# Stress-Energy Tensor of Reissner-Nordstrom solution

I am trying to derive the R-N solution and i am following Blau's notes (to be found here http://www.blau.itp.unibe.ch/newlecturesGR.pdf) pages 677-679. With the same metric ansatz:

$$ds^2 = -A(r)dt^2 + B(r)dr^2 + r^2 d\Omega^2$$

and four potential ansatz:

$$A_{\alpha} = (-\phi(r),0,0,0).$$ i am trying to calculate the energy-momentum tensor:

$$T_{\alpha \beta} = F_{\alpha \kappa}F^{\kappa}_{\beta} - \cfrac{1}{4}g_{\alpha \beta}F^2 .$$

The only non-zero components of the Faraday tensor are:

$$F_{tr} = - F_{rt} = -\phi'(r)$$

where:

$$F_{ab} = \partial_{a}A_b - \partial_{b}A_a.$$ I can calculate the same $$F^2$$:

$$F^2 = F_{\alpha \beta}F^{\alpha \beta} = F_{\alpha \beta}g^{\kappa \alpha}g^{\lambda \beta}F_{\kappa \lambda} = F_{tr}g^{tt}g^{rr}F_{tr} + F_{rt}g^{rr}g^{tt}F_{rt} = -\cfrac{2\phi'(r)^2}{A(r)B(r)}$$

with him (equation 31.5) but i cannot find the same components with him (eq 31.7).

For example for the $$tt$$ component i have:

$$F_{t\kappa}F^{\kappa}_{t} = F_{tr}g^{rr}F_{rt} = \phi '(r) \cfrac{1}{B(r)}\big( -\phi '(r)\big) = -\cfrac{\phi '(r)^2}{B(r)}$$

which of course will not give the correct answer. Can anyone point out what i am missing??

Your last equation (v3) should be $$F_{t\kappa}{F_t}^\kappa=F_{tr}g^{rr}\color{red}{F_{tr}}=\color{red}+\cfrac{\phi '(r)^2}{B(r)}.$$ With this, check that $$T_{tt}=\frac{1}{4\pi}\left[F_{t\kappa}{F_t}^\kappa-\frac{1}{4}g_{tt}F_{\mu\nu}F^{\mu\nu}\right]$$ gives $$T_{tt}=\frac{1}{4\pi}\left[\cfrac{\phi '(r)^2}{B(r)}-\frac{1}{4}\frac{2\phi'(r)^2}{B(r)}{}\right]=\frac{1}{8\pi}\cfrac{\phi '(r)^2}{B(r)}$$ as Blau gets in (31.7).
It is essentially a mistake with index placement. Start with $$F_{t\kappa}{F_t}^\kappa$$, since we are summing over $$\kappa$$ and we know $$F_{tr}$$ is the only term that doesn't vanish, we have $$F_{t\kappa}{F_t}^\kappa=F_{tr}{F_t}^r.$$ Now lower the second index of $${F_t}^r$$ (it is the second index, so keep it in the second place)$${F_t}^r=g^{r\mu}F_{t\mu}$$ and finally sum over $$\mu$$, leaving only the term $$g^{rr}F_{tr}$$. Putting all the steps, we have $$F_{t\kappa}{F_t}^\kappa=F_{tr}{F_t}^r=F_{tr}g^{r\mu}F_{t\mu}=F_{tr}g^{rr}F_{tr}$$
• I agree with you, but i cannot see why this index comes second. I actually tried varying E/M action to see where the index goes myself and i obtained the following: $F_{\mu \kappa}F^{\kappa}_{\nu} - 1/2 g_{\mu \nu}F^2$ where the $\nu$ index comes second. Aug 30, 2020 at 17:24
• Ok, i made a mistake in my calculation indeed index $\nu$ comes first and $\kappa$ second!!! Thank you very much! Aug 30, 2020 at 17:31