A Weinberg's proof of ##{T^{\mu\nu}}_{,\nu}=0## for a perfect fluid

[Kostik]

TL;DR

His proof is hard to follow, can someone help?

Weinberg ("Gravitation and Cosmology") defines the energy-momentum tensor $T^{\mu\nu}$ in equations (2.8.1)-(2.8.2). He proves $${T^{\mu\nu}}_{,\nu}=0$$ on page 44. But:

(1) Why does he have a minus sign at the very beginning; see the equation which starts $$\frac{\partial}{\partial x^i}T^{\alpha i}(x,t) =$$ when there is no such minus sign in (2.8.2)?

(2) How does he do what looks like an integration by parts (third equality) when there is no integration?

What makes his work more confusing is that on pp. 43-44 he alternately uses the notation $({\bf{x}}t)$, $(x)$, $(x,t)$ and $({\bf{x}},t)$.

 

[renormalize]

His proof is hard to follow, can someone help?

(1) Weinberg gets his negative sign by using the identity: $$\frac{\partial}{\partial x^{i}}\delta^3\left(\vec{x}-\vec{x}_{n}\right)\equiv -\frac{\partial}{\partial x_{n}^{i}}\delta^3\left(\vec{x}-\vec{x}_{n}\right)$$
(2) This is just the product rule for differentiation written in the form:$$-u\left(t\right)\frac{\partial v\left(t\right)}{\partial t}=-\frac{\partial}{\partial t}\left(u\left(t\right)v\left(t\right)\right)+\frac{\partial u\left(t\right)}{\partial t}v\left(t\right)$$

 

[Kostik]

Of course; thanks. I think I have found another proof, but I should have seen this.