electromagnetism | Mamuta memuāri

This is how the differential Maxwell’s equations look like using vector notation:

$\dfrac{\partial\mathbf{B}}{\partial t} = - \nabla \times \mathbf{E}$ $\nabla \cdot \mathbf{B} = 0$

$\dfrac{\partial\mathbf{E}}{\partial t} = \nabla \times \mathbf{B}$ $\nabla \cdot \mathbf{E} = 0$

It turns out that these equations can be written in a more compact form:

$d\mathrm{F} = 0$ $d{*}\mathrm{F} = 0$

known as the differential geometric formulation.

The main idea to derive this alternative formulation is as follows. First, we define the electromagnetic tensor which contains the components of both $\mathbf{E}$ and $\mathbf{B}$ :

$F = \begin{pmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0 \end{pmatrix}$

This matrix is skew-symmetric, so we can encode it as a 2-form:

$\mathrm{F} = \mathrm{E}(\mathbf{E}) + \mathrm{B}(\mathbf{B})$

where the 2-forms $\mathrm{E}(\mathbf{E})$ and $\mathrm{B}(\mathbf{B})$ encode the entries of vectors $\mathbf{E}$ and $\mathbf{B}$ in the following way:

$\mathrm{E}(\mathbf{v}) = v_x \, dt \wedge dx + v_y \, dt \wedge dy + v_z \, dt \wedge dz$

$\mathrm{B}(\mathbf{v}) = v_x \, dz \wedge dy + v_y \, dx \wedge dz + v_z \, dy \wedge dx$

Next, one defines an operation known as Hodge dual and checks that

${*}\mathrm{E}(\mathbf{v}) = \mathrm{B}(\mathbf{v})$ ${*}\mathrm{B}(\mathbf{v}) = -\mathrm{E}(\mathbf{v})$

Finally, one has to define the exterior derivative and check that

$d\mathrm{E}(\mathbf{v}) = dt \wedge \mathrm{B}(\nabla \times \mathbf{v})$

$d\mathrm{B}(\mathbf{v}) = dt \wedge \mathrm{B} \biggl( \dfrac{\partial \mathbf{v}}{\partial t} \biggr) - (\nabla \cdot \mathbf{v}) \, dx \wedge dy \wedge dz$

Using these equations we can rewrite $d\mathrm{F} = 0$ and $d{*}\mathrm{F} = 0$ in terms of $\mathbf{E}$ and $\mathbf{B}$ and verify that the resulting equations are equivalent to the original Maxwell’s equations.

For more details see: Differential geometric formulation of Maxwell’s equations.