Conservation of Energy Momentum Tensor

[kent davidge]

Unfortunetly, I found across the web only the case where there's no source, in which case $\partial_\alpha T^{\alpha \beta} = 0$. I'm considering Minkowski space with Minkowski coordinates here.

When there's source, is it true that $\partial_\alpha (T^{\alpha \beta}) = 0$ or is it $\int \partial_\alpha (T^{\alpha \beta}) = 0$? Where now this latter $T^{\alpha \beta}$ is the result of the variation of the complete action (source included).

 

[Ibix]

$\nabla_\alpha T^{\alpha\beta}=0$ (edit: and not $\partial_\alpha T^{\alpha\beta}=0$, as I incorrectly typed originally). Since the stress-energy tensor is the same as the Einstein tensor give or take a constant factor, this turns out to be simply a statement of the Bianchi identity.

 

[PeterDonis]

When there's source, is it true that $\partial_\alpha (T^{\alpha \beta}) = 0$

No, because if there's stress-energy present, spacetime is not flat, so you have to use the correct curved spacetime equation, which is

$$
\nabla_\alpha T^{\alpha \beta} = 0
$$

$\partial_\alpha T^{\alpha\beta}=0$. Since the stress-energy tensor is the same as the Einstein tensor give or take a constant factor, this turns out to be simply a statement of the Bianchi identity.

Careful! If there is a non-zero stress-energy tensor, spacetime isn't flat. If spacetime is flat, then it is true that $\partial_\alpha T^{\alpha \beta} = 0$, but only in the vacuous sense that $T^{\alpha \beta} = 0$.

 

[Ibix]

Careful! If there is a non-zero stress-energy tensor, spacetime isn't flat. If spacetime is flat, then it is true that $\partial_\alpha T^{\alpha \beta} = 0$, but only in the vacuous sense that $T^{\alpha \beta} = 0$.

I knew it looked wrong as I was typing it. I should listen to my instincts more, apparently.