Poisson equation for the field of an electron

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Homework Help Overview

The discussion revolves around the Poisson equation in classical electrodynamics, specifically concerning the scalar field produced by an electron. The original poster is attempting to understand the radial dependence of the field as described by the equation involving the Laplacian operator.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the Laplacian operator in spherical coordinates and the implications of the delta function in the context of the problem. There are suggestions to express the delta function in spherical coordinates and to consider the integration over all space.

Discussion Status

Some participants have provided guidance on the necessary mathematical concepts, including the gradient and divergence, while others have noted the prerequisites for understanding the problem. There is an acknowledgment of the complexity involved for those unfamiliar with three-dimensional calculus.

Contextual Notes

There is a mention of the original poster's lack of familiarity with the Laplacian operator and related concepts, which may hinder their ability to derive the required relationship. Additionally, the notation of the delta function is discussed, indicating a potential misunderstanding in its representation.

Keonn
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Homework Statement


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

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

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.
[itex]\nabla^2 \phi =\frac {\partial^2 \phi}{\partial r^2} + \frac 1r \frac {\partial \phi}{\partial r}[/itex]

The Attempt at a Solution


Plugging in e/r for phi, you get:
[itex]\nabla^2 \phi = \frac {2e}{r^3} - \frac {e}{r^3} = \frac {e}{r^3}[/itex]
which isn't even close to the equation given in the problem. I am so lost, please help.
 
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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 [itex]<\frac{∂}{∂x_1}, \frac{∂}{∂x_2}, ..., \frac{∂}{∂x_n}>[/itex]. 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 = [itex]<\frac{∂ξ}{∂x_1}, \frac{∂ξ}{∂x_2}, ..., \frac{∂ξ}{∂x_n}>[/itex]

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

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) = [itex]\frac{∂^2ξ}{∂x_1^2} + \frac{∂^2ξ}{∂x_2^2} + ... + \frac{∂^2ξ}{∂x_n^2}[/itex]
// as a note, ∇⋅∇ is more commonly written ∇2 or Δ, known as the Laplacian
 
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