2
$\begingroup$

The wave-equation is given by $$\nabla^2E=\frac{1}{c^2}\frac{\partial ^2E}{\partial t^2} $$

And I'm trying to prove that this wave $ E(r,\theta,\phi)= \frac{A_o}{r}sin(\theta) cos(\omega t - \frac{\omega}{c}r) \space\hat{\theta}$ satisfies the wave equation.

I've already performed the laplacian on the LHS and the second derivative of time on the RHS. However I end up with a chunk of algebra which does not seem very elegant to me.

So I'm wondering if there's a better way to prove this?

If anyone could direct me to readings / articles or lay out roughly how I should approach it, that would be great!

EDIT: Added my working

In spherical coordinates on LHS, the laplacian is given by

$$\nabla^2E= \frac{\partial^2E}{\partial r^2} + \frac{2}{r}\frac{\partial E}{\partial r} + \frac{\cot(\theta)}{r^2}\frac{\partial E}{\partial \theta} + \frac{1}{r^2}\frac{\partial ^2E}{\partial \theta^2} + \frac{1}{r^2sin^2(\phi)}\frac{\partial ^2E}{\partial \phi^2}$$

After working out the details, and I let $(\omega t - \frac{\omega}{c}r) = \chi $

$\frac{\partial E}{\partial r}= \frac{\omega }{c}\frac{A}{r}\sin(\theta)\sin(\chi)-\frac{A}{r^2}\sin(\theta)cos(\chi)$

$\frac{\partial^2E}{\partial r^2}= 2\frac{A}{r^3}\sin(\theta)\cos(\chi)-\frac{\omega^2}{c^2}\frac{A}{r}\sin(\theta)\cos(\chi)$

$\frac{\partial E}{\partial \theta} = \frac{A}{r}\cos(\theta)\cos(\chi)$

$\frac{\partial^2 E}{\partial \theta^2}= -\frac{A}{r}\sin(\theta)\cos(\chi)$

$\frac{\partial E}{\partial t} = -\frac{\omega A}{r}\sin(\theta)\sin(\chi)$

$\frac{\partial ^2E}{\partial t^2} = -\frac{\omega^2 A}{r}\sin(\theta)\cos(\chi)$

So putting them all together, $$ LHS = 2\frac{A}{r^2}\frac{\omega}{c}\sin^2(\theta)\sin(\chi) - \frac{A}{r}\frac{\omega^2}{c^2}\sin^2(\theta)\cos(\chi) + \frac{A}{r^3}\cos(\chi)[\cos^2(\theta) - \sin^2(\theta)]$$

$$ RHS = -\frac{1}{c^2}\frac{A\omega^2}{r}\sin(\theta)\cos(\chi)$$

I suppose from here on, I'll have to use some trigonometric trick to get to what I need. But was just wondering if there is any other way.

$\endgroup$
  • $\begingroup$ Show us your calculations. $\endgroup$ – MathematicalPhysicist Jan 27 '16 at 10:00
  • $\begingroup$ this looks like the heat equation $\endgroup$ – Ellya Jan 27 '16 at 11:25
  • $\begingroup$ @ellya I believe looks very similar to it. The heat (diffusion) equation is the first partial derivative w.r.t time. I forgot to add a ^2 to the time derivative. It has now been fixed. But thanks for pointing that out! $\endgroup$ – Candy Man Jan 28 '16 at 8:17

Your Answer

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

Browse other questions tagged or ask your own question.