Conservation of angular momentum and RoS

[SlowThinker]

So I'm starting to see what's going on: the tubes lead through time as well as space, and the photons rotate in a square that is not purely spatial. However, trying to read Peter's math, I found 2 strange things and I can't quite see where it gets wrong.
1.

So, taking the first photon again, its new 4-position will be $X'^a = ( - \gamma u, \gamma, 1 )$, and its new 4-momentum will be $P'^a = ( \gamma k, - \gamma u k, k )$ before the bounce

I find it strange that a photon, that is clearly moving to the right (and up), has a negative momentum along the x-axis.
All the math leading to this seems correct. When I tried to replace $u$ with $-u$, it broke the position of the corner (moved it into the future rather than past).
So I'm not quite sure what's wrong here. (I am?)

2. Out of curiosity, I wanted to find all 8 momenta (before and after each corner) but got stuck pretty fast. Let's follow the photon to the next corner.
Again the front-top corner is ${X'}_{FT}^a = ( - \gamma u, \gamma, 1 )$ and, unless I'm mistaken, the rear-top corner is ${X'}_{RT}^a=(\gamma u, -\gamma, 1)$. The momentum along the top tube is $P'^a = \left[ \gamma \left( 1 + u \right) k, - \gamma \left( 1 + u \right) k, 0 \right]$ . Now let's look at the angular momentum 4-tensors.
The ${M'}_{FT}^{01}$ component is
$${M'}_{FT}^{01}=X_{FT}^0 P^1 - X_{FT}^1 P^0=-\gamma u(-\gamma (1+u) k)\,-\,\gamma \gamma (1+u)k=\gamma^2(1+u)k(u-1)=-k$$
but
$${M'}_{RT}^{01}=X_{RT}^0 P^1 - X_{RT}^1 P^0=\gamma u (-\gamma(1+u)k)\,-\,-\gamma \gamma (1+u)k=\gamma^2 k(1+u)(-u+1)=k$$
That looks like this component of the angular 4-momentum changes during flight
This looks like a pretty basic mistake but again, I can't find it. Other components seem to match.

 

[PeterDonis]

I find it strange that a photon, that is clearly moving to the right (and up), has a negative momentum along the x-axis.

No, in the primed frame, it's moving to the left and up. The frame itself is moving to the right (positive x direction) relative to the CoM frame, so a photon that is moving purely up (positive y direction) in the CoM frame will be moving to the left (negative x' direction) and up (positive y' direction) in the primed frame.

the rear-top corner

It really helps if you don't use words but math; words are ambiguous, math is not. I don't know which corner you think is the "rear-top corner".

In the CoM frame, here are the four mirrors with their 4-positions $X^a = (t, x, y)$ at the instant of the bounce, and the 4-momenta $P^a = (E, p^x, p^y)$ of the photons bouncing off of them at time $t = 0$ in that frame, just before and just after the bounce:

Mirror A: $X^a = (0, 1, 1)$. Photon bouncing off mirror A: $P^a = (k, 0, k)$ before bounce; $P^a = (k, -k, 0)$ after bounce.

Mirror B: $X^a = (0, -1, 1)$. Photon bouncing off mirror B: $P^a = (k, -k, 0)$ before bounce; $P^a = (k, 0, -k)$ after bounce.

Mirror C: $X^a = (0, -1, -1)$. Photon bouncing off mirror C: $P^a = (k, 0, - k)$ before bounce; $P^a = (k, k, 0)$ after bounce.

Mirror D : $X^a = (0, 1, -1)$. Photon bouncing off mirror D: $P^a = (k, k, 0)$ before bounce; $P^a = (k, 0, k)$ after bounce.

In the moving frame, which is moving in the positive $x$ direction at speed $u$ relative to the CoM frame, the above 4-positions and 4-momenta transform to:

Mirror A: $X^a = ( - \gamma u, \gamma, 1)$. Photon bouncing off mirror A: $P^a = ( \gamma k, - \gamma u k, k )$ before bounce; $P^a = ( \gamma (1 + u) k, - \gamma ( 1 + u ) k, 0)$ after bounce.

Mirror B: $X^a = ( \gamma u, - \gamma, 1)$. Photon bouncing off mirror B: $P^a = ( \gamma (1 + u) k, - \gamma ( 1 + u ) k, 0 )$ before bounce; $P^a = ( \gamma k, - \gamma u k, -k)$ after bounce.

Mirror C: $X^a = ( \gamma u, - \gamma, -1)$. Photon bouncing off mirror C: $P^a = ( \gamma k, - \gamma u k, - k)$ before bounce; $P^a = ( \gamma (1 - u) k, \gamma ( 1 - u ) k, 0 )$ after bounce.

Mirror D : $X^a = ( - \gamma u, \gamma, -1)$. Photon bouncing off mirror D: $P^a = ( \gamma (1 - u) k, \gamma ( 1 - u ) k, 0)$ before bounce; $P^a = ( \gamma k, - \gamma u k, k )$ after bounce.

That looks like this component of the angular 4-momentum changes during flight

Your intuition that that shouldn't happen is correct. Hopefully the above will help you to re-check your computations.

 

[PeterDonis]

Let's follow the photon to the next corner.

If by this you mean, let's look at what the angular momentum tensor looks like when the photon has reached its next corner, you have to account for the fact that time has passed, so the 4-position vectors of the mirrors in the CoM frame will now have nonzero $t$ components. Let's look at just mirror B, since that's the next mirror that the photon that bounces off mirror A at $t = 0$ in the CoM frame will hit.

It will take 2 units of time for the photon--call it photon a--to travel from mirror A to mirror B, since the mirrors are two units of distance apart (in this case, two units in the $x$ direction). So when photon a reaches mirror B, the 4-position of mirror B in the CoM frame will be $X^a = (2, -1, 1)$. The 4-momentum of photon a before the bounce will be the same as when it left mirror A, i.e., $P^a = (k, -k, 0)$. If you calculate the 4-angular momentum tensor using these values, you should get the same answer as when you calculated it for photon a just leaving mirror A at time $t =0$ in the CoM frame. And if you then transform the new 4-position vector into the primed frame, and use it to calculate the 4-angular momentum of photon a in that frame just before it hits mirror B, you should get the same answer as what you calculated in the primed frame for photon a just leaving mirror A.

 

[SlowThinker]

It will take 2 units of time for the photon--call it photon a--to travel from mirror A to mirror B, since the mirrors are two units of distance apart (in this case, two units in the $x$ direction). So when photon a reaches mirror B, the 4-position of mirror B in the CoM frame will be $X^a = (2, -1, 1)$. The 4-momentum of photon a before the bounce will be the same as when it left mirror A, i.e., $P^a = (k, -k, 0)$. If you calculate the 4-angular momentum tensor using these values, you should get the same answer as when you calculated it for photon a just leaving mirror A at time $t =0$ in the CoM frame. And if you then transform the new 4-position vector into the primed frame, and use it to calculate the 4-angular momentum of photon a in that frame just before it hits mirror B, you should get the same answer as what you calculated in the primed frame for photon a just leaving mirror A.

I still can't quite make it to add up. Just a quick question:
The speed of the Center of mass is to be ignored, or should I somehow subtract its speed from that of the photon?
I'll check my spreadsheet again and again and hopefully make it work eventually.

A somewhat unrelated question, can we say that a photon with 4-momentum (p,q,r,s) has wavelength hc/p?

 

[jartsa]

You may want to consider the possibility that you are wrong
I'm still digesting Peter's post but
seems to suggest that the vertical photon's energy is $\gamma k$

Vertical photon's energy is proportional to gamma. So energy is Increased not decreased. I don't know what the k includes though.
I'll explain simple Doppler shift in a simple way. Pay attention. Or ignore if you dislike simple stuff:

A moving thing waves, it creates a start of a crest at position p, then the thing either chases or runs away from the start of the crest for a time: (1/frequency of the thing) / 2. Then the thing creates the end of the crest, near the start of the crest if the thing was chasing the wave, far away from the start of the crest if the thing was running away from the wave.

Crest length is: velocity of the wave times the time we calculated, minus closing or separation speed between the thing and the wave multiplied by the time we calculated.

Oh yes, the frequency of the moving thing is: the frequency in the thing's rest frame divided by gamma.

 

[PeterDonis]

The speed of the Center of mass is to be ignored, or should I somehow subtract its speed from that of the photon?

The way I derived quantities in the moving frame was to first derive them in the CoM frame and then Lorentz transform. If you do that, you only have to worry about the 4-vectors that describe each photon (4-position and 4-momentum); you don't have to worry about what the CoM is doing.

can we say that a photon with 4-momentum (p,q,r,s) has wavelength hc/p?

If p has units of energy, yes, the photon's frequency will be p / h (strictly speaking, $\hbar$, but we'll ignore that here), and its wavelength will be hc / p, using conventional units. However, often physicists will use units in which $\hbar = c = 1$, so energy, frequency, momentum, and inverse wavelength all have the same units; in those units, the frequency and wavelength would be p and 1 / p, respectively.

 

[PeterDonis]

Vertical photon's energy is proportional to gamma.

Yes, that' what $\gamma k$ means.

I don't know what the k includes though.

$k$ is just the photon's energy in the CoM frame, as should be obvious from the 4-momentum vectors that I wrote down in that frame. I am using units in which $\hbar = c = 1$, as I described in my response to SlowThinker just now.

the frequency of the moving thing is: the frequency in the thing's rest frame divided by gamma.

You're ignoring 4-vectors again. Frequency and inverse wavelength are components of a 4-vector; in fact, if we take into account quantum mechanics, it is just the 4-momentum vector divided by $\hbar$. So you have to Lorentz transform frequency and wavelength the same way you transform energy and momentum, as components of a 4-vector.

Also, what do you mean by "the frequency of the moving thing", as opposed to the frequency of the light it's emitting? The frequency of the light does not get transformed by $\gamma$; it gets transformed by the Doppler shift factor, which is not $\gamma$. (If you look at the math, you will see that this is because, as I said just now, the frequency-wave number 4-vector has to be transformed as a 4-vector.)

 

[jartsa]

You're ignoring 4-vectors again.

Maybe I have to use 4-vectors then. But I must use them in my own special way. So if I boost 4-positions of two consecutive wave crests, find 3-positions of those boosted 4-positions, subtract those 3-positions, do I get a boosted wave-length?

 

[PeterDonis]

if I boost 4-positions of two consecutive wave crests, find 3-positions of those boosted 4-positions, subtract those 3-positions, do I get a boosted wave-length?

Have you tried it?

You might also want to try boosting the 4-momentum of the light, since that is really what defines the light's frequency and wavelength--see post #56.

 

[pervect]

I still can't quite make it to add up. Just a quick question:
The speed of the Center of mass is to be ignored, or should I somehow subtract its speed from that of the photon?
I'll check my spreadsheet again and again and hopefully make it work eventually.

A somewhat unrelated question, can we say that a photon with 4-momentum (p,q,r,s) has wavelength hc/p?

This is the first time the "center of mass" has come up in this thread. A few remarks were made earlier that I think were helpful.

https://en.wikipedia.org/wiki/Relativistic_angular_momentum
In any frame, the "space-space" components of this 4-tensor will be the ordinary angular momentum in that frame (but as an antisymmetric 3-tensor instead of a pseudovector), and the "time-space" components will be the "mass moment".

[the total angular momentum 4-tensor for all 4 photons before the boost, which is constant in coordinate time t]

$$
M^{ab} = \left[ \begin{matrix}
0 & 0 & 0 \\
0 & 0 & 4k \\
0 & -4k & 0
\end{matrix} \right]
$$

So we see in the original frame, there is no mass-moment, i.e. the origin of the system is at the center of mass-energy. I'll add that I didn't double check all the work here personally, due to laziness.

I believe we do see a mass-moment in the boosted frame, at least that's what I got via a Lorentz boost on the above tensor. It didn't quite match Peter's boosted version in the quoted post. Because Peter's posted version didn't have a factor of 4, I didn't quite trust it.

What I get via the direct boost (using a computer algebra package) is:

$$
\left[ \begin {array}{ccc} 0&0&4\,k{\gamma}\,v\\0&0&4\,k{\gamma}\\-4\,k{{\gamma}}\,v&-4\,k{\gamma}&0
\end {array} \right]
$$

It's not quite what I expected, but - assuming I didn't make an error in my rather hasty calculation, it confirms the idea that the mass-energy moment does change as a result of the boost. In other words, the 3-angular-momentum around the given point increases by a factor of gamma, which we can ascribe as being due to time dilation. Additionally the mass-energy moment changes, which we can ascribe as being due to the relativity of simultaneity. So the angular momentum around the given point doesn't change, but in the boosted frame this point is no longer the "center of mass-energy". I believe this is also what the various diagrams (which I haven't really checked either) are also telling us.

 

[PeterDonis]

I believe we do see a mass-moment in the boosted frame

We certainly should on physical grounds, since the CoM is moving in the boosted frame.

It didn't quite match Peter's boosted version in the quoted post.

My notation in post #45 was rather confusing, as I can see on re-reading. The boosted $M'_{ab}$ tensor I give there is only for the first photon, not for the total of all four. In other words, it corresponds to this:

its value is:
$$
M^{ab} = \left[ \begin{matrix} 0 & -k & -k \\ k & 0 & k \\ k & -k & 0 \end{matrix} \right]
$$

If you do the same computation of the boosted tensor individually for the other three photons, and add them all up to get the counterpart to this:

it turns out to be, in the CoM frame:
$$
M^{ab} = \left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 4k \\ 0 & -4k & 0 \end{matrix} \right]
$$

Then I think you get what you posted, i.e., this:

What I get via the direct boost (using a computer algebra package) is:
$$
\left[ \begin {array}{ccc} 0&0&4\,k{\gamma}\,v\\0&0&4\,k{\gamma}\\-4\,k{{\gamma}}\,v&-4\,k{\gamma}&0 \end {array} \right]
$$

 

[SlowThinker]

Surely I must be stupid or something. I can't figure out the 4-momentum even in the rest frame.

In the CoM frame, ...

Mirror A: $X^a = (0, 1, 1)$. Photon bouncing off mirror A: $P^a = (k, 0, k)$ before bounce; $P^a = (k, -k, 0)$ after bounce.

Mirror B: $X^a = (0, -1, 1)$. Photon bouncing off mirror B: $P^a = (k, -k, 0)$ before bounce; $P^a = (k, 0, -k)$ after bounce.

The 4-momentum of photon leaving mirror A: X x P = (0, 1, 1) x (k, -k, 0):
$M_{01}=0 (-k) - k = -k\text{ ,}\ \ \ \ \ \ \ \ \ \ M_{02}=-k\text{ , }M_{12}=+k$
This same photon arriving at B: X x P = (0, -1, 1) x (k, -k, 0):
$M_{01}=0 (-k) - (-1)k = +k\text{ , }M_{02}=-k\text{ , }M_{12}=+k$
So again, it looks like the angular 4-momentum (component 01) changes in flight.

Your intuition that that shouldn't happen is correct. Hopefully the above will help you to re-check your computations.

What am I doing wrong? This is pretty basic stuff, right?
I've tried to fix it in various ways but none made sense.

The funny thing is that the 4-angular momentum conserves in all 4 corners, even in the moving frame.

 

[PeterDonis]

This same photon arriving at B: X x P = (0, -1, 1) x (k, -k, 0):

No. The same photon arriving at B: X x P = (2, -1, 1) x (k, -k, 0). As I posted before, you have to update the 4-position to account for the travel time of the photon; it isn't at mirror B at time t = 0, it's there at time t = 2, and the time is part of the 4-position.

 

[SlowThinker]

No. The same photon arriving at B: X x P = (2, -1, 1) x (k, -k, 0). As I posted before, you have to update the 4-position to account for the travel time of the photon; it isn't at mirror B at time t = 0, it's there at time t = 2, and the time is part of the 4-position.

The trouble is, at that time, the center of mass is at position (2,0,0) as well, so the "4-distance" from COM to B is (0, -1, 1).
Or, at which point to I advance the COM to new time? If I don't, then after a full cycle, leaving mirror A, I end up with $(8, 1, 1) \times (k, -k, 0) = -9k$ for the 01 component.
It gets even worse in the moving frame.

 

[PeterDonis]

The trouble is, at that time, the center of mass is at position (2,0,0) as well, so the "4-distance" from COM to B is (0, -1, 1).

The 4-position vector is not the "4-distance" from the point to the CoM. (There is no such thing anyway; 4-distance is something that goes between events, not objects or worldlines.) It is the 4-position of the event relative to the spacetime origin of the frame. The very fact that you are getting the wrong answer by taking the "4-distance" to the CoM should be a clue that you are doing it wrong.

at which point to I advance the COM to new time?

You don't. The spacetime origin of the frame is an event, not an object. Remember, this is geometry: the spacetime origin of coordinates does not move. It's a particular point in the geometry, and the 4-position of any other point is taken relative to that particular point. The same goes for 4-angular momentum; it is 4-angular momentum about a particular point in spacetime, not about a particular point in space.

 

[jartsa]

I have now boosted an energy-momentum vector, witch had some momentum to the right, to left. IOW I changed to a frame that moves to the left.

There was more energy in the boosted frame than in the original frame. (Just as I expected)

So right moving light should have extra energy in the picture in post #28.

 

[PeterDonis]

I changed to a frame that moves to the left.

I'm not sure about post #28, but in my analysis in post #45, my "moving" frame was moving to the right (positive x direction) relative to the CoM frame. In this frame, a right-moving photon has less energy, and a left-moving one has more (which is consistent with the analysis I posted).

 

[SlowThinker]

I'm not sure about post #28, but in my analysis in post #45, my "moving" frame was moving to the right (positive x direction) relative to the CoM frame. In this frame, a right-moving photon has less energy, and a left-moving one has more (which is consistent with the analysis I posted).

You seem to love to use unusual coordinates to confuse people
Now that I realized your movement is the opposite of the one in my pictures, your numbers do support Jartsa.
But that, unfortunately, means that my reasoning is wrong in this point as well, although again, I can't see where.

Surely the number of waves between bounces must be invariant in all frames?
It is true for the vertical photons, but not for the horizontal ones, if your 4-momenta are correct. A left-moving photon in a left-moving square has shorter wavelength, and spends more time+distance in flight, so the number of waves it makes is $\gamma^2(1+u)^2$-times higher.

 

[PeterDonis]

You seem to love to use unusual coordinates to confuse people

That wasn't my intent; to be honest, I wasn't sure in which direction things were supposed to be moving in previous scenarios in this thread, so I picked a frame moving to the right because that's the first version of the Lorentz transformation that I learned.

A left-moving photon in a left-moving square has shorter wavelength

No, it doesn't, it has longer wavelength. A left moving photon in a right-moving frame has shorter wavelength.

 

[SlowThinker]

No, it doesn't, it has longer wavelength. A left moving photon in a right-moving frame has shorter wavelength.

Now I'm thoroughly confused.

Mirror A: $X^a = (0, 1, 1)$. Photon bouncing off mirror A: $P^a = (k, 0, k)$ before bounce; $P^a = (k, -k, 0)$ after bounce.

Mirror B: $X^a = (0, -1, 1)$. Photon bouncing off mirror B: $P^a = (k, -k, 0)$ before bounce; $P^a = (k, 0, -k)$ after bounce.

Moving Frame:

Mirror A: $X^a = ( - \gamma u, \gamma, 1)$. Photon bouncing off mirror A: $P^a = ( \gamma k, - \gamma u k, k )$ before bounce; $P^a = ( \gamma (1 + u) k, - \gamma ( 1 + u ) k, 0)$ after bounce.

Mirror B: $X^a = ( \gamma u, - \gamma, 1)$. Photon bouncing off mirror B: $P^a = ( \gamma (1 + u) k, - \gamma ( 1 + u ) k, 0 )$ before bounce; $P^a = ( \gamma k, - \gamma u k, -k)$ after bounce.

So we have, for positive $u$, the whole square moving to the left (-x axis) (cross-check: negative x-component of P before bounce off A).
Also the photon from A to B is moving to the left.
The photon's 4-momentum is $( \gamma (1 + u) k, - \gamma ( 1 + u ) k, 0)$. So the A-B photon's wavelength, as agreed before, is $$\lambda_{AB}=\frac{h}{\gamma(1+u)k}$$
with friends
$$\lambda_{BC}=\frac{h}{\gamma k}$$
$$\lambda_{CD}=\frac{h}{\gamma(1-u)k}$$
Clearly, $\lambda_{AB}$, the wavelength of a left moving photon in left moving square, is the shortest, contradicting the first quote (which I think is true, BTW).

 

[PeterDonis]

Now I'm thoroughly confused.

That appears to be because you are interpreting the word "frame" incorrectly. See below.

So we have, for positive u, the whole square moving to the left (-x axis)

Yes, although I described it as the frame itself moving to the right. In other words, $u$ describes the velocity of the "moving" frame relative to the CoM frame; positive $u$ means the moving frame is moving in the positive $x$ direction relative to the CoM frame. In the moving frame, therefore, the square itself is moving to the left. But "frame" doesn't refer to the square; it refers to the coordinates we are using.

Also the photon from A to B is moving to the left.

Yes. But the frame is moving to the right.

Clearly, $\lambda_{AB}$, the wavelength of a left moving photon in left moving square, is the shortest, contradicting the first quote (which I think is true, BTW).

No, it's a left-moving photon in a right-moving frame, which is what I said.

To see why it's the motion of the frame that matters, consider: if we put an observer at rest on the square, and have him measure the wavelengths of the photons, he will not measure them to be Doppler shifted! But an observer at rest in the right-moving frame will--he will measure the photon moving from A to B, a left-moving photon, to have a shorter wavelength. That is really what we are saying when we say a left-moving photon in a right-moving frame has a shorter wavelength.

 

[pervect]

Surely I must be stupid or something. I can't figure out the 4-momentum even in the rest frame.

The 4-momentum of photon leaving mirror A: X x P = (0, 1, 1) x (k, -k, 0):

You need to anti-symmeterize the product, as per Peter's post, i.e. in abstract index notation

$M^{ij} = r^i p^j - p^i r^j$
Modulo possible sign issues and factors of two, that is - I need to run & don't have time to figure out the details at the moment.

 

[PeterDonis]

You need to anti-symmeterize the product

He's doing that correctly. The issue is that he is using the wrong 4-position for the same photon just before it hits mirror B. See post #63.

 

[PeterDonis]

A left-moving photon in a left-moving square has shorter wavelength, and spends more time+distance in flight

Let me now go back and clear this up. First, I see that I misread "square" as "frame" when I responded; the left-moving square and the right-moving frame refer to the same thing, the square in the "moving" frame. But that's a relatively minor point, which should now be cleared up with post #71.

The more important point is that the wavelength of the photon has nothing to do with the time/distance it spends in flight. The wavelength of the photon has nothing to do with its 4-position; it has to do with its 4-momentum. That is how we derived the wavelength in the moving frame, after all--by looking at the 4-momentum components in that frame. Or consider this: suppose we doubled the side length of the square, so the mirrors were 4 units apart instead of two. That would certainly change the time of flight and distance traveled of the photons; but would it change their wavelength? Obviously not.

Thinking of "the number of waves between bounces" isn't really applicable to the model we've been considering, because that model doesn't actually treat the photon as a wave. It treats it as a massless particle moving at the speed of light. The only reason we sometimes refer to the components of the photon's 4-momentum as "frequency" and "wavelength" instead of "energy" and "momentum" is force of habit; we're so used to associating the word "photon" with the quantum treatment of light that we forget we're using a classical model in this case, where the photon doesn't really have a frequency and wavelength, just an energy and momentum. Saying the left-moving photon's wavelength is shorter in the moving frame than in the CoM frame is really just a sloppy way of saying its momentum is larger, and the Doppler shift in this model doesn't change a photon's frequency and wavelength (since the photon doesn't really have them), but its energy and momentum.

 

[SlowThinker]

That appears to be because you are interpreting the word "frame" incorrectly. See below.

Guilty as charged
I'm still not quite clear on this, I'll start another thread.

Can you comment on this?

Or, at which point to I advance the COM to new time? If I don't, then after a full cycle, leaving mirror A, I end up with $(8, 1, 1) \times (k, -k, 0) = -9k$ for the 01 component.

You don't.

I don't understand why the COM, to which we are relating the angular momentum, must not advance into the future together with the photons. As you can see, it causes trouble down the road.

 

[PeterDonis]

I don't understand why the COM, to which we are relating the angular momentum

No, we are not. As I said before, the angular momentum, as a 4-tensor, is "about" a point in spacetime, not a point in space. That point in spacetime happens to be a particular point on the worldline of the CoM (the point labeled $t = 0$ in the CoM frame). But the point in spacetime, itself, does not "move" with the CoM. It's always the same point. Remember, once more, this is geometry.

 

[SlowThinker]

No, we are not. As I said before, the angular momentum, as a 4-tensor, is "about" a point in spacetime, not a point in space. That point in spacetime happens to be a particular point on the worldline of the CoM (the point labeled $t = 0$ in the CoM frame). But the point in spacetime, itself, does not "move" with the CoM. It's always the same point. Remember, once more, this is geometry.

But, if the 4-angular momentum of the square+photons is always about the spacetime point (0, 0, 0), then it is not constant with time, as shown by the -9k result earlier.

In this document, my "best" attempt is the first tab (v=0, Rel).
What you are suggesting (as I understand it), is the second tab (v=0, Abs).
https://docs.google.com/spreadsheets/d/13kGjH3PPKqe3KkJWyK5wo_REGzSMGCFFZVjdnkTTqPY
Each tab's flow is
- "Corner absolute" + "COM" -> "Corner relative"
- "Photon" + "Corner relative" -> "tx-xt" etc.
The desired goal is that the columns "tx-xt", "ty-yt", "xy-yx" should stay constant at all times... which is not happening.

 

[PeterDonis]

if the 4-angular momentum of the square+photons is always about the spacetime point (0, 0, 0), then it is not constant with time

If you look at times far enough in the future, yes, the "time-space" components of the individual 4-angular momentum tensors do change. I apologize for not being clearer about that before. This is because the "mass moment" of each individual photon changes--if you work out the details, you will see that it changes, in the CoM frame, every time the photon bounces off a mirror at a time other than $t = 0$. But if you add up the 4-angular momentum tensors for all four photons, the components that change with time all cancel, so the tensor of the system as a whole is still constant in time.

In the CoM frame, the total tensor only has the "space-space" components that I gave in a previous post (indicating ordinary spatial angular momentum in the x-y plane); the "mass moment" of the CoM of the system is zero in the CoM frame. In the moving frame, the total tensor has the components pervect gave, which include "time-space" components (indicating that in this frame, the CoM has a nonzero "mass moment"), but the total tensor is still constant with time.

 

[PeterDonis]

if you add up the 4-angular momentum tensors for all four photons, the components that change with time all cancel, so the tensor of the system as a whole is still constant in time.

Let me derive this result explicitly for the CoM frame. We'll consider the 4-angular momentum tensors of each photon before and after it hits a mirror at a time which is now not restricted to $t = 0$. In the CoM frame, these times are given by $t = 2n$, where $n$ is any integer. So the 4-position vectors of the four mirrors are:

Mirror A: $(2n, 1, 1)$

Mirror B: $(2n, -1, 1)$

Mirror C: $(2n, -1, -1)$

Mirror D: $(2n, 1, -1)$

We have already shown that the 4-angular momentum tensor of any photon does not change in flight; so all we need to consider are the bounces. That means we don't have to worry about which photon is hitting which mirror at which time; we just need to use the "before" and "after" 4-momentum of whichever photon is hitting each mirror, and those will be the same for each mirror at each bounce. So the 4-momentum vectors are (just to re-post them for clarity):

Photon hitting Mirror A: $(k, 0, k)$ before, $(k, -k, 0)$ after.

Photon hitting Mirror B: $(k, -k, 0)$ before, $(k, 0, -k)$ after.

Photon hitting Mirror C: $(k, 0, -k)$ before, $(k, k, 0)$ after.

Photon hitting Mirror D: $(k, k, 0)$ before, $(k, 0, k)$ after.

The four individual tensors are then (labeling them by the mirror and "before" or "after"):

$$
M^{ab}_{A-before} = \left[ \begin{matrix}
0 & -k & \left( 2n - 1 \right) k \\
k & 0 & k \\
- \left( 2n - 1 \right) k & -k & 0
\end{matrix} \right]
$$

$$
M^{ab}_{A-after} = \left[ \begin{matrix}
0 & - \left( 2n + 1 \right) k & -k \\
\left( 2n + 1 \right) k & 0 & k \\
k & -k & 0
\end{matrix} \right]
$$

$$
M^{ab}_{B-before} = \left[ \begin{matrix}
0 & - \left( 2n - 1 \right) k & -k \\
\left( 2n - 1 \right) k & 0 & k \\
k & -k & 0
\end{matrix} \right]
$$

$$
M^{ab}_{B-after} = \left[ \begin{matrix}
0 & k & - \left( 2n + 1 \right) k \\
-k & 0 & k \\
\left( 2n + 1 \right) k & -k & 0
\end{matrix} \right]
$$

$$
M^{ab}_{C-before} = \left[ \begin{matrix}
0 & k & - \left( 2n - 1 \right) k \\
-k & 0 & k \\
\left( 2n - 1 \right) k & -k & 0
\end{matrix} \right]
$$

$$
M^{ab}_{C-after} = \left[ \begin{matrix}
0 & \left( 2n + 1 \right) k & k \\
- \left( 2n + 1 \right) k & 0 & k \\
-k & -k & 0
\end{matrix} \right]
$$

$$
M^{ab}_{D-before} = \left[ \begin{matrix}
0 & \left( 2n - 1 \right) k & k \\
- \left( 2n - 1 \right) k & 0 & k \\
-k & -k & 0
\end{matrix} \right]
$$

$$
M^{ab}_{D-after} = \left[ \begin{matrix}
0 & -k & \left( 2n + 1 \right) k \\
k & 0 & k \\
- \left( 2n + 1 \right) k & -k & 0
\end{matrix} \right]
$$

Adding up the "before" and "after" tensors separately, it is evident that in each one, all the $n$ dependent terms cancel, and we are left with just the overall tensor

$$
M^{ab} = \left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 4k \\ 0 & -4k & 0 \end{matrix} \right]
$$

A similar calculation can be done in the moving frame.

 

[SlowThinker]

If you look at times far enough in the future, yes, the "time-space" components of the individual 4-angular momentum tensors do change. I apologize for not being clearer about that before. This is because the "mass moment" of each individual photon changes--if you work out the details, you will see that it changes, in the CoM frame, every time the photon bounces off a mirror at a time other than $t = 0$.

I'm starting to see how it works...
1. If we compute the 4-angular momentum about a constant point in spacetime, it stays constant during flight, for each individual photon.
2. If a photon bounces at the same time, about which we measure the 4-angular momentum, it stays constant as well.
3. It's not constant if we move the "hub" point into the future (which is not surprising, after all). I'm wondering if we're seeing the mysterious spin in action...
4. 4-angular momentum is not constant if a photon bounces at some time in the future or past (hard to imagine, but acceptable).

But this brings me back to the original problem of simultaneity...
It seems I'll have to repeat your full computations for the moving case, before adding them together , which is going to take some time.
I'll refrain from making predictions about the result, until I work it out.
Thanks for your patience, Peter.

 

[PeterDonis]

If we compute the 4-angular momentum about a constant point in spacetime, it stays constant during flight, for each individual photon.

Yes.

If a photon bounces at the same time, about which we measure the 4-angular momentum, it stays constant as well.

In the CoM frame, yes. But remember that we also computed those same bounces in the moving frame, where they don't happen at the same coordinate time, and they still conserved angular momentum; they had to, because we're dealing with tensors, and a tensor that is conserved in one frame is conserved in every frame. So to be precise, we need to specify that "at the same time" here refers to the CoM frame. A more geometric way of stating it would be that angular momentum stays constant at bounce events that are in the plane of simultaneity that passes through the point about which we are computing 4-angular momentum and is orthogonal to the 4-velocity of the CoM at that point. The latter condition is what picks out events that are simultaneous in the CoM frame.

It's not constant if we move the "hub" point into the future

Or the past. In my computation above, $n$ could be positive or negative. A positive $n$ actually corresponds to the "hub" point (the spacetime origin of coordinates) being to the past of the bounce event (i.e., the time $t = 0$ in the CoM frame, which is the time of the "hub" point, is earlier than the time of the bounce event); a negative $n$ corresponds to the "hub" point being to the future of the bounce event.

I'm wondering if we're seeing the mysterious spin in action...

It's actually the "mass moment" in action, as has been pointed out before.

4-angular momentum is not constant if a photon bounces at some time in the future or past

This is really saying the same thing as your #3. See above.

Also, as I said, when you add up the tensors for all four photons, the total 4-angular momentum is conserved, even taking into account what happens at the bounces. This is significant because the point we picked as the "hub" point, about which we measure the 4-angular momentum of all the photons, is on the worldline of the CoM of the system of all four photons. So we would not expect the 4-angular momentum of just one photon, looked at relative to that CoM point, to be conserved. We would only expect the total 4-angular momentum of all the photons to be conserved, since the CoM point is only the CoM for all four photons combined.

this brings me back to the original problem of simultaneity

As you'll see when you repeat the computations in the moving frame, all of the above still applies, with the clarifications I gave above. This is to be expected, since we're dealing with tensor equations, and tensor equations that are valid in one frame are valid in every frame.

There is one other clarification, though. Notice that when we defined the "moving" frame, we put its spacetime origin at the same point as for the CoM frame. In other words, the "hub" point, the point about which we are measuring 4-angular momentum, is at $(0, 0, 0)$ in both frames. If we choose a different "hub" point, i.e., if we pick a different event on the CoM worldline as the event about which we measure 4-angular momentum, then to keep everything working as above, we would need to also redefine the "moving" frame so its spacetime origin is at the new "hub" point.

 

[SlowThinker]

So to be precise, we need to specify that "at the same time" here refers to the CoM frame. A more geometric way of stating it would be that angular momentum stays constant at bounce events that are in the plane of simultaneity that passes through the point about which we are computing 4-angular momentum and is orthogonal to the 4-velocity of the CoM at that point. The latter condition is what picks out events that are simultaneous in the CoM frame.

Actually, this is what I was trying to clarify/point out the whole time
Except you know how to say it "the scientific way"

3. It's not constant if we move the "hub" point into the future (which is not surprising, after all).
4. 4-angular momentum is not constant if a photon bounces at some time in the future or past (hard to imagine, but acceptable).

This is really saying the same thing as your #3.

By #3, I meant that the 4-angular momentum about (0,0,0) is different from 4-angular momentum of that same system/flywheel about (5,0,0). It seems to be different from #4... but I'm not going to delve into this.

I'm wondering if we're seeing the mysterious spin in action...

It's actually the "mass moment" in action, as has been pointed out before.

My physics teacher explained how angular momentum is the rotation in the xy, xz and yz planes, and that spin is very much like rotation in the tx, ty, tz planes. So it seems he described the mass moment instead...

 

[PeterDonis]

By #3, I meant that the 4-angular momentum about (0,0,0) is different from 4-angular momentum of that same system/flywheel about (5,0,0).

Careful. If you're talking about a single photon, yes, its 4-angular momentum about (0, 0, 0) is different from its 4-angular momentum about (5, 0, 0). But if you're talking about the system composed of all 4 photons, that isn't true. The calculation I posted earlier, where I showed that the 4-angular momentum of the system of all 4 photons combined is independent of time, shows that. Moving the "hub" point to some other event on the CoM worldline is equivalent to changing the value of $n$ in the calculation I did, and that calculation shows that the 4-angular momentum of the system as a whole is independent of $n$.

My physics teacher explained how angular momentum is the rotation in the xy, xz and yz planes, and that spin is very much like rotation in the tx, ty, tz planes. So it seems he described the mass moment instead...

Yes, it does. If by "spin" he was trying to describe the intrinsic property that elementary particles like electrons and quarks have, that is a kind of angular momentum, not a kind of mass moment.