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.

 

[PeterDonis]

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.

If the structure were perfectly rigid, this might make sense. But it isn't.

As argued before, these sound waves are exactly the same when the photons move the other way.

That's not the point. The point is that the sound waves can carry angular momentum. Or, rather, you can't just help yourself to the assumption that they don't. It may be that this case is too complicated to analyze here if we have to include the sound waves; but that doesn't mean the sound waves have no effect. It means we can't compute their effect, so this problem is too complicated for us to solve. Which is why pervect is suggesting trying a simpler problem instead.

 

[PeterDonis]

Angular (3-or 4-)momentum = ∑\sum angular moments of parts
without explicitly stating "when".

If you use 4-momentum and 4-angular momentum, there is no "when". You don't have to split spacetime into space and time in order to use 4-vectors, 4-tensors, etc., which means you don't have to adopt any assumptions about "when". That's why 4-vectors, tensors, etc. are such good tools to learn to use.

 

[SlowThinker]

As argued before, these sound waves are exactly the same when the photons move the other way.

That's not the point. The point is that the sound waves can carry angular momentum. Or, rather, you can't just help yourself to the assumption that they don't. It may be that this case is too complicated to analyze here if we have to include the sound waves; but that doesn't mean the sound waves have no effect. It means we can't compute their effect, so this problem is too complicated for us to solve. Which is why pervect is suggesting trying a simpler problem instead.

OK so last try before I give up.
Do you agree that any infinitesimal movements, internal stresses, sound waves and anything else you can think of, in the front-bottom corner, are an exact mirror image of those in the front-top corner?
To me, the answer is "obviously yes". And I'm having trouble seeing how a perfectly symmetric movement could result in a net angular momentum.

 

[PeterDonis]

Do you agree that any infinitesimal movements, internal stresses, sound waves and anything else you can think of, in the front-bottom corner, are an exact mirror image of those in the front-top corner?

In which frame? You're bringing "when" into it again, and "when" is frame-dependent. It's impossible for all these movements to be "exact mirror images" in all frames. If you are content to work only in the center of mass frame, then you can reason more or less the way you have been, and just accept that your reasoning is only valid in that one frame; but if you were content with that, you wouldn't have needed to ask all these questions, would you?

If you're looking for a way to go beyond one frame and understand conservation laws in a more general way, the best approach I know of is to learn about 4-vectors and 4-tensors and how to express conservation laws in terms of them. Those expressions will automatically be valid in any frame, and in fact you often don't even need to pick a frame to understand what they're telling you physically. But you have to give up the urge to constantly try to view things from the viewpoint of a particular frame; the 4-vector 4-tensor approach is fundamentally about geometry and spacetime invariants, not about how things look from any particular frame.

 

[SlowThinker]

Do you agree that any infinitesimal movements, internal stresses, sound waves and anything else you can think of, in the front-bottom corner, are an exact mirror image of those in the front-top corner?

In which frame? You're bringing "when" into it again, and "when" is frame-dependent. It's impossible for all these movements to be "exact mirror images" in all frames.

In any reference frame that's moving horizontally.
Cancellation would also occur in a vertically moving frame, even though between different corners. I'm not sure if that means the cancellation occurs in all inertial reference frames, but it looks possible.
After all, zero angular momentum cannot transform into non-zero, can it? Even if it's angular 4-momentum.

So, are you saying, that angular 4-momentum of this system is always constant, no matter at which time I view it? I could even count some photons twice, if there is no limit on "when" the summation of parts is taking place.

Well I was hoping this system could help me understand how the angular 4-momentum works. Too bad I got stuck on such a silly issue.
Thanks for participation, everyone.

 

[jartsa]

SlowThinker, in your post #28 there is an error in photon energies.

In some frame the photons have equal energies. In that frame let's give everything a boost to the right. What happens is:
Boost to the right increases energy of photons moving to the right.
Boost to the right decreases energy of photons moving to the left.

Another way to say it: An observer of photons accelerates for a while, he sees some photons climbing up in a pseudo-gravity field and redshift, and he sees some photons traveling down in said field and blueshift.

 

[SlowThinker]

SlowThinker, in your post #28 there is an error in photon energies.

In some frame the photons have equal energies. In that frame let's give everything a boost to the right. What happens is:
Boost to the right increases energy of photons moving to the right.
Boost to the right decreases energy of photons moving to the left.

My reasoning goes as follows: in the rest frame, imagine we set the photons to have a wavelength 1000 times shorter than the length of a tube.
This must hold in all frames. So the count of wavelengths in each tube stays constant.
Since the top tube moves against the photon, the photon has to wave much faster than before, to reach that 1000 waves very quickly.
On the other hand, the photons in the bottom tube have a lot of time to make those 1000 waves, so they appear red-shifted.
I agree this is very counter-intuitive, but it seems it must be true. Perhaps you need to boost "to the left" to agree with my scenario.

Another way to say it: An observer of photons accelerates for a while, he sees some photons climbing up in a pseudo-gravity field and redshift, and he sees some photons traveling down in said field and blueshift.

Or, are you saying the photon changes color in flight? That only happens in accelerated motion. My example is inertial.

 

[PeterDonis]

are you saying, that angular 4-momentum of this system is always constant, no matter at which time I view it?

Once more: with angular 4-momentum, there is no time. 4-vectors, 4-tensors, etc. are independent of any splitting of spacetime into space and time. They are geometric objects in 4-dimensional geometry, not "things" that change or don't change with "time".

I was hoping this system could help me understand how the angular 4-momentum works.

If by "this system", you mean 4-vectors, 4-tensors, etc., you haven't even tried to actually use it. You are still using your old system of concepts, and it isn't working.

 

[jartsa]

My reasoning goes as follows: in the rest frame, imagine we set the photons to have a wavelength 1000 times shorter than the length of a tube.
This must hold in all frames. So the count of wavelengths in each tube stays constant.
Since the top tube moves against the photon, the photon has to wave much faster than before, to reach that 1000 waves very quickly.
On the other hand, the photons in the bottom tube have a lot of time to make those 1000 waves, so they appear red-shifted.
I agree this is very counter-intuitive, but it seems it must be true. Perhaps you need to boost "to the left" to agree with my scenario.

Good reasoning, wrong result, therefore there must be an error in some premise.

My reasoning is this: Doppler shift. And I guarantee that there is no error.

The fronts of moving things emit (reflect) blueshifted light.
The rears of moving things emit (reflect) redshifted light.
Right?

 

[PeterDonis]

In the interest of following my own advice, I'm going to at least make a start at analyzing a "photon and mirror" scenario using the angular momentum 4-tensor. I'm going to use a simpler scenario than the one in the OP, along the lines pervect suggested in an earlier post.

The basic setup is as follows: we have four mirrors which, at some instant in the center of mass frame of the mirrors, are all at rest at the four corners of a square. I'll put the four corners at coordinates (in the CoM inertial frame) $(x, y) = (1, 1), (-1, 1), (-1, -1), (1, -1)$. We also have four photons, which, at the same instant in the CoM inertial frame, are just bouncing off the four mirrors; the photons all have energy $k$ in the CoM frame, and you can see that, in that frame just before the bounce, the four photons (in the same order as the mirrors they are just about to bounce off of) have momentum $k$ in, respectively, the positive $y$, negative $x$, negative $y$, and positive $x$ directions. (We are using units where $c = 1$.) Just after the bounce, the photons will have momentum $k$ in, respectively, the negative $x$, negative $y$, positive $x$, and positive $y$ directions.

Let's focus in on just one photon and mirror to start; we'll pick the mirror at $(x, y) = (1, 1)$ and the photon bouncing off of it, which has momentum $k$ in the positive $y$ direction before the bounce and the negative $x$ direction after it.

The angular momentum 4-tensor is $M^{ab} = X^a P^b - X^b P^a$, where $X^a$ is the 4-position and $P^a$ is the 4-momentum. What we want to do is evaluate this tensor just before and just after the bounce for the photon. We assume the bounce happens at $t = 0$ in the CoM frame, so in that frame, we have $X^a = (0, 1, 1)$ both before and after the bounce. (We are leaving out all $z$ components since nothing of interest is happening in the $z$ direction.) Before the bounce, we have $P^a = (k, 0, k)$, and after the bounce, $P^a = (k, -k, 0)$.

If we then calculate $M^{ab}$ before the bounce, we have, for example, $M^{01} = X^0 P^1 - X^1 P^0 = 0 - k = -k$; and after the bounce, we have, for example, $M^{01} = X^0 P^1 - X^1 P^0 = 0 - k = -k$. So that component does not change. Similar calculations show that none of the components change, so $M^{ab}$ is the same both before and after the bounce; its value is:

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

Similar calculations at the other three corners will show that the same is true there; none of the photon bounces transfer any angular momentum from the photons to the mirrors. So SlowThinker's intuition that, in the CoM frame, the bounces do not exchange angular momentum is correct.

As a brief aside, we can also add up the four angular momentum tensors we compute for the four photons to get the total angular momentum of the system; it turns out to be, in the CoM frame:

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

So, in the CoM frame, the angular momentum tensor is indeed purely spatial, as I said in an earlier post.

But what happens if we boost to a different frame? First of all, as I said before, tensor equations are valid in all frames, so if we have shown that a bounce transfers zero angular momentum in one frame, it must transfer zero angular momentum in every frame. That is why the tensor formalism is so powerful; we can pick a frame that makes the component calculation easy, and then, as long as we can write the result in tensor form, we can invoke Lorentz invariance to conclude that the equation is valid in all frames.

But let's go ahead and work through the gory details for this example to see how it goes. We'll boost in the $x$ direction with velocity $u$; that will transform all 4-vectors from $V^a = (v^0, v^1, v^2)$ to $V'^a = \left[ \gamma \left( v^0 - u v^1 \right), \gamma \left( - u v^0 + v^1 \right), v^2 \right]$. 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, and $P'^a = \left[ \gamma \left( 1 + u \right) k, - \gamma \left( 1 + u \right) k, 0 \right]$ after the bounce. Going through the same calculation as before, and remembering that $\gamma^2 \left( 1 - u^2 \right) = 1$, we again see that $M'^{ab}$ is the same before and after the bounce; in this new frame, it is given by

$$
M'^{ab} = \left[ \begin{matrix}
0 & -k & - \gamma \left( 1 + u \right) k \\
k & 0 & \gamma \left( 1 + u \right) k \\
\gamma \left( 1 + u \right) k & - \gamma \left( 1 + u \right) k & 0
\end{matrix} \right]
$$

So each bounce transfers zero angular momentum in the new frame as well. Hence, it doesn't matter that in the new frame, the bounces are happening at different times; each one individually transfers no angular momentum, so there is no need to view them in "pairs" for things to cancel out.

So far, we have not looked at the mirrors at all; but it should be evident, first, that in the CoM frame, before the bounces, each mirror has a vanishing angular momentum tensor (since they have zero momentum, being at rest). After the bounces, each mirror must still have a vanishing angular momentum tensor, since no angular momentum is transferred; and if you compute $M^{ab}$ for each mirror after the bounce, using conservation of linear momentum through the bounce to find $P^a$ for each mirror after the bounce, you will see that $M^{ab}$ still vanishes. And, since $M^{ab}$ is a tensor, if it vanishes in one frame, it vanishes in all frames, so the mirrors must have zero angular momentum in all frames.

So SlowThinker's original intuition was actually correct: the photons cannot transfer any angular momentum to the mirrors at all! But now we can see that this is true individually, at each bounce at each mirror, so there is no problem with relativity of simultaneity, since there is no need to "pair up" bounces at opposite mirrors and do any cancellation to zero out the angular momentum transfer.

 

[jartsa]

Good reasoning, wrong result, therefore there must be an error in some premise.

My reasoning is this: Doppler shift. And I guarantee that there is no error.

The fronts of moving things emit (reflect) blueshifted light.
The rears of moving things emit (reflect) redshifted light.
Right?

Seriously, wavelengths must contract in the tube into which the photons gather, so that all the photons can fit in there.
A water tube analogy: A water tubing moving very fast is length contracted, the water that moves faster than the tubing is even more length contracted, that's how the water fits in the part of tubing where the water accumulates.

 

[SlowThinker]

Seriously, wavelengths must contract in the tube into which the photons gather, so that all the photons can fit in there.
A water tube analogy: A water tubing moving very fast is length contracted, the water that moves faster than the tubing is even more length contracted, that's how the water fits in the part of tubing where the water accumulates.

Length contraction always works in team with time dilation and relativity of simultaneity. You need to take all 3 into account.
I'm pretty sure that my reasoning does that.
You can imagine the standard light clock: the number of waves stays constant between bounces as well, in all frames. Faster moving clock have longer wave, that is, bigger redshift.

 

[jartsa]

Length contraction always works in team with time dilation and relativity of simultaneity. You need to take all 3 into account.
I'm pretty sure that my reasoning does that.
You can imagine the standard light clock: the number of waves stays constant between bounces as well, in all frames. Faster moving clock have longer wave, that is, bigger redshift.

So let's say I have a light clock on my table, where light bounces vertically.

If I boost that light clock to the left, energy of the light increases.
If I boost that light clock to the right, energy of the light increases.

Now our problem is: How do we convince SlowThinker about that.

Hmmm ... $$ E'= \gamma E $$ where E = energy in energy's rest frame, isn't it like that how we transform energy between frames?

 

[SlowThinker]

If I boost that light clock to the left, energy of the light increases.
If I boost that light clock to the right, energy of the light increases.

Now our problem is: How do we convince SlowThinker about that.

You may want to consider the possibility that you are wrong
I'm still digesting Peter's post but

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, and $P'^a = \left[ \gamma \left( 1 + u \right) k, - \gamma \left( 1 + u \right) k, 0 \right]$ after the bounce.

seems to suggest that the vertical photon's energy is $\gamma k$ (green in my picture), while the returning photon's energy is $\gamma(1+u)k$ (violet in my picture). This agrees, both qualitatively and quantitatively, with what I said.

 

[PeterDonis]

Hmmm ..
$$
E'= \gamma E
$$
where E = energy in energy's rest frame, isn't it like that how we transform energy between frames?

No. Energy is the "time" component of the 4-momentum vector; you have to transform the whole vector, and then take the "time" component of the transformed vector to get the energy in the new frame.