How to properly show that : $\sum_{i} \partial_{i} T_{i \mu}=\partial_0 T_{0 \mu} $ knowing that $\partial_{\alpha} T^{\alpha \mu}=0$

Question

We know that : (*) $\partial_{\alpha} T^{\alpha \mu}=0$ for a field following Klein Gordon equation ($T$ is the energy impulsion tensor).

And we say in my QFT course, that because of (*) we have : $\sum_{i} \partial_{i} T_{i \mu}=\partial_0 T_{0, \mu} $.

I know how to "prove" it but I think there is a more automatic way using indices to prove it.

This is what I did :

$$ 0=\partial_{\alpha} T^{\alpha \mu}=\partial^{\alpha} T_{\alpha}^{\mu}=\eta^{\mu \beta}\partial^{\alpha} T_{\alpha \beta}=\eta^{\mu \beta}(\partial_0 T_{0 \beta}-\partial_i T_{i \beta})$$

And as we know that $\eta^{\mu \beta}=\pm \delta^{\mu \beta}$ we thus have :

$$\partial_0 T_{0 \beta}-\partial_i T_{i \beta}=0 $$

But I think there is a better and more "clean" way to prove it because we never say in practice things like $\eta^{\mu \beta}=\pm \delta^{\mu \beta}$ for example.

In fact I need to learn tricks to calculate more efficiently with indices so this is the point of my question : did I do the calculation in the most efficient way or is there a better way to do it ? (In particular I think that to say $\eta^{\mu \beta}=\pm \delta^{\mu \beta}$ is not a good thing in practice even if it is true, but I may be wrong).

Answer 1

There is always a relative minus sign between the temporal and spatial components of the metric - it is this minus sign that allows for the desired equality. Words to this effect is essentially the statement $\eta^{\mu \beta} = \pm \delta^{\mu \beta}$ that you wrote and most authors would assume this is understood rather than write it out explicitly.