Poisson equation for the field of an electron

Click For Summary
SUMMARY

The discussion centers on the application of the Poisson equation in classical electrodynamics, specifically for the scalar field \(\phi(r)\) produced by an electron at the origin. The equation is expressed as \(\nabla^2\phi(r) = -4\pi e\delta(r)\), leading to the radial dependence \(\phi(r) = \frac{e}{r}\). Participants emphasize the necessity of understanding the Laplacian operator \(\nabla^2\) and its application in spherical coordinates, as well as the importance of multi-dimensional calculus concepts such as gradient, divergence, and delta-functions for deriving the solution.

PREREQUISITES
  • Understanding of the Laplacian operator \(\nabla^2\) in three dimensions
  • Familiarity with gradient and divergence theorems in vector calculus
  • Knowledge of delta-functions and their representation in spherical coordinates
  • Completion of Calculus III or equivalent coursework
NEXT STEPS
  • Study the derivation of the Poisson equation in spherical coordinates
  • Learn about the properties and applications of the delta-function in physics
  • Review the concepts of gradient and divergence in multi-dimensional calculus
  • Explore resources on the Laplacian operator and its physical interpretations
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying electrodynamics, as well as educators and anyone seeking to deepen their understanding of vector calculus and its applications in physics.

Keonn
Messages
1
Reaction score
0

Homework Statement


In classical electrodynamics, the scalar field \phi(r) produced by an electron located at the origin is given by the Poisson equation
\nabla^2\phi(r) = -4\pi e\delta(r)

Show that the radial dependence of the field is given by
\phi(r) = \frac er

Homework Equations


I'm not really sure if this is right, but this is what I found on wikipedia. I've never learned about this ∇^2 thing in a math class.
\nabla^2 \phi =\frac {\partial^2 \phi}{\partial r^2} + \frac 1r \frac {\partial \phi}{\partial r}

The Attempt at a Solution


Plugging in e/r for phi, you get:
\nabla^2 \phi = \frac {2e}{r^3} - \frac {e}{r^3} = \frac {e}{r^3}
which isn't even close to the equation given in the problem. I am so lost, please help.
 
Physics news on Phys.org
What you found looks more like the ##\nabla^2## in polar coordinates (2D). In the same encyclopaedia I found (here) something else for three dimensions -- more like the world where electrons manifest themselves. With spherical symmetry you can focus on the r part.

You also want to think what you want to do with the ##\delta## function. It has a nice habit of being integrable (if that's an existing word). And then there is a famous theorem to link E on a surface to charge enclosed (named after a German named Carl Friedrich).

But perhaps there is a more direct route.
 
Keonn said:
I've never learned about this ∇^2 thing in a math class.

Hi. This is a bit of a problem because the Laplacian is usually covered in Calculus III, which would be a pre-requisite for E&M... You would also need to know about the three major theorems of 3-dimensions calculus (gradient, divergence and curl), and finally about multi-dimensional delta-functions and how to express them in non-cartesian coordinates.
If you don't know about these, it's going to be a long way for you to derive the relation you are given. I can outline the first steps:
1 - express the delta-function in spherical coordinates (by the way, your equation should read δ3(r), not δ(r))
2 - express the Laplacian as ∇⋅∇Φ, then use the theorem for gradients to integrate over all space.
The whole derivation takes four lines, but you really need to know about 3-dimensional calculus...
 
2 is equivalent the divergence of the gradient of a scalar function ƒ: ∇⋅∇(ƒ).
I suggest reading about gradient and divergence if you're not familiar with these (plenty of sources online)

To provide a brief intro:
∇ (aka del or nabla operator) can be denoted <\frac{∂}{∂x_1}, \frac{∂}{∂x_2}, ..., \frac{∂}{∂x_n}>. Note there is no function following the ∂ in the numerator.

Let ƒ be a function ƒ(x1, x2, ..., xn) = ξ, where ξ is a scalar
The Gradient ∇ƒ maps the output of ƒ on ℝ to ℝn (Think scalar f multiplied by vector ∇)
∇ƒ = <\frac{∂ξ}{∂x_1}, \frac{∂ξ}{∂x_2}, ..., \frac{∂ξ}{∂x_n}>

Let ϑ be a function ϑ(x1, x2, ..., xn) = <ϑ_1, ϑ_2, ..., ϑ_n>
The Divergence ∇⋅ϑ maps the output of some function on ℝn to ℝ (Think dot product of ∇ and ϑ)
∇⋅ϑ = \frac{∂ϑ}{∂x_1} + \frac{∂ϑ}{∂x_2} + ... + \frac{∂ϑ}{∂x_n}

If we apply both to the first function ƒ(X) = ξ, where ∇(ƒ) = ϑ, we map the output on ℝ, to ℝn and back to ℝ
∇⋅∇(ƒ) = \frac{∂^2ξ}{∂x_1^2} + \frac{∂^2ξ}{∂x_2^2} + ... + \frac{∂^2ξ}{∂x_n^2}
// as a note, ∇⋅∇ is more commonly written ∇2 or Δ, known as the Laplacian
 
Last edited:

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
11
Views
3K
Replies
29
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
Replies
10
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K