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

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

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

- () = k_{n}^{2}() ,
| (21) |

d^{ 3}r()() =
| (22) |

d^{ 3}re^{ - i }e^{ i' } = (2)^{3}( - ')
| (23) |

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

G(,t',,t) = a_{n}(t)()
| (24) |

() = - k_{n}^{2}c^{ 2}a_{n}(t)() - 4c^{ 2}( - )(t - t') .
| (25) |

= - k_{m}^{2}c^{ 2}a_{m}(t) + 4c^{ 2}()(t - t') .
| (26) |

G^{ (R)}(,t,,t')
| = |
4c^{ 2}(t - t')sin(k_{n}c[t - t']) .
| |

G^{ (A)}(,t,,t')
| = |
- 4c^{ 2}(t' - t)sin(k_{n}c[t - t']) .
| (27) |

(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) |

G^{ (R)}(,t,,t')
| = |
(t - t')dksin(k| - |)sin (kc[t - t'])
| |

= |
(t - t')dk
| ||

= |
(t - t')dk
| ||

= | |||

= |
(t - t' - c^{ - 1}| - '|)
| (30) |

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

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^{ 3}r'dt'G(,t,,t')f (',t')
| |

- t'=t_{i}^{t' = tf} + t'=t_{i}^{t' = 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 *t*_{i} allows us to drop the time surface terms. Therefore for
this case, we can write

(,t) = d^{ 3}r' .
| (33) |

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

(,t) = b_{n}(t)()
| (34) |

- 4f (,t)
| (35) |

= - k_{n}^{2}c^{ 2}b_{m}(t) + 4d^{ 3}rf (,t)()
| (36) |