Derivation of the conservation of total energy and momentum

  • #1
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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!
 

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  • #3
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$$T^{\mu \nu}{}_{, \nu} = 0 \Leftrightarrow \frac{\partial T^{\mu 0}}{\partial t} = - \frac{\partial T^{\mu i}}{\partial x^i} \\ \int \frac{\partial T^{\mu 0}}{\partial t} d^3 x = - \int \frac{\partial T^{\mu i}}{\partial x^i}d^3 x$$ 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.
 
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  • #4
Orodruin
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$$T^{\mu \nu}{}_{, \nu} = 0 \Leftrightarrow \frac{\partial T^{\mu 0}}{\partial t} = - \frac{\partial T^{\mu i}}{\partial x^i} \\ \int \frac{\partial T^{\mu 0}}{\partial t} d^3 x = - \int \frac{\partial T^{\mu i}}{\partial x^i}d^3 x$$ 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.
 
  • #5
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$$T^{\mu \nu}{}_{, \nu} = 0 \Leftrightarrow \frac{\partial T^{\mu 0}}{\partial t} = - \frac{\partial T^{\mu i}}{\partial x^i} \\ \int \frac{\partial T^{\mu 0}}{\partial t} d^3 x = - \int \frac{\partial T^{\mu i}}{\partial x^i}d^3 x$$ 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!
 
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  • #6
vanhees71
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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
 

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