that defines (or represents) the curvature of spacetime. For a spacetime to be completely flat at a point means all the components of $R^\rho_{\kappa\mu\nu}$ to identically vanish there. But there are some less strict criteria for differently defined flatness, e.g. if $R^\rho_{\kappa\rho\nu}\equiv R_{\kappa\nu}=0$ then it's called Ricci-flatness and if $R_{\mu\nu}g^{\mu\nu}\equiv R=0$ then it's called scalar-flatness. But such kinds of flatness don't imply that the spacetime is not curved. As long as the Riemann curvature tensor has non-zero components the spacetime is curved - which can be verified by making a vector parallel transport along a closed curve and seeing that it doesn't match with what it was originally.
| {
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Why is the Schwarzschild radius associated with the tiniest micro black hole formed by a Planck mass twice the Plank length? If one calculates the Schwarzschild radius, $r_s$, of a Plank mass $m_p=2,18*10^{-8} (m)$ one gets:
$$r_s=2{\frac{G{m_p}}{c^2}}=1,48*10^{-27}*2,18*10^{-8}=3,22*10^{-35}(m)$$
Now the Planck length $m_p$ is $1,61*10^{-35}(m)$, which is exactly half the Schwarzschild radius of the black hole associated with a Plank mass. Thus $l_p=\frac{G{m_p}}{c^2}$. Filling in the expression for $m_p=\sqrt{\frac{\hbar c} G}$ in the expression for $l_p$ gives the more familiair form for $l_p$, namely $\sqrt{\frac{\hbar G} {c^3}}$.
Why is it that the Schwarzschild radius of a Planck mass micro black hole (the tiniest that exists) is exactly twice the Planck length?
You can of course answer this question by saying that the formulae tell us that this is the case. But is the factor two by which the two lengths are connected just a coincidence, or has this connection some deeper significance?
| Things like the Planck mass, the Planck length, etc. are order of magnitude estimates of when certain effects should be important. You shouldn't take it literally that Quantum gravity becomes important at EXACTLY $\ell_{p}$, and is completely irrelevant at any longer distance. It's a heuristic to help us think. In reality, just like in electromagnetism, we get a factor of $\frac{1}{4\pi}$ in front of the equation, we can expect some sort of pure number factor to modify the various relevant quantum gravity equations.
| {
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Are pointlike massive particles naked singularities? If elementary particles (specifically, those with mass, such as the electron or other leptons) are pointlike particles, wouldn't that mean they are naked singularities?
But these particles have spin- wouldn't that make them naked ring singularities, thus giving them an observed radius, making them non-pointlike?
If I remember correctly, the radius of a ring singularity is given by $a=\frac{J}{Mc}$. If we assume the intrinsic spin property of a particle is equal to $J$ of the corresponding singularity, we get for the electron:
$$r=\frac{\frac{\sqrt{3}\hbar}{2}}{m_ec}≈3.3\cdot10^{-13}>>10^{-22}$$
So this seems utterly nonsensical given the upper bound on the electron radius.
| If you consider only classical physics, an electron can be seen as a Kerr-Newman black hole, that is a rotating charged black hole, violating the naked singularity bound since it's "rotating" too fast.
But this analogy is flawed in many ways, mainly not taking into account quantum physics. You cannot have a black hole with mass smaller than the planck mass, since the quantum fluctuations of the horizon would be of the size of the horizon itself.
Moreover current theories (the standard model!) regard as fundamental entity the fields, while particles can be seen as excitations of this field. Fields live in the whole spacetime, so they are never exactly localized.
You can find some material here:
https://en.wikipedia.org/wiki/Black_hole_electron
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Bounds of Integration (with respect to something that is not time) I have been reading Richard Feynman's lectures and came across an interesting proof regarding the Earth's gravitational force. At one point in the proof, Feynman uses the following the integral:
$\int_{R+a}^{R-a} dr$
(13.18 on http://www.feynmanlectures.caltech.edu/I_13.html)
In this integral, r is the distance between a point in space and the surface of the Earth, R is the distance between that point and the center of the Earth, and a is the radius of the Earth. I interpret this integral as summing up all of the dr's going around the Earth. The proof itself makes sense to me, I am just confused about the bounds of integration.
As $\int_{R+a}^{R-a} dr$, I interpret the integral as summing up the dr's starting on the right side of the Earth and going to the left side. However, in this sense, $\int_{R-a}^{R+a} dr$ should be the sum of all the dr's starting from the left side and going to the right side. Conceptually, I feel as if these should be the same, but mathematically $\int_{R+a}^{R-a} dr = -\int_{R-a}^{R+a} dr$. My question is, how did Feynman choose the ordering of his bounds of integration? It does not appear arbitrary, but I am not sure how the decision was made. Thank you!
| The problem has nothing to do with change of variable. It is just about orientation, where you are choosing to put your zero potential energy. He is putting it at $r=\infty$. This is just a convention, no math. You can put the zero anywhere you like. The integral is then from $r=\infty$ to $0$. So, the integral is really $\int_{\infty}^{0}dW$. There is no mass everywhere in that interval. The mass only counts in the interval from $R+a$ to $R−a$. That is why you end up computing $\int_{R+a}^{R-a}dW$. That's all. Put your zero potential energy at $r=0$ and the order reverses. The sign of the resulting potential energy accounts for the different choice in zero level.
See Why choose a convention where gravitational energy is negative?
| {
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On a derivative of the Poynting vector's derivation I'm studying the Poynting theorem (the mathematical proof). It starts considering the energy held by the electromagnetic field inside a surface $S$, which is:
$$
U_{em} = \int_{\tau}\frac{1}{2}(\vec{E}\cdot\vec{D})\ \mathrm{d}\tau + \int_{\tau}\frac{1}{2}(\vec{H}\cdot\vec{B})\ \mathrm{d}\tau
$$
where d$\tau$ is the infinitesimal volume inside $S$. Now the energy is derived in time, giving the expression:
$$
\frac{\mathrm{d}U_{em}}{\mathrm{d}t} = \int_{\tau}\vec{E}\cdot\frac{\mathrm{d}\vec{D}}{\mathrm{d}t} + \vec{H}\cdot\frac{\mathrm{d}\vec{B}}{\mathrm{d}t}\ \mathrm{d}\tau
$$
So here's what I'm not understanding: why, mathematically speaking
$$
\frac{\mathrm{d}}{\mathrm{d}t}\frac{1}{2}(\vec{E}\cdot\vec{D}) = \vec{E}\cdot\frac{\mathrm{d}\vec{D}}{\mathrm{d}t}
$$
I know is about differential properties of the fields, but I really can't figure out why, also because the derivation a time derivative and I cnnot understand if the fields depend on time this way $\vec{E}\equiv\vec{E}(x,y, z, t)$ (just the magnitude) or this way $\vec{E}\equiv\vec{E}(x(t), y(t), z(t))$...
Thanks for helping, even on a silly question.
| Look to the product rule. First, what is $\mathbf{D}$? $\mathbf{D}(\mathbf{x}, t) = \epsilon(\mathbf{x}, t)\, \mathbf{E}(\mathbf{x}, t)$, right? Thus:
\begin{array}
\ \frac{1}{2} \frac{\operatorname{d}}{\operatorname{d}t} \left(\mathbf{D} \cdot \mathbf{E}\right) & = \frac{1}{2}\frac{\operatorname{d}}{\operatorname{d}t} \left(\epsilon \mathbf{E} \cdot \mathbf{E}\right) \\
& = \frac{1}{2} \left[\mathbf{E} \cdot \mathbf{E}\frac{\operatorname{d}\epsilon }{\operatorname{d}t} + 2\epsilon \mathbf{E} \cdot \frac{\operatorname{d}\mathbf{E}}{\operatorname{d}t} \right].
\end{array}
If $\epsilon$ is constant in time, then the first term vanishes. The $\epsilon$ in the second term can then be brought inside the derivative to change the $\mathbf{E}$ back into a $\mathbf{D}$, giving the desired result.
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What produces an electromagnetic field? A charge that moves, or a charge that changes over time? I've learnt an electromagnetic field is produced by moving charges, i.e. a current.
Is it the case, or is it actually the fact that the charge is changing at a given location?
I mean: imagine I have a charge $q_1$ located at a given point of the space. Then imagine the charge changes and becomes $q_2$, but without moving (this is probably impossible? but let's imagine). Does it produce an electromagnetic field?
Or, without "imagining" a charge changes over time:
Is the EM field produced by the movement of the charge, of by the fact the charge vanishes from its location?
|
I have a charge ... located at a given point of the space. Then imagine the charge changes ..., but without moving (this is probably impossible? ... Does it produce an electromagnetic field?
Elementary charges (electron, proton and their anti-particles) in unbounded state have unchangable charges and by this electric fields of certain size.
Charges emit EM radiation in the case of accelerations. This acceleration can be a spiral path induced by a magnetic field or be this during the deflection from an electric field.
Synchrotron radiation .............................................................Bremsstrahlung
But there is one more detail. During the approach to the nucleus the electron emits EM radiation, called spontaneous emission. The energy for the emission has to come from somewhere. Now your point of a changing electric field come into play.
What if the electric fields of the electron and the proton inside an atom get weaker during the approach of the electron towards the nucleus? The excess energy wil be realized in the form of EM radiation.
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Is there any qualitative difference between the WZW $SO(2)_1$ and the WZW $SU(2)_1$ CFT? Consider the anisotropic spin-$\frac{1}{2}$ Heisenberg chain $$H = \sum_{n=1}^N S^x_n S^x_{n+1}+S^y_n S^y_{n+1} + \Delta S^z_n S^z_{n+1}$$
which for $\Delta = 0$ realizes the Wess-Zumino-Witten (WZW) $SO(2)_1$ conformal field theory (CFT), whereas at the isotropic point $\Delta = 1$ we have the WZW $SU(2)_1$ CFT. Moreover, the line $\Delta \in [0,1]$ is said to be a fixed point line connecting the two critical models.
So on the level of the central charge and scaling dimensions, both models continuously connect to one another, and hence on that level there is no qualitative difference. However, I was just wondering: is there any qualitative difference between them? For example, even though the scaling dimensions are not qualitatively different, I could imagine that (for example) one model has log-like contributions whereas the other doesn't. After all: even though the above model has a line of fixed points connecting both extremes, $\Delta=1$ is the end point of such a line of fixed points, and hence one might expect something funny to happen there.
EDIT: rereading this question I posed two years ago, I regret its poor formulation. The question attempts to ask whether the $\Delta = 1$ point has any singular behavior on the compact boson line, e.g., are there loglike corrections at this point which are not present for $|\Delta|<1$ etc?
| They are very different. First of all the free fermion is fermionic. Then $SU(2)_1$ has non-chiral fields coming from the left and right primary fields with $h=1/4$ (spin 1/2), namely the Hilbert space decomposes as
$(\mathcal{H}_0\otimes\bar{\mathcal{H}}_0)\oplus (\mathcal{H}_{1/4}\otimes\bar{\mathcal{H}}_{1/4})$.
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Quantum systems: three-level vs qutrit Is there a difference between a three-level atomic system and a qutrit? If yes, please tell me what it is and how to relate both quantum systems.
| In a qutrit, every transition between pairs of levels is independent, in the sense that $\vert 1\rangle \leftrightarrow \vert 2\rangle$ is independent from $\vert 1\rangle \leftrightarrow \vert 3\rangle $ which is independent from $\vert 2\rangle\leftrightarrow \vert 3\rangle$.
Thus, not every three-level system is a qutrit. For instance, the angular momentum states with $\ell=1$ do not form a qutrit if we restrict the observables to angular momentum operators, since, for instance the matrix elements $\langle 1,-1\vert L_- \vert 1,0\rangle =
\langle 1,0\vert L_-\vert 1,1\rangle=\sqrt{2}$, i.e. the transitions between adjacent levels are not independent. In addition, a pulse resonant for the $m=-1\to m=0$ transition is also resonant for the $m=0\to m=1$ transition: in the case of $\ell=1$ states, not very transition can be accessed independently so it's NOT a true qutrit.
Unfortunately too many people do confuse a spin triplet for a true qutrit. This is because the states
$\{\vert 1,m\rangle\}$ do span the space, but the difficulty is with the operators: when restricted to angular momentum operators, not every transition can be accessed separately. You need to use the full set $su(3)$ operators acting between basis states to properly described a qutrit.
An important and useful way to see the difference between angular momentum and $su(3)$ operators is that the standard ladder operators in $su(3)$ connect only 1 pair of states, whereas angular momentum operators connect two pairs of states when $\ell=1$.
| {
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First law of thermodynamics as conservation of energy We have that $\Delta U = Q + W$. What I don't see is how this formula relates to the law of conservation of energy. Can someone please clarify?
Does this mean that $\frac{dU}{dt}=\frac{dQ}{dt}+\frac{dW}{dt}=0$, so that $\frac{dQ}{dt}=-\frac{dW}{dt}$?
| The equation of the first law of Thermodynamics does not directly imply the conservation of energy; rather the first law of Thermodynamics is a consequence of the conservation of energy when applied to Thermodynamics( systems involving heat, temperature etc.)
The conservation of energy was known somewhat by scientists like Galileo in 1638 (knew about potential and kinetic energy conservation in a pendulum) and others before him too. This was before 1850 when Rudolf Clausius and Lord Kelvin stated the First Law of Thermodynamics for energy conservation in systems involving heat energy.
To answer your second question, yes if you take time derivatives on both sides that equation results but that does not mean that the rate of change of internal energy($U$) is zero, it's dependent on how the other factors like rate of heat energy supplied($\frac{dQ}{dt}$) or rate of work done on the system($\frac{dW}{dt}$) changes with time. If both of these are zero or they are equal in magnitude and opposite in sign then only can the rate of change of $U$ be zero. It's a special condition when internal energy does not increase not a condition valid for all times. Ofcourse the conservation of energy can hold even if there is a change in $U$, as then the energy comes from the surroundings to add to the system.
| {
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Are special relativity calcs affected by media? Time dilation, increase in mass, Lorentz contraction calcs all involve velocity of light in vacuum. But in optical media light slows down. So what of relativity calcs in media? Do we ever need to adjust the speed of light?
| No, because in a dielectric medium light isn't light, and that's why its speed isn't $c$.
The speed of light in a dielectric medium remains unchanged. You'll find lots of questions discussing exactly what goes on with light in a medium, but basically the EM field of the light interacts with electrons in the medium to form an entangled system that has an effective mass greater than zero and hence travels slower than light. In highly interacting systems like BECs we get a distinct quasiparticle called a polariton, though in usual dielectrics like glass and water the coupling isn't strong enough to make the quasiparticle a useful description.
Though undergraduate descriptions of special relativity focus on the constant speed of light, special relativity is really a geometric theory. It is a consequence of the geometry of flat spacetime, and the constant speed of light is a result of this geometry. The geometry of spacetime is the same inside a dielectric medium as it is outside it.
| {
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How is it that in a car crash, four 8mm bolts can anchor the seat to the car? In a car crash at for example twenty metres per second. I used suvat equations and newtons second law to work out the force as as body accelerates(negatively). I estimated that the distance travelled in the crash by the body would be roughly 0.4 metres.Even using average mass of a human and car seat the force calculated was way too large to be accurate as the tensile strength of steel would be easily exceeded. I concluded that a large portion of energy is transferred by the front of the car before it affects the body. My question is how could I find an accurate but rough figure for the force in newtons acting on each individual bolt and if anybody has any data or estimates.
Thank you
| It isn't uncommon for the breaking load for M8 bolts to be 1800 kg or more. That puts four at 7200 kg, enough to statically support the weight of many vehicles with passengers.
Also important is that the bolts are not the only thing transferring the force. In any good design, much of the force will be transferred across the mating surfaces of the seat and rails to which it is bolted.
| {
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Estimate of the effect of quantum disturbances on a macroscopic object I am self-studying P. Davies, D. Betts, Quantum Mechanics. Exercise 4 of Chapter 1 says:
"A snooker ball of mass $0.1$kg rests on top of an identical ball and is stabilized by a dent $10^{-4}$m wide on the surface of the lower ball. Use the uncertainty principle to estimate how long the system will take to topple, neglecting all but quantum disturbances." (the answer is "About $10^{27}$s.")
I have a hard time thinking about how the uncertainty principle can be applied here. Is the ball to be considered as a "macroscopic particle", or should the principle be applied to a specific particle (or a group of particles) in the ball itself?
I would prefer to not receive the solution straightforwardly. I am looking for suggestions on the correct way to look at this problem.
| If one uses the position-momentum uncertainty relation $\Delta x\Delta p\sim h$, one arrives at a time of $\tau\sim 10^{25}$ s. If, on the other hand, one uses the energy-time uncertainty relation $\Delta E\Delta t\sim h$, one arrives at a time of $\tau\sim 10^{29}$ s.
Let's just take the geometric mean of these answers??
All joking aside, this is a pretty teachable moment. The uncertainty principle shouldn't really be used for anything more than incredibly rough calculations (except for rigorous circumstances where it actually provides useful information as a mathematical statement). Just as in astrophysics, this is a question where getting your answer within the right order-of-magnitude of the correct order-of-magnitude is a pretty okay result.
I hope this helps!
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How is the equation of Mach number derived? Wikipedia states that for a pitot-static tachometer, the mach number for subsonic flow equates to
$$M = \sqrt{5\left[\left(\frac{p_t}{p_s}\right)^\frac{2}{7}-1\right]}.$$
How did they get to that result? Is there a derivation, or is it just from a polynomial fit of a tabulated set of data?
Update
I accepted J.G's answer after glancing at the referenced flight test document (a treasure in itself) and realising that $\frac {7}{5}$ is the same as 1.4, but there remains an issue.
Sadly I don't have my uni books anymore with Bernoulli's equation for compressible flow. The issue is with dynamic pressure: for incompressible flows we can take $p_d = \frac {1}{2} \cdot \rho \cdot V^2$, for compressible flow this is $p_d = \frac {1}{2} \cdot \gamma \cdot p_{static} \cdot M^2$.
Right? If I substitute this I don't get to the equation above. So the answer is unfortunately not accepted anymore.
| If you want to know more about calculating a Mach number, it helps to read Wikipedia's article on Mach number. As explained here, the Mach number for subsonic compressible flow is obtainable from Bernoulli's equation (Wikipedia cites this source). The result you cited then follows from $\gamma=\frac{7}{5},\,p_t=q_e+p$.
| {
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(Why) is it necessary to invoke Newtonian Gravity to fix normalization constants in General Relativity? I'm studying GR on Carroll's book. In Chapter 4 the constant appearing in Einstein's equation is fixed by requiring to obtain Poisson's equation in a Newtonian framework.
In chapter 5 the constant of the Schwarzschild metric is fixed invoking the weak field metric.
I understand that Einstein's equation is built, not derived or proved, and hence we somehow do not have a real ground to build on, but we have to work the other way round. Still, one would expect GR to be self contained, so what is going on here? Is it actually necessary to use Newtonian results?
|
Is it actually necessary to use classical results?
Of course. Any new theory in physics must replicate current theories in those areas where those current theories already do a very good job of predicting behaviors. This is true in quantum mechanics as well.
Newtonian gravity predicts behaviors quite nicely in the limit of small densities, large distances, and small velocities, with a very small discrepancy from observations with regard to the orbit of Mercury. General relativity has to explain the near-Newtonian behavior of the outer planets and explain that small discrepancy in Mercury's orbit. It does both. The explanation of Mercury's anomalous precession is a natural consequence of general relativity. But matching the near-Newtonian behavior of the outer planets was done by design.
| {
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Two-point correlation function in Peskin's book I am reading Peskin's book on QFT and I reached a part (in chapter 4) where he is analyzing the two-point correlation function for $\phi^4$ theory. At a point he wants to find the evolution in time of $\phi$, under this Hamiltonian (which is basically the Klein-Gordon - $H_0$ - one plus the interaction one). Anyway, when he begins his derivation he says that for a fixed time $t_0$ we can still expand $\phi$ in terms of ladder operators in the same way as we did in the free (non-interaction) case (this is on page 83), i.e. $$\phi(t,\mathbf{x})=\int{\frac{d^3p}{(2\pi)^3}\frac{1}{E_\mathbf{p}}(a_\mathbf{p}e^{i\mathbf{x}\mathbf{p}}+a_\mathbf{p}^\dagger e^{-i\mathbf{x}\mathbf{p}})}.$$ I am not sure I understand why can we do this, for a fixed time. When we wrote this in term of ladder operators for the free case, we used the KG equation in the free case, which resembled to a harmonic oscillator in the momentum space, and hence we got the ladder operators. But now, the equation of motion is different (it has $\phi^3$ term, instead of 0, as before), so can someone explain to me why we can still use the same formula as before, even if the equation of motion is different?
| I think the question is related with the answer given by Valter Moretti from
Here
Clearly, you cannot say that the field $\phi(x)$ can be decomposed as creating and annihilation part.
| {
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Will overlapping two different beams of coherent light with different wavelength cause interference? If I use two different wavelength lasers to transmit light into a single mode optic fiber will they interfere with each other? If so, how much will be that interference.
| Yes, light from the two lasers will interfere, but the light will interfere temporally, not spatially. The result, when the light is incident on a detector, is a beat frequency: a signal whose frequency is equal to the difference between the frequencies of the two lasers.
The phase of the signal will wander relatively slowly, at a rate that depends on the phase stability of the two lasers.
The beat frequency between two lasers having slightly different frequencies is temporal interference, and is actually exactly the same phenomenon as spatial interference. A fringe pattern that is stationary in one reference frame is moving in another. The frequency of both beams is the same in the first frame, but different (because of Doppler shift) in the second frame, the rate at which the fringe peaks go by in the second frame is the beat frequency between the beams: one Doppler shifted up, and the other Doppler shifted down.
| {
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Is there a case when it is better to use the integral form of the Maxwell equations rather than the differential form? I was wondering if there is a case where the integral form of the Maxwell equations is preferred over the differential form?
If you could provide with an example for each one of the equations I would really appreciate that.
| The integral form of Maxwell's equations is the most general form.
You may ask, why? Aren't both of them equivalent statements. No, they are not, the integral form is more powerful, yet it's underappreciated.
The first condition for a function to be differentiated is differentiable, if there's a single point which is non differentiable within the interval, the whole function cannot be differentiated.
For example, the differential form is useless when you deal with boundaries.
If you have a metal ball, you can't just setup a function and differentiate it to get your result, metal boundaries make the function discontinuous.
The integral form is not limited to continuous function, as long as the function is piecewise continuous, and integrable, you can integrate it easily with Gauss's surface and Ampèrian loop.
Stoke's theorem and Divergence Theorem can only applied if the loop/surface is simply connected, which means, you can use the Integral form to recover differential form if the function is continuous, or the other way round.
However, If the function is discontinuous, you must use integral form, and you can't switch to differential form.
The integral form really is the "true" and most general Maxwell Equation.
| {
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Determining mass radius and charge radius of electrons First the mass radius problem:
Why can't the mass radius of electrons be determinded by shooting neutral particles on it. Similar to Rutherford's gold model only a bit more sophisticated.
Secondly the charge radius problem:
I often hear the term charge radius and the charge radius for protons has been calculated with accuracy for example described in this article:
https://phys.org/news/2016-08-deuterium-nucleus-proton-radius-puzzle.html
Why can't the charge radius of electrons be determinded in a similar way?
| Rutherford used $\alpha$-particles as projectiles. Compared to the gold foil they have "little" mass and they interact "heavily" via the electro-magnetic force. Furthermore, compare to the size of a gold atom (incl. the electronic shells) the $\alpha$-particles are small. All three properties were important:
*
*small mass, so that the gold foil will not move when it interacts with the $\alpha$-particles. Otherwise, we would not find back-scattered particles.
*"strong" interactions via the electro-magnetic force, so that we can find some reflected particles in the forward direction.
*small size, so that it forms a local probe.
So what neutral particles would you use in the electronic Rutherford experiment?
| {
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Non-local field redefinition and $S$-matrix It is known that for local field redefinitions for which the LSZ formula is valid:
$$\langle 0|\phi(x)|p\rangle \neq 0$$ field redefinitions don't change the S-matrix.
(See QMechanic's answer to Equivalence Theorem of the S-Matrix)
But is it true for non-local field redefintions? For instance if I take a field redefinition of the form:
$$\psi(x)=e^{-l\Box} \phi(x)$$
Will the S-matrix be invariant under this?
From QMechanic's explanation linked above and the answer by AccidentalFourier Transform here it would seem that the answer should be yes.
Edit: My main concerns are
*
*Is such a transformation invertible? Is this a concern?
*Any boundary condition on $\psi$ would translate to infinitely many boundary conditions on $\phi$. Is this relevant?
| Claim 1. If $\psi(x)$ is an arbitrary operator that satisfies
\begin{equation}
\langle 0|\psi(x)|p\rangle\neq 0\tag1
\end{equation}
then it is a valid interpolating field, and as such, it can be used in the LSZ formula. The proof can be found in any introductory text, such as Weinberg.
Claim 2. If we assume that
\begin{equation}
\langle 0|\phi(x)|p\rangle\neq 0\tag2
\end{equation}
and define
\begin{equation}
\psi(x)\overset{\mathrm{def}}=\mathrm e^{-\ell\partial^2}\phi(x)\tag3
\end{equation}
then we have
\begin{equation}
\langle 0|\psi(x)|p\rangle\neq 0\tag4
\end{equation}
The proof is straightforward. One just need to use $\langle0|\phi(x)|p\rangle=c\mathrm e^{ipx}$ for some non-zero constant $c$, and the trivial identity $\mathrm e^{-\ell\partial^2}\mathrm e^{ipx}=\mathrm e^{\ell p^2}\mathrm e^{ipx}$.
Conclusion: the non-local redefinition $(2)$ is a valid redefinition, and the $S$ matrix is invariant under it.
| {
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Does stream function hold only for steady flow? This is just a basic question about streamline, equation of streamline and stream function. I am used to perceiving that the stream fucntion is just another form of the equations of streamlines. Since streamlines can be obtained at an "instant time" of an unsteady flow field, then the stream function should be the same, it could be obtained at an instant time of an unsteady flow field.
However, I stuck at this point: consider an unsteady, compressible, 2D - flow field at an instant time, it has stream function $\bar {\psi}$. From textbooks, we already know:
$$\rho u = \frac{\partial \bar {\psi}}{\partial y}\space\space\space (1) $$
$$\rho v = -\frac{\partial \bar {\psi}}{\partial x}\space\space\space (2) $$
Then I do some maths:
$$(1)\Rightarrow \frac{\partial (\rho u)}{\partial x} = \frac{\partial^2 \bar {\psi}}{\partial y \partial x}$$
$$(2)\Rightarrow -\frac{\partial (\rho v)}{\partial y} = \frac{\partial^2 \bar {\psi}}{\partial x \partial y}$$
Since $\frac{\partial^2 \bar {\psi}}{\partial y \partial x} = \frac{\partial^2 \bar {\psi}}{\partial x \partial y} $, hence:
$$\frac{\partial (\rho u)}{\partial x} = -\frac{\partial (\rho v)}{\partial y} \iff \frac{\partial (\rho u)}{\partial x} + \frac{\partial (\rho v)}{\partial y}=0 \iff \nabla.(\rho\vec V)=0\space\space\space (3)$$
But continuity equation says:
$$\frac{\partial \rho}{\partial t}+ \nabla.(\rho\vec V)=0\space\space\space (4)$$
Here you can see the flow field is unsteady, then $\frac{\partial \rho}{\partial t}$ is not equal zero and the equation (4) is inconsistent with (3). Here is where I stuck... if I am wrong, so does stream function hold only for steady flow?
| Generalizing your mathematics (which are all correct), the 2D stream function automatically satisfies continuity for any 2D case where $\partial\rho / \partial t$ is zero. In flows other than these, you must independently confirm that your results satisfy continuity.
| {
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Deformation of a self-gravitating sphere from two forces I have a fluid sphere (say a gas or a liquid of uniform density, under its own gravity) on which forces is applied to its surface. I would like to find its approximate shape (most probably an oblate ellipsoid), from the forces applied on its (initialy) spherical surface. Using spherical coordinates, the radial (pressure) and tangential (shear) forces are these :
\begin{align}\tag{1}
F_r(\vartheta) &= C \sin^3 \vartheta, \\[12pt]
F_{\vartheta}(\vartheta) &= 4C \, \sin^2 \vartheta \, \cos{\vartheta}, \tag{2}
\end{align}
where $C$ is an arbitrary positive constant. In vectorial form :
\begin{equation}\tag{3}
\vec{\boldsymbol{\mathrm{F}}} = C \sin^3 \vartheta \, \vec{\boldsymbol{\mathrm{u}}}_r + 4C \, \sin^2 \vartheta \, \cos{\vartheta} \, \vec{\boldsymbol{\mathrm{u}}}_{\vartheta}.
\end{equation}
There's an axial symetry around the $z$ axis. The deformation may be considered "weak", as a first approximation ($C$ may be "small", compared to the gravitationnal force on the surface : $C \ll G M^2/R^2$).
Note that the pressure force is 0 at the poles, and maximal at the equator, so it tends to "squash" the sphere to an oblate ellipsoid (of unknown ellipticity). The shear force is 0 at the poles and at the equator.
Any idea about how to find the deformation's ellipticity?
| If the deformation is minimal, I think you can approximate the problem by just increasing the radius of the sphere by the amount that it is deformed outwards (of course on the poles this gives the wrong results then).
Then you can require the increased gravitational force to be equal to the force acting on the equator due to rotation.
This approximation should for example work on earth.
| {
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How can we show Power = $\mathbf{F}\cdot \mathbf{v}$? How can we say that $$\text{Power} = \mathbf{F}\cdot \mathbf{v}$$
We know that small work done by a force $\mathbf{F}$ to displace an object by '$\mathbf{x}$' is
$$W = \mathbf{F}\cdot \mathbf{x}$$
So derivating wrt time, we get
$$\begin{align}
P=\dfrac{dW}{dt}&=\frac{d\mathbf{F}}{dt}\cdot \mathbf{x}+\mathbf{F}\cdot\dfrac{d\mathbf{x}}{dt}\\
&=\frac{d\mathbf{F}}{dt}\cdot \mathbf{x}+\mathbf{F}\cdot\mathbf{v}
\end{align}$$
We get this wrong result. How actually can we show $P=\mathbf{F}\cdot\mathbf{v}$ ?
Edit
Actually I know that total work $W$ is $\int \mathbf{F}\cdot d\mathbf{x}$.
Infinitesimal work done by $\mathbf{F}$ to displace body by $d\mathbf{x}$ will be $dW = \mathbf{F}\cdot d\mathbf{x}$, so dividing by $dt$ on both sides gives $$P =\dfrac{dW}{dt} = \mathbf{F}\cdot \frac{d\mathbf{x}}{dt}$$
But I wanted a proper proof not involving differentials!
| The work $W$ is not equal to $F\cdot x$ generally. The correct form is
$$W=\int F \cdot dx$$
So $dW=F\cdot dx$ and $P=\dfrac{dW}{dt}=F\cdot\dfrac{dx}{dt}=F\cdot v$
| {
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When short circuits are not exactly short circuits?
Here $R1 = 2 \Omega$, $R2= 4 \Omega$ and $R3= 4 \Omega$
Though there looks to be a short circuit in this diagram, my teachers say that this circuit can easily be redrawn into simple parallel circuit. As far as I know a short circuit is an alternate way for current to move where there is no resistance whatsoever. So all of the current will flow through a short circuit if found.
Here if the current flows through that part of the wire ACB then after that R2 and R3 being the same it will get confused which way to go and even if it goes both sides then some part of the current is bouncing back and moving from the negative side of the cell to the positive. This is just a conjecture.
So can anyone please describe why there is no short circuit in the circuit though it seems to be? And also how to understand by looking at any circuit that there is actually no short circuit though there seems to be one.
| The wire which joins the points $A$ and $B$ is essentially a short circuit. Assuming the wire to be ideal (no resistance), the electric potential at point $A$ is equal to the electric potential at point $B$.
Therefore, the circuit is equivalent to the circuit where $R_2$ and $R_3$ were connected to point $A$.
With similar reasoning, you can do the same for the rest of the short circuit wires and obtain a simple parallel resistors combination.
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Why does the Sun appear larger during the sunrise and sunset? Why does the Sun appear larger during sunrise and sunset compared to its size during midday even though its distance from the Earth remains fixed all the day? It is only during the Winter when the Sun is away from the Earth and that too is due to the motion of Earth around the Sun, not due to the diurnal motion of the Earth. So is it some optical effect? If yes, what is it really? I apologize if the question has already been asked.
| I spent over 30 years at sea, and I can tell you that when you're out in the middle of the ocean the setting and rising sun sometimes seems to be much larger. Atmospheric conditions seem to influence this. If you wake up to a red sky and the sun looks huge on the horizon you can expect some nasty weather.
I think that the water vapor in the atmosphere, coupled with the oblique angle of the radiation from the rising sun, magnifies the apparent size.
| {
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Why can a regular infrared camera not show temperature (thermography)? There are a lot of questions here dealing with infrared cameras and thermographic cameras. I think I understand the reason why a thermographic camera is able to retrieve the temperature values from any object and convert them to a falsecolor representation, but why is a "regular" infrared camera not able to retrieve this information? What are the differences between these cameras? Is it just the sensor within the camera?
| It depends on the sensor. If all that makes a camera "infrared" is that it is capable of detecting light with a lower frequency than red light, that is not enough to identify temperature. Objects of different temperature emit different frequencies of infrared light at different amounts. In order to accurately determine the temperature of an object, you need to be able to accurately measure the distribution of frequencies, not just the amount of light. Thermographic cameras are tuned to multiple infrared frequencies and possibly a few visible ones just like how ordinary cameras are tuned to red, green, and blue.
| {
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Renormalization and canonical commutation relations My question is whether canonical commutation relations hold for renormalized quantum fields. Below I show reasoning which caused by doubts.
Consider a relativistic scalar QFT. We have spectral decomposition of two-point function
$$ \langle \Omega | \phi(x_1) \phi(x_2) | \Omega \rangle = \int \frac{\mathrm d m^2}{2 \pi} \rho(m^2) \Delta_+(x_1-x_2,m^2), $$
where $\rho \geq 0$ is called spectral density function and distribution $\Delta_+$ is defined as
$$ \Delta_+ (x,m^2) = \int \frac{\mathrm d ^3 p}{(2 \pi)^3 2p^0} e^{-ipx}, $$
with integral evaluated over the positive frequency ($p^0 \geq 0$) mass-shell $p^2=m^2$. I assumed above that field $\phi$ has no vacuum expectation value. If we take the difference of the first formula with itself with $x_2$ and $x_1$ interchanged, set $x_2 = 0$, take derivative with respect to $x_1^0$ and set $x_1^0=0$ we get canonical commutator on the left hand side. By comparing with the right hand side one obtains the Weinberg sum rule for the spectral density:
$$ \int \frac{\mathrm d m^2}{2 \pi} \rho(m^2) = 1. $$
What bothers me is that value of this integral depends on the values of finite parts of renormalization constants. Hence it is not renormalization scheme and scale independent. I checked some simple examples and it turned out to be possible to enforce this relation as renormalization condition and fix the value of wavefunction renormalization. However, I don't think this is what is usually done.
| The commutation relations for renormalized fields are different then those of the bare fields by factors of the wavefunction renormalization. As an example, consider a complex scalar field, $\phi$. The bare fields obey, e.g.,
$$
\left[ \phi (x) , \phi (y) ^\dagger \right] = \int \frac{ d^3p }{ (2\pi)^3 } e ^{ i p \cdot x }
$$
while the renormalized fields ($ \phi _r \equiv \phi /\sqrt{ Z} $) obey,
$$
\left[ \phi _r (x) , \phi _r (y ) ^\dagger \right] = Z \int \frac{ d^3p }{ (2\pi)^3 } e ^{ i p \cdot x }
$$
Should we be troubled by this? I don't think so. The important conclusion with regards to the commutation relations is they vanish for space-like points in order to be consistent with special relativity. Other than that they don't play a significant role here.
As a side point, the time-ordered product of fields is also different for the bare and renormalized fields. This leads to a modification of the propagator as I suspect you are already aware.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/348328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Buoyancy Force and Density I am doing a practice problem about buoyancy force. If there are two objects, (one made of wood and one made of metal), and we hold them underwater, will the buoyancy force of both objects stay the same? That's what I believe at least.
| The buoyant force will be same if the fluid density is the same and the volume of liquid displaced by both objects is the same. The formula to calculate upward buoyant force is ρ × V × g , where ρ is the density of the liquid the object is immersed in, V is the volume of the displaced liquid and g is the gravitational acceleration.
I have created this Buoyancy Simulation which will help you understand the concept better.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/348430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Current experimental evidence of lepton flavour violation other than neutrino oscillation Is Lepton Flavour Violation (LFV) experimentally established in processes other than neutrino oscillation? This answer by Luboš Motl points out that
Interestingly, CMS has detected a 2.5 sigma excess of Higgs bosons that seemingly decay to
$$ h \to \mu^\pm \tau^\mp $$
This is an example of a process that would violate the flavor numbers. A muon is created with an anti-tau, or vice versa. Note that this process is compatible with the charge (and energy, momentum etc.) conservation laws, the truly important ones. But it changes $L_\mu$ and $L_\tau $ by $\pm 1$ and $\mp 1$, respectively.
*
*Can it taken as the experimental evidence for lepton flavour violation independent from neutrino oscillation?
*Did the LHC observe other signatures for LFV? If yes, what is the confidence level of the observations?
| The answer is no :)
(At least thus far.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/348613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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