Field stress energy tensors

[JonathanApps]

Hey! I have a question about spin-1 fields.

 

Symmetry of the stress-energy tensor comes from conservation of orbital angular momentum, ie:

 

\partial_{\mu} (T^{\mu}^{\nu} x^{\rho} - T^{\mu}^{\rho} x^{\nu}) = 0

 

But for a spin-1 e.m. field, orbital angular momentum isn't conserved because of the spin. Like for spin-1/2 you need a spin term added to get the conserved angular momentum.

 

So my question is, why is the stress-energy tensor symmetric? OK this could just be a fluke but in Itzykson and Zuber I think page 24 they actually make a statement which seems to imply there would be a problem if it wasn't, i.e. they seem to _demand_ symmetry.

 

For spin-1/2 it isn't symmetric, and shouldn't be as far as I can see. The asymmetric part corresponds exactly to the spin part of the angular momentum tensor as I've worked out (can give details if needed). So why should the stress tensor be symmetric for spin-1?

 

Thanks for any answers.

Jonathan

Sorry just trying out the latex here - I'm new :

 

[latex]

\partial_{\mu} (T^{\mu}^{\nu} x^{\rho} - T^{\mu}^{\rho} x^{\nu}) = 0

[/latex]

[elfmotat]

There are really two types of stress-energy tensor: the Hilbert SET and the canonical SET. The Hilbert SET is defined by a functional derivative of the Lagrangian, and is always symmetric. This is the SET that appears in the Einstein field equations. The canonical SET is defined as the conserved Noether current that comes from spacetime translation symmetry. The canonical SET is not symmetric in general. Since Noether currents remain conserved with the introduction of a divergence term, both tensors can be related to each other by introducing another tensor (called the Belinfante tensor):

 

[math]T^{\mu \nu}= \Theta^{\mu \nu} +\partial_\sigma B^{\sigma \mu \nu}[/math]

 

where theta is the canonical tensor. Integrating T over spacetime is identical to integrating theta over spacetime, because the two only differ by a total divergence. So you can think of B as something that contains all of the information about the intrinsic angular momentum in the field.

 

Explicitly we have:

[math]T^{\mu \nu} \equiv \frac{2}{\sqrt{-g}} \frac{\delta ( \mathcal{L} \sqrt{-g})}{\delta g_{\mu \nu}}[/math]

 

This is how T is defined (the definition comes from varying the Hilbert action to get the EFE's).

 

If you work out the conserved current under [math]x^\mu \mapsto x^\mu +a^\mu[/math] ,[math]~~~[/math] [math]\phi (x) \mapsto \phi (x+a)=\phi(x) + a^\mu \partial_\mu \phi(x)[/math] (with infinitesimal a) for some field phi (could be scalar, vector, tensor, etc.), you get the quantity:

 

[math]\Theta^{\mu \nu} \equiv \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \partial^\nu \phi - \eta^{\mu \nu} \mathcal{L}[/math]

 

Now we define the tensor B such that [math]T^{\mu \nu}= \Theta^{\mu \nu} +\partial_\sigma B^{\sigma \mu \nu}[/math]. If you go through the work, you get B to be of the form:

[math]B^{ \alpha \mu \nu } = \frac{1}{2} \left (\Pi^{\alpha} \Sigma^{\mu \nu} + \Pi^{\mu} \Sigma^{\nu \alpha} - \Pi^{\nu} \Sigma^{\alpha \mu} \right ) \phi[/math]

 

where [math]\Pi^\mu = \partial \mathcal{L}/\partial (\partial _\mu \phi )[/math] and the sigmas are the spin matrices. So, for example, with a spin-1 field like the EM field, we can plug in the EM Lagrangian to find the canonical SET, and we get:

[math]\Theta^{\mu \nu} =F^\mu_{~\lambda} \partial^\nu A^\lambda-\frac{1}{4}\eta^{\mu \nu} F^{\alpha \beta } F_{\alpha \beta }[/math]

 

Indeed this is not symmetric due to the presence of intrinsic spin in the field. If we use the vector spin matrices and plug everything into B, we get [math]B^{\alpha \mu \nu }=-F^{\alpha \mu } A^\nu[/math]. When we take its divergence and add it to the canonical SET, we get:

[math]T^{\mu \nu}=F^{\mu}_{~ \lambda} F^{\lambda \nu}-\frac{1}{4} \eta^{\mu \nu} F^{\alpha \beta }F_{\alpha \beta }[/math]

 

This tensor is symmetric, and it corresponds to what you would get with the above definition. This is the tensor that couples to curvature in the EFE's.

Edited September 12, 2014 by elfmotat

[JonathanApps]

This is EXTREMELY helpful. Thankyou!

This is awesome. I'm printing it out and taking it home to pore over in detail. An ex particle-physics lecturer of mine couldn't provide this answer, but I don't think he's well-versed in general relativity.

Edited September 12, 2014 by JonathanApps

[elfmotat]

No problem. Good question.

[JonathanApps]

Hi, I assume B is antisymmetric in the first two indices? That immediately gives conservation of T.

 

I think I can derive your explicit formula for B in terms of \Pi \Sigma \phi by demanding that the Hilbert angular momentum density (T cross x) is equal to the canonical angular momentum density (\Theta cross x + spin bit) plus a total divergence....

[elfmotat]

Yes, B is antisymmetric in the first two indices. I should have mentioned that as part of its definition. If I were going to try to derive B, the way I would do it would be to show that if [math]\Theta^{\mu \nu}-\Theta^{\nu \mu}=-\partial_\sigma C^{\sigma \mu \nu}[/math], then the form of B must be:

[math]B^{\sigma \mu \nu}=\frac{1}{2} (C^{\sigma \mu \nu}+C^{\mu \nu \sigma}-C^{\nu \sigma \mu })[/math]

 

If you then take the angular momentum tensor (the kinetic part plus the spin) and impose Lorentz invariance as a symmetry of the field (i.e. you impose conservation of angular momentum), then you must have:

 

[math]\Theta^{\mu \nu} - \Theta^{\nu \mu}=\partial_\sigma C^{\sigma \mu \nu}= -\partial_\sigma (\Pi^\sigma \Sigma^{\mu \nu} \phi)[/math]

Edited September 15, 2014 by elfmotat

[JonathanApps]

Thanks again.

 

One quibble: I'm not sure that adding a total divergence to \Theta to get T is meaningful in itself. I'm not sure what you get by integrating over all spacetime (d^4 x). Yes the integrals of the two SETs over all spacetime might be the same (although I'm not even sure of this - B has got to vanish at the time boundaries, in other words the field has to eventually disappear, violating conservation of matter) but I don't know if that really gives you anything useful.

 

What IS useful AFAICS is that \partial_\mu \partial_\nu B^{\mu \nu \rho} = 0 so that T is conserved if \Theta is (which it is!).

So we're allowed to add the div B term because it's divergenceLESS rather than because it's a divergence in itself (even though it is)...?

To put it another way, we couldn't just add any old total divergence.

[elfmotat]

Thanks again.

 

One quibble: I'm not sure that adding a total divergence to \Theta to get T is meaningful in itself. I'm not sure what you get by integrating over all spacetime (d^4 x). Yes the integrals of the two SETs over all spacetime might be the same but I don't know if that really gives you anything useful.

 

The following must be true for both SET's to represent the same physical observable (the energy-momentum):

 

[math]\int d^3x T^{0 \mu}= \int d^3 x \Theta^{0 \mu}[/math]

 

In my first post I said "spacetime" where I should have said "space," but what I meant was the above. Sorry for the confusion.

 

 

 

 

What IS useful AFAICS is that \partial_\mu \partial_\nu B^{\mu \nu \rho} = 0 so that T is conserved if \Theta is (which it is!).

So we're allowed to add the div B term because it's divergenceLESS rather than because it's a divergence in itself (even though it is)...?

To put it another way, we couldn't just add any old total divergence.

 

Yes, div(B) must also be divergence free, as mandated by B's definition.