Question: Can a boosted frame remove unwanted terms from a transformed metric?

[Mentz114]

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
<br /> G_{\mu\nu}= \left[ \begin{array}{cccc}<br /> A &amp; B &amp; 0 &amp; 0\\<br /> B &amp; p1 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; p2 &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; p3<br /> \end{array} \right]<br />
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,
<br /> \Lambda^\mu_\rho=\left[ \begin{array}{cccc}<br /> 1 &amp; -\frac{B}{p1} &amp; 0 &amp; 0\\<br /> 0 &amp; 1 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; 1 &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array} \right]<br />
This seems to be transforming t into T=t-h\ r where h=B/p1. This can be used to give the differential transformation
<br /> dT=dt -hdr-rdh=dt-hdr-r(\partial_t h\ dt + \partial_r h\ dr)<br />
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]

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]

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]

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]

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.