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

[yungman]

Let \;\vec r\;'\; be position vector pointing to the source. \;\vec r\; be the position vector pointing to a field point. Therefore the distance from the source and field point is:

\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}

Where \;\theta\; is the angle between the two vector.

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

\vec {\eta} =\vec r -\vec r\;&#039; \;\hbox { and }\; \eta = \sqrt{r^2+r&#039;^2 - 2rr&#039;cos\theta}= r \sqrt{1+\left (\frac{r&#039;}{r}\right )^2 -2\frac{r&#039;}{r} cos \theta} = r \left (1-\frac {r&#039;}{r} cos \theta\right )

I know for r'<<r, \left ( \frac {r&#039;}{r}\right )^2\approx \;0. But how can you remove the square root.

 

[RedX]

To first order, for |x|<1:
\sqrt{1+x}=1+\frac{x}{2}+O[x^2]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.

 

[yungman]

Thanks for the response, can you show me how to derive this?

Alan

 

[vanhees71]

The Taylor series for the square root reads

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

Thus to order r&#039;/r 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.

 

[RedX]

Thanks for the response, can you show me how to derive this?

d/dx(\sqrt{1+x})=\frac{1}{2}\frac{1}{\sqrt{1+x}}
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+...

 

[RedX]

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.

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.

 

[yungman]

I thought Bi-Nomial is:

\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 { &lt;&lt;1. }

How can you show me how to does the other work?

 

[yungman]

Thanks for all your help. I have to go back and read up infinite/Taylor series to refresh my memory.

 

[RedX]

I thought Bi-Nomial is:

\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 { &lt;&lt;1. }

How can you show me how to does the other work?

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.

 

[yungman]

Thanks for all the help.

Alan

 

[yungman]

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