Re: [Phys-l] spherical waves

[David Bowman <David_Bowman@georgetowncollege.edu>]

Regarding Kyle's question:

> Hi;
> I have mightily confused myself, maybe someone can straighten
> me out. A spherical wave spreads out so obviously the energy
> per area (intensity) decreases as 1/r2, we expect that just
> from geometric considerations.

Yep.

> But does the amplitude also decrease?

Yep.  The energy density and flux is typically related to the square of the amplitude, so as the energy density and flux decay as 1/r^2 the amplitude must typically decay as 1/r in the far field.

> A simple y(r,t)=Asin(kr-wt) with r being a polar coordinate
> says no (and a plane wave in a perfect medium does not).

The function you wrote here does not a solution or even asymptotic solution of a typical wave equation (i.e. one whose spatial deriviatives are a Laplacian).  If you divided the above function by r the quotient would be such a solution (at least a spherically symmetric L=0 one which would correspond to monopole radiation, say, from a point-like source of sound).

> My intuition is that it seems like if you are far away from a 
> source the amplitude can't be the same but amplitude squared is 
> proportional to energy. So is the intensity also decreasing
> because amplitude is decreasing?

You have almost answered your own question.

> But that would mean the 1/r2 dependence for energy/area is wrong.

That part is correct.  What is wrong is your hypothesis that the y(r,t) function you wrote above was a solution of the appropiate wave equation.

> Several wave books (Berkeley physics waves p 372) I have show
> that the Poynting vector of the radiation from an accelerating
> charge is proportional to charge squared, acceleration squared
> and 1/r^2.

In the nonrelativistic limit for the source's motion.

> But how does that relate to amplitude of the wave?

For EM waves the monopole spherically symmetric L=0 solution is not allowed because they are vector waves, not scalar waves, and have no spherically symmetric solution.  (Think of trying to comb the hair on a hairy sphere).  The lowest order multipole solution has L = 1 (i.e. the angular symmetry of a chemist's p-orbital).  But regardless of the spherical harmonics defining the angular dependence of the wave, their amplitudes all fall off with distance as 1/r in the far field.  For scalar spherical waves they typically are made of linear combinations of spherical harmonics for the angular dependence times spherical Bessel functions for the radial dependence times trig functions or (complex exponentials) for the temporal dependence.  For spherical vector waves they are made of linear combinations of *vector* spherical harmonics, (and for spherical spinor waves they are made of linear combinations of spinor spherical harmonics).  In the far field the radial dependences of all these kinds of wave functions each asymptotically fall off as 1/r with distance from the source.

> kyle

David Bowman