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I Taylor expansion of 1/distance

  1. Sep 1, 2017 #1
    upload_2017-9-1_22-11-7.png
    Hi, I would like to express that r and r' are vectors in the attachment and let's say that r is observer distance vector r' is source distance vector. By the way I know this is taylor expansion (for instance if there was only x component (scalar form) I would not any ask question ). But I do not understand when we include vectors, why there is a dot product of r and r'???????????? Also I know the taylor expansion of multiple scalar variables but I do not know how to derive taylor expansion of multiple vector variables ( I think 1/distance is a scalar function comprised of 2 different vectors here )??????????
     
  2. jcsd
  3. Sep 1, 2017 #2
    You can write the vectors in component form, although this isn't a very elegant solution. It's very straightforward though.
    ##\frac{1}{|\mathbf{r}-\mathbf{r}'|} = \frac{1}{\sqrt{\sum_i(r_i - r'_i)^2}}##
    If you do a taylor expansion around ##r_x'=r_y'=r_z'=0##
    you get ##\frac{1}{r} + \sum_i{\left[\partial_i \left(\frac{1}{\sqrt{\sum_j (\mathbf{r}-\mathbf{r}')^2}} \right) r_i' \right] } ## (partial operates only on the next term in paren)
    ##= \frac{1}{r} + \sum_i{\left[\frac{-1}{2} \left(\sum_j (\mathbf{r}-\mathbf{r}')^2 \right)^{-3/2} \partial_i \left(\sum_j (\mathbf{r}-\mathbf{r}')^2 \right) r_i' \right] } ## (chain rule)
    ##= \frac{1}{r} + \sum_i{\left[\frac{-1}{2} \left(\sum_j (\mathbf{r}-\mathbf{r}')^2 \right)^{-3/2} (-2) \left(r_i-r_i' \right) r_i' \right] } ## (chain rule)
    The derivative was supposed to be evaluated at r' = 0, so it becomes
    ##= \frac{1}{r} + \sum_i{\left[\left(\sum_j (\mathbf{r}-\mathbf{r}')^2 \right)^{-3/2} \left(r_i \right) r_i' \right] } ## (chain rule)
    Let ##r=\sqrt{\sum_i(r_i - r'_i)^2}##
    ##= \frac{1}{r} + \frac{1}{r^3} r_i r_i' ##
    That's the same as ##\frac{1}{r} + \frac{1}{r^3} \mathbf{r} \cdot \mathbf{r}_i'##
     
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