I am a bit confused as to how we obtain regular wave equations from Maxwell's equations when the vector Laplacian is defined the way it is. We have the differential equation for waves in the form: $$ \nabla^{2} \mathbf{E}=\mu \epsilon \frac{\partial^{2} \mathbf{E}}{\partial t^{2}}+\mu \sigma \frac{\partial \mathbf{E}}{\partial t} $$ and solutions of the form: $$\tilde{\mathrm{E}}(z, t)=\tilde{\mathrm{E}}_{0} e^{i(\bar{k} z-\omega t)}$$ (this is from Maxwell's equations for a conductor).

When we expand the vector Laplacian we get: $$\nabla^{2} \mathbf{E}= \left(\nabla^{2} E_{x}, \nabla^{2} E_{y}, \nabla^{2} E_{z}\right)$$ but then we have the issue of having to expand these laplacians out to find for example in the x directions: $$\frac{\partial^{2} E_x}{\partial x^{2}}+\frac{\partial^{2} E_x}{\partial y^{2}}+\frac{\partial^{2} E_x}{\partial z^{2}}=\mu \epsilon \frac{\partial^{2} E_x}{\partial t^{2}}+\mu \sigma \frac{\partial E_x}{\partial t}$$ which is not the standard wave equation.

Is this the real form and when we solve the wave equation we are assuming some components are zero or have I messed up somewhere/am not understanding it correctly?

  • 1
    $\begingroup$ Rob Jeffries' answer is incorrect. There is nothing wrong with your defintion of the Laplacian. $\endgroup$ – Milan Nov 8 at 17:05
  • $\begingroup$ What do you mean by “the standard wave equation”? $\endgroup$ – G. Smith Nov 8 at 17:06
  • $\begingroup$ Well in my class it is simplified to $\frac{\partial^{2} E_{x}}{\partial x^{2}}=\mu \epsilon \frac{\partial^{2} E_{x}}{\partial t^{2}}+\mu \sigma \frac{\partial E_{x}}{\partial t}$ or similar. I am trying to understand the reason for neglecting the other terms. And what would it mean for them to be present. How does that change the solutions? $\endgroup$ – user244685 Nov 8 at 17:07
  • $\begingroup$ Your class has simplified it for some particular case. That equation applies only when the field doesn’t depend on $y$ or $z$. The equation with $x$, $y$, and $z$ derivatives is standard and more general. $\endgroup$ – G. Smith Nov 8 at 17:26

If you are dealing with E-field in vacuum or in a neutral medium then Gauss's law tells you that $$\nabla \cdot {\bf E} = 0$$

For any wave solution of the form ${\bf E} = {\bf E_0} f({\bf k}\cdot {\bf r} - \omega t)$ then this mean that ${\bf E_0}\cdot {\bf k} = 0$.

Thus if your E-field is in the x-direction then $k$ must be perpendicular to the x-direction and ${\bf k} \cdot {\bf r} = k_y y + k_z z$ (in general).

That means your final equation has one term that is zero $$\frac{\partial^2 E_x}{\partial x^2} = 0$$

Thus $$\frac{\partial^2 E_x}{\partial y^2} + \frac{\partial^2 E_x}{\partial z^2} = \frac{\partial^{2} E_x}{\partial t^{2}}+\mu \sigma \frac{\partial E_x}{\partial t}$$

This is still a wave equation, that allows for the wave to be travelling in any direction perpendicular to the $x$-axis. The final term is just a damping term that makes the amplitude of the wave die away.

  • $\begingroup$ That is the same form as $\nabla^{2} \mathbf{A}=\left(\nabla^{2} A_{x}, \nabla^{2} A_{y}, \nabla^{2} A_{z}\right)$ right? In that case, wont I still end up with what I stated? $\endgroup$ – user244685 Nov 8 at 16:44
  • $\begingroup$ But when you expand it out and add the vectors they come out identical. $\endgroup$ – user244685 Nov 8 at 17:01
  • $\begingroup$ Yes, I agree. @user244685 $\endgroup$ – Rob Jeffries Nov 8 at 17:25
  • $\begingroup$ So the other terms are usually omitted to make for easier calculations I am assuming? $\endgroup$ – user244685 Nov 8 at 17:34
  • 1
    $\begingroup$ @user244685 If working in Cartesian coordinates it is usual to assume that the wave is travelling along one of the axes. If that is the case then (e.g.) $k_y =0$ for a wave travelling along the z-axis and $\partial^2 E_x/\partial y^2=0$. $\endgroup$ – Rob Jeffries Nov 8 at 17:36

In general, a first order derivative in a wave equation represents a dissipative term (think of ODE's: friction for a damped harmonic oscillator or resistance in an RLC circuit).

This term is present in your wave equation because you are dealing with a conducting medium. The amplitude of your resulting EM wave will decay exponentially.

  • $\begingroup$ Thank you for answering but this is not actually what I was asking. What I meant was that I do not understand why the second spatial derivatives in both y and z are missing (well, two of the three at least) such that the usual form of $\frac{\partial^{2} E_{x}}{\partial x^{2}}=\mu \epsilon \frac{\partial^{2} E_{x}}{\partial t^{2}}+\mu \sigma \frac{\partial E_{x}}{\partial t}$ is seen. Why do we omit those terms and what would they add? $\endgroup$ – user244685 Nov 8 at 17:22
  • $\begingroup$ I'm sorry I misread. This is the standard wave equation: $\nabla^2E_x=\mu \epsilon\dfrac{\partial^2E_x}{\partial t^2}+\mu \sigma\dfrac{\partial E_x}{\partial t}$. Only when the second order derivates wrt $y$ and $z$ of $E_x$ disappear, does the equation take on the form you posted. $\endgroup$ – Milan Nov 8 at 17:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.