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fluidistic
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#7
Mar29-12, 02:51 PM
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Quote Quote by sunjin09 View Post
The δ(θ-pi/2) part of your expression is missing a 1/r factor, if you think about the difference between dθ and dz. What you want is an infinitesimal displacement in the z direction, i.e., dz=r*dθ at pi/2.

BTW in cylindrical system δ(r,r')=[itex]\delta(r-r')\frac{\delta(\phi-\phi')}{r}\delta(z-z') [/itex], where the 1/r is associated with the delta of [itex]\phi[/itex] coordinate, for the same consideration, but z coordinate is unaffected.
What I don't understand is why do I have to use "δ(r,r')"?
For instance in cylindrical coordinates, I made the ansatz that the solution to the problem is [itex]\rho (\vec x)=C \Theta (r-R) \delta (z)[/itex]. When integrated over all the space, this gave me [itex]C=\frac{Q}{\pi R^2}[/itex] leading to [itex]\rho (r,z)=\frac{Q}{\pi R^2}\Theta (r-R) \delta (z)[/itex] which coincides with the answer found on the Internet. I tried the exact same method in spherical coordinates but for some unknown reason to me, the answer found on the Internet differs from mine.
In either cases, I didn't use the expression for δ(r,r').