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10.1 - The Potential Formulation

10.0.1 Presentation/Paper notes

A few words about the paper/presentation

• It's supposed to be on a topic related to the class (classical electromagnetism)
• The topic can be wide-ranging. Some sample topics will be put it. It should be thematically related to what we've been talking about, and you should be making connections to things we've been talking about (waveguides, waves, generation of waves, etc.)
• There will be two class periods in the last week which are reserved for presentations of about 15 minutes each.

Examples of possible topics:

• Applications of X-ray radiation
  • Advanced light source
  • Advanced photon source
• Waveguiding
  • Optical fibers
  • Applications of coaxial cables
• Microwave cavities
  • ADMX experiment
  • Atom-photon entanglement with Rydberg atoms in high finesse cavities
• Optical interferometry
  • Fabry-Perot interferometer / cavity and applications
  • LIGO and gravitational waves
• Microcavities, Photonic crystal cavities

10.1.1 Scalar and Vector Potentials

Ultimately the question of electromagnetism is given some sources $\rho(\vec r, t)$ and $\vec J ( \vec r, t)$, what are the resulting fields $\vec E( \vec r, t)$ and $\vec B ( \vec r, t)$? In the static case, Coulomb's law and the Biot-Savart law provide deterministic answers, so how do we then generalize to time-dependent configurations?

It turns out that once again, it will pay to represent the fields in terms of potentials. Just like in electrostatics, $\curl \vec E = 0$ allowed us to write $\vec E$ as the gradient of a scalar potential. We can't do that any more, but we do still have a divergenceless $\vec B$, so $$\vec B = \curl \vec A$$ is still valid, as in magnetostatics. Plugging into Faraday's law gives us $$\curl \vec E = - \pdv{\vec B}{t} \\ = - \pdv{}{t} \left( \curl \vec A \right) \\ = - \curl \left( \pdv{\vec{A}}{t} \right) \\ \rightarrow \curl ( \vec E + \pdv{\vec{A}}{t} ) = 0 \\ \rightarrow \vec E + \pdv{\vec{A}}{t} = - \grad V$$

In terms of a scalar and a vector potential, then, we can write $\vec E$ as $$\vec E = - \grad V - \pdv{\vec{A}}{t}$$

What happens with Gauss' law and the Ampere-Maxwell law? $$\div \vec E = \rho / \epsilon _0 \\ \rightarrow \div ( - \grad V - \pdv{\vec A}{t}) = \rho / \epsilon_0 \\ \rightarrow \laplacian V + \pdv{}{t} \left( \div \vec A \right) = - \rho / \epsilon_0$$

$$\curl \vec B = \mu_0 \vec J + \mu_0 \epsilon_0 \pdv{\vec E}{t} \\ \curl ( \curl \vec A) = \mu_0 \vec J + \mu_0 \epsilon_0 \pdv{}{t} \left(- \grad V - \pdv{\vec A}{t} \right) \\ \laplacian \vec A - \mu_0 \epsilon_0 \pdv{^2 \vec A}{t^2} - \grad ( \div \vec A + \mu_0 \epsilon_0 \pdv{V}{t} )= - \mu_0 \vec J$$

Well now what! We've got a fairly complicated differential equation on our hands... how can we make it simpler? The fields actually only care about the curl of the vector potential, and that allows us some freedom in the gauge of $\vec A$. We can add any curl-less function to modify the divergence $\div \vec A$ and nothing at all about the real fields will change. In particular, wouldn't it be nice if $$\div \vec A = 0 \quad \text{(Coulomb gauge)}$$

so that solving for $V$ just amounts to the Poisson equation:

$$\laplacian V + \pdv{}{t} (\div \vec A) = - \rho / \epsilon_0 \\ \rightarrow \laplacian V = - \rho / \epsilon_0 \quad \text{(Coulomb Gauge)}$$

But the Coulomb gauge still makes it very difficult to solve for $\vec A$. If we want to solve for the vector potential more easily, it's pretty obvious that we should set

$$\div \vec A = - \mu_0 \epsilon_0 \pdv{V}{t} \quad \text{(Lorenz gauge)}$$

so that $$\laplacian \vec A - \mu_0 \epsilon_0 \pdv{^2 \vec A}{t^2} - \grad ( \div \vec A + \mu_0 \epsilon_0 \pdv{V}{t} )= - \mu_0 \vec J \\ \rightarrow \laplacian \vec A - \mu_0 \epsilon_0 \pdv{^2 \vec A}{t^2} = - \mu_0 \vec J \quad \text{(Lorenz gauge)}$$

The result is an inhomogeneous wave equation with a "source" term on the right. It's quite similar to the wave equations we've been solving for the past couple of chapters, but the sources have been re-introduced, and it's important to remember that now the sources $\rho$ and $\vec J$ are allowed to vary in both space and time.