Writing up the charge distribution with Dirac's delta

Click For Summary
SUMMARY

The discussion focuses on representing charge distributions using Dirac's delta function in spherical coordinates, particularly for point charges located along the z-axis. The user initially struggles with defining the charge density function, ##\rho (\vec x)##, in spherical coordinates due to the ambiguity of the angles phi and theta at the origin and along the z-axis. The user explores several formulations and ultimately concludes that the correct representation is ##\rho (\vec x) = q \cdot k \frac{\delta (\theta) \delta (r - d_1)}{r^2 \sin \theta}##, where k is determined by the integral of the Jacobian, specifically k = 1/(2π) for their example.

PREREQUISITES
  • Understanding of electrostatics and charge distributions
  • Familiarity with Dirac's delta function
  • Knowledge of spherical coordinates and their Jacobians
  • Basic calculus and integration techniques
NEXT STEPS
  • Study the properties and applications of Dirac's delta function in physics
  • Learn about the Jacobian transformation in coordinate systems
  • Explore charge distributions in different coordinate systems, including cylindrical coordinates
  • Investigate the Green's function method in electrostatics
USEFUL FOR

Students and professionals in physics, particularly those focusing on electrostatics, mathematical physics, and anyone working with charge distributions and coordinate transformations.

fluidistic
Gold Member
Messages
3,932
Reaction score
283

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:

Similar threads

Replies
5
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K