Maxwell’s Equations and the Helmholtz Wave Equation

There are four Maxwell equations, which you can find in many places.  I’ll repeat them here, but I want to give you some feeling for what the equations mean.  After all, we’re not mathematicians, interested in equations for their own sake.  We’re interested in what equations tell us about the physical world.

The first Maxwell equation is called Ampere’s Law:

$latex \displaystyle \nabla \times \vec{H} = \vec{J} + \frac{\partial \vec{D}}{\partial t} $

where H is the magnetic field, J is the electrical current density, and D is the electric flux density, which is related to the electric field.   In words, this equation says that the curl of the magnetic field equals the electrical current density plus the time derivative of the electric flux density.  Physically, this means that two things create magnetic fields curling around them: electrical current, and time-varying (not static) electric fields.  The typical example of this is a vertical current-bearing wire with magnetic field lines looping around it:

(insert picture)

The second Maxwell equation is called Faraday’s Law:

$latex \displaystyle \nabla \times \vec{E} = – \frac{\partial \vec{B} }{\partial t} $

where E is the electric field and B is the magnetic flux density, which is related to the magnetic field.  This is called Faraday’s Law, and similar to Ampere’s Law, it tells us that time-varying magnetic fields create electric fields curling around them.

What does ‘related’ mean, anyways?  Well, it turns out that

$latex \displaystyle \vec{D} = \epsilon \vec{E} $


$latex \displaystyle \vec{B} = \mu \vec{H} $

which don’t really count as Maxwell equations – they’re called ‘constitutive relations’ – but they’re still very important.  $latex \epsilon $ is the permittivity, and $latex \mu $ is the permeability, both of which are properties of whatever material you’re in (air, glass, water, plastic, metal, etc.)

If we consider Ampere’s law and Faraday’s law (only two Maxwell equations – there are two more, we’ll get to them in a second!) we see that time-varying electric fields create magnetic fields curling around them, and time-varying magnetic fields create electric fields curling around them.  This is why electromagnetic waves can exist, and can carry energy far away from their source (billions of light-years, in the case of distant galaxies):  the electric and magnetic fields can support one another.

The third Maxwell equation is Gauss’ Law:

$latex \displaystyle \nabla \cdot \vec{D} = \rho $

where $latex \rho $ is the electric charge density.  This equation tells us that charge creates electric fields diverging from it.  Think of that charged metal sphere you grabbed as a kid to make your hair stand up.  The electric field lines were radiating outward from it.

And finally, the fourth Maxwell equation, which is nameless:

$latex \displaystyle \nabla \cdot \vec{B} = 0$

which tells us that magnetic fields don’t diverge from anything, they only curl around.  Or equivalently: there is no such thing as magnetic charge, at least not that we’ve found so far.

Since we’re mostly interested in electromagnetic waves here, and in particular light waves, we have to convert the Maxwell equations into a form that easily yields wave-like solutions.  To accomplish this, we will derive the Helmholtz wave equation from the Maxwell equations.

We’ve discussed how the two ‘curl’ equations (Faraday’s and Ampere’s Laws) are the key to electromagnetic waves.  They’re tricky to solve because there are so many different fields in them: E, D, B, H, and J, and they’re all interdependent.  So our goal will be to combine those two equations into a single equation with a single field in it.

First, let’s assume we’re in a uniform material, so that the permittivity epsilon and the permeability mu are constants – they don’t change in space or in time.  Of course, that can’t be true for the entire universe, but it can be approximately true in some limited-size region, which is where we’ll solve Maxwell’s equations for now.  We’ll also pick a region that has zero conductivity, and therefore zero electrical current density J.  That gets rid of one of our fields right away.  Of course our solution won’t be entirely general, because it won’t necessarily apply to regions with nonzero conductivity, but we can fix that up later.

Our next goal will be to somehow get rid of the magnetic field on the right hand side of Faraday’s law, and replace it with an expression involving the electric field.  That way we’d have an equation with only one field – E – on both sides of the equation.  How can we accomplish this?  Well, we know that Ampere’s law relates the curl of the magnetic field to the electric field, so we’re going to take the curl of both sides of Faraday’s law:

$latex \displaystyle \nabla \times \nabla \times \vec{E} = – \frac{\partial ( \nabla \times \vec{B}) }{\partial t} $

I’ve brought the curl inside the time derivative, but that’s ok – it’s just interchanging the order of differentiation.  We’ve certainly made Faraday’s law look messier, how does it help us?  Well, let’s rewrite Ampere’s law using our constitutive relations, and getting rid of J:

$latex \displaystyle \frac{1}{\mu} \nabla \times \vec{B} = \epsilon \frac{\partial \vec{E}}{\partial t} $


$latex \displaystyle \nabla \times \vec{B} = \epsilon \mu \frac{\partial \vec{E}}{\partial t} $

OK, so now I have an expression that allows me to replace the curl of B with an expression involving E.  Let’s swap that into our modified Faraday’s law:

$latex \displaystyle \nabla \times \nabla \times \vec{E} = – \epsilon \mu \frac{\partial^2 \vec{E} }{\partial t^2} $

Mission accomplished!  We’ve condensed the two Maxwell curl equations down into a single equation involving nothing but E.  This is one form of the Helmholtz wave equation, although not necessarily the nicest form to solve, since it has the curl of a curl on the left hand side.  We can use some vector identities to simplify that a bit.  One useful vector identity is the following:

$latex \displaystyle \nabla \times \nabla \times \vec{A} = \nabla ( \nabla \cdot \vec{A} ) – \nabla ^2 \vec{A} $

where $latex \vec{A} $ is any vector field.  One tedious but reliable way of deriving this relatively wacky-looking vector identity is to write out all of the vector components and derivatives; in the time-honored words of many distinguished textbook writers, ‘we leave this as an exercise for the reader’.

The vector identity doesn’t really seem to simplify things much – it just allows us to replace the curl of the curl with a different complicated-looking expression – but we can improve things by putting one more restriction on the region where we’re solving these equations; namely, that there is no ‘free’ (unbound) electrical charge in that region, or $latex \rho = 0$.  In that case, Gauss’ Law becomes

$latex \displaystyle \nabla \cdot \vec{D} = 0 = \nabla \cdot (\epsilon \vec{E}) = \epsilon \nabla \cdot \vec{E} $

where we’ve assumed that $latex \epsilon$ doesn’t depend on position, allowing us to take it outside of the derivative.  Dividing both sides by $latex \epsilon$ finally gives us

$latex \displaystyle \nabla \cdot \vec{E} = 0 $

You’d be excused for wondering what the point is of all this.  We’re getting to it!  Let’s go back to our vector identity and replace generic field $latex \vec{A}$ with electric field $latex \vec{E}$ :

$latex \displaystyle \nabla \times \nabla \times \vec{E} = \nabla ( \nabla \cdot \vec{E} ) – \nabla ^2 \vec{E} $

You see that $latex \nabla \cdot \vec{E}$ in there?  Now we know it’s zero, as long as we’re in a region with no charge, and as long as the permittivity $latex \epsilon $ is constant with position.  So now we have a pretty nice simplification; namely

$latex \displaystyle \nabla \times \nabla \times \vec{E} = -\nabla ^2 \vec{E} $

I’m going to put that back into the Helmholtz equation, to give me.. uh, still the Helmholtz equation:

$latex \displaystyle -\nabla^2 \vec{E} = – \epsilon \mu \frac{\partial^2 \vec{E} }{\partial t^2} $

usually we gather everything on one side:

$latex \displaystyle \nabla^2 \vec{E} – \epsilon \mu \frac{\partial^2 \vec{E} }{\partial t^2}  = 0 $

Whew!  And that is the Helmholtz wave equation.


Solution of Helmholtz equation on separate page