I Conservation of Energy Momentum Tensor

kent davidge
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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).
 
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##\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 said:
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

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

Ibix said:
##\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##.
 
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PeterDonis said:
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.
 
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