Derivation of the conservation of total energy and momentum

[MathematicalPhysicist]

I want to derive from $T^{\mu \nu}_{,\nu}=0$ the equation: $\int T_{0\mu}d^3 y=constant$, I don't see how exactly.

From the derivative I know that $T^{0\mu}_{,\mu}=0$, but I don't see how to integrate this equation, it's $T^{00}_0+T^{0i}_i=0$.
But how to proceed from here?

Thanks in advance, great forums!

 

[Orodruin]

Gauss’ theorem.

 

[kent davidge]

$$T^{\mu \nu}{}_{, \nu} = 0 \Leftrightarrow \frac{\partial T^{\mu 0}}{\partial t} = - \frac{\partial T^{\mu i}}{\partial x^i} \\ \int \frac{\partial T^{\mu 0}}{\partial t} d^3 x = - \int \frac{\partial T^{\mu i}}{\partial x^i}d^3 x$$ Right hand side is a divergence. Integrate over all space to get $\int \frac{\partial T^{\mu 0}}{\partial t} d^3 x = - \int T^{\mu i}$ over the boundary at infinity. Demand that the fields go to zero there $\Longrightarrow \int T^{\mu 0} d^3 x$ is conserved.

 

[Orodruin]

$$T^{\mu \nu}{}_{, \nu} = 0 \Leftrightarrow \frac{\partial T^{\mu 0}}{\partial t} = - \frac{\partial T^{\mu i}}{\partial x^i} \\ \int \frac{\partial T^{\mu 0}}{\partial t} d^3 x = - \int \frac{\partial T^{\mu i}}{\partial x^i}d^3 x$$ Right hand side is a divergence. Integrate over all space to get $\int \frac{\partial T^{\mu 0}}{\partial t} d^3 x = - \int T^{\mu i}$ over the boundary at infinity. Demand that the fields go to zero there $\Longrightarrow \int T^{\mu 0} d^3 x$ is conserved.

This is fine, but obscures the beauty of using Gauss’ theorem in 4 dimensions.

 

[MathematicalPhysicist]

$$T^{\mu \nu}{}_{, \nu} = 0 \Leftrightarrow \frac{\partial T^{\mu 0}}{\partial t} = - \frac{\partial T^{\mu i}}{\partial x^i} \\ \int \frac{\partial T^{\mu 0}}{\partial t} d^3 x = - \int \frac{\partial T^{\mu i}}{\partial x^i}d^3 x$$ Right hand side is a divergence. Integrate over all space to get $\int \frac{\partial T^{\mu 0}}{\partial t} d^3 x = - \int T^{\mu i}$ over the boundary at infinity. Demand that the fields go to zero there $\Longrightarrow \int T^{\mu 0} d^3 x$ is conserved.

$$T^{\mu \nu}{}_{, \nu} = 0 \Leftrightarrow \frac{\partial T^{\mu 0}}{\partial t} = - \frac{\partial T^{\mu i}}{\partial x^i} \\ \int \frac{\partial T^{\mu 0}}{\partial t} d^3 x = - \int \frac{\partial T^{\mu i}}{\partial x^i}d^3 x$$ Right hand side is a divergence. Integrate over all space to get $\int \frac{\partial T^{\mu 0}}{\partial t} d^3 x = - \int T^{\mu i}$ over the boundary at infinity. Demand that the fields go to zero there $\Longrightarrow \int T^{\mu 0} d^3 x$ is conserved.

Thanks for your illuminating answer!

 

[vanhees71]

The important point to remember is that only if $\partial_{\nu} T^{\mu \nu}=0$, the naive calculation of the total four-momentum density leads to a four-vector,
$$g^{\mu}=\int_{mathbb{R}^3} \mathrm{d}^3 \vec{x} T^{\mu 0}.$$
The proof follows again from Gauss's theorem in four dimensions. For the proof (for the completely analogous case of a conserved four-current) see p. 19 in

https://itp.uni-frankfurt.de/~hees/publ/kolkata.pdf