Conservation of angular momentum in GR

[TrickyDicky]

Total angular momentum is not conserved due to lack of spacetime spherical symmetry, it is precisely this fact that causes the angular momentum of the quadrupole moment to have to be radiated away as gravitational radiation. (see Schutz, chapter 9: exercises 39,40 and 47).

In this context, there is something I don't understand about an example also found often in GR textbooks: in the Earth-moon system there is a tidal torque due to the moon's influence and also the sun's, that changes the Earth spin angular momentum by acting on the equator bulge and that slows down the Earth's spin, however in this case the total angular momentum is effectively conserved by correcting the orbit angular momentum thru its enlarging of about 4.5 cm/year.
What makes the total angular momentum to be conserved in this particular setting? Is this small system considered practically spherically symmetric?

 

[Bill_K]

Angular momentum IS conserved in any asymptotically flat spacetime, even if radiation is present.

 

[TrickyDicky]

Angular momentum IS conserved in any asymptotically flat spacetime, even if radiation is present.

Consider the Kerr spacetime, I think it is asympotically flat but it has axisymmetric metric, so it is not spherically symetric, this according to various GR textbooks means total angular momentum is NOT conserved, for instance Hobson's GR in the chapter about the Kerr metric in page 313: "Note, however, that the total angular momentum of a particle is not a conserved quantity, since the spacetime is not spherically symmetric about any point."

On the other hand the total angular momentum in a spherically symmetric spacetime is conserved and therefore the quadrupole moment's angular momentum is absent and can not produce gravitational radiation. I believe all this to be basic stuff with no much room for disagreement.

So apparently your statement is not corect but in the context of the moon-earth system I'm not sure what you mean by it anyway, are you saying that the moon-earth system can be considered asymtotically flat? Please explain.

 

[Q-reeus]

Total angular momentum is not conserved due to lack of spacetime spherical symmetry, it is precisely this fact that causes the angular momentum of the quadrupole moment to have to be radiated away as gravitational radiation. (see Schutz, chapter 9: exercises 39,40 and 47).

In this context, there is something I don't understand about an example also found often in GR textbooks: in the Earth-moon system there is a tidal torque due to the moon's influence and also the sun's, that changes the Earth spin angular momentum by acting on the equator bulge and that slows down the Earth's spin, however in this case the total angular momentum is effectively conserved by correcting the orbit angular momentum thru its enlarging of about 4.5 cm/year.
What makes the total angular momentum to be conserved in this particular setting? Is this small system considered practically spherically symmetric?

TrickyDicky, I have no access to that textbook and probably couldn't make much use of it if I did, but just a thought that may be off the mark. Sometimes authors confuse owing to failure to specify context adequately. Maybe 'Total angular momentum is not conserved' Schutz refers to above is that of the orbiting masses only ('radiation reaction' couple slowing the bodies down)? And we are supposed to assume the balance is in the GW's? In the case of the Earth-Moon system, my assumption is probably that GW's are too weak to consider - ie tidal transfer is way larger in effect and taken to be momentum conserving. I think Clifford Will covers calcs on Earth-Moon tidal coupling in http://relativity.livingreviews.org/Articles/lrr-2006-3/ but maybe wrong there.

In #3

...Hobson's GR in the chapter about the Kerr metric in page 313: "Note, however, that the total angular momentum of a particle is not a conserved quantity, since the spacetime is not spherically symmetric about any point."

Just maybe another example of context not properly specified by author - an unstated assumption the particle's angular momentum is coupling to that of a much larger spinning mass via it's Kerr metric - ie a GR variant of Earth-Moon tidal exchange?
On the other hand, would be most interested if none of my comments above hit the mark!

 

[TrickyDicky]

Sometimes authors confuse owing to failure to specify context adequately. Maybe 'Total angular momentum is not conserved' Schutz refers to above is that of the orbiting masses only ('radiation reaction' couple slowing the bodies down)? And we are supposed to assume the balance is in the GW's? In the case of the Earth-Moon system, my assumption is probably that GW's are too weak to consider - ie tidal transfer is way larger in effect and taken to be momentum conserving.

In #3

Just maybe another example of context not properly specified by author - an unstated assumption the particle's angular momentum is coupling to that of a much larger spinning mass via it's Kerr metric - ie a GR variant of Earth-Moon tidal exchange?
On the other hand, would be most interested if none of my comments above hit the mark!

In this case, this seems to be not controversial common knowledge in GR, I can assure you my citations are not out of context, I just didn't quote the whole paragraphs.
See Wikipedia Angular momentum page: "Angular momentum in relativistic mechanics:
In modern (late 20th century) theoretical physics, angular momentum is described using a different formalism. Under this formalism, angular momentum is the 2-form Noether charge associated with rotational invariance (As a result, angular momentum is not conserved for general curved spacetimes, unless it happens to be asymptotically rotationally invariant)."

It is precisely the no conservation of total angular momentum in GR that allows the quadrupole moment to exist.

I grant you my own comparison with the moon-earth system might be not valid here and thus my question in post #1 but I think Bill-k made a not exact assertion AFAICS in his answer.

 

[Q-reeus]

In this case, this seems to be not controversial common knowledge in GR, I can assure you my citations are not out of context, I just didn't quote the whole paragraphs...

OK sorry I wasn't giving you much credit there.

See Wikipedia Angular momentum page: "Angular momentum in relativistic mechanics:
In modern (late 20th century) theoretical physics, angular momentum is described using a different formalism. Under this formalism, angular momentum is the 2-form Noether charge associated with rotational invariance (As a result, angular momentum is not conserved for general curved spacetimes, unless it happens to be asymptotically rotationally invariant)."

Well it makes sense if energy is ill-defined in curved spacetime, momentum too. While asymptotically flat has for me a clear interpretation not so asymptotically rotationally invariant. This specifies whether there is some nonzero metric coupling to the spin of a test particle at infinite distance from some source of angular momentum (say a notional Kerr BH), or something else? This is new to me.

It is precisely the no conservation of total angular momentum in GR that allows the quadrupole moment to exist.

That surprises me. Two masses connected by a spring undergoing linear oscillation have a nonzero mass quadrupole moment, right? Or are we restricted to the quad moment of say two co-orbiting bodies where angular momentum is inherently present?

 

[Bill_K]

Consider the Kerr spacetime, I think it is asympotically flat but it has axisymmetric metric, so it is not spherically symetric

Of course Kerr is not spherically symmetric in the near zone, and that's not what I said. It is spherically symmetric in the asymptotic region, which is all you need. Asymptotically flat means it approaches Minkowski space as you get far away.

this according to various GR textbooks means total angular momentum is NOT conserved... "Note, however, that the total angular momentum of a particle is not a conserved quantity, since the spacetime is not spherically symmetric about any point."

This quote is referring to the angular momentum of a test particle, not the total angular momentum. The angular momentum of the particle is not conserved but the total angular momentum is conserved.

On the other hand the total angular momentum in a spherically symmetric spacetime is conserved and therefore the quadrupole moment's angular momentum is absent and can not produce gravitational radiation. I believe all this to be basic stuff with no much room for disagreement.

I agree completely with this statement, it is quite basic and there should be no disagreement.

I'm not sure what you mean by it anyway, are you saying that the moon-earth system can be considered asymtotically flat? Please explain.

Yes certainly, the gravitational field of the Earth-Moon system is asymptotically flat. The combined gravitational field goes down as M r^-3, where M is the total mass, and the spacetime asymptotically approaches Minkowski space.

 

[TrickyDicky]

Well it makes sense if energy is ill-defined in curved spacetime, momentum too. While asymptotically flat has for me a clear interpretation not so asymptotically rotationally invariant. This specifies whether there is some nonzero metric coupling to the spin of a test particle at infinite distance from some source of angular momentum (say a notional Kerr BH), or something else? This is new to me.

In fact that term "asymptotically rotationally invariant" seems to be not very commonly used, normally just rotational invariance is mentioned.

That surprises me. Two masses connected by a spring undergoing linear oscillation have a nonzero mass quadrupole moment, right? Or are we restricted to the quad moment of say two co-orbiting bodies where angular momentum is inherently present?

You are right. I'm actually referring only to binary systems of co-orbiting bodies configurations.

 

[Cleonis]

[...] in the Earth-moon system there is a tidal torque due to the moon's influence and also the sun's, that changes the Earth spin angular momentum by acting on the equator bulge and that slows down the Earth's spin, however in this case the total angular momentum is effectively conserved by correcting the orbit angular momentum thru its enlarging of about 4.5 cm/year.
What makes the total angular momentum to be conserved in this particular setting?

There are two distinct effects. (Probably there a bit of cross-influence, but afaik that's negligable.)
The two have in common that they each relate to a way of the Earth not being perfectly spherical.

1. Moon and Sun act upon the Earth's equatorial bulge, giving rise to a torque upon the Earth, hence a corresponding gyroscopic precession. For the Earth that gives the precession of the equinox.

2. The moon acts upon the tidally distorted Earth. Due to the internal friction the Earth's tidal distortion lags behind. Given that lag there is a gravitational effect that increases the Moon's orbital energy, and it decreases the Earth's rotational energy. The number of 4.5 cm/year increase of Moon orbit altitude is the increase of Moon orbital energy.

(As I understand it: when the Moon first formed it was rotating somewhat faster than 1 rotation per month. The Moon was tidally distorted by the Earth, and rotating more than once a month. Internal friction makes such distortion lag behind. This lag gave the Earth opportunity to slow the Moon's rotation down. At some point in time the Moon reached a state of tidal lock. )

Anyway, the precession of the equinox and the Earth rotation slowing down arise both from gravitational distortion effects, but it's two different effects.
Neither of them are relativistic effects, which makes it unlikely that any relativity textbook author would mention them.

 

[TrickyDicky]

Of course Kerr is not spherically symmetric in the near zone, and that's not what I said. It is spherically symmetric in the asymptotic region, which is all you need. Asymptotically flat means it approaches Minkowski space as you get far away.

Ok, that solves the doubt about asymptotically rotational invariance in the above post too.

This quote is referring to the angular momentum of a test particle, not the total angular momentum. The angular momentum of the particle is not conserved but the total angular momentum is conserved.

Ok, I see.

Yes certainly, the gravitational field of the Earth-Moon system is asymptotically flat. The combined gravitational field goes down as M r^-3, where M is the total mass, and the spacetime asymptotically approaches Minkowski space.

I see, I didn't consider the fact that it could be modeled that way.

Thanks for the clarifications

 

[PAllen]

My understanding of some of the issues touched on here is as follow:

A pair of orbiting bodies in GR does not exactly conserve angular momentum considering just the two bodies. Neither do a mutually orbiting pair of opposite charges, considering mass to be gravitationally negligible, and only EM force significant. The deviation in the gravity case is much smaller (due to existence only of quadrupole radiation), versus dipole for the EM case.

HOWEVER, the total angular momentum of bodies plus gravitational radiation; or charges plus EM radiation *does* conserve angular momentum exactly.

Issues like the rotational symmetry or assymptotic flatness of the universe relate to whether you can define a universally conserved angular momentum. As long as, within some large region, you have sufficient overall flatness, there is conservation of angular momentum within the large region (e.g. the milkyway's mass, energy, plus gravitational radiation conserve angular momentum).

For me, a good source on some of these issues is (in passing):

http://arxiv.org/abs/gr-qc/9909087

 

[pervect]

I can't find the reference in Schutz that says that angular momentum isn't conserved. Is this "A First course in General Relativity?" Perhaps there's some change with editions.

There is mention in Held, "General Relativity and Gravitation", about various ways to define angular momentum in an asymptotically flat space-time, in a chapter by Winicour. My version of Schutz also cites Held:

The measurement of the mass and angular momentum of a source by looking at its distant gravitational field is discussed in Misner et al(1973), Ashtekar(1980), and Winicour (1980)

the Winicour reference is found in Held's collection of papers, "General Relativity and Gravitation, 100 yeras after the birth of Einstein".

[add]
I"m also reasonably certain that there is a conserved orbital angular momentum for a particle orbiting a spining black hole. As I said, I couldn't find the section of Schutz that you were referring to - it's not my favorite textbook.

Wiki seems to back me up on this:

http://en.wikipedia.org/w/index.php?title=Carter_constant&oldid=420908353

The Carter constant is a conserved quantity for motion around black holes in the general relativistic formulation of gravity. Carter's constant was derived for a spinning, charged black hole by Australian theoretical physicist Brandon Carter in 1968. Carter's constant along with the energy, axial angular momentum, and particle rest mass provide the four conserved quantities necessary to uniquely determine all orbits in the Kerr-Newman spacetime (even those of charged particles).

As none of the metric coefficients is a function of $\phi$, I think you should have rotational symmetry as is required by Noether's theorem. The space-time will be "stationary" rather than static because of the $dt \, d\phi$ term, but that shouldn't spoil the conservation laws - it does prevent you from having "hypersurface orthogonality" though.

 

[Q-reeus]

Here's something I've hastily put together as purely hand-waving argument (no numbers, no equations) you folks might like to pick over while I get to a very late sleep.
Take an ideally 'incompressible' hollow right circular cylinder, spinning at constant angular velocity about it's major vertical axis and supported on frictionless bearings. Now apply gradually pressure to the cylinder ends, placing the sides under purely axial compression. Unlike the case of an ideally incompressible fluid so compressed, there is no resulting induced circumferential stress. A relativistic transformation into the frame of a moving element in the cylinder wall will therefore notice no circumferential stress gradient during the time changing squeeze phase, and therefore no relativistic force in that direction. So, to the extent elastic energy is absent (and we have idealized to an 'incompressible' solid), our cylinder undergoes no change in either spatial dimensions or angular velocity as a result of the stressing process. Ho hum you might say, but here's the rub. Pressure/stress is a source of gravity (active mass) in GR, and by the equivalence principle that means equally also a source of both passive and inertial mass. The latter is key here. Stressing the cylinder has increased it's inertial mass 'for free' and therefore it's angular momentum and rotational kinetic energy. So in principle we can endlessly cycle this process and generate angular momentum and energy - spin up in unstressed state, spin down in stressed state, over and over. Of course in the real world case friction and finite elastic energy contributions and cross couplings to the 'pure pressure' effect will be overwhelmingly greater. But these will be material dependent factors and hence not fundamental. Can it be shown there is no such effect at all though? Cheers!

 

[pervect]

Wincour's article in Held (pg 74, eq 2-10) gives the expression for the total angular momentum as a volume intergal as:

$$\int_{\Sigma} \xi^{b} \left(T^{a}{}_{b} - \frac{1}{2} \, \delta^{a}{}_{b} T \right) dS_{a}$$

Here $\xi^{b}$ is the appropriate killing vector - for angular momentum, it would be a phi-translation vector.

I *think* dS_a is just a unit future (i.e. normal to the volume element sigma) - and I *think* this implies that pressure doesn't contribute to angulalr momentum (though you are correct that it contributes to energy) - because the kronecker delta factor vanishes when a is not equal to b, so the dependence on T is present only when you have a time-like killing vector and are computing energy, it doesn't contribute when you have a space-like killing vector and are computing linear or angular momentum.I'm only 50/50 on this...

 

[Q-reeus]

Wincour's article in Held (pg 74, eq 2-10) gives the expression for the total angular momentum as a volume intergal as:

$$\int_{\Sigma} \xi^{b} \left(T^{a}{}_{b} - \frac{1}{2} \, \delta^{a}{}_{b} T \right) dS_{a}$$

Here $\xi^{b}$ is the appropriate killing vector - for angular momentum, it would be a phi-translation vector.

I *think* dS_a is just a unit future (i.e. normal to the volume element sigma) - and I *think* this implies that pressure doesn't contribute to angulalr momentum (though you are correct that it contributes to energy) - because the kronecker delta factor vanishes when a is not equal to b, so the dependence on T is present only when you have a time-like killing vector and are computing energy, it doesn't contribute when you have a space-like killing vector and are computing linear or angular momentum.
I'm only 50/50 on this...

I'm assuming pervect your last entry refers to mine in #13, in which case thanks for your interest in the problem, but it was posed as a potential counterexample to the kind of generalized theorems employing the style of high end maths you have used. Sorry but I can't really follow it at that level; could you please translate it back to the specifics of the setup? What I expected was an argument that because the cylinder axial stress is normal to the motion, v.sigma = 0 (v the local velocity, sigma the axial stress) for any given moving element and thus no actual 'boost' to momentum and KE. My objection in turn would be that implies a breakdown in the principle of equivalence - if the pressure generated inertial mass is directional in nature, why not the same for the active and passive mass? But no-one would argue the gravitational properties of the latter two would be anything but isotropic, surely. Nearly all cases of astrophysical interest seem to be limited to fluids under pressure, and here isotropy hides the issue - ie you just consider rho + 3p, the 3 factor signifying that all three orthogonal components of pressure in a fluid are equally contributing to the effective mass density. But is it consistent? If m_a = m_p = m_i here, it automatically implies that for any given acceleration a, there will be a full contribution F = m_ia = 3pa owing to fluid pressure inertial mass contribution. Which necessarily implies a full contribution from the two components of p orthogonal to a. Which in turn brings us back to the case of our spinning cylinder - either the principle of equivalence fails for pressure, or we have a genuine perpetuum mobile (we should use v|sigma|, not v.sigma). What's wrong with this line of reasoning?

 

[TrickyDicky]

Which in turn brings us back to the case of our spinning cylinder - either the principle of equivalence fails for pressure, or we have a genuine perpetuum mobile What's wrong with this line of reasoning?

Hi, Q-reeus
Probably I don't understand your thought experiment very well but I would say that if in a experiment about conservation of momentum-energy you leave out friction and compressibility you can get all kind of perpetuum mobiles.

 

[Q-reeus]

Hi, Q-reeus
Probably I don't understand your thought experiment very well but I would say that if in a experiment about conservation of momentum-energy you leave out friction and compressibility you can get all kind of perpetuum mobiles.

Firstly TrickyDicky I feel a little guilty now, having jumped in with a different angle altogether - even though it has kinda stuck to the title, didn't really stop to think if this was really a 'highjacking' of your thread. If you feel that I will put a stop to this now - not that there's much to stop mind you!
On the matters you have raised, I agree the natural instinct is to assume this has got to be wrong if one stipulates physically impossible idealizations. But in this context it follows a long line of similar examples. To isolate the essential features, remove any facors that merely complicate but are not germane. Friction is a fairly obvious one - not claiming a perfect flywheel or anything like that. Finite compressibility is less obvious but is standard procedure for what seems like an endless succession of papers in AJP etc that keep rolling out on 'hidden momentum' and 'stored field momentum' for instance, and many of these are actually quite close to this scenario. As explained in #13, these two factors are both material dependent and thus not of fundamental significance to the principle in question. One caveat there though; SR dictates an upper theoretical limit to rigidity (Born Rigid), but I believe even there it does not fundamentally touch on the principles in question. And I have not really claimed a PMM as such - merely pointed out this is implied imho if stress as source of inertial mass has the properties I think is so. So if you would like to add anything, here or by my starting a new thread, be my guest. Same old problem here though, an almost deathly silence!

 

[TrickyDicky]

... If you feel that I will put a stop to this now

No problem at all.

So if you would like to add anything, here or by my starting a new thread, be my guest. Same old problem here though, an almost deathly silence!

After reading pervect's post I would say his answer makes sense, in GR the conservation laws are encoded in symmetries of the system, one way to look at these symmetries uses Killing vectors, where energy conservation is related to the time-like Killing vector and angular momentum is related to space-like kiling vector, if the formula pervect writes applies here it means (I think, if not please pervect explain) that the cylinder might gain energy (in a static setting), but not angular momentum and therefore no spin cycle and no free energy.

 

[bcrowell]

I can't find the reference in Schutz that says that angular momentum isn't conserved. Is this "A First course in General Relativity?" Perhaps there's some change with editions.

The argument is pretty elementary: angular momentum isn't a scalar. No non-scalar quantity can be globally conserved for all spacetimes in GR, because parallel transport is path-dependent, but you can't add up the quantity without parallel-transporting to a single point first. (A related but slightly different way of putting it is that GR doesn't have global frames of reference, so there is no frame of reference that could be used in order to express the vector representing the total.)

 

[pervect]

Q-reeus. I thought your experiment was interesting. My tentative conclusion is simply that the angular momentum does not change when you compress the cylinder, as one would expect.

It is true that if you have a cylinder with a bolt through it, and that if you tighten up the bolt to put the cylinder under pressure, that the Komar mass of the cylinder increases. While the Komar mass, of the cylinder increases because of the pressure, the Komar mass of the bolt decreases, because it's under tension. Unless something actually deforms, doing work, the whole closed system doesn't gain or lose mass, it's just distributed differently.

However, (assuming my understanding is correct) even though rotating cylinder is more massive, it appears to be the case that it doesn't have any more angular momentum. (Again, using the Komar formula, which can be generalized to handle angular and linear momentum.) The pressure terms, which do contribute to the Komar mass, simply don't matter to the angular momentum.

That's assuming I've understood the equations properly. It all appears to make sense, but I don't have a lot of "worked problems" in this area to alert me to potential confusion on my part, and I've learned to be cautious about that situation.

 

[Sam Gralla]

Some of the confusion in this thread is a result of the fact that angular momentum can be defined either at spatial infinity or at null infinity. Angular momentum defined at spatial infinity is conserved, for every asymptotically flat spacetime, period. This is because even if you radiate angular momentum the radiation never makes it out to spatial infinity, and the angular momentum of the radiation is included when you calculate the angular momentum, no matter what time (slice) you choose. Angular momentum defined at null infinity is not conserved, precisely due to the radiation loss that Schutz talks about.

 

[Q-reeus]

No problem at all.

Thanks - quite a relief!

After reading pervect's post I would say his answer makes sense, in GR the conservation laws are encoded in symmetries of the system, one way to look at these symmetries uses Killing vectors, where energy conservation is related to the time-like Killing vector and angular momentum is related to space-like kiling vector, if the formula pervect writes applies here it means (I think, if not please pervect explain) that the cylinder might gain energy (in a static setting), but not angular momentum and therefore no spin cycle and no free energy.

Fair enough but my rather simple approach is to test general theorems by way of a specific setup, and just see if all situations have been fully and accurately accounted for by said theorem(s). See my comments to pervect's post in #20.

 

[Q-reeus]

Q-reeus. I thought your experiment was interesting. My tentative conclusion is simply that the angular momentum does not change when you compress the cylinder, as one would expect.
It is true that if you have a cylinder with a bolt through it, and that if you tighten up the bolt to put the cylinder under pressure, that the Komar mass of the cylinder increases. While the Komar mass, of the cylinder increases because of the pressure, the Komar mass of the bolt decreases, because it's under tension. Unless something actually deforms, doing work, the whole closed system doesn't gain or lose mass, it's just distributed differently.

Agreed, no argument there. However it matters much that the mass is distributed differently when a rotating system is under consideration. If for instance the bolt as assumed lies on the axis of rotation, it's contribution to angular momentum will be relatively tiny, even though it cancels the overall system mass change from the cylinder. I had actually envisaged a stationary external agent in applying pressure (eg. G-clamp) which could be quite compressible itself but an irrelevant factor - only changes to the rotating mass is important imo.

However, (assuming my understanding is correct) even though rotating cylinder is more massive, it appears to be the case that it doesn't have any more angular momentum. (Again, using the Komar formula, which can be generalized to handle angular and linear momentum.) The pressure terms, which do contribute to the Komar mass, simply don't matter to the angular momentum.

Well that would be extremely problematic from my pov given points raised in #15. How could this be explained? I argued in #13 there would be no circumferential forces of relativistic origin, owing to the purely axial nature of the induced stress. Therefore no physical mechanism to reduce angular velocity. Now if cylinder Komar mass increase is 'really real', how on Earth can there not be an increase in angular momentum? My suspicion is that conservation of energy/momentum is a priori built into the Komar 'proof' of conservative behavior. Hence the specific counterexample. I have argued in #15 one cannot have it both ways - if m_a = m_p = m_i (WEP) holds for pressure induced mass, transverse components of pressure must contribute to inertial mass m_i just as for the component in line with any acceleration a. That leaves no way around the conclusion angular momentum must increase for the setup of #13. The other alternative is there is indeed a directional nature to pressure generated m_i (but not for m_a or m_p!). Hence WEP must fail here, which would still be news imo. My bet is though WEP holds. Thoughts?

 

[TrickyDicky]

Some of the confusion in this thread is a result of the fact that angular momentum can be defined either at spatial infinity or at null infinity. Angular momentum defined at spatial infinity is conserved, for every asymptotically flat spacetime, period. This is because even if you radiate angular momentum the radiation never makes it out to spatial infinity, and the angular momentum of the radiation is included when you calculate the angular momentum, no matter what time (slice) you choose. Angular momentum defined at null infinity is not conserved, precisely due to the radiation loss that Schutz talks about.

So I guess in the Hulse-Taylor binary pulsar system angular momentum is defined at null infinity, not asymptotically flat situation, and thus radiates GW, and angular momentum is not conserved.
However in the earth-moon orbit system angular momentum is conserved because it is defined at spatial infinity, and it is an asymptotically flat modeled situation.

Is this right according to your explanation?
How do we choose to define angular momentum either at null infinity or at spatial infinity for a given orbital system?

 

[PAllen]

So I guess in the Hulse-Taylor binary pulsar system angular momentum is defined at null infinity, not asymptotically flat situation, and thus radiates GW, and angular momentum is not conserved.
However in the earth-moon orbit system angular momentum is conserved because it is defined at spatial infinity, and it is an asymptotically flat modeled situation.

Is this right according to your explanation?
How do we choose to define angular momentum either at null infinity or at spatial infinity for a given orbital system?

No, I don't think this is right. As I explained earlier, the Hulse Taylor system *including its gravitational radiation* conserves angular momentum. The fact that the system excluding radiation does not is completely trivial and non-relativistic and is analagous to the loss of angular momentum in a pair of co-orbiting charges in EM + SR; including the EM radiation, angular momentum is conserved.

In a radiating system, it seems not very sensible to use a null-infinity definition, because you want to include the radiation at all times.

I think the Hulse-Taylor system is normally analyzed with an asymptotic flatness condition. It is not static, or spherically symmetric, but can be asymptotically flat.

 

[TrickyDicky]

No, I don't think this is right. As I explained earlier, the Hulse Taylor system *including its gravitational radiation* conserves angular momentum. The fact that the system excluding radiation does not is completely trivial and non-relativistic and is analagous to the loss of angular momentum in a pair of co-orbiting charges in EM + SR; including the EM radiation, angular momentum is conserved.

In a radiating system, it seems not very sensible to use a null-infinity definition, because you want to include the radiation at all times.

This doesn't agree with Sam Gralia's statement "Angular momentum defined at null infinity is not conserved, precisely due to the radiation loss...".
The fact is we include the radiation precisely because angular momentum is not conserved.
Precisely the reason dipole moment is not radiated as GW is because linear momentum is conserved.

I think the Hulse-Taylor system is normally analyzed with an asymptotic flatness condition. It is not static, or spherically symmetric, but can be asymptotically flat.

This I could agree with. Edit: In this case angular momentum is defined at spatial infinity and therefore conserved because the GW can't reach spatial infinity, is this right?

 

[Q-reeus]

TrickyDicky: In the other recent thread, I mentioned an Angelo Loinger, who is a strictly GR theorist (of sorts), and claims full GR does not admit to GW's - if true then coupled to binary pulsar data, automatically means failure of angular momentum conservation in toto. For sure 'controversial' but anyway the paper is relatively short, and if you do a search at arXiv.org will not be hard to find - you could no doubt make more of than I could. Just a thought.

 

[Sam Gralla]

So I guess in the Hulse-Taylor binary pulsar system angular momentum is defined at null infinity, not asymptotically flat situation, and thus radiates GW, and angular momentum is not conserved.
However in the earth-moon orbit system angular momentum is conserved because it is defined at spatial infinity, and it is an asymptotically flat modeled situation.

Is this right according to your explanation?
How do we choose to define angular momentum either at null infinity or at spatial infinity for a given orbital system?

I think probably all that is going on here is that the earth-moon system analysis you read about is in a non-relativistic (i.e., Newtonian) limit, where there is no gravitational radiation. In reality it will radiate a bit of angular momentum, but my guess is this is tiny and simply being neglected in whatever you read about. In a binary pular, on the other hand, there is a lot of gravitational radiation.

Whether to use angular momentum defined at null or spatial infinity is just a choice. If you want to see how much angular momentum escapes as radiation, you use null infinity. If you want to see that angular momentum is always conserved, as long as the angular momentum in the radiation is included, you use spatial infinity. Both are well defined for any asymptotically flat spacetime (although the definitions of asymptotic flatness people give in the respective cases can be a little bit different). I didn't really make this comment in response to your original question; it's just I noticed some people were saying "angular momentum is conserved" and some people were saying "no, you can radiate it away", and the different definitions of angular momentum are probably the cause there.

 

[PAllen]

TrickyDicky: In the other recent thread, I mentioned an Angelo Loinger, who is a strictly GR theorist (of sorts), and claims full GR does not admit to GW's - if true then coupled to binary pulsar data, automatically means failure of angular momentum conservation in toto. For sure 'controversial' but anyway the paper is relatively short, and if you do a search at arXiv.org will not be hard to find - you could no doubt make more of than I could. Just a thought.

I don't know his credentials, but he says many things all other GR researcher's disagree with. For example, in the intro to one of his papers 'refuting' gravitational waves he says:

"The exact (non-approximate) formulation of general relativity (GR)
does not allow the existence of physical gravitational waves (GW’s). I have
given several proofs of this fact [1]. Quite simply, we can observe, e.g.,
that bodies which interact only gravitationally describe geodesic lines, and
therefore – as it is very easy to see – they do not generate any GW."

Except that going all the way back to Einstein and Infeld (in the 1940s), and everyone else who derives equations of motion directly from the field equations, finds that bodies follow geodesics only in the limit of point particles. Real, massive bodies closely approximate geodesics, but the difference cannot be ignored for a system like two orbiting pulsars. Thus I immediately suspect the whole body of work.

 

[Q-reeus]

...Except that going all the way back to Einstein and Infeld (in the 1940s), and everyone else who derives equations of motion directly from the field equations, finds that bodies follow geodesics only in the limit of point particles. Real, massive bodies closely approximate geodesics, but the difference cannot be ignored for a system like two orbiting pulsars. Thus I immediately suspect the whole body of work.

I'm in no position to argue that - you are probably quite right. I never noticed him referring his findings to the binary pulsar results which seemed odd, but owing to how short his arguments were, I figured someone with a good grasp of GR (not me) could sort out his logic without too much sweat

 

[bcrowell]

Loinger is a kook with academic credentials. In addition to the kookery discussed by PAllen, Loinger and his coauthor Marsico believe that there is something wrong with the standard analysis of black holes: http://arxiv.org/abs/1011.2600 Basically they make a big deal out of the fact that an observer at infinity sees infalling matter take an infinite amount of time to reach the event horizon. They treat this as having Dramatic Physical Implications, and they criticize other authors: "Now, if we take into account the decisive role of the Hilbertian gravitational repulsion, which is neglected by [Broderick et al.][...]" This is stupid and misleading, since Broderick hasn't made a mistake by ignoring some relevant effect. The "repulsion" isn't really a repulsion; it's just a *coordinate* acceleration with no intrinsic physical significance. An observer at infinity sees infalling material as taking infinite time to reach the event horizon, and since the Schwarzschild t coordinate is the time measured by an observer at infinity, the coordinate acceleration d2r/dt2 has an outward direction.

 

[TrickyDicky]

I think probably all that is going on here is that the earth-moon system analysis you read about is in a non-relativistic (i.e., Newtonian) limit, where there is no gravitational radiation. In reality it will radiate a bit of angular momentum, but my guess is this is tiny and simply being neglected in whatever you read about. In a binary pular, on the other hand, there is a lot of gravitational radiation.

Yes, I also think that's all.
I guess in a Newtonian-limit non-relativistic analysis of the binary pulsar, you do have total angular momentum conservation with no gravitational radiation, with the orbit angular momentum (shortening of the orbital period) compensated by the spin angular moment of the pulsar (acceleration of spin) just like in the earth-moon system Newtonian treatment the slowing of the Earth spin is compensated by the enlarging of the orbit and therefore total angular momentum is conserved.

Whether to use angular momentum defined at null or spatial infinity is just a choice. If you want to see how much angular momentum escapes as radiation, you use null infinity. If you want to see that angular momentum is always conserved, as long as the angular momentum in the radiation is included, you use spatial infinity. Both are well defined for any asymptotically flat spacetime (although the definitions of asymptotic flatness people give in the respective cases can be a little bit different). I didn't really make this comment in response to your original question; it's just I noticed some people were saying "angular momentum is conserved" and some people were saying "no, you can radiate it away", and the different definitions of angular momentum are probably the cause there.

Yeah, there is some confusion about asymptotically flat spacetimes, it is not an easy theme for non-experts at least. Part is derived from the fact that this and many other common themes in GR have a lot to do with the concepts of "conformal infinity" and "null infinity" that are not usually treated with depth in introductory texts but that are vital to give a sound theoretical basis to GW, black holes etc.
Here is a pretty informative site about this: http://relativity.livingreviews.org/Articles/lrr-2004-1/

Also as I read angular momentum in GR is a thorny subject, or it has been until relatively recently.

 

[TrickyDicky]

Real, massive bodies closely approximate geodesics, but the difference cannot be ignored for a system like two orbiting pulsars. Thus I immediately suspect the whole body of work.

Most likely this guy's work is flawed, but not for this reason.
If we are limiting the analysis to bodies trajectories is perfectly licit to talk about geodesics for anybody trajectory, the fact that in the surface of a massive body test particles at rest are not following geodesic motion is not relevant in this case. The fact is curvature of spacetime (gravity) manifests thru the convergence-divergence of geodesic trajectories.

 

[bcrowell]

Real, massive bodies closely approximate geodesics, but the difference cannot be ignored for a system like two orbiting pulsars. Thus I immediately suspect the whole body of work.

Most likely this guy's work is flawed, but not for this reason.
If we are limiting the analysis to bodies trajectories is perfectly licit to talk about geodesics for anybody trajectory, the fact that in the surface of a massive body test particles at rest are not following geodesic motion is not relevant in this case. The fact is curvature of spacetime (gravity) manifests thru the convergence-divergence of geodesic trajectories.

No, PAllen was correct, and he wasn't saying anything about surfaces. Bodies with nonnegligible mass don't follow geodesics because they radiate gravitational waves.

 

[Q-reeus]

If we are limiting the analysis to bodies trajectories is perfectly licit to talk about geodesics for anybody trajectory, the fact that in the surface of a massive body test particles at rest are not following geodesic motion is not relevant in this case. The fact is curvature of spacetime (gravity) manifests thru the convergence-divergence of geodesic trajectories.

Can't argue the specifics, but agree 100% it's not proper to dismiss anyone who makes a serious argument without presenting their entire rationale first. While no doubt PAllen in #29 meant only the best, it would have been a bit kinder to A.Loinger to have quoted the remainder of the intro "..If we add non-gravitational forces, the conclusion remains the same, because the new trajectories do not possesses kinematical elements (velocity, acceleration,
time derivative of the acceleration, etc.) different from those of the geodesic motions."
Not saying the final conclusion in the full course of time would be different, but, inadvertently or not, excising aspects of his pov paints a particularly simplistic 'straw man' picture. Likewise, bcrowell's costic assessment in #31 to my mind doesn't address the fact that while 'Hilbertian repulsion' may be a poor way of labelling things, is it truly 'of no physical significance' that in Schwarzschild coords a distant observer finds infalling matter will decelerate to zero velocity on approach to the nominal EH of a nominal stationary BH? Isn't this a legitimate 'physically meaningful' perspective, even if the terminology 'repulsion' is somewhat misguided? I'm not the only one to have pointed out that 'infinite redshift' is more than mere 'optical illusion' in this setting -it has a genuine physical meaning, imho at least. And honestly, it may be true Loinger is ultimately off the mark, but is it right to label such folks with terms like 'kook', 'nutter'. 'fringe', 'weirdo', [insert your own pejorative and demonizing term(s) here..]? More reasonable surely to just say 'mistaken', 'minority view' etc. Can't help but see parallels with a particularly intolerant and 'huge vote of hands' religious movement that has just one word for dissenters - 'infidel'.
One last thing here. Notwithstanding TrickyDicky's accepting stance, fact is I have created a 2-stream thread here, and am officially 'pulling' my part of that at this point. Will be in due time be reformulating and starting as new thread - who knows, I might even get some decent feedback (exempting two participants from that observation!).