Writing up the charge distribution with Dirac's delta

fluidistic
Gold Member
Messages
3,928
Reaction score
272

Homework Statement


In electrostatics it's useful to have ##\rho (\vec x )## written with Dirac's delta so that we can know the total charge by integrating the charge distribution over a region of space.
Many problems/situations deal with point charges. In Cartesian coordinates for example, ##\rho (\vec x ) = q \delta (x+3) \delta (y ) \delta (z)## means there's a charge q situated at (-3,0,0).
My question is, how do you write up the charge distribution of a point charge in spherical coordinates ##(r, \theta , \phi )## when the charge lies over the z-axis? (or at the origin for example).
Because in such a coordinate system, the angle "phi" is not well defined for the z-axis. And at the origin both phi and theta are not well defined. So I don't know how to write the charge distribution in such cases.
Thank you!
 
Physics news on Phys.org
I've tried a few things, like if a charge q is at ##(d_1,0,0)## in Cartesian coordinates. In spherical coordinates I tried
1)##\rho (\vec x ) =q \frac{\delta (\theta ) \delta (r-d_1 ) \delta ( \phi )}{r^2 \sin \theta }##. In this case I reached what I should, namely that ##q=\int \rho (\vec x ) d ^3x##.
2)##\rho (\vec x ) =q \frac{\delta (\theta ) \delta (r-d_1 ) \delta ( \phi - \text {any value between 0 and 2 \pi} )}{r^2 \sin \theta }## which also works well.
3)##\rho (\vec x ) =q \frac{\delta (\theta ) \delta (r-d_1 ) }{r \sin \theta }## which doesn't yield ##q= \int \rho (\vec x ) d^3 x## so that's really bad.
4)Other variants that didn't work.

Overall, only when I chose an arbitrary value for phi, I could get a sensical result. However I'm not satisfied that any value for phi work, because when I use ##\rho (\vec x )## in other equations, it makes a big difference what the value of phi is, and not any value will work.
Example: For the Green function in the case of a conducting sphere at potential 0, the potential is given by ##\Phi (\vec x ) = \int G(x,x') \rho (x' )d^3x'## where ##G(x,x')## does depend on phi and therefore ##\Phi## (the potential) will depend on the value I choose for phi in the Dirac's delta for rho, the charge density.
Since the potential is uniquely determined, only 1 value for phi work. I just don't know how to find it. Thus my question, how to write rho (x) in spherical coordinates when there's a charge on the z-axis or on the origin.

Edit: I found out the solution.
It's to write up ##\rho (\vec x ) =q*k \frac{\delta (\theta ) \delta (r-d_1 ) }{r^2 \sin \theta }## where k=1/(2 pi) in my example. In general it's just evaluating the integral of the Jacobian of the coordinate system and integrate with respect to the "missing" Dirac's delta. That's worth the denominator. So in my example it's just 2 pi.
 
Last edited:
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top