In my lectures, we considered the conserved stress energy tensor $T^{\mu \nu}$ and noted that we could always add a conserved tensor to it such that $T^{\mu \nu}$ is symmetric.

As a consequence, a tensor of higher rank, $M^{\lambda \mu \nu}= x^\mu T^{\lambda \nu} - x^{\nu}T^{\lambda \nu}$ is always conserved in that $\partial _{\lambda} M^{\lambda \mu \nu}=0$.

What I don't understand is that the claim is that the conservation of $M^{\lambda \mu \nu}$ results in the conservation of the total angular momentum tensor:

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

In essence, it is claimed that symmetry of stress energy tensor -> conservation of $M^{\lambda \mu \nu}$ -> Conservation of $J^{\mu \nu}$.

I don't see how this last step holds. I have tried the algebra, but I don't see how you can deduce anything about $\partial _{\mu} M^{0 \mu \nu}$, which you would need to vanish for $\partial _{\mu} J^{\mu \nu}$ to vanish i.e. for the angular momentum tensor to be conserved, based on $\partial _{\lambda} M^{\lambda \mu \nu}=0$ where $\lambda$ has not been set to 0, and is a sum over $\lambda$.


The quantity $J^{\mu\nu}(t)$ isn't a conserved current, it's a conserved quantity. Unlike $M^{\lambda \mu\nu}(\mathbf{x}, t)$, it doesn't have spatial dependence; at each time it is a tensor rather than a tensor field. The statement is that it doesn't depend on time at all.

The proof of this statement is just the same as the proof for a rank one tensor, since the extra indices just come "along for the ride". If we know $\partial_\mu J^\mu(\mathbf{x}, t) = 0$, then we define $$Q(t) = \int J^0(\mathbf{x}, t) \, d^3x.$$ Then $Q(t)$ is conserved because $$\frac{dQ}{dt} = \int \partial_0 J^0(\mathbf{x}, t) \, d^3x = - \int \nabla \cdot \mathbf{J} \, d^3x = - \int \mathbf{J} \cdot d\mathbf{S} = 0$$ where the last integral is at spatial infinity, and we assume $\mathbf{J}$ vanishes there. The same proof works for $M^{\lambda \mu \nu}$ since the extra two indices don't interfere. (For the case of curved spacetime, see here.)

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  • $\begingroup$ Perfect! Thank you! I've been mulling over that point for a week already! $\endgroup$ – Meep Dec 25 '18 at 20:51

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