Solving Maxwell Equations for the Electric Field: Is it Possible? - A Discussion

[Gavroy]

hi

i asked myself whether it is possible to solve the partial differential equation rot E=-dB/dt
for the electric field.
i assumed that at least for a few right hand sides this should be possible, but i have never seen anybody doing this.

 

[Polyrhythmic]

You want to know if you can solve that single equation? The Maxwell equations can be combined so that one arrives at wave equations for the electric and magnetic field, which have wave solutions.

 

[vanhees71]

It depends also on what you mean by "solve" the equation. If you want the electric field as function of a given magnetic field, this is not sufficient.

You should look for Helmholtz's Theorem in vector calculus. It explains how to construct a given vector fields when its curl and its sources are given. According to this theorem, the field can be split, under certain assumptions on boundary conditions uniquely, in a potential field and a solenoidal field, i.e.,

\vec{V}=\vec{V}_1+\vec{V}_2 \quad \text{with} \vec{\nabla} \times \vec{V}_1=0, \quad \vec{\nabla} \cdot \vec{V}_2=0.

 

[Gavroy]

what do you exactly mean by "not suffiecient"? Let me say, if I have a magnetic field B=At/r

where A is some real number. Is there a possibility now to calculate the electric field?

 

[Bill_K]

You can find a solution to this equation very easily.
∇ x E = - ∂B/∂t. Take the curl of both sides:

∇ x (∇ x E) ≡ ∇ ∇·E - ∇²E = - ∂(∇ x B)/∂t

Let's look for a particular solution with ∇·E = 0. We have

∇²E = ∂(∇ x B)/∂t with solution given by Poisson's integral,

E(x) = ∫(1/|x - x'|) ∂(∇ x B(x'))/∂t d³x'

The resulting solution does satisfy ∇·E = 0.

 

[vanhees71]

Well, who says that

\vec{\nabla} \cdot \vec{E}=0?

That's what I meant, it's not sufficient to have only the induction Law and a given magnetic field to calculate the electric field. In addition, you also need the sources of the electric field, which is the charge distribution.

 

[Gavroy]

@ Bill_k
sorry, but I have some doubts about your solution.

my problem is, that there are always when you solve a differential equation some conditions that the solution must satisfy, but this integral gives only one solution, so:

How can you be sure that this is the right one?
or:
are you sure that your solution is always applicable?

 

[cragar]

you could integrate both sides over the boundary and use stokes theorem to get it to a line integral on one side. But you probably already know this.

 

[Gavroy]

yes i do, but thank you too.

my intention was to get the electric field. bill k already answered my question, but i am not sure whether his equation is always meaningful, as it is just A solution and maybe not always the correct one.

so now maybe someone can tell me a bit more about the meaning of bill's equation