Daniel Berwick Evans. Maxwell’s equations and electromagnetism II. Example: Current through a wire. E = 0. From this data we can derive Ampere’s law: ∫ B · ds =

j_{e}/(ϵ_{0}c^{2}) = |B| · 2π. Hence |B| = |j|/(2πϵ_{0}c^{2}).

Example: Solenoid. We assume that B_{o} = 0. ∫ B · ds = j_{e}/(ϵ_{0}c^{2}) = N|j|/(ϵ_{0}c^{2}). Hence |B| = n|j|/(ϵ_{0}c^{2}).

Potentials are a nice way to cook up solutions to Maxwell’s equations. Since div B = 0 we can write B = curl A for some non-unique vector potential A if there are no topological obstructions. Also curl(E+∂_{t}A) = 0, hence E + ∂_{t}A = − div ϕ for some non-unique scalar potential ϕ.

Gauge transformations: A ↦ A + div ψ and ϕ ↦ ϕ − ∂_{t}ψ. By choosing a particular A and ϕ we are “fixing our gauge”. For example, div A = −c^{−2}∂_{t}ϕ is called

the Lorentz gauge.

Relativistic formulation. Motivation. Consider two wires with current. Lorentz’ law: F = q(E + v × B) (force on a test particle

with velocity v and charge q). (We cannot derive it from Maxwell’s equations.) Hence the wires are attracted to each other if the have the same direction of current and repelled otherwise.

Now we move into a frame of reference in which electrons in wires are at rest. We observe that now we have electric field instead of magnetic.

We introduce some forms: B = B_{x}dy ∧ dz + B_{y}dz ∧ dx + B_{z}dx ∧ dy, E = E_{x}dx + E_{y}dy + E_{z}dz, and F = B + E ∧ dt (electromagnetic field). J = ρdt + j_{x}dx + j_{y}dy + j_{z}dz. Maxwell’s equations: dF = 0 and ∗d ∗ F = J, where ∗ is the Hodge star.

In the vacuum, given a solution F, ∗F is another solution. In particular we have self-dual and anti-self-dual solutions: F = ± ∗ F.

Self-dual solutions are left circularly polarized wave solutions and anti-self-dual are right circularly polarized (for given orientation on R^{3+1} determined by ∗.

Lagrangian formulation of Maxwell’s equations: fields are closed 2-forms. Lagrangian (vacuum case): ∫ F ∧ ∗F.

We can also take A ∈ Ω^{1}(R^{3+1}) and L = ∫ dA ∧ ∗dA + ∗J ∧ A. Aharonov-Bohm effect: S_{q} = ∫ exp(ih̄^{−1}q ∫_{γ} A).