other independent solutions $(\lambda=1,2)$, which represent physical degrees of freedom, i.e., which satisfy the field equations and $G=0$ but are not pure gauge, will be chosen to have $A_t = 0$. Those with $A_r \neq 0$ are given as $$A_r^{(1n;\,\omega lm)} = R_{\omega l}^{(1n)}(r)Y_{lm}e^{-i\omega t}\,,$$ where $$\left[ \frac{\omega^2}{f}-\frac{l(l+p-1)}{r^2}\right] R_{\omega l}^{(1n)}(r)
+ \frac{1}{r^2}\frac{d}{dr}
\left[ \sqrt{\frac{f}{h}\,}r^{2-p}\frac{d}{dr} \left(
\frac{r^p}{\sqrt{fh}} R_{\omega l}^{(1n)}(r) \right)\right]
= 0\,.
\label{R^1}$$ Note here that the condition $G=0$ cannot be solved if $l=0$. Hence, we have $l \geq 1$. The corresponding angular components can be found by solving the condition $G=0$ as $$A_i^{(1n;\,\omega lm)} = \frac{r^{2-p}}{l(l+p-1)}\sqrt{\frac{f}{h}}
\frac{d}{dr} \left[ \frac{r^p}{\sqrt{fh\,}}R_{\omega l}^{(1n)}
\right]\partial_i Y_{lm}\; e^{-i\omega t} \,.$$ We call these modes “physical modes I". The other set of physical solutions can be obtained by letting $A_t= A_r =0$ and $$A_i^{(2n;\,\omega lm)} = R_{\omega l}^{(2n)}(r)Y_i^{(lm)}e^{-i\omega t}\,,$$ where $$\left[ \frac{\omega^2}{f} + \frac{1}{\sqrt{fh\,}r^{p-2}}\frac{d}{dr}
\left( \sqrt{\frac{f}{h}\,} r^{p-2}\frac{d}{dr} \right)
- \frac{(l+1)(l+p-2)}{r^2}\right]R_{\omega l}^{(2n)}(r) = 0\,.
\label{R^2}$$ Here, the $Y_i^{(lm)}$ are divergence-free vector spherical harmonics on the unit $p$-sphere satisfying $$\tilde{\nabla}^k (\tilde{\nabla}_k Y_i^{(lm)} - \tilde{\nabla}_i Y_k^{(lm)})
= - (l+1)(l+p-2)Y_i^{(lm)}$$ and $$\int d\Omega_p \tilde{\eta}^{ij}\overline{Y_i^{(lm)}}Y_j^{(l'm')} = 
\delta_{ll'}\delta_{mm'}\,.$$ (See, e.g., Refs. [@CM; @CH].) We call these modes “physical modes II". Physical modes I and II obtained here (and restricted to four dimensions) 