# Poisson equation for the field of an electron

1. Feb 5, 2015

### Keonn

1. The problem statement, all variables and given/known data
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
$\phi(r) = \frac er$

2. Relevant 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.
$\nabla^2 \phi =\frac {\partial^2 \phi}{\partial r^2} + \frac 1r \frac {\partial \phi}{\partial r}$

3. 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.

2. Feb 5, 2015

### BvU

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.

3. Feb 5, 2015

### Goddar

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...

4. Feb 5, 2015

### Brian T

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: Feb 5, 2015