# Multipole expansion. Problems with understanding derivatives

1. Jun 26, 2013

### Lindsayyyy

Hi everyone

1. The problem statement, all variables and given/known data

I want to find the multipole expansion of

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

2. Relevant equations

Taylor series

3. 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?

2. Jun 26, 2013

### TSny

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'.

3. Jun 26, 2013

### Marioeden

Use the vector form of Taylor Expansion i.e.

where x and h are vectors, grad is the usual gradient operator and "." indicates the dot product.

4. Jun 27, 2013

### Lindsayyyy

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.

5. Jun 27, 2013

### andrien

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: Jun 27, 2013
6. Jun 27, 2013

### Marioeden

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.