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:

$$T^{\mu\nu} = (\epsilon + p) \frac{dx^{\mu}}{ds} \frac{dx^{\nu}}{ds} -p g^{\mu\nu}$$

since in cylindrical coordinates
$$x^0 =t \qquad x^1 =\rho \qquad x^2 = \phi \qquad x^3 =z$$

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

I obtain
$$T_{11}=T_{33}$$

but from the definition of the einstein tensor I obtain
$$G_{11}=-G_{33}$$

but einstein equation says
$$G_{\mu\nu}=8\pi k T_{\mu\nu}$$

which in this case then implies
$$T_{11}=-T_{33}$$

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.