Pre-Quantum Electrodynamics

For a completely generic situation, we can more formally compute the divergence and curl of \({\bf B}\) directly from Biot-Savart (here taking for example the volume current density version BiotSavart_v)

\[ {\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' ~\frac{{\bf J}({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \] Applying the divergence gives \[ {\boldsymbol \nabla} \cdot {\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' ~{\boldsymbol \nabla} \cdot \left(\frac{{\bf J}({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \right) \label{Gr(5.46)} \] Note that we can write (using div1or) \(\frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} = -{\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|}\). Substituting this in and using product rule div_xprod, we get

\begin{align} {\boldsymbol \nabla} &\cdot \left(\frac{{\bf J}({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \right) = -{\boldsymbol \nabla} \cdot \left( {\bf J}({\bf r}') \times {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} \right) \nonumber \\ &= -{\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} \cdot \bigl({\boldsymbol \nabla} \times {\bf J} ({\bf r}') \bigr) + {\bf J} ({\bf r}') \cdot \left({\boldsymbol \nabla} \times {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|}\right) \label{Gr(5.47)} \end{align}

But \({\bf J}\) depends only on \({\bf r}'\) so \({\boldsymbol \nabla} \times {\bf J} ({\bf r}') = 0\), and since the curl of a gradient always vanishes, we obtain

The divergence of a magnetic field is always zero. Said equivalently (and thinking of the divergence theorem and the magnetic equivalent of Gauss's law): there are no magnetic charges.