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Mentz114
May26-11, 06:50 PM
A certain metric gives an Einstein tensor that has the form below. The coordinate labelling is
x^0=t,\ x^1=r,\ x^2=\theta,\ x^3=\phi

G_{\mu\nu}= \left[ \begin{array}{cccc}
A & B & 0 & 0\\
B & p1 & 0 & 0\\
0 & 0 & p2 & 0\\
0 & 0 & 0 & p3
\end{array} \right]

where A,B,C,p1,p2,p3 are functions of t and r. A transformation \Lambda so \Lambda^\mu_\rho\ \Lambda^\nu_\sigma\ G_{\mu\nu} is diagonal is easily found,

\Lambda^\mu_\rho=\left[ \begin{array}{cccc}
1 & -\frac{B}{p1} & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{array} \right]

This seems to be transforming t into T=t-h\ r where h=B/p1. This can be used to give the differential transformation

dT=dt -hdr-rdh=dt-hdr-r(\partial_t h\ dt + \partial_r h\ dr)

so we can find dt^2 and substitute into the original metric to get a transformed one written in coordinates T,r,\theta,\phi.

Question: will the Einstein tensor obtained from the transformed metric be \Lambda^\mu_\rho\ \Lambda^\nu_\sigma\ G_{\mu\nu} ?

I think it will be but I haven't convinced myself.

bcrowell
May26-11, 07:26 PM
Question: will the Einstein tensor obtained from the transformed metric be \Lambda^\mu_\rho\ \Lambda^\nu_\sigma\ G_{\mu\nu} ?

Isn't this true regardless of the specifics of the problem, just because that's how a rank-2 tensor transforms?

Mentz114
May26-11, 10:19 PM
Isn't this true regardless of the specifics of the problem, just because that's how a rank-2 tensor transforms?
That is what I think - but I have some doubts.

It would be true if \Lambda were a frame field, i.e. a transformation from the coordinate basis to a frame basis - but it's not.

In the untransformed Einstein tensor, the terms I've called p1 etc do come out as isotropic pressure, i.e. p1=Pg_{11}, with the same P in all three. So if the off-diagonal terms were absent, it might be a static perfect fluid with pressure. Can one change an unphysical metric to a physical one with a coordinate transformation ? It seems too easy.

PAllen
May26-11, 11:17 PM
I could be all wet here, but it looks like your lambda is impossible as a coordinate transform. I think the (0,0) component being 1 is saying you must have T=t + f(r). Your solution would then be possible if B and p1 depended only on r. But you've said they depend on r and t. Contradiction.

Mentz114
May27-11, 09:28 AM
I could be all wet here, but it looks like your lambda is impossible as a coordinate transform. I think the (0,0) component being 1 is saying you must have T=t + f(r). Your solution would then be possible if B and p1 depended only on r. But you've said they depend on r and t. Contradiction.

I think you're right. It's a bust.

But I have found a boosted frame that removes the unwanted terms, which is what I should have done in the first place.