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Photons

Maxwell's equations of classical electrodynamics in vacuum are


$\displaystyle \vec{\nabla} \cdot \vec{E} = \rho$ $\textstyle ,$ $\displaystyle \vec{\nabla} \times \vec{E} +
\frac{\partial \vec{B}}{\partial t} = 0 ,$ (7.88)
$\displaystyle \vec{\nabla} \cdot \vec{B} = 0$ $\textstyle ,$ $\displaystyle \vec{\nabla} \times \vec{B} -
\frac{\partial \vec{E}}{\partial t} = \vec{j} ,$  

where we are using Heaviside-Lorentz (rationalized Gaussian) units.

Defining $j^\mu = (\rho,\vec{j})$ and $A^\mu = (\phi,\vec{A})$, the four-vector potential is related to the electric and magnetic fields by


\begin{displaymath}
\vec{E} = -\frac{\partial A}{\partial t} - \vec{\nabla} \phi \quad
,\quad \vec{B} = \vec{\nabla} \times \vec{A} .
\end{displaymath} (7.89)

Further, we can show that in terms of the antisymmetric field strength tensor


\begin{displaymath}
F^{\mu\nu} \equiv \partial^\mu A^\nu - \partial^\nu A^\mu
\end{displaymath} (7.90)

that Maxwell's equations now take the compact form


\begin{displaymath}
\partial_\mu F^{\mu\nu} = j^\nu
\end{displaymath} (7.91)

and current conservation, $\partial_\nu j^\nu = 0$, follows as a natural compatibility condition.

These equations are equivalent to the following covariant equation for $A^\mu$.


\begin{displaymath}
\Box A^\mu - \partial^\mu (\partial_\nu A^\nu) = j^\mu ,
\end{displaymath} (7.92)

which is general and $A^\mu$ is in the presence of charge and currents.

$\vec{E}$ and $\vec{B}$ are unchanged by the gauge transformation


\begin{displaymath}
A_\mu \rightarrow A_\mu^\prime = A_\mu + \partial_\mu \chi ,
\end{displaymath} (7.93)

where $\chi$ is any scaler function of $x$. We use this freedom to pick $\chi$ so that we can write Maxwell's equations in the form


\begin{displaymath}
\Box A^\mu = j^\mu \quad\textrm{with}\quad \partial_\mu A^\mu = 0 .
\end{displaymath} (7.94)

The requirement $\partial_\mu A^\mu=0$ is known as the Lorentz condition. The process of choosing a particular condition on $A^\mu$ so as to define it uniquely is called ``choosing a gauge''. Potentials satisfying $\partial_\mu A^\mu=0$ are said to be ``in the Lorentz gauge''. However, even after choosing a gauge, there is still some residual freedom in the choice of the potential $A^\mu$. We can still make another gauge transformation


\begin{displaymath}
A_\mu \rightarrow A^\prime_\mu = A_\mu + \partial_\mu \Lambda
\end{displaymath} (7.95)

where $\Lambda$ is any function that satisfies


\begin{displaymath}
\Box \Lambda = 0 .
\end{displaymath} (7.96)

This last equation ensures that the Lorentz condition is still satisfied.



Subsections
next up previous contents index
Next: Polarization Vectors Up: QED Processes Previous: Coulomb Scattering of Positrons
Douglas M. Gingrich (gingrich@ ualberta.ca)
2004-03-18