BELIEVE   ME   NOT!    - -     A   SKEPTIC's   GUIDE  

Spherical Waves

The utility of thinking of $\Vec{k}$ as a "ray" becomes even more obvious when we get away from plane waves and start thinking of waves with curved wavefronts. The simplest such wave is the type that is emitted when a pebble is tossed into a still pool - an example of the "point source" that radiates waves isotropically in all directions. The wavefronts are then circles in two dimensions (the surface of the pool) or spheres in three dimensions (as for sound waves) separated by one wavelength $\lambda$ and heading outward from the source at the propagation velocity $c$. In this case the "rays" $k$ point along the radius vector $\Hat{r}$ from the source at any position and we can once again write down a rather simple formula for the "wave function" (displacement $A$ as a function of position) that depends only on the time $t$ and the scalar distance $r$ from the source.

A plausible first guess would be just   $A(x,t) \; = \; A_{_0} \; e^{i(k r - \omega t)}$,  but this cannot be right! Why not? Because it violates energy conservation. The energy density stored in a wave is proportional to the square of its amplitude; in the trial solution above, the amplitude of the outgoing spherical wavefront is constant as a function or $r$, but the area of that wavefront increases as $r^2$. Thus the energy in the wavefront increases as $r^2$? I think not. We can get rid of this effect by just dividing the amplitude by $r$ (which divides the energy density by $r^2$). Thus a trial solution is

$$\mbox{ \fbox{ \rule[-1.25\baselineskip]{0pt}{3\baselineskip . . . . . . \; = \; A_{_0} \; { e^{i(k r - \omega t)} \over r } . }$~ }}$$ (14.38)

which is, as usual, correct.^14.11The factor of $1/r$ accounts for the conservation of energy in the outgoing wave: since the spherical "wave front" distributes the wave's energy over a surface area $4\pi r^2$ and the flux of energy per unit area through a spherical surface of radius $r$ is proportional to the square of the wave amplitude at that radius, the integral of $\vert f\vert^2$ over the entire sphere (i.e. the total outgoing power) is independent of $r$, as it must be.

We won't use this equation for anything right now, but it is interesting to know that it does accurately describe an outgoing^14.12spherical wave.

The perceptive reader will have noticed by now that Eq. (38) is not a solution to the WAVE EQUATION as represented in one dimension by Eq. (10). That is hardly surprising, since the spherical wave solution is an intrinsically 3-dimensional beast; what happened to $y$ and $z$? The correct vector form of the WAVE EQUATION is

$$\mbox{ \fbox{ \rule[-1.25\baselineskip]{0pt}{3\baselineskip . . . . . . c^2} \; {\partial^2 A \over \partial t^2} \; = \; 0 }$~ }}$$ (14.39)

where the LAPLACIAN operator  $\nabla^2$  can be expressed in Cartesian^14.13coordinates ($x,y,z$) as^14.14

$$\nabla^2 \; = \; {\partial^2 \over \partial x^2} \; + \; . . . . . . over \partial y^2} \; + \; {\partial^2 \over \partial z^2} .$$ (14.40)

With a little patient effort you can show that Eq. (38) does indeed satisfy Eq. (39), if you remember that   $r = \sqrt{x^2 + y^2 + z^2}$. Or you can just take my word for it . . . .

Footnotes

. . . correct.^14.11

I should probably show you a few wrong guesses first, just to avoid giving the false impression that we always guess right the first time in Physics; but it would use up a lot of space for little purpose; and besides, "knowing the answer" is always the most powerful problem-solving technique!

. . . outgoing^14.12

One can also have "incoming" spherical waves, for which Eq. (38) becomes

$$A(x,t) \; = \; A_{_0} \; { e^{i(k r + \omega t)} \over r } .$$

. . . Cartesian^14.13

The LAPLACIAN operator can also be represented in other coordinate systems such as spherical ($r,\theta,\phi$) or cylindrical ($\rho,\theta,z$) coordinates, but I won't get carried away here.

. . . as^14.14

The LAPLACIAN operator can also be thought of as the inner (scalar or "dot") product of the GRADIENT operator  $\Vec{\nabla}$  with itself:   $\nabla^2 = \Vec{\nabla} \cdot \Vec{\nabla}$,  where

$$\Vec{\nabla} \; = \; \Hat{\imath} {\partial \over \partial x} . . . . . . over \partial y} \; + \; \Hat{k} {\partial \over \partial z}$$

in Cartesian coordinates. This VECTOR CALCULUS stuff is really elegant - you should check it out sometime - but it is usually regarded to be beyond the scope of an introductory presentation like this.