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.

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....

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.

Spin is always derived from the tensor rank of the field, as ajb says. If it's a scalar field (rank 0), the spin is zero. If it's a spinor (electrons) then the spin is 1/2; I guess a spinor is kind of a 1/2-rank tensor. Electromagnetic fields are vectors (1st rank tensors) so the spin is 1. The derivation comes from the field Lagrangian being a scalar. There is a good rundown of the derivation in Mandl & Shaw "Quantum Field Theory", Chapter 2 I think. ...although the gravity field is the metric tensor (not the stress - energy tensor; the latter is the _source_ of the field, not the field itself). But anyway, the metric tensor is 2nd rank, so - Spin =2.

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]