Electromagnetic Field Energy Kirk T. McDonald Joseph Henry Laboratories, Princeton University, Princeton, NJ 08544 (April 3, 2002) 1 Problem The (time-dependent) electromagnetic field energy in vacuum (or in a medium where D = E and B = H) is given as U EM =  E 2 + B 2 8π dVol, (1) in Gaussian units. Here, D is the electric displacement vector, E is the electric field, B is the magnetic induction, and H is the magnetic field. In static situations the electromagnetic energy can also be expressed in terms of sources and potentials as U EM = 1 2   ρφ + J · A c  dVol, (2) where ρ is the charge density, J is the current density, φ is the scalar potential and A is the vector potential. Time-dependent electromagnetic fields include radiation fields that effectively decouple from the sources. Verify that the expression (1) cannot in general be transformed into expression (2). The Lorentz invariant quantity E 2 − B 2 vanishes for radiation fields. Hence the integral  (E 2 − B 2 )dVol (3) excludes the contribution from the radiation fields, and remains related to the sources of the fields. Show that eq. (3) can be transformed into an integral of the invariant (j · A)= cρφ − J · A plus the time derivative of another integral. The quantity E 2 − B 2 has the additional significance of being the Lagrangian density of the “free” electromagnetic field [1], while ρφ − J · A/c is also considered to be the interaction term in the Lagrangian between the field and sources. The above argument indicates that the “free” fields retain a kind of memory of their sources, as they must. 2 Solution The aspects of this problem related to E 2 − B 2 were suggested by J.D. Jackson. To bring the potentials into the field energy (1), we recall that since ∇· B = 0 always, the magnetic induction can always be related to a vector potential A according to B = ∇× A. (4) 1