statistical mechanics - Definition of quantum microcanonical ensemble in Landau & Lifshitz

I'm reading the first chapters of Landau & Lifshitz's Statistical Physics and I don't understand the definition of the quantum microcanonical ensemble.

The microcanonical distribution for a quantum isolated system with energy $E_0$ is defined in the text as:

$$\text d w = \text{const}\times \delta (E-E_0) \cdot \prod _a \text d \varGamma_a.$$

where $\text d w$ stands for “probability” and $\text d \varGamma_a\!$ is the differential of the function $\varGamma_a(E_a)$, the number of quantum states of a subsystem with with $\text {energies}\leq E_a$ (which is treated as a smooth function).

I see two problems here: first of all, the distribution should be given in terms of a density operator $\rho$: I don't immediately see how to relate this density function to a matrix. However, forgetting for a second about the operator, what does $\text d w = ...$ mean? It is the probability density of what variables and how do these variables identify the state of the ensemble?

Finally, the text goes on asserting that, if the system is subdivided in $n$ small (quasi-)independent subsystems of the original system, then $\text d \Gamma$ can be written as the product $\text d \Gamma_1\cdots \text d \Gamma _n$. This fact is used later to derive the second law of thermodynamics: the joint probability density for the values $E_1 , \dots ,E_n$ of the subsystems energy is written as

$$\qquad\qquad\qquad\qquad\qquad \text d w = \text{constant} \times \delta (E-E_0)\cdot\prod_a \dfrac{\text d \varGamma_a}{\text d E_a} \text d E_a. \quad\qquad\qquad(7.15)$$

This one looks like the meaningful equation (to express the joint probability density of energies in the microcanonical ensemble). Is Landau's notation just a short-hand for the above one?

Answer

The first thing to point out is that there are two equivalent ways to describe a quantum statistical distribution: the density matrix, and a probability distribution on the results of measurements of a "complete" set of observables. (It is remarkable that one should think of givin such a probability distribution on the classical phase space of the quantum system, but let's not go into this here. It is remarkable since it seems to go against the whole principle of quantum mechanics to assign a probability to a point with simultaneously assigned values of both position and momentum. But it works somehow anyway.)
Landau explains this equivalence between $w$ as a probability distribution and $w$ as a density matrix. In his formula he wants to point out the analogy with classical mechanics and for that reason interprets $w$ as a probability distribution.

The microcanonical ensemble consists only of states with a given fixed energy $E_0$: any other value of the energy has a zero probability of resulting from a measurement. But of course there are other commuting observables, and they will take on possibly different values when measured, so $w$ is a distribution on those values. But it is, by definition, somehow "constant". Other statistical distributions on this same set of states would be possible, but would not be the microcanonical one.

The second edition of Landau--Lifschitz explicitly says that $dw$ is the probability of finding the system in one of the states that "belongs" to one of the energy levels infinitesimally close to $E_0$. He also says explicitly that $d\Gamma$ is the number of quantum states with energies in that same infinitesimal range around $E_0$. It is very hard to make rigorous sense out of this, although it has been done by other authors. That is why the density matrix is often preferred.

In fact the multiplication by the delta function means all states with a non-zero probability have the same energy. So $w$ on that ensemble is not a function of $E_0$, since $E_0$ is a parameter. But if you look at that formula, there are no other variables except $E$. So $w$ is constant on that ensemble, i.e. every state has the same probability. So we don't really need a formula at all: the constant $d\Gamma$ doesn't affect relative probabilities, and if you normalise this to make them genuine probabilities you just divide by $d\Gamma$.... Landau was a great physicist but he was not a very logical writer. Nor a very clear one. And throwing a junior co-author into the mix just makes things worse...

Varying the parameter and integrating $w$ over $E_0$ makes no real sense for the micro-canonical distribution.

The point is if you wish to calculate the entropy of the microcanonical distribution, as a function of the parameter $E_0$ (which is what Landau does in the very next section). Because the normalised density matrix has to have trace one, and be diagonal, and since every probability is the same, the value of the diagonal elements is just the reciprocal of the dimension of the space of states with that energy, which is the number or elements in a basis, which is what Landau really means by "the number of quantum states that 'belong' to an energy in that infinitesimal range." (Literally, of course, the number is uncountably infinite, not finite.)

I never recommend reading Landau unless you already understand the basic concepts. Landau is the kind of chess grandmaster who can be champion of the world but could not explain to you in writing how a knight moves.

electromagnetism - What is a gauge in a gauge theory?

As I study Jackson, I am getting really confused with some of its key definitions. Here is what I am getting confused at. When we substituted the electric field and magnetic field in terms of the scalar and vector potential in the inhomogeneous Maxwell's equations, we got two coupled inhomogeneous wave equations in terms of $\mathbf{A}$ and $\phi$. So, the book states that to uncouple them, which definitely makes our equations simpler to solve, we introduced gauge transformations as adding a gradient to $\mathbf{A}$ and adding a constant to $\phi$ would not affect their meaning. My question is which one is a gauge and why in the expression for a gauge transformation $$\mathbf{A'}=\mathbf{A+\nabla \gamma}.$$ Somewhere in the internet, I read that $\gamma$ is a gauge function. So, is $\gamma$ a gauge, if yes, then why?

Basically: What is a gauge?

Answer

In normal usage, a gauge is a particular choice, or specification, of vector and scalar potentials $\mathbf A$ and $\phi$ which will generate a given set of physical force fields $\mathbf E$ and $\mathbf B$.

More specifically, a physical situation is specified by the electric and magnetic fields, $\mathbf E$ and $\mathbf B$. A set of potentials $\mathbf A$ and $\phi$ generates the force fields if it obeys the equations \begin{align} \mathbf B & =\nabla\times\mathbf A \\ \mathbf E & = -\nabla\phi-\frac{\partial \mathbf A}{\partial t}. \end{align} As you know, for a given set of force fields, the potentials are not unique. A gauge is a specific, additional requirement on the potentials. One good example of a gauge is the Coulomb gauge, which is mostly embodied by the requirement that $\mathbf A$ also be divergenceless, $$\nabla \cdot\mathbf A=0.$$ "The Coulomb gauge" refers to the set of potentials which satisfy this.

Gauges are usually thought of as specifying the potentials uniquely. This is not really true, but they do tend to specify the potentials "uniquely up to reasonable physical assumptions". The Coulomb gauge is a good example of this: the gauge transformation to \begin{align} \mathbf A'&=\mathbf A+\nabla \chi(\mathbf r)\\ \phi'&=\phi \end{align} preserves the physical fields, and if $$\nabla^2 \chi(\mathbf r)=0$$ then it also preserves the gauge condition that $\nabla \cdot\mathbf A'=0$. This is not great for unicity, because there are a lot of harmonic functions that satisfy the above condition. However, for a function to really be harmonic throughout all of space - with no exceptions and no singularities - then it must diverge at infinity, which is not really palatable in most cases. Because of that, saying that $\mathbf A$ is the vector potential in the Coulomb gauge usually means that $\nabla \cdot\mathbf A=0$ and that such 'infinite-self-energy' terms have been set to zero; this is usually a unique set of potentials in situations where the energy in the physical fields themselves is not infinite.

It is worth noting that, in certain situations, the word gauge can be naturally free of this ambiguity. In my field, strong-field physics, the words 'length gauge' and 'velocity gauge' are taken to mean that the total energy of an electron interacting with a laser field, at position $\mathbf r$ and with momentum $\mathbf p$, is of the form $$E=\tfrac1{2m}\mathbf p^2-e\mathbf r\cdot \mathbf E$$ and $$E=\tfrac1{2m}\left(\mathbf p-e\mathbf A\right)^2,$$ respectively. For a uniform field (i.e. in the 'dipole approximation') the two energies are equivalent via a gauge transformation. However, here the word 'gauge' is completely unambiguous except for a total constant energy which can very safely be ignored.

---

Thus far for technical matters. I think, though, that a lot of what worries you is the word 'gauge' itself, which is indeed a weird choice. In everyday usage, a gauge is a generic form of meter or dial. The phrase 'gauge invariance' seems to have come into physics via German, in Hermann Weyl's use of the word 'Eichinvarianz', which loosely means 'scale invariance' or 'gauge invariance' (in the sense that a choice of measuring instrument (gauge) determines the measured physical values in a given setting, i.e. determines the scale).

This invariance under changes of scale is exactly (part of) the (technical) gauge invariance in general relativity, which is invariant under coordinate transformations.

Note, though, that my source for this history is Wikipedia, so if someone can chime in with a better source it would be fantastic.

classical electrodynamics - Problem with Maxwell's theory

What exactly is the problem with classical Maxwell theory and the blowing up of energy at $r=0$? Does it have any other problems on the classical level?

Answer

There is nothing wrong with Classical Electrodynamics. Electromagnetism is an effective theory in the sense that it provides an almost exact description of physics in our everyday life energy scales, but has a few technical problems like the ones you mention. This is due to the fact that classical electromagnetism is just a very good approximation to a more deeper theory, one we now realise as Quantum Electrodynamics(QED), which takes into account the relativistic quantum mechanical description of maxwell's equations to tackle these problems.

For the above problem physicists have realised that in the proper quantum mechanical description of charged particles that interact via the electromagnetic force is through the emission and absorption of virtual gauge bosons, in the case of QED, photons. This means the separation between two particles cannot physically go to $0$ because they will interact and exchange momentum (via the gauge boson) much before they physically occupy the same spatial state. There are of course other ways this could be avoided like the exclusion principle preventing this from happening to two interacting fermions, or the discrete energy states of electrons in atoms (and thus a stable ground state at a distance $r > 0$) in the case of electrostatics.

homework and exercises - Kinetic energy stored in a triangular wave

I usually used to solve questions based on plane progressive harmonic wave. Recently I encountered a triangular plane progressive wave. Here is a snapshot of the wave.

I am provided with mass/unit length of the string and the tension in the string. I had to find out the kinetic energy of the wave pulse travelling in the taut string as in the figure above.

Here is my attempt:

The triangular wave seems to me like $\sin^{-1}(\sin x)$ with some modifications. I am struggling finding out correct expression of $y$ as expression of $x$ and then integrate to find out the required energy. Any help is appreciated.

Answer

Outline of solution:

• Find the total amount of potential energy stored. This is simply $T \Delta \ell$ where $T$ is the tension and $\Delta \ell$ is the change in length of the rope.


• Don't bother doing integrals, just compute $\Delta \ell$ with the Pythagorean theorem.


• Note that for waves on a string, the kinetic and potential energies are equal.

astrophysics - Effect of relativity on observers of two different inertial frames observing an event taking place very far away

I am new to relativity ,So please correct me If I am wrong. In the case I have mentioned, In a gravity free space, Let us assume a stationary observer and an observer moving with a velocity 'V' w.r.t to the stationary observer.

The stationary observer looks at an event taking place very far away('X') at time 't' So the moving observer's reference (the event is taking place far away for the moving observer as well) of space time can be calculated using Lorentz transformation. if 'V' is considerably small compared to speed of light, then the transformation for space will be similar to Galilean transformation, but since the distance of the event taking place is very large the time transformation will not be Galilean. ( because of the term (V*X)/C^2)

A light year being of the order 10^16 and all objects in universe being beyond this distance, does that mean an observer moving with a speed of 10 m/s will have relativistic effects in significant form recording an event like star explosion, wobbling effects, light variation etc?

quantum mechanics - Plotting $psi$ for finite square well potential

Lets say we have a finite square potential well like below:

This well has a $\psi$ which we can combine with $\psi_I$, $\psi_{II}$ and $\psi_{III}$. I have been playing around and got expressions for them, but they are not the same for ODD and EVEN solutions but lets do this only for ODD ones.

---

ODD solutions:

$$ \boxed{\psi_{I}= Ae^{\mathcal{K} x}~~~~~~~~\psi_{II}= - \dfrac{A e^{-\mathcal{K}\tfrac{d}{2}}}{\sin\left( \mathcal{L} \tfrac{d}{2} \right)}\, \sin\left(\mathcal{L} x\right)~~~~~~~~ \psi_{III}=-Ae^{-\mathcal{K} x}} $$

When i applied boundary conditions to these equations i got transcendental equation which is:

\begin{align} &\boxed{-\dfrac{\mathcal{L}}{\mathcal{K}} = \tan \left(\mathcal{L \dfrac{d}{2}}\right)} && \mathcal L \equiv \sqrt{\tfrac{2mW}{\hbar^2}} && \mathcal K \equiv \sqrt{\tfrac{2m(W_p-W)}{\hbar^2}} \\ &{\scriptsize\text{transcendental eq.} }\\ &\boxed{-\sqrt{\tfrac{1}{W_p/W-1}} = \tan\left(\tfrac{\sqrt{2mW}}{\hbar} \tfrac{d}{2} \right)}\\ &{\scriptsize\text{transcendental eq. - used to graph} } \end{align}

Transcendental equation can be solved graphically by separately plotting LHS and RHS and checking where crosssections are. $x$ coordinates of crossections represent possible energies $W$ in finite potential well.

So i can theoreticaly get values for possible energies $W$ and when i get these i can calculate $\mathcal L$ and $\mathcal K$. But i am still missing constant $A$.

---

ORIGINAL QUESTION:

I would like to plot $\psi_I$, $\psi_{II}$ and $\psi_{III}$ but my constant $A$ is still missing. How can i plot these functions so the normalisation will be applied?

---

EDIT:

After all of your suggestions i decided to work on a speciffic case of an electron with mass $m_e$ which i put in a finite well. So the constants i know are:

\begin{align} d &= 0.5nm\\ m_e &= 9.109\cdot 10^{-31} kg\\ W_p &= 25eV\\ \hbar &= 1.055 \cdot 10^{-34} Js {\scriptsize~\dots\text{well known constant}}\\ 1eV &= 1.602 \cdot 10^{-19} J {\scriptsize~\dots\text{need this to convert from eV to J}} \end{align}

I first used constants above to again draw a graph of transcendental equation and i found 2 possible energies $W$ (i think those aren't quite accurate but should do). This looks like in any QM book (thanks to @Chris White):

Lets chose only one of the possible energies and try to plot $\psi_I$, $\psi_{II}$ and $\psi_{III}$. I choose energy which is equal to $0.17\, W_p$ and calculate constants $\mathcal K$ and $\mathcal L$:

\begin{align} \mathcal K &= 2.3325888\cdot 10^{10}\\ \mathcal L &= 1.5573994\cdot 10^{10}\\ \end{align}

Now when the picture above looks like in a book i will try to use constants $\mathcal K$, $\mathcal L$ and $\boxed{A \!=\! 1}$ (like @Chris White sugested) to plot $\psi_I$, $\psi_{II}$ and $\psi_{III}$. Even now the boundary conditions at $-\tfrac{d}{2}$ and $\tfrac{d}{2}$ are not met:

EXTENDED QUESTION:

It looks like boundary conditions are not met. I did calculate my constants quite accurately, but i really can't read the energies (which are graphicall solutions to the first graph) very accurately. Does anyone have any suggestions on how to meet the boundary conditions?

Here is the GNUPLOT script used to draw 2nd graph:

set terminal epslatex color colortext size 9cm,5cm
set size 1.5,1.0
set output "potencialna_jama_6.tex"

set style line 1 linetype 1 linewidth 3 linecolor rgb "#FF0055"
set style line 2 linetype 2 linewidth 1 linecolor rgb "#FF0055"
set style line 3 linetype 1 linewidth 3 linecolor rgb "#2C397D"
set style line 4 linetype 2 linewidth 1 linecolor rgb "#2C397D"

set style line 5 linetype 1 linewidth 3 linecolor rgb "#793715"
set style line 6 linetype 2 linewidth 1 linecolor rgb "#793715"
set style line 7 linetype 1 linewidth 3 linecolor rgb "#b1b1b1"
set style line 8 linetype 3 linewidth 1 linecolor rgb "#b1b1b1"

set grid

set samples 7000

set key at graph .70, 0.4

set key samplen 2
set key spacing 0.8

m = 9.9109*10**(-31)
d = 0.5*10**(-9)
U = 25 * 1.602*10**(-19)
h = 1.055*10**(-34)

K = 2.3325888*10**10
L = 1.5573994*10**10

A = 1

f(x) = A*exp(K*x)
g(x) = -( A*exp(-L*(d/2)) )/( sin(L*(d/2)) )*sin(L*x)
h(x) = -A*exp(-K*x)

set xrange [-d:d]
set yrange [-8*10**(-2):8*10**(-2)]

set xtics ("$0$" 0, "$\\frac{d}{2}$" (d/2), "$-\\frac{d}{2}$" -(d/2))

set ytics ("$0$" 0)

set xlabel "$x$"

plot [-1.5*d:1.5*d] f(x) ls 1 title "$\\psi_{I}$", g(x) ls 3 title "$\\psi_{II}$", h(x) ls 5 title "$\\psi_{III}$" 

Answer

Wavefunctions are found by solving the time-independent Schrödinger equation, which is simply an eigenvalue problem for a well-behaved operator: $$ \hat{H} \psi = E \psi. $$ As such, we expect the solutions to be determined only up to scaling. Clearly if $\psi_n$ is a solution with eigenvalue $E_n$, then $$ \hat{H} (A \psi_n) = A \hat{H} \psi_n = A E_n \psi_n = E_n (A \psi_n) $$ for any constant $A$, so $\psi$ can always be rescaled. In this sense, there is no physical meaning associated with $A$.

To actually choose a value, for instance for plotting, you need some sort of normalization scheme. For square-integrable functions, we often enforce $$ \int \psi^* \psi = 1 $$ in order to bring the wavefunction more in line with the traditional definition of probability (which says the sum of probabilities is $1$, also an arbitrary constant).

In your case, $$ \psi(x) = \begin{cases} \psi_\mathrm{I}(x), & x < -\frac{d}{2} \\ \psi_\mathrm{II}(x), & \lvert x \rvert < \frac{d}{2} \\ \psi_\mathrm{III}(x), & x > \frac{d}{2}. \end{cases} $$ Thus choose $A$ such that $$ \int_{-\infty}^\infty \psi(x)^* \psi(x) \ \mathrm{d}x \equiv \int_{-\infty}^{-d/2} \psi_\mathrm{I}(x)^* \psi_\mathrm{I}(x) \ \mathrm{d}x + \int_{-d/2}^{d/2} \psi_\mathrm{II}(x)^* \psi_\mathrm{II}(x) \ \mathrm{d}x + \int_{d/2}^\infty \psi_\mathrm{III}(x)^* \psi_\mathrm{III}(x) \ \mathrm{d}x $$ is unity.

If you happen to be in the regime $E > W_p$, then $\mathcal{K}$ will be imaginary, $\psi_\mathrm{I}$ and $\psi_\mathrm{III}$ will be oscillatory rather than decaying, and the first and third of those integrals will not converge. You could pick an $A$ that conforms to some sort of "normalizing to a delta function," but there are many different variations on this, especially for a split-up domain like this. In that case I would recommend picking an $A$, if you really have to do it, based on some other criterion, such as $\max(\lvert \psi \rvert) = 1$ or something.

gravity - Is the graviton hypothetical?

Wikipedia lists the graviton as a hypothetical particle. I wonder whether graviton is indeed hypothetical or does its existence directly follow from modern physics? Does observation of gravitational waves amount to the discovery of graviton or could there be gravitational waves not composed of gravitons?

Answer

Gravitons are hypothetical, but they're far less hypothetical than most of the other particles which theorists speculate about (such as axions, magnetic monopoles, strings, sterile neutrinos, and the like).

That probably sounds a little strange. Let me explain.

We don't have a complete theory of quantum gravity. But we do actually have an extremely good incomplete theory of quantum gravity. This theory is an "effective quantum field theory". It doesn't work for physics at distances smaller than the Planck scale, but it works just fine above that scale. In other words, it's good enough for all practical purposes. You get this theory -- maybe we should call it 'effective quantum general relativity' to distinguish it from classical GR -- by introducing a Planck scale cutoff into general relativity and then doing the path integral or canonical quantization or whatever your favorite procedure is.

Such effective theories are widely used in physics. One of the more famous ones was Fermi's theory of the weak interaction. It was broken in ways that drive theorists crazy -- non-renormalizable and non-unitary -- and it eventually had to be replaced with the Glashow-Weinberg-Salam theory. But it worked great. You can do essentially all of nuclear physics with just Fermi's theory. Hans Bethe even used it to explain how the sun produces energy.

So we have this theory, and it's ugly, but it works just fine at energies below the Planck scale. It gives correct answers; for example, it predicts that quantum gravity phenomena are practically impossible to detect in normal terrestrial physics. Moreover, this theory is mathematically pretty much unique: any theory of quantum gravity which reduces to classical general relativity at low energies must reduce to this effective QFT along the way. You really can't avoid it.

So we should take this theory pretty seriously, since it works well and doesn't require a lot of extra assumptions. And since this theory describes all gravitational phenomena we know about in terms of gravitons, it's probably fair to say that the existence of gravitons is a fairly good bet. (You might have to wait a long time to collect on that bet though.)

What is the square root of the Dirac Delta Function?

What is the square root of the Dirac Delta Function? Is it defined for functional integrals? Can it be used to describe quantum wave functions?

\begin{align} \int_{-\infty}^{\infty} f(x)\sqrt{\delta(x-a)}dx \end{align}

Answer

You can think of the Dirac distribution as being an element in some operator valued Banach algebra. One can then use the Riesz calculus (http://en.wikipedia.org/wiki/Holomorphic_functional_calculus) to define an arbitrary holomorphic function of this Dirac delta (or more general distributions).

In particular (much like the Cauchy-Integral Formula) we have, for a Banach algebra $\mathscr{A}$ and holomorphic function $f$, for $a \in \mathscr{A}$:

$f(a) = \frac {1} {2 \pi i} \oint \frac {f(z)} {z - a} d z$ ;

This can in principle be evaluated. For example, you can represent the Dirac delta as a matrix (if you are thinking about finite dimensions). The integration is then fairly straightforward (for the square-root you would set $f(z) = z^{1/2} = e^{\frac {1} {2} \log(z)}$). You will have to be careful about branch cuts for this particular case.

As for quantum wave functions, I could not say. However, this tool is very useful to define Dirac operators. For example, you could consider the algebra $\mathscr{A}$ to consist of differential operators of some dimension. Taking the Klein-Gordon operator $\nabla^2 + m^2$ one can attempt to define square-roots of this operator by evaluating the above integral.

Corpuscular theory of light and Double slit experiment

Physicists would initially have attempted to explain Young's double slit experiment's results using the concept of light as a stream of particles, ryt? Can somebody tell me what these attempts were and how they failed?

Answer

Every phenomenon has a level of interpretation. The photoelectric effect, Planck's law of Black body radiation, the emission of photons from excited electrons, all these phenomena are interpreted as light to consist of quanta, later called photons. Only the quantized deflection of EM radiation on edges and colors in thin films were not explainable in Young’s time. Wikipedia states:

... diffraction effects at edges or in narrow apertures, colours in thin films and insect wings, and the apparent failure of light particles to crash into one another when two light beams crossed, could not be adequately explained by the corpuscular theory ...

So the emission as well as the absorption of EM radiation is explained by photons. The deflection of EM radiation withe a swelling intensity distribution behind edges and the deflection in transparent media (lenses for example) are interpreted as coming from the wave characteristics of light in analogy to water waves:

But such an interpretation has some weaknesses:

1. Comparing the both sketches you should agree that in the upper sketch of water waves the drawn points of minimum intensity C and D are moving “upwards” an E and F are moving “downwards”. This is not the case for the intensity distribution of EMR behind slits. How Young could not point this out? The analogy is missing an important point.


2. An intensity distribution from single emitted photons occurs after a while even behind single edges. The explanation of a path difference from a coherent light source is obsolet in this experiment.

what these attempts were and how they failed?

Do these last phenomena failed the corpuscular nature of light? I think non. Simply in Young’s time where was no knowledge about the quantized interaction between EMR and the surface electrons from edges and transparent media as well as from crystal lattices.

Anyway, these different interpretations don’t change anything in everyday life except less mystery of the double slit experiment. Ernst Abbe made the right calculations for lenses based on observations, not care about a wave or corpuscular behavior of light.

cosmology - Is Dark Matter evenly spread out in the universe?

Is Dark Matter evenly spread out? If no, could we ever find a correlation between the amount of dark matter and matter in a specific place?

What gonna happen to beer on the moon?

Let's assume that we have bottle of beer on the Moon. What would happen to it in sunlight? Some cases:

1. 
   

   Beer can be dark or lager.

   
   


2. 
   
   

   It can be contained in glass bottle (transparent or dark) or aluminium can.

   
   

Will it freeze or vaporize or what?

Answer

According to NASA, the temperature on the Moon in direct sunshine can reach 400K and this is above the boiling point of beer. So the beer will boil and possibly burst it's container depending on how strong the container is.

The same article tells us the the temperature falls to only 100K at night, so at night the beer would freeze.

The colour of the beer and/or the bottle may change the rate at which it boils or freezes because a dark beer and/or bottle will absorb more sunlight. However the end result will be the same.

optics - Linearizing Quantum Operators

Possible Duplicate:
Linearizing Quantum Operators

I was reading an article on harmonic generation and came across the following way of decomposing the photon field operator. $$ \hat{A}={\langle}\hat{A}{\rangle}I+ \Delta\hat{a}$$

The right hand side is a sum of the "mean" value and the fluctuations about the mean. While I understand that the physical picture is reasonable, is this mathematically correct? If so what are the constraints this imposes? In literature this is designated as a "linearization" process.

My understanding of a linear operator is that it is simply a homomorphism. I have never seen anything done like this and I'm having a hard time finding references which justify this process.

I would be grateful if somebody can point me in the right direction!

Answer

The operation you mentioned $$\Delta \hat{a} = \hat{A}-\langle \hat{A}\rangle \hat{I}$$ is just shifting of the annihilation operator. Typically people even drop the identity and write $$\hat{a}_{new}=\hat{a}_{old}-\alpha_0,$$ where $\alpha_0$ is a complex number. In your question the shift is such that $\langle \hat{a}_{new} \rangle = 0$, which may be convenient for calculations. In particular when the pump beam has many photons, the quantum part (i.e. related to $\hat{a}_{new}$) may be neglected.

Additionally, $\hat{a}_{new}$ is an annihilation operator with the same anti-commutation relations as $\hat{a}_{old}$.

From mathematical point of view it is a perfectly legit operation. Moreover, both operators have the same domain, and spectrum only shifted by $\alpha_0$. To see that domain is the same take any $|\psi\rangle \in \text{dom}(\hat{a}_{old})$. Then denoting $|\phi\rangle := \hat{a}_{old} |\psi\rangle$, we check that $\hat{a}_{new} |\psi\rangle = |\phi\rangle-\alpha_0|\psi\rangle$ is a well-defined vector.

quantum mechanics - Does a normal torch emit entangled photons?

I was reading a sciencenews.org post about three photons being entangled. My question here is, why is the chance of producing an entangled pair once in a billion times? Isn't every particle produced in pairs, each of them being entangled to the other one formed in the pair? So shouldn't every photon that is produced, have an entangled photon? Why is the probability so low?

Shouldn't entanglement happen immediately during the production of the particle?

Note Since I am a little new to this topic, a little background of entanglement would be appreciated as I might be wrong in my conceptualization.

quantum mechanics - How many of which particles are in Hawking radiation?

My understanding is that a black hole radiates ~like an ideal black body, and that both photons and massive particles are emitted by Hawking radiation. So for a low temperature black hole, photons are emitted according to Planck's law, but the peak of this spectrum shifts to higher frequencies as temperature increases.

For sufficiently large temperature, does it follow that massive particles of equivalent energy are also emitted, without preference to other particle properties? If not, I'd be interested to know what physics are involved in determining the form in which the black hole's energy is radiated away.

astrophysics - Why did the gamma ray burst from GW170817 lag two seconds behind the gravitational wave?

The ABC, reporting on the announcement of gravitational wave GW170817, explained that for the first time we could identify the precise source of a gravitational wave because we also observed the event in the electromagnetic spectrum. It notes however that the gamma ray burst detected by the FERMI space telescope was observed nearly two seconds later than the gravitational wave.

1. 
   

   Did the gamma ray burst actually arrive at Earth two seconds after the gravitational wave, or is this time delay just some kind of observational artefact?

   
   


2. 
   

   If the delay is real, what is its cause? Is the delay due to the gamma ray burst somehow being slowed, or was it 'emitted' at a different time in the merger event?

   
   


3. 
   

   What does a delay even mean when the gravitational wave was detected for over 100 seconds? Is it a delay from peak GW to peak gamma ray?

   
   

astrophysics - What mechanism is responsible for the creation of these dunes on Comet 67P/Churyumov-Gerasimenko?

What mechanism is responsible for the creation of these dunes on Comet 67P/Churyumov-Gerasimenko?

The following high resolution picture from ESA's Rosetta mission shows the dunes:

At a distance of 7.8 km from the surface, the image scale is about 66.5 cm/pixel, so each 1024 x 1024 pixel frame is about 680 m across (although if we assume the furthest point away is an additional ~1 km further from the centre, the image scale is about 92 cm/pixel).

So the scale of the dunes is on the order of tens of meters along the crest. And each crest are probably just a few meters apart.

When a charged particle absorbs a photon, does the particle get 'pulled' or 'pushed'?

When a charged particle absorbs a photon, does the particle accelerate towards where the photon was coming from, or where the photon was going?

quantum field theory - Feynman $ivarepsilon$-prescription in path integral by adding an imaginary part to time

It is known that the well-definiteness of the path integral leads to the Feynman's $i\varepsilon$-prescription for the field propagator. I've found many ways of showing this in the literature, but it is precisely the way that I have learned in my QFT course (and which I have not found in literature) that I do not understand.

Context of the problem

Considering the case of a real scalar field for simplicity, one has that the following path integral (evaluated at asymptotic times)

\begin{equation} \lim_{T \rightarrow \infty}\int_{\phi(-T, \vec{x})}^{\phi(T,\vec{x})} \mathcal{D}\phi \ \text{exp} \left( i \int^T_T dt \int d³ x \ ( \mathcal{L}+J \phi) \right)\tag{1} \end{equation}

can be expressed as

\begin{equation} \lim_{T \rightarrow \infty} \sum_{m, n} e^{-i\left(E_n+E_m \right)T} <\phi, T|n, T>_J _J \tag{2} \end{equation}

where $|n>_J$ are eigenstates of the hamiltonian $H$ in the pressence of the source $J$. In order to make this oscillatory exponential converge (and properly define the path integral) one adds to $T$ a small imaginary part $T \rightarrow T(1-i\varepsilon)$. With this, one writes the vacuum persistence amplitude as \begin{equation} <0|0>_J = \frac{1}{N} \lim_{\varepsilon \rightarrow 0} \ \lim_{T \rightarrow \infty(1-i\varepsilon)} \int \mathcal{D} \phi \ \text{exp} \left( i \int_{-T}^T dt \int d³x (\mathcal{L}+J\phi)\ \right) \equiv \frac{1}{N} Z[J]\tag{3} \end{equation} where the constant $N$ is typically taken to be $N=Z[0]$.

My problem

In order to relate $Z[J]$ with the Feynman propagator $D_{F}(x-y)$, one typically writes the argument of the exponential in the Fourier space, then makes the change of variable $$\hat{\phi}(p)'=\hat{\phi}(p)+(p²-m²)^{-1} \hat{J}(p)\tag{4}$$ (which leaves $\mathcal{D}\phi'=\mathcal{D}\phi$) to get

\begin{equation} Z[J]=Z[0] \text{exp} \left(-\frac{i}{2} \int \frac{d⁴ p}{(2\pi)⁴} \hat{J}(p) \frac{1}{p²-m²} \hat{J}(-p)\right).\tag{5} \end{equation} Here I'm using the $(+,-,-,-)$ Minkowski sign convention.

Now here I've been told that the fact of replacing $T\rightarrow (1-i\varepsilon)T$ to define the path integral is equivalent in Fourier space as $p⁰ \rightarrow (1+i\varepsilon)p⁰$. With this, one gets the correct $i \varepsilon$ Feynman prescription for the propagator $D_F(x-y)$

\begin{equation} Z[J]=Z[0] \text{exp} \left(-\frac{i}{2} \int \frac{d⁴ p}{(2\pi)⁴} \hat{J}(p) \frac{1}{p²-m²+i\varepsilon} \hat{J}(-p) \right) = Z[0] \text{exp} \left(-\frac{1}{2} \int d⁴y \int d⁴x \ J(x) D_F(x-y) J(y) \right). \tag{6} \end{equation}

And this is the part that I don't get at all, I've tried but I don't see how the fact that $T\rightarrow (1-i\varepsilon)T$ leads in the previous approach to $p⁰ \rightarrow (1+i\varepsilon)p⁰$ and therefore to the Feynman prescription. I'm having nightmares with this, any help would be really appreciated.

NOTE: I use the following convention for the Fourier transform \begin{equation} \hat{\phi}(p)=\int d⁴ x \ \phi(x) e^{-ip \cdot x}\tag{7} \end{equation}

so that \begin{equation} \phi(x)= \int \frac{d⁴p}{(2 \pi)⁴} \ \hat{\phi}(p) e^{+ip \cdot x}.\tag{8} \end{equation}

gravity - Gravitational lensing or cloud refraction?

My current understanding of gravitational lensing follows

When a star or other massive body passes between us and another star, the phenomenon generally labeled 'gravitational lensing' occurs. The term 'gravitational lensing' refers to the bending of light rays due to the gravitational influence of the massive body. Since the bends concentrate light for the observer, they act like a magnifying lens.

OTOH

But since I was young I have pondered other explanations for such observations. One thought about what (else) might cause the observed effect is a cloud of gas around the massive body itself. The assumption being that any change in density between the thin gas of interstellar space and the thicker gas surrounding a massive object should form a refractive boundary. In the case of an active star, that gas & might be the heliosphere of the star, and for any massive object that is not creating an heliosphere, it could be due to clouds of gas that the massive object is sucking in.

I raised this idea with a physics teacher in high school & he assured me that any possible refraction from clouds of gas had been ruled out in observations of gravitational lensing. I never got to see papers on the subject, so I cannot confirm that (and am not convinced).

My questions are:

1. Does anybody know of papers on gravitational lensing that discuss the refraction from clouds of gas around the lensing body?


2. How could an observer/experimenter go about quantifying the likely refractive index of such a gas cloud.

Answer

One point in addition to the other answers. Refraction is almost always chromatic (i.e., different wavelengths refract differently). Gravitational lensing is achromatic. Some studies of gravitational lensing, particularly the MACHO-like microlensing studies, look specifically for achromaticity to test that what they're seeing really is gravitational.

general relativity - Difference of connections in the Killing vector equation

For the Killing vector equation, I sometimes see it written in terms of spin connection $\omega$ and other times in terms of the affine connection $\Gamma$.

More clearly $$\nabla_{\mu}V_{\nu}+\nabla_{\nu}V_{\mu}=0=\partial_{\mu}V_{\nu}+\Gamma_{\mu\nu\lambda}V^{\lambda}+\partial_{\nu}V_{\mu}+\Gamma_{\nu\mu\lambda}V^{\lambda}$$

Another time, you find it satisfying this $$\partial_A V_B+\partial_B V_A=\omega_{A,CB}V^C+\omega_{B,CA}V^C$$

What is the difference between the two descriptions?

Answer

Comments to the question (v3):

1. 
   

   The vielbein $e^A{}_{\mu}$ in the Cartan formalism is an intertwiner $$\tag{1} g_{\mu\nu}~=~e^A{}_{\mu} ~\eta_{AB} ~e^B{}_{\nu} $$ between the curved (pseudo) Riemannian metric $g_{\mu\nu}$ and the corresponding flat metric $\eta_{AB}$. Here Greek indices $\mu,\nu,\lambda, \ldots,$ are so-called curved indices, while capital Roman indices $A,B,C, \ldots,$ are so-called flat indices. See also this related Phys.SE post.

   
   


2. 
   

   The vielbein $e^A{}_{\mu}$ translates the vector field $$V~=~V^{\mu}\partial_{\mu}$$ with curved components $V^{\mu}$ into flat components $$V^A~:=~e^A{}_{\mu} V^{\mu}.$$

   
   



3. 
   

   The christoffel symbols $\Gamma^{\lambda,\mu\nu}$ and the spin connection $\omega^{AB}_{\mu}$ can be written in terms of the vielbein $e^A{}_{\mu}$, see Wikipedia for details.

   
   


4. 
   

   TL;DR: Using the vielbein $e^A{}_{\mu}$ as an intertwiner, it is possible to show that OP's various conditions for a Killing vector field are equivalent.

   
   

experimental physics - How would you go about finding the natural frequencies (resonance frequency)

How would you go about finding the natural frequencies of solid materials like wood (e.g., teak, pine), stone (e.g., marble, granite) liquid, etc?

quarks - Ratio of electrons and protons in Universe

What is the relation between amount of electrons and protons in Universe? I expect that the Universe should not be charged, so the estimation is 1:1.

But then, why there should be the same amount of leptons and quarks/3 ? (if I generalize a bit the question) I remember, that lepton and baryon numbers are conserved, which sounds to me like we speak about two completely different species. That have the same number of members by chance...

Answer

The Universe is indeed electrically neutral at the cosmological length scales which means that the total charge of the positively charged particles is equal to (minus) the total charge of the negatively charged particles.

However, one must be more careful what these particles are. Electrons and protons are two dominant charged particle species. However, the Universe also contains other charged particles including antiprotons, positrons, and, less importantly, some unstable particles.

But if one ignored all charged particles except for protons and electrons, $N_e=N_p$ would really arise from the neutrality of the Universe, and it is approximately obeyed by the Universe around us, anyway. (Most of electrons and protons in the Universe exist in the form of hydrogen atoms, anyway.)

If one talks about the number of quarks, the counting is different. A proton contains 3 (valence) quarks so the number of (valence) quarks is $N_q=3N_p+\dots $. However, protons aren't the only particles that contain quarks. There are lots of neutrons, so neglecting all other hadrons, $N_q=3N_p+3N_n\gt 3N_p\approx 3N_e$.

However, I have mentioned that most of the electrons and protons in the Universe come in the form of hydrogen-1 which has no neutrons, so the number of neutrons in the Universe is much smaller than the number of the protons, and this term may be approximately neglected. However, heavy elements – like those on Earth – actually contain a greater number of neutrons than protons. And neutron stars are full of neutrons. One must be careful about this counting.

electromagnetism - What would be the force law for the electric force between two electric charges in higher dimensions in the case of photons that have rest mass?

I was reading about if photons have rest mass https://galileospendulum.org/2013/07/26/what-if-photons-actually-have-mass/, and found that if photons have mass the force law for the electric force between two electric charges, in 4+1 dimensions, would be $$F=\frac{Qq\left(1+{\mu}r\right)}{4{\pi}{\varepsilon_0}r^2e^{{\mu}r}}$$ and $$\mu=\frac{m_{\gamma}c}{\hbar}$$ with $Q$, and $q$ being the two electric charges, $r$ being the distance between the two electric charges, $\varepsilon_0$ being the electric constant, $m_{\gamma}$ being the rest mass of the photon, $c$ being the speed of massless particles, $\hbar$ being the reduced planks constant, and $F$ being the electric force between the two electric charges.

This equation can also be expressed as $$F={D_r}\left(\frac{Qq\left(sinh\left({\mu}r\right)-cosh\left({\mu}r\right)\right)}{4{\pi}{\varepsilon_0}r}\right)$$ $$F=\frac{Qq\left(D_r\left(sinh\left({\mu}r\right)-cosh\left({\mu}r\right)\right)r-\left(sinh\left({\mu}r\right)-cosh\left({\mu}r\right)\right)D_r(r)\right)}{4{\pi}{\varepsilon_0}r^2}$$ $$F=\frac{Qq\left(\left({\mu}cosh\left({\mu}r\right)-{\mu}sinh\left({\mu}r\right)\right)r-\left(sinh\left({\mu}r\right)-cosh\left({\mu}r\right)\right)\right)}{4{\pi}{\varepsilon_0}r^2}$$ $$F=\frac{Qq\left({\mu}rcosh\left({\mu}r\right)-{\mu}rsinh\left({\mu}r\right)-sinh\left({\mu}r\right)+cosh\left({\mu}r\right)\right)}{4{\pi}{\varepsilon_0}r^2}$$ $$F=\frac{Qq\left({u}rcosh\left({\mu}r\right)-sinh\left({\mu}r\right)-{\mu}rsinh\left({\mu}r\right)+cosh\left({\mu}r\right)\right)}{4{\pi}{\varepsilon_0}r^2}$$

I'm not sure if the force law in higher dimensions would be $$F=D_r\left(\frac{Qq\left(sinh\left({\mu}r\right)-cosh\left({\mu}r\right)\right)}{S{\varepsilon_{0}}r^{n-3}}\right)$$ or if it would be $$F=\frac{Qq\left(1+{\mu}r\right)}{S{\varepsilon_{0}}r^{n-2}e^{{\mu}r}}$$ or if it would be something else with, $n$ being the total number of dimensions and $S$, being what the radius of a sphere is multiplied to get the hyper-surface area in the number of dimensions minus one. Also the units of $\varepsilon_0$ would depend on the number of dimensions.

What would be the force law for the electric force between two electric charges if photons are massive?

Answer

$$[\ WARNING : This\ answer\ uses\ natural\ units\ $$ $$\ Use\ dimensional\ analysis\ to\ bring\ back\ any\ constants\ you\ need]$$

Let's jump right into the workings - which is basically an assumption and some integrations.

In $3+1$ dimension (without that time component, we wouldn't have a magnetic force, which will be bad and so we will talk about d+1 dimensions - although magnetism becomes complicated for higher dimensions, but that's a story for another day), Coulomb's law takes the form of an inverse-square law, $$f = \frac{1}{r^2}$$ with proper definition of the source and distance.

Jumping dimensions :

Time for our assumptions! We assume that the Lagrangian stays the same form in $d+1$ dimension.

$ -\ Massless\ calculations\ :\ $

It means, the Maxwell equations hold; or equivalently, the Poisson equation holds $$\nabla^2 \varphi(\mathbf{x}) = 0$$

Solving $\varphi$ in free space will produce the potential and hence the force. By doing a Fourier transform, $$\varphi(\mathbf{x}) = \int \frac{\mathrm{d}^d k}{(2\pi)^d} \frac{ e^{i \mathbf{k}\cdot \mathbf{x}}}{k^2}$$

Solving $\varphi(r)$ $$\varphi = \frac{1}{(2\pi)^d}\int \mathrm{d} k \; k^{d-3} \mathrm{d}^{d-1}\Omega {e^{i k r \cos \theta_1}}$$ where $\mathrm{d}^{d-1} \Omega$ is the $(d-1)-D$ angular element. Now this integral involves the integral of one azimuthal angle $\theta$. We can parametrize the coordinate in $d-D$ spherical coordinate as: $$x_1 = r \cos\theta_1$$ $$x_2 = r \sin\theta_1 \cos\theta_2\cdots$$ $$x_d = r \sin\theta_1 \sin\theta_2\cdots \sin\theta_{d-2}\cos\phi$$ Then the surface element becomes: $$\mathrm{d}^{d-1} \Omega = \sin^{d-2}\theta_1 \sin^{d-3}\theta_2 \cdots \sin \theta_{d-2} \mathrm{d}\theta_1 \mathrm{d}\theta_2 \cdots \mathrm{d}\theta_{d-2} \mathrm{d} \phi$$ The integral over angles except $\theta_1$ is just the surface area of a $(d-2)-D\space hypersphere$ (Which I will assume you know, and so I am just stating the formula) $$S_{d-2} = \frac{2 \pi^{\frac{d-1}{2}}}{\Gamma\left( \frac{d-1}{2} \right) }$$

So $$\varphi = \frac{ S_{d-2}}{ (2\pi)^d r^{d-2}} I_{d}$$ where $I_d = \int_0^\infty \mathrm{d}\xi \; \int_0 ^\pi \mathrm{d}\theta \; \xi^{d-3} \sin^{d-2}\theta \exp\left[ i \xi \cos\theta \right]$

We can do the $\theta$ integral first(the other way around results in a tan integral with an argument of $\pi /2$, and is very very big and bad), which yields (I am using mathematica, please forgive me...): $$I_d = \int_0^\infty \mathrm{d}\xi \; \sqrt{\pi} \Gamma\left( \frac{d-1}{2} \right) \frac{{ }_0F_1\left( \frac{d}{2}, -\frac{\xi^2}{4} \right)}{\Gamma\left( \frac{d}{2} \right)} \xi^{d-3} = 2^{d-3} \sqrt{\pi} \Gamma\left( \frac{d-2}{2} \right) \Gamma\left( \frac{d-1}{2} \right).$$

Therefore, $$\varphi = \frac{ S_{d-2}}{ (2\pi)^d r^{d-2}} 2^{d-3} \sqrt{\pi} \Gamma\left( \frac{d-2}{2} \right) \Gamma\left( \frac{d-1}{2} \right) = \frac{\Gamma\left( \frac{d-2}{2}\right)}{4 \pi^\frac{d}{2} }\frac{1}{r^{d-2}}$$

$-\ Massful\ calculations\ :\ $

Now that we have easily solved for an intermediary massless boson, time to give it some mass. The Poisson equation becomes $$(\nabla^2 - m^2) \varphi(\mathbf{x}) = 0$$

By doing Fourier transform, $$\varphi(\mathbf{x}) = \int \frac{\mathrm{d}^d k}{(2\pi)^d} \frac{ e^{i \mathbf{k}\cdot \mathbf{x}}}{k^2+m^2}$$ Apply the same technique as the massless counterpart
$$\varphi(r) = \frac{ (m r)^{\frac{d}{2}-1} K_{\frac{d}{2}-1} (mr)}{(2 \pi)^\frac{d}{2}}\frac{1}{ r^{d-2}}$$ where $K_n(x)$ is the Bessel function of the second kind.

To go from the potential to the force, you will need to differentiate this equation. This shouldn't be much of a problem, except the Bessel function.

The differentiation of the Bessel function yields :

$$K'_n(x) = - \frac{1}{2} [ K_{n-1} (x) + K_{n+1} (x)].$$

Putting this in should give you the answer you seek.

Cheers!!

semiconductor physics - In solar cells, do photons break apart electron-hole pairs, or create them?

Some sources say that when a photon hits the PV cell, it breaks apart electron-hole pairs. Other sources say that photons create electron-hole pairs. Can anyone explain which one is right? I've read several explanations of what goes on in the solar cell, but they don't seem very clear. To me, who has little prior knowledge, a few aspects of the many different explanations seem to contradict each other. I don't know if it's because some of the explanations said things a certain way for simplicity's sake, but this part seems too significant to ignore.

For example, http://www.solarenergyexperts.co.uk/buyersguides/photovoltaic-glass-how-does-it-work/ : "When photons (light particles) from the sun hit the cell, the energy breaks up the paired particles. The freed electrons go into the n-type layer, while the holes go down into the p-type layer."

http://science.howstuffworks.com/environmental/energy/solar-cell3.htm : "When light, in the form of photons, hits our solar cell, its energy breaks apart electron-hole pairs."

http://solarjourneyusa.com/bandgaps.php has a hole section called "Generation of electron-hole pairs"

This one basically seems to be talking about the same thing, but doesn't really mention electron-hole pairs, and the explanation seems more confusing to me, but anyway, http://www2.pv.unsw.edu.au/nsite-files/pdfs/UNSW_Understanding_the_p-n_Junction.pdf: "...broken bonds created by the light act as holes ... and these holes are also free to move throughout the material. Electrons and holes created in this way are physically near each other: for every electron excited by the light there is a corresponding hole generated. These electrons and holes can remain excited only for a short period of time. In a process called recombination, excited electrons stray too close to holes and the two fall back into bonded positions." This one seems to be saying that photons break apart the bonds between the electron and its atom, which creates a free electron and a hole, i.e. another one saying that an electron-hole pair is created.

http://solarcellcentral.com/junction_page.html : "When photons hit the solar cell, free electrons attempt to unite with holes in the p-type layer."

http://www.solarenergy.net/Articles/how-photovoltaic-cells-work.aspx : "When enough photons are absorbed by the negative layer of the photovoltaic cell, electrons are freed from the negative semiconductor material."

So what actually happens when light hits a solar cell?

statistical mechanics - Partition function of lattice without kinetic energy

Consider a single particle on a cubic lattice with $N$ sites each with a site length of $a$. The particle has no kinetic energy but only feels a spatially varying continuous potential $v(\boldsymbol{r})$.

One can directly write down the partition function as sum over possible states (particle at different sites):

$$Z=\sum_i^N\mathrm{e}^{-\beta v(\boldsymbol{r_i})}$$

But what happens if we want to evaluate the partition function from the continuous representation, i.e.

$$Z=\frac{1}{h^3}\left(\int \mathrm{d}\boldsymbol{p}\right)\left(\int \mathrm{d}\boldsymbol{r} \mathrm{e}^{-\beta v(\boldsymbol{r})}\right)=\frac{1}{h^3}\left(\int \mathrm{d}\boldsymbol{p}\right)a^3\sum_i^N \mathrm{e}^{-\beta v(\boldsymbol{r_i})}$$

Does that mean that $\frac{a^3}{h^3}\int\mathrm{d}\boldsymbol{p} = 1$?

---

Edit: As suggested by AccidentalFourierTransform we could integrate over the first Brilloin zone, this would exactly be in agreement with the above statement:

$\int\limits_{-\infty}^\infty \mathrm{d} \boldsymbol{p} = \hbar^3\int\limits_{-\infty}^\infty \mathrm{d} \boldsymbol{k} = \hbar^3\int\limits_{-\pi/a}^{\pi/a}\mathrm{d}\boldsymbol{k} = \frac{h^3}{a^3}$

Could someone comment on that or give an explanation? Is this a valid approach? Can e.g. atoms sitting on the lattice sites be considered as waves? And why do we not need to integrate over higher momenta as the Brilloin zone?

astronomy - How to explain the Moon halo phenomenon?

Today, here in Brazil, I have observed (and is still observing) an interesting phenomenon.

The Moon is near to a big star in the sky, but this is normal. The interesting part is what's around them.

A huge circle in which the moon and the star is inside. Why does this circle appear? What is it?

(This isn't a cloud. It looks like more something like a light reflection of a lamp.)

Unfortunately, I hadn't a good camera to catch this. I would appreciate having an image posted for this.

Answer

You might be describing a moon aura.

It's caused by diffraction on tiny ice crystals in the atmosphere: When the moonlight hits the ice, you get interference effects that depend on the angle of the incident light. The idea is similar to powder diffraction in crystallography.

string theory - Same partition functions, different theories

I am reading the book "Basic Concepts of String Theory" by Blumenhagen, Lust and Theisen and in page 290 they say:

"It follows that the $E8\times E8$ and the $SO(32)$ heterotic string theories have the same number of states at every mass level which are however differently organized under the internal gauge symmetries. So, even though the partition functions are identical, the theories are nevertheless different. The differences show up in correlation functions."

I am trying to understand how these differences would show up in correlation functions, ie what would be an example of two correlation functions that are different in these two theories. Any references where this is done/explained would be greatly appreciated!

Answer

The differences will show up in the correlation functions because the correlation functions "know" about the group under which the states transform.

For example, the first excited level of both CFTs, one with $E_8\times E_8$ (HE) and one with $SO(32)$ (HO), contains $248+248=32\times 31/(2\times 1)=496$ states (and therefore the corresponding operators $K_i$, $i=1,2,\dots, 496$, assigned by the state-operator correspondence) that transform as the adjoint of the gauge group.

These $K_i(z)$ operators may become factors in the vertex operators in string theory (the operators encoding the external gauge bosons and their superpartners). The scattering amplitudes involving two gauge bosons (and something else) are integrals of integrands that are proportional to the correlators $$\langle K_i(z_1) K_j(z_2) \cdot \cdots \rangle $$ perhaps with some additional operators. But these correlators may be computed if you replace the $KK$ according to the OPEs (operator product expansions). These OPEs will also contain terms proportional to $K_k$ operators themselves: $$ K_i(z_1) K_j(z_2) \sim \frac 1{z_1-z_2} f_{ij}{}^k K_k(z_2) $$ where $f_{ij}{}^k$ are the structure constants of the Lie group. The structure constants of $E_8\times E_8$ and $SO(32)$ groups are different from each other – regardless of the basis of the Lie algebra you choose (for example, it's because the $E_8\times E_8$ algebra splits into two decoupled pieces while the $SO(32)$ algebra does not) so if you study the OPEs of the 496 operators $K_i$, you will be able to extract the structure constants $f_{ij}{}^k$ and therefore determine which of the two gauge groups is involved, too.

The structure constants may be extracted not only from the OPEs but also from the correlators of three vertex operators such as $$ \langle K_i(z_1) K_j(z_2)K_k(z_3) \rangle \sim f_{ij}{}^k $$ so these correlators "know" about the representations under which the operators $K_i(z)$ transform. Although these operators may look like a collection of 496 operators with the same dimension in both cases, when you switch to the interacting theory, the difference between the HE and HO theories appears through the correlators (and structure constants in them).

The formulae above were just examples for the basic operators in the adjoint. The formulae for more complicated correlators possibly involving more than 3 operators and perhaps some operators that are more excited (higher dimensions) are more involved but they display differences between HE and HO, too. At any rate, the correlators of the simplest states above (the gauge bosons) are enough to show that the theories are different at the interacting, perturbative level.

One should emphasize that after one bosonic dimension of the heterotic string is compactified on a circle, the theories actually become equivalent – T-dual to each other – but the Wilson lines and other moduli must be carefully adjusted on both sides for the theories to match. It's because the even self-dual lattices of signature 17+1 are all isometric to each other. In particular $$\Gamma^{16}_{Spin(32)/Z_2} +\Gamma^{1,1} \equiv \Gamma^8_{E_8} + \Gamma^8_{E_8} + \Gamma^{1,1}$$ after an appropriate $SO(17,1)$ "Lorentz" transformation of the lattices on both sides.

quantum mechanics - Ground state of Spherical symmetric potential always have $ell=0$?

I was given a problem where I have a spherically symmetric potential (the exact form is not relevant to this question, I think - but anyway it is 0 for $r\in[a,b]$ and $\infty$ everywhere else) and I was asked to find the ground state energy. Note: I am actually able to solve the problem, but I first assumed $\ell=0$. Now that I think about it, I was just making the assumption blindly and I would like a justification for it.

Is it always true that the ground state of any spherically symmetric potential function has zero orbital angular momentum? [The case of hydrogen atom seems to fall out from algebra, and a physical explanation that works in general would be way nicer]

Answer

Here's a paper with a proof that the ground state must be l=0 for spherically symmetric potentials for a single particle, assuming there's a bound state. Abstract:

The variational principle is used to show that the ground-state wave function of a one-body Schrödinger equation with a real potential is real, does not change sign, and is nondegenerate. As a consequence, if the Hamiltonian is invariant under rotations and parity transformations, the ground state must have positive parity and zero angular momentum.

Essentially,

1. 
   
   

   It can be proven that the ground state must be real and non-negative everywhere, assuming a bound state exists and assuming zero spin. The argument in the paper is abstract, but the idea is that, after making the wavefunction into a product of amplitude and phase factors which may vary with position, after calculating the potential and kinetic energy parts of the expectation value $$, the only part of the energy which depends on the phase is a kinetic energy term which is minimized by making the phase constant. If the phase is constant, the wavefunction can be assumed to be real and can't change sign.

   
   


2. 
   

   From there, it can be proved that the ground state is unique. Basically, if it weren't, then there would be a second ground state wavefunction (which also can't change sign), but they can't be orthogonal since you can't integrate the product of two real non-sign-changing wavefunctions and get zero.

   
   

The spherical symmetry of the Hamiltonian demands that if there's an $l \neq 0$ state with an energy, there must also be a $-l$ state with the same energy, so the only possibility is to have $l=0$.

(Having said that, the paper cites this paper, saying that if we have a collection of two particles with strong spin-orbit coupling between them, the overall Hamiltonian can have spherical symmetry, but the system will have spontaneous symmetry breaking and end up with $l \neq 0$.)

quantum mechanics - Action of Parity operator on Impulse representation

Is my derivation of the action of the parity operator $\mathbb{P}$ on the $|p\rangle$ representation correct?

$$\left( \mathbb{P}\tilde\psi \right)(p)= - \tilde\psi (p).$$

Obtained from

$$\left( \mathbb{P}\tilde\psi \right)(p) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} dx \, e^{-\frac{i}{\hbar}px}(\mathbb{P}\psi)(x)= $$

$$ = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} dx \, e^{-\frac{i}{\hbar}px}\psi(-x)=$$

$$= -\frac{1}{\sqrt{2\pi\hbar}}\int_{\infty}^{-\infty} dx \, e^{-\frac{i}{\hbar}(-p)x}\psi(x)=-\tilde\psi(p).$$

$$= \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} dx \, e^{-\frac{i}{\hbar}(-p)x}\psi(x)=\tilde\psi(-p).$$

quantum field theory - Spin Up with Indefinite Helicity

Imagine we are studying the spin quantization along the same axis as the momentum. What if I have a Dirac spinor with a spin up but no definite helicity ($\psi_L,\psi_R\neq0$):

$$ u(p)= \left(\begin{array}{c} \sqrt{p ·\sigma}\xi_{\uparrow} \\ \sqrt{p ·\bar\sigma}\xi_{\uparrow} \end{array}\right) = \left(\begin{array}{c} \psi_L \\ \psi_R \end{array}\right)$$

How should I understand this? If the spin along the $p$-direction is positive, shouldn’t the particle have definite right-hand helicity and have 0 left-hand component? In other words, shouldn’t we have $ u(p) \sim \left(\begin{array}{c} 0 \\ \rm something \end{array}\right)$ ?

Answer

So the problem is that I did not understand that $\psi_L$and $\psi_R$ are chirality eigenstates not helicity eigenstates.

In the high energy limit where helicity and chirality are basically the same thing, the state does in fact become an eigenstate (I did the math but it’s too long to post here unless someone asks for it.)

newtonian mechanics - Relation and difference between work and kinetic energy

I don't really understand the difference and the relation between work and kinetic energy. When you move an object a distance you do work (or does the object work?), what's the object's kinetic energy? Is the kinetic energy the work you've put in to the box all gathered up at the end of the distance given the box keeps moving when you stopped pushing the box? The box can't have kinetic energy if's stationary after I've pushed the box can it?

Sorry if it's unclear.

Answer

The kinetic energy is indeed the work you have done (assuming no change in potential energy), however this is also neglecting the effect of friction. In real life, the kinetic energy of the box is going to be constant if its moving at a constant velocity even if you are exerting a force on the box and hence doing work on the box. This is because in this case, the work you are doing is being converted into heat and sound due to friction. If we neglect friction, your box would keep accelerating as you applied a force and hence the kinetic energy would keep increasing.

quantum mechanics - Does time reversal symmetry hold for (kitaev model) 1D spinless $p-$ wave superconductor?

The hamiltonian 1D spinlesss p wave superconductor can be written as $$ H=\sum_k \phi_k^\dagger \begin{pmatrix} \xi(k) & 2i\Delta \sin(k)\\ -2i\Delta \sin(k ) & -\xi(k)\end{pmatrix}\phi_k $$ and by using $h(-k)=h(k)^*$ we can show that it holds time reversal symmetry Where as in real it shouldn't hold this symmetry so what I am missing here?

Answer

Obviously if one starts from spin-1/2 physical electrons and want to get effective spinless fermions, one has to break time-reversal symmetry.

But let us imagine that we just have spinless electrons to begin with, and time-reversal symmetry is just complex conjugation(since now electrons do not have any internal degrees of freedom). Assume that $\xi_k=-2t\cos k-\mu$, corresponding to a uniform nearest-neighbor hopping with strength $t$ and a chemical potential $\mu$. Then the Hamiltonian as written is time-reversal invariant and there is nothing wrong about it. So if one keeps this definition of time-reversal symmetry, the Hamiltonian belongs to the BDI class. However, one can add time-reversal breaking terms (the simplest thing is probably adding some imaginary next-nearest-neighbor hopping) to the Hamiltonian. The universal physics of the Hamiltonian (e.g. Majorana zero modes on the edge) is completely unaffected by the addition of such terms, provided one does not close the excitation gap. Then the Hamiltonian belongs to class D.

quantum mechanics - Can the Berry Connection be derived from a metric?

The Berry Connection is $$A_\mu(R)=-i \langle \Psi(R) |\partial_\mu \Psi(R) \rangle$$ which allows us to parallel transport a state indexed by $R$. We can integrate the Berry Connection to get the Berry Phase, and we can differentiate the Berry Connection to get the Berry Curvature.

Can the Berry Connection be derived from a metric? As a prototypical example, I'm thinking about how the Christoffel Symbols in General Relativity (GR) can be derived from the metric tensor. Also, I think for every connection, there exists a metric for which the connection is a Levi-Civita connection. However, is there a natural and physical metric that induces the Berry Connection?

In this presentation, the great and powerful Haldane relates "quantum distance" to the Berry Curvature, but it doesn't look like you can derive the Berry Curvature from the quantum distance in the same way curvature and the metric are related in GR.

soft question - Your favorite Physics/Astrophysics blogs?

What are the Physics/Astrophysics blogs you regularly read? I'm looking to beef up my RSS feeds to catch up with during my long commutes. I'd like to discover lesser-known gems (e.g. not well known blogs such as Cosmic Variance), possibly that are updated regularly and recently.

Since this is a list, give one entry per answer please. Feel free to post more than one answer.

gravity - Atmospheric escape of gas molecules

Most of the objects in space are likely to have an atmosphere. Since space is void, the gases in atmosphere should have either dissolved or emptied into space. But, some of the objects still seem to be revolving with their atmosphere. I'm curious why the molecules can't just escape into space?

Answer

Like any other form of matter gas molecules feel the gravitational pull of the planet they surround, so they're attracted to the planet by gravity. At any temperature above absolute zero the atoms/molecules in a gas have a velocity distribution known as the Maxwell-Boltzmann distribution. As long as their velocities remain well below the escape velocity of the planet the atmosphere will be bound to it.

Although this gives the basic idea it's an oversimplification for several reasons. For example the escape velocity decreases as you move up through the atmosphere, however the temperature changes as well so the average gas atom/molecule velocity also changes with height. Also even if the average is well below the escape velocity a small fraction of molecules will have high enough velocities to escape. However even then only molecules near the top of the atmosphere are likely to escape as the mean free path near the ground is too short for even an energetic molecule to escape. Finally radiation from the Sun is an important factor in removing gas molecules from planets. On Earth the magnetic field keeps most of the radiation out, but on Mars gas loss due to solar radiation is important.

For more info you might want to have a look at the Wikipedia article on atmosphere loss.

I'm not sure what your second question is asking as it doesn't seem relevant to atmosphere loss. If you're asking why planets generally have a non-zero tilt this should be posted as a separate question. In brief, for most planets the tilt is chaotic and varies contnuously, and the planet may even flip over completely. Earth's tilt is stabilised by the moon and varies only slightly with time.

Re the revised question: At any given temperature the average velocity of lighter gas molecules is greater than heavier gas molecules, so it's the ligher gas molecules that escape most easily. For example the Earth loses a few kg of hydrogen per second but almost no oxygen or nitrogen.

However if the rate of loss of a light gas is very fast it can carry molecules of heavier gases along as well by colliding with them and transferring momentum to the heavier molecules. This is known as hydrodynamic escape. Note that this only happens when there is a rapid loss rate of the lighter gas, so it isn't happening to any significant extent on any of the planets in the solar system.

vectors - Metric tensor and imaginary time

I just started a re-reading of the Conformal Field Theory yellow book by Di Francesco et al. In chapter two, after defining imaginary time $\tau$ as $t=-i\tau$, the authors state that the metric tensor, being in real time, e.g., $g_R=\mbox{diag}(1,-1,-1,-1)$, becomes in imaginary time $g_I=\mbox{diag}(1,1,1,1)$.

How can this statement be proven? I know it is a maybe stupid question, but I can't figure it out. For instance, if I consider the covariant four-vector $x^\mu_R=(t,x_0,x_1,x_2)$, the contravariant vector becomes, by application of $g_R$, $x_{R,\mu}=(t,-x_0,-x_1,-x_2)$. Performing Wick rotation, they become $x^\mu_I=(-i\tau,x_0,x_1,x_2)$ and $x_{I,\mu}=(-i\tau,-x_0,-x_1,-x_2)$, that are not transformed into each other by $g_I$! Where does my argument breaks?

solar system - What would happen if the Earth was tidally locked with the Sun?

I'm thinking of writing a short story set on a version of Earth that is tidally locked to the Sun. I'm not exactly sure how to research the topic. Here's a number of questions about what would happen:

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  How hot would the light side get? Are we talking ocean-boiling levels? I imagine that life would eventually flourish, given the massive constant energy source. Is this accurate?

  
  
  
  
  • On that note, I imagine massive thunderstorms along all the coasts due to increased evaporation. How bad would they get? Would the ground ever see the Sun, or only rainfall?
  
  
  
  


• How cold would the dark side get? Is it conceivable that any life could still exist there? (Life has proven itself quite versitile in the past, i.e. life at the bottom of the ocean.)


• What wind speed would the twilight zone experience? I imagine the atmosphere would transfer heat from one side to the other, but would the wind speeds be bearable? In what direction would air flow?


• I hear that the oceans would recede into disjoint northern and southern oceans if the world stopped spinning. Would this also happen if the Earth became tidally locked?



• Would the Sun create a 'tidal' bulge in the ocean at the apex of the light side? Would this or the above dominate ocean behavior?


• Would we completely lose the magnetic field? Would life be able to survive without such shielding from magnetic radiation?


• Would the Moon eventually unlock the Earth? What state would the Moon have to be in for there to be both a locking between the Sun and the Earth as well as the Earth and the Moon?


• What other radical differences would exist between our Earth and a tidally locked alternative?

Answer

There are a lot of questions here, and I'm more sure about the answers to some of them than I am about others. But nevertheless, I'll give it a go.

The first and most important question is what would the temperatures be like. This is the one I'm least sure about. The reason I'm unsure is that it's highly dependent on a lot of things that would probably be quite different on a tidally locked Earth. These include the greenhouse effect (which is highly dependent on water vapour), albedo (which depends on how many clouds there are, as well as on ice), and most importantly on the wind speeds.

If there was no heat transport from the hot side to the cold side then we could expect the temperatures to be similar to the temperature range on the Moon - around -150 °C on the cold side, and more than 100 °C on the hot side. However, if the planet has an atmosphere then it will transport heat from the hot side to the cold side, and this might make the temperature difference much more moderate. This could only work if the rate of heat transport (and hence the wind speeds) were quite substantial, because if it gets too cold on the cold side then the atmosphere will start to freeze. This will leave less air to transport heat, leading to more freezing, and so on in a feedback loop that would result in an airless world. This is probably the most likely outcome of a tidally locked Earth-like planet - but it's not so interesting from the point of view of fiction, so I'll focus on how an inhabitable world might be plausible.

A high enough heat transfer rate might be possible if the hot side were very hot, perhaps because of a large greenhouse effect in addition to permanently facing the Sun. I'm not sure whether this would mean the hot side would have to be so hot that the ocean would boil, but I suspect so. In this case we would expect to find most of the water in the form of ice on the cold side - but perhaps some of it could be found in a liquid state near the boundary. One would hope there could be some kind of water cycle - I'm imagining something like glaciers being continually melted by the warm air blowing in from the hot side, with the meltwater flowing in huge rivers (or even flowing seas) to the hot side, where it evaporates and cycles back around to fall as snow on the cold side.

Next we need to consider what the air flow patterns would be like. The situation on the present-day Earth is complicated, but, oversimplifying a bit, the basic principle is that air heats up at the equator, then rises, then travels towards the poles at high altitudes, then cools (due to emitting thermal radiation into space), then sinks again and travels back toward the equator. So the prevailing winds at ground level tend to blow towards the equator. On a tidally locked Earth I'd expect a more extreme version of this phenomenon. The flow patters in the middle of the hot side would probably be crazily chaotic - there'd be a lot of energy there to drive storms, especially if there was a water cycle providing moisture - but in the region near the boundary I would expect a fairly stable convection cell, with hot air blowing to the cold side at high altitude, sinking, and blowing back to the hot side at low altitude. So a person standing on the surface near the boundary would experience very strong prevailing winds towards the sunlit side. Exactly how strong these winds would be, I can't say, and it's also very hard to say how much cloud cover or rainfall there would be. However, in general a higher temperature difference means more energy goes into weather systems, and on this world the temperature difference would be bigger than on Earth, so we'd generally expect the atmosphere's dynamics to be stronger and more violent.

Something to note is that heat can be transferred as latent heat rather than simply hotter and cooler air. Water absorbs heat when it evaporates and releases it when it condenses. So if there is a strong water cycle, with lots of cool liquid water flowing towards the hot side and lots of warm water vapour being transported by the atmosphere toward the cold side, then you can have a higher heat transport without needing such strong winds. In this scenario, someone on the surface would experience a prevailing wind towards the sunlit side, but looking up at the sky they would see clouds moving in the opposite direction.

An extra complication on Earth is the Coriolis force, which causes the prevailing surface winds to blow around the Earth instead of just from pole to equator, and also allows hurricanes to form. It also causes the jet streams. But a tidally locked Earth would hardly be rotating, so there wouldn't be a Coriolis force and these phenomena wouldn't happen - the prevailing winds near the boundary would blow directly towards the Sun, and rotating hurricane-type storm systems would be extremely rare if they existed at all.

The next question is about whether there would be a distinct Northern and Southern ocean. I don't think this would be the case. This idea comes from the fact that the Earth's rotation causes the oceans to bulge out around the equator - but it also causes all the rock to bulge out, too, which is why we don't have just a big ocean around the equator. I guess you can imagine that if the Earth somehow stopped spinning then the oceans would flow towards the poles, but the rock would move much more slowly, so that there would be a period of time where there would be oceans near the poles but land around the equator. But over longer geological time scales the rocky part of the Earth would change its shape as well, and you'd end up with roughly evenly distributed oceans again.

The same goes for the question about the Sun causing a permanent tidal bulge in the oceans. It would, but it would also bulge the shape of the planet, so you wouldn't necessarily have all the ocean on the sides facing toward and away from the Sun. I think the fact that the water would tend to freeze on the cold side would have a much bigger effect on water distribution than anything else. Of course, the Moon would still cause tides, if there was a moon.

Next is the magnetic field - I don't feel qualified to answer this one. I've always wanted to understand how Earth's magnetic field is generated, but I've never found a good explanation, so I don't know how plausible it is to imagine it working without the Earth's rotation. The importance of the magnetic field is that prevents the atmosphere from being slowly blown away by the solar wind, so it would be needed for this planet's atmosphere to persist in the long term.

I'm also not sure how to answer the question about the Moon. It seems hard to imagine how the Moon could be in orbit yet the Earth still be tidally locked to the Sun, but I'm no expert.

Another potential big difference is plate tectonics. If the oceans are frozen on the cold side then this will put a lot of weight on the plates there, which will probably change the dynamics of the whole system - but unfortunately I'm not currently able to imagine what the result would be. Plate tectonics are quite important for recycling elements over very long time scales. The water cycle on this planet would tend to transport nutrients to the hot side and deposit them there, so you'd need geological activity to recycle them in the long term.

One more really important thing we have to think about is photosynthesis. It will be very difficult for plants to persist on this planet, because the hot side will be very very hot, and the cold side has no light. So probably there would be much less life on this planet than there is on Earth, with most of it existing towards the edges of the hot side. (The prevailing winds and water currents here will be from the cold side, so these could keep things relatively cool even though the sun is always in the sky.) If the cloud cover is 100% then this will also make photosynthesis harder due to reducing the available light. This is important because it might mean there wouldn't be enough oxygen to support human life, so I guess in designing this fictional world there is a tradeoff between having a strong water cycle to transport lots of heat, versus having it weak enough that enough light reaches parts of the surface.

You mention the possibility of life on the dark side. This isn't impossible of course, but you'd have to consider the question of what it would eat. Most of the life on the sea floor, for example, eats goo composed of dead stuff that drips down from the surface - so its power source is ultimately sunlight, which is converted into chemical free energy by photosynthetic plankton. So while there could be some life on the dark side (feeding off geothermal energy from hydrothermal vents under the ice, for example) there couldn't be very much of an ecosystem there unless there was a constant source of chemical energy being transported somehow from the light side. But it's hard to imagine how this could happen without making the composition of the atmosphere very different from Earth's.

I hope this provides some food for thought. It is of course all very speculative, and there may well be important things that I haven't thought of, or where I've got something wrong due to lack of expertise - so caveat emptor.

Does the force of kinetic friction increase with the relative speed of the objects involved? If not, why not?

Does the force of kinetic friction increase with the relative speed of the objects involved?

I have heard and read that the answer is no. This is counter intuitive, and is a big part of why the "Airplane on a Treadmill" question is so interesting.

What phenomena are at work with kinetic friction, and why does it not increase with relative speed?

Answer

At the simple level of approximation where you talk about kinetic friction, it doesn't depend on speed. It's not a great approximation (the coefficients of kinetic friction that you find for materials tend to have huge uncertainties), though.

The reason we use the approximation (other than that it makes for good intro mechanics problems) is that the microscopic physics is pretty complicated. At a very small scale, all objects are somewhat rough (at the atomic scale, if not before), and friction is the result of trying to drag one corrugated surface over another. Larger projections from the surfaces will snag against each other and require some force to dislodge, and the sum of all those microscopic snags and drags is the force we see as friction. As it's impossible to keep track of all those interactions in detail for any reasonable size object, we approximate the total force using the kinetic friction model.

Kinetic friction has nothing to do with the airplane-on-a-treadmill problem, though. Kinetic friction involves two surfaces sliding across one another. In a situation involving rolling, however, there is no sliding. At the point where the wheels of the airplane come in contact with the surface of the ground (or a treadmill), the wheel surface is not moving relative to the ground. The relative speed of the bit of the wheel touching the ground and the bit of ground that it is touching is always zero, no matter how fast the wheel is moving relative to the surface.

A cute way to see this is to place a ruler on top of a couple of soda cans (or other convenient round objects), and roll it some distance along the surface of a table next to another ruler. You'll find that the distance covered by the ruler on top of the cans is double that covered by the center of one of the cans. This happens because the point where the cans touch the table is stationary relative to the table. The centers of the cans move, though, which means that the top part of the rolling can must be moving at twice the speed of the center for the average speed of the can to work out. You can think of the contact point, center, and top being (at some instant) like points connected by a stick that pivots about the contact point-- the point at the far end of the stick will move twice as fast as the point at the halfway mark (obviously, this only holds exactly for the infinitesimal amout of time between when a given bit of can hits the table and when it lifts off again as the roll continues, but it gets you the right idea). Thus, the ruler along the top moves twice as far as the center of the cans.

electromagnetism - Derivation of the Eikonal Equation - Neglecting Derivative of $epsilon$?

In "Principles of Optics" by Max Born and Emil Wolf, the eikonal equation is derived from Maxwell's equations. In there, the authors use the approximation of low wavelengths / high wavenumbers, to neglect some terms in the wave equation and arrive at the Eikonal equation. One of those neglected terms is the term $\nabla \log \epsilon$, and the reason for that is that it is smaller than the "wavenumber" $k$ of the solution. We then proceed to derive the Eikonal equation and fermat's principle and it is happily ever after used in geometric optics.

I may show some steps in the derivation: In general linear inhomogenous media, the wave equations that follow from Maxwell's equations are given as: $$ \Delta \vec{E} - \frac{\epsilon \mu}{c^2} \ddot{\vec{E}} + (\nabla \log \mu ) \times (\nabla \times \vec{E}) + \nabla (\vec{E} \nabla \log \epsilon) = 0 $$ Plugging in $\vec{E} = \vec{E}_0(\vec{r})e^{i(k\chi(\vec{r} - \omega t)}$, looking only at the imaginary part of the equation and neglecting everything that isn't of order $k^2$, you arrive at the Eikonal equation. In a similar way (see my other question on that), you make an approximation to arrive at the commonly used "approximated wave equation" (I will for now call it like that) in a medium:

$$ \Delta \vec{E} - \frac{\epsilon (\vec{x}) \mu (\vec{x})}{c^2} \ddot{\vec{E}} = 0 $$

So here is my question: Think of the case of a sharp material change in surface (that would be a discontinuity of $\epsilon$), and of a plane wave that is refracted at this plane (we all know Snell's law of refraction is applicable here). All in all, the whole solution in this case (incident wave and refracted wave and reflected wave) do have a discontinuity at the plane.

This solution was obtained by solving the wave equation for both materials and then glueing them together, using the right continuity conditions. Similar, one could solve the given general wave equation given at first. Our "approximated wave equation" is not able to describe this solution, it will not yield a discontinuity. So far everything's fine, we made an approximation (neglecting the terms containing derivations of $\epsilon$), we can't expect the equation to still be as powerful as it has been before.

But then (and that really buffles me, and is the core of my question): Making stronger assumptions and neglecting even more terms, we arive at a theory (geometrical optics) that can describe refraction at a surface perfectly well.

Why is it that geometrical optics can describe something like refraction better than the "approximated wave equation", when geometrical optics are even a bigger approximation of said wave equations? And why do geometrical optics describe surfaces of edges of different media, when in there you have discontinuities in the refractive index, and thus can't make those said approximations?

experimental physics - How does LIGO remove the effects of environmental noise?

Since LIGO is dealing with readings at nanometers, events such as vehicles driving nearby, and constant (but extremely minor) tremors of the earth can cause movement with the mirrors at nanometers.

Such tiny movements are extremely difficult to measure, but can easily cause errors when using such a small scale.

How does LIGO remove these factors, and take precautions to prevent false readings due to noise?

Answer

This is indeed a problem, which is dealt with using seismic isolation via Internal Seismic Isolation and Hydraulic External Pre-Isolators. The original LIGO isolation systems were passive - described here as shock absorbers - and included a single pendulum system, but the Advanced LIGO (aLIGO) upgrades added actuators that counteract vibrations in various components, as well as a quadruple pendulum system:

Hild (2011) talks about noise issues in general, including several graphs of noise severity at different frequencies:

It is clear that seismic noise is problematic mainly in lower frequencies, while quantum noise dominates at high frequencies. As such, other adjustments must be made (e.g. raising the power of the lasers, which aLIGO does, although it creates some radiation pressure that can be troublesome). However, these effects are not environmental, per se, as much as seismic activity is.

Related information is available in this presentation and this paper.

I'm not aware of what is done theoretically to compensate that; this is just how the apparatus reduces the impact of noise.