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.
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.