word - A DJ at a radio station?

What is the significance of this gentleman:

J. Jason, DJ, FM/AM

?

Answer

The significance is:

His initials spell the months of the year.

Specifically:

June .

July
August

September
October
November ,

December
January , February
March /

April
May

What does the magnetic field of the (quantum-mechanical) electron look like?

While a treatment of electron spin can be found in any introductory textbook, I've noticed that the electron's magnetic field seems to be treated classically. Presumably this is because a quantum treatment of the electromagnetic field would venture into the much more difficult topic of quantum electrodynamics. However, treating the magnetic field classically also seems to create conceptual difficulties. How can we write something like

$$\mathbf{\mu} = \frac{g_e \mu_b}{\hbar} \mathbf{S}$$

and treat the left-hand side as a vector, while treating the right-hand side as a vector-valued operator?

So, what does really happen when we measure the magnetic field around an electron? For simplicity imagine that the electron is in the ground state of the hydrogen atom, where it has zero orbital angular momentum. It seems to me that we can't observe what looks like a classical dipole field, because such a field would have a definite direction for the electron's magnetic moment, which would appear to contradict the quantum-mechanical properties of spin.

My guess is that measuring any one component of the magnetic field at a point near an electron would collapse the spin part of the electron wave function, and in general the three components of the magnetic field will fail to commute so we cannot indeed obtain a definite direction for the electron magnetic dipole moment. However, I'm not even sure how to begin approaching this problem in a rigorous fashion without breaking out the full machinery of QED. For an electron in a magnetic field we have the Dirac equation. For the magnetic field of the electron I wasn't able to find an answer online or in the textbooks I have at hand.

Answer

The magnetic moment of the electron is a magnetic moment, so the right magnetic field around it is $$ \mathbf{B}({\mathbf{r}})=\nabla\times{\mathbf{A}}=\frac{\mu_{0}}{4\pi}\left(\frac{3\mathbf{r}(\mathbf{\mu}\cdot\mathbf{r})}{r^{5}}-\frac{{\mathbf{\mu}}}{r^{3}}\right). $$ The world is quantum mechanical – and so is any viable description of the spin – so we have to respect the postulates of quantum mechanics. In particular, the magnetic field above corresponds to a "state" (e.g. spin up and spin down) and one may construct complex linear superpositions of such states. It is important to realize that the quantum mechanical superposition of states in the Hilbert space in no way implies that the corresponding magnetic fields are being added according to the classical electromagnetic field's superposition principle.

Indeed, they're linear superpositions of states that contain different profiles of the magnetic field.

The magnetic field around the electron is so weak that indeed, it can in no way be thought of as a classical magnetic field, in the sense that the classical fields are "large". But the classical formula for the magnetic field is still right! This formula defines the magnetic moment. The quantum effects are always important, however. Also, if you try to measure this very weak magnetic field, it will unavoidably influence the state of the measured system, including the electron's spin itself.

It is of course completely wrong to imagine that we could measure such a weak magnetic field by a big macroscopic apparatus, like a fridge magnet. The effect of one electron's magnetic field on such a big object would be nearly zero, of course. In fact, quantum mechanics guarantees quantization of many "phenomena", so instead of predicting a very tiny effect on the fridge magnet, it predicts a finite effect on the fridge magnet that occurs with a tiny probability.

You may "measure" the electron's magnetic field by creating a bound state with another magnet in the form of an elementary particle. For example, the electron and the proton in a hydrogen atom are exerting the same kind of force that you would expect from the usual "classical" formulae – but it's important to realize that all the quantities in the equations are operators with hats.

Let me show you an example of a simple consistency check implying that there is no contradiction. Calculate the expectation value of the magnetic field (an operator) $\vec B(\vec r)$ at some point for the state $c_{up}|up\rangle + c_{down} |down\rangle$. The column vector of the amplitudes $c$ is normalized. The check is that you get the same expectation value of $\vec B(\vec r)$ for each $\vec r$ if you first compute it for the "up" component and "down" component separately, and then you add the terms, or if you first realize that it's a spin state "up" with respect to a new axis $\vec n$, and compute $\vec B$ from that.

It's a nice exercise. The point is that the expectation value of $\vec B(\vec r)$ is a bilinear expression in the bra-vector and ket-vector $|\psi\rangle$, much like the direction $\vec n$. And indeed, $\vec B$ is linear in the direction $\vec n$, according to the formula above, so things will agree. You may insert anything else to the expectation value, in fact, so the check works for all linear expressions in $\vec B$. The higher powers of $\vec B$ also have expectation values but they will behave differently than in classical physics because there will be extra contributions from the "uncertainty principle", analogous to zero-point energies of the harmonic oscillator in quantum mechanics.

It's extremely important to realize that the field $\vec B(\vec r)$ is also an operator – so it has nonzero off-diagonal matrix elements with respect to the "up" and "down" spin states of the electron. In fact, if you just write $\vec \mu$ in the formula for the magnetic field I started with as a multiple of the Pauli matrices (electron's spin), you will exactly see how the "key" term in the magnetic field behaves with respect to the up- and down- spin states. The off-diagonal elements do not contribute to the expectation value in the up state or in the down state but they do impact the "mixed matrix elements between up and down, and those affect the expectation values in spin states polarized along non-vertical axes.

BTW I added a semi-popular blog version of my answer here

http://motls.blogspot.com/2014/07/does-electrons-magnetic-field-look-like.html?m=1

astronomy - What does the sky look like to human eyes from orbit?

There are numerous pictures, obviously, of the blackness of space from the shuttle, the space station, and even the moon. But they all suffer from being from the perspective of a camera, which is not sensitive enough to pick up the stars in the background when compared to the bright foreground objects (the limb of the Earth, the station, moon, etc). I've seen some photos that show a few of the brightest stars, but nothing special.

Are there any photos or eye witness accounts from astronauts of what it looks like to a human with night-adjusted vision? If I were in an orbit similar to the space station and looked away from the Earth, would I be able to see more stars than I ever could on Earth, or would it only be marginally better than the best terrestrial night viewing?

Answer

I haven't been to space :( and don't know of any accounts to point you to but I suspect that the view would be marginally better than that on earth.

First, the "black of space" would be really black i.e. no light. Even in dark skies there is a bit of scattered light in the atmosphere (even if just from scattered starlight) so you'd have higher contrast.

Additionally, you wouldn't have any "seeing" effects, the bluring of the starlight by the shifting atmosphere so the stellar images would be more concentrated. Not that you'd probably be able to consciously notice. The net effect would be to sharpen the stellar image on you eyes and increase the contrast a bit more.

I suspsect you'd see a few more stars for the above reasons, especially the latter as you eyes would have a better chance to detect light from the sharper stellar images. However, it wouldn't be a huge amount more as the stars are intrinsically faint and your eyes are only so senstive.

quantum mechanics - Is the wavefunction a real physical wave or only a mathematical abstraction?

Does interaction with physical slits in a double slit experiment indicate that the wavefunction is a real physical wave, as opposed to a mathematical abstraction? This question pertains to the psi ontology debate as described here https://arxiv.org/pdf/1409.1570.pdf. In summary, the debate is about whether the wavefunction is a real physical wave as the psi-ontologists claim, or simply a mathematical abstraction that provides information in the form of probabilities as the epistemologists claim. Why doesn't interaction with physical slits prove the wavefunction is a real physical wave, i.e., an ontic entity?

quantum mechanics - Fock Space and fermionic annihilation & creation operators

I have been trying very hard to understand, I am reading Ballentine's book on this topic, but I need help:

I realized that I don't understand how many particle states work with the creation & annihilation operators $ C_a $ and $C_a^\dagger $ while trying to calculate $\{C_a,C_b^\dagger\}$.

I will illustrate my problem starting with $C_a C_b^\dagger |a...(\sim b)\rangle $ where Ballentine uses $ \sim b$ to mean the state b is not occupied.

Here is my confusion. If I do what seems sensible: $C_a C_b^\dagger |a... (\sim b)\rangle =C_a |a... b\rangle=|(\sim a)... b\rangle $ but $C_b^\dagger C_a |a... (\sim b)\rangle= C_b^\dagger |(\sim a)... (\sim b)\rangle = |(\sim a)... b\rangle $

This is obviously wrong but from the definition I don't get what to do in the above case: $ C_a^\dagger C_b^\dagger |0\rangle=|ab \rangle $

Can someone explain how exactly one can relate a general fock state to the nice but confusing: $ | a,b,c...\rangle$. And how formally one can make sense of just a row of operators $C_a^\dagger C_b$ so I can transfer this to other situations?

I would be really glad for help. If my problem is unclear please comment.

Answer

So, the problem is that you've got to enforce Fermionic antisymmetry, but Fock space tries to make things easier by making that invisible.

So if we've got two electrons in a box in a definite Fock state, the electrons definitively occupy some single-particle states which we can just call $1, 2$. The actual state that is being occupied is therefore:

$|\psi\rangle = |12\rangle - |21\rangle$

where the "first electron" (arbitrarily chosen) is in the first numbered state, etc.

Looking at your $C_a^\dagger$ and $C_a$ operators, it is somewhat clear that they are not capturing this distinction completely. Let us say that we're looking at $C_3^\dagger$ and $C_1$. Perhaps the action of $C_3^\dagger$ will look like:

$|123\rangle - |213\rangle - |132\rangle + |231\rangle - |321\rangle + |312\rangle$

Here I am associating the $+$ sign with appending onto the end, a $-$ sign with appending one before that. This means that $C_1$ should probably have a + sign for deleting from the end, a $-$ sign for deleting from the one before that, etc. This sign convention leads to the state:

$|23\rangle - |32\rangle $

But if we reverse these for $C^\dagger_3 C_1$ then the very same sign convention would force us first into the state $ -|2\rangle $ thus generating $ -|23\rangle + |32\rangle$. So you see that the results you get are negatives of each other, but this result is hidden by a naive Fock space solution.

We can focus on the orders which are associated with a + sign and phrase all of this simply as:

1. For $C_1 C^\dagger_3$ I started with [12], prepended a 3 to get [312], swapped 1 to the front to get -[132], then removed the 1 from the front to get -[32].


2. For $C^\dagger_3 C_1$ I started with [12], removed the 1 from the front, prepended with 3, got +[32].

Similarly with a starting point of three states, you start with [123] having a + sign associated with it:

1. For $C_3 C^\dagger_4$ I started with [123], prepended a 4 to get [4123], swapped 3 to the front with 3 swaps to get -[3412], then removed it from the front to get -[412].


2. For $C^\dagger_4 C_3$ I started with [123], swapped the 3 to the front with 2 swaps to get [312], removed the 3 from the front, prepended with 4, got +[412].

Now you can maybe see why they will always be negatives of each other: in the first case you will do $k$ swaps to get that number to the start of the permutation. In the second case you will do $k - 1$ because the 4 will not be there. So you'll do an odd number of swaps total.

general relativity - Are time loops real?

I'm new to physics and I was wondering if time loops (like those seen in the movies Groundhog Day or Edge of Tomorrow) possible?

condensed matter - Holes in a P-type semiconductor under external force E

Basically in almost every semiconductor texts, there will be all these concepts concerning electrons, holes, dopants, fermi-levels.

However, I have been always confused about the picture of hole transport in semiconductor device, say, a simple PN junction.

With a specific acceptor level and dopant concentration, we have some "holes" in the valence bands, which are in fact the absence of some electrons having gone to the acceptor levels, then all theses books seem to assume in the valence bands of the p-type region, there will be only holes that conduct current?

My questions:

• 
  

  aren't there still many electrons in the valence band? though they have some negative effective masses, but still do contribute to the transport?

  
  



• 
  

  under a certain external force(say,E), the electron and holes in the valence band are moving in the same directions since electrons have negative effective mass and holes have positive one, so the corresponding currents cancel with each other in a P-type semiconductor?

  
  

Can I swap quantum mechanical ground state for some classical trajectory distribution and have it sit still after the swap?

Suppose that I have a single massive quantum mechanical particle in $d$ dimensions ($1\leq d\leq3$), under the action of a well-behaved potential $V(\mathbf r)$, and that I let it settle on the ground state $|\psi_0⟩$ of its hamiltonian, $$ \left[\frac{\mathbf p^2}{2m}+V(\mathbf r)\right]\left|\psi_0\right> = E_0\left|\psi_0\right>. $$ Suppose, because I'm a hater of QM or whatever reason, that I want to model this state of the system as an ensemble of classical trajectories. (More realistically, I might be interested in doing classical trajectory Monte Carlo (CTMC) calculations of some interaction that is hard to model using the full TDSE.) As such, I want to find a classical equivalent to my ground state that I can then use as starting condition for whatever simulation I want to do.

This wavefunction can be examined in phase space in a number of ways, such as using Wigner functions or the Sudarshan $P$ and Husimi $Q$ representations, all of which offer different views into the state and different quasi-classical ways to understand it. My precise question is as follows:

• Is there a way to translate wavefunctions $|\psi⟩$ of a quantum system into probability distributions $\rho(\mathbf r,\mathbf p)$ over classical phase space, in such a way that the eigenstates of a given quantum hamiltonian will be stationary states of the Liouville equation for the corresponding classical system?

To be fully explicit, I want a map such that produces a classical density with the correct position and momentum distributions, i.e. $\int \rho(\mathbf r,\mathbf p)\mathrm d\mathbf p = |⟨\mathbf r|\psi⟩|^2$ and $\int \rho(\mathbf r,\mathbf p)\mathrm d\mathbf r = |⟨\mathbf p|\psi⟩|^2$, and ideally also for all possible quadratures at any angle. The classical $\rho(\mathbf r,\mathbf p)$ should remain stationary under Liouville's equation with a classical hamiltonian $H(\mathbf r,\mathbf p)$ that connects to the quantum hamiltonian via a classical limit or canonical quantization, in the general case, but I'm happy to restrict this to hamiltonians of the form $H(\mathbf r,\mathbf p) = \frac{\mathbf p^2}{2m}+V(\mathbf r)$, in which the correspondence is obvious.

More intuitively, I know that after the translation I will get a counterfeit $\rho( \mathbf r, \mathbf p)$ that doesn't actually describe what's going on, but at least I would like it to sit still once I let classical mechanics take over.

condensed matter - Spontaneous Time Reversal Symmetry Breaking?

It is known that you can break P spontaneously--- look at any chiral molecule for an example. Spontaneous T breaking is harder for me to visualize. Is there a well known condensed matter system which is uncontroversial example where T is broken spontaneously?

I remember vaguely articles of Wen, Wilczek, and Zee from 1989 or so on standard High Tc hopping models, electrons which singly-occupy lattice sites, double-occupation repulsion, small amount of p-doping (holes running around), where they made the claim that T is spontaneously broken. Unfortunately I didn't understand how this happened or if it actually happened. If somebody understands the Zee example, that's good, but I would be happy with any example.

I am not looking for explicit T breaking, only spontaneous T breaking. I would also like an example where the breaking is thermodynamically significant in the large system limit, so mesoscopic rings with permanent currents caused by electron discreteness is not a good example.

Answer

The simplest example in condensed matter physics that spontaneously breaks time reversal symmetry is a ferromagnet. Because spins (angular momentum) change sign under time reversal, the spontaneous magnetization in the ferromagnet breaks the symmetry. This is a macroscopic example.

The chiral spin liquid (Wen-Wilczek-Zee) mentioned in the question is a non-trivial example that breaks time reversal but with out any spontaneous magnetization. Its order parameter is the spin chirality $E_{123}=\mathbf{S}_1\cdot(\mathbf{S}_2\times\mathbf{S}_3)$, which measures the Berry curvature (effective magnetic field) in the spin texture. Because $E_{123}$ also changes sign under time reversal, so the T symmetry is broken by spontaneous development of the spin chirality. Chiral spin liquid can be consider as a condensation of the skyrmion which carries the quantum of spin chirality but is spin neutral as a whole.

In fact, within the spin system, one can cook up any order parameter consisting of odd number of spin operators ($\mathbf{S}_1$ for ferromagnets and $E_{123}$ for chiral spin liquid are both examples of such constructions). Then by ordering such order parameter, the time reversal symmetry can be broken spontaneously.

Beyond the spin system, it is still possible to break time reversal symmetry by the development of orbital angular momentum (loop current) ordering. Just think of spins and loop currents are both angular momenta, what can be done with spins can also be done with loop currents. Indeed, the spinless fermion system can break the time reversal symmetry using the loop current (Note the word "spinless", so there is no spin SU(2) nor spin-orbit coupling involved in the following discussion). Simply consider the spinless fermion $c_i$ on a square lattice coupling to a U(1) gauge field $a_{ij}$, the Hamiltonian reads $$H=-t\sum_{\langle ij\rangle}e^{ia_{ij}}c_i^\dagger c_j+g\sum_\square \prod_{\langle ij\rangle\in\partial\square}e^{ia_{ij}}+h.c.$$ With zero flux per plaquette and with the filling of 1/2 fermion per site, the system has a fermi surface and the fermi level rest on a Van Hove singularity, which is very unstable energetically. The fermions wish to develop any kind of order as long as a it helps to open a gap at the fermi level, such that the fermi energy can be reduced. It is found that the stagger flux is a solution, in which the U(1) flux $\pm\phi$ goes through the plaquette alternately following the checkboard pattern. The corresponding gauge connection is $a_{i,i+x}=0, a_{i,i+y}=(\phi/2)(-)^{i_x+i_y}$. One can show that the energy dispersion for the fermion is given by $$E=\pm\sqrt{\cos^2k_x+\cos^2k_y+2\cos\frac{\phi}{2}\cos k_x\cos k_y},$$ which removes the Van Hove singularity and opens up a pseudo gap (like Dirac cones) as long as $\phi\neq 0$. Therefore driven by the fermi energy, $\phi$ wishes to grow toward the maximum flux $\pi$. However due to the $g$ term in the Hamiltonian, the development of stagger flux consumes magnetic energy (the energy of orbital angular momentum), which grows as $\phi^2$ for small $\phi$. The competition between the fermi energy $t$ and the magnetic energy $g$ will eventually agree on a saddle point value for $\phi$ which is between 0 and $\pi$ and its specific value can be tuned by the $t/g$ ratio. In terms of fermions, the stagger flux $\phi$ is interpreted as loop currents alternating between clockwise and counterclockwise around each plaquette following the check board pattern. Such a state is also call the orbital antiferromagnet (an antiferromagnetic arrangement of orbital angular momentum) or d-wave density wave (DDW) in high-Tc context.

Here $\phi$ serves as the order parameter of the stagger flux state. Because $\phi$ changes sign under time reversal symmetry (like any other magnetic flux), the spontaneous development of the stagger flux pattern in the spinless fermion system will break the time reversal symmetry. In solid-state materials, such phenomenon has not been observed due to the too small $t/g$ ratio which is unable to drive $\phi$ away from 0. However considering the fast development of cold atom physics, the spontaneous time reversal symmetry broken in spinless fermion system may be realized in the future in the optical lattice.

pattern - Color Logic - Find the rule

Find the rule specified in the image below.

---

_{Taken from Caterpillar Logic, now the app is gone.}

Answer

I think I found the rule.

There must be exactly one green tile per row, and there must be more tiles to the left of the green than to the right.

fourier transform - Edge states of Kitaev chain

I am reading paper about Kitaev chain of electrons, which can exhibit famous Majorana fermions at ends of wire. The Hamiltonian (his Eq. (6)) reads

$H = \frac{i}{2} \sum_j - \mu c_{2j-1}c_{2j} +(w+|\Delta|)c_{2j}c_{2j+1} +(-w+|\Delta|)c_{2j-1}c_{2j+2}$

in terms of the Majorana operators $c_{2j-1}$ and $c_{2j}$ and constants $\mu, w, |\Delta|$ can be thought of as parameters controlling Fermi level, hopping and gap.

Kitaev shows there are solutions of Hamiltonian $H$ with zero-energy. He gives the operators associated to zero-energy solutions in ansatz form (his Eq. (14)):

$b' = \sum_j \left( \alpha_{+}' x_{+}^{j} + \alpha_{-}' x_{-}^{j}\right) c_{2j-1}$

$b'' = \sum_j \left( \alpha_{+}'' x_{+}^{-j} + \alpha_{-}'' x_{-}^{-j}\right) c_{2j}$

where all $\alpha$ are constants and $x_{\pm}$ are unknowns to be found.

Q1) How to find $x_{\pm}$?

Q2) How to show $x_{\pm} = \frac{-\mu \pm \sqrt{\mu^2 - 4 w^2 + 4 |\Delta|^2}}{2 (w+|\Delta|)}$?

I attempted to show it by computing $[b', H]=[b'', H]=0$, but it did not give correct answer.

Any help appreciated.

lagrangian formalism - Constrained Hamiltonian systems: spin 1/2 particle

I am trying to apply the Constrained Hamiltonian Systems theory on relativistic particles. For what concerns the scalar particle there is no issue. Indeed, I have the action \begin{equation} S=-m\int d\tau \sqrt{-\dot{x}^\mu \dot{x}_\mu}\tag{1} \end{equation} and computing the momentum \begin{equation} p_\mu=\frac{m\dot{x}_\mu}{\sqrt{-\dot{x}^2}}\tag{2} \end{equation} I see that it satisfies the constraint $p_\mu p^\mu+m^2=0$. I then proceed to quantize the system with the Dirac method.

I am finding issues with the relativistic massless spin 1/2 particle. Indeed, it is described by the space-time coordinates $x^\mu$ and by the real grassmann variables $\psi^\mu$, according to my notes. The action should take the form \begin{equation} S=\int d\tau \space \dot{x}^\mu\dot{x}_\mu+\frac{i}{2}\psi_\mu\dot{\psi}^\mu\tag{3} \end{equation} which exhibits a supersymmetry on the worldline, the supersymmetric conserved charge being $Q=\psi^\mu p_\mu$. According to the lecturer I should find the constraints \begin{equation} H=\frac{1}{2}p^2, \quad Q=\psi^\mu p_\mu\tag{4} \end{equation} i.e. the dynamics on the phase space should take place on hypersurfaces $$H=0, Q=0.\tag{5}$$

My question: How can I derive these constraints? They should arise simply with the definition of momenta, but, having no constants to work with, I'm left with \begin{equation} p_\mu=\dot{x}_\mu\\ \Pi_\mu=\frac{i}{2}\dot{\psi}_\mu\tag{6} \end{equation} and I don't know what to do with them. I see that, in principle, the first constraint is obtained by setting $m=0$ in the constraint of the scalar particle for example, but what if I want to derive it without the previous knowledge? And what about $Q$?

Edit: By intuition, knowing that the model exhibits a ${\cal N}=1$ supersymmetry, I may understand that the dynamics must take place on a surface such that $H=const$ and $Q=const$ (then I could set the constant to zero without lack of generality?), being $Q$ and $H$ conserved charges. Is it the only way to find these constraints? Should I need this previous knowledge about supersymmetry to study the model? I think I should be able to find these constraints just by looking at the Lagrangian itself.

Answer

1. 
   

   We consider here the massless case $m=0$. Let us start from the Lagrangian$^1$ $$L_0~=~\frac{\dot{x}^2}{2e} +\frac{i}{2}\psi_{\mu}\dot{\psi}^{\mu} \tag{A}$$ with an einbein field $e$, cf. e.g. this Phys.SE post. If we introduce the momentum $$ p_{\mu}~=~\frac{\partial L_0}{\partial \dot{x}^{\mu}} ~=~\frac{\dot{x}_{\mu}}{e},\tag{B}$$ the corresponding Legendre transformation $\dot{x}^{\mu}\leftrightarrow p_{\mu}$ yields a first-order Lagrangian $$ L_1~=~p_{\mu}\dot{x}^{\mu} +\frac{i}{2}\psi_{\mu}\dot{\psi}^{\mu}-eH, \qquad H~:=~\frac{p^2}{2}. \tag{C}$$ This explains OP's first constraint $H\approx 0$, which is indirectly due to world-line (WL) reparametrization invariance, cf. this Phys.SE post.

   
   
   


2. 
   

   It is unnecessary to introduce momentum for the fermions $\psi^{\mu}$ as the Lagrangian $L_1$ is already on first-order form, cf. the Faddeev-Jackiw method.

   
   


3. 
   

   The Lagrangian $L_1$ has a global super quasisymmetry. The infinitesimal transformation $$ \delta x^{\mu}~=~i\varepsilon\psi^{\mu}, \qquad \delta \psi^{\mu}~=~-\varepsilon p^{\mu}, \qquad \delta p^{\mu}~=~ 0 , \qquad \delta e~=~ 0, \tag{D}$$ changes the Lagrangian with a total derivative $$ \delta L_1~=~\ldots~=~i\dot{\varepsilon}Q+\frac{i}{2}\frac{d(\varepsilon Q)}{d\tau}, \qquad Q~:=~p_{\mu}\psi^{\mu}, \tag{E}$$ for $\tau$-independent Grassmann-odd infinitesimal parameter $\varepsilon$.

   
   


4. 
   

   OP's other constraint $Q\approx0$ arises by gauging the SUSY, i.e. $\delta L_1$ should be a total derivative for an arbitrary function $\varepsilon(\tau)$. On reason to do this is given in Ref. 2 below eq. (3.3):

   
   

   
   
   

   Because of the time component of the field $\psi^{\mu}$ there is a possibility that negative norm states may appear in the physical spectrum. In order to decouple them we require an additional invariance and, inspired by the Neveu-Schwarz-Ramond model, it seems natural to demand invariance under local supergauge transformations.

   
   

   
   


5. 
   

   Concretely, we impose $Q\approx0$ with the help of a Lagrange multiplier $\chi$. This leads to the Lagrangian $$ L_2~=~L_1-i \chi Q~=~p_{\mu}\dot{x}^{\mu} +\frac{i}{2}\psi_{\mu}\dot{\psi}^{\mu}-eH - i \chi Q .\tag{F}$$

   
   


6. 
   

   Let us mention for completeness that in order to have gauged super quasisymmetry of the new Lagrangian $L_2$, the previous transformation $\delta e= 0$ needs to be modified into $$ \delta e~=~2i\chi\varepsilon, \qquad \delta \chi~=~\dot{\varepsilon}.\tag{G}$$

   
   


7. 
   
   

   An alternative perspective is the replacement $$L_2~=~ L_1|_{\dot{x}\to Dx}\tag{H}$$ of the ordinary derivative $$\dot{x}^{\mu}\quad\longrightarrow\quad Dx^{\mu}~:=~\dot{x}^{\mu} -i\chi \psi^{\mu}\tag{I}$$ with a gauge-covariant derivative $Dx^{\mu}$. Here $\chi$ is a compensating gauge field. The gauge-covariant derivative transforms as $$ \delta Dx^{\mu}~=~i\varepsilon(\dot{\psi}^{\mu}-\chi p^{\mu}).\tag{J}$$

   
   

References:

1. 
   

   F. Bastianelli, Constrained hamiltonian systems and relativistic particles, 2017 lecture notes; Section 2.2.

   
   


2. 
   

   L. Brink, P. Di Vecchia & P. Howe, Nucl. Phys. B118 (1977) 76; Below eq. (3.3).

   
   
   


3. 
   

   C.M. Hull & J.-L. Vazquez-Bello, arXiv:hep-th/9308022; Chapter 2, p. 7-8.

   
   

--

$^1$ Conventions: We use the Minkowski sign convention $(-,+,+,+)$ and we work in units where $c=1$.

special relativity - Does SR treat time and space symmetrically? Then why does $t$ dilate while $ell$ contracts?

Can you resolve this paradox: Time and space are supposed to be treated symmetrically in SR, yet time dilates while length contracts. I've been working on this for weeks, and I can't resolve it.

The mirror experiment shows that time and space both expand for the moving observer, in contradiction to the results one derives from the LT, yet a well-known text, Eisberg's Fundamentals of Modern Physics (pp. 20-21) incorrectly concludes the usual asymmetrical results from that, by changing the definition of time dilation! He defines it as meaning that time measured in the moving frame is longer, when tau is actually well-known to be shorter. Time dilation refers to the mover's time UNITS LOOKING longer to the stationary observer because the total number of time units is FEWER, the opposite situation from the mirror experiment.

electromagnetism - Why does the Earth not lose its magnetism?

A magnet loses its magnetic properties when subjected to heat, the Earth's core is also hot but still it has magnetic properties, Why?

mathematics - Crack the Code #3

You return to the bench after cracking the last code, and you notice an envelope on the ground directly under it. You put down your envelope with the previous code and pick up the new one, anticipating another challenge.

Crack the Code #2

_{Digits are referred to as A-B-C-D in the clues. "A + B" is the sum of the first and second digit. All math follows the standard order of operations}

Clues

• The number is prime.



• No digits are repeated.


• $A\cdot B=C\cdot D$


• The first digit is greater than 3.

What four digit number matches these criteria? Also, if you want, post your methodology for finding the correct answer, as this will help me in the future.

_{Note: I am pretty sure that only one number matches all these clues. However, I may have miscalculated. Please correct me in the comments. If you find the answer, put it in a spoiler so the question is not ruined for those who want to solve it.}

Answer

Do not hover over the below unless you wish to know the answer...

6329

I cheated and wrote a small program to do the hard work for me :p which I'll post in a moment - I'm a lazy software developer and we never do any un-necessary maths.

/* package whatever; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Ideone

{

    public static boolean isPrime(long n)
    {
        if(n < 2) return false;

        if(n == 2 || n == 3) return true;

        if(n%2 == 0 || n%3 == 0) return false;


        long sqrtN = (long)Math.sqrt(n)+1;

        for(long i = 6L; i <= sqrtN; i += 6)
        {
                if(n%(i-1) == 0 || n%(i+1) == 0) return false;
        }

        return true;
    }


    public static void main (String[] args) throws java.lang.Exception
    {
        int a;
        int b;
        int c;
        int d;
        int combined;

        for(a = 4; a < 10; a++)
        {

            for(b = 0; b < 10; b++)
            {
                for(c = 0; c < 10; c++)
                {
                    for(d = 0; d < 10; d++)
                    {
                        if (a != b && a != c && a !=d && b != c && b != d && b != d && c != d)
                        {
                            if(a*b == c*d)
                            {

                                combined = (((a * 10 + b) * 10 + c) * 10) + d;
                                if(isPrime(combined))
                                {
                                    System.out.println(combined);
                                }
                            }

                        }
                    }
                }

            }
        }
    }
}

particle physics - Why is quark flavor just a SU(N) group?

In the standard model one has U(1) for electromagnetism, SU(2) for the weak sector and SU(3) for the color sector. One could say that in the quark part of the fermions, there are $$ \underbrace{6}_\text{flavors} \cdot \underbrace{3}_\text{colors} \cdot \underbrace{(2 + 1)}_\text{left and right}$$ “different” spinor particles. Those states can be grouped into color triplets and weak-isospin doublets and singlets depending on their chirality. For massive quarks, the weak-isospin symmetry is only approximate, of course. The Lagrangian stays invariant under the unitary transformations, therefore those are the gauge symmetries.

The weak-isospin doublets made up from up-down, charm-strange and top-bottom are just the same thing in different generations. Their only difference I think they have is the mass. Neglecting the CKM-matrix and the masses, I would think that there is no way to distinguish a down-quark from a strange-quark or bottom-quark. With masses, there is just this one difference.

One then generalizes the flavor symmetry to include the strange quark. The weak-isospin group is extended from SU(2) to just SU(3). This bugs me: The (approximate) SU(2) weak-isospin symmetry groups the (left-handed) up- and down-quarks together in a doublet. They are two states that differ by their electromagnetic charge. When adding the strange quark I would consider this to be another down-type quark (which is distinct from an up-type quark) from the second generation. Why does one put the strange quark onto equal footing with the other ones by including it with one $\mathrm{SU}(N_\mathrm F = 3)$ group? Shoudn't the strange-quark be treated similarly to the down-quark but different from the up-quark?

For any given force, it only sees part of it. So for the weak-isospin gauge symmetry, all the color states appear to be on equal footing. And the strong force does not care about the electric charge or the weak-isospin $z$-component. Is my problem in understanding that I distinguish up-type and down-type by quantum numbers that belong to other symmetry groups and the flavor would not see at all? Or is there something more to this?

electromagnetic radiation - How do we know that these radio bursts are from billions of light-years away?

NASA just announced that they detected the first radio bursts from outside of our galaxy.

Astronomers, including a team member from NASA's Jet Propulsion Laboratory in Pasadena, Calif., have detected the first population of radio bursts known to originate from galaxies beyond our own Milky Way. The sources of the light bursts are unknown, but cataclysmic events, such as merging or exploding stars, are likely the triggers.

The new radio-burst detections -- four in total -- are from billions of light-years away, erasing any doubt that the phenomenon is real.

If we don't know what the source is, then how do we know how far away the source is? How can we tell how far the light waves have traveled?

Answer

The relevant Science summary is "Radio Bursts, Origin Unknown" by Cordes, abstract here. Briefly, it notes that a burst of radio waves will undergo dispersion in the interstellar and intergalactic medium (ISM and IGM), the amount of which is indicative of how much matter the signal has passed through.

Some further background not given in that summary: In general, the ISM/IGM consists largely of free protons and electrons. As a plasma, this medium has a frequency-dependent plasma frequency, so different frequency components will propagate at slightly different speeds (just under the speed of light). (In fact, radio waves that are too low in frequency cannot propagate at all through the "emptiness" of space.)

The more material the signal has passed through, the more spread-out the pulse will be, with lower frequencies arriving after higher ones. Assuming the original shape of the pulse is known (perhaps it is just a very sharp spike, at least compared to how broad it is by the time we detect it), then one can take the shape of the detected pulse and figure out how much dispersion has occurred. Then one asks, "How much of this is due to material in our own galaxy?" which is answerable based on maps astronomers have constructed of the interstellar (intra-galactic) medium along various sight lines. Next one guesses (in an educated fashion) how much is due to the host galaxy of the source. The rest is attributed to the IGM, which, assuming some uniform density, yields a distance. A summary of this technique can also be found on Wikipedia; see Dispersion in pulsar timing

As it turns out, this is indeed the method used in the Science paper by Thornton et al., "A Population of Fast Radio Bursts at Cosmological Distances," abstract here, to which the NASA press release was referring. These particular events were found at high galactic latitude, meaning they were seen looking "up" or "down" out of the plane of the galaxy, rather than through the bulk of the disk, so not much of the measured dispersion can be attributed to the ISM in the Milky Way.

That second Wikipedia article defines dispersion measure. The Thornton paper reports dispersion measures for these four objects are $553$, $723$, $944$, and $1104\ \mathrm{pc/cm^3}$. After subtracting the effect of our own ISM, they conclude the extragalactic dispersion measures (including any contribution from host galaxies) are $521$, $677$, $910$, and $1072\ \mathrm{pc/cm^3}$. They then assume host dispersion measures are $100\ \mathrm{pc/cm^3}$ in all cases, subtract that off, and divide by a value for the IGM number density of electrons (in $\mathrm{cm^{-3}}$, to get a distance in parsecs). Actually, this last step is a little more complicated due to the fact that the universe has expanded over the long time those radio waves have been traveling, but the authors take that into account.

In summary, radio astronomy uses the fact that the space between galaxies is not completely empty, and that radio waves, like all forms of light, slow down in various ways when traveling through matter. This is particularly useful when the signal doesn't have sharp spectral features from which to obtain redshifts (this technique being common in optical, UV, and IR astronomy).

statistical mechanics - Literature recommendation for classical density functional theory (DFT) and fundamental measure theory (FMT)

I'm very much interested in properly learning about density functional theory calculations (DFT) in classical settings, for example as used in the theory of liquids. Apart from the success of DFT applied to many-body QM systems, for classical​ systems it remains the main theoretical approach of the statistical physics of liquids and solids. Similarly, but more recently, applications of fundamental measure theory (FMT, which is a more geometric approach) can be noticed more and more.

In most current liquid state theory books, these approaches are only briefly introduced and almost never at depth (as they are often assumed known by the authors). Although I have the basics of statistical mechanics, I am very new to classical DFT calculations, and would be very much interested in any piece of literature, be they review papers, lecture notes or textbooks, that would softly and slowly introduce these techniques.

Answer

1. 
   

   An excellent introduction to the field of classical DFT is Robert Evan's article Density functionals in the theory of nonuniform fluids (Fundamentals of inhomogeneous fluids 1 (1992): 85-176). It is a bit older, but very accessible and yet thorough. Some proofs are omitted, but references are provided for further information.

   
   
   


2. 
   

   The standard textbook for liquid state theory, Theory of Simple Liquids by Jean-Pierre Hansen and Ian McDonald has a section on DFT and also discusses Fundamental Measure Theory. It is a great book but the presentation is quite terse, and makes numerous references to preceding sections that you will have to go through first. Nevertheless it a good starting point if you have the book lying around (as everyone interested in liquid theory should).

   
   


3. 
   

   DFT is typically presented in terms of correlation functions of simple fluids, which means that one has to deal with complicated-looking integral equations. The article An introduction to inhomogeneous liquids, density functional theory, and the wetting transition by Hughes, Thiele and Archer presents an alternative introduction to DFT in terms of a simple (Ising) lattice model, which might be a good approach for those who have a more general statistical mechanics background.

   
   

kinetic theory - How much do we know about the characteristic time for electron-ion recombination for a dense plasma?

As mentioned, how much do we know about the characteristic time for electron-ion recombination for a dense (noble gas) plasma (i.e. density of electrons at $\approx 10^{21} \textrm{cm}^{-3}$) at $10000 \textrm{K}$ (i.e. at around $\approx 1\textrm{eV}$)?

The dominant mechanism for electron-ion recombination at this temperature and density for a thermalised plasma seems to be three-body recombination (or dielectronic recombination). The mentioned process is the backward process as represented in the equation, $ X + e^- \rightarrow X^{+1} + e^- + e^- $, where $X$ denotes the participating atom. The cross section of its inverse process, electron impact ionization, has been measured experimentally (probably at a low density of $X$), and the rate of this inverse process can be obtained from the cross section (i.e. $\approx $, where $<\cdots>$ denotes a thermal average over the energy of the impacting electron, $v$ the thermal velocity of the impacting electron and $\sigma$ the cross section). It seems that the rate of three-body recombination can be obtained from the principle of detailed balance.

The problem is that the density effect of $X$ has been ignored. It seems that we are extrapolating the cross sections at low densities of $X$ to obtain the cross sections at high densities of $X$. Of course the obtained cross sections for Xenon for example, are about (PRA, 65, 042713, doi:10.1103/PhysRevA.65.042713) $10^{-16} \textrm{cm}^2$, which suggests a length $\approx 10^{-10} \textrm{m}$. The average separation between the atoms seems to be greater than this value at $21\times 10^{21} \textrm{cm}^{-3}$, and so the density effect may be unimportant. Yet the argument is ad-hoc.

Is there any reason to believe that the electron-ion recombination time (or rate) will be significantly different from the one obtained using the above procedure? ("significantly" means the correct time will be several orders larger/smaller than the calculated time). Thanks.

particle physics - How do we know Dark Matter isn't simply Neutrinos?

What evidence is there that dark matter isn't one of the known types of neutrinos?

If it were, how would this be measurable?

Answer

Dark matter can be hot, warm or cold. Hot means the dark matter particles are relativistic (kinetic energy on the order of the rest mass or much higher), cold means they are not relativistic (kinetic energy much less than rest mass) and warm is in between. It is known that the total amount of dark matter in the universe must be about 5 times the ordinary (baryonic) matter to explain the CMB as measured by WMAP.

However, cold dark matter must be a very significant component of the universe to explain the growth of structures from the small fluctuations in the early universe that grew to become galaxies and stars (see this reference). Thus cold dark matter is also required to explain the currently measured galactic rotation curves.

Now, the neutrino oscillation experiments prove that neutrinos have a non-zero rest mass. However, the rest masses must still be very small so they could only contribute to the hot dark matter. The reason they can only be hot dark matter is because it is assumed that in the early hot, dense universe, the neutrinos would have been in thermal equilibrium with the hot ordinary matter at that time. Since the neutrino's rest mass is so small, they would be extremely relativistic, and although the neutrinos would cool as the universe expands, they would have still been very relativistic at the time of structure formation in the early universe. Thus, they can only contribute to hot dark matter in terms of the early growth of structure formation. [Because of the expansion of the universe since then, the neutrinos should have cooled so much that they are non-relativistic today.]

According to this source:

Current estimates for the neutrino fraction of the Universe’s mass–energy density lie in the range 0.1% <∼ ν <∼ a few %, under standard assumptions. The uncertainty reflects our incomplete knowledge of neutrino properties.

So most cosmic neutrinos are probably less than 10% of the total dark matter in the universe. In addition most of the rest (of the non-neutrino) 90% of dark matter must also be cold dark matter - both in the early universe and even now.

quantum mechanics - Proving which QM interpretation is correct

Let's assume that the existence of gravitons is theoretically proven or they are detected by LHC could one interpretaion be proven correct?

quantum field theory - Ultraviolet behaviour in dimensional regularization

In dimensional regularization, we introduce an arbitrary energy scale $\mu$. Naively, it plays the role of another parameter of the theory that needs to be fixed experimentally, but actually it is not since the RG-flow equations allow a change in $\mu$ to be compensated by a change in renormalized couplings.

The RG-flow is then to be analyzed, and based on its behavior in the UV we can either say that in the UV limit our theory reaches a fixed point and therefore is equivalent to some conformal field theory (like in QCD), or some couplings blow up in which case perturbative expansions make no sense anymore (like in QED).

My question is: why can't we just fix some $\mu$ and go from there?

In cut-off regularization scheme, for example, it is obvious that $\Lambda$ defines a "bounding energy scale" beyond which the regularized theory can not work (since the corresponding degrees of freedom simply do not exist). It is therefore natural to seek a higher $\Lambda$ whenever our current $\Lambda$ is not good enough, i.e. whenever we need to explain some higher-energy phenomena.

In contrast, in dimensional regularization we have already taken the $d \rightarrow 4$ limit, so the theory with $\mu$ and renormalized couplings evaluated at $\mu$ is already exact (well, at least perturbatively) and I can not imagine why one would need a higher $\mu$ after all.

Whilst I completely understand that in both schemes the resulting RG-flow equations are the same and define a beautiful and concise mathematical structure, I would nevertheless like to understand this conceptual point.

UPDATE: I also understand that $\Lambda$ plays a role of regularizer in the cut-off scheme, just like $d$ does in dimensional regularization (and not $\mu$). But it does not answer my question: why can't we simply take some $\mu$ and forget about the RG flow?

UPDATE 2: in case the answer to my question turns out to be "sure, why not, take some $\mu$ and go from there" $-$ then I have another one. How come that our theory can have different regimes in the UV and in the IR if I could just take some $\mu$? Maybe it has something to do with the validity of perturbative expansions?

Answer

We have to change the renormalization scale to a typical scale of the process we are considering because otherwise perturbation theory becomes rather useless, since process far away from the renormalization scale get large higher order corrections due to the "large log" $\ln(-p^2/\mu^2)$, where $p$ is the process scale and $\mu$ the renormalization scale.

The standard perturbative expansion is not only an expansion in $\lambda$, but also in $\ln(-p^2/\mu^2)$. As an illustrative example, consider massless 4D $\phi^4$ theory, where we might fix the renormalization scale at any $\mu$ by imposing $$ \mathrm{i}\mathcal{M} = -\mathrm{i}\lambda$$ for the 2-to-2 scattering $p_1 p_2\to p_3 p_4$ at $s=t=u=-\mu^2$ where $s,t,u$ are the usual Mandelstam variables. It's clear that we defined $\lambda(\mu)$ by this to be the relevant coupling at the scale $\mu$ - and to see that it becomes a bad choice for the coupling at scales far away from $\mu$, we have to consider the full amplitude $$ \mathrm{i}\mathcal{M} = -\mathrm{i}\lambda(\mu) - \mathrm{i}\frac{\lambda(\mu)^2}{32\pi^2}\left(\ln(-s/\mu^2) + \ln(-t/\mu^2) + \ln(-u/\mu^2)\right) + \mathcal{O}(\lambda(\mu)^3)$$ where the 1-loop contribution of order $\lambda(\mu)^2$ grows larger as the ratio of the Mandelstam variables (which are good measures for the energy scale of the process) to the renormalization scale goes away from unity.

It's not that it is not possible to compute the amplitudes of processes far away from the renormalization scale, but the perturbation series becomes inefficient, and the coupling $\lambda(\mu)$ is no longer really the "physical coupling" which should be pretty straightforwardly related to the vertices. "Running the coupling" is essentially restoring the coupling to a value where it is again a good parameter to consider the Feynman diagram perturbation series as being expanded in. This also explains why one would speak of the breakdown of perturbation theory as soon as the coupling becomes of the order of unity - you can't get around this by staying at a lower scale and perturbing that, since then the logarithms appear large and blow up the series, anyway.

newtonian mechanics - Deriving D'Alembert's Principle

The wiki article states that D'Alembert's Principle cannot derived from Newton's Laws alone and must stated as a postulate. Can someone explain why this is? It seems to me a rather obvious principle.

Answer

Superficially, D'Alembert's Principle

$$\tag{1} \sum_{j=1}^N ( {\bf F}_j^{(a)} - \dot{\bf p}_j ) \cdot \delta {\bf r}_j~=~0 $$

may look like a trivial consequence of Newton's 2nd law, but the devil is in the detail. Here the detail is the superscript $(a)$ on the force, which stands for applied forces. The term applied forces refers to that we have divided all forces (such as, e.g., gravity force, constraint force, etc.) into two bins:

1. 
   

   The applied forces, and

   
   


2. 
   

   the rest.

   
   
   

It is important to realize that D'Alembert's principle may be true or false, depending on how the above division is made. See also this Phys.SE post. The standard example of a force that one cannot put into the second bin is a sliding friction force.

In particular, if one puts all forces into the first bin, then indeed, D'Alembert's Principle would be a trivial consequence of Newton's 2nd law. But in practice one would not like to do that. One would instead like to minimize the type of forces that one puts in the first bin to simplify the equations as much as possible.

electromagnetism - Why the electric bulb turns on almost instantly when the switch is closed?

The electron drift speed is estimated to be very low.How could there is current almost the instant a circuit is closed??

By the discussions it is known that The information about beginning of the flow of current is transmitted through the propagation of electromagnetic waves(electric impulse)and not with the drift velocity of the electrons.
But I want any one to explain how this process takes place.CURIE:)

homework and exercises - How do I Derive the Green's Function for $-nabla^2 + m^2$ in $d$ dimensions?

What is the solution to this equation in $d$ dimensions: $$(-\nabla_d^2 + m^2)G(\mathbf{x}, \mathbf{x}') = A \delta(\mathbf{x} - \mathbf{x}'),$$ with the boundary condition that $\lim_{|\mathbf{x} - \mathbf{x}'|\rightarrow \infty} G(\mathbf{x}, \mathbf{x}') = 0$?

Answer

The first step is to recognize that equation is invariant under $d$-dimensional rotations around $\mathbf{x} - \mathbf{x}' = \mathbf{0}$ and simultaneous identical translations of $\mathbf{x}$ and $\mathbf{x}'$, so we can make the following step: $$\begin{align}(-\nabla_d^2 + m^2)G(\mathbf{x}, \mathbf{x}') &= A \delta(\mathbf{x} - \mathbf{x}') \\ & \mathrm{let: }\ r \equiv |\mathbf{x} - \mathbf{x}'| \Rightarrow \\ \frac{A}{\Omega_d r^{d-1}} \delta(r) &= - \frac{1}{r^{d-1}} \frac{\partial}{\partial r} \left[r^{d-1} \frac{\partial G(r)}{\partial r}\right] + m^2 G(r),\end{align}$$ where the factor $\Omega_d r^{d-1}$ comes from the volume element $\operatorname{d} V = \Omega_d r^{d-1} \operatorname{d}r$ and the derivatives on the right hand side (rhs) are the radial term of $\nabla_d^2$ in $d$-dimensional spherical coordinates (wiki link).

The next step is to integrate both sides of the equation over a spherical volume centered at the origin with radius $r$, then take the limit as $r \rightarrow 0$. This yields the normalization condition for $G$: $$\lim_{r\rightarrow 0} \left[r^{d-1} \frac{\partial G}{\partial r}\right] = -\frac{A}{\Omega_d},$$ and handles the part of the equation where the delta function is non-zero.

The region where the delta function is zero, the homogeneous region, becomes: $$0 = \frac{\partial^2 G}{\partial r^2} + \frac{d-1}{r} \frac{\partial G}{\partial r} - m^2 G.$$ The equation in the homogeneous region can be brought into a more familiar form by the function substitution $G(r) = f(r) r^{-(d/2 - 1)}$ giving: $$0 = r^2 \frac{\partial^2 f}{\partial r^2} + r \frac{\partial f}{\partial r} - \left(\frac{d}{2} - 1\right)^2 f - m^2 r^2 f.$$ The familiar form to this equation is the modified Bessel's equation. The most general solution to this equation is: $$f(r) = C K_{d/2-1}(mr) + D I_{d/2-1}(mr),$$ with $I_{d/2-1}$ and $K_{d/2-1}$ modified Bessel functions of the first and second kind, respectively, and $C$ and $D$ constants fixed by the boundary conditions.

The boundary condition at $r\rightarrow \infty$ requires $D=0$, giving the following form for $G$: $$G(r) = \frac{C}{r^{d/2-1}} K_{d/2-1}(mr).$$ Plugging our solution for $G$ into the left hand side (lhs) of the $r \rightarrow 0$ boundary condition derived above gives: $$\lim_{r\rightarrow0} \left[ r^{d-1} \frac{\partial G}{\partial r}\right] = -\Gamma\left(\frac{d}{2}\right) 2^{d/2-1} m^{1-d/2} C,$$ after application of the small argument limit form of $K_\nu$. This implies that: $$\begin{align}C &= \frac{A m}{2^{d/2-1}\Gamma(d/2) \Omega_d} \\ & = \frac{A m^{d/2-1}}{2^{d/2} \pi^{d/2}},\end{align}$$ where the explicit form of $\Omega_d = S_{d-1}$ has been inserted.

Finally, replacing $C$ gives: $$G(r) = \frac{A}{(2\pi)^{d/2}} \left(\frac{m}{r}\right)^{d/2-1} K_{d/2-1}(mr).$$

enigmatic puzzle - True blueblood made disturbance twice (6, 7)

I found this note the other day...

Captured confidante without it returning (6)

Capricious committee accommodation after rats return (4, 7)
Absolute leader, remove fractions nearby (6)
Precious drink messed up hearing aid (7)
Blind tiger talk; naive (9)
Dark times clothing sounds of valiant struggle (10)
Yet means eternal, lose permanence (7)
Forcibly print mail contract (11)

What did this cheerful puzzler say?

Answer

They weren't so cheerful - they said:

"I'm scared, send help"!

Clue explanations:

Title: DOUBLE TROUBLE*
1. IN(-ti<)MATE
2. STAR<+CHAMBER
3. A(-fraid)+ROUND

4. (EARD+RUM)* [indirect anagram of DEAR+RUM: note that indirect anagrams are generally not allowed]
5. SPEAK+EASY ["blind tiger" is slang for a speakeasy]
6. NIGHTDRESS ("knight duress" homophone?)
7. HOW+EVER(-lasting)
8. LETTER+PRESS

Getting the message:

The italicised letters spell: "It came nearer; I hear it rave; I'm torn apart".

The last line is the key: each of these answers is made up of two words, and can be torn apart into those separate words. IN+MATE, STAR+CHAMBER, A+ROUND, EAR+DRUM, SPEAK+EASY, NIGHT+DRESS, HOW+EVER, LETTER+PRESS. The first letters of each of these, in order, spell I'M SCARED SEND HELP.

optics - Keep the light beam in a closed room, is it possible?

I mean if I am in a room totally closed to light. If I switch on a torch for a second then switch it off. So will the inside of room be always bright?

Answer

No mirror can be perfectly reflective due to quantum tunneling so that already answers your question. But even if it could be done, you would never be able to check the situation because when you look inside, the light almost instantly leaves through the peephole. This also poses a problem for your initiation method, which John M already touched on: you need to be very, very quick to insert the light beam and remove your insertion device/hole before it can reabsorb the light.

In any case, let's consider what would happen if you had been able to pull of this insertion of a light beam. The light would definitely be inside for a short but finite period of time. The problem is that even classically perfect mirrors are impossible with current technology; I believe the highest reflectivity that we're able to achieve is about 99.999% if the wavelength is just right. Alas, even this kind of amazing reflectivity means that a light beam will lose 90% of its intensity in less than a second in a spherical container with a diameter of $1\,\mathrm{km}$!

You can do the maths of this yourself: use that the reflectivity $R$ is the factor by which the intensity $I$ of the light beam is reduced when it reflects off a wall. So

$$I_{after} = R I_{before}$$

Now give the intensity of the light an index corresponding to the number of interactions with the wall. So the original beam has intensity $I_0$. After one reflection, it still has $I_1 = RI_0$ left. After two reflections, it has $I_2 = RI_1 = R^2I_0$ left. There is a pattern here. After $n$ reflections, it has $I_n = R^nI_0$ left. Now, you want to find out for which $n$ the intensity has dropped to 10 percent of the original $I_0$. So the question is: when is $R^n < 0.1$ (with $n$ a natural number)?

You can probably work that out by yourself. Now consider a spherical container of diameter $1\,\mathrm{km}$. Then the light will always have to travel $1\,\mathrm{km}$ or less before it hits the wall again. In other words, it has to travel (at the highest) $n$ times $1\,\mathrm{km}$ before it has lost 90 percent of its original intensity. Using that the speed of light is approximately $300000\,\mathrm{km/s}$ you should find that it only takes the light about $0.77\,\mathrm{s}$ to do this.

homework and exercises - Scalar invariance under Lorentz-transformation

With $u^\alpha v^\beta$ the components of two vector fields, is

$$u^\alpha v^\alpha=u^1v^1+u^2v^2+u^3v^3+u^4v^4$$

a scalar invariant under a Lorentz-transformation? And why?

Answer

First note that your expression of $u^{\alpha}v^{\alpha}$ is wrong. Einstein summation convention tell you that you sum if an index appears twice - once up and once down. Thus, in fact

$$u^{\alpha}v_{\alpha}=u^1v_1+u^2v_2+u^3v_3+u^4v_4$$

The quantity $u^\alpha v^\beta$, on the other hand, is a tensor of rank $(2,0)$, which can be represented by a $4\times 4$ matrix

$$\begin{pmatrix}u^1v^1&u^1v^2&u^1v^3&u^1v^4\\u^2v^1&u^2v^2&u^2v^3&u^2v^4\\u^3v^1&u^3v^2&u^3v^3&u^3v^4\\u^4v^1&u^4v^2&u^4v^3&u^4v^4\end{pmatrix}$$

so by writing $u^\alpha v^\alpha$ you refer to its diagonal elements.

Second, you are confusing upper and lower indices. The location of the index determines its transformation properties. Take for example a vector $v^\alpha$, it transform in the following manner

$$v^{\alpha^\prime}=\frac{\partial x^{\alpha^\prime}}{\partial x^\alpha}v^\alpha$$

On the other hand, a co-vector $u_{\alpha}$ transforms differently

$$u_{\alpha^\prime}=\frac{\partial x^{\alpha}}{\partial x^{\alpha^\prime}}u_{\alpha}$$

In your case, the quantity $u^\alpha v^\beta$ transform like this

$$u^{\alpha^\prime} v^{\beta^\prime}=\frac{\partial x^{\alpha^\prime}}{\partial x^\alpha}\frac{\partial x^{\beta^\prime}}{\partial x^\beta}u^\alpha v^\beta$$

and in particular, the diagonal elements

$$u^{\alpha^\prime} v^{\alpha^\prime}=\frac{\partial x^{\alpha^\prime}}{\partial x^\alpha}\frac{\partial x^{\alpha^\prime}}{\partial x^\beta}u^\alpha v^\beta$$

In general, it is not invariant under the Lorentz transformations. However, the quantity $u^\alpha v_\alpha$ has no free indices since you sum over $\alpha$, and thus it remains invariant under transformations. You can also see it by transforming $u^\alpha$ and $v_\alpha$ separately

$$u^{\alpha^\prime}v_{\alpha^\prime}=\frac{\partial x^{\alpha^\prime}}{\partial x^\beta}u^\beta\frac{\partial x^{\gamma}}{\partial x^{\alpha^\prime}}v_\gamma=\frac{\partial x^{\gamma}}{\partial x^\beta}u^\beta v_\gamma=\delta^{\gamma}_{\beta}u^\beta v_\gamma=u^\beta v_\beta$$

enigmatic puzzle - Read. Learn. Live. Escape

You wake lying face down on a hard stone floor. Your head hurts. As your eyes find focus, you see a haiku scratched into the ground beside you:

A wrong step brings death!
But, a wrong guess brings knowledge…
Read. Learn. Live. Escape.

Death? Escape? You leap up and scan your surroundings... You're in a large smooth-walled cavern, whose floor is made entirely of a grid of seemingly floating stone platforms, suspended somehow over an immeasurable drop. The only way out appears to be a doorway on the far side of the room. As you stare, you notice that each platform has a single letter carved into it. You quickly note them down:

$$ \begin{matrix} \fbox{R} & \fbox{U} & \fbox{Y} & \fbox{E} & \fbox{R} & \fbox{O} & \color{blue}{\fbox{▲}} & \fbox{L} & \fbox{N} & \fbox{R} & \fbox{A} & \fbox{E} & \fbox{Y} \\ \fbox{A} & \fbox{R} & \fbox{D} & \fbox{Y} & \fbox{E} & \fbox{E} & \fbox{R} & \fbox{U} & \fbox{O} & \fbox{F} & \fbox{M} & \fbox{O} & \fbox{T} \\ \fbox{C} & \fbox{U} & \fbox{L} & \fbox{A} & \fbox{A} & \fbox{T} & \fbox{Y} & \fbox{L} & \, & \fbox{S} & \fbox{A} & \fbox{I} & \, & \\ \fbox{B} & \fbox{T} & \fbox{W} & \fbox{O} & \fbox{T} & \fbox{H} & \fbox{E} & \fbox{N} & \fbox{D} & \fbox{I} & \fbox{I} & \fbox{W} & \fbox{N} \\ \fbox{E} & \fbox{E} & \fbox{D} & \fbox{E} & \fbox{O} & \fbox{R} & \fbox{M} & \fbox{F} & \fbox{D} & \fbox{A} & \fbox{D} & \fbox{N} & \fbox{E} \\ \fbox{N} & \fbox{L} & \fbox{N} & \fbox{I} & \fbox{R} & \fbox{S} & \fbox{R} & \fbox{A} & \fbox{E} & \fbox{N} & \fbox{M} & \fbox{I} & \, \\ \fbox{O} & \fbox{A} & \fbox{E} & \fbox{R} & \fbox{O} & \fbox{M} & \fbox{T} & \fbox{B} & \fbox{U} & \fbox{A} & \fbox{R} & \fbox{O} & \fbox{T} \\ \fbox{K} & \fbox{A} & \fbox{E} & \fbox{T} & \, & \fbox{E} & \fbox{C} & \fbox{N} & \fbox{O} & \fbox{D} & \fbox{A} & \fbox{E} & \fbox{R} \\ \fbox{A} & \, & \fbox{E} & \fbox{A} & \fbox{P} & \fbox{T} & \fbox{H} & \fbox{A} & \fbox{N} & \, & \fbox{T} & \fbox{O} & \fbox{P} \\ \fbox{T} & \fbox{A} & \fbox{F} & \fbox{S} & \fbox{Y} & \fbox{E} & \, & \fbox{O} & \fbox{S} & \fbox{A} & \fbox{G} & \fbox{I} & \fbox{O} \\ \fbox{O} & \fbox{H} & \fbox{E} & \fbox{P} & \fbox{R} & \fbox{F} & \fbox{R} & \fbox{O} & \fbox{M} & \fbox{T} & \fbox{H} & \fbox{E} & \fbox{T} \\ \fbox{U} & \fbox{Y} & \fbox{O} & \fbox{N} & \fbox{O} & \fbox{W} & \fbox{E} & \fbox{S} & \fbox{E} & \fbox{A} & \fbox{M} & \fbox{T} & \fbox{I} \\ \fbox{D} & \fbox{N} & \fbox{A} & \color{green}{\fbox{▲}} & \fbox{C} & \fbox{S} & \fbox{P} & \fbox{E} & \fbox{A} & \fbox{E} & \fbox{S} & \fbox{O} & \fbox{G} \\ \end{matrix} $$

You're standing on the platform represented by the green arrow ($\color{green}{▲}$) in the bottom row. To reach the door, and thus freedom, you need to cross the room to the platform on the far side indicated by the blue arrow ($\color{blue}{▲}$). You take a breath to calm yourself and make the following observations:

• The platforms are large, but if necessary, you estimate that you could safely jump completely over one to another, as well as jump diagonally between them, though not completely over one on the diagonal (I.e. from $\color{green}{▲}$, you could safely reach any of the nearby $\text{N}$, $\text{A}$, $\text{C}$, $\text{S}$, $\text{O}$, $\text{N}$, $\text{O}$, and $\text{P}$ platforms - [see addendum at the bottom of this post for an example])


• The platforms are too spaced out to spread your weight between more than one at a time, so when jumping you're going to have to commit yourself fully and assume, based on the haiku, that leaping to any invalid platform, will result in it collapsing, leading to death


• There are some missing platforms in the grid, which can obviously be jumped over if necessary, but are otherwise impassible

What platforms can you safely traverse in order to escape?

You're only going to get one shot at this, so you better be damn sure of your path before you take that first leap...

---

_{For convenience, here's an ascii/csv version of the floor tiles, since MathJax seems to render matrices in columns:}

R,U,Y,E,R,O,▲,L,N,R,A,E,Y,
A,R,D,Y,E,E,R,U,O,F,M,O,T,
C,U,L,A,A,T,Y,L, ,S,A,I, ,
B,T,W,O,T,H,E,N,D,I,I,W,N,
E,E,D,E,O,R,M,F,D,A,D,N,E,
N,L,N,I,R,S,R,A,E,N,M,I, ,
O,A,E,R,O,M,T,B,U,A,R,O,T,
K,A,E,T, ,E,C,N,O,D,A,E,R,
A, ,E,A,P,T,H,A,N, ,T,O,P,
T,A,F,S,Y,E, ,O,S,A,G,I,O,

O,H,E,P,R,F,R,O,M,T,H,E,T,
U,Y,O,N,O,W,E,S,E,A,M,T,I,
D,N,A,▲,C,S,P,E,A,E,S,O,G,

---

Edit: To be 100% clear about "legal" moves (i.e. jumps that you're physically capable of making), here's a diagrammatic version of the explanation:

$$ \begin{matrix} \color{red}{\fbox{A}} & \color{red}{\fbox{B}} & \color{blue}{\fbox{C}} & \color{red}{\fbox{D}} & \color{red}{\fbox{E}} \\ \color{red}{\fbox{F}} & \color{blue}{\fbox{G}} & \color{blue}{\fbox{H}} & \color{blue}{\fbox{I}} & \color{red}{\fbox{J}} \\ \color{blue}{\fbox{K}} & \color{blue}{\fbox{L}} & \color{green}{\fbox{●}} & \color{blue}{\fbox{M}} & \color{blue}{\fbox{N}} \\ \color{red}{\fbox{O}} & \color{blue}{\fbox{P}} & \color{blue}{\fbox{Q}} & \color{blue}{\fbox{R}} & \color{red}{\fbox{S}} \\ \color{red}{\fbox{T}} & \color{red}{\fbox{U}} & \color{blue}{\fbox{V}} & \color{red}{\fbox{W}} & \color{red}{\fbox{X}} \\ \end{matrix} $$

So if you were standing on the green dot ($\color{green}{●}$) in the centre, you would be able to safely jump to any of the blue letters ($\text{C}$, $\text{G}$, $\text{H}$, $\text{I}$, $\text{K}$, $\text{L}$, $\text{M}$, $\text{N}$, $\text{P}$, $\text{Q}$, $\text{R}$ and $\text{V}$). If any of those platforms were missing, you obviously couldn't land there, however, if $\text{H}$, $\text{L}$, $\text{M}$, or $\text{Q}$ were missing, you could still reach $\text{C}$, $\text{K}$, $\text{N}$, or $\text{V}$.

Answer

Final answer

Continuing where Rand left off, we

Read off the letters, moving the value of each letter. So if we hit an A, we go forward 1 to reach the next; if B, we skip one and go to the second, and so on. Starting with the R in the top-left corner and going right (wrapping around back to the left side when we hit the right edge), this gives us:
"REMAINDER ARE ANAGRAMS".

Now if we look at each section delimited by the path, holes, and yellow squares, we can anagram them to get a word. These words form the message: "You're nearly ready to actually win freedom, and nil remains but to take a path to safety, so I hope you see it and escape. Go!"

The phrase "PATH TO SAFETY" spells out a valid way to get to the exit!

thermodynamics - Black hole entropy versus entropy of normal matter

It has been established that the entropy of a black hole is equal to:

$$ \frac{1}{4} \frac{c^3 A}{ G \hbar} k$$

Which if one substitutes for A the surface area of the event horizon:

$$16\pi\frac{G^2M^2}{c^4}$$

One gets for the entropy:

$$8\pi^2\frac{GM^2}{hc}k$$

My question is the following:

Entropy for ordinary matter very roughly scales with the number of particles in the system. Which for stars mostly composed of fully ionized hydrogen is proportional to the mass of the system. However for black holes this seems to scale with the mass squared.

• 
  

  Why is this?

  
  


• 
  

  Does this seem to imply that matter falling into a black hole is heated to such an extreme that more particles are created (from energy gained by gravity or by decomposition of nuclei and nucleons into constituent particles) in such high numbers that entropy seems to have this proportionality to the mass squared?

  
  


• 
  

  Do neutron stars have a similar relationship?

  
  

optics - Why do metals only glow red, yellow and white and not through the full range of the spectrum?

Why don't metals glow from red to yellow to green to blue etc.? Why only red, then yellow and then white? Shouldn't all wavelengths be emitted one by one as the temperature of the metal increases?

If some metals do glow at with different colours, could you give me examples of such metals and the reason why this happens in specific cases?

differential equations - How are Fourier transforms of any dynamical system different to traditional ones?

When projecting a vector in Hilbert space into its (closed?) subspace, its best approximation is its Fourier series. The technique has been using in many traditional problems (heat, wave, Schrödinger) and in other low dimensional dynamical systems by finding $\lambda$ in the characteristic polynomial $det(A-\lambda I)$.

However, in general, are there any differences when applying this to any given dynamical system compared to the traditional ones? Not all systems have nice or symmetrical equations, and they may involve more variables/higher dimensions, and I think it might be stuck to find the characteristic polynomial. Is there ever such a thing, and how to solve it?

Answer

The current version (v3) of the question seem to describe a particular linear approximation to the system.

If that's the case, then

• 
  

  no, there's no difference in the application of the method; and

  
  
  


• 
  

  it's a valid analysis, but with all limitations of local approximations.

  
  

quantum field theory - Does dilation/scale invariance imply conformal invariance?

Why does a quantum field theory invariant under dilations almost always also have to be invariant under proper conformal transformations? To show your favorite dilatation invariant theory is also invariant under proper conformal transformations is seldom straightforward. Integration by parts, introducing Weyl connections and so on and so forth are needed, but yet at the end of the day, it can almost always be done. Why is that?

magnetic fields - Permanent Magnet Lines & Currents:

A permanent magnet produces lines of magnet flux that we call a "magnetic field". Those lines come from inside the magnet, come out of the N pole, loop outside the magnet, & return back into the S pole to complete a magnetic circuit. Using ferromagnetic materials, other magnets, &/or test equipment, we feel & measure forces that are caused by a magnet/s when other materials are placed within close enough proximity to the magnet/s (i.e. within their magnetic field/s). We therefore conclude that the magnetic lines flow in a specific direction & exert definite forces (i.e. magnitudes/strength) on some materials (ferromagnetic & other magnets), depending on where those materials are placed within the magnetic field of a magnet/s.

We label those forces as "H Field" & "B Field".
We measure H in terms of Amperes/meter. We measure B in terms of Newtons/meter/Ampere (or Tesla). We also commonly defined B in terms of the force that it exerts on moving electric charges (i.e. the Lorentz force).

An "Ampere" is basic unit of electrical current in the SI system, which = 1 Coulomb per second--which is formally defined to be the constant current which if maintained in two straight parallel conductors of infinite length, of negligible circular cross section, and placed one meter apart in vacuum, would produce between these conductors a force equal to 2 × 10 −7 newton per meter of length.

A "Coulomb" is the standard unit of quantity of electricity in the SI system, which = the quantity of charge transferred across a conductor in which there is a constant current of 1 Ampere/Second.

Consequently, those "magnetic lines" are currents (or flows) of charges that don't appear to need any external source of energy to continue generating their currents.

We also know those magnetic lines exist inside a vacuum, so they are independent of air molecules to flow outside of a magnet.

So now my questions:

1. 
   

   Exactly what charged particles are flowing outside (and inside) a permanent magnet that create the magnetic "lines"?

   
   


2. 
   

   Do those particles come from something inside the magnet or does the magnet do something outside of it to affect unknown particles to make the lines?

   
   


3. 
   

   If there is a current (i.e. a continuous flow of charged particles), then why don't we harness that current like a water wheel (instead of 'using energy' to rapidly move copper wires through magnetic fields--like we do with electricity generators?) Shouldn't we be able to get the line currents to charge a capacitor (or or other device) & then otherwise discharge the capacitor for the energy that we want?

   
   
   

nuclear physics - Why does an atom remain uncharged after emission of an alpha particle?

When an alpha particle is emitted, two protons and two neutrons leave the nucleus but the electrons remain the same in number. Why does the atom remain uncharged although it appears it should have a net charge of $-2e$?

quantum field theory - Several stationary points of the action functional

In QFT the principle of stationary action states that we choose fields that will make the action stationary but what if the action has many stationary points (for a fixed choice of boundary conditions)? What's the significance of these other solutions?

enigmatic puzzle - Five-Minute Comics: Part 1

_{Part of the Fortnightly Topic Challenge #35: Restricted Title 1}

---

Puzzles made in 5 minutes or less.

brought to you by having like an hour, but too committed to break my streak now

Answer

These are all clues with the subscripts enumerating the answers:
-- / -- BE
C / O / AC
A / BED / D
There are two of each letter, A-E. Read these as flag semaphore to get MERGE.

The three categories are "words that can come before BOARD", "words that can come before CRAFT" , and "words that can come before TEXT". The middle word is HOVER.

This is a cryptic clue: "Trump supporter, debatably" is PUTIN, and switching sides makes that into INPUT (with definition "gave advice to").

The shapes of these objects are letters spelling out MOUSE.

This is a rebus: MINER + VENTS - NERVE = MINTS.

A LARGE INTESTINE is a "colon", and an ERA is a "period" (the bottom half of a colon, as in punctuation marks). The second part is a "semicolon" since it's half of the clue for colon, so the corresponding word is COMMA.

Each puzzle has five numbers below it. Match each five-letter answer to five numbers, then read off the numbers in order from 1 to 30 to get "UMMM SOME CREATIVE PUN OR SOMETHING".

Steve's crossword

^{Congrats! Niranj Patel for finding the answer of Steve makes a riddle!}

$\bullet$ Yesterday Bangladesh won against South Africa by 21 runs.

Here's a crossword made by Steve. Fill it with the names of Cricket players.

ACROSS

2 Mr. Fantastic
7 Hitman
12 Kung fu
13 Watto
14 The Champion and Music
15 Superman

DOWN

1 Steady the Ship
3 The Slinga
4 Express + Leg-spin
5 Boom Boom !!!
6 Enigma
8 The Boss
9 The Big Show
10 The Untold Story
11 New Balance and me...
16 Moggie

Answer

ACROSS

2 Mr. Fantastic

Brendon McCullum

7 Hitman

Rohit Sharma

12 Kung fu

Hardik Pandya

13 Watto

Shane Waston

14 The Champion and Music

DJ Bravo

15 Superman

AB DE Villiers

DOWN

1 Steady the Ship

Kane Williamson

3 The Slinga

Lasith Malinga

4 Express + Leg-spin

Imran Tahir

5 Boom Boom !!!

Shahid Afridi

6 Enigma

Faf Du Plessis : Using IXU bat

8 The Boss

Chris Gayle

9 The Big Show

Glenn Maxwell

10 The Untold Story

M S Dhoni

11 New Balance and me...

Steve smith : Using NB bat

16 Moggie

Eoin Morgan

homework and exercises - What Exactly is a Shock Wave?

The Wikipedia defintion of a shock wave pretty much sums up all I've found online about what a shock wave is:

A shock wave is a type of propagating disturbance. Like an ordinary wave, it carries energy and can propagate through a medium (solid, liquid, gas or plasma) or in some cases in the absence of a material medium, through a field such as an electromagnetic field. Shock waves are characterized by an abrupt, nearly discontinuous change in the characteristics of the medium. Across a shock there is always an extremely rapid rise in pressure, temperature and density of the flow.... A shock wave travels through most media at a higher speed than an ordinary wave.

To me, however, this doesn't seem to provide a very rigorous definition that would allow me to look at a bunch of propagating disturbances and be able to clearly classify it as being a shock wave or (as Wikipedia puts it) a "normal" wave. Although this definition provides a qualitative definition of what sets a shock wave apart from a normal wave, I am wondering if there is a definite difference between a shock wave and normal waves that would allow me to definitively classify a wave as one or the other or if there is a continuous spectrum of wave properties between normal waves and shock waves with no clear boundary between the two (like the electromagnetic spectrum, with only arbitrary boundaries being drawn between the various classes of EM waves).

Answer

The first thing that distinguishes a shock wave from an "ordinary" wave is that the initial disturbance in the medium that causes a shock wave is always traveling at a velocity greater than the phase velocity of sound (or light) in the medium. Notice that I said light - that is because there is also a kind of electromagnetic analogue to a shock wave known as Cherenkov radiation (Wikipedia article is here )that is created when a charged particle travels through a medium at a velocity faster than that of the phase velocity of light in the medium (which for many media is some fraction of c).

So getting back to acoustic waves in a gas, the main characteristic that divides a shock wave from an ordinary wave is the thermodynamics of the changes in pressure and temperature due to the wave. For ordinary waves (disturbance less than the phase velocity of sound), the compression and rarefaction of the gas does not entail a change in entropy of the gas - thus an ordinary wave is a reversible process thermodynamically speaking.

For shock waves, this is not the case. The process of compression and rarefaction caused by a shock wave is an irreversible process - it leads to a change in entropy of the gas.

Why is this the case ? Without going too deep into the mathematics, it relates to your question as to how distinct shock waves are from ordinary sound waves. The zone of discontinuity is quite sharp between the disturbance and the shock waves, and the changes in pressure, temperature, and density are large enough that dissipative effects like heat transfer and gas friction come into play.

The boundary conditions involved in analyzing shock waves are known as the Rankine-Hugoniot conditions. The Wikipedia article on Rankine-Hugoniot conditions is actually more detailed about explaining shock waves than the Wikipedia article on shock waves itself.

homework and exercises - Divergence of $frac{hat{r}}{r^2}$

In David J. Griffiths's Introduction to Electrodynamics, the author gave the following problem in an exercise.

Sketch the vector function $$ \vec{v} ~=~ \frac{\hat{r}}{r^2}, $$ and compute its divergence, where $$\hat{r}~:=~ \frac{\vec{r}}{r} , \qquad r~:=~|\vec{r}|.$$ The answer may surprise you. Can you explain it?

I found the divergence of this function as $$ \frac{1}{x^2+y^2+z^2} $$ Please tell me what is the surprising thing here.

Answer

Pretty sure the question is about $\frac{\hat{r}}{r^2}$, i.e. the electric field around a point charge. Naively the divergence is zero, but properly taking into account the singularity at the origin gives a delta-distribution.

electrostatics - Difference between electric field $mathbf E$ and electric displacement field $mathbf D$

$$\mathbf D = \varepsilon \mathbf E$$ I don't understand the difference between $\mathbf D$ and $\mathbf E$. When I have a plate capacitor, a different medium inside will change $\mathbf D$, right? $\mathbf E$ is only dependent from the charges right?

Answer

$\mathbf E$ is the fundamental field in Maxwell equations, so it depends on all charges. But materials have lots of internal charges you usually don't care about. You can get rid of them by introducing polarization $\mathbf P$ (which is the material's response to the applied $\mathbf E$ field). Then you can subtract the effect of internal charges and you'll obtain equations just for free charges. These equations will look just like the original Maxwell equations but with $\mathbf E$ replaced by $\mathbf D$ and charges by just free charges. Similar arguments hold for currents and magnetic fields.

With this in mind, you see that you need to take $\mathbf D$ in your example because $\mathbf E$ is sensitive also to the polarized charges inside the medium (about which you don't know anything). So the $\mathbf E$ field inside will be $\varepsilon$ times that for the conductor in vacuum.

quantum electrodynamics - Most trivial neutral pion decay

Literature states neutral pion decay by QED cannot occur directly because the pion is uncharged.

However, I cannot see why Photons are not allowed to interact with the Quarks directly.

Please elaborate which statement is false and why.

wordplay - Venetian word pairs (i.e., sectioned word reversals)

It's fun to think about reverse word pairs. For example, STRESSED spelled backward yields DESSERTS. Or, my personal favorite: EVIAN spelled backward is NAIVE!

Reverse pairs have a long history and are well documented. Let's explore a more sophisticated reversal mechanism.

We are going to reverse words in sections:

1.  Take a word
2.  Break the word into sections
3.  Reverse each section
4.  Recombine the sections to form a new word

For example, PARROT can become RAPTOR:

1.  PARROT
2.  PAR ROT
3.  RAP TOR

4.  RAPTOR

I'm calling these "Venetian blinds" word pairs because all the sections flipping over simultaneously reminds me of Venetian blinds being turned. Or perhaps a Jacob's ladder toy. Maybe you can come up with a better analogy.

There's no reason to restrict the sections to having the same number of letters. Allowing for sections of different sizes gives us new possibilities. For example, OCELOTS can become COOLEST:

1.  OCELOTS
2.  OC ELO TS
3.  CO OLE ST
4.  COOLEST

Some word pairs behave in surprising ways. For example, NAMELESS can become MANELESS:

1.  NAMELESS
2.  NAM ELE SS
3.  MAN ELE SS
4.  MANELESS

Other words contain enough internal palindromes to yield themselves again, given the appropriate sectioning. For example, DIFFIDENTNESS:

1.  DIFFIDENTNESS
2.  DIFFID ENTNE SS
3.  DIFFID ENTNE SS
4.  DIFFIDENTNESS

In order to keep it fun, here are the rules:

• There must be more than one section


• Each section must have more than one letter


• Each section must participate in the reversal

Word pairs involving long and common words are preferred over word pairs involving short or obscure words.

There are many such "Venetian blinds" word pairs. Below are some whimsical hints pointing you to one or both words in a pair.

Your _____ in judgment was bad, but _____ in comparison to his!

These little flies are giving me anxiety!

Who has ____ to wait in the checkout line when you only have one ____?

Topmost space in a house

Name of one of your fingers

What does an active volcano do?

The elements of music

What does Archie Bunker shout at Edith?

The driveway was paved a while ago, but it needed it again

It's unlikely that anyone feels this strongly about their supervisors!

See more exploration of Venetian word pairs in part 2.

Answer

Rest of the spoiler text, besides what @ExcitedRaichu already filled in:

These little flies are giving me anxiety! GNATS -> ANGST
Topmost space in a house: ATTIC -> TACIT
Name of one of your fingers: INDEX -> NIXED
What does an active volcano do? ERUPTS -> PUREST

The elements of music: NOTES -> ONSET
What does Archie Bunker shout at Edith? STIFLE -> ITSELF
The driveway was paved a while ago, but it needed it again: REPAVED -> PERVADE
It's unlikely that anyone feels this strongly about their supervisors! BOSSES -> OBSESS

That was a fun puzzle by itself! For the challenge of coming up with the longest/least obscure Venetian Blind, I'll leave it to someone smarter :-)