electromagnetism - Isn't the front edge of a wave, kind of "information" which travels faster than light?

Considering the definition of phase and group velocities, We know group velocity can't exceed C but phase velocity can be infinitely high.

Assume a monochromatic electromagnetic wave traveling with a phase velocity greater than C in the space,

1. 
   

   In terms of existence or inexistence of a wave, why don't we consider "changing the wave from being absent to being present" as a kind of "information" in an area of space at which wave arrives?

   
   
   


2. 
   

   How much is the value of velocity at which the leading edge of the wave progresses in the space? Isn't it equal to the phase velocity of the wave? Why/why not?

   
   


3. 
   

   Assume a monochromatic source of sinusoidal electromagnetic wave (like $A\left( {t,x} \right){\rm{ }} = {\rm{ }}{A_0}cos(kx - wt)$ ) in vaccum, located at the position of ${x_0} = 0$ and a receiver at ${x_1}$ , $\,3 \times {10^8}\,\,meters\,$ distant from the source. We turn the source on at the time ${t_0}$ , calculate the time interval between ${t_0}$ and the first instant (${t_1}$) at which the receiver feels an electromagnetic wave from the source.(assume $\,\omega = 6 \times {10^8}\,{{rad} \over {\sec }}\,$ , $\,k = 1\,{{rad} \over m}\,$ )

   
   

Answer

Let me clarify two things here:

a. A perfect monochromatic electromagnetic wave has a single frequency, which means it is a purely sinusoidal wave. For a purely sinusoidal wave, the phase velocity is equal to the group velocity which is c.

b. Now think of having a source of perfect monochromatic wave and assume by default it is off. The moment you switch it on say at time 0, you create a wave packet (or a signal) that has zero amplitude for time t < 0 and sinusoidal varying amplitude at t > 0. That is NO LONGER a chromatic wave because the Fourier transform of this wave packet is not a delta function. A chromatic wave has a delta Fourier transform in the frequency domain. Such a thing can be obtained only if you have a sinusoidal wave that started from minus infinity in time.

With respect to your questions:

1. 
   

   The absence/presence of the wave is information, that is correct. But because of the wave packet you created by switching on the source, the group velocity is not necessarily equal to phase velocity. In such a case the group velocity is the velocity of propagation of information, which is always less than or equal to c naturally.

   
   


2. 
   

   If the medium of propagation is not dispersive (vacuum and air are non-dispersive media), the phase velocity and the group velocity are equal to c regardless of the nature of the signal (whether it is purely sinusoidal or a wave packet). So the velocity of the leading edge is c. If the medium is dispersive or has any special property related to propagation, I can't give a general answer. Some experiments showed that the group velocity can be greater than c or even negative. Nevertheless, it was proven that the real velocity of the photons is always less than or equal to c. For more information on those special media please read "Group and phase velocity" section at http://en.wikipedia.org/wiki/Dispersion_(optics)

   
   



3. 
   

   Since vacuum is non-dispersive the group velocity is equal to the phase velocity with is equal to omega divided by k which is 2 seconds (I believe you made a mistake in typing omega or k).

   
   

I don’t know how familiar you are with Fourier transform. Think of it as follows, any signal that has computable Fourier transform can be expresses as a summation of infinite number of sinusoidal signals, each one of those signals has its own amplitude, frequency, wave vector and phase shift compared to other signals. In that sense any arbitrary signal (as long as its Fourier transform exists) can be expressed in a summation of harmonic waves. More accurately, continuous spectrum of harmonic signals.

The propagation speed of one of those harmonic signals is called the phase velocity. The speed of propagation of the whole packet is the group velocity. For the very special case of monochromatic signal, the Fourier transform is a delta function, which means it has only one amplitude, frequency , wave vector and phase shift, so the phase velocity is equal to the group velocity (because it is the only signal in the packet). So for monochromatic waves, the definition of phase velocity and group velocity is meaningless. It has a meaning only for non-monochromatic signals, just as the one you described in your question.

The signal you described has phase velocity, group velocity and a frequency spectrum instead of single frequency. The mathematical expression you used DOESN'T REPRESENT the signal you described. The mathematical expression you used represents an ideal monochromatic wave. The signal you described can be described mathematically using the same expression you used multiplied by a Heaviside step function. That step function changes everything.

Briefly, in air or vacuum you don't need to worry about the phase velocity or group velocity, they are the same and have a value of c.

lagrangian formalism - How to show that $partial S/partial q=p$ without variation of $S$?

I'm trying to get some understanding in treating action $S$ as a function of coordinates. Landau and Lifshitz consider $\delta S$, getting $\delta S=p\delta q$, thus concluding that

$$\frac{\partial S}{\partial q}=p.$$

I'm now trying to understand this result, and consider the definition of action. I suppose that action as a function of coordinates $x$ will have $q=q(x,t)$ and $\dot q=\dot q(x,t)$, so $S(x)$ will look like:

$$S(x)=\int_{t_1}^{t_2}L(q(x,t),\dot q(x,t),t)dt.$$

Now I take the partial derivative with respect to $x$ to get $p$:

$$\frac{\partial S}{\partial x}=\int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}\frac{\partial q}{\partial x}+\frac{\partial L}{\partial \dot q}\frac{\partial \dot q}{\partial x}\right)dt.$$

And now I'm stuck. I see the momentum $\frac{\partial L}{\partial \dot q}=p$ and force $\frac{\partial L}{\partial q}=F$ inside the integral, but I can't seem to get, how to extract the momentum, so that all the other things cancelled.

Am I on the right track? What should be the next step?

Answer

Thanks to Qmechanic's answer, I've understood that $\partial S/\partial x=p(t_2)$, while I was under the illusion that it would somehow equal $p(t)$.

Now follows the finalization of my attempts.

First, consider the second part of the expression inside the integral for $\partial S/\partial x$ in the OP,

$$A=\int_{t_1}^{t_2}\frac{\partial L}{\partial \dot q}\frac{\partial \dot q}{\partial x}dt.$$

Integrating by parts, namely using $u=\frac{\partial L}{\partial\dot q}$, $dv=\frac{\partial\dot q}{\partial x}dt$, we get, with $du=\frac d{dt}\frac{\partial L}{\partial\dot q}dt=\frac{\partial L}{\partial q}dt$ and $v=\frac\partial{\partial x}\int\dot q dt=\frac{\partial q}{\partial x}$,

$$A=\left.p\frac{\partial q}{\partial x}\right|_{t_1}^{t_2}-\int_{t_1}^{t_2}\frac{\partial q}{\partial x}\frac{\partial L}{\partial q}dt.$$

Substituting this into the integral in the OP, we get the integrals cancel, thus

$$\left.\frac{\partial S}{\partial x}=p\frac{\partial q}{\partial x}\right|_{t_1}^{t_2}=p(t_2)\frac{\partial q(x,t_2)}{\partial x}-p(t_1)\frac{\partial q(x,t_1)}{\partial x}.$$

But by definition of $x$, $q(x,t_2)\equiv x$, and because first point is fixed, we have $q(x,t_1)=q^{(1)}=\text{const}(x)$, thus the result is what was to be proved:

$$\frac{\partial S}{\partial x}=p(t_2).$$

electrostatics - What happens to half of the energy in a circuit with a capacitor?

For a simple circuit with a battery supplying a voltage $V$ to a capacitor, let us assume that the charge on the capacitor is $Q$. Now, the work done by the battery or the energy supplied is given by the relation:

$$W=QV$$

But the energy stored in the capacitor is given by:

$$U = \tfrac12 QV$$

The value of $Q$ as well as that of $V$ should be the same in both the equations.

Now my question is, where is the other half of the energy that the battery supplied?

homework and exercises - Coupled Harmonic Oscillator (Forced Vibration)

I derived two equations for a 2DOF harmonic oscillator system, declared state variable equations, and placed them into matrix form: $Ax' + Bx = C$. I have a Matlab script to determine the constants ($K$'s, $m$'s, & $R$'s). I'll be seeing how the system responds from 40 - 1000 Hz.

How can I manipulate these matrices to find the solutions to $x'$: \begin{array}{c} \dot{x}_s , \dot{x}_m , \dot{v}_s , \dot{v}_m \end{array}

$$ \begin{split} m_s \ddot{x}_s + R_{s1}\dot{x}_s + R_{2s}\left( \dot{x}_s - \dot{x}_m\right) + K_{s1}x_s + K_{s2} \left( x_s - x_m \right) &=& P_0 \\ m_m \ddot{x}_m + R_{m}\dot{x}_m - R_{2s}\left( \dot{x}_s - \dot{x}_m\right) + K_{m}x_m - K_{s2} \left( x_s - x_m \right)&=&0 \\ \dot{x}_s - v_s&=& 0 \\ \dot{x}_m - v_m &=& 0 \end{split} $$

$$ \left[ \begin{array}{cccc} 1 &0 & 0 &0 \\ 0 &1 & 0 & 0 \\ 0 & 0 & m_s &0 \\ 0 &0 & 0 & m_m \\ \end{array} \right] \left[ \begin{array}{c} \dot{x}_s \\ \dot{x}_m \\ \dot{v}_s \\ \dot{v}_m \end{array} \right] + \left[ \begin{array}{cccc} 0 &0 & - 1& 0 \\ 0& 0 &0 & -1 \\ K_{s1} + K_{s2} & -K_{s2} & R_{s1} + R_{s2} & -R_{s2} \\ -K_{s2} & K_{m} + K_{s2} & - R_{s2} & R_m + R_{s2} \end{array} \right] \left[ \begin{array}{c} {x}_s \\ {x}_m \\ {v}_s \\ {v}_m \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \\ 1 \\0 \end{array} \right] e^{i\omega t} $$

quantum mechanics - Do Bell inequality violations appear instantly when the source is turned on, or do they increase over time?

This experimental Question is a result of reading a particular article on Bell violations. I addressed the e-mail below to the corresponding authors —because who knows, they might reply— but it is not specific to them, it's just that their article crystallized the Question, and perhaps someone on Physics SE can tell me what the results of the experiment would be without doing it (although I think these time-dependent properties are not part of the usual data sheets for off the shelf components).

There is a subsidiary question: have such experiments been done, and I haven't seen them in the literature?

Dear Johannes and Anton,

  I've read your PNAS article "Violation of local realism with freedom of choice" as a result of a blog posting by Sabine Hossenfelder at http://backreaction.blogspot.com/2011/06/nonlocal-correlations-between-canary.html [Physics SE readers can find PNAS and arXiv links there]. I was impressed by your three key remarks on the second page, which seem to me nicely balanced.

  Reading your article I wondered as I have many times whether Bell inequality violations appear instantly when the source is turned on, or to what extent they increase over time. I am partly led to the questions below because it seems that Bell violating experiments run continuously, whereas it seems that technological applications may well be intermittent.

  I imagine, specifically, an experiment based on an experiment in which we observe Bell violations routinely (presumably in CHSH form). The crucial physical modification is to block both paths from the source to the two detectors physically, at a point near the source, on a time scale of, say, about a second, so that light travels from the source only for half a second at a time. A steady state condition of the source, the quantized electromagnetic field, and the detectors will presumably not be instantly established, but, I suppose, would be established in half a second, so that if we considered data only from the last quarter second of the on-phase we would see the usual violation of Bell inequalities.

  The crucial modification of the analysis is to consider the way Bell violation changes over time, microsecond by microsecond (or more or less finely time-sliced as experience indicates), after the moment that the physical block is removed. Clearly this is a statistical analysis, since we would expect to see approximately one photon pair every four microseconds (at a local production rate comparable to the 250,000 photon pairs per second that you report in your PNAS article).

  I suppose that a number of questions emerge, some of which may be characteristics of the source and detectors; others might be characteristic of the experimental apparatus more generally considered. Firstly, most elementarily, how does the rate of single photon detections change over time in the microseconds after the physical block is removed? We would expect that there would be a rapid approach to the steady state, but not instantaneous.

  Secondly, how does the rate of photon pair detections change over time in the microseconds after the physical block is removed? Does the approach to the steady state have the same relaxation rate as the relaxation rate for the single photon approach to the steady state?

  Thirdly, how does the violation of Bell inequalities vary over time in the microseconds after the physical block is removed? Again, does this have the same relaxation rate for the approach to the steady state?

  It seems a matter of technological interest as well as of foundational interest whether relaxation of Bell inequality violations towards the steady state immediately after a light source is exposed is the same as the relaxation rates of single photon detections and/or double photon detections, or whether we have to wait longer for usable Bell violations to emerge.

  Because I suppose these questions to be of general interest, I have also posted them at http://physics.stackexchange.com. I don't imagine the results of such an experiment will go against quantum theoretical expectations, but I'm curious whether characterizing these time-dependent properties of the sources and detectors might give some surprises.

word - The road to El Dorado reconstructed from fragmented accounts

You are seeking the route to the lost city of El Dorado. You have assembled several historical accounts from explorers who claim to have reached it. Unfortunately, the accounts are incomplete, fragmented, or evasive. None provide all the milestones necessary to chart the route to El Dorado. Your only certainty is that the milestones which were reported are correct and in the proper order. Can you reconstruct a complete route from the fragmented accounts?

We're going to do something similar, but through a word puzzle.

RETAIN, HERO, STATION, CHEST, and ORCA are all examples of "fragments" taken from the source word ORCHESTRATION. Notice that letter ordering in the source word is preserved in the fragments.

I'm going to give you a set of "fragments" and we'll see if you can reconstruct the source word. I have designed it so that the given words are sufficient to determine the source word, but not easily.

For example:

Given words:


CRAM
CUTLET
TIN


-----------------------

Solution:


CURTAILMENT

If you're having difficulty seeing it, here is a graphical demonstration:

C   R   A     M   
C U   T     L   E   T
      T   I       N 
---------------------
C U R T A I L M E N T

It sort of looks like word Tetris!

As always, I construct my puzzles in such a way that they can be solved using only well-known words, so if you find yourself conjuring up increasingly obscure words, you may be overthinking it.

See if you can figure out the source word for each of these eight sets of "fragments" below:

1.  AUTO
    STRAIN

2.  CAKE
    FIERCER
    FREAK


3.  COURT
    MINUTES
    RUINS

4.  COLAS
    SCOTCHES
    SHOOTERS

5.  SEND

    SPROUT
    SUCCOR

6.  CANTON
    CONTAIN
    OCEAN

7.  LEGIT
    PORES
    TRESS


8.  PRUDE
    SEANCE
    URBANE
    (okay, maybe the source word for this one is just a little bit... professorial)

Answer

EDIT: I think I got them all!

1)

SATURATION

2)

FIRECRACKER

3)

MICRONUTRIENTS

4)

SCHOOLTEACHERS

5)

SUPERCONDUCTOR

6)

CONCATENATION

7)

POLTERGEISTS

8)

SUPERABUNDANCE

general relativity - Parallel Transport of a 4-vector

Why does the parallel transported $4$-vector change from $X^a(x)$ to $X^a(x) + \bar{\delta}X^a(x)$ ?

This is also discussed in D'Inverno's relativity book [page - 72]; but the reason is not clear.

Answer

I really don’t like that diagram. No, REALLY. I think it conveys a bad intuition that may confuse you. I don’t like what it does with the connexion coefficients (Christoffel symbols). Here’s why.

In a general manifold, tangent spaces of course are not comparable at different points. This is in contrast with the situation in Euclidean space, which can be taken as isomorphic to the tangent space to any of its points.

So to define a “change” in shifting between tangent spaces at different points, we need to define some mapping which defines how our basis vectors in the first tangent space are mapped to those in the second. Once we define that mapping, we have something that we can measure “change” relative to. We take an original vector in the tangent space and we find its image under this mapping and declare this image to be “the same as” the original vector for the purposes of measuring change and thus defining for example a derivative.

It turns out, however, in general manifold that this mapping, i.e. “parallel transport” depends on the path. Refer to Aman-Abhishk’s neat summary of parallel transport around a closed loop to understand intuitively why this is so. More abstractly, although we have a great deal of freedom in defining our connection and covariant derivative, the latter has to fulfil the Leibniz rule. This forces restrictions that mean that the “no change mapping” of parallel transport in general depends on the path (see footnote, though). What this means is that there is no set of vector fields on the manifold whose tangents we can use as reference vectors. Wouldn’t that be easy? We’d just measure how vector changes relative to these tangents. But of course, this would imply that the parallel transport were independent of path.

So here is what I believe to be a better diagram.

I’ve drawn the tangent spaces to three points on a two-dimensional manifold. The directions marked $x$ and $y$ in each case are tangent vectors to two orthogonal vector fields which we might hope to use some kind of reference. We can indeed choose a connexion such that the connexion coefficients vanish at one point: let this be the point $p_1$ in my diagram. So in this case parallel transported versions of the vector $X$ stay “aligned” to two reference vector fields. So I’ve worked out through the covariant derivative what the big green vector at point $p_2$ must be if we work out the change in this vector (independent of coordinates of course): our parallel transport mapping leaves $X$ aligned to the $y$-axis, and then we just add on $\delta X$. Likewise our parallel transport leaves $X$ aligned to the $y$-axis when we shift to point $p_3$. However, as Aman-Abhishk’s neat summary of parallel transport around a closed loop shows, we can’t in general pull this trick off if we begin with the vector at point $p_3$ and work out its change through the covariant derivative in shifting to $p_2$. Our parallel transported $X$ around the loop is in general nonzero. So there must in general be some component $\Gamma X$ so that our parallel transported $X$ in going straight from $p_1$ to $p_2$ is different from that in going from $p_1$ to $p_2$ to $p_3$. Here $\gamma\in GL(N, V)$ is the square $N\times N$ matrix (here $N$ is the manifold's dimension, in this case two) of the parallel transport map. The “no-change” parallel transported vector $\Gamma X$ is different this time, and this is what the connection coefficients measure. The “absolute change” in going from $p_2$ to $p_3$ is shown as $\delta X^\prime$.

Footnote: The noneuclideanhood (i.e. deviation of a manifold's geometry from fulfilling the Euclid parallel postulate) is in general encoded in two quantities, the curvature and torsion tensors. It is the curvature that measures how vector changes when parallel transported around a loop. General Relativity, as with theory on a psuedo-Riemannian or Riemannian manifold, makes use of one of the co-called "fundamental theorems of Riemannian geometry" that one can always define a connexion to absorb the tortion, i.e. set it to nought. In this case, parallel transport always depends on the path. However, it is still possible to have non-Euclidean geometry without curvature but nonzero torsion: as with Christoph's comment below, Einstein's teleparallelism is one theory that is curvature free. A more wonted example is that of a Lie group: one can define parallel transport by the idea of left (or right) translation of vectors. Since the translation is effected by the group member that is also the point of the manifold in question, parallel transport cannot depend on path! However likewise most Lie groups have nonzero torsion as in general they are most decidedly "not flat"!

logical deduction - The Expedition Into The Ancient Cave

You are a archeologist and have taken 8 of your students on a expedition trip to a ancient cave to try and find ancient treasure. After you and your team entered the cave, you travel as a group exploring the entire cave. There is only one tunnel left in the cave to explore. As you reach the end of the tunnel one of your students accidentally leans on a panel that opens up a secret entrance before you.

As you enter you see ancient writing on the wall and a giant gem sitting on a pedestal. After sometime you and your students deciphered the ancient writing, it says "Beware of the curse! The darts will control whether you lie or tell the truth and the green gas will turn you into a hideous beast. The only way to return back to normal is to escape alive". Some of your students laugh and call it a bluff designed to scare and prevent theft of the treasure hidden in the cave.

You decided to take a chance and take the gem off of the pedestal. As you do, two darts fly out of nowhere, barely missing you, and hit two of your students. You jumped away from the pedestal just in time as green gas sprays all 8 of the students. After the gas clears, everyone but you, look exactly the same. You can't tell them apart. One of the students points behind you and says "What's that?!". You turn around to see purple gas starting to fill the room, naturally you and your students start running back to the entrance.

After escaping the tunnel you came from, you have to make a choice between 4 other tunnels. From what you recall it took around 4 minutes to travel from one end of the tunnel to the other. The gas will reach where you are in about 10 minutes. You decide to split up into groups, check each tunnel, and report back here to see which tunnel is the correct way out. Which means there is only have 2 minutes to discuss and make a decision before the gas reaches the groups location.

Remember, two of the students can't control if they lie or not, and they all look the same so you can't tell them apart. How many groups should there be when you all split up? How many people per group? and how can you tell which is the correct tunnel to escape through?

Answer

EDIT: Handled the unpredictable nature of the affected students this time.

Thinking quickly, you decide to send

3 students down one tunnel, 3 down another, 2 down a third, and you check the forth.

In 8 minutes all return back and you make your decision.

You know for certain the status of your own tunnel, so if it is the exit gather them up and leave.

Otherwise, that was not the exit, so it must be in the other tunnels. Listen to what the students have to say. The possibilities are:

* Both 3 person groups are unanimous in their report
* One 3 person group is in disagreement and the 2 person group is unanimous
* One 3 person group is in disagreement and the 2 person group is in disagreement
* Both 3 person groups are in disagreement

But these are all easily resolvable:

* For the first case, both 3 person groups have a truth teller and thus are not lying. If neither indicate the exit, the 2 person tunnel will be the exit, regardless of how they answered.
* For the second case, we don't know if there are two liars in this group, or if there are two truth tellers. Regardless, we know all the other groups are telling the truth since they have truth tellers and are in agreement. Thus, infer the truth value of this group based on the other two groups. For example, if the other two groups didn't find the exit, then we know this group did find the exit regardless of how they answer.
* For the third case, we know that there are 2 truth tellers in the 3 person group and only one liar since the other liar must be in the 2 person group. So use the majority decision of both 3 person groups to infer the status of the two person tunnel.
* For the last case, if both 3 person groups are in disagreement, then they both have a single affected person. But that means the other two people in the group are unaffected, so simply use the majority decision of all three groups.

Thus, you can easily infer the status of the tunnels and can safely make your escape!

thermodynamics - How do I prove that the Carnot cycle is the most efficient graphically?

I have to prove that the Carnot cycle is the most efficient by comparing it with a arbitrary cycle working within the temperature range of $T_L$ to $T_H$ in a ST-diagram. How do I do this?

Answer

How do I prove that the Carnot cycle is the most efficient graphically?

See the diagrams below.

Fig 1 shows an arbitrary cycle on a T-S diagram.

Fig 2 shows a Carnot cycle on a T-S diagram operating between the same high and low temperature extremes and having the same range of entropy (the Carnot graph can be superimposed on top of the arbitrary cycle to show this).

We will assume that both the arbitrary and Carnot cycles are reversible.

For any such arbitrary cycle we can see that area enclosed by the T-S diagram of the Carnot cycle is greater than that for any arbitrary cycle operating between the same maximum and minimum temperatures and range of entropies. This means the net work of the Carnot cycle is greater than the arbitrary cycle operating between the two temperature extremes an range in entropy difference.

In order to show that, in addition to their being more work done, the Carnot cycle is the most efficient we need to take into consideration entropy.

First, we know that for both heat engine cycles shown in the figure we have the following:

$$W_{Net}=Q_{IN}-Q_{OUT}$$

Where $W_{Net}$ is the net work done in the cycle, $Q_{IN}$ is the total heat added during the cycle, and $Q_{OUT}$ is the total heat rejected during the cycle.

The efficiency for both heat engine cycles is given by:

$ζ=$ Net work done ($W_{NET})$ / Gross heat added ($Q_{IN})$

Combining the last two equations we get, for both cycles

$$ζ=1-\frac {Q_{OUT}}{Q_{IN}}$$

So now we need to show that $\frac {Q_{OUT}}{Q_{IN}}$ for the arbitrary cycle is greater than that for the Carnot cycle in order to show that $ ζ_{Carnot}> ζ_{Arbitrary}$

To do this we start with the definition of a differential change in entropy, $dS$, which is defined as, for a reversible transfer of heat:

$$dS=\frac {dQ}{T}$$

Or,

$$dQ=TdS$$

Where $dS$ is a differential change in entropy, $dQ$ is a differential reversible transfer of heat and $T$ is the temperature at which the heat transfer occurs.

We now return to the diagram for the arbitrary cycle, Fig 1.

Heat in, $Q_{IN}$:

Note that with the exception of heat transfer in at the maximum temperature $T_H$, for a given $dS$ each differential transfer of heat occurs at a temperature $T$ less than $T_H$. That means each $dQ_{IN}$ for the arbitrary cycle is less than $dQ_{IN}$ for the Carnot cycle except where it occurs at the maximum temperature of the arbitrary cycle. Adding them all up, $Q_{IN}$ for the arbitrary cycle is less than $Q_{IN}$ for the Carnot cycle.

Heat out, $Q_{OUT}$:

Note that with the exception of heat transfer out at the minimum temperature $T_L$, for a given $dS$ each differential transfer of heat occurs at a temperature $T$ greater than $T_L$. That means each $dQ_{OUT}$ for the arbitrary cycle is greater than $dQ_{OUT}$ for the Carnot cycle except where it occurs at the minimum temperature of the arbitrary cycle. Adding them all up, $Q_{OUT}$ for the arbitrary cycle is greater than $Q_{OUT}$ for the Carnot cycle.

Summing it up, since $Q_{IN}$ for the arbitrary cycle is less than $Q_{IN}$ for the Carnot cycle, and since $Q_{OUT}$ for the arbitrary cycle is greater than $Q_{OUT}$ for the Carnot cycle, we conclude that

$\frac {Q_{OUT}}{Q_{IN}}$ for the arbitrary cycle is greater than $\frac {Q_{OUT}}{Q_{IN}}$ for the Carnot cycle, and therefore $ζ_{Carnot}>ζ_{Arbitrary}$.

Hope this helps

quantum mechanics - Supercharges for N=1 supersymmetry of a super Hamiltonian

I currently have determined a super Hamiltonian for a non-relativistic electron in an external electromagnetic field using an appropriate Lagrangian and use of the Euler-Lagrange equations:

$$H=\frac{1}{2m}\left(\overrightarrow{p}-\frac{q}{c}\overrightarrow{A}\right)^2+q\phi$$

I then considered an electron in a purely magnetic field finding that there is an additional interaction such that

$$H=\frac{1}{2m}\left(\overrightarrow{p}+\frac{q}{c}\overrightarrow{A}\right)^2+\frac{g}{2}\frac{e\hbar}{2mC}\overrightarrow{B}\cdot{\overrightarrow{\sigma}}.$$

By setting $g=2$ I can analyse the Hamiltonian in $\mathcal N=1$ supersymmetry by considering the anti-commutation relations between supercharges and the Hamiltonian:

$$H=2Q_i^2=\{Q_i,Q_i\}$$

where $[H,Q_i]=0$. My struggle is that this closely follows the Fred Cooper book on Supersymmetric Quantum mechanics and he achieves a supercharge as shown below

$$Q_i=\frac{1}{\sqrt{4m}}\left(\overrightarrow{p}+\frac{e}{c}\overrightarrow{A}\right)\cdot{\overrightarrow{\sigma}}.$$

I am lost as to how this is found by simple factorisation. I am unsure, how is this done?

Answer

If you are unsure of how to verify it, start with $$Q^2=(p+A)^i\sigma_i\,(p+A)^j\sigma_j,$$ where I'm summing over both indices and I'm getting rid of the constants for simplicity. This splits into a term where both $p$ are acting on the wavefunction, and a term where the left $p$ is acting on the right $A$. $$=(p+A)^i(p+A)^j\sigma_i\sigma_j-i\partial^iA^j \sigma_i\sigma_j$$ Now use the identity $\sigma_i\sigma_j=\delta_{ij}+i\epsilon_{ijk}\sigma_k.$ The first term in $Q^2$ is symmetric in $i,j$ so only the Kronecker delta matters, in the second term the Kronecker delta leads to the divergence of $A$ which vanishes in the Coulomb gauge, so only the Levi-Civita symbol matters. $$=(p+A)^2+\epsilon_{ijk}\partial^iA^j \sigma_k=(p+A)^2+(\nabla\times \vec A)\cdot\vec\sigma=(p+A)^2+\vec B\cdot\vec\sigma.$$

If you are wondering how they figured out $Q$ in the first place, I imagine they were aware of the identity $\vec{V}\cdot{\vec\sigma}\,\vec{V}\cdot{\vec\sigma}=V^2$, for an ordinary (non-operator) $\vec V$, so they just looked at the kinetic term and tried out an operator $Q=(\vec p+\vec A)\cdot \vec\sigma$. The maybe unexpected thing is this also gives you the magnetic field term with $g=2$.

thermodynamics - What exactly is fire?

What is fire? Is it a wave or is it matter?

Where does fire come from?

Does everything burn with fire? (for example: water and some metals don't burn).

Answer

I'll let the far greater minds than me at Minute Physics give the bulk of the answer. Turns out that the way you perceive fire is the result of numerous chemical and physical processes. I'll discuss and expand on just two major ones that were mentioned in the video I linked:

1. Atomic Transitions- The distinct colors that you see in a flame are the result of electrons changing energy levels within at atom. When an excited electron drops to a lower energy level, a photon is released, in order to conserve energy. The frequency of the photon released depends on the exact transition the electron makes. In fact, the energy levels of a particular element are somewhat unique, and so the exact colors emitted by electrons changing energy levels are somewhat unique to the element. In chemistry class you may have done a flame ion lab at some point to see that different metals produce flames of different colors, as a result of the different transitions taking place, unique to each element. In the second video I just linked, you can see a variety of colors such as green, blue, red, or orange depending on the element.


2. Thermal Radiation- However, the red/orange colors you typically see from most combustion reactions in real life are actually the result of a different process. As mentioned in the Minute Physics video, impure flames usually burn reddish orange because the soot and ash particles around the flame are so hot that they emit visible light. This is a different process than the atomic transitions I mentioned. All objects, when heated, emit radiation in the form of electromagnetic waves and the temperature of the object determines the exact wavelengths emitted. You are actually emitting "light" as we speak, but you are so cool that the radiation you emit is in the infrared range, and therefore not visible.

quantum field theory - Inverse Green's Function identity in derivation of Hedin's equations

I'm trying to work through a derivation of Hedin's Equations in Effect of Interaction on One-Electron States by Hedin and Lundqvist (1969) and I've come across an identity that is given without much explanation that I'd like to understand.

We start with an equation of motion with single particle Green's functions G:

$\left[ i\frac{\partial}{\partial t_1} - h(1) - \phi(1) - V_H(1)\right] G(1,2) - i\int v(1^{+},3) \frac{\delta G(1,2)}{\delta \phi(3)} d3 = \delta(1,2)$

where $1=(x_1,t_1)$, $h$ is the one-electron kinetic energy and ion interaction, $\phi$ is a small perturbing field that is set to zero at the end, $V_H$ is the Hartree potential, and $v$ is the Coulomb interaction.

They then insert the identity

$\frac{\delta G(1,2)}{\delta\phi(3)} = -\int G(1,4) \frac{\delta G^{-1}(4,5)}{\delta\phi(3)} G(5,2) d(4,5)$

which is then used to define the self energy

$\Sigma(1,2) = -i \int v(1^+,3) G(1,4)\frac{\delta G^{-1}(4,2)}{\delta\phi(3)} d(3,4)$.

I know this identity is related to the definition of the inverse Green's function

$ \int G(1,3) G^{-1}(3,2) d(3) = \delta(1,2)$

but I can't figure out how to get from this definition to the identity that is inserted into the equation of motion. Any ideas?

Thank you!

Answer

Take the functional derivative of the functional identity $$\mathbf{\int d(3)G_{}^{}(1,3)G_{}^{-1}(3,2)=\delta_{}^{}(1,2)} \tag{1}$$ with respect to $\phi(4)$ to get $$\int d(3)\frac{\delta G_{}^{}(1,3)}{\delta\phi_{}^{}(4)}G_{}^{-1}(3,2)+\int d(3)G_{}^{}(1,3)\frac{\delta G_{}^{-1}(3,2)}{\delta\phi_{}^{}(4)}=0. \tag{2}$$ In going from $(1) \rightarrow (2)$, chain rule for functional differentiation and independence of $\delta_{}^{}(1,2)$ on $\phi$ are used.

Right multiply Eq.$(2)$ on both sides with $G_{}^{}(2,5)$ and integrating over variables $(2)$ to get $$\int d(2)\int d(3)\frac{\delta G_{}^{}(1,3)}{\delta\phi_{}^{}(4)}G_{}^{-1}(3,2)G_{}^{}(2,5)+\int d(2)\int d(3)G_{}^{}(1,3)\frac{\delta G_{}^{-1}(3,2)}{\delta\phi_{}^{}(4)}G_{}^{}(2,5)=0. \tag{3}$$ Finally use Eq.$(1)$ to get the desired identity $$\mathbf{\frac{\delta G_{}^{}(1,5)}{\delta\phi_{}^{}(4)}=-\int d(2)\int d(3)G_{}^{}(1,3)\frac{\delta G_{}^{-1}(3,2)}{\delta\phi_{}^{}(4)}G_{}^{}(2,5)}. \tag{4}$$

electromagnetism - Is the Lorentz force expression valid for magnetic field created by a magnetic monopole?

Will the Lorentz force expression be valid for a magnetic field created by a magnetic monopole?

I haven't seen any derivation of Lorentz force expression yet and I don't know whether it was derived applying Lorentz transformations to field created by charge particles.

But if this is the case then magnetic field created by moving charges of course has zero divergence ,as we know (not same as magnetic field of a monopole) and a magnetic monopole can create magnetic field even when it is at rest. So should the Lorentz force expression be applied in the case of magnetic field by a monopole?

Edit:I was talking about the force on a charge particle due to magnetic field created by a monopole.

$$F=q_e(v \times B)$$

special relativity - Do photons have relativistic mass?

I am conducting research on photons and was wondering if they have relativistic mass. I already know that they they have zero rest mass. Any answers are welcome!

thermodynamics - Proving (instead of discovering) the laws of quantum mechanics

A single toss of a fair coin cannot be predicted. But if we observe a large number of tosses, we can prove mathematically the law that roughly half of them will show up heads.

The movements of individual molecules in a gas cannot be predicted and can be assumed to be random. But if we observe some macroscopic phenomena such as temperature or pressure, we can prove mathematically that some laws are satisfied.

Individual quantum events are random. But if we observe a large number of such events, we discover experimentally that they satisfy the laws of quantum mechanics. Could the laws of quantum mechanics be proved mathematically as in the examples above?

wordplay - Seven-words (-------|||||||)

Seven-words:

My first is what the bully may show
My second describes some spouses post-nuptials
My third is when something is like something else

My fourth are people who tell you what to pay
My fifth are microscopic and may move through you
My sixth will reduce your quality
My seventh is something that Homer wrote

---

In the neo-tradition of Four-words, Five-words, and Six-words.

Answer

BRAVADO 

RENAMED 
ANALOGY 
VALUERS 
AMOEBAS 
DEGRADE 
ODYSSEY

eigenvalue - What is the meaning of pre-tension for a stiff membrane?

On one hand, I know that the tighter a drum head is stretched, the higher its natural frequencies. This relation is given by:

$$f_{ij}=\frac{k_{ij}}{2\pi R}\sqrt{\frac{T_0}{h\rho}}$$

where $k_{ij}$ are roots of the Bessel function of the first kind, $R$ is the membrane radius and $T_0$ is the pre-tension.

On the other hand, if my membrane is a stiff solid that is clamped to edges of a cavity, no pre-tension is required to produce frequency-response when a disturbance is introduced. Even if we neglect the induced pre-tension by gravity, we would still expect our stiff membrane to respond to an impulse, because tension will be generated "internally" upon deflection due to the impulse (see also: "Chladni plate", which has free boundary conditions).

However, I cannot use $T_0=0$ in the equation above, since it provides only the trivial solution. So what is the meaning of pre-tension for a stiff membrane?

Answer

You cannot use that formula for a stiff plate - you have a fundamentally different situation.

In the stiff plate, the restoring elastic force comes from the stress arising from the strain induced elastic deformation of the plate's material. This is related to the materials' modulusses (Young's, Shear and Bulk Modulusses) and Poisson ratio. See how these ratios do NOT appear in your formula.

In the stretched membrane, on the other hand, the material is assumed to be infinitely flexible and flimsy. At rest, all the tension is in the membrane's plane. As the membrane is deformed sideways, there is a component of this roughly constant tension opposing the deformation: a two dimensional analogy of the stretched, infinitely flimsy string. Other than its being infintely flimsy, no other material properties matter in this analysis.

I can't give you a formula for what you need, but you would have to work out orfind the analogy for Euler-Bernoulli beam theory or Timoshenko beam theory. Timoshenko Beam theory applies to the stiff beam, and this is quite a different theory to that of the Stretched flimsy string.

wormholes - Can humans make white holes

I have a question that i could not find on any site on the internet. do humans have the ability to create white holes, and if so could it be used to stabilize a wormhole.

quantum field theory - Self energy, 1PI, and tadpoles

I'm having a hard time reconciling the following discrepancy:

Recall that in passing to the effective action via a Legendre transformation, we interpret the effective action $\Gamma[\phi_c]$ to be the generating functional of 1-particle irreducible Green's functions $\Gamma^{[n]}$. In particular, the 2-point function is the reciprocal of the connected Green's function,

$$\tilde \Gamma^{[2]}(p)=i\big(\tilde G^{[2]}(p)\big)^{-1}=p^2-m^2-\Sigma(p)$$

which is the dressed propagator.

But, the problem is this: in the spontaneously broken $\phi^4$ theory, the scalar meson (quantum fluctuations around the vacuum expectation value) receives self energy corrections from three diagrams:

$-i\Sigma(p^2)=$ + +

Note that the last diagram (the tadpole) is not 1PI, but must be included (see e.g. Peskin & Schroeder p. 361). In the MS-bar renormalization scheme, the tadpole doesn't vanish.

If the tadpole graph is included in $\Sigma$, and hence in $\tilde{G}$ and $\tilde\Gamma$, then $\tilde\Gamma$ cannot be 1PI. If the tadpole is not included, then $\tilde G$ is not the inverse of the dressed propagator (that's strange, too). What's going on?

Answer

I'm going to give an explanation at the one loop level (which is the order of the diagrams given in the question).

At one loop, the effective action is given by $$ \Gamma[\phi]=S[\phi]+\frac{1}{2l}{\rm Tr}\log S^{(2)}[\phi],$$ where $S[\phi]$ is the classical (microscopic) action, $l$ is an ad hoc parameter introduced to count the loop order ($l$ is set to $1$ in the end), $S^{(n)}$ is the $n$th functional derivative with respect to $\phi$ and the trace is over momenta (and frequency if needed) as well as other indices (for the O(N) model, for example).

The physical value of the field $\bar\phi$ is defined such that $$\Gamma^{(1)}[\bar\phi]=0.$$ At the meanfield level ($O(l^0)$), $\bar\phi_0$ is the minimum of the classical action $S$, i.e. $$ S^{(1)}[\bar \phi_0]=0.$$ At one-loop, $\bar\phi=\bar\phi_0+\frac{1}{l}\bar\phi_1$ is such that $$S^{(1)}[\bar \phi]+\frac{1}{2l}{\rm Tr}\, S^{(3)}[\bar\phi].G_{c}[\bar\phi] =0,\;\;\;\;\;\;(1)$$ where $G_c[\phi]$ is the classical propagator, defined by $S^{(2)}[\phi].G_c[\phi]=1$. The dot corresponds to the matrix product (internal indices, momenta, etc.). The second term in $(1)$ corresponds to the tadpole diagram at one loop. Still to one-loop accuracy, $(1)$ is equivalent to $$ S^{(1)}[\bar \phi_0]+\frac{1}{l}\left(\bar\phi_1.\bar S^{(2)}+\frac{1}{2}{\rm Tr}\, \bar S^{(3)}.\bar G_{c}\right)=0,\;\;\;\;\;\;(2) $$ where $\bar S^{(2)}\equiv S^{(2)}[\bar\phi_0] $, etc. We thus find $$\bar \phi_1=-\frac{1}{2}\bar G_c.{\rm Tr}\,\bar S^{(3)}.\bar G_c. \;\;\;\;\;\;(3)$$

Let's now compute the inverse propagator $\Gamma^{(2)}$. At a meanfield level, we have the meanfield propagator defined above $G_c[\bar\phi_0]=\bar G_c$ which is the inverse of $S^{(2)}[\bar\phi_0]=\bar S^{(2)}$. This is what is usually called the bare propagator $G_0$ in field theory, and is generalized here to broken symmetry phases.

What is the inverse propagator at one-loop ? It is given by $$\Gamma^{(2)}[\bar\phi]=S^{(2)}[\bar\phi]+\frac{1}{2l}{\rm Tr}\, \bar S^{(4)}.\bar G_{c}-\frac{1}{2l}{\rm Tr}\, \bar S^{(3)}.\bar G_{c}. \bar S^{(3)}.\bar G_{c}, \;\;\;\;\;\;(4)$$ where we have already used the fact that the field can be set to $\bar\phi_0$ in the last two terms at one-loop accuracy. These two terms correspond to the first two diagrams in the OP's question. However, we are not done yet, and to be accurate at one-loop, we need to expand $S^{(2)}[\bar\phi]$ to order $1/l$, which gives $$\Gamma^{(2)}[\bar\phi]=\bar S^{(2)}+\frac{1}{l}\left(\bar S^{(3)}.\bar\phi_1+\frac{1}{2}{\rm Tr}\, \bar S^{(4)}.\bar G_{c}-\frac{1}{2}{\rm Tr}\, \bar S^{(3)}.\bar G_{c}. \bar S^{(3)}.\bar G_{c}\right). \;\;\;\;\;\;$$ Using equation $(3)$, we find $$\bar S^{(3)}.\bar\phi_1= -\frac{1}{2}\bar S^{(3)}.\bar G_c.{\rm Tr}\,\bar S^{(3)}.\bar G_c,$$ which corresponds to the third diagram of the OP. This is how these non-1PI diagrams get generated in the ordered phase, and they correspond to the renormalization of the order parameter (due to the fluctuations) in the classical propagator.

astronomy - Is there an algorithm to find the time when the sun is X degrees above the horizon for a given latitude B at date C?

Is there an accurate algorithm / method to determine the precise time of day/night when the sun is X degrees above (or below) the horizon for a given latitude Y at date Z?

Is this the same question here? Reverse Sun position algorithm? I'm having a hard time plugging in specifics.

Answer

First, you need to know the Right Ascension and Declination of the Sun on that day. Then there are some angular relations connecting all of these. Everything is available in the RASC Observer's Handbook. Another way would be to interpolate the table produced by this program provided by the US Naval Observatory: http://aa.usno.navy.mil/data/docs/AltAz.php Hope this helps!

riddle - I am described by temperature, and never by looks. I might be found in your camera, as well as children's books

Riddle me this:

I am described by temperature, and never by looks.

I might be found in your camera, as well as children's books.

I described what you observed, but perhaps not at night.

I am swallowed by darkness, but am an offspring of light.

I have a pair of spellings, with and without;

An infinitude of relatives, though every eye has its doubt.

All of us can be found where seven rule the family.

I have not seen it, but there is a pot of gold, apparently.

My gender is defined by my shade and/or tone.

Unravel these clues, and the answer you will own.

---

I made up this riddle for fun. Hope you enjoy!

Hints for the answer are below.

---

Hint:

I can be light or dark and might be found on a wheel.
I am, quite literally, what your eyes may reveal.

---

Edit:

In line $3$, I changed the word "saw" to "observed" because it means the same thing; has more syllables; and sounds better with the word "darkness".

Answer

proposed solution

color / colour

I am described by temperature, and never by looks.

colors are warm/cold

I might be found in your camera, as well as children's books.

cameras capture color. children's books are full of them!

I described what you saw, but perhaps not at night.

it's hard to see color when it's dark..

I am swallowed by darkness, but am an offspring of light.

white light in a prism produces color

I have a pair of spellings, with and without;

color / colour

An infinitude of relatives, though every eye has its doubt.

the human eye can perceive approximately 10 million colors

All of us can be found where seven rule the family. I have not seen it, but there is a pot of gold, apparently.

rainbow

My gender is defined by my shade and/or tone.

traditional pink=girl blue=boy

Hint

color wheel

visible light - Why does the moon sometimes appear giant and a orange red color near the horizon?

I've read various ideas about why the moon looks larger on the horizon. The most reasonable one in my opinion is that it is due to how our brain calculates (perceives) distance, with objects high above the horizon being generally further away than objects closer to the horizon.

But every once in a while, the moon looks absolutely huge and has a orange red color to it. Both the size and color diminish as it moves further above the horizon. This does not seem to fit in with the regular perceived size changes that I already mentioned.

So what is the name of this giant orange red effect and what causes it?

electromagnetism - The definition of the Lorenz gauge condition

The inner product of two vectors in space-time is:

$$(x_1, y_1, z_1, t_1) \cdot (x_2, y_2, z_2, t_2) = x_1 x_2 + y_1 y_2 + z_1 z_2 - t_1 t_2$$

So

$$(\frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z}, \frac 1c \frac{\partial }{\partial t}) \cdot (A_1, A_2, A_3, \phi) = \text{div}(\vec A) - \frac 1c \frac{\partial \phi}{\partial t}$$

is Lorentz invariant, where $\vec A=(A_1, A_2, A_3)$. But the [Lorenz gauge condition] (https://en.wikipedia.org/wiki/Lorenz_gauge_condition) is defined by $\text{div}(\vec A) + 1/c\ \partial_t \phi=0$. Why has the minus changed into plus? So there is apparently no longer invariance.

newtonian mechanics - Types of Collisions and velocity

I know that there are three different collisions elastic, inelastic and perfectly inelastic. I wanted to know how the three velocities would rank respective to the type of collision. For example, if a ball were to collide with marble, which type of collision would maximize and/or minimize the velocity of the marble.

From my understanding, the velocities from greatest to least should go like- Elastic, Inelastic, Perfectly Inelastic.

topological order - FQH Edge Theory as decoupled chiral bosons

The action describing the edge theory of the Fractional Quantum Hall effect is given by \begin{equation} S = \frac{1}{4\pi} \int \mathrm{d}x \ \mathrm{d}t \left[ K_{IJ} \ \partial_{t}\phi_{RI} \partial_{x}\phi_{RJ} - V_{IJ} \partial_{x}\phi_{RI}\partial_{x}\phi_{RJ} \right] \end{equation} for scalar fields $\phi_{RI}$ with $I=1....dim(K)$ and some symmetric,invertible matrix K and some positive definite, symmetric "velocity matrix" V.

The equations of motion for the fields read \begin{equation} \partial_{t}\partial_{x} K_{IJ}\phi_{RJ} - \partial_{x}^{2}V_{IJ}\phi_{RJ} = 0. \end{equation} This smells already a lot like the chiral boson theory in 1+1D Minkowski space which this action is supposed to be equivalent to. However I do not see the transformation that "decouples" the fields. Can anyone help?

Thanks in advance for your responses!

general relativity - Violation of Lorentz symmetry on cosmological distances

This question is about the domain of validity of Lorentz symmetry.

As far as I know, general relativity is a generalization of special relativity. Does that mean that Lorentz symmetry is violated on cosmological distances, i.e in curved spacetime?

I am just curious, since people and the media are crazy about the OPERA result, yet it seems to me that there is already a violation of Lorentz symmetry in curved spapcetime that nobody mentioned.

Answer

There are two senses in which Lorentz invariance is preserved in general relativity, local and asymptotic.

In GR, if you look at the neighborhood of a given point, the local space-time looks like the space-time of special relativity. In such a little patch, you can't violate special relativity without violating General Relativity. So there is a local limit in which special relativity holds.

But on larger scales, curvature does violate special relativity. In particular, you can't find a global time which dilates or contracts, because the slices of constant time don't fit together globally to make a flat spacetime.

But on super-large scales, in a mostly empty patch, the gravitational field dies away. If the gravitational field is localized, the asymptotic space is that of special relativity. In this case, you have a global notion of special relativity--- you can boost or translate these solutions arbitrarily, according to the rules of special relativity. The transformations are symmetries of the asymptotic solution.

The result of this symmetry is that special relativity continues to hold at very large scales. This leads to stringent constraints on solutions of General Relativity, for example, the constraint that they should not be able to get something to move faster than light globally (see this question, and the linked question and answer: Does a Weak Energy Condition Violation Typically Lead to Causality Violation?).

The OPERA results violate Lorentz invariance in a way that is not allowed by standard General Relativity. Gubser has argued that it requires a violation of the Weak energy condition, the condition that gravity always is attractive on light rays, and Motl has argued further that it leads to causality violations, meaning that if you can go faster than light in a gravitational background, you can go back in time (his cogent suggestion for a physical argument appears here: ergosphere treadmills )

This is one aspect of the asymptotic special relativity restrictions still present in General Relativity. There are more, of course, as any asymptotically flat gravitational background obeys all of special relativity. String theory, for example, is a special relativistic theory which reproduces General Relativity, and it can do this because it's special relativity symmetry is acting on asymptotic regions, on far-past and far-future states which define the string S-matrix.

electrostatics - Whats the meaning of this equation relating electric field and potential energy?

My electrodynamics teacher was explaining that it is required to apply some energy (or do some work) on charges in order to arrange them in space. When he stated an example about energy in a capacitor the result was

$$U = \frac{1}{2}\frac{Q^2}{\epsilon_0 A / d}$$

Where

• 
  

  $Q$ is Capacitor charge

  
  


• 
  

  $A$ is area of capacitor

  
  


• 
  
  

  $d$ is distance between plates

  
  


• 
  

  $\epsilon_0$ is the permittivity of space

  
  

He later stated that it was equivalent to write it like this:

$$U = \frac{\epsilon_{0}}{2}E^2(Ad)$$

Where $E$ is the electric field inside the capacitor.

Then, as an analogy, he told us that for any other charge set, we could find something similar, and told us that it could be calculated as

$$U = \frac{\epsilon_{0}}{2} \int_v E^2 \, dV$$

• 
  

  How do I get to the second equation?

  
  


• 
  

  What does the integral in the third equation mean? How do I interpret it physically?

  
  

Answer

You have

$${U = \frac{1}{2}\frac{Q^2}{\frac{\epsilon_{0}A}{d}}}$$

$${\frac{\epsilon_{0}A}{d}}=C\equiv\textrm{ the capacitance of the capacitor.}$$

Hence

$${U=\frac{1}{2}\frac{Q^2}{C}}$$

Now $\displaystyle{Q=CV}$, where V is the p.d. between the capacitor plates.

So, $${U=\frac{1}{2}CV^2}$$

or $${U=\frac{1}{2}\frac{\epsilon_{0}A}{d}V^2}$$

Now the V (scalar potential) and the electric field are related as:

$\mathbf E= -\mathbf\nabla V$, which is valid for static electric fields as in the case of electric field between capacitor plates

where $\displaystyle{\mathbf \nabla V=\frac{\partial V}{\partial x}\hat{\textbf{x}}+\frac{\partial V}{\partial y}\hat{\textbf{y}}+\frac{\partial V}{∂z}\hat{\textbf{z}}}$

called the gradient of $V$ which gives you the maximum rate of change of V in all the three dimensions.

$\therefore$ $\displaystyle{V=-\int {\textbf{E}}.d{\textbf{r}}}$

Substituting this value of V in the equation for $U$, we get

$$\displaystyle{U=\frac{1}{2}\frac{\epsilon_0A}{d}\int (\textbf{E}.d{\textbf{r}})^2}$$

or

$$\color{red}{\displaystyle{ U=\frac{\epsilon_0}{2}\int_\textrm{all space} E^2d\tau}}$$

where $d\tau$ is the elemental volume

This is the electrical energy stored in the electric field of a charge configuration. This equation is valid for any charge configuration. This equation tells you that the electric energy is stored in the electric field when integrated over all space. The all space has certain importance because the electric field drops of as square of distance and hence it will be zero only if the distance from the charge is infinity. So, to get the entire energy of a charge configuration, you need to integrate the square of electric field taken over a small volume over the entire space. This means that the electric field itself exists as a manifestation of electric field which is the property of an electric charge. Greater the magnitude of the charge, greater will be the electric field and the energy stored in the field. The energy stored in the field drops off as you move farther away from the charge. Since the electric field drops off by square of distance from the charge, the energy drops off by the fourth power of distance from the charge. So this means the electric energy is well confined near the charge while the electric field could spread out to some distance. That region of space is called the sphere of influence of the charge.

Now, in the case of the parallel plate capacitor, the electric field exists only in between the plates and so the energy stored by a capacitor appears in between the plates. But if you treat the plates individually, the electric energy is present around the plate, but when you reconfigure it by placing another plate at some finite distance to it in a parallel plate capacitor arrangement, the electric field modifies and there is no single field of the plates, but only the superposition of field due to the combination of the plates. In such a configuration, no field exist outside the plates. So the entire energy of the modified field appears in between the plates. However, don't confuse this energy as the potential difference between the plates as you can easily verify from the equation. The energy in between the plates is the work done to assemble the charged plates configuration. But potential difference is the work done in order to move a unit positive charge from one plate to the other. The energy required for that also comes from the energy stored in the field between the plates. Remember that the energy stored in the field contains an $E^2$ term, while the potential difference contain only $E$ term.

rotational kinematics - Moment of inertia of a football and its angular momentum

What are the ways to create a mathematical model for the moment of inertia of a football? Can the moment of inertia of the football be simplified to two cones stack against each other?

I'm trying to find the angular momentum of the football by using $L=I \omega$

Is the angular momentum of football actually conserved?

What are the ways one can obtain the angular velocity on a rotating football?

general relativity - Speed of light when accelerating

I'm studying special relativity and saw the 2 postulates of Einstein. The most remarkable one for me is the universal speed of light. Einstein postulated that the speed of light in vacuum is the same for all inertial observers, regardless of the motion of the source. But what if we are accelerating and we are changing from one inertial frame to another, how will we observe the speed of light? Is it still the same? I suggest it's not? Because I alread saw a small introduction to general relativity where the light will bend in a gravitational field and therefore also when you accelerate. So does the speed of light change because it bends?

Answer

To understand this you need to understand what we mean by speed.

If I want to measure positions and times I need to set up a coordinate system. For example I can take my stopwatch and my metre rule and construct some Cartesian axes $t$, $x$, $y$ and $z$, then I can describe every point in spacetime by its position in my coordinates $(t,x,y,z)$. Once I have done this I can calculate the velocity of some object by watching how its position measured using my coordinates changes with the time measured using my coordinates. So for example if something is moving along my $x$ axis the speed is just:

$$ v = \frac{dx}{dt} $$

All very well, but why did I repeatedly use the phrase measured using my coordinates in the paragraph above? Well, it's because my coordinate system is just one way of measuring out spacetime and it doesn't necessarily have any fundamental physical significance. That means the speeds determined by my coordinates don't necessarily have any fundamental physical significance either.

To look into this a bit farther let's stick to flat spacetime so we don't have the complications introduced by general relativity. The obvious coordinates to choose are those of an inertial frame, and you'll meet the phrase inertial frame over and over again in studying relativity. In these coordinates the speed of light is always $c$.

But now suppose I'm accelerating with some acceleration $a$. There's nothing to stop me choosing coordinates where I remain at the origin i.e. I measure all distances and times relative to me. This would be a non-inertial frame, and as you'd expect if I use a non-inertial frame all sorts of weird things can happen e.g. Newton's laws no longer apply.

To make this concrete suppose I am accelerating along the $x$ axis and I measure the velocity of a light beam travelling along the $x$ axis using my non-inertial coordinates. I get the result:

$$ v_\text{light} = c\,\left(1 + \frac{a}{c^2}x \right) \tag{1} $$

I won't go thorough the derivation, but the accelerating coordinates are known as Rindler coordinates and the speed of light is calculated using an equation called the Rindler metric.

Anyhow, what I find is that the velocity of light now changes with the distance along the $x$ axis away from me. When the product $ax$ is positive (i.e. ahead of me) the velocity of light is greater than $c$ and when $ax$ is negative (behind me) the velocity of light is than than $c$. In fact when $ax = -c^2$ the velocity of the light slows to zero and there is an event horizon there (called the Rindler horizon).

But it's the same light in the same spacetime, just described using two different sets of coordinates. So what then is the real speed of the light. And here we reach the key point: there is no real speed. Relativity tells us that any set of coordinates is as valid as any other set of coordinates - you cannot say the inertial coordinates are right and the accelerating coordinates are wrong because they are equally valid ways of describing the same spacetime.

There is one last point to make. Suppose we take my equation (1) for the speed in the accelerating coordinates and calculate the speed at my position i.e. at $x=0$. The value is:

$$ v_\text{light} = c\,\left(1 + \frac{a}{c^2}\,0 \right) = c $$

So even though the speed of light is variable, at my position the speed is equal to $c$. And this is another absolutely key point: although the speed of light may vary in some coordinate systems, if you measure the speed of light at your position you will always get the value $c$. So the speed of light is always locally constant, it's just not globally constant.

A blue, white and red maze

First, let's give to Caesar what belongs to Caesar: this maze comes from the mind of Dave Phillips, a brilliant maze designer ¹.

So, for the maze. It is a fairly easy one:

           

Your goal is to exit by the blue tile on top left, after having entered the maze by the red tile at the bottom right. And here are the rules:

• Keep to the path;


• Do not turn around;



• Alternate between red and blue tiles (i.e. don't pass over two red/blue tiles in a row).

---

¹ _{Check his website!}

Answer

If going over the same path twice is allowed, this is a probable solution. I just started at the end and worked my way backwards to the start since there is only one possibility for the second-last tile (the other red tile would have resulted in an infinite loop)

The line colour is the colour of the last tile passed through

thermodynamics - Joule-Kelvin coefficent what does $p$ and $T$ represent?

In the Joule-Kelvin expansion we take a gas at one pressure $p_1$ and throttle it through a valve to another pressure $p_2$ where these two pressures are taken to be constant. The Joule-Kelvin coefficient is given by: $$\mu_{JK} = \left(\frac{\partial T}{\partial p}\right)_H$$ So what temperature does $T$ represent and what pressure does $p$ represent?

optics - Real and apparent depth not looking perpendicular to the interface

What is the formula for Real vs Apparent Depth, when not looking perpendicular to the interface?

I know the formula $$\frac{\text{real depth}}{\text{apparent depth}}=n$$ but I'm not looking for this.

mathematics - 2016 coins and a balance

On the table, there is a balance with two pans. The display on the balance tells the difference between the weight in the left ban and the weight in the right pan (measured in gram).

There are also $2016$ coins on the table. Cosmo tells Fredo: "There are exactly $99$ fake coins among these $2016$ coins. All genuine coins have the same weight. Some fake coins weigh one gram less than the genuine coins, and the other fake coins all weigh one gram more than the genuine coins."

Cosmo points at the leftmost coin $x$ and asks Fredo to determine whether $x$ is fake.

Question: Can Fredo decide whether coin $x$ is fake by using the balance at most twice?

Answer

Fredo should weigh the single coin against nothing, then all 2016 coins together against nothing. If the coin is real, then 2016 times its weight will be within 99 grams of the total weight. Otherwise, it will be at least 1917 grams off.

special relativity - Derivation of rest energy in Landau & Lifshitz

In Landau & Lifshitz The Classical Theory Of Fields there's a statement:

$$\mathscr E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}\tag{9.4}$$ This very important formula shows, in particular, that in relativistic mechanics the energy of a free particle does not go to zero for $v=0$, but rather takes on a finite value $$\mathscr E=mc^2.\tag{9.5}$$ This quantity is called the rest energy of the particle.

But this is the direct result of choosing Lagrangian as $$L=-mc^2\sqrt{1-\frac{v^2}{c^2}}.$$

We could as easily subtract $-mc^2$ from the Lagrangian to have $L^\prime=L+mc^2$ and have nothing changed in equations of motion, but now having the energy $$E^\prime=mc^2\left(\frac1{\sqrt{1-\frac{v^2}{c^2}}}-1\right),$$ which would give rest energy $E^\prime_0=0$.

So my question is what is the physical significance of using $(9.4)$ and corresponding Lagrangian to define rest energy? Why is such a seemingly special-case result makes the authors say "in relativistic mechanics the energy of a free particle does not go to zero" as if it's a general result?

Answer

So my question is what is the physical significance of using (9.4) and corresponding Lagrangian to define rest energy?

If the energy is defined as in 9.4, then this energy together with the three components of the momentum transform by Lorentz transformation in the same way as spacetime displacement four-vector, which is nice and preferable. If the energy was defined only as the kinetic energy, this could not be regarded as a component of such four-vector.

There are other reasons why the rest energy $mc^2$ is included, for example Einstein wrote many papers about the fact that change of energy of system in its rest frame leads to corresponding change in the rest mass according to relation $\Delta \mathscr{E} = \Delta mc^2$. This also suggests the above definition of energy.

symmetry - What happens to the Lagrangian of the Dirac theory under charge conjugation?

Consider a charge conjugation operator which acts on the Dirac field($\psi$) as
$$\psi_{C} \equiv \mathcal{C}\psi\mathcal{C}^{-1} = C\gamma_{0}^{T}\psi^{*}$$ Just as we can operate the parity operator on the Lagrangian, and we say that a theory has a symmetry if $$\mathcal{P}\mathcal{L}(t,x^{i})\mathcal{P}^{-1} = \mathcal{L}(t,-x^{i})$$

Suppose we operate $\mathcal{C}$ on the Dirac Lagrangian what should we get? $$\mathcal{C}\mathcal{L}_{Dirac}(x^{\mu})\mathcal{C}^{-1} = \mathcal{L}^{*}_{Dirac}(x^{\mu}) ?$$ in analogy to the transformation of the scalar field $\phi$ under charge conjugation.

On a same note one can ask what equation should $\psi_{C}$ satisfy? Should it satisfy the conjugated Dirac equation as $$(i\gamma^{\mu}\partial_{\mu} + m)\psi_{C} = 0 ?$$ If so can someone give me the physical interpretation for it.

I am asking this question as I want to explicitly use $\psi_{C}$ and check whether it keeps the Dirac Lagrangian invariant. I have done a calculation by substituting $\psi_{C}$ in the Dirac equation and have found it is not satisfying as shown below.

$$(i\gamma_{\mu}\partial_{\mu} - m)\psi_{C} = i(\gamma_{\mu}C\gamma_{0}^{T})\partial^{\mu}\psi^{*} - (C\gamma_{0}^{T})m\psi^{*}$$ We will use $C^{-1}\gamma_{\mu}C = - \gamma_{\mu}^{T}$ and $\{\gamma_{\mu}^{T},\gamma_{\nu}^{T}\} = 2g_{\mu\nu}$.
Consider \begin{align} \gamma_{\mu}C\gamma_{0}^{T} &= CC^{-1}\gamma_{\mu}C\gamma_{0}^{T} \\ &= -C\gamma_{\mu}^{T}\gamma_{0}^{T} \\ &= C\gamma_{0}^{T}\gamma_{\mu}^{T} \end{align} Hence substituting back we will get
\begin{align} (i\gamma_{\mu}\partial_{\mu} - m)\psi_{C} &= C\gamma_{0}^{T}(i\gamma_{\mu}^{T}\partial^{\mu}\psi^* - m\psi^*) \\ &= C\gamma_{0}^{T}[(i\gamma_{\mu}^{T}\partial^{\mu}\psi^* - m\psi^*)^{T}]^{T} \\ &= C\gamma_{0}^{T}(i\psi^{\dagger}\gamma_{\mu}\partial^{\mu} - m\psi^{\dagger})^{T} \\ &= C\gamma_{0}^{T}[(i\bar{\psi}\gamma_{0}\gamma_{\mu}\partial^{\mu} - m\bar{\psi}\gamma_{0})]^{T} \\ \end{align} Now \begin{align} \gamma_{0}\gamma_{\mu}\partial^{\mu} &= (\gamma_{0}\partial^{t} + \gamma_{i}\partial^{i})\gamma_{0} \end{align} If we substitute back we will get \begin{align} (i\gamma_{\mu}\partial_{\mu} - m)\psi_{C} &= C\gamma_{0}^{T}[\{i\bar{\psi}(\gamma_{0}\partial^{t} + \gamma_{i}\partial^{i}) - m\bar{\psi}\}\gamma_{0}]^T \\ &\neq 0 \end{align}

Answer

The short answer to the question, "What happens to the Lagrangian of the Dirac theory under charge conjugation?" is, "Nothing." It is invariant with respect to charge conjugation.

Before getting to the longer exposition, I'd like to point out a potential misunderstanding about the nature of invariance of the equations of motion under symmetry transformations that arises about your statement regarding the parity ($\mathbf{x}\to-\mathbf{x}$). Your equation \begin{align} \mathcal{P}\mathcal{L}(t,x^{i})\mathcal{P}^{-1} = \mathcal{L}(t,-x^{i}) \end{align} is correct. But this, in and of itself, doesn't "say that a theory has a symmetry." In fact, it places a restriction on the theory -- that the theory must be an even function of the position vector. For example, for a real scalar field, $\phi(t,\mathbf{x})$, the kinetic energy operator in the Lagrangian density, $\partial_\mu \partial^\mu \phi(x)$ is invariant under parity. In fact, it is only the action ($S=\int d^4x \mathcal{L}$) that need be invariant under the symmetry transformation. (Often this reduces to the invariance of the Lagrangian density.) So each term of the Lagrangian (density) need not be invariant (though this is often the case).

Returning to the Dirac equation -- the full statement, in words, of invariance of the theory of the interaction of electrons with light (QED) under the discrete transformations ($\mathcal{P},\mathcal{C}$, & $\mathcal{T}$) is that the theory is invariant under each of them separately or in any combination. (QED is less "interesting" than the electroweak theory in this regard since the electroweak theory appears to violate all three of these separately -- but perhaps not all simultaneously.)

We have to remember that the invariance under $\mathcal{C}$ requires the transformation not only of the wave function, $\psi_{C} \equiv \mathcal{C}\psi\mathcal{C}^{-1} = C\gamma_{0}^{T}\psi^{*}$ but also the charge, $q\to -q$. In the case of the free Dirac equation/Lagrangian considered above, the charge does not feature, so it's not directly relevant to the present discussion but it's important to keep in mind.

Now the direct answers to your questions. (I won't do the algebra since, if you can carry out the calculations for the Dirac equation itself, then the transformation of the Lagrangian should be straightforward.)

"Suppose we operate C on the Dirac Lagrangian what should we get?" The corrected relation is:

\begin{align}\mathcal{C}\mathcal{L}_{Dirac}(x^{\mu})\mathcal{C}^{-1} = \mathcal{L}_{Dirac}(x^{\mu})\end{align}

(Incidentally, your relation turns out to be all right since the Lagrangian (density) must be a Hermitian, scalar operator, so $\mathcal{L}^* = \mathcal{L}^\dagger = \mathcal{L}$. EDIT: Thanks to Omkar for pointing out that this is wrong. $\mathcal{L}^*\ne\mathcal{L}$.)

On a same note one can ask what equation should ψC satisfy? Should it satisfy the conjugated Dirac equation as \begin{align} > (i\gamma^{\mu}\partial_{\mu} + m)\psi_{C} = 0? \end{align}

If you're using the "West Coast" metric $(+1,-1,-1,-1)$ then the equation that $\psi_C$ should satisfy is the one above with $m\to -m$. That is, the free Dirac equation is the same for $\psi$ and $\psi_C$. This is because the masses of the particle and anti-particle are identical, as first divined by Dirac. (If you're using the metric of the opposite sign, then you're equation is correct.)

Is Newtonian gravity consistent with an infinite universe?

Let us assume that we have have an infinite Newtonian space-time and the universe is uniformly filled with matter of constant density (no fluctuations whatsoever), all of it at rest. By symmetry, the stuff in this universe should not collapse or change position in any way if gravity is the only force acting on it. Now, consider Gauss theorem. It says that within a spherical system (it says more that that but this will suffice), the gravitational force felt by any point will be the same as if all matter between the center and the point were concentrated at the center. The matter outside the sphere does not contribute any force). Thus, in such a system, the matter (stuff, I do not say gas so we can consider it continuous) will collapse towards the center of the sphere. We can apply this argument to any arbitrary point in our previous infinite homogeneous universe, and conclude that matter will collapse towards that point (plus the point is arbitrary). So, why is Gauss's theorem is not valid in this case?

I was signaled that this question is a duplicate and has been answered, however: The best I could get from the redirected question is this quote: "However, the mass can't be negative and the energy density is positive. This would force a violation of the translational symmetry in a uniform Newtonian Universe". It still doesn't give a satisfactory answer. For instance: how is that symmetry broken if we assume that there is no noise nor small density fluctuations in the system? How can you choose then the absolute "origin" that will break the symmetry? Still doesn't make sense to me.

word - When you give it out in the wrong way

To know the secrets of the universe and hence attain ultimate power, you just need to enter a four-letter password. All you have is a sheet of paper with the following written on it, which is supposed to be a clue. Can you read between the lines and make out the password?

To You, Dear!

Try it just deviously; this is easy, Mr. Puzzler Omniscient. Knowledge about all languages aids. Don't enter nonsense, BS or rubbish answers. Zero experience in this? Hang on, read again this epistle, minding peculiar unusual specialities. Simple afterthoughts may assist you.

Answer

The password is

TIME

Because

The first letters of each sentence/clause spell a word that translates to time in another language

To You, Dear! -> Tyd (Afrikaans)
Try it just deviously -> Tijd (Dutch)
this is easy, Mr. Puzzler Omniscient -> tiempo (Spanish)
Knowledge about all languages aids -> kaala (Sanskrit)
Don't enter nonsense, BS or rubbish answers -> denbora (Basque)
Zero experience in this? -> zeit (German)
Hang on, read again -> hora (Galician)
this epistle, minding peculiar unusual specialities -> tempus (Latin)

Simple afterthoughts may assist you -> samay (Hindi)

And the title...

When you give it out the wrong way = EMIT backwards = TIME

quantum mechanics - I don't understand the relationship between electron indistinguishability and the Pauli exclusion principle

I know I'm wrong but this is my line of thought: If electrons are indistinguishable, then why do we have an exclusion principle? If we have two electrons in an s orbital, the Pauli exclusion principle says that they can't have the same set of quantum numbers, but then what does that say about electrons being indistinguishable?

So we have these two electrons that are supposed to be indistinguishable, but then we say, no they can't have the same set of quantum numbers, isn't this making them distinguishable then?

Answer

The indistinguishability of particles is expressed by imposing certain symmetry constrains on the state functions and on the observables. As you may know, there can be symmetric and antisymmetric state functions as you interchange two particle coordinates, and all the observables must be invariant under such operations. And this postulate agrees with the experimental data. The particles being identical can be explained as saying that the physical system is unchanged if the particles are interchanged. This formulation can be expressed mathematically in the following way

$\lvert\psi(p(x_1,\dots,x_n))\rvert^2 = \lvert\psi(x_1,\dots,x_n)\rvert^2$

where $p$ is the permutation of the N particle coordinate. However, from the above you can see that the word interchange, here has no physical meaning. The l.h.s and the r.h.s have no separate meaning and it shows the redundancy in the notation. Putting it in other words, the same particle configuration can be expressed in different ways.

As for Pauli exclusion principle, you know that it says that no two identical fermions may occupy the same quantum state. This is because the total wave function for two identical fermions is anti symmetric with respect to the interchange of the particles.

Hope it helps a bit, but for a far better understanding you should look up the Occupation Number Representation. The starting point of this formalism is the notion if indistinguishability.

experimental physics - Are random errors necessarily Gaussian?

I have seen random errors being defined as those which average to 0 as the number of measurements goes to infinity, and that the error is equally likely to be positive or negative. This only requires a symmetric probability distribution about zero. However typing this question into Google, I did not find a single source that suggested random errors could be anything other than gaussian. Why must random errors be gaussian?

Answer

Are random errors necessarily gaussian?

Errors are very often Gaussian, but not always. Here are some physical systems where random fluctuations (or "errors" if you're in a context with the thing that's varying constitutes an error) are not Gaussian:

1. 
   

   The distribution of times between clicks in a photodetector exposed to light is an exponential distribution.$^{[a]}$

   
   


2. 
   

   The number of times a photodetector clicks in a fixed period of time is a Poisson distribution.

   
   


3. 
   

   The position offset, due to uniformly distributed angle errors, of a light beam hitting a target some distance away is a Cauchy distribution.

   
   

I have seen random errors being defined as those which average to 0 as the number of measurements goes to infinity, and that the error is equally likely to be positive or negative. This only requires a symmetric probability distribution about zero.

There are distributions that have equal weight on the positive and negative side, but are not symmetric. Example: $$ P(x) = \left\{ \begin{array}{ll} 1/2 & x=1 \\ 1/4 & x=-1 \\ 1/4 & x=-2 \, . \end{array}\right.$$

However typing this question into Google, I did not find a single source that suggested random errors could be anything other than gaussian. Why must random errors be gaussian?

The fact that it's not easy to find references to non-Gaussian random errors does not mean that all random errors are Gaussian :-)

As mentioned in the other answers, many distributions in Nature are Gaussian because of the central limit theorem. The central limit theorem says that given a random variable $x$ distributed according to a function $X(x)$, if $X(x)$ has finite second moment, then given another random variable $y$ defined as the average of many instances of $x$, i.e. $$y \equiv \frac{1}{N} \sum_{i=1}^N x_i \, ,$$ the distribution $Y(y)$ is Gaussian.

The thing is, many physical processes are the sums of smaller processes. For example, the fluctuating voltage across a resistor is the sum of the voltage contributions from many individual electrons. Therefore, when you measure a voltage, you get the underlying "static" value, plus some random error produced by the noisy electrons, which because of the central limit theorem is Gaussian distributed. In other words, Gaussian distributions are very common because so many of the random things in Nature come from a sum of many small contributions.

However,

1. 
   

   There are plenty of cases where the constituents of an underlying error mechanism have a distribution that does not have a finite second moment; the Cauchy distribution is the most common example.

   
   


2. 
   

   There are also plenty of cases where an error is simply not the sum of many small underlying contributions.

   
   

Either of these cases lead to non-Gaussian errors.

$[a]$: See this other Stack Exchange post.

general relativity - Would it be possible for a human to simply "step" through a traversable wormhole?

(For the purposes of a science-fiction story where the author is attempting to have as much basis in fact as possible) if one were to create or make use of a Morris-Thorne or similar traversable wormhole and accelerate one end to modify it to be a "time machine", would one simply be able to step through? Would the transition, in theory, be safe or deadly? Would some kind of life-support gear be in order? Would this apply also if one stepped back to the younger end (from the point of view of the person travelling through).

For the purposes of this I'm assuming that both ends of the wormhole are on the Earth's surface and at points in time where the atmospheric, surface, etc. conditions are suitable for a human to live normally.

I apologise if this should be in the Science Fiction site, but I wish to look at this from the point of view of a scientist with the sense of wonder of, say, Hawking. Any information and thoughts on this would be amazingly helpful.

EDIT: I don't feel this question is a duplicate of the question: Would there be forces acting on a body as it traverses space near a wormhole?; as I seek to understand the effects of passing through the wormhole to a human's life and health by "stepping" through, not an ambiguous body in space.

Answer

There are quite a variety of concerns here. For a full overview you can check "Lorentzian wormholes" by Visser. For now we'll consider wormholes defined by the thin-shell formalism, where the matter propping up the wormhole is on a very thin shell on the wormhole mouth.

First, there's the tidal forces. As the tidal stress induced by gravitational forces depends on the Riemann tensor, we can consider the Riemann tensor of a thin-shell spacetime :

$$R_{abcd} = - \delta(r) \tau_{abcd} + \Theta(r) R^+_{abcd} + \Theta(-r) R^-_{abcd}$$

with $\tau_{abcd}$ a tensor depending on the discontinuity of the extrinsic curvature $K$. Since this is the most important part in propping the wormhole open, let's assume (without any good reasons) that we can neglect the curvature outside of the wormhole mouth.

The extrinsic curvature tensor locally looks roughly like

\begin{equation} K = \begin{pmatrix} 1/R_t & 0 & 0 & 0 \\ 0 & 1/R_1 & 0 & 0 \\ 0 & 0 & 1/R_2 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \end{equation}

$R_t$ is the radius of the timelike curve described by that point (it's related to the wormhole's acceleration), while $R_1$, $R_2$ are the principal radii of the surface. Hence by making the radii as large as possible (that is, making the surface as flat as possible), we can make the tidal effects arbitrarily small.

In other words, if the wormhole mouth has at least one part of it that is flat, then one can pass through it unmolested by tidal forces.

Then there's the matter of the matter itself propping the wormhole open. As there isn't really any realistic model of matter that can prop up a people-sized wormhole (or even one large enough to get one electron to go through), it's hard to say exactly what are the possible risks (the matter involved might not even couple to ordinary matter), but considering a rough estimate of the order of magnitude for the (absolute) energy is about $10^{27} kg$, in an area of a few cubic meter, it's safe to say that it's probably not a great idea to approach it too much.

There again, flat faces are a benefit : the stress-energy tensor for the thin-shell part is again zero if the surface is flat. This does not mean still that this is a good idea : such very dense matter is likely to radiate in the vicinity.

Another issue is the throat length and time : if we drop the thin-shell assumption, which isn't terribly realistic, there is still some way to travel between the two mouthes. The length of the throat is related to the amount of negative energy required (long throats require less negative energy), which means that a more reasonable worhole may have very long travel times to actually cross. The perceived time is also an issue, as the travel time could be very short for the person crossing it but very long from the outside perspective (this is related to the redshift function which brings up a whole bunch of other potential problems).

Then if we actually assume a time machine scenario, things get much much worse. To simplify matters somewhat, you might be aware that you can describe the quantum vacuum as a gallore of virtual particles (wrong yes, but let's go with that for now). As a wormhole approaches time machine formation, there are more and more virtual particles forming almost-loops, which has the bad side effect of blue-shifting them, increasing their energy. Those become actual loops when the time machine is actually formed, which has the result of making the stress-energy tensor diverge : the quantum vacuum has infinite energy, which probably bad results if this was possible.

It's quite likely that the wormhole collapses before this happens.