Conservation of relativistic angular momentum

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  • #1
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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
"Angular momentum has not been forgotten in this universal conjugation of physical quantities. It’s partner is the mass moment (mass × distance), which is proportional to the center of mass, or centroid."
. Again, what is the "mass moment" (in practical terms) or centroid?
 
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  • #2
WannabeNewton
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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
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yuiop said:
... under what circumstances is angular momentum conserved in General Relativity? What does Noether's thereom tell us.
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 !]
 
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  • #4
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Thanks for the quick response, WN and Mentz!
Are you talking about the angular momentum of a test particle/matter field or of a space-time?
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
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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}}##
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.
 
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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.
As far as I can tell there are no circular geodesics except in an equatorial plane...
 
  • #7
PAllen
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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
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As far as I can tell there are no circular geodesics except in an equatorial plane...
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
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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.
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
PAllen
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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?
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
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No, asymptotically flat is a feature of the geometry at infinity (specifically spatial infinity for a conserved quantity in GR).
From Wikipedia http://en.wikipedia.org/wiki/Constant_of_motion we have
Another approach is to recognize that a conserved quantity corresponds to a symmetry of the Lagrangian. Noether's theorem provides a systematic way of deriving such quantities from the symmetry. For example, conservation of energy results from the invariance of the Lagrangian under shifts in the origin of time, conservation of linear momentum results from the invariance of the Lagrangian under shifts in the origin of space (translational symmetry) and conservation of angular momentum results from the invariance of the Lagrangian under rotations. The converse is also true; every symmetry of the Lagrangian corresponds to a constant of motion, often called a conserved charge or current.
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
WannabeNewton
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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
PAllen
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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.
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
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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?
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
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... 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.
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
Bill_K
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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?
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
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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?
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
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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 ...
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?
 

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