Maxwell’s equations

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.


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