Showing that an equation satisfied the helmholtz equation

[warfreak131]

Homework Statement

Show that \epsilon(r)=\frac{A}{r}e^{ikr} is a solution to \nabla^{2}\epsilon(r)+k^{2}\epsilon(r)=0

Homework Equations

The Attempt at a Solution

Is \nabla^{2} in this case equal to \frac{\partial^2}{\partial r^2} or \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}?
I know that using r simplifies things rather than using x, y, z, but I am not sure if I am doing it correctly.

 

[dextercioby]

Well, of course you need to express the 3-d Laplacian in spherical coordinates, because the calculations look simpler than in Cartesian ones.

 

[warfreak131]

but the equation only uses the vector R, not phi or theta. e is only a function of r, so wouldn't it be like saying take the derivative of this function w.r.t. r, then take the derivative of a constant with respect to theta, then the derivative of a constant with respect to phi?

 

[ideasrule]

but the equation only uses the vector R, not phi or theta. e is only a function of r, so wouldn't it be like saying take the derivative of this function w.r.t. r, then take the derivative of a constant with respect to theta, then the derivative of a constant with respect to phi?

Yes. The Laplacian is equivalent to <br /> (\frac{\partial^2}{\partial r^2},0,0)<br /> in spherical coordinates.

 

[Troponin]

Yes. The Laplacian is equivalent to <br /> (\frac{\partial^2}{\partial r^2},0,0)<br /> in spherical coordinates.

Because there is no \theta or \phi dependence.


Just wanted to clarify because I've seen students make the conclusion that the laplacian is always <br /> (\frac{\partial^2}{\partial r^2},0,0)<br /> after doing their first "make use of the spherical symmetry" lapacian in spherical coordinates...