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?

  • 3
    $\begingroup$ Split into real and imaginary parts $\endgroup$ Dec 5, 2014 at 11:52

1 Answer 1


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.

  • $\begingroup$ thank you for your nice answer. I have one question though, why would one do that? I mean why combine these two equations with one another? And why did we use the imaginary number "i"? Is their some physical interpretation behind this? $\endgroup$ Dec 5, 2014 at 12:57
  • $\begingroup$ @Fluctuations you should continue reading the paper to find out why one would do this... $\endgroup$
    – Danu
    Dec 5, 2014 at 12:57

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