- #1
Jimmy Snyder
- 1,127
- 21
This time I am asking for help on a problem. I think I have an answer for problem 23, part (a), but I would like verificiation. I don't have a solution for part (b).
Problem 23: Use the identity [itex]T^{\mu \nu}{}_{,\nu} = 0[/itex] to prove the following results for a bounded system (i.e. a system for which [itex]T^{\mu \nu} = 0[/itex] outside a bounded region of space).
(a)
[tex]\frac{\partial}{\partial t}\int T^{0 \alpha}d^3x = 0[/tex]
My solution is:
[tex]
\frac{\partial}{\partial t}\int T^{0 \alpha}d^3x
[/tex]
[tex]
= \int T^{\alpha 0}{}_{,0}d^3x
[/tex]
[tex]
= -\int T^{\alpha j}{}_{,j}d^3x[/tex] (by the identity)
[tex]
= -\oint \bold{n} \cdot T^{\alpha j}dS[/tex] (Gauss's theorem)
[tex]
= 0[/tex](by boundedness, using a sphere that surrounds the support of T)
I think all this is OK, but when I try to extend this to part (b), the tensor virial theorem, I don't get the right answer:
(b)
[tex]\frac{\partial^2}{\partial t^2}\int T^{00} x^i x^j d^3x = 2\int T^{ij} d^3x[/tex]
The problem for me is that after I convert the first partial in Gauss's theorem, the volume integral turns into a surface integral and I can't do that a second time for the second partial.
Problem 23: Use the identity [itex]T^{\mu \nu}{}_{,\nu} = 0[/itex] to prove the following results for a bounded system (i.e. a system for which [itex]T^{\mu \nu} = 0[/itex] outside a bounded region of space).
(a)
[tex]\frac{\partial}{\partial t}\int T^{0 \alpha}d^3x = 0[/tex]
My solution is:
[tex]
\frac{\partial}{\partial t}\int T^{0 \alpha}d^3x
[/tex]
[tex]
= \int T^{\alpha 0}{}_{,0}d^3x
[/tex]
[tex]
= -\int T^{\alpha j}{}_{,j}d^3x[/tex] (by the identity)
[tex]
= -\oint \bold{n} \cdot T^{\alpha j}dS[/tex] (Gauss's theorem)
[tex]
= 0[/tex](by boundedness, using a sphere that surrounds the support of T)
I think all this is OK, but when I try to extend this to part (b), the tensor virial theorem, I don't get the right answer:
(b)
[tex]\frac{\partial^2}{\partial t^2}\int T^{00} x^i x^j d^3x = 2\int T^{ij} d^3x[/tex]
The problem for me is that after I convert the first partial in Gauss's theorem, the volume integral turns into a surface integral and I can't do that a second time for the second partial.
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