Pre-Quantum Electrodynamics

Suppose we have a piece of material with known magnetization \({\bf M}\). What is the field produced by this object? For a single dipole: refer to \ref{Gr(5.83)} (vector potential of single dipole): \[ {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \frac{{\bf m} \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \label{Gr(6.10)} \] For a chunk of material with local magnetization \({\bf M} ({\bf r})\), by the principle of superposition we thus have:

In principle, this is all that is needed. As in electric case however, a more illuminating version of this equation can be given by using some simple identities: using \({\boldsymbol \nabla}' \frac{1}{|{\bf r} - {\bf r}'|} = \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3}\), \[ {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' {\bf M} ({\bf r}') \times \left( {\boldsymbol \nabla}' \frac{1}{|{\bf r} - {\bf r}'|} \right), \] we can further integrate by parts and use product rule: \({\boldsymbol \nabla} \times (f {\bf A}) = f ( {\boldsymbol \nabla} \times {\bf A}) - {\bf A} \times ({\boldsymbol \nabla} f)\), we get \[ {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \left\{ ∫_{\cal V} dτ' \frac{{\boldsymbol ∇}' × {\bf M} ({\bf r}')}{|{\bf r} - {\bf r}'|}

• ∫_{\cal V} dτ' {\boldsymbol ∇}' × \left( \frac{{\bf M} ({\bf r}')}{|{\bf r} - {\bf r}'|} \right)

\right\} \] Problem 1.61 b) (p.56): leads to \[ {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} ∫_{\cal V} dτ' \frac{{\boldsymbol ∇}' × {\bf M} ({\bf r}')}{|{\bf r} - {\bf r}'|}

• \frac{\mu_0}{4\pi} \oint_{\cal S} \frac{{\bf M} ({\bf r}') × d{\bf a}'}{|{\bf r} - {\bf r}'|}

\label{Gr(6.12)} \] Reinterpretation: first term: potential from volume current,

second term: potential from surface current,

With these definitions,

so the field produced by the material is the same as that produced by {\bf bound currents} in the volume and surface of the material.