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How do I derive the step on the formula of distance of source to field point.

  1. Jun 23, 2011 #1
    Let [itex]\;\vec r\;'\; [/itex] be position vector pointing to the source. [itex]\;\vec r\;[/itex] be the position vector pointing to a field point. Therefore the distance from the source and field point is:

    [tex] \vec {\eta} =\vec r -\vec r\;' \;\hbox { and }\; \eta = \sqrt{r^2+r'^2 - 2rr'cos\theta}= r \sqrt{1+\left (\frac{r'}{r}\right )^2 -2\frac{r'}{r} cos \theta}[/tex]

    Where [itex]\;\theta\;[/itex] is the angle between the two vector.

    For r'<<r, Why is the book than say

    [tex] \vec {\eta} =\vec r -\vec r\;' \;\hbox { and }\; \eta = \sqrt{r^2+r'^2 - 2rr'cos\theta}= r \sqrt{1+\left (\frac{r'}{r}\right )^2 -2\frac{r'}{r} cos \theta} = r \left (1-\frac {r'}{r} cos \theta\right )[/tex]

    I know for r'<<r, [itex]\left ( \frac {r'}{r}\right )^2\approx \;0 [/itex]. But how can you remove the square root.
     
  2. jcsd
  3. Jun 23, 2011 #2
    To first order, for |x|<1:
    [tex]\sqrt{1+x}=1+\frac{x}{2}+O[x^2] [/tex]so basically the squared term you have under the radical it too small to bother with.

    You can derive this from expanding (1+x)1/2 with a Taylor expansion about x=0, or just use the binomial theorem.
     
  4. Jun 23, 2011 #3
    Thanks for the response, can you show me how to derive this?

    Alan
     
  5. Jun 23, 2011 #4

    vanhees71

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    The Taylor series for the square root reads

    [tex]\sqrt{1+x}=1+x/2+O(x^2)[/tex]

    Thus to order [itex]r'/r[/itex] you immediately obtain the approximation given by your book. That's one way to obtain the multipole expansion of the electrostatic field in terms of its sources. It's not a very clever way, but it works.:wink:
     
  6. Jun 23, 2011 #5
    [tex]d/dx(\sqrt{1+x})=\frac{1}{2}\frac{1}{\sqrt{1+x}} [/tex]
    so evaluted at x=0 gives 1/2.

    So the Taylor expansion is f(x)=1+x/2 since f(0)=sqrt(1+0)=1, and f(x)~f(0)+f'(0)x+...
     
  7. Jun 23, 2011 #6
    That's the only way I can think of! I think I can get the second order terms but higher terms would be tedious because of that squared term in the radical, but a computer could probably get the higher order terms without trouble.
     
  8. Jun 23, 2011 #7
    I thought Bi-Nomial is:

    [tex]\frac 1 {\sqrt {1+\epsilon}}\approx (1-\frac 1 2 \epsilon + \frac 3 8 \epsilon^2 -\frac 5 {16} \epsilon^3.......)\;\hbox { for }\;\epsilon \;\hbox { <<1. } [/tex]

    How can you show me how to does the other work?
     
  9. Jun 23, 2011 #8
    Thanks for all your help. I have to go back and read up infinite/Taylor series to refresh my memory.
     
  10. Jun 23, 2011 #9
    Here is the Wikipedia article on the binomial theorem for arbitrary exponents:

    http://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalized_binomial_theorem

    Just use formula (2) and substitute x=1, and you have an expansion for (1+y)^n. So formula (2) gives you the first four terms of a binomial expansion.
     
  11. Jun 23, 2011 #10
    Thanks for all the help.

    Alan
     
  12. Jun 23, 2011 #11
    I am glad I run across this. I studied infinite/taylor series so long ago, I forgot they are like fourier, bessels etc. that you can approx a function with the power series.

    Thanks

    Alan
     
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