Spin Density and Non-symmetric Stress-Energy Tensor

[stevendaryl]

In General Relativity, the assumption is made that the stress-energy tensor T^αβ is symmetric. However, if there are particles with intrinsic spin, then this assumption is false, as described here: http://en.wikipedia.org/wiki/Spin_tensor

The spin tensor S^αβμ satisfies:

∂_μ S^αβμ = T^βα - T^αβ

Here's what I don't understand about this: Surely, there should be not much observational difference between (A) a particle with intrinsic spin, and (B) a composite particle made of much smaller spinless subparticles, which have orbital angular momentum.

My intuitive feeling is that at a gross level of description, there should not be much difference between particles with spin and composite particles with orbital angular momentum. It seems like one should be some kind of limit of the other. (The article here http://en.wikipedia.org/wiki/Einstein–Cartan_theory#.CF.89-consistency describes a continuum limit of many tiny black holes.) But how can a symmetric T^αβ become non-symmetric in the limit?

 

[Bill_K]

stevendaryl, The confusion results because there are several definitions for the energy momentum tensor, and unfortunately many references use the oldest and simplest, the "canonical" energy momentum tensor. As long ago as 1939, Belinfante pointed out this tensor is unsymmetrical for a field with spin, and defined a modified "symmetrical" energy momentum tensor.

The simplest definition of all: in general relativity the energy momentum tensor is defined to be the source of the gravitational field, and may be calculated as

T^μν = (2/√-g) δS/δg_μν

where S is the action. (The √-g factor is required to make the result a tensor rather than a tensor density.) This energy momentum tensor is always symmetrical.

 

[stevendaryl]

stevendaryl, The confusion results because there are several definitions for the energy momentum tensor, and unfortunately many references use the oldest and simplest, the "canonical" energy momentum tensor. As long ago as 1939, Belinfante pointed out this tensor is unsymmetrical for a field with spin, and defined a modified "symmetrical" energy momentum tensor.

The simplest definition of all: in general relativity the energy momentum tensor is defined to be the source of the gravitational field, and may be calculated as

T^μν = (2/√-g) δS/δg_μν

where S is the action. (The √-g factor is required to make the result a tensor rather than a tensor density.) This energy momentum tensor is always symmetrical.

I'm not exactly sure I understand what you're saying. You're certainly right, that in GR, the energy momentum tensor is always symmetrical. The question is whether it should be symmetrical, in the presence of intrinsic spin. The claim made in the article:
http://en.wikipedia.org/wiki/Einstein-Cartan_theory
is that GR cannot correctly describe particles with intrinsic spin (in particular, the spin-orbit coupling---The article is actually ambiguous as to what that means; it makes reference to quantum mechanics. But in QM, the coupling is not gravitational, it is electromagnetic. What I assume to be the case is that spin-orbital coupling in GR is about gravitational coupling of some sort.)

The point about the canonical stress-energy tensor being nonsymmetric may or may not have anything to do with the derivation in http://en.wikipedia.org/wiki/Spin_tensor. In the latter case, they derive that a non-vanishing spin density should lead to a nonsymmetric stress-energy tensor. It doesn't assume any particular form for fields--it's basically non-quantum.

 

[stevendaryl]

The point about the canonical stress-energy tensor being nonsymmetric may or may not have anything to do with the derivation in http://en.wikipedia.org/wiki/Spin_tensor. In the latter case, they derive that a non-vanishing spin density should lead to a nonsymmetric stress-energy tensor. It doesn't assume any particular form for fields--it's basically non-quantum.

Well, in another Wikipedia article, here: http://en.wikipedia.org/wiki/Belinfante–Rosenfeld_stress-energy_tensor
the Belinfante approach to the stress-energy tensor is described. It basically amounts to distinguishing between

T^αβ, which is not symmetric, and
T_B^αβ, which is, where

T_B^αβ = T^αβ + 1/2 ∂_λ [S^αβλ + S^βαλ - S^λβα]

So in regular GR, T_B^αβ is the contribution to the curvature, not T^αβ. Fine. But that doesn't really clear everything up. Are you claiming that the article about Einstein-Cartan (here http://en.wikipedia.org/wiki/Einstein-Cartan_theory) is wrong in claiming that GR doesn't handle spin-density correctly?

 

[Bill_K]

In general relativity the energy momentum tensor is symmetric, even for particles or fields with intrinsic spin. Einstein-Cartan theory is (or was) a rival theory that introduced torsion. The initial success Einstein had in explaining gravitation through geometry (Riemannian manifold) encouraged many workers to explore more general geometrical environments, and this was one of those experiments.

All the experimental evidence to date points to general relativity as being the correct theory. Einstein-Cartan theory predicted gravitational spin-orbit coupling, which does not occur in general relativity.

 

[Ben Niehoff]

GR and Einstein-Cartan theories are equivalent (in EC, torsion is a non-propagating field). The only difference is that in GR, you symmetrize the SE tensor via the Belinfante construction, and in EC, you don't. The end result is mathematically equivalent.

 

[Ben Niehoff]

At least, I am pretty sure they are equivalent. The Wiki article contains at least one glaring mistake: It defines the momentum tensor via

$$\frac{\delta \mathcal{L}_\mathrm{G}}{\delta g^{ab}} -\frac{1}{2}P_{ab}=0$$
and then claims that $P_{ab}$ is not symmetric! Obviously $P_{ab}$ defined this way is symmetric, because $g_{ab}$ is symmetric.

I don't see any mention in the article of attempting to use a non-symmetric metric tensor, but even if so, this is equivalent to coupling ordinary GR to a 2-form gauge potential B with 3-form field strength (which in turn is equivalent to allowing some amount of torsion, but under some constraints).

EC theory has the exact same action as GR,

$$S = \frac{1}{2 \kappa^2} \int \big( R \sqrt{|g|} d^4 x + \mathcal{L}_M \big),$$
with the difference that the Ricci scalar is considered to be a function of both the metric and the connection separately. One of the equations of motion of EC theory is the one that relates the connection to the metric in the standard way, so it has to be equivalent to standard GR, as far as I can tell.

 

[tom.stoer]

GR and Einstein-Cartan theories are equivalent (in EC, torsion is a non-propagating field). The only difference is that in GR, you symmetrize the SE tensor via the Belinfante construction, and in EC, you don't. The end result is mathematically equivalent.

GR and EC are not mathematical identical but indistinguishable experimentally; the reason is that (as Ben says correctly) torsion is non-propagating (in contrast to curvature) and therefore torsion vanishes identically in vacuum; effects due to spin-density and non-vanishing torsion are restricted to matter distributions; but within matter spin-density effects are highly supressed compared; one may expect spin effects to be important in quantum gravity theories; one may expect the semi-classical limit of LQG coupled to spinning matter to be EC, not GR.

 

[Ben Niehoff]

I'll have to think about that more, Tom. Like I pointed out, GR and EC have exactly the same action. Whether you end up with torsion depends both on what matter you couple to, as well as whether you vary the action with respect to g only, or with respect to g and \Gamma separately.

Can you point to a case where GR and EC make different predictions? I know the Wiki article mentions "spin-orbit coupling", but they don't give a concrete example. I have a feeling the torsion effects in EC can be accounted for by force effects in GR, and that the physics will be the same.

The article does contain other inaccuracies, such as suggesting that geodesics "twist around each other" in the presence of torsion. No such thing happens; the torsion does not enter the geodesic equation. (Torsion does cause parallel-transported vectors to "twist around", but this is not the same thing).

 

[tom.stoer]

Like I pointed out, GR and EC have exactly the same action.

The variables and the actions are different; the EC action reduces to EH action only if you apply the constraints + the condition of vanishing torsion to reduce the vierbein and connection variables to the metric formalism - which you can't do when torsion is present.

Have a look at

http://arxiv.org/abs/gr-qc/0606062
Einstein-Cartan Theory
Authors: Andrzej Trautman
(Submitted on 14 Jun 2006)
Abstract: The Einstein--Cartan Theory (ECT) of gravity is a modification of General Relativity Theory (GRT), allowing space-time to have torsion, in addition to curvature, and relating torsion to the density of intrinsic angular momentum. This modification was put forward in 1922 by Elie Cartan, before the discovery of spin. Cartan was influenced by the work of the Cosserat brothers (1909), who considered besides an (asymmetric) force stress tensor also a moments stress tensor in a suitably generalized continuous medium.
Comments: 7 pages, uses amsmath.sty, amssymb.sty
Journal reference: Encyclopedia of Mathematical Physics: Elsevier, 2006, vol. 2, pages 189--195

 

[cosmik debris]

I think theories with torsion predict that identical masses with opposite chirality fall at different rates in a Gravitation field.

 

[Ben Niehoff]

The variables and the actions are different; the EC action reduces to EH action only if you apply the constraints + the condition of vanishing torsion to reduce the vierbein and connection variables to the metric formalism - which you can't do when torsion is present.

As far as I can tell, both actions are simply the integral of the Ricci scalar, just written in terms of different fundamental fields. Did you have something else in mind?

I'm not concerned with whether there is torsion or not. I'm more concerned with whether the effects of torsion in the EC theory can be achieved by coupling to additional matter fields in standard GR. The geometry is a model of the physics; there may be more than one way to model the same physics.

I'm open to the idea of the two theories being different, but I still think it may be possible to get the same physics as EC theory by coupling GR to a B field, or something like it. I'll have to look at it in more detail when I have some time.

 

[Muphrid]

I'm not sure if it's useful for this discussion, but the geometric/clifford algebra-based Gauge Theory Gravity formalism neatly handles GR as a subset of EC theory with vanishing torsion. See http://arxiv.org/abs/gr-qc/0405033, but basically, it requires a field $\bar h(a)$ such that $\overline h(e^\mu) \cdot \overline h(e^\nu) = g^{\mu \nu}$. But with that, you can get

$$\begin{align*} \underline G(a) &= \kappa \underline T(a) \\ \mathcal D \wedge \bar h(a) &= \kappa S \cdot \bar h(a) \end{align*}$$

where $S$ comes from the spin involved. The second equation defines the torsion tensor, and when it is constrained to be zero everywhere, all the usual GR results about, for example, $\underline T(a)$ being symmetric follow (albeit with...much math).

 

[tom.stoer]

As far as I can tell, both actions are simply the integral of the Ricci scalar, just written in terms of different fundamental fields. Did you have something else in mind?

The action is identical for vanishing matter density = in vacuum; more precisely, the EC e.o.m. reduce to the GR e.o.m. in vacuum, but this need not be the case in general b/c neither the stress-energya-tensor nor the Ricci-tensor are symmetric. In addition the constraint to reduce vierbein + connection to metric differ in both theories.

there may be more than one way to model the same physics.

You are right; there is a special formulation of teleparallelism using a curvature-free geometry with torsion only which is identical to GR.

I'm not concerned with whether there is torsion or not. I'm more concerned with whether the effects of torsion in the EC theory can be achieved by coupling to additional matter fields in standard GR.

b/c in EC you start with more geometric d.o.f. and w/o a torsion constraint you have more choices to couple matter to it; therefore there are matter couplings which can't exist in GR (which is restricted to vanishing torsion which modifies the field equations)

... but I still think it may be possible to get the same physics as EC theory by coupling GR to a B field, or something like it.

You should look at a spin 1/2 field which should make the difference most explicit.

Besides Trautman's paper

http://arxiv.org/abs/gr-qc/0606062
Einstein-Cartan Theory
Authors: Andrzej Trautman
(Submitted on 14 Jun 2006)
Abstract: The Einstein--Cartan Theory (ECT) of gravity is a modification of General Relativity Theory (GRT), allowing space-time to have torsion, in addition to curvature, and relating torsion to the density of intrinsic angular momentum. This modification was put forward in 1922 by Elie Cartan, before the discovery of spin. Cartan was influenced by the work of the Cosserat brothers (1909), who considered besides an (asymmetric) force stress tensor also a moments stress tensor in a suitably generalized continuous medium.
Comments: 7 pages, uses amsmath.sty, amssymb.sty
Journal reference: Encyclopedia of Mathematical Physics: Elsevier, 2006, vol. 2, pages 189--195

Hehl is a good reference:

http://arxiv.org/abs/gr-qc/9712096
Alternative Gravitational Theories in Four Dimensions
Authors: Friedrich W. Hehl (University of Cologne)
(Submitted on 26 Dec 1997)
Abstract: We argue that from the point of view of gauge theory and of an appropriate interpretation of the interferometer experiments with matter waves in a gravitational field, the Einstein-Cartan theory is the best theory of gravity available. Alternative viable theories are general relativity and a certain teleparallelism model. Objections of Ohanian and Ruffini against the Einstein-Cartan theory are discussed. Subsequently we list the papers which were read at the `Alternative 4D Session' and try to order them, at least partially, in the light of the structures discussed.

http://arxiv.org/abs/gr-qc/9602013
On the Gauge Aspects of Gravity
Authors: F. Gronwald, F.W. Hehl
(Submitted on 8 Feb 1996)
We give a short outline, in Sec.\ 2, of the historical development of the gauge idea as applied to internal ($U(1),\, SU(2),\dots$) and external ($R^4,\,SO(1,3),\dots$) symmetries and stress the fundamental importance of the corresponding conserved currents. In Sec.\ 3, experimental results with neutron interferometers in the gravitational field of the earth, as inter- preted by means of the equivalence principle, can be predicted by means of the Dirac equation in an accelerated and rotating reference frame. Using the Dirac equation in such a non-inertial frame, we describe how in a gauge- theoretical approach (see Table 1) the Einstein-Cartan theory, residing in a Riemann-Cartan spacetime encompassing torsion and curvature, arises as the simplest gravitational theory. This is set in contrast to the Einsteinian approach yielding general relativity in a Riemannian spacetime. In Secs.\ 4 and 5 we consider the conserved energy-momentum current of matter and gauge the associated translation subgroup. The Einsteinian teleparallelism theory which emerges is shown to be equivalent, for spinless matter and for electromagnetism, to general relativity. Having successfully gauged the translations, it is straightforward to gauge the four-dimensional affine group $R^4 \semidirect GL(4,R)$ or its Poincar\'e subgroup $R^4\semidirect SO(1,3)$. We briefly report on these results in Sec.\ 6 (metric-affine geometry) and in Sec.\ 7 (metric-affine field equations (\ref{zeroth}, \ref{first}, \ref{second})). Finally, in Sec.\ 8, we collect some models, currently under discussion, which bring life into the metric-affine gauge framework developed.

 

[stevendaryl]

At least, I am pretty sure they are equivalent. The Wiki article contains at least one glaring mistake: It defines the momentum tensor via

$$\frac{\delta \mathcal{L}_\mathrm{G}}{\delta g^{ab}} -\frac{1}{2}P_{ab}=0$$
and then claims that $P_{ab}$ is not symmetric! Obviously $P_{ab}$ defined this way is symmetric, because $g_{ab}$ is symmetric.

Hmm. There's something weird about that, because the conclusion

R_αβ - 1/2 R g_αβ = P_αβ

is only possible if both R_αβ and P_αβ are symmetric, or neither are.

 

[stevendaryl]

This is all very interesting, and it seems that the Wikipedia articles that I used as the basis for my original post are mistaken in certain ways (I'm assuming that the article isn't original research, that the authors are misquoting something published?)

My original questions haven't really been answered, though. Is it true that the presence of a nonzero spin-density leads to a nonsymmetric canonical stress-energy tensor? (Whether or not the canonical tensor is the full stress-energy).

If so, I would like to understand how that's possible--my intuition is that nonvanishing spin-density should be obtainable through a "coarse-graining" of ordinary orbital angular momentum. That seems to be what the Wikipedia article is saying about the fluid of many, tiny black holes. But how can a limiting process turn a symmetric stress-energy tensor into a nonsymmetric one?

There is another puzzle about this stuff, which is that I remember reading an argument for why the stress-energy tensor must be symmetric, which showed that a nonsymmetric tensor would lead to some anomalous situation. It's described here:
http://www.worldscibooks.com/etextbook/4134/4134_chap03.pdf

So I'm assuming that the argument breaks down if there is nonzero spin density. Is that correct?

 

[Ben Niehoff]

Hmm. There's something weird about that, because the conclusion

R_αβ - 1/2 R g_αβ = P_αβ

is only possible if both R_αβ and P_αβ are symmetric, or neither are.

I think the answer is that

$$\frac{\delta \mathcal{L}_\mathrm{G}}{\delta g^{ab}} -\frac{1}{2}P_{ab}=0$$
is not the correct definition of the stress-energy tensor. I would define the stress-energy tensor as "whatever is the source of gravity", which in this case will be a non-symmetric tensor.

Actually, I've seen many statements in various Wiki articles about GR and differential geometry where the authors have been sloppy about the what assumptions have been used; for example, statements purported to apply in all generality, when they only apply in the case of vanishing torsion.

 

[Ben Niehoff]

There is another puzzle about this stuff, which is that I remember reading an argument for why the stress-energy tensor must be symmetric, which showed that a nonsymmetric tensor would lead to some anomalous situation. It's described here:
http://www.worldscibooks.com/etextbook/4134/4134_chap03.pdf

So I'm assuming that the argument breaks down if there is nonzero spin density. Is that correct?

I haven't thought about this too hard, but it is possible that "the thing that sources gravity", and "the thing that measures local stress-energy density" are two different objects in EC theory.

 

[Ben Niehoff]

Tom,

Check out this draft of Van Proeyen and Freedman's "Supergravity":

http://itf.fys.kuleuven.be/~toine/SUGRA_DoctSchool.pdf

Specifically, section 7.3, in the chapter on first- and second-order formulations of GR. Here they explicitly show that the torsion terms in EC theory can be accounted for by adding four-fermi terms to GR theory. This is the kind of contact interaction I suspected, given the non-propagation of torsion.

Therefore one does not have to consider the presence of torsion; it is enough to stick with torsion-free geometry, but instead add four-fermion interaction terms to the action. In more generality, I think one would also have to add a three-form field strength.

However, using the first-order formulation and allowing for torsion is clearly a more elegant way of organizing these extra terms; the point is that it accomplishes nothing new.

See also these lecture notes by the same authors,

http://arxiv.org/abs/1106.1097v2

where in section 8 the above is described in words, but not done explicitly. The authors do call the theories with and without torsion as "distinct", but the only distinction is the extra four-fermi term.

 

[tom.stoer]

The four-fermion interaction is derived by solving the constraint i.e. integrating out unphysical d.o.f. The underlying geometry is Riemann-Cartan w/ torsion, not Riemann w/o torsion. So the theories are different geometrically.

This difference can be seen both via non-vanishing of torsion and via fact that there is no longer a unique geodesic equation for test particles; in Riemann geometry geodesics are both straightest and shortest lines between two given spacetime points; in Riemann-Cartan spacetime the definition of 'straightest' and 'shortest' are no longer equivalent.

Regarding your statement "Here they explicitly show that the torsion terms in EC theory can be accounted for by adding four-fermi terms to GR theory": this is not precise in a mathematical sense. Unfortunately in physics we often make no difference between GR which is a specific theory and the underlying Riemannian geometry on which it is defined. It is correct that you can construct a specific Einstein-Cartan theory by adding fermions to GR, but the geometric difference i.e. the presence of torsion is much more fundamental. You can have a variety of different "EC theories" using the same Riemann-Cartan geometry. 'Supergravity with spin 3/2 gravitinos' is different from 'GR with spin 1/2 fermions', but the underlying Riemann-Cartan geometry is identical.

That means then when going from Einstein to Einstein-Cartan one has to enlarge the geometrical setup in a very precise sense, i.e. to allow for the presence of torsion - which is strictly speaking not 'GR + something' b/c
1) the geodesic equation for test particles and therefore the equivalence principle of GR is lost and
2) 'GR + something' is not Riemann-Cartan geometry but only a very specific model using Riemann-Cartan geometry.

(it would be interesting to understand whether there can be models in which both curvature and torsion as propagating d.o.f. and what that would mean physically)

 

[Ben Niehoff]

The four-fermion interaction is derived by solving the constraint i.e. integrating out unphysical d.o.f. The underlying geometry is Riemann-Cartan w/ torsion, not Riemann w/o torsion. So the theories are different geometrically.

I understand that. I'm saying that the equations of motion cause the torsion to fall within a certain class that can be represented by other means, such as 3-form field strengths and 4-fermi interactions.

This difference can be seen both via non-vanishing of torsion and via fact that there is no longer a unique geodesic equation for test particles; in Riemann geometry geodesics are both straightest and shortest lines between two given spacetime points; in Riemann-Cartan spacetime the definition of 'straightest' and 'shortest' are no longer equivalent.

I'll have to ask you for a reference on this. It is well-known that two connections differing only by their torsion have exactly the same geodesics; the torsion does not enter the geodesic equation. I pointed out earlier that the Wikipedia page on torsion contains some inaccuracies (specifically they claim that torsion causes geodesics to "twist around each other", which it certainly does not!).

Regarding your statement "Here they explicitly show that the torsion terms in EC theory can be accounted for by adding four-fermi terms to GR theory": this is not precise in a mathematical sense. Unfortunately in physics we often make no difference between GR which is a specific theory and the underlying Riemannian geometry on which it is defined. It is correct that you can construct a specific Einstein-Cartan theory by adding fermions to GR, but the geometric difference i.e. the presence of torsion is much more fundamental. You can have a variety of different "EC theories" using the same Riemann-Cartan geometry. 'Supergravity with spin 3/2 gravitinos' is different from 'GR with spin 1/2 fermions', but the underlying Riemann-Cartan geometry is identical.

I know. But the EoM of EC theory do not allow the torsion to be generic; they restrict it to a certain specific class that can just as well be represented by GR with additional matter fields and interactions.

That means then when going from Einstein to Einstein-Cartan one has to enlarge the geometrical setup in a very precise sense, i.e. to allow for the presence of torsion - which is strictly speaking not 'GR + something' b/c
1) the geodesic equation for test particles and therefore the equivalence principle of GR is lost and

The geodesic equation is identical. Torsion only affects spinning test particles, because it changes the definition of parallel propagation; i.e., it will cause additional precession.

2) 'GR + something' is not Riemann-Cartan geometry but only a very specific model using Riemann-Cartan geometry.

Sure, but EC theory also only describes very specific models using Riemann-Cartan geometry; specifically where the torsion tensor is non-propagating and satisfies certain algebraic relationships.

(it would be interesting to understand whether there can be models in which both curvature and torsion as propagating d.o.f. and what that would mean physically)

Yes it would.

 

[Ben Niehoff]

Actually, I think the non-propagation of torsion is exactly what saves the equivalence principle (if by "equivalence principle" we mean "observers falling on a geodesic in empty space feel no local forces"). Torsion is only present in non-empty space, where any observer can reasonably interpret torsion forces as being due to interaction with the medium they are traveling through.

 

[tom.stoer]

you are right; torsion vanishes in vacuum and in vacuum GR and EC are strictly equivalent; therefore the geometrical difference between GR and EC is restricted to non-vacuum regions; sorry for the confusion

 

[stevendaryl]

you are right; torsion vanishes in vacuum and in vacuum GR and EC are strictly equivalent; therefore the geometrical difference between GR and EC is restricted to non-vacuum regions; sorry for the confusion

If you have a point-mass with spin, will its trajectory depend on the orientation of the spin? Specifically, if there is a massive rapidly-rotating star, then is there a difference in the trajectory between the case where the spins are aligned and anti-aligned? If so, is that a violation of the equivalence principle, or not?

 

[tom.stoer]

If you have a point-mass with spin, will its trajectory depend on the orientation of the spin? Specifically, if there is a massive rapidly-rotating star, then is there a difference in the trajectory between the case where the spins are aligned and anti-aligned? If so, is that a violation of the equivalence principle, or not?

In Riemann-Cartan geometry trajectories will depend on torsion. Refer e.g. to the last equation of section 3.1 in

http://www.slimy.com/~steuard/teaching/tutorials/GRtorsion.pdf

As you can see even in flat space and for zero spin (!) the geodesic equation is changed in the presence of torsion.

But in Einstein-Cartan theory this effect will not show up in vacuum b/c there torsion is exactly zero. Therefore the equivalence principle holds in vacuum but is violated geometrically within matter

(the fact that torsion couples to particles of zero spin is sometimes regarded as severe deficiency of EC theory - but I cannot see why this should be problematic; it's a new effect predicted by EC theory; unfortunaterly we will never be able to prove / disprove this experimentally, I guess)

Regarding you question of rotating stars: they will follow geodesics in vacuum, torsion is zero in vacuum, therefore their geodesics will be identical with the usual ones derived in GR.

 

[TrickyDicky]

Regarding you question of rotating stars: they will follow geodesics in vacuum

In an old thread about geodesics most people were arguing that to comply with gravitational radiation, massive stars didn't follow geodesics, for instance this was evident neutron stars in systems like the Hulse-Taylor binary, do you agree?

 

[tom.stoer]

Yes, I agree, if you do no longer consider massive stars to be pointlike spinning test bodies (which was my interpretation of the assumption in stevendaryl's post) but extended objects with back-reaction on spacetime (which is the realistic assumption for massive binaries with extremely short orbital periods) then one has to take gravitational radiation and deviations from geodesic motion into account.

But I guess what we should discuss here are effects due to spin and torsion restricted to pointlike test bodies in leading order.

 

[TrickyDicky]

Yes, I agree, if you do no longer consider massive stars to be pointlike spinning test bodies (which was my interpretation of the assumption in stevendaryl's post) but extended objects with back-reaction on spacetime (which is the realistic assumption for massive binaries with extremely short orbital periods) then one has to take gravitational radiation and deviations from geodesic motion into account.

But I guess what we should discuss here are effects due to spin and torsion restricted to pointlike test bodies in leading order.

Ok, I figured you would. But then EC and GR should make different predictions about massive stars (I mean realistic ones, not pointlike) trajectories. Because in EC torsion effects affect non-vacuum and in the realistic case (where as you say one no longer considers massive stars pointlike and thus gravitational radiation must be taken into account) we are no longer talking vacuum (gravitational radiation and vacuum are incompatible).

 

[Ben Niehoff]

OK, I see how the torsion affects the geodesic equation; or in general, how in the presence of torsion, length-minimizing curves and autoparallels will not necessarily be the same. Interesting. Which kind of trajectory do free test particles follow?

 

[TrickyDicky]

Which kind of trajectory do free test particles follow?

I don't think that is defined in general in GR. I guess it will depend in the particular geometry of the chosen solution.

 

[stevendaryl]

Regarding you question of rotating stars: they will follow geodesics in vacuum, torsion is zero in vacuum, therefore their geodesics will be identical with the usual ones derived in GR.

Hmm. I thought that the Thirring effect causes the trajectory of a spinning object to be different from a nonspinning one. In the Wikipedia article on "Gravitoelectromagnetism":

"For instance, if two wheels are spun on a common axis, the mutual gravitational attraction between the two wheels will be greater if they spin in opposite directions than in the same direction. This can be expressed as an attractive or repulsive gravitomagnetic component."

Does this a nonlinear effect that vanishes as the mass of one of the "wheels" goes to zero (that is, becomes a test particle that has negligible effect on spacetime curvature)?

 

[tom.stoer]

@TrickyDicky, regarding you statement that this is not "... defined in ... GR." and that "it will depend in the particular geometry": in GR pointlike test particles follow geodesics and b/c GR uses Riemannian geometry the geodesics are both straightest and shortest worldlines, so the two definitons coincide; it is defined in GR.

I do not know how to define the trajectories in EC; one could consider either 'straightest' or 'shortest' world lines and I guess physically one should use shortest world lines; but this is not a mathematical question, it depends which actions you chose and whether a certain choice makes sense physically.

Pointlike particles are limiting cases; I do not know whether it is possible to formulate a limit for "realistic massive objects" which converges to a 'geodesic' in a certain limit, especially when backreaction is supressed. I would expect that in this limit the difference between GR and EC should vanish b/c for pointlike particles the torsion must vanishes everywhere. Otherwise I would say that something is wrong either with EC or with the limit.

@stevendaryl, a short remark regarding the Thirring effect in GR; this effect has nothing to do with spin or torsion. It is simply the non-straight motion of a piece of matter and therefore should be covered by the standard energy-momentum tensor.

 

[stevendaryl]

@stevendaryl, a short remark regarding the Thirring effect in GR; this effect has nothing to do with spin or torsion. It is simply the non-straight motion of a piece of matter and therefore should be covered by the standard energy-momentum tensor.

I'm diverging a little from the topic of torsion, but asking a related question, which is whether point-masses with intrinsic spin follow geodesics. You say that the Thirring effect has nothing to do with spin, but in the quote that I gave (which I assume is based on some actual calculation---you can never be certain with Wikipedia articles, I know) it was said that two massive spinning wheels will be slightly more or less attracted to each other depending on whether their spins are aligned or not. I would think that the effect would persist if one took the limit as the size of one of the wheel goes to zero, provided that you keep the angular momentum constant. In this limit, you would have a point-mass with intrinsic angular momentum, and it would be affected by the gravity of another massive rotating "wheel" in a way that depends on the orientation of the angular momentum of the point-mass.

 

[stevendaryl]

@TrickyDicky, regarding you statement that this is not "... defined in ... GR." and that "it will depend in the particular geometry": in GR pointlike test particles follow geodesics and b/c GR uses Riemannian geometry the geodesics are both straightest and shortest worldlines, so the two definitions coincide; it is defined in GR.

I do not know how to define the trajectories in EC; one could consider either 'straightest' or 'shortest' world lines and I guess physically one should use shortest world lines; but this is not a mathematical question, it depends which actions you chose and whether a certain choice makes sense physically.

As I understand it from Misner, Thorne and Wheeler's book on gravity, the fact that test particles follow geodesics is not an additional assumption in GR, but is actually provable from the field equations. I don't have it handy, so I can't actually verify that, but I seem to remember such a claim.

I would assume that the same is true for EC, that one need not postulate that test particles follow geodesics (of either variety), but that it would follow from whatever the field equations are.

 

[tom.stoer]

As I understand it from Misner, Thorne and Wheeler's book on gravity, the fact that test particles follow geodesics is not an additional assumption in GR, but is actually provable from the field equations. I don't have it handy, so I can't actually verify that, but I seem to remember such a claim.

Interesting. I cannot remember such a statement and I have to get the book as well from a library.

I would assume that the same is true for EC, that one need not postulate that test particles follow geodesics (of either variety), but that it would follow from whatever the field equations are.

That is a reasonable assumption based on what you are saying regarding GR and MTW.