# 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).

2022 Award
##\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.

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• kent davidge
Mentor
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