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How do I generate a system of equations that describe an electromagnetic wave? Maxwell's equations are very generalized, which is good, but I want to create a concrete example (with specific numbers and functions and vector fields) of a mathematical system of an electromagnetic wave in free space from scratch.

From digging around in Wikipedia I think I could somehow use electromagnetic potentials and the Lorenz gauge condition to satisfy Maxwell's equations, but I'm not really sure where to begin. Admittedly, the math is a little over my head, but I'm pretty sure I can pick up most of it, and it's the reason I'd like to learn, anyhow.

Here's what I've found.

Maxwell's equations in free space

$$\nabla \cdot \mathbf{E} = 0$$

$$\nabla \cdot \mathbf{B} = 0$$

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

$$\nabla \times \mathbf{B} = \frac{1}{c^{2}} \frac{\partial \mathbf{E}}{\partial t}$$

Electric and magnetic potentials (Lorenz gauge)

$$\mathbf{B} = \nabla \times \mathbf{A}$$

$$\mathbf{E} = - \nabla \phi - \frac{\partial \mathbf{A}}{\partial t}$$

$$\nabla \cdot \mathbf{A} + \frac{1}{c^{2}} \frac{\partial \phi}{\partial t} = 0$$

If $\phi_{1}$ and $\mathbf{A}_{1}$ are another solution of the same electromagnetic wave, then there should exist a scalar function $\lambda$ of position and time such that

$$\phi_{1} = \phi - \frac{\partial \lambda}{\partial t}$$

and

$$\mathbf{A}_{1} = \mathbf{A} + \nabla \lambda$$

Using the Lorenz gauge condition means that $\lambda$ must satisfy

$$\nabla^{2} \lambda - \frac{1}{c^{2}} \frac{\partial^{2} \lambda}{\partial t} = 0$$

which is the wave equation.

Quantum electrodynamics

If $\psi = \lambda, \phi, \mathbf{A}, \mathbf{E}$ or $ \mathbf{B}$, then $\psi$ satisfies the wave equation.

$$\nabla^{2} \psi - \frac{1}{c^{2}} \frac{\partial \psi}{\partial t} = 0$$

The solutions to $\psi$ in one dimension are arbitrary functions $f$ and $g$, such that

$$\psi = f( x - ct) + g( x + ct)$$

In three dimensions, $\psi$ would be the superimposition of the one-dimensional solution such that

$$\psi = f( x - ct) + g( x + ct) + f( y - ct) + g( y + ct) + f( z - ct) + g( z + ct)$$

(not sure about this)

In conclusion,

Any directions on how I should proceed? Do I just pick arbitrary functions $f$ and $g$ as solutions to $\lambda$ and then work up from there? Are there restrictions/conditions I should be aware of? Any more information I would need?

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closed as too broad by AccidentalFourierTransform, ZeroTheHero, Cosmas Zachos, user191954, hyportnex Nov 19 '18 at 0:13

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    $\begingroup$ Question is too broad and you should say what standard treatments of this problem you have looked at. $\endgroup$ – Rob Jeffries Oct 28 '18 at 20:48
  • $\begingroup$ You can safely remove lots of this, including the Maxwell equations. They're well-known enough. $\endgroup$ – user191954 Oct 29 '18 at 12:30
  • $\begingroup$ If I understand you correctly, are you aware of plane waves? $\endgroup$ – Ofek Gillon Oct 29 '18 at 22:25
  • $\begingroup$ @Chair I am sure that everyone else on this site knows perfectly well the stuff written here. I just wanted to show what I found to give an approximation of what I have an understanding of. $\endgroup$ – Ilyankor Oct 30 '18 at 2:13
  • $\begingroup$ @OfekGillon I am not. I've read somewhere about them, but I don't understand them very well. $\endgroup$ – Ilyankor Oct 30 '18 at 2:15
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Uh, I'm not sure I understood you correctly. If you want an example of how you get the wave equation from electromagnetic equations, it's pretty simple. LEt's choose vacuum, or any linear isotropic homogeneous medium, but vacuum is the easiest one.

You ONLY need Maxwell's equations, so forget about the rest.

Then, the key is recalling this vector identity:

$$\vec{\nabla}\times(\vec{\nabla}\times\vec{v})=\vec{\nabla}\cdot(\vec{\nabla}\cdot\vec{v})-\vec{\nabla}^2\vec{v}, \qquad\forall\vec{v}$$

In particular, for the electric field:

$$\vec{\nabla}\times(\vec{\nabla}\times\vec{E})=\vec{\nabla}\cdot(\vec{\nabla}\cdot\vec{E})-\vec{\nabla}^2\vec{E}$$

But now, you plug MAxwell's equatiosn inside. You see that the divergence of $\vec{E}$ would be 0 in absence of charges (vaccuum)

$$\vec{\nabla}\times(\vec{\nabla}\times\vec{E})=-\vec{\nabla}^2\vec{E}$$

But, on the other Hand, replacing another Maxwell equation yields:

$$\vec{\nabla}\times(\vec{\nabla}\times\vec{E})=\vec{\nabla}\times\left(-\mu \frac{\partial}{\partial t}\vec{H}\right)=-\mu \frac{\partial}{\partial t} (\vec{\nabla}\times\vec{H})$$

$$ =-\mu\frac{\partial}{\partial t} \left(\epsilon \frac{\partial\vec{E}}{\partial t}\right)=-\mu\epsilon \frac{\partial^2 \vec{E}}{\partial t}$$

So, you get that

$$\nabla^2 \vec{E}=\mu\epsilon \frac{\partial^2 \vec{E}}{\partial t}$$ The wave equation with velocity $\dfrac{1}{\sqrt{\epsilon\mu}}$, which is $c$ in case of vacuum.

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  • $\begingroup$ Yes. I understand most of that. But what is a concrete example of a solution to that partial differential equation? And $\vec{B} $ (or $\vec{H} $)? Like, idk, make something up: $\vec{E} = t^2 sin(xy) \vec{i} + y \cos (3z^2)e^{4x} \vec{j} + \left \cos^{-1} (t - yx) - t + e^{izy-x} \right \vec{z} $? It's a system of 8 unknowns, it's hard to keep track of all the variables and conditions. Is there a way to simplify this process? $\endgroup$ – Ilyankor Nov 4 '18 at 9:09
  • $\begingroup$ The equation $\nabla^2 f = \frac{1}{v^2}\frac{\partial^2 f}{\partial t^2}$ is the wave equation, the famous "wave equation", you can look for information just searching for "wave equation". And, going to your point, the solutions depend on the boundary conditions. In case of vacuum, the solution is a superposition of plane waves of the form $\vec{E}=\vec{A}\cdot e^{i(\omega t - \vec{k}\cdot\vec{r})}$. You can check by yourself. $\endgroup$ – FGSUZ Nov 4 '18 at 11:29

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