quantum field theory - Propagator of gauge boson

Propagator for W boson in unitary gauge is given by $\left(-g_{\mu \nu}+\frac{q_{\mu}q_{\nu}}{m_w^2}\right)$ which can be written as $$ \left(-g_{\mu \nu}+\frac{q_{\mu}q_{\nu}}{q^2}-\frac{q_{\mu}q_{\nu}}{q^2}(1-\frac{q^2}{m_W^2})\right)$$ In a paper (1402.2787) I read that the first half of the propagator i.e, $\left( -g_{\mu \nu}+\frac{q_{\mu}q_{\nu}}{q^2}\right)$ is identified as Spin-1 part and remaining part corresponds to spin-$0$ contribution.

I understand that the first half part is transverse in nature as it vanishes when contacted with $q_{\mu}$ while other half part is longitudinal but I am not able to relate them by spin-$0$ and spin-$1$ combinations. Please explain.

Answer

The propagator of an arbitrary vector field is [ref.1] \begin{equation} \langle A_\mu A_\nu\rangle=\frac{-\eta_{\mu\nu}+p_\mu p_\nu/m_1^2}{p^2-m_1^2}-\frac{p_\mu p_\nu/m_1^2}{p^2-m_0^2}\tag1 \end{equation} for a pair of masses $m_0,m_1$. This propagator is usually called the Stückelberg propagator, and $A_\mu$ a Stückelberg field.

The Stückelberg field corresponds to an irreducible representation of the Lorentz group, but it corresponds to a reducible representation of the orthogonal group, as given by the decomposition $A=(j=0)\oplus(j=1)$. In other words, a typical Stückelberg field creates both spin $j=0$ and $j=1$ particles.

This can easily be seen in the propagator itself: it has two poles, at $p^2=m_0^2$ and $p^2=m_1^2$, which means that $A_\mu$ creates two particles, with masses $m_0$ and $m_1$. Therefore, the correct decomposition is as follows: \begin{equation} \underbrace{\frac{-\eta_{\mu\nu}+p_\mu p_\nu/m_1^2}{p^2-m_1^2}}_{\text{spin $j=1$}}-\underbrace{\frac{p_\mu p_\nu/m_1^2}{p^2-m_0^2}}_{\text{spin $j=0$}}\tag2 \end{equation}

In other words, the paper is wrong. The propagator of the $W$ boson is already the propagator of a spin $j=1$ field, and there is no need to decompose it any further: it is already irreducible as it stands: \begin{equation} \langle W_\mu W_\nu\rangle=\frac{-\eta_{\mu\nu}+p_\mu p_\nu/m_W^2}{p^2-m_W^2}\equiv\text{spin $j=1$ propagator}\tag3 \end{equation}

The structure \begin{equation} \frac{q_{\mu}q_{\nu}}{q^2}\left(1-\frac{q^2}{m_W^2}\right)\tag4 \end{equation} is not identified with a spin $j=0$ particle. Indeed, a scalar particle has \begin{equation} \langle \phi \phi\rangle=\frac{1}{p^2-m_0^2}\tag5 \end{equation} or, upon taking two derivatives, \begin{equation} \langle \partial_\mu\phi \partial_\nu\phi\rangle=\frac{p_\mu p_\nu}{p^2-m_0^2}\tag6 \end{equation} in agreement with $(2)$ (the factor of $1/m_1^2$ is due to the fact that $\partial_\mu\phi$, the scalar part of $A_\mu$, is not typically canonically normalised).

Similarly, the structure \begin{equation} \left( -g_{\mu \nu}+\frac{q_{\mu}q_{\nu}}{q^2}\right)\tag7 \end{equation} is not identified with a spin $j=1$ particle. Indeed, if we take the limit $m_0,m_1\to 0$ of the Stückelberg propagator while keeping $\xi=m_0^2/m_1^2$ fixed, we get \begin{equation} \langle A_\mu A_\nu\rangle\to \frac{-\eta_{\mu\nu}+(1-\xi)p_\mu p_\nu/p^2}{p^2}\tag8 \end{equation} where the spin $j=1$ and $j=0$ particles have mixed into a single term. The structure mentioned in the paper is obtained by further taking $\xi=0$, known as the Landau gauge. This structure is clearly not the structure of a pure spin $j=1$ particle, but it contains a spin $j=0$ part.

To sum up: the structures the paper claims to correspond to spin $j=0$ and $j=1$ are incorrect. The correct structures are those given by $(2)$.

References

1. Quantum Field Theory, by Itzykson and Zuber.

relativity - Degrees of freedom in the infinite momentum frame

Lenny Susskind explains in this video at about 40min, as an extended object (for example a relativistic string) is boosted to the infinite momentum frame (sometimes called light cone frame), it has no non-relativistic degrees of freedom in the boost direction. Instead, these degrees of freedom are completely determined by the (non-relativistic) motions in the plane perpendicular to the boost direction.

I dont see why this is, so can somebody explain to me how the degrees of freedom are described in this infinite momentum frame?

Answer

Without seeing the quote/context I can only imagine that it means something like: if you take, say, a cube moving at close to c in the z direction, then (in the frame in which it's moving) its z extent gets Lorentz contracted to virtually zero, so it is effectively now a square in the xy plane and has only the degrees of freedom that a square in the xy plane has.

grid deduction - Help! Can you get my puzzle back?

Puzzling has a lot of really complicated stuff these days, so I just wanted to post a simple Boggle grid.

You know that game, right? You form words by starting from one letter, then taking adjacent letters moving in any of the eight directions, but you can't use the same letter twice to form a single word.

Here are the words you are looking for, they should be familiar to most of you:

time termination tetris nim animals real tangram pattern pangram game anagram flags nonogram

And here is the grid:

Um... that was not what the grid was supposed to look like at all! Have I got one of those newfangled viruses or something?

Since we're already here, can you help me recover my puzzle? I didn't take any backups while I was making this, and I'm really desperate to remember what it was like!

---

^{Luckily, the virus or whatever it was seems to have picked colorblind-friendly colors for the image, but if that somehow is still an issue here is a text transcription, in which different letters represent different colors:}

ABCBB

CDEFG
HGABI
JKGGC
LMFMF

Answer

The answer:

rtitt
imena
sartp

lgaai
fonon

Explanation:

Here's the initial grid:

ABCBB
CDEFG
HGABI
JKGGC
LMFMF

Start by counting the number of occurrences of each colour (which I'll represent using capital letters):

• A×2, B×4, C×3, D×1, E×1, F×3, G×4, H×1, I×1, J×1, K×1, L×1, M×2

and the minimum number of times we know each letter occurs in the final grid (I'll use lowercase letters for the letters in the eventual grid):

• a×3 (anagram), i×2 (termination), n×2 (termination), o×2 (nonogram), t×2 (pattern), other letters could appear once

There are a total of thirteen letters in the tags (aefgilmnoptrs), and thirteen colours on the grid, so there must be a 1-to-1 correspondence.

I started off by looking at t, o, and n. t must clearly be B or G, as no other colour allows for the adjacent ts in pattern. Looking at all the places we could possibly spell nono (from nonogram), they must either be in B and C (some way round), or else two of F/G/M.

The next letter to check was a. The requirement to spell anagram means we can't have a=C (which would force n into B with a as C, or into D which isn't a valid location for it). We also can't have a=F, as that quickly leads to a contradiction:

ABgBB    ABgBB
gDmao    gDman
HoABr or HnABr
JKoog    JKnng
Lnana    Loaoa

In other words, we've found our first letter position, a=G (immediately forcing t=B to spell pattern).

AtCtt

CDEFa
HaAtI
JKaaC
LMFMF

no now must be MF (one way round or the other). This means that p=I:

AtCtt
CDEFa
HaAtp
JKaaC
LMFMF

If M=n, then we can't spell pangram. Thus, F=n and M=o:

AtCtt
CDEna
HaAtp
JKaaC
Lonon

To be able to spell nonogram, anagram, and pangram, we need K=g:

AtCtt
CDEna
HaAtp

JgaaC
Lonon

To spell termination, we need C=i:

Atitt
iDEna
HaAtp
Jgaai
Lonon

and ADE need to spell erm in some order (for termination); to be able to fit tetris in too we need A=r, E=e, H=s:

rtitt

imena
sartp
Jgaai
Lonon

animals requires J=l, flags requires L=f:

rtitt
imena
sartp
lgaai
fonon

We can find all the words in this grid, so we know it's correct.

thermodynamics - How to find the power required to maintain object temperature in a different-temperature environment?

For example, if I'd like to maintain a 1kg sphere of water at 50$\unicode{xb0}$C in a room full of air at 25$\unicode{xb0}$C, how much power must I put into the water as heat? Assume the room is large compared to the water.

I've found explanations and calculators for some specific cases. Most of what I'm finding has to do with the energy transfered as the temperature reaches equilibrium, but I don't know how to transfer that information to this question. I'm not even sure what search terms to use in a case like this where energy is being constantly put into the system.

I'd like to get enough of an understanding to understand the cases in which:

• The materials are different (e.g. steel ball in water, instead of water in air)


• There is a third material at the boundary (e.g. water in a rubber membrane in air)


• The shape is different (e.g. a planar interface, or an arbitrary shaped object)

classical mechanics - What is Maupertuis' principle good for?

The strength of Hamilton's principle is obvious to me and I see the advantage. Now, for conservative systems we also have Maupertuis' principle that says:

$$ \delta \int p dq =0$$

and I am not sure how to derive an equation of motion from this? Is this of any use in practical computations? So, can one apply this principle for example to the harmonic oscillator?- I have never seen anybody using it.

Further, I read in Goldstein's classical Mechanics that the variation in Maupertuis' principle is not the one in Hamilton's principle, since we have constant Hamiltonian and changing time, whereas Hamilton's principle has constant time and varying Hamiltonian (in general).

I am a little bit wondering about this, since you could easily get Maupertuis' principle from Hamilton's principle: $$ \delta \int L dt = \delta \int p \dot{q} - H dt = \delta \int p \dot{q} dt = \delta \int p dq =0,$$ if $H$ is constant. Can anybody here explain to me, why we have to use a different variation and how one can use this principle?

Answer

In fact, Maupertuis or Euler principles of least action are historically the first formulation of a least action principle, but one have to wait Lagrange and Hamilton to have a modern version, with the so-called Euler-Lagrange equations, which allow us to obtain the equations of movement.

If I am not mistaken, you cannot deduce directly the equations of movement from the Maupertuis/Euler principles. The problem I see is that you cannot know the dependence of the potential energy $V$ in $x$, in seeing only the kinetic energy $T$.

Now, as you state, but written differently, for a movement with conservative energy, one may see that the variation of the Maupertuis' action is equivalent to the variation of the Lagrange/Hamilton action, for instance, starting with Maupertuis' principle :

$$ \delta\int (2T) dt = 0$$

We have : $2T = T + (E - V)$, so Maupertuis' principle can be written :

$$ \delta\int (E + (T-V)) dt = 0$$

But $E$ being a constant, this is not useful in the variation, so finally, we have :

$$ \delta\int (T-V) dt = 0$$ which is the usual Lagrange/Hamilton action.

But, to really have the equations of movement, you have to use a functional, that is :

$$ \delta\int L(x,\dot x,t) dt = 0$$ and this gives you, thanks to the Euler/Lagrange equations, the equations of movement.

homework and exercises - Work done changes between reference frames?

(This is not homework; a friend shared with me this puzzler and neither of us can figure it out.) Suppose you are in a plane traveling at velocity $v_1$ relative to the ground. The flight attendent pushes a cart of mass, say, $m$, accelerating it from rest to $v_2$, relative to the plane (so relative to the ground, its velocity goes from $v_1$ to $v_1+v_2$). From your perspective, the work done is then $$W = \Delta E = \dfrac{1}{2}mv_2^2$$ but from the perspective of the ground, the work done is $$W = \Delta E = \dfrac{1}{2}m(v_1+v_2)^2 - \dfrac{1}{2}mv_1^2 = \dfrac{1}{2}mv_2^2 + mv_1v_2.$$ Why are these quantities unequal?

I've seen some similar problems, like a body colliding with a wall in different reference frames, but I don't have the physics experience to tell if that question is analogous to this one. So, my apologies if this is a duplicate. :)

Answer

There is a subtlety involved here. Let's phrase the problem this way: Say that the flight attendant has an amount of energy E to accelerate his cart to a speed $v_2$ with respect to the plane. If he does that while the plane is sitting on the ground, the total energy of the cart is $\frac{\text{mv}_2^2}{2}$. If he does that when the plane is in the air flying at speed $v_1$, then the total energy of the cart is $\frac{\text{mv}_2^2}{2}+\text{mv}_1 v_2$ which is greater than the total energy of the cart in the first case even though the attendant did the same amount of work in both cases. So we seem to have a paradox.

The key to unraveling the paradox is to recognize that when the flight attendant pushes the cart forward, he is pushing the rest of the plane backwards by some small velocity. That small decrease in the plane's velocity would seem to be so infinitesimal that it could be ignored, but when you consider the fact that the kinetic energy of the plane is proportional to its (very large) mass, even a small decrease in the plane's velocity can play a significant role here when considering the changes in energy. You'll find that the decrease in the kinetic energy of the plane cancels out the problematic $\text{mv}_1 v_2$ term, so that the total energy imparted to the entire system (plane+cart) by the attendant is the same in both cases.

Or, another way to look at it is if you require that the plane maintains the exact same velocity $v_1$ despite the fact that the flight attendant suddenly decides to push his cart forward at a velocity $v_2$, then the plane's engines are required to work a little harder briefly in order to maintain velocity $v_1$ against the force exerted by the flight attendant's feet in trying to push the rest of the plane slightly backwards when he pushes the cart forward. That additional energy provided by the plane's engines works out to be $\text{mv}_1 v_2$ which is the additional energy that you were wondering about in your originally stated problem.

Try working out the problem yourself in full and taking into account that the plane has some large (but finite) mass M and the cart has a smaller mass m. Assume that the plane has some high velocity $v_1$ and that an amount of energy is suddenly provided which propels the cart forward at a velocity $v_2$ with respect to the cart+plane center of mass. Using conservation of momentum, see how much the plane slows down as a result. Then calculate the total before and after energies and see how they compare.

perturbation theory - Perturbative Quantum Mechanics

I am, in full generality, confused about perturbation theory in quantum mechanics.

My textbook and Wikipedia have the same general approach to explaining it: given some Hamiltonian $H=H^{(0)} + H^\prime$, we can break down each eigenfunction $\left\vert n \right\rangle$ into a power series in an invented constant $\lambda$ and the eigenenergies likewise:

$\left\vert n \right\rangle = \sum\lambda^i\left\vert n^{(i)}\right\rangle$

$E_n = \sum \lambda^i E_n^{(i)}$

$\left(H^{(0)} + H^\prime\right) \left(\left\vert n^{(0)}\right\rangle + \lambda \left\vert n^{(0)}\right\rangle + \cdots \right) = \left(E^{(0)}+ \lambda E^{(1)} + \cdots\right) \left(\left\vert n^{(0)}\right\rangle + \lambda \left\vert n^{(1)}\right\rangle + \cdots \right)$

... and then they take $\lambda\to1$.

My question is - what's the logic here? Where did this come from? What purpose does $\lambda$ serve, given that the actual size of each contribution will be determined by the $E^{(i)}$'s and $\left\vert n^{(i)}\right\rangle$'s?

Answer

Firstly, I refer you to Prof. Binney's textbook (see below) which covers perturbation theory in quantum mechanics in explicit detail. When doing perturbation theory, we perturb the Hamiltonian $H^{(0)}$ of a system which has been solved analytically, i.e. the eigenstates and eigenvalues are known. Specifically,

$$H^{(0)}\to H^{(0)} + \lambda H'$$

where $H'$ is the perturbation, and $\lambda$ is a coupling constant. Why include such a constant? As Binney says, it provides us a 'slider' which when gradually increased to unity increases the strength of the perturbation. When $\lambda = 0$, the system is unperturbed, and when $\lambda=1$ we 'fully perturb the system.'

Introducing a coupling constant $\lambda$ also provides us with a manner to refer to a particular order of perturbation theory; $\mathcal{O}(\lambda)$ is first order, $\mathcal{O}(\lambda^2)$ is second order, etc. As we increase in powers of the coupling constant, we hope the corrections decrease. (The series may not even converge.)

A caveat: the demand that a coupling $\lambda \ll1$ may not be sufficient or correct to ensure that the coupling is small; this is only the case when the coupling is dimensionless. For example, if the coupling, in units where $c=\hbar=1$, had a mass (or equivalently energy) dimension of $+1$, then to ensure a weak coupling we would need to demand, $\lambda/E \ll 1$, where $E$ had dimensions of energy. Such couplings are known as relevant as at low energies they are high, and at high energies the coupling is low.

classical mechanics - Amplitude-phase decomposition as a canonical transformation

I am studying a classical dynamical system defined on $\mathbb{CP}^2$: the phase space is parametrized in terms of three complex coordinates $\psi_i$ ($i=1,2,3$) and Hamilton's equations of motion take the form,

$$\imath \frac{d\psi_i}{dt} = \frac{\partial H}{\partial \psi_i^*},\quad \imath \frac{d\psi_i^*}{dt} = -\frac{\partial H}{\partial \psi_i}.$$

I would like to make an amplitude-phase decomposition, replacing the three complex coordinates and their conjugates, $\{\psi_i, \psi_i^*\}$, with six real ones $\{n_i, \phi_i\}$, with

$$\psi_i = \sqrt{n_i} \exp(\imath \phi_i)$$

But this transformation appears not to be canonical: instead of the usual,

$$\nabla_\xi \Theta \cdot \Omega \cdot \nabla_\xi \Theta = \Omega,$$

where $\nabla_\xi \Theta$ is the Jacobian of the transformation and $\Omega$ is the symplectic block matrix, I get,

$$\nabla_\xi \Theta \cdot \Omega \cdot \nabla_\xi \Theta = \frac{1}{\imath}\Omega.$$

Is the amplitude-phase decomposition not a canonical transformation? Or did I make a mistake?

I'm sure this is a standard problem, but I am very new to the idea of classical dynamics on complex manifolds and haven't gotten my bearings yet. Any reference suggestions would be welcome!

Answer

For simplicity consider the 1-d case, with $\psi =\sqrt{n} e^{2i\phi}$, then

$$i \psi_t =\frac{i}{2} \frac{\dot{n}}{\sqrt{n}} e^{2i\phi} -\sqrt{n} e^{2i\phi} 2\dot{\phi}.$$

Similarly

$$ \frac{\partial H}{\partial \psi^*} = \frac{\partial H}{\partial n}\frac{\partial n}{\partial \psi^*} + \frac{\partial H}{\partial \phi}\frac{\partial \phi}{\partial \psi^*} = 2\sqrt{n} e^{2i\phi} \frac{\partial H}{\partial n} + \frac{i}{2\sqrt{n}} e^{2i\phi} \frac{\partial H}{\partial \phi}.$$

Equating the real and imaginary parts (with H real), we have

$$\frac{d n}{dt} = \frac{\partial H}{\partial \phi}; \quad \quad \frac{d \phi}{dt} = -\frac{\partial H}{\partial n}.$$

The other governing equation for $d \psi^*/dt$ gives the same information. Hence, $(n,\phi)$ are canonical variables.

$\textbf{EDIT}$: As Ted Pudlik correctly pointed out, the above reasoning is incorrect. Why? Well, it's because I was being sloppy and got bit. Let's try this again.

As usual, we need to work at the order of the action in order to get coherent results.

Consider $$ S = \int i\dot{\psi}\psi - H dt.$$

Hamilton's principle states the dynamics of the system are given when $S$ is stationary, and indeed this yields the set of Hamilton's equations you originally stated.

Next, we consider a different action, $S'$ defined as

$$S' = \int -2n\dot{\phi}- 2H' dt$$

for some undetermined $H'$. Hamilton's principle yields (1) with $H\to H'$.

For $S$ and $S'$ to give the same dynamics, they must differ by a constant, ie

$$S-S' =\int \frac{d f}{dt} \ dt$$

for some function $f$. Now, when we substitute in our two actions, we find

$$ \int -2n\dot{\phi} +\dot{n} -H +2n\dot{\phi} + 2H' \ dt $$

The perfect derivative integrates to 0 (we assume the wave is compact in time) and we are left with the requirement that for the transformation to be canonical, $2H' \equiv H$, as you pointed out in your comment.

tic tac toe - What is the smallest tic-tac-toe board to have a winning strategy?

There is not winning strategy for 9cell 3x3 board. There is a winning strategy for 16cell 4x4 board, or even for 10cell "3x3 + 1_cell_near_a_corner" board (see M.Gardner "Mathematical games"):

$$ \begin{array}{|c|c|} \hline & & & - \\ \hline & & & - \\ \hline \space\space & \space\space & X & \\ \hline \end{array} $$

here "$-$" states for unused cells. "$X$" - for the first winning move.

I wonder is there a board with 9 or less cells where a player has a winning strategy? So the question:

What is the smallest tic-tac-toe board to have a winning strategy for first or second player?

P.S. win means the usual thing: to put 3 signs ("x" or "o") consecutively in a row (horizontally, vertically or diagonally). So one obviously needs at least 5 cells to perform the moves until the winning one.

Answer

6 or fewer cells will never make a board in which either player can force a win.

On a 6 cell board, each player moves exactly thrice. To win, $X$ must pick 3 cells which happen to lie adjacent to each other on a line, meaning that his only chance to win is on move 5. It's a standard strategy-stealing argument to show that $O$ can never force a win on any board (because making a move is always advantageous); either $X$ can force a win or the board is a draw with perfect play.

To be able to win move 5 on a board of any size, he must have at least two potential winning cells after move 3, since $O$ will be able to block one with move 4. But he only has selected two cells after his second move. These two cells determine a line, so both of his winning cells must also be on that line. Hence, he must have picked adjacent cells along the same line.

Now suppose there were only one line of 4 or more cells passing through $X$'s first move. $O$ knows $X$ can only force a win along this line with his second move (if at all). With move 2, she can pick either cell along this line adjacent to move 1. As we've already said, $X$'s only chance with move 3 is to play in the other square adjacent to move 1 along this line. But now $O$ already has one of the (at most) 2 cells $X$ could complete this line with blocked, and she can pick the other one with move 4. So with move 5, $X$ can't win since he's already committed to this line and $O$ has stopped him there.

So now we just have to show that a board with 6 cells or fewer can't possibly have two distinct lines of 4 or more. But this is actually obvious. If it did, by Pidgeonhole principle, those two lines must overlap on at least two cells. But if they overlap on two cells, they're each the line determined by those two cells; hence actually the same line. So 7 cells is required for a board to be a forced win for $X$. The proof also suggests the example given in this answer, which indeed is a forced win for $X$ on move 5. Interestingly, although the board has 7 cells, $X$ still only needs 3 moves to win.

---

You'll note that the fact that two points on a plane uniquely determine a line was used repeatedly here. This is necessary; it's what tells you that the board is on Euclidean space as opposed to something else. You might ask what happens if we loosen this and allow boards on other sorts of spaces. This isn't exactly a precise question, though there are several ways one can make it precise. If you look at tic-tac-toe on a cube (where the faces are cells), for example, in some sense the "diagonal" lines correspond to picking 3 vertices which meet at a corner. So in fact, any collection of 3 faces of a cube are adjacent along a "line" of some sort. Thus, for either the full cube or a subset of 5 faces, $X$ automatically wins turn 5 regardless of the moves chosen by either player.

optics - How does infrared light 'erase' phosphorescence on zinc sulfide?

I found some sheets of zinc sulfide in my basement that phosphoresce green for up to 24 hours or so after exposure to bright light in the violet range or shorter. One of the first things I tried was drawing on it with a violet laser pointer (405nm, 5mW) and as expected it draws bright lines. What I found more surprising was that I can 'erase' the phosphorescent lines if I focus my green laser onto them. I've determined with some color filters that it's infrared at 1064nm leaking from the green laser causing the erasure. This erasure does not appear to damage the panel in any way, once erased the erased spot can be lit back up to full brightness normally.

How is this infrared light erasing the panel, and how effective would other infrared wavelengths be at doing this?

Answer

Thanks to @Manishearth for the edit

In normal phosphorescence, the atoms are in a "metastable" state--where electrons are in a higher energy level, but do not immediately come to ground state due to partial stability. The electrons come down slowly, giving rise to the (relatively) long lasting glow.

The IR light frees away the electrons from the shallow metastable trap. It's like the usual chemical reaction, where the IR light provides enough activation energy to overcome the barrier. Basically, the IR light promotes the electrons to a higher, non-metastable "energy level" (virtual state). From here, the electron jumps down nearly immediately--dumping all its energy. So, instead of the electrons trickling down like in normal phosphescence, they all come down in a torrent.

The same experiment was performed by this guy: http://ajp.aapt.org/resource/1/ajpias/v29/i3/pxxv_s2

The effect has been explained by the assumption that the long-wave radiation frees electrons from shallow trapping centers into which they have fallen after being excited to the conduction band by the ultraviolet light. The freed electrons recombine quickly with ionized luminous centers with the emmision of light. More recent investigations indicate that probably other processes also are involved.

visible light - Why is not everything transparent?

There is a related question on this site here: Why glass is transparent? Which explains that glass is transparent because the atoms in glass have very large energy differences between energy levels and photons of visible light do not have enough energy to excite electrons from one energy level to another. Whereas, electrons in atoms of most other substances can be excited so the photon is absorbed. But my question is, why don't these excited electrons return to their original energy level and release a photon in the direction the original photon was travelling, hence allowing the light to pass through the object? Edit: I had not realised earlier that this exact same question had been asked before on this site here: Why aren't all objects transparent? So, I shall clarify my question a bit more. The answers to the linked question say that the energy of the excited electron is lost so the light is re emitted as waves with longer wavelengths which we cannot see. I'd like to know how exactly the electron loses this energy. One answer to the linked question states that the energy is lost to lattice vibrations, but I'd like to know how exactly an excited electron still bound to the atom can transfer its energy to lattice vibrations.

Answer

When an atom or molecule absorbs a photon, it enters an excited state; each excited state has a mean lifetime.

When the atom or molecule returns to the ground state it may emit a phonon (vibrations), or it may decay through multiple levels; in this case there are multiple photons, with different wavelengths.

In the case where the absorbed and emitted photons have the same wavelength, the new photon is emitted at a random time and random direction.

So there are four things going on that break up the image: loss of photons which are transformed into vibrations (heat), or which change color (wavelength) or become invisible (infrared), delays in timing which breaks up the coherence of the image (similar to a wavy mirror or water), and random directions.

The last, the random directions, rapidly destroys the intensity of the transmitted image, generating a random background.

For those curious as how a transparent medium transmits an image, and why light slows down inside (but resumes speed when it leaves), I've repeated my previous answer to this question:

Transparent materials (glass, air) transmit images; if the image is distorted or indistinct, we know that the material is altering the coherence of the optical information. That is, what started out at the beginning has not arrived all at the same time. With enough distortion the image is completely lost.

So what is required for a transparent medium to successfully transmit an image? Since light is a physical wave, the transparent medium must preserve the coherence of the phase information of the light. In a typical glass the phase front is slightly slowed while traveling through the glass; this slowing is encoded in the index of refraction, $n = c/v$.

If the material absorbs some frequencies, the material will appear to be colored; a photon that is absorbed (depending on the energy level structure) can be re-emitted, but this will be at (a) a random time later, and (b) in a random direction. No image for this color! There is an exception: stimulated emission, which is the key to building a laser. But this is not how images are transmitted in a passive material.

The process that transmits images can be summed up as Coherent Forward Scattering: Coherent, because otherwise the image integrity is reduced; Forward, because the image is transmitted in this direction, through the material; and Scattering, the remaining available generalized mechanism at the quantum level.

The result is quite like the Huyghen's wavelet model for light transmission: the photons are the waves that are scattered coherently, and because it is coherent, they are able to interfere both constructively and destructively to maintain the coherence of the overall phase front.

It is the interference that slows the phase velocity through through the material; the individual photons continue to "move" at the speed of light, $c$, but the effective motion of the phase front is slowed.

Richard Feynman devotes some time to this in his lectures on QED: The Strange Theory of Light and Matter

electromagnetism - Does a Faraday cage block outgoing signals?

I see lots of discussions about how a Faraday cage can block EM signals, but almost no one addresses the possible difference between incoming signals versus outgoing signals. It seems to me that those are different situations.

For sake of argument, lets say the cage is a hollow superconducting sphere.

I can see how EM fields outside the sphere could re-distribute the charge on the surface of the sphere in a way that there would be no EM field inside the sphere.

However, in the simple case where a point charge is placed in the center of the sphere, the charge on the surface redistributes in such a way that the EM field outside the sphere is as if the sphere was not there.

Similarly, it seems like if the charge inside the sphere were rapidly moved up and down along the z-axis, it would create an EM wave outside the sphere as if the sphere was not there.

Am I missing something here?

**Edit: I used the term "Faraday Cage" because it is in common use and it makes this discussion easier for others to find. But I am asking about a highly idealized case of a superconducting spherical shell with a charge inside but not connected to the shell. For me it provides a good starting point for understanding real Faraday cages.

By the way: I was once told by a physics professor that this is one of the few things Feynman gets wrong in the Feynman lectures. Which shows how easy it is to get this wrong. Either a physics professor at a major university gets this wrong or Feynman.

Answer

An "RF cage" is commonly used to keep signals IN as well as OUT (see for example the mesh door of a microwave oven.)

The short answer is - reciprocity says "if it works in one direction, it works in the opposite direction". The induced charges on the sphere (in the case of a charge inside it) are just enough to cancel the field outside exactly - because that's how superconductivity works. The E field must be normal to the surface at every point. The charge redistributes to make it so. Any field due to the charge inside is exactly cancelled by the induced charge on the surface. Note - this does require the cage to be grounded: that is especially important at low frequencies (where the wave length of the radiation is long compared to the size of the cage). Grounding ensures there is no net charge on the sphere-plus-contents, and should eliminate your concern.

So yes, a Faraday cage blocks outgoing signals.

condensed matter - How does the notion of topological order relate to the Landau-Ginzburg theory of phase transitions?

As per Landau-Ginzburg (LG) theory, we write down a theory (Hamiltonian) with all possible interactions/operators (in terms of some order parameter) that respects certain symmetries. The ground state (which varies with the tunable couplings/parameters in the theory) might partially break some of the symmetry spontaneously, and the behaviour of correlation functions (observables) depends on the spectrum of fluctuations about the ground state. So, we characterize the ground state and the spectrum of fluctuations by some local order parameter which tells us about the qualitative behaviour of the system (aka "phase").

How does the notion of "topological order" and "quantum states of matter at zero temperature" fit into this picture of matter and phases? I would appreciate if someone could place this in context. What are the observables we use to characterize states/phases? Is it talking about a different understanding of the same phenomena as LG, or does is aim to explain completely different phenomena? If it has a broader scope than LG, then does LG fit into this theory in some manner? Is there some overarching principle here, like my description above for LG theory?

general relativity - Time dilation in a gravitational field and the equivalence principle

A clock near the surface of the earth will run slower than one on the top of the mountain. If the equivalence principal tells us that being at rest in a gravitational field is equivalent to being in an accelerated frame of reference in free space, shouldn't the clock near the earth run increasingly slower than the other clock over time? If those two clocks are considered to be in two different frames of acceleration, the one near the earth will have a greater acceleration than the one on the mountain, and thus over time, their relative velocity will increase over time which will increase the time dilation. Or is this not a proper use of the equivalence principle?

Answer

A clock near the surface of the earth will run slower than one on the top of the mountain.

Rather: the geometric (and kinematic) relations between two (or more) given, distinct, separated clocks must be determined and taken into consideration in order to compare intervals (from any one indication to any other indication) of each clock to each other, on this basis to compare the "proper rates" (or frequencies) at which these given clocks "ran" (or "ticked"), and thus to find out whether they "ran equally", or which "ran slower" than the other (in terms of their individual "proper rates").

If the equivalence principal tells us that being at rest in a gravitational field is equivalent to being in an accelerated frame of reference in free space

Certainly it is possible to consider participants in a flat region which move with constant proper acceleration;
and certainly it is possible to consider pairs of such participants which are rigid to each other (i.e. either one of them finding constant ping duration to the other), just as "the surface of the earth" and "the top of the mountain" are rigid to each other.

Some relevant calculations concerning such geometric/kinematic relations between two participants in a flat region are shown in my answer there.

the one near the earth will have a greater acceleration than the one on the mountain

Yes; as shown in the indicated answer, the two participants which remain rigid to each other in a flat region move at ("slightly") different constant proper accelerations, too;
namely "the top" moving with proper acceleration $ k \, e^{(\frac{-k}{c^2} L )} $,
compared to "the bottom" moving with proper acceleration $k$,
where $\frac{2 L}{c}$ is the constant ping duration of "the bottom" to "the top" and back.

general relativity - Might local gaps in spacetime suffice to leave its infinity and eternality incomplete?

At https://arxiv.org/pdf/1403.1599.pdf, Vilenkin, in 2014, used the Borde-Guth-Vilenkin theorem's premise that the "bouncing" local universes of an expanding or contracting multiverse would necessarily be geodesically incomplete either to the past or to the future, using General Relativity without the one assumption added to it by Einstein and Cartan, which was an assumption that fermions have spatial extent. That extent would be submicroscopic, but nevertheless greater than the Planck length, and the addition of it is said to have left relativity more complicated mathematically, but still able to meet all of its experimental proofs. 

The math and physics notations in Vilenkin's paper are beyond my comprehension, but I'm assuming that they reveal some gap in space, and, consequently, in the time that's integral with it. It appears to me that the totality of that space which is occupied by fermions might correspond to that gap, if there is one.

One factor that might bear on the answer would be the use of "parallel transport" in Einstein-Cartan theory, which seems to be required in circumstances resembling some of those where geodesics, or affine geodesics, are used in GR.

quantum field theory - Representations of the Poincare group

Which type of states carry the irreducible unitary representations of the Poincare group? Multi-particle states or Single-particle states?

Answer

Essentially by definition (due to Wigner), one-particle Hilbert spaces of elementary particles support unitary strongly continuous irreducible representations of Poincaré group.

Conversely, any multi-particle Hilbert space, with either fixed or undefined number of particles either identical or distinguishable, cannot be irreducible under the action of the associated representation of Poincaré group.

Proof. A multi-particle representation is the tensor product of the representations in each factor one-particle subspace. If $P_\mu$ denotes the total four-momentum operator of the system of particles, the bounded unitary operator $e^{i a P_\mu P^\mu}$ ($a\in \mathbb R$) commutes with all the unitary operators of the tensor representations and it is not proportional to the identity operator (as it instead happens for a one-particle space). In view of Schur's lemma the representation cannot be irreducible.

An invariant closed subspace is, evidently, the subspace of state vectors where the squared mass $M^2= P_\mu P^\mu$ assumes values (in the sense of spectral decomposition) inside a fixed interval $[a,b]$.

fluid dynamics - Kinematic Viscosity

How would you define kinematic viscosity? What does it physically represent? Around the Internet I've found it defined as just a ratio, and that's it.

I saw in an answer that I can think of it as "diffusivity of momentum". What does this mean?

Answer

In mechanics, "kinematics" means describing the motion mathematically, so, for example, if the acceleration is known I can integrate to find the velocity and position. "Dynamics" means analyzing motion due to the influence of forces. The two are related through Newton's second law: F = ma (dynamics version) is the same as a = F/m (useful for kinematics).

In both cases, "coefficient of viscosity" refers to the effect on one part of the flow due to neighboring flow with a different speed, i.e. flow shear. That is, if you imagine yourself as one parcel of fluid and the next parcel over is moving in the same direction as you but faster, it will pull you along due to the effect of viscosity. "Dynamic viscosity" gives the force on your parcel (per unit volume), while "kinematic viscosity" gives the acceleration (force per unit mass).

(Diffusivity of momentum is also a valid way to think of it that is equivalent to this, but if you don't have any intuition about diffusivity to begin with that's not very helpful.)

ADDENDUM: Air and water make an interesting comparison: air has much lower dynamic viscosity than water (by a factor of 50), but due to its low density, it has much higher kinematic viscosity (by a factor of 17):

$\mu_\text{air} = 2\times 10^{-5}$ Pa s; $\;\mu_\text{water} = 1\times 10^{-3}$ Pa s

$\nu_\text{air} = 0.17 \text{cm}^2/\text{s}$; $\;\nu_\text{water} = 0.01 \text{cm}^2/\text{s}$

potential - When to use Poisson's equation in electrostatics

When does when elect to use Laplace's equation when dealing with charge distributions. For example, if I had a metallic sphere of radius $R$ and charge $Q$, then

$$\mathbf E = \begin{cases} 0, & \text{for } r < R \\ \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}, & \text{for } r \ge R \end{cases}$$

The potential at all regions of space can be calculated in a straight-forward way then.

For $r \ge R$,

$$V = - \frac{Q}{4 \pi \epsilon_0}\int_{\infty}^r 1/r^2 \ dr$$

And for $r < R$

$$\implies V = 0$$

However, I could've just as easily tried to work it out by saying for $r > R$ and $r:

$$\nabla ^2 V = 0$$

As all charge resides on the surface, and then say that

$$V(r,\theta) = \sum_{l=0}^\infty \left(A_l r^l + \frac{B_l}{r^{l+1}}\right) P_l (\cos{\theta})$$

Wouldn't this be an equally valid way to derive the field? If so, when is it better to derive things this way rather than the way I did with Gauss's Law?

classical mechanics - Why do non-Newtonian fluids go hard when having a sudden force exerted on them?

You can dip your hands into a bowl of non-Newtonian fluid but if you are to punch it, it goes hard all of a sudden and is more like a solid than anything else.

What is it about a non-Newtonian fluid that makes it go hard when having a force suddenly exerted on it? How does it go from being more like a liquid to a solid in such a short amount of time? Does it change its state as soon as the force has made contact with it?

Answer

A quick comment on your terminology. The description "non-Newtonian" just means the stress/flow rate graph is not linear i.e. there isn't a single constant viscosity coefficient. The fluid you describe is what we colloid scientists call "dilatant", and it is certainly non-Newtonian. However there are lots of other non-Newtonian fluids such as tomato ketchup and shampoo that behave in different ways. See Are there good home experiments to get a feel for the behavior of yield-stress liquids? for a related question.

Anyhow, kleingordon has explained why the dilatant effect occurs, but let me try a slightly different approach to the explanation.

Oobleck is a suspension of solid (starch) particles in water. Suppose you had a very dilute suspension i.e. lots of water and a little starch. In this case the spacing between the starch grains is large so the grains can flow around without hitting each other, and the suspension just behaves like water. As you increase the amount of starch the spacing between the grains decreases, until at some point the spacing between the grains becomes less than the size of a grain. At this point, when you try apply a large force to suspension the starch grains bump into each other and lock together to form a framework. The water in the suspension now has to flow through the small pores in the starch grain "framework" and this requires a lot of force. Hence you can stand on the suspension for a moment. If the apply a small force the water/starch grains move slowly and this gives time for the starch grains to slide around between each other so they will flow. This is why the chap in the white shirt could run on the oobleck, but when he stood still he gradually sank.

homework and exercises - Correct formula for Mass Defect / Binding Energy?

I'm a web developer and I have to change an online course. The course teaches Advanced Nuclear Theory. In the 'Mass Defect and Binding Energy' chapter, it has this formula:

$$\Delta M = Z(m_p) + N(m_n) - \frac{A}{Z}(M).$$

In the formula above, the letter 'p' and 'n' are written in subscript, and for the 'AZM' part, the letter A (written in superscript) is above the letter Z (which is in subscript). Above this equation, this is written

1/1(p) + 1/0(n) -> 2/1(D)


mass of proton = 1.007272 u
mass of neutron = 1.008665 u
mass of deuterium nuclear = 2.013553 u

. Below the equation which I wrote above (delta M = Z(mp) + N(mn) - A/Z(M)), the worker / user is asked to fill in the blank:

∆M = __ x __ + __ x __ - __

B.E. of nucleaus = 931.5 x __ MeV

When the neutron and proton combine, the nuclear force pulling them together does __ MeV of work.

Now, can anyone tell me what the correct answers are to the blanks? I'm just a web developer and I have to write a program which checks that the user filled in the correct answers, but I do not know the correct answers. I have some knowledge of physics and I'm guessing mp = mass of proton, mn = mass of neutron and then M = mass of deuterium nucleus, but what about the other blanks?

Answer

On the Wikipedia page for the semi-empirical mass formula (based on the Gamow liquid drop model of nuclei) it basically says that the mass defect $\Delta M$ is given by the difference in the masses of the unbound protons and neutrons $Zm_p+Nm_n$ minus the actual mass of the nucleus, $^A_ZM$ (the rest of the semi-empirical formula is not needed here).

The only other bit of trivia you need to know is that 1 amu of mass is the equivalent of 931.5 MeV energy.

So in order, the blanks should be typeset as $Z,m_p,N,m_n,^A_ZM,\Delta M,$ and 2.22 MeV.

integration - Asymptotic behaviour of the propagator for a scalar field

When discussing causality in Chapter 2 of Peskin & Schroeder a couple of equations giving the asymptotic behaviour of the propagator for a scalar field appear:

$$ \text{If} \,\, x^0-y^0=t, \, \, \mathbf{x-y} = 0 \Rightarrow D (x-y) = \frac{1}{4 \pi^2} \int_{m}^{\infty} dE \sqrt{E^2-m^2} e^{-iEt} \underset{t \to \infty}{\sim} e^{-imt} $$ $$ \text{If} \,\, x^0-y^0=0, \, \, \mathbf{x-y} = \mathbf{r} \Rightarrow D (x-y) = \frac{1}{4 \pi^2 r} \int_{m}^{\infty} d\rho \frac{\rho\, e^{-\rho r}}{\sqrt{\rho^2 - m^2}} \underset{r \to \infty}{\sim} e^{-mr} $$

I can't see how you derive these asymptotic behaviours (I have no problem deriving the integral exact expressions, but then I get stuck). All I could do was to rewrite the first integral as follows:

$$ D (x-y) = \frac{1}{4 \pi^2} \int_{m}^{\infty} dE \sqrt{E^2-m^2} e^{-iEt} = \frac{m}{4 \pi^2 i t} K_1(imt) $$

using this article on modified Bessel functions of the second kind. But checking with Mathematica, this vanishes for $t \to \infty$. For the second integral I don't have any clue, so any help would be more than welcome!

Extra (but related) question:

In the first discussion of the chapter something similar appears

$$ U(t) = \frac{1}{2 \pi^2 | \mathbf{x - x_0} | } \int_{0}^{\infty}dp\,p\, \sin (p | \mathbf{x - x_0} | ) e^{-it \sqrt{p^2 + m^2}} $$

Is this also obtained through a similar procedure?

quantum mechanics - Can an electron absorb a photon even if it exceeds the energy required for the electron to jump up an energy level?

To clarify my question, consider the hydrogen atom. An electron on the ground state needs 10.2 eV to be excited to the second state (first excited state). If a photon with less than 10.2 eV strikes the electron, it will not be absorbed-- okay. My question is: what happens if the photon has an energy larger than 10.2 eV? Is the photon once again not absorbed, or is it partially absorbed?

I know that the same question was posted here (Can an electron jump to a higher energy level if the energy is insufficient or exceeds the $\Delta E$?) but I didn't understand the most upvoted answer, and the only other answer seemed to be conflicting. It was also posted by some other users but again, I didn't understand the answers because I'm really a beginner and they went into (what for me) is complicating QM equations. I've looked in other places too, but the answers are conflicting (some people saying photon does get absorbed, some saying it doesn't, etc).

So yeah, please use very simple language, and thank you for your time (I really appreciate it!) :)

Answer

It depends on how much higher. Electronic transitions are not infinitesimally thin, and they all have a certain linewidth $\Delta f$, i.e. a span of frequencies below and above the central frequency $f_0$ at which the photon can be absorbed. If the photon energy is so high that it falls above the transition linewidth, then it will not be absorbed.

Generally speaking, transition linewidths are rather narrow, at least by human standards, though their precise width depends on the conditions; there are several different mechanisms which underlie them, and those produce widths on different scales.

• 
  

  For all transitions, there is an intrinsic linewidth $\Delta f = 1/\tau$ produced by the lifetime $\tau$ of the excited state, which is basically an instance of the Heisenberg uncertainty principle: you can't specify the frequency of a wave to something sharper than $\Delta f$ if its substrate is changing on timescales of order $\tau$.

  
  

  For transitions like the Lyman alpha line, $\tau$ is in the order of a few nanoseconds, giving a $\Delta f$ of the order of about a gigahertz. Here the central frequency is of the order of $f_0 \sim 1 \:\rm PHz$, which means that the line can look quite narrow: $\Delta f/f_0 \sim 10^{-9}$. (I say "look" because, by atomic precision spectroscopy standards, this is extremely wide. Everyday fare in precision spectroscopy uses lines which are narrower by many orders of magnitude, reaching down to $\Delta f/f_0 \sim 10^{-18}$.)

  
  


• 
  

  There are also external causes for nonzero linewidths, often called "inhomogeneous broadening", from a range of sources such as e.g. collisions between different atoms, or Doppler shifts coming from the different velocities in the thermal distribution of a gas. Unless you do something specific to counteract it, then inhomogeneous broadening will almost certainly dominate, giving you linewidths in the tens or hundreds of GHz or even higher. But again, since the central frequency is of the order of a petahertz, $\Delta f/f_0$ will still be quite small.

  
  

The bottom line is that photons typically do have some leeway in getting absorbed even if they have a photon energy that's a bit higher than the transition, but for most electronic transitions in gaseous atomic samples, that leeway will generally be rather limited. And, if the photon exceeds that leeway, then it won't be absorbed.

variational principle - When can we add a total time derivative of $f(q, dot{q}, t)$ to a Lagrangian?

The other day, I was listening to this lecture on the Lagrangian for a charged particle in an electromagnetic field, and at one point in the video, the lecturer mentions that we can add any total time derivative of a function $f(q, t)$ to the Lagrangian without altering its equations of motion.

This is nothing new to me, and I understand it fully, but shortly afterwards (approximately two minutes after the linked starting point), he goes on to say that you can, in fact, add a total time derivative of a function $f(q, \dot{q}, t)$, given certain conditions. This definitely surprised me, and I would love to know more about it, but the lecturer quickly moves on, so my question is as follows: under what conditions can one add the total time derivative of a function which depends on the particle's generalized velocities in addition to its generalized coordinates and time without affecting the particle's equations of motion?

Answer

I) In general, it is true that if we plug a local Lagrangian

$$\tag{1} L\quad \longrightarrow \quad \tilde{L}~=~L+\frac{df}{dt}$$

modified with a total derivative term into the Euler-Lagrange expression

$$\tag{2} \sum_{n} \left(-\frac{d}{dt}\right)^n \frac{\partial \tilde{L}}{\partial q^{(n)}}~=~\sum_{n} \left(-\frac{d}{dt}\right)^n \frac{\partial L}{\partial q^{(n)}}, $$

it would lead to identically the same Euler-Lagrange expression without any restrictions on $L$ and $f$.

II) The caveat is that the Euler-Lagrange expression (2) is only$^1$ physically legitimate, if it has a physical interpretation as a variational/functional derivative of an action principle. However, existence of a variational/functional derivative is a non-trivial issue, which relies on well-posed boundary conditions for the variational problem. In plain English: Boundary conditions are needed in order to justify integration by parts. See also e.g. my related Phys.SE answers here & here.

III) A Lagrangian $L(q,\dot{q},\ldots, q^{(N)},t)$ of order $N$ leads to equation of motion of order $\leq 2N$. Typically we require the Lagrangian $L(q,\dot{q},t)$ to be of first order $N=1$. See e.g. this and this Phys.SE posts.

IV) Concretely, let us assume that we are given a first-order Lagrangian $L(q,\dot{q},t)$. If one redefines the Lagrangian with a total derivative

$$\tag{3} \tilde{L}(q, \dot{q}, \ddot{q}, t)~=~L(q, \dot{q}, t)+\frac{d}{dt}f(q, \dot{q}, t), $$

where $f(q, \dot{q}, t)$ depends on velocity $\dot{q}$, then the new Lagrangian $\tilde{L}(q, \dot{q}, \ddot{q}, t)$ may also depend on acceleration $\ddot{q}$, i.e. be of higher order.

V) With a higher-order $\tilde{L}(q, \dot{q}, \ddot{q}, t)$, we might have to impose additional boundary conditions in order to derive Euler-Lagrange equations from the principle of a stationary action by use of repeated integrations by parts.

VI) It seems that Prof. V. Balakrishnan in the video has the issues IV and V in mind when he spoke of 'putting further conditions' on the system. Finally, OP may also find this Phys.SE post interesting.

--

$^1$ Here we ignore derivations of Lagrange equations directly from Newton's laws, i.e. without the use of the principle of a stationary action, such as e.g. this Phys.SE post, because they usually don't involve redefinitions (3).

Matlab package: graphical calculus for quantum operations (esp. linear optics)

I need a matlab package that will make my life easier. I have quantum circuits with optical beam splitters, polarizing beam splitters and photodetectors. These circuits are getting very complicated and I am wondering if anyone knows of a matlab package that would make my life easier. A graphical package would be excellent and I suppose any software will do.

Answer

Beyond the qotoolbox, which is showing its age but still very useful, I'm not sure what to suggest. A graphical package does not exist (as far as I'm aware), and is not likely to unless you make it yourself since this not in high demand.

word - Captain Pun's new boat

My good friend Captain Pun has bought a new boat.

"And this is a picture of its sail!" he said, showing me this:

"Solve the nonogram and you'll soon know what type of boat it is!"

Can you solve the two-colour nonogram to help me work out what type of boat he bought?

_{This puzzle can be solved without guessing. Note that cells filled with different colours do not necessarily need to have a gap between them. Numbers marked in bold are red, while those in normal weight are blue.}

Answer

Finished nonogram:

These are

The flags of Finland, Laos, Austria, Georgia, Somalia, Honduras, Iceland and Peru.

Meaning the captain's new boat is literally a

FLAGSHIP, as spelt by the first letters of the country names.

logical deduction - Reverse Puzzling 2

This puzzle was inspired by Dr Xorile's first reverse puzzle! Kudos to him for coming up with such a fun format.

Intrigued, I gave George yet another famous puzzle to ponder over. Once again, I was surprised that he had not seen this before either.

George frowned, paused for a few moments, and then picked up his trusty marker and started drawing a diagram.

My eyes widened in horror as I realized that George would take a long time if he kept up this strategy. I didn't have time for him to complete his diagram.

"Wait, George! Let's... let's try an easier version of the same puzzle. A much easier version."

I gave George the revised puzzle and he eventually came up with the following diagram for it.

"Ah! The answer is obvious now!" George exclaimed. I eyed George warily. I was starting to think he was more of a 'tedious' puzzler than a 'great' one.

What was/were the puzzle(s)?

Answer

This puzzle is the Towers of Hanoi with three disks. A, B, and C are the pegs, and the positions from left to right are the disks; the lines connect positions that are valid to move between.

The initial puzzle is probably the Towers of Hanoi with more disks - the area needed to draw the diagram is quadrupled with every new disk added, so a larger diagram would take up much more space and time.

wordplay - I behave and am armed

Starting with I behave and am armed, do all the following in the correct order to reveal a famous American history quote.

1. Remove one word


2. Remove two letters from a word


3. Replace one word with its anagram


4. Remove one letter from another word

Answer

The answer is

I have a dream!

I behave and am armed
Remove 1 word

I behave am armed

Remove 2 letters from a word

I have am armed

Replace one word with its anagram

I have am dream

Remove one letter from another word

I have a dream

astrophysics - Why does a supernova explode?

This is really bugging me. When you look up some educational text about stars life, this is what you find out:

1. Gravity creates the temperature and pressure to start fusion reactions.


2. The fusion proceeds to heavier and heavier cores ending with iron, which remains in the centre of the star.



3. One moment, all light cores are depleted and the gravity wins over the power of the fusion reactions, now absent.


4. The core of the star collapses into high density object, which may vary depending on the star mass.


5. And the top layers of the star explode.

And I just cannot find clear explanation why. According to what I imagine, the top layers of the star should just fall into the collapsing core.

Is that because of the 3rd Newtons rule?

Or do the stars have some need to end with a cool boom?

Answer

There are lots of possible ways that stars can end their life, even in the subset of cases where the end is violent. Eloff has given an excellent answer, but I wanted to add a few points.

Summary (tl;dr):

You need the right conditions (mass, angular momentum, metallicity, etc) to produce a proto-neutron-star which is able to resist complete collapse to a black-hole. The bounce from hitting that proto-neutron-star surface, and the heating from neutrinos, is what drives the explosion of material. Radioactivity is eventually the source of the light we see from supernovae.

---

The basic picture for producing a supernova from a massive star¹:

1. 
   

   The star burns progressively heavier elements on shorter timescales until producing iron (Fe) on the timescale of seconds.

   
   


2. 
   

   After iron, fusion in the core ceases, and pressure support is lost. Gravity is unhindered, and the star begins dynamical collapse.

   
   



3. 
   

   As the Fe-core contracts, electron-capture begins to convert protons + electrons into neutrons, emitting MeV neutrinos.

   
   


4. 
   

   The Fe-core, now largely composed of neutrons is stabilized to further collapse by neutron degeneracy pressure at nuclear densities.

   
   


5. 
   

   Material further out, which is still collapsing, hits the incredibly hard proto-neutron-star surface - causing a bounce (see video analog): the launch of a powerful shockwave outwards through the star.

   
   


6. 
   
   

   Because the neutrinos produced from electron-capture are so energetic (as dmckee points out), and because the densities are so high - the neutrinos are able to deposit significant amounts of energy into the outer-material, accelerating it beyond escape velocities. This is the supernova explosion.

   
   


7. 
   

   Due to the hot, dense, nucleon-rich nature of the ejecta, r(apid)-process nucleosynthesis produces radioactive Nickel (Ni) and Cobalt (Co).

   
   


8. 
   

   After roughly 10's of days, the expanding supernova ejecta becomes optically thin - allowing the radiation produced by Ni and Co decay to escape - this causes the optical emission we call a supernovae.

   
   

from http://arxiv.org/abs/astro-ph/0612072

Why does a supernova explode?

All massive stars are not believed to produce supernovae when they explode. In the following figure (which is intended to convey the basic idea - but not necessarily the quantitative aspects), regions titled 'direct black-hole formation' are regions of initial mass where the neutron-degeneracy pressure (stage '4' above) is insufficient to halt collapse. The Fe core is massive enough that it continues collapsing until a black-hole is formed, and most of the material further out is rapidly accreted.

The region in this plot between about 8 and 35 solar masses is where the vast majority of observed supernovae are believed to come from.

To answer why supernovae explode: Consider the schematic process outlined above. The reason why some deaths-of-massive-stars explode and others don't, is that you need the right conditions (mass, angular momentum, metallicity, etc) to produce a proto-neutron-star which is able to resist complete collapse. The bounce from hitting that proto-neutron-star surface, and the heating from neutrinos is what drives the explosion of material. Radioactivity is eventually the source of the light we see from supernovae.

from http://rmp.aps.org/abstract/RMP/v74/i4/p1015_1

---

Footnote

1: This discussion is constrained to 'core collapse' supernovae - the collapse of massive stars, observed as type Ib, Ic, and type II supernovae

Additional References

Basically any paper by or with Stan Woosley, e.g.
Woosley & Janka 2006 - The Physics of Core-Collapse Supernovae

Lecture Notes by Dmitry A. Semenov - "Basics of Star Formation and Stellar Nucleosynthesis"

quantum field theory - Why do we say that irreducible representation of Poincare group represents the one-particle state?

Only because

1. 
   

   Rep is unitary, so saves positive-definite norm (for possibility density),

   
   


2. 
   

   Casimir operators of the group have eigenvalues $m^{2}$ and $m^2s(s + 1)$, so characterizes mass and spin, and

   
   


3. 
   
   

   It is the representation of the global group of relativistic symmetry,

   
   

yes?

Answer

First, note that in physics, we consider unitary representations $U$ of the Poincare group acting on the Hilbert space $\mathcal H$ of the theory because we are interested in a precise formulation of the concept of Poincare transformations acting on the quantum mechanical states of the theory as symmetries (since the laws of physics should be inertial frame-invariant); and by Wigner's theorem, we choose these symmetries to be realized by unitary operators. These observations are related to your #1 and #3 and I think they should be kept conceptually distinct from the notion of a state that represents a single particle state.

Second, since such quantum field theories are supposed to allow for the emergence of states of particles, and in particular should account for states in which there is a single elementary particle, we expect that there is some subset $\mathcal H_1$ of the Hilbert space of the theory corresponding to states "containing" a single elementary particle.

Given these observations, let's rephrase your question as follows:

What properties do we expect that the action of the representation $U$ will have when its domain is restricted to the subspace $\mathcal H_1$?

In particular, we would like to justify the following statement

The restriction of the unitary representation $U$ acting on $\mathcal H$ to the single-particle subspace $\mathcal H_1$ is an irreducible representation of the Poincare group acting on $\mathcal H_1$.

This requires justifying two things:

1. The restriction maps $\mathcal H_1$ into itself.


2. The restriction is irreducible.

I think that the justification of the first property is pretty intuitive. If all we are doing is applying a Poincare transformation to the state of the system, namely we are just changing frames, then the number of particles in the state should not change. It would be pretty strange if you were to, for example, boost or rotate from one inertial frame into another and find that there are suddenly more particles in our system.

The irredicibility requirement means that the only invariant subspace of the single particle subspace $\mathcal H_1$ is itself and $\{0\}$. The physical intuition here is that since we are considering a subspace of the Hilbert space in which there is a single elementary particle, expect that there is no non-trivial subspace of $\mathcal H_1$ in which vectors of this subspace are simply "rotated" into one another. If there were, then the particle would not be "elementary" in the sense that the non-trivial invariant subspace would represent the states of some "more elementary" particle. When it really comes down to it, however, I'm not sure if there is some more fundamental justification for why the restriction of $U$ to $\mathcal H_1$ is irreducible aside from the decades of experience we've now had with particle physics and quantum field theory.

Why is the strong nuclear force > electrostatic repulsion?

In a nucleus there is a gravitational force between the nucleons and also electrostatic repulsion between the protons, and since electrostatic repulson >> gravitational attraction, it follows that there must be an additional attractive force acting on the nucleons or else there is nothing stopping them from flying apart.

So if we let the gravitational and electrostatic forces be $g$ and $e$, respectively, and denote the additional attractive force by $x$, then we would need $g+x=e$ (because the attractive and repulsive forces must balance).

This gives $x=e-g

But in my revision guide, it says that the strong nuclear force is more than electrostatic repulsion, which seems counter-intuitive according to the above.

Please explain!

Answer

Consider the Earth-Moon system. They are subject to an attractive force (gravitation) and to no repulsive forces (neglecting solar tides, anyway), yet they stay at a nearly constant distance from one another because of their dynamics.

A a static analysis of this system would prompt us to postulate some repulsive force holding the bodies apart (and you can find it by using a non-inertial frame of reference: it is the centrifugal pseudoforce). The lesson is that static analysis will break when applied to dynamic systems.

You are trying to analyze the nucleus in terms of statics when it is a dynamic system (and moreover a dynamic quantum system). As nuclear particles are confined to a limited region in space they necessarily acquire a larger range of momenta as a consequence of the commuter between positions and momentum (we can wave our hands and say "Heisenberg Uncertainty Principle" if you want a shorter label for this effect).

kinematics - Rotational physics of a playing card

A playing card leaves a dealers hand with some angular velocity. As the card slides across a table the friction of the table causes the rotation to slow. How is the friction coefficient between the card and the table related to the angular acceleration of the card? IE how can I calculate angular acceleration?

Answer

I think this is a good question, but difficult to answer.

Frictional coupling between sliding and spinning motion includes a mathematical treatment of this situation. The solution involves elliptic integrals. The rotating and translating motions are coupled through friction, so that the disk always stops rotating and translating at the same time. The same thing happens with playing cards - the phenomenon is not peculiar to disks. There is a more accessible description in the Focus section of the American Physical Society website.

Calculating the effect of friction on a disk which is translating but not rotating, or a disk which is rotating but not translating, are both relatively straight-forward. But the combination is much more difficult to analyze.

For example, a disk of radius $R$ and mass $M$ translates on a flat horizontal surface with dry friction $\mu$. The friction force is $F=\mu Mg$. The linear deceleration is $\frac{F}{M}=\mu g$.

If the disk is rotating without translating, the friction force creates a torque. $F$ is distributed evenly across the disk but the torque it exerts depends on distance $r$ from the axis of rotation. The area of an annulus of thickness $\delta r$ at radius $r$ is $2\pi r \delta r$, the friction force on it is $\frac{2\pi r \delta r}{\pi R^2}F=\frac{2r\delta r}{R^2}F$ and the torque is $\frac{2r^2\delta r}{R^2}F$. Integrating from $r=0$ to $R$, the torque on the whole disk is $\tau=\frac23RF$. The moment of inertia of the disc is $I=\frac12MR^2$ so the angular deceleration is
$\alpha=\frac{\tau}{I}=\frac{\frac23R\mu Mg}{\frac12MR^2}=\frac{4\mu g}{3R}$.

classical mechanics - How to formulate variational principles (Lagrangian/Hamiltonian) for nonlinear, dissipative or initial value problems?

Although this questions is very much math related, I posted it in Physics since it is related to variational (Lagrangian/Hamiltonian) principles for dynamical systems. If I should migrate this elsewhere, please tell me.

Often times, in graduate and undergraduate courses, we are told that we can only formulate the Lagrangian (and Hamiltonian) for "potential" systems, where in the dynamics satisfy the condition that: $$ m\ddot{\mathbf{x}}=-\nabla V $$ If this is true, we can formulate a functional which is stationary with respect to the system as: $$ F[\mathbf{x}]=\int^{t}_0\left(\frac{1}{2}m\dot{\mathbf{x}}(\tau)^2-V(\mathbf{x}(\tau))\right)\,\text{d}\tau $$

Taking the first variation of this functional yields the dynamics of the system, along with a condition that effectively states that the initial configuration should be similar to the final configuration (variation at the boundaries is zero).

Now, given the functional: $$ F[\mathbf{x}]=\frac{1}{2}[\mathbf{x}^{\text{T}} * D(\mathbf{x})]+\frac{1}{2}[\mathbf{x}^{\text{T}} * \mathbf{Ax}]-\frac{1}{2}\mathbf{x}'(0)\mathbf{x}(t) $$ With $\mathbf{A}$ symmetric and $\mathbf{x}(0)$ being the initial condition, and: $$ [\mathbf{f}^{\text{T}} * \mathbf{g}]=\int^{t}_0 \mathbf{f}^{\text{T}}(t-\tau)\mathbf{g}(\tau)\,\text{d}\tau $$

If we take the first variation and assume only that the initial variation is zero, the functional is stationary with respect to: $$ \frac{d\mathbf{x}(t)}{dt}= \mathbf{Ax}(t) $$

This is a functional derived by Tonti and Gurtin, it represents a variational principle for linear initial value problems with symmetric state matrices and shows, as a proof of concept, that functionals can be derived for non-potential systems, initial value or dissipative systems.

My question is, is it possible to derive these functionals for arbitrary nonlinear systems which do not have similar initial and final configurations (and cannot have similar initial and final configurations due to dissipation)?

What sorts of conditions would exists on the dynamics of these systems?

In this example, $\mathbf{A}$ must be symmetric which already implies all of it's eigenvalues are real and thus it is a non-potential system, but there is still a functional which can be derived for it.

Any related sources, information, or answers regarding specific cases would be appreciated. If anyone needs clarification, or a proof of any result I presented here, let me know.

Edit: Also, a related question anyone seeing this: I'm currently just interested in the abstract aspect of the problem (solving/investigating it for the sake of it), but why are functional representations such as these useful? I know there are some numerical application, but if I have a functional which attains a minimum for a certain system, what can I do with it?

Answer

Comments to the question (v3):

I) The Gurtin-Tonti bi-local method [which OP mentions in an example; see also Section II below] of pairing opposite times $t\leftrightarrow (t_f-t_i)-t$ (hidden inside a convolution) is an artificial trick from a fundamental physics point of view, unless further justified. Why would such correlations into the past/future take place?

In fact, it may have non-local quantum mechanical consequences if such non-local action is supposed to be used in a path integral formalism.

Also the Gurtin-Tonti convolution method does not work for a non-compact time interval $[t_i,t_f]$, i.e. if $t_i=-\infty$ or $t_f=\infty$.

Most fundamental physics models typically obey locality, but there are various non-local proposals on the market.

The question of whether a certain set of equations of motions $E_i(t)$ has an action principle (or not!) can be very difficult to answer, and is often an active research area, cf. e.g. this Phys.SE post.

Also what constitutes an acceptable action principle? E.g. can we just introduce some Lagrange multipliers $\lambda^{i}(t)$ and an action $S=\int\! dt ~\lambda^i(t) E_i(t)$ so that $\delta S/ \delta\lambda^i(t) = E_i(t)$, and call it a day? Or are we not allowed to introduce auxiliary variables or non-locality? Should it satisfy a minimum principle rather than a stationary principle? And so forth.

II) Example. Let us for simplicity consider the unit time interval $[t_i,t_f]=[0,1]$. A symmetrized version of the Gurtin-Tonti model is the following bi-local action

$$S[q]~:=~ \frac{1}{4}\iint_{[0,1]^2} \!dt~du~\left\{ q^i(t) \left(\frac{dq^i(u)}{du}- A_{ij}(t,u) q^j(u)\right)+(t\leftrightarrow u) \right\}\delta(t+u-1) $$ $$~=~\frac{1}{2}\int_{[0,1]} \!dt~\left\{\frac{1}{2} q^i(1\!-\!t) \frac{dq^i(t)}{dt}-\frac{1}{2}q^i(t) \frac{dq^i(1\!-\!t)}{dt}- q^i(1\!-\!t)A_{ij}(1-t,t) q^j(t) \right\} $$ $$~=~\frac{1}{2}\int_{[0,1]} \!dt~\left\{ q^i(1\!-\!t) \frac{dq^i(t)}{dt}- q^i(1\!-\!t)A_{ij}(1-t,t) q^j(t) \right\} \tag{1}$$

with symmetric matrix

$$\tag{2} A_{ij}(t,u) ~=~A_{ji}(u,t) .$$

Interestingly, the boundary contributions in the variation $\delta S$ cancel without imposing any boundary conditions (BC). In other words, as far as the finding stationary solutions, we may assume that the variables $q^i$ are free at both end points. (However, there might be other reasons to impose BCs.)

The functional derivative

$$\tag{3} \frac{\delta S[q]}{\delta q^i(t)}~=~\left.\left\{\frac{dq^i(u)}{du}- A_{ij}(t,u) q^j(u)\right\}\right|_{u=1-t}. $$

Hence the equations of motion become

$$\tag{4} \frac{dq^i(t)}{dt}~\approx~A_{ij}(1\!-\!t,t) q^j(t). $$

References:

1. V. Berdichevsky, Variational Principles of Continuum Mechanics: I. Fundamentals, 2009; Appendix B.

electromagnetism - Expanding electromagnetic field Lagrangian in terms of gauge field

The electromagnetic field tensor is given by $F_{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$, and it appears in the Lagrangian as $$L = -\frac{1}{4}F_{\mu\nu}^2 - A_{\mu}J_{\mu}.$$ Schwartz's QFT textbook says that (chapter 8, page 116) $$F_{\mu\nu}^2 = 2(\partial_{\mu}A_{\nu})^2 - 2(\partial_{\mu}A_{\mu})^2,$$ but I do not see how that is correct.

The first step of the expansion gives $F_{\mu\nu}^2 = (\partial_{\mu}A_{\nu})^2 - \partial_{\mu}A_{\nu}\partial_{\nu}A_{\mu} - \partial_{\nu}A_{\mu}\partial_{\mu}A_{\nu} + (\partial_{\nu}A_{\mu})^2$, and I see that the first and last term add up to produce $2(\partial_{\mu}A_{\nu})^2$. However, the second term $\partial_{\mu}A_{\nu}\partial_{\nu}A_{\mu}$ when expanded has a term like $\partial_{0}A_{1}\partial_{1}A_{0}$, and I don't see how this is present in $(\partial_{\mu}A_{\mu})^2 = (\partial_{0}A_{0} + \partial_{1}A_{1} + \partial_{2}A_{2} + \partial_{3}A_{3})^2$.

wordplay - Cryptic Five-words (-----|||||)

Inspired by Aggie Kidd's redux of the original "Four-words" puzzle by Prem, I decided to create a similar "Five-words" puzzle.

Some clues are intentionally somewhat vague and/or misleading, to keep it from being too easy, but they should all make sense when the solution is determined.

My first's what ant calls a cricket, no doubt
My second is third when you really zoom out
My third will protect you from villain or foe

My fourth won't get verdant when it's on the go
My fifth's what you did with the game that you lost
My whole is a square with five words that are crossed

Answer

Here is my answer:

B E A S T
E A R T H
A R M O R
S T O N E
T H R E W

Explanation:
My first's what ant calls a cricket, no doubt

A cricket would be a beast to an ant.

My second is third when you really zoom out

The earth is the third planet if we zoom out on the solar system.

My third will protect you from villain or foe

Armor protects you from everyone.

My fourth won't get verdant when it's on the go

Verdant is green, moss is green, rolling stone gathers no moss.

My fifth's what you did with the game that you lost

You threw the game.

rotational dynamics - Acceleration of body rolling down inclined plane

Acceleration of a body rolling down an inclined plane is given by:

$$\frac{g\sin\theta}{1+\frac{k^2}{r^2}}$$

$g$=acceleration due to gravity

$\theta$=angle of inclined plane

$k$=radius of gyration

$r$=radius of abject

How come acceleration is independent of coefficient of friction of plane when friction itself is causing the torque? I am not able to understand this although I am able to derive the formula on my own.

english - Longest Word without repeating character-pairs

What's the longest English word that doesn't contain the same two letters in a row more than once?

Good word : too contains to once and oo once.

Good word : tot contains to once and ot once.

Bad word : aardvark contains most pairs once but ar appears twice.

Handy JavaScript test function :

function test_word(w){
    for(var l=[],i=1; i        var p=w[i-1]+w[i];
        if( l[p] ) return p;
        l[p]=true;
    }
    return true;
}

To test your word read the link above then copy and paste the test function hit enter then use the command:

test_word('yourword');

It will return true if it is good and the repeated characters otherwise.

You can also check the word in Excel by entering the word into the cell A1 and entering this formula elsewhere as an array formula using Ctrl+Shift+Enter: (It will return TRUE if it is good and FALSE if it is not.)

=MAX(LEN($A$1)-LEN(SUBSTITUTE(UPPER($A$1),MID(UPPER($A$1),ROW(OFFSET($A$1,0,0,LEN($A$1)-1)),2),"")))<=2