# Conservation of relativistic angular momentum

1. Sep 12, 2013

### yuiop

Is several other threads some arguments depend on what circumstances angular momentum is conserved in relativity and that is what I would like to establish here.

This Wikipedia article http://en.wikipedia.org/wiki/Relativistic_angular_momentum describes relativistic angular momentum as being coupled with a vector quantity called the "time varying moment of mass" in a similar manner to the way energy and linear momentum couple to form the invariant four momentum. What exactly is the "moment of mass" in layman's terms? Is the four angular momentum conserved under a Lorentz transformation? Is the angular momentum part (p) of the exterior product x^p defined as $p=m*r*u*\gamma (u) = m*r^2\omega*\gamma (\omega r)$ ?

As I understand it the the quantity $p=m*r*u*\gamma(u)$ is conserved for a system in a given reference frame (eg reduced r causes a corresponding increase in u maintaining p constant) but is not invariant under a transformation.

Also, under what circumstances is angular momentum conserved in General Relativity? What does Noether's thereom tell us.

I also found this paper http://panda.unm.edu/Courses/Finley/P495/TermPapers/relangmom.pdf [Broken] which states
. Again, what is the "mass moment" (in practical terms) or centroid?

Last edited by a moderator: May 6, 2017
2. Sep 12, 2013

### WannabeNewton

Are you talking about the angular momentum of a test particle/matter field or of a space-time? The former is trivial but the latter is very complicated.

If a space-time has an axial killing field $\psi^a$ then along the worldline of a freely falling test particle, one has $u^b \nabla_b (u^a\psi_a ) = 0$ so $u^a\psi_a$ is conserved along its worldline and the quantity $\psi^a u_a$ can be associated with the reduced angular momentum of the freely falling test particle. Similarly, if $T_{ab}$ is a matter field then $\nabla_a (T^{ab}\psi_b) = 0$ so $T^{ab}\psi_b$ defines a conserved current, in particular the angular momentum density of the matter field.

For the angular momentum of a space-time, the situation is not nearly as simple. See here for the easiest case (section 2.2): http://arxiv.org/pdf/1001.5429.pdf and Wald chapter 11 (in particular problem 11.6).

3. Sep 12, 2013

### Mentz114

If a spacetime possesses a spatial Killing vector field $\psi^a$ then a geodesic $u^a$ will conserve the quantity $\psi^a u_a$. I'm looking for an example.

In the Scwarzschild vacuum, $\psi^a=\partial_\phi$ is a KVF and the circular orbit ( r=constant, θ=π/2) the geodesic covector is $u_a=-\frac{r-2\,m}{\sqrt{r}\,\sqrt{r-3\,m}}\partial_t+\frac{\sqrt{m}\,r}{\sqrt{r-3\,m}}\partial_\phi$, giving a conserved angular momentum $L=\frac{\sqrt{m}\,r}{\sqrt{r-3\,m}}$

[edit: too slow to beat the WN machine !]

Last edited: Sep 12, 2013
4. Sep 12, 2013

### yuiop

Thanks for the quick response, WN and Mentz!
I was talking about the former. In particular I am looking for some simple examples, such as a thin disc with angular velocity W that lies in the x,y plane and rotates around the z axis. Will the angular momentum be conserved if the disc is accelerated in the z direction?

5. Sep 12, 2013

### yuiop

Is the angular momentum conserved if the particle circles around the 'North pole' rather than around the great circle at the equator? See post #20 in another thread https://www.physicsforums.com/showthread.php?t=709197&page=2 where I am discussing the issue with PeterDonis.

Last edited: Sep 12, 2013
6. Sep 12, 2013

### Mentz114

As far as I can tell there are no circular geodesics except in an equatorial plane...

7. Sep 12, 2013

### PAllen

The established results are the angular momentum is exactly conserved in SR, and also fully defined and exactly conserved in GR for asymptotically flat spacetimes (unfortunately, not ours). Otherwise, the GR situation for angular momentum falls into the same morass of quasilocal methods sometimes used for mass/energy.

8. Sep 12, 2013

### yuiop

Ah yes, I was aware that the equatorial orbit is a geodesic and can described by a particle in freefall, while the small 'orbit' around the the pole cannot be. Nevertheless, can we define angular momentum and conservation of that quantity for the non free fall case?

9. Sep 12, 2013

### yuiop

Cannot a small enough local region be considered asymptotically flat in the same way that that a small enough region can be considered essentially Minkowskian?

10. Sep 12, 2013

### PAllen

No, asymptotically flat is a feature of the geometry at infinity (specifically spatial infinity for a conserved quantity in GR). The energy and momentum and angular momentum of gravitational radiation can only be defined at asymptotically flat spatial infinity, it cannot be described locally. In an expanding universe, there is no accepted way to define total angular momentum at all.

11. Sep 12, 2013

### yuiop

From Wikipedia http://en.wikipedia.org/wiki/Constant_of_motion we have
and by extension these conservation properties can be used to predict the conservation of angular momentum for a free falling particle falling deep into the curved spacetime around a black hole, so things are not as dire as you describe. See http://www.fourmilab.ch/gravitation/orbits/ Here the angular momentum is conserved in terms of proper time.

12. Sep 12, 2013

### WannabeNewton

As I stated earlier, there is a difference between conserved angular momentum/angular momentum densities of test particles/matter fields propagating on space-time and the conserved angular momentum of a space-time itself. PAllen is referring to the latter; as far as is I know, angular momentum of a space-time only has a natural extension/generalization of Komar integrals for asymptotically flat space-times. What you are referring to are conserved currents of a matter field and conserved quantities along the worldline of a test particle which can always be defined in the simplest case when killing vector fields (isometries of space-time) are present in the sense described in post #2; more complicated conserved quantities (e.g. approximately conserved quantities) can also be defined but that's more advanced than needed.

13. Sep 12, 2013

### PAllen

As WBN noted, these statements only apply to test masses - considered gravitationally insignificant against a fixed background. Where a rotating body is the the primary subject of interest, a factor in GR is that it will radiate GW and angular momentum will not be conserved unless that carried by the GW is included. That can only be done in asymptotically flat spacetime.

However, I think this is all irrelevant to the issues of we are stuck on in the rotating disc scenarios, where we have assumed SR, and the case of uniform acceleration of a disc along its spin axis. There is no point worrying about GR until the SR problem is fully understood. In SR, angular momentum can be locally defined and is exactly conserved, always.

14. Sep 12, 2013

### Mentz114

One can write the worldline of a test body on the 'small' orbit and predict what accelerations will be required to maintain the orbit. I suppose there must be a conserved angular momentum if r=constant.

15. Sep 12, 2013

### yuiop

OK, for the SR case, in what form is angular momentum conserved under a Lorentz boost? Do we take a clue from pervect's linear momentum example https://www.physicsforums.com/showthread.php?t=701257 and use the classic Newtonian formula and use proper time instead of coordinate time? i.e. is L = m*r*dx/dtau where dx/dtau is the instantaneous proper tangential velocity of a point on the perimeter invariant under a boost?

Also no one has addressed one of my main questions in the OP. What is the "mass moment" (in practical terms) in the articles I linked to?

16. Sep 12, 2013

### Bill_K

It's the mass dipole moment, analogous to the electric and magnetic dipole moments in electromagnetism. For a discrete set of particles, it's Σ mi xi. In Newtonian gravity and linearized gravity, the gravitational field obeys Poisson's equation, and the external field of a mass distribution therefore depends only on the mass multipole moments and has an easy expression in terms of them.

17. Sep 12, 2013

### Mentz114

Suppose we have a worldline of a circular orbit $u^\mu=\dot{t}\partial_t+\dot{\phi}\partial_\phi$ where the overdots indicate differentiation wrt proper time $\tau$. Our coordinates are $t,z,r,\phi$, and if we boost in the z-direction by $\beta_z$ the worldline becomes $\gamma \dot{t}\partial_t + \beta\gamma \dot{t}\partial_z+\dot{\phi}\partial_\phi$ and the angular momentum is not changed ( this is a long-winded way to say something obvious). Naturally, a boost in the $\phi$-direction will change the angular momentum. I'm not sure I understood the question now ...

18. Sep 15, 2013

### yuiop

Yes, I meant a boost in the z direction, and yes it is obvious that the momentum remains constant if we do a 'soft' boost. For example consider an observer (A) that is initially at rest with respect to the centre of the rotating disc.This observer accelerates to a new velocity in the -z direction and the disc appears to move in the +z direction and its rotation rate appears to slow down. Another observer (B) that remains at the centre of the disc considers the rotation rate of the disc to be constant because nothing has happened to the disc or to te observer (he feels no proper acceleration.

Now we repeat the exercise but this time with a 'hard' boost. The disc itself is actually accelerated in the positive z direction relative to observer A (who remains inertial this time), such that observer B that remains at the centre of the disc experiences proper acceleration. Will observer B agree that the angular velocity of the disc remains constant over time, if the radius and invariant mass of the disc remain constant?

Assuming that the thickness of the disc is negligible, are 'soft' and 'hard' boosts equivalent for this situation, for the same relative velocity between observers A and B?