I am reading the paper "Shortcuts to high symmetry solutions in gravitational theories" written by Deser and Tekin, http://cds.cern.ch/record/625743/files/0306114.pdf

and I am having troubles to understand how they reach formula (9) from the Maxwell term \begin{equation}\label{eq.form1} -\frac{1}{4}\int{\sqrt{-g}F^2_{\mu\nu}} \end{equation} in presence of a potential $A_{\mu}=(A_0(r),0,...,0)$.

As far as I understand I should simply add the previous term to the results they got in Eq. (6), but this would give, after ignoring the overall constant that would come by integrating over the ($D-2$) sphere element and time : \begin{equation} I-\frac{1}{4}\int{dr r^{D-2}b g^{\mu\rho}g^{\nu\sigma}F_{\mu\nu}F_{\rho\sigma}}=I-\frac{1}{4}\int{dr r^{D-2}b2g^{tt}g^{rr}(\partial_{r}A_0)^2}=I+\frac{1}{2}\int{dr\frac{r^{D-2}}{b}(A'_0)^2}, \end{equation} where in the last equality I use the fact that $g^{tt}g^{rr}=-\frac{a}{ab^2}=-1/b^2$.

Comparing my result with Eq. (9) in the paper:

1) I find $+1/2$ as scale factor in the second addend, while they find a $-1/2$

2) I don't understand how could I get rid of the ($D-2$)-scale they find in (6). I can understand it could be absorbed into a new definition of the radial coordinate, but I don't see where the ($D-2$)-term should come from the kinetic term of the e.m. field.

3) Actually, it is not even clear to me why they use a Maxwell term with a factor $-1/4$ in the front: they ignored the normalization when they defined the action I in Eq. (4), so I believe I should ignore the $1/4$ in the Maxwell term as well. If I do so, I would not even get the scale factor of the second addend right in my result

Am I having problem in the generalization of the Maxwell theory to $D$-dimensions?

  • $\begingroup$ Remark on point (2). It is not a good idea to rescale $r$, because the Maxwell term contains several functions that depends on $r$ $\endgroup$ – Voltrack Jun 9 at 9:33

I believe this problem can be solved in the following way. I can multiply the second addend by $(D-2)/(D-2)$ and ignore the overall factor $(D-2)$. Then, the second addend picks up a factor $1/(D-2)$. However, I can define a new scalar field $\tilde{A}\equiv A_0/(D-2)$. Well, then the minus sign difference, which I described in point (1), and the argument about the normalization factor, which I described in point (3), doesn't bother me anymore because I can define $\tilde{A}\equiv \alpha A_0/(D-2)$, where $\alpha$ is a scale that takes care of that. In few words, Eq. (9) in the paper meant to be defined for $\tilde{A}$ rather than for $A_0$.

At the end of the day, all they want to show is that by varying the action in Eq. (9), of the paper, with respect to $A_0$, I get $A_0\propto 1/(r^{D-3})$ (note that $b$ is actually a constant), which provides the Coulomb potential for $D=4$. This argument would work also for $\tilde{A}$, because it brings into play only one more constant, what I called $\alpha$ before.


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