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Stress-Energy Tensor of Reissner-Nordstrom solution

Physics Asked by VladimirA on October 27, 2020

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:

$$ ds^2 = -A(r)dt^2 + B(r)dr^2 + r^2 dOmega^2 $$

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 $F^2$:

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

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:

$$F_{tkappa}F^{kappa}_{t} = F_{tr}g^{rr}F_{rt} = phi ‘(r) cfrac{1}{B(r)}big( -phi ‘(r)big) = -cfrac{phi ‘(r)^2}{B(r)}$$

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

One Answer

Your last equation (v3) should be $$F_{tkappa}{F_t}^kappa=F_{tr}g^{rr}color{red}{F_{tr}}=color{red}+cfrac{phi '(r)^2}{B(r)}.$$ With this, check that $$T_{tt}=frac{1}{4pi}left[F_{tkappa}{F_t}^kappa-frac{1}{4}g_{tt}F_{munu}F^{munu}right]$$ gives $$T_{tt}=frac{1}{4pi}left[cfrac{phi '(r)^2}{B(r)}-frac{1}{4}frac{2phi'(r)^2}{B(r)}{}right]=frac{1}{8pi}cfrac{phi '(r)^2}{B(r)}$$ as Blau gets in (31.7).

It is essentially a mistake with index placement. Start with $F_{tkappa}{F_t}^kappa$, since we are summing over $kappa$ and we know $F_{tr}$ is the only term that doesn't vanish, we have $$F_{tkappa}{F_t}^kappa=F_{tr}{F_t}^r.$$ Now lower the second index of ${F_t}^r$ (it is the second index, so keep it in the second place)$${F_t}^r=g^{rmu}F_{tmu}$$ and finally sum over $mu$, leaving only the term $g^{rr}F_{tr}$. Putting all the steps, we have $$F_{tkappa}{F_t}^kappa=F_{tr}{F_t}^r=F_{tr}g^{rmu}F_{tmu}=F_{tr}g^{rr}F_{tr}$$

Correct answer by Urb on October 27, 2020

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