Líenard-Weichert potential satisfies the Lorenz gauge

Question

It can be shown in 3-vector notation that the Líenard-Weichert potential satisfies the Lorenz gauge condition $\frac{\text{d}\phi}{\text{d}t}+\vec{\nabla}\vec{A}=0$; I am interested in showing this in covariant notation. $$A^{\mu}=\frac{qU^{\mu}(\tau_0)}{U_{\nu}(\tau_0)[x-r(\tau_0)]^\nu}\to\partial_{\mu}A^{\mu}=0$$ Here, $\tau_0$ depends implicitly on $x$ as $(x-r(\tau_0))^2=0$. My attempt is as follows: $$\partial_{\mu}A^{\mu}=\frac{q}{U_{\nu}(\tau_0)[x-r(\tau_0)]^\nu}\frac{A_\alpha(\tau_0)}{U_{\alpha}(\tau_0)}-\frac{\partial_{\mu}[U_{\nu}(\tau_0)[x-r(\tau_0)]^\nu]}{U_{\nu}(\tau_0)U^{\nu}(\tau_0)[x-r(\tau_0)]^\nu[x-r(\tau_0)]_\nu}$$ Firstly, I have no idea when the use of additional indices is required ($\alpha$ in this case), and secondly what does it even mean to write $\partial_{\mu}[U_{\nu}(\tau_0)[x-r(\tau_0)]^\nu]$ given that the indices differ?

Answer 1

For anyone interested: $$\partial_\mu U_\nu(\tau_0)[x-r(\tau_0)]^\nu=\partial_\mu U^{\alpha}(\tau_0)g_{\alpha\nu}[x-r(\tau_0)]^\nu=\frac{\psi^\alpha(\tau_0)}{U^\alpha(\tau_0)}\delta^{\alpha\mu}g_{\alpha\nu}[x-r(\tau_0)]^\nu=\frac{\psi^\mu(\tau_0)}{U^\mu(\tau_0)}[x-r(\tau_0)]_\mu$$ \begin{align} \partial_\mu([x-r(\tau_0)]^\alpha [x-r(\tau_0)]_\alpha) &=\partial_\mu [x-r(\tau_0)]^\alpha [x-r(\tau_0)]_\alpha+\partial_\mu [x-r(\tau_0)]_\alpha [x-r(\tau_0)]^\alpha\\ &=\partial_\mu [x-r(\tau_0)]^\alpha [x-r(\tau_0)]_\alpha+\partial_\mu [x-r(\tau_0)]^\delta g_{\alpha\delta} [x-r(\tau_0)]^\alpha\\ &=\partial_\mu [x-r(\tau_0)]_\alpha [x-r(\tau_0)]^\alpha+\partial_\mu [x-r(\tau_0)]^\delta [x-r(\tau_0)]_\delta\\ &=2\partial_\mu [x-r(\tau_0)]^\alpha [x-r(\tau_0)]_\alpha\to \partial_\mu [x-r(\tau_0)]_\alpha=0 \end{align} Where in the final line we used that $[x-r(\tau_0)]^\alpha [x-r(\tau_0)]_\alpha=0$: $$\partial_\mu A^\mu(x)=\frac{q}{U_\nu(\tau_0)[x-r(\tau_0)]^\nu}\frac{\psi^\mu(\tau_0)}{U^\mu(\tau_0)}-\frac{q}{U^\nu(\tau_0)[x-r(\tau_0)]_\nu}\frac{\psi^\mu(\tau_0)}{U^\mu(\tau_0)}=0$$

Edit: Forgot to mention that $\psi^\mu(\tau_0)$ denotes the 4-acceleration in order to avoid confusion with the potential $A^\mu(x)$