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MA3431/3432 Classical Field Theory and Classical Electrodynamics

Lecturer: Dr. Nigel Buttimore

Website: Link

Notes by Eoin Curran
Notes by Chris Blair

[edit]

Alternative Derivation of the EM field from the Liénard Potentials

Given the Liénard - Wiechert potentials

$A^{\nu}=\frac{e V^{\nu}}{V_\sigma R^\sigma}$

where the expression is to be evaluated at retarded time defined by

$R^\sigma R_\sigma=0$

and where

$R^\mu=x^\mu-x_e^\mu(\tau)$

$V^\mu=\frac{\mathrm{d}x^\mu_e(\tau)}{\mathrm{d}\tau}=-\frac{\mathrm{d}R^\mu}{\mathrm{d}\tau}$

We wish to find the electromagnetic fields $F^{\mu\nu}$ .

We begin by taking finding how the retarded time is related to the change in the observation point

$(\partial^{\mu} R^{\sigma})R_{\sigma}=0$

now regarding $R^{\sigma}=R^{\sigma}(x^\rho,\tau)$ and since the path of the charge e $x_\mathrm{e}^\mu(\tau)$ intersects the past lightcone of the event $x^\rho$ uniqely we can regard $\tau=\tau(x^\rho)$ . And so we find

$\left(\frac{\partial R^\sigma}{\partial \tau}\partial^\mu \tau+g^{\mu \sigma}\right)R_\sigma=0$

$\left(-V^{\sigma} \partial^{\mu} \tau+g^{\mu \sigma}\right)R_{\sigma} =0$

$-V^{\sigma} R_{\sigma} \partial^\mu \tau+R^\mu=0$

$\partial^{\mu} \tau=\frac{R^\mu}{V^\sigma R_\sigma}$

Now taking $\partial^\mu$ of $A^\nu$ , regarding $\tau=\tau(x^\rho)$ and using the expression above we find

$\partial^\mu A^\nu=\frac{e \partial^\mu V^\nu}{V^\sigma R_\sigma}-\frac{e V^\nu}{\left(V^\sigma R_\sigma\right)^2}\left( \left(\partial^\mu V^\rho\right)R_\rho+V^\rho \partial^\mu R_\rho\right)$

$=\frac{e \dot{V}^\nu}{V^\sigma R_\sigma}\frac{R^\mu}{V^\rho R_\rho} -\frac{eV^\nu}{(V^\sigma R_\sigma)^2} \left( \dot{V}^\rho R_\rho \frac{R^\mu}{V^\lambda R_\lambda} +V^\rho \left(\dot{R}_\rho\frac{R^\mu}{V^\lambda R_\lambda}+\delta^\mu_\rho \right) \right)$

$=\frac{e }{(V^\sigma R_\sigma)^2}R^\mu \dot{V}^\nu-\frac{e \dot{V}^\rho R_\rho}{(V^\sigma R_\sigma)^3}R^\mu V^\nu +\frac{e V^\rho V_\rho}{(V^\sigma R_\sigma)^3} R^\mu V^\nu -\frac{e}{(V^\sigma R_\sigma)^2}V^\mu V^\nu$

where $\dot{X}$ means $\frac{\mathrm{d}}{\mathrm{d}\tau}X$

Noting that

$V^\sigma V_\sigma =\gamma^2(c^2-\overrightarrow{v}^2)$

$=\gamma^2 c^2 \gamma^{-2}$

$=c^2$

and using our expression for $\partial^\mu A^\nu$ we find

$F^{\mu \nu}=\partial^\mu A^\nu-\partial^\nu A^\mu$

$=F_{acc}^{\mu \nu}+F_{vel}^{\mu \nu}$

where

$F_{acc}^{\mu \nu}=\frac{e}{(V^\sigma R_\sigma)^2}(R^\mu \dot{V}^\nu-R^\nu \dot{V}^\mu)-\frac{e \dot{V}^\rho R_\rho}{(V^\sigma R_\sigma)^3}(R^\mu V^\nu-R^\nu V^\mu)$