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Can somebody help show me how a pulsed spherical wave has a wavefunction of the form U(r,t) = (1/r)a(t-r/c), where a(t) is an arbitrary function, r is the radius of the spherical wave, t is time, and c is the speed of light?

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up vote 3 down vote accepted

It follows from the Laplacian in the spherical coordinates $$ \Delta f = {1 \over r} {\partial \over \partial r} \left( {\partial (r f) \over \partial r} \right) + {1 \over r^2 \sin \theta} {\partial \over \partial \theta} \left(\sin \theta {\partial f \over \partial \theta} \right) + {1 \over r^2 \sin^2 \theta} {\partial^2 f \over \partial \phi^2}. $$ I could have changed the $1/r^2$ to $1/r$ and similarly $r^2$ to $r$ as long as I correctly sandwiched two copies of $\partial / \partial r$ in between instead of one. It's the same operator.

The angular derivatives vanish for a spherically symmetric wave so the angular part of the Laplacian is $$ \frac{1}{r} \frac{\partial}{\partial r} \frac{\partial}{\partial r} r $$ and when we include the time derivative term from the box operator, both nonvanishing terms may be combined to $$ \frac{1}{r} \frac{\partial}{\partial r} \frac{\partial}{\partial r} r - \frac{1}{c^2} \frac{1}{r} \frac{\partial}{\partial t} \frac{\partial}{\partial t} r $$ up to an overall scaling that doesn't matter. The time-derivative commutes with $r$ or $1/r$ so I could add the redundant $1/r$ and $r$ to the term with the time derivatives, too. It's also equal to $$ \frac{1}{r} \left( \frac{\partial}{\partial r}+\frac{1}{c}\frac{\partial}{\partial t} \right) \left( \frac{\partial}{\partial r}-\frac{1}{c}\frac{\partial}{\partial t} \right) r $$

Now, when this operator acts on $$ U(r,t) = \frac{1}{r} a(t-r/c), $$ what happens is that first, the $1/r$ cancels against the $r$ at the end of the Laplacian. then the factor $a$ is annihilated by one of the parentheses because it only depends on $t-r/c$. The most general solution of this equation would be a combination of an incoming and outgoing wave, i.e. there could be a similar piece in $U$ that depends on $t+r/c$, i.e. $(1/r)b(t+r/c)$.

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Thanks for the answer. I'm just a bit confused by your very last step. How does the cancellation of the U(r,t) function by the operator prove what we set out to prove? – John Roberts Sep 22 '12 at 15:02
I suppose you were solving the equation $\Box U = 0$. ;-) A pulsed spherical wave has to be a solution to some equation, so if you didn't know what the equation is, although it may have been a mandatory part of the formulation of the question, I just told you what the right equation in the question was. ;-) – Luboš Motl Sep 22 '12 at 15:44

Here are my thoughts on a solution:

If we know that a pulsed planar wave can be expressed as U(r,t) = A(t-z/c)*exp(j*2*pi*v0*(t-z/c)) (where A(t-z/c) is the complex envelope and v0 is frequency), then a pulsed spherical wave can be equivalently expressed as U(r,t) = (A0/radius)*exp(j*2*pi*v0*(t-radius/c)), making the proper substitutions. If we then set A0*exp(j*2*pi*v0*(t-radius/c)) = a(t-radius/c), we will arrive at our required expression.

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