# How to correctly differentiate the Lienard-Wiechert four-vector potential to get the EM tensor?

The retarded 4-vector potential for a moving charge is given by $$A^\alpha = \left. \frac {eV^\alpha(\tau)}{V\cdot[x-r(\tau)]} \right|_{\tau = \tau_0}$$ where $$e$$ is the charge, $$V$$ the four-velocity, $$x$$ the field point, $$r$$ the retarded position of the charge at its retarded proper time $$\tau = \tau_0$$.

Using the light-cone condition, I can convert $$dx^\alpha$$ to $$d\tau$$: $$[x - r(\tau_0]]^2=0\\ dx^{\alpha}(x-r)^{\alpha}-dr\cdot(x-r) = 0\\ dx^{\alpha}(x-r)^{\alpha}-d{\tau}(x_0-r_0)= 0\\ dx^{\alpha}=\frac{d\tau(x_0-r_0)}{(x-r)^{\alpha}}$$

For the EM tensor I hence get: $$F^{\alpha\beta}= \partial^{\alpha}A^{\beta}-\partial^{\beta}A^{\alpha}= \frac{e(x-r)^{\alpha}}{(x-r)^0}\frac{d}{d\tau} \left [\frac{V^{\beta}}{V\cdot(x-r)} \right] - \frac{e(x-r)^{\beta}}{(x-r)^0}\frac{d}{d\tau} \left [\frac{V^{\alpha}}{V\cdot(x-r)} \right]$$

By convention $$\frac{d}{d\tau}[\;]$$ is taken with the field point $$x$$ held constant which isn't the case here. I therefore need to add $$\frac{e(x-r)^{i}}{(x-r)^0}\frac{dx^i}{d\tau}\frac{\partial}{\partial x^i}[\;]=\frac{e\partial}{\partial x^i}[\;]$$ to the above expression:

\begin{align} F^{\alpha\beta}&= \frac{e(x-r)^{\alpha}}{(x-r)^0}\frac{d}{d\tau} \left [\frac{V^{\beta}}{V\cdot(x-r)} \right] + \frac{e\partial}{\partial x^\alpha} \left [\frac{V^{\beta}}{V\cdot(x-r)} \right]\\ &- \frac{e(x-r)^{\beta}}{(x-r)^0}\frac{d}{d\tau} \left [\frac{V^{\alpha}}{V\cdot(x-r)} \right] - \frac{e\partial}{\partial x^\beta} \left [\frac{V^{\alpha}}{V\cdot(x-r)} \right] \end{align}

However, Jackson$$^1$$ calculates the EM tensor as $$F^{\alpha\beta}= \partial^{\alpha}A^{\beta}-\partial^{\alpha}A^{\beta}= \frac{e}{V\cdot(x-r)}\frac{d}{d\tau} \left [\frac{(x-r)^{\alpha}V^{\beta}- (x-r)^{\beta}V^{\alpha}}{V\cdot(x-r)} \right]$$

What do I need to do to get this expression also?

1 Classical Electrodynamics, Jackson, 3rd edition page 663

• Use: $A(x)^\alpha=2e\int d\tau V^\alpha(\tau)\theta[x_{0}-r_{0}(\tau)]\delta\{[x-r(\tau)]^2\}$ instead - then $\partial^{\alpha}A(x)^\beta=2e\int d\tau V^\beta(\tau)\theta[x_{0}-r_{0}(\tau)]\partial^{\alpha}\delta\{[x-r(\tau)]^2\}$. Using $\partial^{\alpha}\delta[f]=\frac{(x-r)^{\alpha}}{V\cdot(x-r)}\frac{d}{d\tau}\delta[f]$ where $f=[x-r(\tau)]^2$ integrate by parts and $\theta$ will vanish.Result will be first term of $F^{\alpha\beta}$. You will need to use the result $\frac{d}{d\tau}[x-r(\tau)]^2=-2[x-r(\tau)]_{\beta}V^{\beta}(\tau)$ – Cinaed Simson Apr 28 at 3:04
• @CinaedSimson yes, there are other ways of doing it but I'm interested in someone pointing out the flaw in my working. – Physiks lover May 2 at 23:42
• $[x - r(\tau_0)]^2=0$ is the light cone constraint which enforces the retardation $x_{0} > r_{0}(\tau_{0})$. Since you wrote down the constraint, I presumed you were trying to evaluate the integral form - which contains the term $\delta {\{[x-r(\tau)]^2\}}$. So take your expression for $A^\alpha =\left. \frac {eV^\alpha(\tau)}{V\cdot[x-r(\tau)]}\right|_{\tau = \tau_0}$ and calculate $\partial^{\beta}A^{\alpha}$. – Cinaed Simson May 3 at 21:53