- #1
Lindsayyyy
- 219
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Hi everyone
I want to find the multipole expansion of
[tex] \Phi(\vec r)= \frac {1}{4\pi \epsilon_0} \int d^3 r' \frac {\rho(\vec r')}{|\vec r -\vec r'|}[/tex]
Taylor series
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 [tex] \frac {1}{\sqrt{(1-x)}}[/tex]
Which I can use I guess for this problem, where my x is:
[tex]2 \frac {\vec r\vec r'}{r^2} -\frac {r'^2}{r^2}[/tex]
so I get
[tex] \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}}} [/tex]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.
Homework Statement
I want to find the multipole expansion of
[tex] \Phi(\vec r)= \frac {1}{4\pi \epsilon_0} \int d^3 r' \frac {\rho(\vec r')}{|\vec r -\vec r'|}[/tex]
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 [tex] \frac {1}{\sqrt{(1-x)}}[/tex]
Which I can use I guess for this problem, where my x is:
[tex]2 \frac {\vec r\vec r'}{r^2} -\frac {r'^2}{r^2}[/tex]
so I get
[tex] \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}}} [/tex]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.