- #1
kent davidge
- 933
- 56
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).
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).