that $F_{tr}$ is $t$-independent and $(r^p/\sqrt{fh})F_{tr}$ is $r$-independent. Hence, if in addition the modes are assumed to be proportional to $e^{-i\omega t}$ with $\omega \neq 0$, they must be pure gauge because $F_{tr} = 0$. (A time-independent solution has $F_{tr} \propto\sqrt{fh}/r^p$. This will represent a static Coulomb field, which we will disregard here.) To find the solutions which are not spherically symmetric we let $A_t, A_r \propto Y_{lm}e^{-i\omega t}$. From the equation $\nabla^\mu F_{\mu i}=0$ we find $$-i\omega A_t = \sqrt{\frac{f}{h}}\frac{1}{r^{p-2}}
\partial_r \left[ \sqrt{\frac{f}{h}} r^{p-2}A_r\right]\,. \label{Atmode}$$ By substituting this and $A_i =0$ in $\nabla^\mu F_{\mu r} = 0$, we find that $A_r^{(1'n;\,\omega l m)}= R^{(1'n)}_{\omega l}(r)Y_{lm}e^{-i\omega t}$ (here we use primed indices to refer to the spherical Coulomb gauge), where $$\left[ \frac{\omega^2}{f} -\frac{l(l+p-1)}{r^2}\right] R^{(1'n)}_{\omega l}(r)
+ \frac{1}{f}\frac{d}{dr}\left[ \sqrt{\frac{f}{h}}\frac{1}{r^{p-2}}
\frac{d}{dr} \left( 
\sqrt{\frac{f}{h}}r^{p-2}R^{(1'n)}_{\omega l}(r)\right)\right]
= 0\,.$$ The component $A_t^{(1'n;\,\omega lm)}$ can be found from Eq. (\[Atmode\]). (Then, one can readily show that $\nabla^\mu F_{\mu t}=0$.) Indeed one can verify that these modes are related to the ones in the modified Feynman gauge by a gauge transformation: $$A_\mu^{(1 n;\,\omega lm)} \to A_\mu^{(1'n;\,\omega lm)}
= A_\mu^{(1 n;\,\omega lm)} - \nabla_\mu \Phi^{(1 n;\,\omega lm)}\,,$$ where $$\Phi^{(1 n;\,\omega lm)} = \frac{r^{2-p}}{l(l+p-1)}
\sqrt{\frac{f}{h}}\frac{d}{dr} \left( \frac{r^p}{\sqrt{fh}}R^{(1n)}_{\omega l}
\right) 