# Conservation of angular momentum and RoS

1. Nov 2, 2015

### SlowThinker

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

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.

Last edited by a moderator: Nov 6, 2015
2. Nov 2, 2015

### Staff: Mentor

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.

3. Nov 3, 2015

### SlowThinker

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?

4. Nov 3, 2015

### Staff: Mentor

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 ]

Last edited: Nov 3, 2015
5. Nov 3, 2015

### Staff: Mentor

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 .

6. Nov 3, 2015

### SlowThinker

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.

7. Nov 3, 2015

### Staff: Mentor

It can be exchanged between frame and photon, but the whole system cannot change its angular momentum. Every collision conserves angular momentum.
The photons don't have the same energy, and you transfer momentum and therefore energy through your structure at a finite speed.

8. Nov 3, 2015

### SlowThinker

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

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.

9. Nov 3, 2015

### A.T.

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?

10. Nov 3, 2015

### Staff: Mentor

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

11. Nov 3, 2015

### Staff: Mentor

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.
What is a moving reference frame? How can we find a non-moving one, and why is it special?
The structure will behave differently, but some photons probably don't break our structure, right?

12. Nov 3, 2015

### SlowThinker

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.

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

13. Nov 3, 2015

### Staff: Mentor

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.

14. Nov 3, 2015

### SlowThinker

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.
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?

15. Nov 3, 2015

### SlowThinker

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.

16. Nov 3, 2015

### Staff: Mentor

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

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.

17. Nov 3, 2015

### Staff: Mentor

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.)

18. Nov 3, 2015

### SlowThinker

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.
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

Last edited: Nov 3, 2015
19. Nov 3, 2015

### Staff: Mentor

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.

20. Nov 4, 2015

### Staff: Mentor

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.