Conservation of angular momentum and RoS

[SlowThinker]

[Moderator's note: thread spun off from a previous one on a related but different topic.]

But I was thinking about RoS, since events may not happen simultaneously in another frame, will 3-momentum and energy always be conserved with respect to time?

Well that's another question that has been puzzling me for quite some time and would love to know the answer to it...
Imagine 4 tubes of equal length, making a square, with 4 identical photons running around in the same direction, and hitting the corners at the same time. So we have an object with some angular momentum, which is constant with time.
But when the square is viewed from a moving coordinate system, the photons no longer hit the corners in sync. It means that the angular momentum will oscillate around some average value. Also, the center of mass of the object is moving periodically, so the momentum and energy can not even be defined. (Imagine the square standing on one of its sides, moving to the right. Then the center of mass is certainly moving up and down).

This leads me to the conclusion that 3-angular momentum and energy is only defined -and conserved- in a frame co-moving with the center of mass of the system under consideration, or some similar condition. Is that not correct? It seems to contradict Mfb's answer.

 

[Nugatory]

Imagine 4 tubes of equal length, making a square, with 4 identical photons running around in the same direction, and hitting the corners at the same time. So we have an object with some angular momentum, which is constant with time.

The angular momentum of the system is constant, but the angular momentum of the square structure is not. Every time that a circulating flash of light hits a corner, it speeds up the rotation of the structure a bit more while the light loses a bit of energy and momentum so is red-shifted. Thus over time there is a net transfer of angular momentum from the circulating light to the structure and the system reaches equilibrium when the light has been completely absorbed and all the angular momentum has been transferred to the structure.

You also have to remember that the impulse from the flash of light hitting the corner does not immediately accelerate the entire structure - instead the impulse has to travel at the speed of sound through the material. You cannot assume that all parts of the structure are always changing speed simultaneously, nor that the frame is rigid under conditions where relativistic effects matter.

Take all of these effects into consideration, and you will find that:
a) This system is really hard to analyze correctly in a frame that is moving relative to the center of mass (that center of mass is, however, moving in a straight line at a constant velocity, as behooves the center of mass of a system not subject to external forces).
b) Momentum, angular momentum, and energy are all conserved.

 

[SlowThinker]

The angular momentum of the system is constant, but the angular momentum of the square structure is not. Every time that a circulating flash of light hits a corner, it speeds up the rotation of the structure a bit more while the light loses a bit of energy and momentum so is red-shifted.

That's strange. If the photon was a ball, the force needed to turn it 90 degrees would point directly to the center of the structure, so it should not transfer any torque to the structure.
Are photons somehow different from ideal microscopic rubber balls in this respect?

This system is really hard to analyze correctly in a frame that is moving relative to the center of mass

Too bad...

(that center of mass is, however, moving in a straight line at a constant velocity, as behooves the center of mass of a system not subject to external forces).
b) Momentum, angular momentum, and energy are all conserved.

...but if you're sure about this, I'll probably take your word for it.

 

[Nugatory]

That's strange. If the photon was a ball, the force needed to turn it 90 degrees would point directly to the center of the structure, so it should not transfer any torque to the structure.
Are photons somehow different from ideal microscopic rubber balls in this respect?

No. I was imagining a slightly different setup. The other stuff about the non-rigidity of the structure and the motion of the center of mass still hollers holds.

[Edited: "still hollers"? Score one for iPad autocorrect ]

 

[mfb]

It means that the angular momentum will oscillate around some average value.

Why? The collisions don't change angular momentum of the whole system, why would their timing matter? Same for the center of mass: why do you expect an oscillation?

The non-rigidity is a serious issue .

 

[SlowThinker]

Why? The collisions don't change angular momentum of the whole system, why would their timing matter? Same for the center of mass: why do you expect an oscillation?

Angular momentum can only be exchanged at moments when photons hit the corners. But in the moving frame, photons don't hit all 4 corners at the same time: the two ends of the front tube are hit together, but not in sync with the rear tube.
I spent some time trying to figure out how to define the angular momentum, so that it would be conserved at all times in this scenario, but could not find a way.
Same goes for the center of mass. In the moving frame, there is a time interval where there are 2 photons in the bottom tube, while a photon goes from front to top to rear tube. I can't imagine how the center of mass could keep its height.
I was reading https://en.m.wikipedia.org/wiki/Relativistic_angular_momentum but could not quite understand the concept of "moment of mass", which is why I came up with the photon square as a means to think about the concept.

 

[mfb]

Angular momentum can only be exchanged at moments when photons hit the corners.

It can be exchanged between frame and photon, but the whole system cannot change its angular momentum. Every collision conserves angular momentum.

Same goes for the center of mass. In the moving frame, there is a time interval where there are 2 photons in the bottom tube, while a photon goes from front to top to rear tube. I can't imagine how the center of mass could keep its height.

The photons don't have the same energy, and you transfer momentum and therefore energy through your structure at a finite speed.

 

[SlowThinker]

It can be exchanged between frame and photon, but the whole system cannot change its angular momentum. Every collision conserves angular momentum.

Yes but... in the stationary frame, the square construction is not rotating, not even wobbling. How could that transform to a periodically changing angular momentum in the moving frame, so that it could compensate the varying angular momentum of the photons?

My conclusion is, I have no problem with the conservation of "proper angular momentum" (if there is such a thing) but it seems ill-defined in a moving reference frame. So there's no point in debating its conservation.
If you're saying that the ordinary angular-3-momentum is well defined and conserved, I have difficulty seeing it.
But again, if you're sure we are talking about the same thing and what you say is indeed correct, I won't cause trouble

you transfer momentum and therefore energy through your structure at a finite speed.

In my opinion it only adds trouble to the definition and measurement of angular momentum, reinforcing my idea of ill-definiteness.
If the forces cancel in one situation, and I change the material to one with same weight and double speed of sound, will they cancel *again*? That's somewhat hard to grasp.

 

[A.T.]

If the photon was a ball, the force needed to turn it 90 degrees would point directly to the center of the structure, so it should not transfer any torque to the structure.

Yes but... in the stationary frame, the square construction is not rotating, not even wobbling. How could that transform to a periodically changing angular momentum in the moving frame, so that it could compensate the varying angular momentum of the photons?

If the individual collisions exert no torque around the COM, why would the angular momentum of the square vary in any frame? Why should it matter whether zero angular momentum is transferred in sync or not in sync? Can you draw a picture of the scenario you envision?

 

[PeterDonis]

If the photon was a ball, the force needed to turn it 90 degrees

Does the photon turn by 90 degrees in the moving frame?

 

[mfb]

Yes but... in the stationary frame, the square construction is not rotating, not even wobbling.

Sure it does wobble. The corners constantly oscillate outwards and inwards. They have to, momentum conservation and the finite speed of light force them to do so.

My conclusion is, I have no problem with the conservation of "proper angular momentum" (if there is such a thing) but it seems ill-defined in a moving reference frame.

What is a moving reference frame? How can we find a non-moving one, and why is it special?

If the forces cancel in one situation, and I change the material to one with same weight and double speed of sound, will they cancel *again*?

The structure will behave differently, but some photons probably don't break our structure, right?

 

[SlowThinker]

If the individual collisions exert no torque around the COM, why would the angular momentum of the square vary in any frame? Why should it matter whether zero angular momentum is transferred in sync or not in sync? Can you draw a picture of the scenario you envision?

The angular momenta of the photons simplycan't add to a constant at all moments of time.
I've tried to draw something although I agree it's not Mona Lisa:

Leftmost is the situation in co-moving frame. Center is the situation in a moving frame. To the right top is the path of one photon, as I imagine it, obviously not to scale. Right center is the combined path of all 4 photons.
While there *might* be a speed where the knots cancel out, it certainly won't work for all speeds.

Does the photon turn by 90 degrees in the moving frame?

No, as the path of one photon shows. Are you implying the square will wobble in the moving reference frame?

 

[PeterDonis]

No, as the path of one photon shows. Are you implying the square will wobble in the moving reference frame?

I'm saying that if the photon does not change direction by exactly 90 degrees, your argument that the force needed to redirect it is purely radial, so no torque is exerted on the square, no longer holds. There will be a tangential component to the force, and that component will exert a torque on the square.

 

[SlowThinker]

Sure it does wobble. The corners constantly oscillate outwards and inwards. They have to, momentum conservation and the finite speed of light force them to do so.

Good point, but again, it seems to complicate the problem, not solve it. The situation can obviously be described in a moving reference frame, but I see no way to define the angular momentum, so that it would be constant throughout the different phases of the photon cycle.
Unless it's defined in the frame where the square is at rest, which is my original point.

What is a moving reference frame? How can we find a non-moving one, and why is it special?

A moving frame is any where the center of mass of the square is changing coordinates. Let's exclude rotating frames of reference for now, OK?

 

[SlowThinker]

I'm saying that if the photon does not change direction by exactly 90 degrees, your argument that the force needed to redirect it is purely radial, so no torque is exerted on the square, no longer holds. There will be a tangential component to the force, and that component will exert a torque on the square.

Another good point, but how about this:
Let's replace the photons with another 4 that are running in the other direction. In the stationary frame, the square's wobbling is exactly the same as before, so it must be the same in the moving frame as well.
But now it smoothes peaks in a very different movement pattern (the peaks in the sum of the angular momenta of the 4 photons).
To me it implies that the square's angular momentum equals its negative at all times, meaning it's zero, meaning it can't cancel the peaks.

 

[PeterDonis]

Let's replace the photons with another 4 that are running in the other direction. In the stationary frame, the square's wobbling is exactly the same as before

By "replace" you mean we still have 4 photons, just going around in the opposite direction, correct?

But now it smoothes peaks in a very different movement patter

I'm not sure what you mean by this. The situation is symmetric, so I would expect this scenario to look just like the first one as far as time dependence is concerned.

 

[PeterDonis]

I see no way to define the angular momentum, so that it would be constant throughout the different phases of the photon cycle.

Angular momentum has to be defined about an axis. What axis would you use in the moving frame?

You might also want to look at this Wikipedia page:

https://en.wikipedia.org/wiki/Relativistic_angular_momentum

One key idea that is sort of discussed there is this: suppose we consider the angular momentum to be purely spatial in the stationary frame. That is, we define it as a purely spatial vector (actually a pseudovector, as the Wiki page discusses, but we can put that aside for the moment), pointing in the direction perpendicular to the plane of the square and the photons. "Purely spatial", in 4-d spacetime, means a 4-vector with zero time component, i.e., $L^{\mu} = (0, L^x, L^y, L^z)$.

Now transform this vector to the moving frame. It will no longer be purely spatial! That has to be the case just from looking at how a vector Lorentz transforms. So you can't think of the angular momentum in the moving frame the way you are used to; it isn't the same kind of object. (Strictly speaking, it isn't in the stationary frame either, but in the stationary frame you can get away with thinking it is because of the zero time component.)

 

[SlowThinker]

I'm not sure what you mean by this. The situation is symmetric, so I would expect this scenario to look just like the first one as far as time dependence is concerned.

Let me recap how I understand your view:
1) You agree that the sum of angular momenta of the CCW running photons themselves is not constant over time
2) The total angular momentum of the system is constant
3) Thus the square's wobbling must cancel the irregularities in the photon sum (1).

If I make the pattern in (1) negative, and the square's wobbling is the same, it can't add to a constant again.

Now transform this vector to the moving frame. It will no longer be purely spatial!

So, coming back to the title, do you agree that angular-3-momentum is not conserved when Relativity of Simultaneity is considered?

Edit: forgot the "angular" momentum

 

[PeterDonis]

do you agree that 3-momentum is not conserved when Relativity of Simultaneity is considered?

I assume you mean "3-angular momentum"? Relativistic conservation laws never involve 3-vectors. They always involve 4-vectors, or tensors, or other 4-dimensional quantities.

I haven't had the time to analyze the scenario under discussion in detail, so I'll defer further comment on the specifics of it until I have. I suspect that there are things that are being missed in the discussion so far.

 

[mfb]

Good point, but again, it seems to complicate the problem, not solve it.

It is a complication you cannot avoid, that is the point. You focus on one part of the system and ignore the other. The conservation laws only apply to the sum of the two, "this part is too complicated so I'll ignore it" does not work.

 

[Nugatory]

I suspect that there are things that are being missed in the discussion so far.

At a minimum:
- The non-rigidity of the rectangular structure (expecting an unbalanced impact at one corner to uniformly displace the entire structure is a variant of the "pushing a rigid rod" paradox).
- The correct transformation of the reflection angles from frame to frame.
- The center of mass of the rectangular structure is different from the center of mass of the entire system because the latter must include the momentum of the circulating light flashes.
- Relativity of simultaneity affects not only when the circulating light flashes strike the structure but also the relative motions of various parts of the structure.
- The position of the center of gravity of the system is itself frame-dependent.
- And of course energy and momentum are frame-dependent even though they are conserved within a frame.

And I've probably left something out...

 

[SlowThinker]

It is a complication you cannot avoid, that is the point. You focus on one part of the system and ignore the other. The conservation laws only apply to the sum of the two, "this part is too complicated so I'll ignore it" does not work.

- The non-rigidity of the rectangular structure (expecting an unbalanced impact at one corner to uniformly displace the entire structure is a variant of the "pushing a rigid rod" paradox).
- The correct transformation of the reflection angles from frame to frame.
- Relativity of simultaneity affects not only when the circulating light flashes strike the structure but also the relative motions of various parts of the structure.

All these can be ignored, since these effects are the same for photons circulating counter-clockwise and clockwise. Thus none of them can have any effect on the total angular momentum, or vertical momentum, or vertical position of the center of gravity/mass/stress-energy/whatever.

- The center of mass of the rectangular structure is different from the center of mass of the entire system because the latter must include the momentum of the circulating light flashes.

In the stationary frame, the center of the square is the center of mass, at all times, photons included.
In the moving frame, it is a very complicated mess that, I claim, cannot sum to a constant, unless the relativity of simultaneity is explicitly accounted for.

- The position of the center of gravity of the system is itself frame-dependent.

Agreed. But I add that the vertical position is not constant with time, unless, again, the relativity of simultaneity is explicitly accounted for.

- And of course energy and momentum are frame-dependent even though they are conserved within a frame.

This, I'm not so sure that is correct (the conservation part).
In my opinion, the correct approach is to find the energy, momentum and angular momentum in the rest frame, and, somehow, transform them to the moving frame.
But if you try to find the energy "at a time seen by a moving observer", you're in trouble.

 

[mfb]

All these can be ignored, since these effects are the same for photons circulating counter-clockwise and clockwise.

They are not. You can mirror the whole system, sure, but then you mirror the photons and the structure and get the same system back.

 

[PeterDonis]

In my opinion, the correct approach is to find the energy, momentum and angular momentum in the rest frame, and, somehow, transform them to the moving frame.

You can do this, but not if you treat energy as a scalar and momentum and angular momentum as 3-vectors.

The correct approach is to construct an energy-momentum 4-vector and an angular momentum pseudo-4-vector (or antisymmetric 4-tensor) in the rest frame, and then transform those to the moving frame.

 

[SlowThinker]

They are not. You can mirror the whole system, sure, but then you mirror the photons and the structure and get the same system back.

The forces and timing exerted by the photons on the square construction are exactly the same, whether the photons rotate CCW or CW. The construction is not mirrored, only the photons.
This only holds in the rest frame, obviously.

 

[SlowThinker]

You can do this, but not if you treat energy as a scalar and momentum and angular momentum as 3-vectors.

The correct approach is to construct an energy-momentum 4-vector and an angular momentum pseudo-4-vector (or antisymmetric 4-tensor) in the rest frame, and then transform those to the moving frame.

As far as I can tell, I agree with you.
Nugatory and Mfb seem to insist that angular-3-momentum can be used as well, and I'm trying to show they are wrong. Unless I misunderstand their arguments and we are arguing about something else...

 

[PeterDonis]

Nugatory and Mfb seem to insist that angular-3-momentum can be used as well

It can be used if you treat it as a component of a pseudo-4-vector and transform it appropriately between frames. In each frame, the pseudo-4-vector separates into a pseudoscalar and a pseudo-3-vector (the same way that the energy-momentum 4-vector splits into an energy scalar and a momentum 3-vector). In the rest frame, the pseudoscalar part of the angular momentum pseudo-4-vector is zero; but in the moving frame, it isn't. At least, that's the way it looks to me, but as I said before, I have not worked through the details of the calculation.

 

[SlowThinker]

OK now that I got my own thread I can show a bit of "analysis" I prepared before.
The image shows 2 moments:
1. Before and after 2 photons bounce off the 2 front mirrors,
2. Before and after 2 photons bounce off the 2 rear mirrors.

The positions of photons are guessed but should be approximately correct.
My computation says that the time spent in the top tube is $T_T=\frac{\gamma}{c}(1-v/c)$.
Time in front and rear tube is $T_F=T_R=\frac{\gamma}{c}$.
Time in bottom tube is $T_B=\frac{\gamma}{c}(1+v/c)$.
Then, it seems that the photon's energy has to adjust accordingly, to keep constant number of waevs in each tube. This is shown by the colors of photons in the above image.

At the moment of the front bounce, each of the photons gain the same amount of energy ($\frac{\gamma}{c}(v/c)$), so the center of mass should be at the middle, or it will jump.
On the other hand, if angular momentum was to be conserved, the center of rotation would have to be a bit below the middle. I haven't tried to figure out the exact position, partially because it seems like nonsense.

Another issue: at the time of front bounce (left image), the total energy of the photons increases, while the construction is accelerated overall, so energy is not conserved.
This is compensated during the read bounce.

 

[jartsa]

Let's remember that the tubing's longitudinal mass is larger than its transverse mass.

When light hits a mirror, it gives lot of longitudinal momentum to a large longitudinal mass, which results such an acceleration of the mirror that an observer traveling with the tubes can observe the mirror gaining equal amount of horizontal and vertical momentum.

 

[SlowThinker]

Let's remember that the tubing's longitudinal mass is larger than its transverse mass.

When light hits a mirror, it gives lot of longitudinal momentum to a large longitudinal mass, which results such an acceleration of the mirror that an observer traveling with the tubes can observe the mirror gaining equal amount of horizontal and vertical momentum.

My problem is not with "how it all works". Clearly energy, momentum and angular momentum are conserved, and should be constant with time, as the tubing + photons make an isolated system.
The question is, how do you define energy, momentum and angular momentum, so that they are constant with time? It seems we need to take a snapshot of the system in a specific spacetime slice (the rest frame of center of mass), rather than any spacelike slice. But I've never seen that done. For center of mass, you just get $\mathbf{x}_{COM}=\sum_i{m_i \mathbf{x}_i}/\sum_i{m_i}$, with no specific condition on time.
Perhaps I've been reading wrong books

Originally I was trying to understand the concept of "moment of mass" that Wikipedia mentions, but I got stuck on this simultaneity issue.

 

[PeterDonis]

how do you define energy, momentum and angular momentum, so that they are constant with time?

For energy and momentum, you define an energy-momentum 4-vector for the system, whose invariant norm is the system's rest mass. In any inertial frame, the "time" component of this 4-vector gives the energy, and the spatial components give the 3-momentum. Since we're assuming the system is isolated, these will be constant with time in any inertial frame.

For angular momentum, you need to pick a point about which to define it; the simplest choice is the object's center of mass. Then you can define an antisymmetric 4-tensor that describes the object's angular momentum about that point; it's basically the 4-dimensional analogue of the 3-vector $L = r \times p$. See here:

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".

 

[pervect]

My problem is not with "how it all works". Clearly energy, momentum and angular momentum are conserved, and should be constant with time, as the tubing + photons make an isolated system.
The question is, how do you define energy, momentum and angular momentum, so that they are constant with time? It seems we need to take a snapshot of the system in a specific spacetime slice (the rest frame of center of mass), rather than any spacelike slice. But I've never seen that done. For center of mass, you just get $\mathbf{x}_{COM}=\sum_i{m_i \mathbf{x}_i}/\sum_i{m_i}$, with no specific condition on time.
Perhaps I've been reading wrong books

Originally I was trying to understand the concept of "moment of mass" that Wikipedia mentions, but I got stuck on this simultaneity issue.

My perception of your problem is rather different than yours. My perception is that the root of your issue is that you want to treat the assembly of mirrors as a rigid object. And this just doesn't work, there aren't any rigid objects in SR. Ignoring this fact is leading you into paradoxes, which is where you're at now.

There are a couple of possible routes. The seemingly straightforwards route is to consider what happens when a photon bounces off a mirror in a non-rigid object. The mirror moves, and a displacement wave moves through the supporting structure at the speed of sound in a material.

Most existing models of this process (a material model) are not (as far as I know) relativistic. So it's a rather challenging approach to do in detail, but if you skip some of the detail required to model the relativistic propagation of sound waves through a media, you can get some good insights into the problem. Basically, if you tap one end of a bar, the other end does not move instantaneously, and you get paradoxes if you try to assume it does. Similarly, when you transfer momentum to the frame, that momentum does not instantaneously move through the entire frame - rather it causes a displacement wave to creep through the material (at a very low non-relativistic speed for any known material we can do experiments on).

There is a way to change the problem to make it easier to analyze. Put the mirrors on little tracks and let them move. The mirror isn't really rigid either, but you can think of it as a point particle. Then you need some constant force "spring" to slowly transport the momentum at a constant rate back to the mirror.

To carry through this project, first you need to figure out how you'd do it with three-vectors, then you need to change the3-vectors to 4-vectors. From some of your earlier questions, I get the impression that you might also have to learn (or refresh your memory) about 4-vectors (?).

Once you have the momentum in terms of 4-vectors, you can use Lorentz transforms on them to change frames. But then you still have the question - "how do we do angular momentum with 4-vectors?". Well the short answer is we don't write r (x) p, (x) being the cross product, instead we write r ^ p, ^ being what's called the "wedge product". So we just replace the cross product with a wedge product. But to learn about the wedge product, you need to learn about bi-vectors, because the wedge product of two vectors isn't another vector, it's a bi-vector. A bit of Clifford algebra comes in really handy here, but it's not typically something you'd get in an introductory course :(. If you are curious enough to go this route, there's a rather nice introduction to the topic, entitled "Imaginary Numbers are not Real — the Geometric Algebra of Spacetime" - which you can find either with a google search. If you don't want to learn about Clifford Algebras, you still need to learn about the Hodges Dual operator :(. The hodges dual operator basically says that there is a map from bi-vectors to vectors in 3 dimensions, the idea is that the bi-vectors are rather like planes, and the dual vector is rather like a vector perpendicular to the plane.

Anyway, no matter which route you take, you get into a lot of mathematical byways. But the fundamamental issue(in my opinoin) starts with the lack of the rigid bodies you're used to using in classical mechanics in special relativity. If you think of your rigid frame as being made out of the opposite of rigidity, say wet noodles, and just what it would take to analyze the physics of these "wet noodles", it will be helpful in getting rid of the "rigid object" paradigm. Unfortunately, there is still the issue of learning enough physics to handle distributed systems like the "wet noodle", replacing differential equations with partial differential equations, and all the related issues you need to handle non-rigid objects.

 

[SlowThinker]

My perception of your problem is rather different than yours. My perception is that the root of your issue is that you want to treat the assembly of mirrors as a rigid object. And this just doesn't work, there aren't any rigid objects in SR. Ignoring this fact is leading you into paradoxes, which is where you're at now.

I have to disagree. While the movement of the mirrors, and sound waves in the tubing, are certainly a nightmare to compute, I'm pretty sure that no such movement can add to the angular momentum of the structure. As argued before, these sound waves are exactly the same when the photons move the other way. So, when computing the angular momentum of the system, all we need is to focus on the photons.

To carry through this project, first you need to figure out how you'd do it with three-vectors, then you need to change the3-vectors to 4-vectors. From some of your earlier questions, I get the impression that you might also have to learn (or refresh your memory) about 4-vectors (?).

I do have a general understanding of 4-vectors and tensors, although it certainly is not "second nature".
Is the Geometric Algebra making its way to mainstream? I tried to find math that would not be obfuscated with philosophical babble, but it seems that Mr. Hestenes is not getting a large following.

But the fundamamental issue(in my opinoin) starts with the lack of the rigid bodies you're used to using in classical mechanics in special relativity. If you think of your rigid frame as being made out of the opposite of rigidity, say wet noodles, and just what it would take to analyze the physics of these "wet noodles", it will be helpful in getting rid of the "rigid object" paradigm.

Maybe you're right and I don't see it, but any definitions I've found, seem to say something along the lines
Angular (3-or 4-)momentum = $\sum$ angular moments of parts
without explicitly stating "when". Clearly, we need to make a "snapshot" of the system, and perform the summation on this snapshot.
It seems that the snapshot cannot be just any space-like slice. It has to be a slice where the time depends on position, something like
$$t(\vec x)=t_0-\gamma(\frac{\vec v\,\vec x}{c^2})$$
(the standard Lorentz transformation to the rest frame of the system being analyzed)

 

[pervect]

I have to disagree. While the movement of the mirrors, and sound waves in the tubing, are certainly a nightmare to compute, I'm pretty sure that no such movement can add to the angular momentum of the structure. As argued before, these sound waves are exactly the same when the photons move the other way. So, when computing the angular momentum of the system, all we need is to focus on the photons.

Well, unless we can compute them, we can't really tell. Because it is, as you say, a nightmare, I would propose changing the problem. Rather than dealing with the sound waves in the bar, let's compute the behavior of a mirror mounted on a sliding track, which seems much easier. So at each corner, we have a mirror on a sliding track, which oscillates around some equilibrium position. By making the force on the mirror constant, we replace any sound waves with static stresses.

A rather key elemen to make this approach work is that the force exerted by the constant force spring on the moving mirror must be radial in both frames of reference so as not to change to the angular momentum of the mirror in either frame. I believe this to be the case, but I suppose it needs to be checked. The Lorentz transform should "squish" the track in the same way as it squishes the frame, so the track should still point towards the center of the frame.

We can resolve the issue by a fairly simple calculation with the track approach. Before the collision of the photon with the mirror, the linear momentum p of the mirror is some vector $pm_1$ and the linear momentum of the photon is $pp_1$. The angular momentum is $(pm_1 + pp_1) \ times r$. After the collision, the linear momentum of the mirror is $pm_2$ and the photon $pp_2$.

We know that $pm_1 + pp_1 = pm_2 + pp_2$ by the conservation of linear momentum. So we know that the total angular momentum $(pm_1 + pp_1) \times r = (pm_2 + pp_2) \times r$

We additionally not that until the collisions, $pp_1$ and $pp_2$ stay constant, and we argue that because the force is radial, while $pm_1$ and $pm_2$ are not constant, the angular momentum only changes when there is a collision, it doesn't change due to the action of the spring.

The question is, is there an exchange of angular momentum between the photon and the mirror? I believe that the answer is no in the rest frame and yes in the moving frame, so it is dependent. Note that $pm_1$ and $pm_2$ lie along the track in the rest frame, and $pp_1$ is vertical, and $pp_2$ is horizontal. For definiteness, I'm writing as if I'm analyzing the collision in the upper right corner.

While $pp_1$ is vertical in the stationary frame, in the moving frame, $pp_1$ has a rightwards components, the photon in this frame is always centered on the frame, but the frame is moving, so the photon has some rightwards momentum. $pp_2$ will be horizontal in both frames, bu it's magnitudet will change. We can regard this as being due to the doppler shift of the photon by the motion of the frame, which affects its energy and momentum values. Similar remarks apply to $pm_1$ and $pm_2$, $pm_1$ and $pm_2$ will be along the track in the rest frame, but will have an additional righwards component in the moving frame. Given that we assume that the track remains pointed towards the origin where we are calculating the angular momentum L, the fact that $pm_1$ and $pm_2$ are not pointing along the same direction as the track suggests that $pm_1$ and $pm_2$ will not point towards the center of the frame. This implies that they will contribute a non-zero term to the angular momentum, so we see an exhchange of angular momentum between the light and the mirror in the moving frame.

I do have a general understanding of 4-vectors and tensors, although it certainly is not "second nature".
Is the Geometric Algebra making its way to mainstream? I tried to find math that would not be obfuscated with philosophical babble, but it seems that Mr. Hestenes is not getting a large following.

You don't have to learn about Geometric Algebra if you don't want to, but if you want to understand the 4-vector formulation of angular momentum, to go along with the 4-vector approach to linear momentum, you do need to understand bi-vectors. So if you want to avoid geometric algebra it's fine, but we can't avoid talking about bi-vectors, which are also known as rank 2 antisymmetric tensors. MTW has some discussion of this 4-vector form of angular momentum in "Gravitation", I'm sure other textbooks will as well. The other alternative is to just ignore the issue and convert the 4-vectors back to 3-vectors and use the 3-momentum, which might be just as easy.

Maybe you're right and I don't see it, but any definitions I've found, seem to say something along the lines
Angular (3-or 4-)momentum = $\sum$ angular moments of parts
without explicitly stating "when".

When will always be frame dependent. .

Clearly, we need to make a "snapshot" of the system, and perform the summation on this snapshot.
It seems that the snapshot cannot be just any space-like slice. It has to be a slice where the time depends on position, something like
$$t(\vec x)=t_0-\gamma(\frac{\vec v\,\vec x}{c^2})$$
(the standard Lorentz transformation to the rest frame of the system being analyzed)

Using moving mirrors, we can take such snapshots. I haven't done the calculations in detail, but I expect to find that the total value of L in the whole system (photons + mirrors) is, indeed, a constant, but that the division of what part of L is due to the photons and what part of L is due to the mirrors varies, only in the non-moving frame will the motion of the mirrors not contribute to the angular momentum.

To actually do the calcuations, we need to set up some coordinates like (t,x,y,z), set up 4-vectors like (1,1,0,0) for motion of the photons in the x direction and (1,0,1,0) for motions of the photons in the y direction, and $(1/\sqrt{1-2*v^2}, v, v, 0)$ for a typical diagonal mirror motion. Then perform the appropriate lorentz boost to convert these 4-vectors in the rest frame to the 4-vectors in the moving frame. Then we either need to use the 4-vector formalism to compute L with bi-vectors, or convert the 4-vectors back to 3-vectors if we want to use the 3-vector approach.

 

[SlowThinker]

First, thanks for taking the time to analyse the problem.

You don't have to learn about Geometric Algebra if you don't want to, but if you want to understand the 4-vector formulation of angular momentum, to go along with the 4-vector approach to linear momentum, you do need to understand bi-vectors.

I'm OK with the use of G.A. Some time ago I tried to follow the GA Primer but it became hard to read with the scrambled TeX, such as this page Keplerian motion so I put it away for later reading. Now I'm reading the document you mentioned.

We know that $pm_1 + pp_1 = pm_2 + pp_2$ by the conservation of linear momentum. So we know that the total angular momentum $(pm_1 + pp_1) \times r = (pm_2 + pp_2) \times r$

The math looks right but I don't feel quite comfortable with this reasoning.
Should not this be performed with momentum 4-vectors and the wedge or geometric product?
Perhaps there is a time component in the movement of mirrors that I'm missing

I haven't done the calculations in detail, but I expect to find that the total value of L in the whole system (photons + mirrors) is, indeed, a constant, but that the division of what part of L is due to the photons and what part of L is due to the mirrors varies, only in the non-moving frame will the motion of the mirrors not contribute to the angular momentum.

I still believe that the front-top and front-bottom mirrors exactly cancel each other's angular momentum at all times, even though the reasoning above looks convincing. Same goes for the two rear mirrors.

Before doing any calculations, I want to make sure it even makes sense. That's actually my whole point, because we obviously know the final result, via analysis in the rest frame and Lorentz transformation.