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Simple Taylor or Multipole Expansion of Potential

  1. Jun 17, 2008 #1
    Hullo,

    Somehow, I couldn't get the TeX to come out right.

    I have been trying to learn scheme theory (algebraic geometry) and completely forgotten how to do this simple calculus type stuff...

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

    Let V be a potential of the form

    [tex]V = \left(\frac{1}{r} + \left(\frac{1}{\left|\vec{r} - \left(\vec{r_{1}} - vec{r_{2}}\right)\right|} - \left(\frac{1}{\left|\vec{r} + \vec{r_{2}}\right|} - \left(\frac{1}{\left|\vec{r} - \vec{r_{1}}\right|} \right)[\tex]

    where r = [tex]\left|\vec{r}\right|[\tex].

    For large r >> 1 I am supposed to expand in powers of [tex]\frac{\vec{r_{i}}}{r}[\tex] to obtain the expression
    [tex] V \sim \frac{1}{r^3}\left(x_{1}x_{2} + y_{1}y_{2} - 2z_{1}z_{2} \right)[\tex] where higher order terms have been neglected.

    2. Relevant equations


    3. The attempt at a solution

    I tried using the expansion [tex]\left(\frac{1}{\left|\vec{r} + \vec{r'}\right|} = \frac{1}{r} + \frac{\vec{r}\vec{r'}}{r^3}[\tex] + terms of higher order, but somehow end up with V = 0 (up to order [tex]r^-3[\tex]. I have completely forgotten how to do this...

    Would be thankful for any help...

    Greetings from Germany,

    bavaji
     
    Last edited: Jun 17, 2008
  2. jcsd
  3. Jun 18, 2008 #2

    dynamicsolo

    User Avatar
    Homework Helper

    While you do use "backslashes" for the special TeX characters, you need to terminate the TeX command with regular "slash" [ / tex ]:

    [tex]V = \left(\frac{1}{r} + \left(\frac{1}{\left|\vec{r} - \left(\vec{r_{1}} - vec{r_{2}}\right)\right|} - \left(\frac{1}{\left|\vec{r} + \vec{r_{2}}\right|} - \left(\frac{1}{\left|\vec{r} - \vec{r_{1}}\right|} \right)[/tex]

    It looks like the expression may be missing something...
     
    Last edited: Jun 18, 2008
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