# Combination of Maxwell equations and other form of Maxwell equations

In reference to this paper on arXiv, page three, we have the following:

We know that the Bianchi Identites are $\partial_{[\alpha F_\beta\gamma]} = 0$ and are equivalent to

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

We also know that given a Lagrangian $L$ one may define $G^{\mu\nu}$ by: $$G^{\mu\nu} = -2 \frac{\partial L}{\partial F_{\mu\nu}}$$

and equivalently,

$$\nabla \cdot D =0$$ $$\nabla \times H = +\frac{\partial D}{\partial t}$$

The field equations and Bianchi identites may be combined in the form

$$\nabla \cdot (D +iB) =0$$

$$\nabla \times (E +iH) = i\frac{\partial}{\partial t} (D+iB)$$

My question is the last line, how were these combined based on what's before?

• Split into real and imaginary parts – Holographer Dec 5 '14 at 11:52

Like Holographer's comment points out, what one does is the following: First, multiply the equations with $B$ and $H$ on the left hand side by $i$: $$i\nabla\cdot B=\nabla\cdot(iB)=0,\hspace{1cm} \text{and}\hspace{1cm}\nabla\times (iH)=\frac{\partial( iD)}{\partial t}$$ The key observation is that the divergence and curl do not mix real and imaginary components. Now, we simply sum the equations pairwise to obtain the required result:
$$\nabla \cdot (D +iB) =0$$
$$\nabla \times (E +iH) = \frac{\partial}{\partial t} (iD-B)=i\frac{\partial}{\partial t}(D+iB)$$ Note that these two equations are equivalent to Maxwell's equations: Because the derivatives don't mix the real and imaginary parts, they must vanish independently.