# What does it mean to decouple Maxwell's equations with the curl operator?

I'm reading Introduction to Electrodynamics 4th Edition by D.J. Griffiths where after listing the Maxwell equations in empty space (i.e. $$\rho = 0, \mathbf{J} = 0, \mu = \mu_0, \epsilon = \epsilon_o$$) he says on p.g. 393:

[The equations] constitute a set of coupled, first-order, partial differential equations for E and B. They can be decoupled by applying the curl to (iii) and (iv):

where equations (iii) and (iv) are:

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

$$\nabla \times \mathbf{B} = - \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}$$

Griffiths then goes on to apply the curl operator to both functions and arrives to the wave equation for electromagnetic waves.

I don't understand why we use the curl operator. It seems arbitrary to me without understanding what Griffiths means by coupled PDEs and decoupling them by using the curl operator. I know that the two equations rely on each other; one field affects the other and that is reflected in the equations, maybe that is what is meant by coupled.

My questions is made of two parts:

1. What does it mean for equations to be coupled, and why must we decouple them?

2. Why does using the curl operator decouple them?

• Take, e.g., the first equation: it has the time derivative of B, but the second one tells you that the curl of B is related to E. So, if you want an equation with ONLY the field E, you have to take the curl of the first eq and substitute the relation for curl B provided by the second one. That's why Griffiths takes the curl. – Quillo Nov 2 '20 at 14:45
• But why specifically the curl? – MrMineHeads Nov 2 '20 at 14:46
• It may sound "silly", but the reason is because the equations already contain the curl: how do you obtain an equation for E only given the two listed? ( obtaining a single equation for E, or B, only = decoupling the equations). – Quillo Nov 2 '20 at 14:48
• Okay so is it just that because the equations have a curl operator, using another curl decouples them, that is to say, we obtain equations for the fields that do not depend on the other field? – MrMineHeads Nov 2 '20 at 14:51
• Exactly, nothing "extraordinary" behind the use of the curl, it is just the natural way to decouple the equations. Try this game: put another operator (say a generic operation "L" that commutes with the time derivative) in the two Maxwell equations in place of the curl.. and try to decouple them. You will see that you have to apply "L". – Quillo Nov 2 '20 at 14:55

## 1 Answer

Coupling means simply that both equations involve both fields, so that you can't solve them separately.

Now, both equations involve a relation between the curl of one field and the other field. So if we take the first equation and apply curl, the right hand side will contain $$\nabla \times \mathbf{B}$$, which we can write in terms of $$\mathbf{E}$$ and get an equation written only in terms of the electric field: that's what decoupling means.

The curl only complicates a simple idea: if we have a pair of first order equations

\begin{align}\dot{f} &= a g \\ \dot{g} &= bf,\end{align}

how do we decouple them? Just take derivatives! After a bit of algebra, that will turn them into the two second order equations

\begin{align}\ddot{f} &= ab f \\ \ddot{g} &= ab g.\end{align}

As a final note, the curl is not the only option: you can also take time derivatives, go through the same steps, and arrive at the same result.