# Solutions To Spherical Wave Equation

1. Sep 18, 2013

### RESolo

If the solution to the electric part of the spherical wave equations is:

E(r, t) = ( A/r)exp{i(k.r-ωt)

What happens when t=0 and the waves originates at the origin, i.e. r=0 ... which I assume can't be right as you of course cannot divide by zero.

Thanks!

2. Sep 18, 2013

### mikeph

The short answer is: don't try to evaluate a wave due to a point source at the location of the point source!

The wave equation contains a Laplacian operator, which is undefined in spherical polar coordinates at the origin, so if the domain of the wave equation doesn't include the origin, you shouldn't assume that a solution exists there.

3. Sep 18, 2013

### BruceW

well, we could always use Cartesian coordinates instead. So I wouldn't say that the wave equation doesn't include the origin. More that this particular solution diverges at the origin.

4. Sep 18, 2013

### dauto

That must be old news for anybody that has seen Coulombs law (which also diverges at the origin). That's the way it is...

5. Sep 19, 2013

### mikeph

But the detail here is that there is no physically divergent quantity at the origin, because the wave equation is derived from the homogeneous free space Maxwell equations.

The wave equation in Cartesian coordinates is isotropic and homogeneous.

But somewhere between converting to spherical polars and solving for some exp(i(k.r-wt)) type wave, the mathematics have become inhomogeneous in that there is no solution at r=0.

6. Sep 19, 2013

### BruceW

the physically divergent quantity arises when we choose a solution that is a/r exp(i(k.r-wt)), we are effectively saying there is a point source of the wave. (the solution agrees with the free-space solution everywhere except at r=0, which makes sense if the source is at r=0). This is similar to Coulomb's law, where we say that we have a point source at the origin.

7. Sep 19, 2013

### mikeph

Ok, that makes sense, but surely even writing the wave equation using a spherical polar form of the Laplace operator will generate unavoidable 1/r terms? The damage is already done by the coordinate transform stage.

8. Sep 19, 2013

### WannabeNewton

I don't get the issue here. Would you also object to the Coloumb potential $\varphi = \frac{q}{r}$ being a solution to $\nabla^{2}\varphi = -4\pi \rho$, where $\rho = q\delta^{3}(r)$, because of the point charge source? Clearly not. In this situation, the Laplacian at the point charge is handled using distributions. That is $\nabla^{2}\varphi = -4\pi q\delta^{3}(r) = -4\pi \rho$.

9. Sep 20, 2013

### mikeph

I'm not saying the spherical polar wave equation can't be solved, I'm saying it should be undefined at the origin. It's valid for r>0, no? The derivative d(phi)/dr at r=0 is undefined because the r coordinate has no direction.

I understand how delta functions work, but I'm not even discussing the solution, just the mathematics of restricting your domain when you do a coordinate transform. Plus the wave equation is homogeneous, so you don't have any localised source term embedded in the equations like you do with Poisson's equation. r=0 essentially becomes a place where boundary conditions can be specified.

10. Sep 20, 2013

### BruceW

you can tell that the solution a/r exp(i(k.r-wt)) has a source at the origin, by integrating the Poynting vector over a spherical surface. (you might want to take a time-average also), and then you see that energy is flowing away from the origin, so there must be something physically going on at the origin.

11. Sep 20, 2013

### mikeph

Yep, your Poynting vector integral will be nonzero, but you'll never actually evaluate the divergence of S at r=0 because you never solved the equation at that point.

12. Sep 20, 2013

### BruceW

hmm... there's two different things that I think you might be saying, and I'm not sure which one it is.
1) the divergence in spherical polar coordinates has 1/r terms, so we should not be allowed to take the divergence at r=0 using spherical polar coordinates.
2) with a non-zero r, as we take the limit of r tends to zero, the divergence of the Poynting vector will diverge to infinity.

now, 2) is only true when we have a point source. This is the case for the solution in the O.P. This is a wave created by a point source, so close to the point source, there is an infinite amount of energy transmitted per volume (in loose terms). But 2) is not true when we have some finite charge distribution that causes spherical waves. In this case, the divergence of the Poynting vector is always finite, even when we allow r to tend to zero.

As for 1) yeah, I guess that is true. But this is not a fundamental problem. We can just switch to cartesian coordinates for the case r=0. So if we have a spherical wave that is created by a finite charge distribution, this will work, and we can calculate the divergence of the Poynting vector at r=0. And if our spherical wave is created by a point source at the origin, then again, quantities will diverge to infinity when we approach r=0.

13. Sep 21, 2013

### Claude Bile

Coordinates are undefined at the origin in the spherical coordinate system (since the angular terms are undefined).

So the op's point is moot.

Claude.

14. Sep 21, 2013

### BruceW

really? Can't we just define something arbitrary, like $\theta =\pi$ and $\phi =\pi$ when $r=0$ ?

15. Sep 21, 2013

### dauto

You can, but that choice is not unique

16. Sep 21, 2013

### BruceW

we don't need a unique choice. we just make whatever choice we want to.

edit: I am not a mathematician, so I am not certain about this stuff. But I swear I read somewhere that when using spherical polar coordinates, we can just give $\theta$ and $\phi$ arbitrary values at the origin, so that our co-ordinate system does not have any 'missing points'.