Spherical Waves

Imagine tossing a stone into a pond on a calm day. The surface disturbance that propagates out in all direction are circular waves. To extend this description to three dimensions we imagine a small pulsating sphere surrounded by a liquid. As the pulsating sphere expands and contracts it emits spherical waves that propagate out in all directions.

We use this as an example for a point source of light. This point source emits light in all directions uniformly. This type of source is said to be isotropic, and the wave fronts it generates are spherical and expand as the source propagates in all directions. In some cases it is more convenient to describe waves in curvilinear coordinates rather than Cartesian. Because this particular situation displays a high degree of symmetry we find it most convenient to describe it in spherical coordinates. To this end we introduce the Laplace operator in spherical coordinates

$LaTeX: \nabla ^2\equiv \frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial }{\partial r}\right)+\frac{1}{r^2 (\theta \sin )}\frac{\partial }{\partial \theta }\left(\theta \sin \frac{\partial }{\partial \theta }\right)+\frac{1}{r^2 (\theta \sin )^2}\frac{\partial ^2}{\partial \phi ^2}$ $\nabla ^2\equiv \frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial }{\partial r}\right)+\frac{1}{r^2 (\theta \sin )}\frac{\partial }{\partial \theta }\left(\theta \sin \frac{\partial }{\partial \theta }\right)+\frac{1}{r^2 (\theta \sin )^2}\frac{\partial ^2}{\partial \phi ^2}$

where again r, θ, and φ are defined in the figure

Spherical Polar Coordinates

and we have

$LaTeX: x=r \sin (\theta ) \cos (\phi ),\\ y=r \sin (\theta ) \sin (\phi ),\\$ $x=r \sin (\theta ) \cos (\phi ),\\ y=r \sin (\theta ) \sin (\phi ),\\$

and

$LaTeX: z=r \cos (\theta )$ $z=r \cos (\theta )$

In the present discussion we are looking for waves that show spherical symmetry i.e., ones that do not depend on θ or φ or

$LaTeX: \psi \left(\overset{\rightharpoonup }{r}\right)=\psi (r,\theta ,\phi )=\psi (r)$ $\psi \left(\overset{\rightharpoonup }{r}\right)=\psi (r,\theta ,\phi )=\psi (r)$

so that the Laplacian takes the simplified form

$LaTeX: \nabla ^2 \psi(r) \equiv \frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial }{\partial r} \psi(r) \right)$ $\nabla ^2 \psi(r) \equiv \frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial }{\partial r} \psi(r) \right)$

this particular form can be derived from the Cartesian representation by operating on the spherically symmetric wave function and then converting to polar coordinates for instance, in the x-direction

$LaTeX: \frac{\partial \psi (r)}{\partial x}=\frac{\partial \psi (r)}{\partial r} \frac{\partial r}{\partial x}$ $\frac{\partial \psi (r)}{\partial x}=\frac{\partial \psi (r)}{\partial r} \frac{\partial r}{\partial x}$

and the second derivative simplifies to

$LaTeX: \frac{\partial ^2\psi (r)}{\partial x^2}=\frac{\partial ^2\psi (r)}{\partial r^2} \left(\frac{\partial r}{\partial x}\right)^2+\frac{\partial ^2r}{\partial x^2} \frac{\partial \psi (r)}{\partial r}$ $\frac{\partial ^2\psi (r)}{\partial x^2}=\frac{\partial ^2\psi (r)}{\partial r^2} \left(\frac{\partial r}{\partial x}\right)^2+\frac{\partial ^2r}{\partial x^2} \frac{\partial \psi (r)}{\partial r}$ .

Because we have assumed spherical symmetry i.e.,

$LaTeX: \psi \left(\overset{\rightharpoonup }{r}\right)=\psi (r)$ $\psi \left(\overset{\rightharpoonup }{r}\right)=\psi (r)$

we have

LaTeX: r^2=x^2+y^2+z^2 $r^2=x^2+y^2+z^2$

and writing the partial derivatives of r w.r.t. x

$LaTeX: \frac{\partial r}{\partial x}=\frac{x}{r}$ $\frac{\partial r}{\partial x}=\frac{x}{r}$

so that we can now write

$LaTeX: \left(\frac{\partial r}{\partial x}\right)^2=\left(\frac{x}{r}\right)^2$ $\left(\frac{\partial r}{\partial x}\right)^2=\left(\frac{x}{r}\right)^2$

and

$LaTeX: \frac{\partial ^2r}{\partial x^2}=\frac{\partial }{\partial x}\frac{x}{r}=\left(x\frac{\partial }{\partial x}\frac{1}{r}+\frac{1}{r}\right)=\frac{1}{r}-\frac{x^2}{r^3}=\frac{1}{r}-\frac{x^2}{r^3}=\frac{1}{r}(1-\frac{x^2}{r^2})$ $\frac{\partial ^2r}{\partial x^2}=\frac{\partial }{\partial x}\frac{x}{r}=\left(x\frac{\partial }{\partial x}\frac{1}{r}+\frac{1}{r}\right)=\frac{1}{r}-\frac{x^2}{r^3}=\frac{1}{r}-\frac{x^2}{r^3}=\frac{1}{r}(1-\frac{x^2}{r^2})$

Note: if you are unsure of

$LaTeX: x\frac{\partial }{\partial x}\frac{1}{r}=-\frac{x^2}{r^3}$ $x\frac{\partial }{\partial x}\frac{1}{r}=-\frac{x^2}{r^3}$

remember that

$LaTeX: \frac{1}{r}=\frac{1}{\sqrt{x^2+y^2+z^2}}=(x^2+y^2+z^2)^{-1/2}$ $\frac{1}{r}=\frac{1}{\sqrt{x^2+y^2+z^2}}=(x^2+y^2+z^2)^{-1/2}$

inserting these into the second partial derivative of the wave function w.r.t. x

$LaTeX: \frac{\partial ^2 \psi(r) }{\partial x^2}=\frac{x^2}{r^2}\frac{\partial ^2 \psi(r) }{\partial r^2}+\frac{1}{r}(1-\frac{x^2}{r^2})\frac{\partial \psi(r) }{\partial r}$ $\frac{\partial ^2 \psi(r) }{\partial x^2}=\frac{x^2}{r^2}\frac{\partial ^2 \psi(r) }{\partial r^2}+\frac{1}{r}(1-\frac{x^2}{r^2})\frac{\partial \psi(r) }{\partial r}$

In order to form the Laplacian we need $LaTeX: \frac{\partial ^2\psi(r) }{\partial x^2}+\frac{\partial ^2 \psi(r)}{\partial y^2}+\frac{\partial ^2 \psi(r) }{\partial z^2}$ $\frac{\partial ^2\psi(r) }{\partial x^2}+\frac{\partial ^2 \psi(r)}{\partial y^2}+\frac{\partial ^2 \psi(r) }{\partial z^2}$ , adding these together and with a significant amount of simplification we get

$LaTeX: \nabla ^2 \psi(r) \equiv \frac{\partial ^2 \psi(r)}{\partial x^2}+\frac{\partial ^2 \psi(r)}{\partial y^2}+\frac{\partial ^2 \psi(r)}{\partial z^2}=\frac{\partial ^2 \psi(r)}{\partial r^2}+\frac{2}{r}\frac{\partial \psi(r)}{\partial r}$ $\nabla ^2 \psi(r) \equiv \frac{\partial ^2 \psi(r)}{\partial x^2}+\frac{\partial ^2 \psi(r)}{\partial y^2}+\frac{\partial ^2 \psi(r)}{\partial z^2}=\frac{\partial ^2 \psi(r)}{\partial r^2}+\frac{2}{r}\frac{\partial \psi(r)}{\partial r}$

in the case where

$LaTeX: \psi \left(\overset{\rightharpoonup }{r}\right)=\psi (r,\theta ,\phi )=\psi (r)$ $\psi \left(\overset{\rightharpoonup }{r}\right)=\psi (r,\theta ,\phi )=\psi (r)$

which is exactly what we had just from stating the Laplacian in spherical coordinates and throwing out terms that depended on angular derivatives of ψ(r). This can be written in a slightly different form as:

$LaTeX: \nabla ^2 \psi(r) =\frac{\partial ^2 \psi(r)}{\partial r^2}+\frac{2}{r}\frac{\partial \psi(r)}{\partial r}=\frac{1}{r}\frac{\partial ^2}{\partial r^2}(r \psi(r))$ $\nabla ^2 \psi(r) =\frac{\partial ^2 \psi(r)}{\partial r^2}+\frac{2}{r}\frac{\partial \psi(r)}{\partial r}=\frac{1}{r}\frac{\partial ^2}{\partial r^2}(r \psi(r))$

Now--using this in the wave equation

$LaTeX: \nabla ^2\psi =\frac{1}{v^2}\partial _t^2\psi$ $\nabla ^2\psi =\frac{1}{v^2}\partial _t^2\psi$

we have

$LaTeX: \frac{1}{r}\frac{\partial ^2}{\partial r^2}(r \psi) =\frac{1}{v^2}\partial _t^2\psi$ $\frac{1}{r}\frac{\partial ^2}{\partial r^2}(r \psi) =\frac{1}{v^2}\partial _t^2\psi$

multiplying both sides by r, we have

$LaTeX: \frac{\partial ^2}{\partial r^2}(r \psi) =\frac{1}{v^2}\partial _t^2(r \psi)$ $\frac{\partial ^2}{\partial r^2}(r \psi) =\frac{1}{v^2}\partial _t^2(r \psi)$

If we make the definition

$LaTeX: \Psi (r)\equiv r \psi (r)$ $\Psi (r)\equiv r \psi (r)$

we see that

$LaTeX: \frac{\partial ^2}{\partial r^2}\Psi =\frac{1}{v^2}\partial _t^2\Psi$ $\frac{\partial ^2}{\partial r^2}\Psi =\frac{1}{v^2}\partial _t^2\Psi$

we arrive at the one dimensional wave equation. The solution is then

$LaTeX: \Psi (r)=f(r+v t)+g(r-v t)$ $\Psi (r)=f(r+v t)+g(r-v t)$

and inserting $LaTeX: \Psi (r)=r \psi (r)$ $\Psi (r)=r \psi (r)$

$LaTeX: r \psi (r)=f(r+v t)+g(r-v t)$ $r \psi (r)=f(r+v t)+g(r-v t)$

dividing both sides by r,

$LaTeX: \psi (r)=\frac{f(r+v t)}{r}+\frac{g(r-v t)}{r}$ $\psi (r)=\frac{f(r+v t)}{r}+\frac{g(r-v t)}{r}$

represents a sum of two spherical waves, one expanding outward (r - v t) and one converging toward the origin (r + v t). The fact that these solutions approach infinity as r->0 means that they cannot represent a physical quantity there. This is due to the fact at least in optics, that light that appears to be a spherical wave very far from the source is in fact generated by a different physical mechanism than a small sphere contracting and expanding at the origin. The physical source of spherical light waves (to first order) is known as dipole radiation and will be discussed in the next module.

A special case of the general solution is

$LaTeX: \psi (r)=\frac{\mathcal{A} \cos (k r\pm k v t)}{r} ,$ $\psi (r)=\frac{\mathcal{A} \cos (k r\pm k v t)}{r} ,$

which is a harmonic spherical wave. The complex representation takes the form

$LaTeX: \psi (r)=\frac{\mathcal{A} e^{i k(r\pm v t)}}{r}$ $\psi (r)=\frac{\mathcal{A} e^{i k(r\pm v t)}}{r}$

As a spherical wave propagates outward, its radius increases. Far enough away from the source, a small area of the wavefront will very closely approximate a plane wave.