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# 6 Application to the Wave Equation

The wave equation with a source is

 (,t) = - 4f (,t) . (19)

and the Green's function is the solution with a delta function source,

 (',t';,t) = - 4( - r')(t - t') (20)

where, as before, the Green's function is generally the solution of the adjoint equation, but the wave equation is self adjoint.

To use the harmonic oscillator result, we want to eliminate the spatial operators. We can do this exactly as we did for the Green's function for Poisson's Equation. Expanding in the eigenfunctions of . We write

 - () = kn2() , (21)

where the eigenfunctions are determined by the spatial boundary conditions that we enforce at the boundaries of our system. For free space, the eigenfunctions are plane waves, and the transformation to eigenfunctions is the same Fourier transform that we used for the Coulomb case. Eigenfunctions in a finite volume such as a cavity, are typically normalized to one,

 d 3r()() = (22)

while the free space Fourier transform is typically normalized to Dirac delta functions using

 d 3re - i e i' = (2)3( - ') (23)

which was derived in class by taking the limit of the periodic box normalized solutions.

The general solution of the wave equation is given by expanding the Green's function in the complete set of eigenfunctions,

 G(,t',,t) = an(t)() (24)

where an(t) are the expansion coefficients which also depend on and t' , but to keep the notation simple we will not display that dependence explicitly. Substituting the expansion into the wave equation gives

 () = - kn2c 2an(t)() - 4c 2( - )(t - t') . (25)

We multiply this by () and integrate using the orthogonality to get

 = - km2c 2am(t) + 4c 2()(t - t') . (26)

This equation is identical with our harmonic oscillator Green's function equation, with chosen appropriately. As before, we can choose any boundary conditions we like. We will choose either advanced or retarded boundary conditions. Substituting our harmonic oscillator result, the advanced and retarded Green's functions are
 G (R)(,t,,t') = 4c 2(t - t')sin(knc[t - t']) . G (A)(,t,,t') = - 4c 2(t' - t)sin(knc[t - t']) . (27)
where we have used the reality of the solution to put the complex conjugate on the other eigenfunction. We have also enforced the retarded boundary and advanced conditions by using the theta function defined by

 (x) = . (28)

Let's work out the result for free space. The eigenfunctions of are plane waves, (that is you can simply Fourier transform the space part of the wave equation.) The result is

 G (R)(,t,,t') = 4c(t - t')e i ( - ') . (29)
Integrating over the angular part gives
 G (R)(,t,,t') = (t - t')dksin(k| - |)sin (kc[t - t']) = (t - t')dk = (t - t')dk = = (t - t' - c - 1| - '|) (30)
where in the last line, the theta function restricts the time such that the second delta function can never contribute. The calculation for the advanced delta function goes through identically, except in that case the theta function restricts the time so the first delta function never contributes. The minus signs cancel and the result is

 G (A)(,t,,t') = (t - t' + c - 1| - '|) . (31)

Notice these agree with Jackson Eq. 6.44.

Let's calculate the solution of the wave equation given that (,t) is zero at all times before the source f (,t) is turned on. We repeat the same mathematics as before, where we take the wave equation with a source and the Green's function equation, cross multiply, subtract, and use Green's theorem (i.e. the divergence theorem or integration by parts) to arrive at

 (,t) = d 3r'dt'G(,t,,t')f (',t') - t'=tit' = tf + t'=tit' = tf + dt'da'. (32)

By taking infinite space, with the initial field zero everywhere, the disturbance never makes it to the edge'' of the volume, so we can drop the spatial surface terms. If we were calculating the response in a cavity where was specified (e.g. zero) on the cavity surface, we would want to choose our Green's function to go to zero on the surface so that we could drop the surface terms proportional to the normal derivative of , just as in the Poisson case. Similarly, as in the harmonic oscillator, if we use the retarded Green's function the upper limits of the time surfaces give zero, while our boundary condition that the field is zero at ti allows us to drop the time surface terms. Therefore for this case, we can write

 (,t) = d 3r' . (33)

A related but alternative method to using the Green's function is to use the eigenfunction expansion directly for (,t) ,

 (,t) = bn(t)() (34)

where bn(t) are time dependent coefficients. Substituting into the wave equation gives

 - 4f (,t) (35)

multiplying by () and integrating gives the equations of motion for the coefficient bm(t) ,

 = - kn2c 2bm(t) + 4d 3rf (,t)() (36)

which shows that each mode amplitude bm(t) behaves exactly like a driven harmonic oscillator. Any method to solve these differential equations can be used to produce a solution, and by summing over the modes a complete solution is obtained.

Next: 7 Quantum Harmonic Oscillator Up: Green's functions for the Previous: 5 A Simple Example