Poincare conserved currents : Energy-momentum and Angular-momentum tensors

[crackjack]

Not sure if this is the right place to ask, but this doubt originated when reading on string theory and so here it goes...

The general canonical energy-momentum tensor (as derived from translation invariance), T^{\mu\nu}_{C} is not symmetric. Also, the general angular momentum conserved current (as derived from lorentz invariance) consists of two parts to it - the orbital angular momentum component and the spin angular momentum component...
j^{\mu\nu\rho} = T^{\mu\nu}_{C} x^\rho - T^{\mu\rho}_{C} x^\nu + S^{\mu\nu\rho}

But, by taking clues from the above angular momentum expression, we can append a suitable term to the generally non-symmetric canonical energy-momentum tensor and modify it (without breaking its conservation) into a symmetric Belinfante tensor...
T^{\mu\nu}_{C} \to T^{\mu\nu}_{B}


Further, the angular momentum tensor can also be modified to absorb the spin term in its orbital momentum term by rewriting it as...
j^{\mu\nu\rho} = T^{\mu\nu}_{B} x^\rho - T^{\mu\rho}_{B} x^\nu

Now, there is no spin operator at all now - at least not explicitly. And we haven't really modified the physics at all. What is then the spin of the system now?

In light cone gauge of bosonic strings, the above (original) spin operator is used to calculate the spins of the massless fields (photon, graviton)...
So, what would the above disappearance of the spin operator mean then? If you say that its not done away with but is just hidden in the last expression above, what rule do we follow in separating the orbital and spin components of angular momentum?

[samalkhaiat]

[QUOTE=crackjack;2346218]Not sure if this is the right place to ask,

No, this is not the right place! Don't mention string theory in this part of the PF :smile:
People in here are too busy, with the LOOPY "quantum gravity" stuff of Rovelli and Smolin, to reply to your question! Any way let us do it.

Let us write

J^{\rho \mu \nu} = L^{\rho \mu \nu} + S^{\rho \mu \nu}

where


L^{\rho\mu\nu} = x^{\mu}T_{C}^{\rho\nu} - x^{\nu}T_{C}^{\rho\mu} \ \ \ (1)


and


S^{\rho\mu\nu} = \frac{\partial \mathcal{L}}{\partial \partial_{\rho}\phi^{r}} (\Sigma^{\mu\nu})^{r}{}_{s}\phi^{s}


The object L is called by Belinfate the orbital angular momentum density "tensor" because its space components have the form of non-relativistic orbital momentum density. Because the object S clearly expresses the transformation of the internal degree of freedom of the field, it is called (by Belinfate) the "tensor" of internal angular momentum density or spin density.
Notice that S^{\rho\mu\nu} does not have the typical form of a classical angular momentum like L^{\rho\mu\nu}. In order to show that it can be written in such a form, Belifate (and independently Rosenfeld) introduced the so-called symmetric energy-momentum tensor;

T_{B}^{\mu\nu} = T_{C}^{\mu\nu} + t^{\mu\nu}

with


t^{\mu\nu} = (1/2) \partial_{\rho}( S^{\mu\nu\rho} + S^{\rho\mu\nu} + S^{\nu\mu\rho})


Because of the following properties

T_{B}^{\mu\nu} = T_{B}^{\nu\mu}

\partial_{\mu}T_{B}^{\mu\nu} = \partial_{\mu}T_{C}^{\mu\nu} = 0

P^{\mu} = \int T_{B}^{0\mu} \ d^{3}x = \int T_{C}^{0\mu} \ d^{3}x ,

T_{B}^{\mu\nu} appears as an equivalent rnergy-momentum tensor. The corresponding new total angular momentum tensor is defined by;

J_{B}^{\rho\mu\nu} = x^{\mu}T_{B}^{\rho\nu} - x^{\nu}T_{B}^{\rho\mu}

This can also be written as

J_{B}^{\rho\mu\nu} = L^{\rho\mu\nu} + s^{\rho\mu\nu}

with

s^{\rho\mu\nu} = x^{\mu}t^{\rho\nu} - x^{\nu}t^{\rho\mu}

having the required form of an angular momentum.

Again

\partial_{\rho}J_{B}^{\rho\mu\nu} = 0

M^{\mu\nu} = \int J_{B}^{0\mu\nu}\ d^{3}x = \int J^{0\mu\nu} \ d^{3}x

and

S^{\mu\nu} = \int d^{3}x s^{0\mu\nu} = \int d^{3}x S^{0\mu\nu}

consequently, the tensor s^{\rho\mu\nu} could equivalently be considered as the spin density instead of S^{\rho\mu\nu}; it is the angular momentum of a "spin" energy-momentum density t^{\mu\nu}, which does not contribute to the total energy-momentum vector:

\int d^{3}x \ t^{0\mu} = 0.

In this way, the canonical energy-momentum tensor would then represent an "orbital" energy-momentum tensor.
Although it does not change the physics ( M^{\mu\nu} and P^{\mu} remain the same and generate the same Poincare algebra) the above Blinfate-Rosenfeld method has at least two weaknesses. The 1st one lies in the fact that the total spin and the total orbital angular momentum are, in general, not covariant quantities. Indeed, the quantities

S^{\mu\nu} = \int d^{3}x \ s^{0\mu\nu}

and

L^{\mu\nu} = \int d^{3}x \ L^{0\mu\nu}

are tensorial only if the corrsponding densities s^{\rho\mu\nu} and L^{\rho\mu\nu} are conserved. This is however the case, if and only if T_{C}^{\mu\nu} is symmetric. That is, the Belinfate spin and orbital angular momentum are sensible quantities only in those cases where the entire symmetrization procedure seems to be superfluous!
The 2nd weakness of the B-R method is that it leads some people (I believe you are included) to believe (incorrectly) that a field can only posses non-zero spin if T_{C}^{\mu\nu} is non-symmetric.

regards

sam

[crackjack]

Thanks Sam, for the detailed explanation!


The 2nd weakness of the B-R method is that it leads some people (I believe you are included) to believe (incorrectly) that a field can only posses non-zero spin if T_{C}^{\mu\nu} is non-symmetric.

Ya, my confusion is kind of related to this...

Why should the orbital and spin angular momentum components be separately covariant?

[Mentz114]

You may find this interesting


Symmetric energy-momentum tensor in Maxwell, Yang-Mills, and Proca theories
obtained using only Noether’s theorem

Merced Montesinos and Ernesto Flores

(Dated: February 1, 2008)

Abstract:
The symmetric and gauge-invariant energy-momentum tensors for source-free Maxwell and Yang-
Mills theories are obtained by means of translations in spacetime via a systematic implementation
of Noether’s theorem. For the source-free neutral Proca field, the same procedure yields also the
symmetric energy-momentum tensor. In all cases, the key point to get the right expressions for
the energy-momentum tensors is the appropriate handling of their equations of motion and the
Bianchi identities. It must be stressed that these results are obtained without using Belinfante’s
symmetrization techniques which are usually employed to this end.

arXiv:hep-th/0602190v1 20 Feb 2006

The emphasis is mine.

[crackjack]

Thanks for the link! That is a neat way of deriving the symmetric stress tensor using Bianchi identity.

But I actually don't find Belinfante's procedure to be too hotch potch. In Belinfante's procedure, the spin momentum component can be totally done away with, leaving only the orbital momentum component (in the procedure employed in the above paper, we don't ever arrive at any spin component).
But, from what I have read, it is this spin momentum component that is used to determine the (integer) spin in bosonic string theory. This is what is confusing me.

Additional note...
The above paper also shows that we could arrive at one 'pure orbital' angular momentum tensor without first going through one which has both orbital and spin components (as with Belinfante's procedure). So then, we dont have any rule to split the 'pure orbital' angular momentum tensor to get 'back' the spin component (as samalkhaiat did above).

[Ed Rex]

These papers, arXiv:0905.4529 by Andrew Randono and Dave Sloan and the follow-up paper arXiv:0906.1385 by Andrew Randono, may help if you are interested in the way spin is encoded in the gravitational field. The first emphasizes some algebraic properties of intrinsic spin which are not usually covered, and shows how these algebraic properties are encoded in the gravitational field (tetrad) at asymptotic infinity. The algebraic properties reveal how the two tensors are separately covariant, as you asked. The second paper uses the example of a spinor in linearized gravity and shows explicitly how the intrinsic spin is buried in the symmetric stress energy tensor. To unbury it, the paper uses a "gravitational Gordon decomposition" which is a direct analogue of the procedure to extract the spin in electromagnetism.

[crackjack]

Thanks Ed Rex. I will take a look at them.
They seem to be specialized to the case of gravitational fields, but I will see what all generic information I can decouple from this.