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Taylor expansion of 1/|r-r'|

  1. Sep 7, 2010 #1
    I am wroking through an electrodynamics textbook and there is this Taylor expansion to do later a multipole expansion. But I can't figure out how the author does it. Please any help?

    the expansion:

    [tex] \frac{1}{|\vec{r}-\vec{r'}|} = \frac{1}{r} - \sum^3_{i=1} x'_i \frac{\partial}{\partial x_i} \frac{1}{r} + \frac{1}{2} \sum^3_{i,j=1} x'_i x'_j \frac{\partial}{\partial x_i} \frac{\partial}{\partial x_j}\frac{1}{r} + \mathellipsis [/tex]

    And he writes "the occuring differenciations were changed using:"

    [tex] (\frac{\partial}{\partial x'_i} \frac{1}{|\vec{r}-\vec{r'}|})_{r'=0} = - (\frac{\partial}{\partial x_i} \frac{1}{|\vec{r}-\vec{r'}|})_{r'=0} = - \frac{\partial}{\partial x_i} \frac{1}{r} [/tex]

    I just can't follow his argument..
     
  2. jcsd
  3. Sep 7, 2010 #2

    G01

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    The first line is just the statement of the taylor expansion (of a function of 3 variables) about r'.

    It may not seem familiar because here he is expressing everything in terms of components.

    As for the second line, all he is doing is taking the derivative of [itex]\frac{1}{|\vec{r}-\vec{r'}|}[/itex] with respect to the coordinates of r'.

    The he say, "Oh this will be the same as the negative derivative of the same function with respect to r." (The negative sign appears because of the chain rule and the negative in front of r' inside the function.)

    If you don't believe this you can work out the derivatives explicitly and prove the relation. Use:

    [tex]\frac{1}{|\vec{r}-\vec{r'}|}=\frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}[/tex]
     
  4. Sep 13, 2010 #3
    Is he doing the taylor expansion around r taking it as a constant and r' being the variable?

    But all in all I got you argument! Thanks!
     
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