Multipole expansion. Problems with understanding derivatives

Click For Summary

Homework Help Overview

The discussion revolves around the multipole expansion in the context of electrostatics, specifically focusing on the expression for the potential \(\Phi(\vec r)\) derived from a charge distribution \(\rho(\vec r')\). The original poster is attempting to apply the Taylor series to simplify the term \(\frac{1}{|\vec r - \vec r'|}\) under the assumption that \(r \gg r'\).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use the Taylor series expansion but expresses uncertainty about handling derivatives, particularly whether to differentiate both the \(r' \cdot r\) term and \(r'^2\). Some participants suggest considering partial derivatives of both terms with respect to the coordinates.

Discussion Status

Participants are providing guidance on the use of Taylor expansion and the application of the gradient operator (nabla) in this context. There is an acknowledgment of the need to use vector forms of expansion and the binomial expansion, but no consensus has been reached on the specific steps to take next.

Contextual Notes

There is a mention of the relationship between higher-order terms in the expansion and their physical significance, such as monopoles, dipoles, and quadrupoles. The discussion also reflects some confusion regarding the terminology used for the gradient operator.

Lindsayyyy
Messages
215
Reaction score
0
Hi everyone

Homework Statement



I want to find the multipole expansion of

\Phi(\vec r)= \frac {1}{4\pi \epsilon_0} \int d^3 r' \frac {\rho(\vec r')}{|\vec r -\vec r'|}

Homework Equations



Taylor series

The Attempt at a Solution



My attempt at a solution was to use the Taylor series. I tried to approach 1/|r-r'| around r'/r (that's what the task told me to, because r>>r').

I found the taylor series for \frac {1}{\sqrt{(1-x)}}

Which I can use I guess for this problem, where my x is:
2 \frac {\vec r\vec r'}{r^2} -\frac {r'^2}{r^2}

so I get

\frac {1}{| \vec r - \vec r'|}= \frac {1}{r} \frac {1}{\sqrt{ 1- 2 \frac {\vec r \vec r'}{r^2} + \frac {r'^2}{r^2}}}But now I'm stuck. I don't know how to handle the derivates. Do I only have to derive the vector r' with nabla or r'^2 aswell?

Thanks for your help in advance.
 
Physics news on Phys.org
If I understand your question, you will need to take partial derivatives of the r'2 term as well as the r##\cdot##r' term with respect to x', y', and z'. Just think of the argument of the square root as some function of x', y', and z'.
 
Use the vector form of Taylor Expansion i.e.
f(x+h) = f(x) + (h.grad)f(x) + [(h.grad)^2]f(x) + ...

where x and h are vectors, grad is the usual gradient operator and "." indicates the dot product.
 
Thanks for your help. I know that this has something to do with Nabla, but I don't understand why I have to use Nabla here actually.
 
you just have to use the binomial expansion for the term which you got in op.Different higher order terms in r'/r represents monopoles,dipoles,quadrupoles etc.
 
Last edited:
Lindsayyyy said:
Thanks for your help. I know that this has something to do with Nabla, but I don't understand why I have to use Nabla here actually.

I think what you're referring to as nabla is what I call grad.

Just do the vector taylor expansion as I mentioned, this was the box standard thing to do back in electrodynamics exams. Oh, and use summation convention to make life easier.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
995
  • · Replies 13 ·
Replies
13
Views
6K