Stress-energy tensor in static cylindrical case

[ArthurB]

I have some problems using this definition, maybe because it's not valid in every coordinate system:

<br /> T^{\mu\nu} = (\epsilon + p) \frac{dx^{\mu}}{ds} \frac{dx^{\nu}}{ds} -p g^{\mu\nu}<br />

since in cylindrical coordinates
<br /> x^0 =t \qquad x^1 =\rho \qquad x^2 = \phi \qquad x^3 =z<br />

using weyl metric
<br /> g_{00}= e^{2u} \qquad g_{11}=-e^{2v-2u} \qquad g_{22}=-\rho^2 e^{-2u} \qquad g_{33}=-e^{2v-2u} \qquad u=u(\rho,z) \qquad v=v(\rho,z)<br />

I obtain
<br /> T_{11}=T_{33}<br />

but from the definition of the einstein tensor I obtain
<br /> G_{11}=-G_{33}<br />

but einstein equation says
<br /> G_{\mu\nu}=8\pi k T_{\mu\nu}<br />

which in this case then implies
<br /> T_{11}=-T_{33}<br />

in contraddiction with the result obtained from the first formula.
can anyone explain?

 

[bcrowell]

To make the math show up correctly, you need tex and /tex (or itex and /itex for inline equations), not latex and /latex.

 

[ArthurB]

thanks, corrected. Do you have any clue about the problem?

 

[Mentz114]

I find G³³ =-G²² also.

I would not expect this Einstein tensor to be that of a perfect fluid.