The Ricci scalar of the four-dimensional Reissner–Nordström metric is equal to zero. In the case of the five-dimensional Reissner–Nordström metric, the Ricci scalar is different to zero?


1 Answer 1


Yes. The electromagnetic field is nonzero, so the stress-energy tensor is equal to $\frac{1}{2}F_{ac}F^{c}{}_{b}- \frac{1}{8}g_{ab}F^{cd}F_{cd}$. And, of course, $8\pi G T_{ab} = R_{ab} -\frac{1}{2}Rg_{ab}$

EDIT: to clarify, in the riessner-Nordstrom spacetime, you have $g_{ab} = {\rm diag} (-\Delta, \frac{1}{\Delta}, {\rm sphere\; metric})$ and $F_{ab} = f(r)\left(dtdr - drdt\right)$. Meanwhile, you contract Einstein's equation, and you get $$R - (d/2)R = 8\pi G\left(\frac{1}{2}F^{ab}F_{ab} - (d/8)F^{ab}F_{ab}\right)$$

Therefore, you get

$$R = \frac{2\pi G(4-d)}{2-d}F^{ab}F_{ab}$$

So, ALL electrovac solutions have a vanishing ricci scalar in four dimensions only. In higher dimensions $F_{ab} \neq 0$, so $R\neq 0$

  • $\begingroup$ @JuanOspina, then you need to calculate it. You certainly shouldn't expect a zero Ricci scalar in either case. The stress-energy tensor is nonzero. $\endgroup$ Oct 2, 2014 at 21:23
  • $\begingroup$ And seriously, your question just involves going and doing some math. Just do the math. $\endgroup$ Oct 2, 2014 at 22:01
  • $\begingroup$ but, there, see the edit, @JuanOspina $\endgroup$ Oct 2, 2014 at 22:10
  • $\begingroup$ @JuanOspina: why? the exterior derivative is definied in all dimensions, and it adds only one index in all dimensions. $F_{ab}$ is an antisymmetric 2-tensor in all dimensions. $\endgroup$ Oct 3, 2014 at 15:51
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    $\begingroup$ @ JerrySchirme: now all is correct. Again, many thanks. All the best. $\endgroup$ Oct 3, 2014 at 17:57

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