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 outgoing14.12spherical 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 and ?
The correct vector form of the WAVE EQUATION is