The utility of thinking of 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 and heading outward
from the source at the propagation velocity .
In this case the "rays" point along
the radius vector from the source
at any position and we can once again write down a rather simple
formula for the "wave function"
(displacement as a function of position)
that depends only on the time and the *scalar*
distance from the source.

A plausible first guess would be just
,
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 , but the *area*
of that wavefront increases as .
Thus the energy in the wavefront increases as ?
I think not. We can get rid of this effect by just dividing
the amplitude by (which divides the energy density by ).
Thus a trial solution is

We won't use this equation for anything right now,
but it is interesting to know that it does accurately describe
an outgoing^{14.12}spherical 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 and ?
The correct *vector* form of the WAVE EQUATION is

With a little patient effort you can show that Eq. (38) does indeed satisfy Eq. (39), if you remember that . Or you can just take my word for it . . . .

- . . . 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

- . . . Cartesian
^{14.13} - The LAPLACIAN operator can also be represented in other coordinate systems such as spherical () or cylindrical () 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 with itself:
, where

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.

Jess H. Brewer - Last modified: Sun Nov 15 18:06:25 PST 2015