## Spin Density and Non-symmetric Stress-Energy Tensor

In General Relativity, the assumption is made that the stress-energy tensor Tαβ 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 Sαβμ satisfies:

μ Sαβμ = Tβα - Tαβ

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/Einstei...89-consistency describes a continuum limit of many tiny black holes.) But how can a symmetric Tαβ become non-symmetric in the limit?
 PhysOrg.com science news on PhysOrg.com >> King Richard III found in 'untidy lozenge-shaped grave'>> Google Drive sports new view and scan enhancements>> Researcher admits mistakes in stem cell study
 Blog Entries: 1 Recognitions: Science Advisor 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 Tμν = (2/√-g) δS/δgμν 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.

 Quote by 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 Tμν = (2/√-g) δS/δgμν 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.

## Spin Density and Non-symmetric Stress-Energy Tensor

 Quote by 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/Belinfa...-energy_tensor
the Belinfante approach to the stress-energy tensor is described. It basically amounts to distinguishing between

Tαβ, which is not symmetric, and
TBαβ, which is, where

TBαβ = Tαβ + 1/2 ∂λ [Sαβλ + Sβαλ - Sλβα]

So in regular GR, TBαβ is the contribution to the curvature, not Tαβ. 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?
 Blog Entries: 1 Recognitions: Science Advisor 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.
 Recognitions: Science Advisor 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.
 Recognitions: Science Advisor 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.

Recognitions:
 Quote by 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.
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.
 Recognitions: Science Advisor 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).

Recognitions:
 Quote by Ben Niehoff 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
 I think theories with torsion predict that identical masses with opposite chirality fall at different rates in a Gravitation field.

Recognitions:
 Quote by tom.stoer 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.
 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).

Recognitions:
 Quote by Ben Niehoff 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.

 Quote by Ben Niehoff 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.

 Quote by Ben Niehoff 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)

 Quote by Ben Niehoff ... 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.

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

Rαβ - 1/2 R gαβ = Pαβ

is only possible if both Rαβ and Pαβ are symmetric, or neither are.
 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/etextbo...134_chap03.pdf So I'm assuming that the argument breaks down if there is nonzero spin density. Is that correct?

Recognitions:
$$\frac{\delta \mathcal{L}_\mathrm{G}}{\delta g^{ab}} -\frac{1}{2}P_{ab}=0$$