We pick up a couple things for reference, that will come in handy later. They both relate to this vector derivatives (grad, div, curl) that we’ve been talking about.

Important point: I’m going on and on about the divergence and the curl, because Maxwell’s equations specify the divergence and the curl of the E- and the B- fields.

The Helmholtz Theorem

So one question you could ask is….

If we know the divergence and curl of a field, do we know the field?

No.

If we know the divergence and curl of a field, and the value of the field at a boundary, do we know the field?

Yes.

In other words, we need boundary conditions in order to specify the field in many cases. We’ll get to boundary conditions officially in chapter 3 when we talk about potentials more.

Scalar Potential

If the curl of a field (say E) vanishes everywhere, then E can be written as the gradient of a scalar potential (V):

\[\nabla \times \vec{E} = \vec{0} ~~~ \rightarrow ~~~~ \vec{E} = -\nabla V.\]

Griffiths wants to make sure you know this is true for ANY curl-less vector field, not just for \(\vec{E}\) although \(\vec{E}\) is certainly the thing upon which we’re going to employ it..

Is \(E\) always curl-less (i.e. is \(\nabla \times \vec{E}\) always 0?)

No, but most of the time it is. Let’s see it two ways.

1) Consider a point charge. An electron. Which way does the field point? Towards it. Does it have any curl? No, there’s no “swirliness” to the E-field. It’s all divergence, and no curl (sounds like a hair-care ad.)

Okay, but that was for a point charge. Is that generally true? Yes! Any charge distribution can be made up of a sum of charges, and the E-field is the sum of the field due to each charge. So the E-field is a sum of curl-less fields. Is there any way to add a bunch of curl-less fields together and get a curl-ful field?

No, because the curl of a sum is the sum of the curls, so if the curl of each of the individual fields that you’re adding together is 0 then the curl of their sum is also zero.

2) Look at Maxwell’s 4th equation, aka Faraday’s law

\[\nabla \times \vec{E} = -\frac{\partial B}{\partial t}\]

So \(\nabla \times \vec{E} = 0\) as long as \(\vec{B}\) isn’t changing in time.

(That’s pretty interesting - I knew that a changing \(\vec{B}\) would induce an electric field, but I don’t think I appreciated that a changing \(\vec{B}\) would produce a curl-full E-field!! I’m looking forward to learning more about that.)

Vector Potential

AKA the potential that gives you \(\vec{B}\).

If a vector field (say \(\vec{B}\) for example) is divergence-less, i.e. \(\nabla \cdot \vec{B} = 0\) then \(B\) can be expressed as the the curl of some function, i.e.

\[\vec{B} = \nabla \times \vec{A}\]

where \(\vec{A}\) is called the vector potential.

Is \(\vec{B}\) always divergence-less? Yes, that’s Maxwell’s 2nd equation:

\[\nabla \cdot \vec{B} = 0\]

(Also called the “no magnetic monopoles” formula. If a magnetic monopole is ever found, that formula will change. It makes sense, right? that without a magnetic “charge” you can’t get that perfect diverging field that we think of for an electric charge.)

Two things to notice about the above

In both cases, i.e. \(\vec{E} = \nabla V\) and \(\vec{B} = \nabla \times \vec{A}\) the field is the derivative of some other function, which means that there’s an arbitrary element that can be added to \(V\) or \(\vec{A}\) the will not affect the field. In the case of \(V\) it’s any constant. In the case of $\vec{A}$ it’s the gradient of any scalar (because the curl of a gradient is mathematically (and therefore without exception) zero.)

Notice why you need that caveat that these things can only be expressed as the gradient (curl) of a scalar (vector)… if the curl (divergence) is zero…

E-field case

We need \(\nabla \times \vec{E} = 0\) (most of the time, by physics) and the curl of a gradient is 0 (all of the time, by math). So if \(\nabla \times \vec{E} \ne 0\) but \(\nabla \times (\nabla V) = 0\) then we can’t express \(E\) as \(\nabla V\).

B-field case

Similarly we need \(\nabla \cdot \vec{B} = 0\) (all of the time, by Maxwell’s equations, by physics) and the divergence \(\nabla \cdot (\nabla \times \vec{A}) = 0\) (all of the time, by math). So \(\nabla \cdot \vec{B}\) isn’t 0, then it’s a problem if \(\nabla \cdot (\nabla \times \vec{A})\) is 0, and the whole idea of a vector potential falls apart.