Showing that an equation satisfied the helmholtz equation

AI Thread Summary
The discussion focuses on demonstrating that the function ε(r) = A/r * e^(ikr) satisfies the Helmholtz equation, ∇²ε(r) + k²ε(r) = 0. Participants clarify that in spherical coordinates, the Laplacian simplifies to ∂²/∂r² when there is no dependence on the angles θ or φ. This simplification is crucial for correctly applying the Laplacian to the given function. Additionally, there is a caution against assuming the Laplacian is always in the form of (∂²/∂r², 0, 0) without considering the specific context of spherical symmetry. The conversation emphasizes the importance of understanding the application of the Laplacian in different coordinate systems.
warfreak131
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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.
 
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Well, of course you need to express the 3-d Laplacian in spherical coordinates, because the calculations look simpler than in Cartesian ones.
 
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?
 
warfreak131 said:
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.
 
ideasrule said:
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...
 
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