Electrodynamics/Tutorials/3/4/1/Part 2
Video Intro
Hi, this is Jonathan Gardner.
We're covering [section reference] of Griffiths Introduction to Electrodynamics.
I'm going to move fast, but you can always rewind.
Thumbs up and share if you appreciate my effort.
As always, questions in a video response or comments.
Let's get started.
Intro
I'm going to show you how to coax out the dipole, quadrupole, octopole, etc... terms given a volume charge density and a point P sufficiently far away.
First a picture:
(Draw the picture)
- r is the vector to P
- r' is the vector to the bit we are integrating
- <math>\mathfrak{r}</math> is the vector from the bit we are integrating to P.
The potential is:
- <math>V(P) = {1 \over 4 \pi \epsilon_0}\int{1\over \mathfrak{r}} \rho\ d\tau\;</math>
Binomial Expansion
Using the law of cosines,
- <math>\mathfrak{r}^2 = r^2 + (r')^2 - 2rr'\cos \theta\;</math>
where theta is the angle between r and r'.
If we pull out the r^2, in preparation for the binomial expansion,:
- <math>\mathfrak{r}^2 = r^2\left(1 + \left({r' \over r}\right)^2 - 2\left({r'\over r}\right)\cos \theta\right)\;</math>
We'll pull out the naughty bits into epsilon:
- <math>\mathfrak{r}^2 = r^2 (1 - \epsilon)\;</math>
- <math>\epsilon = \left({r' \over r}\right)^2 - 2\left({r'\over r}\right)\cos \theta = \left({r' \over r}\right)\left({r' \over r} - 2\cos \theta \right)\;</math>
This is not epsilon-naught! Confusing the two would be bad.
So, square root:
- <math>\mathfrak{r} = r (1 - \epsilon)^{1/2}\;</math>
Invert:
- <math>{1 \over \mathfrak{r}} = {1 \over r} (1 - \epsilon)^{-1/2}\;</math>
As long as we keep P far away, then epsilon will be small, between 1 and -1, so we can do our binomial expansion.
- <math>{1 \over \mathfrak{r}} = {1 \over r} \left[
1
- {1 \over 2} \epsilon + {3 \over 8} \epsilon^2 - {5 \over 16} \epsilon^3 + ...
\right]\;</math>
Algebra
Plug it back in:
- <math>{1 \over \mathfrak{r}} = {1 \over r} \left[
1
- {1 \over 2} \left({r' \over r}\right)\left({r' \over r} - 2\cos \theta \right) + {3 \over 8} \left({r' \over r}\right)^2\left({r' \over r} - 2\cos \theta \right)^2 - {5 \over 16} \left({r' \over r}\right)^3\left({r' \over r} - 2\cos \theta \right)^3 + ...
\right]\;</math>
Expand:
- <math>{1 \over \mathfrak{r}} = {1 \over r} \left[
1
- {1 \over 2} \left({r' \over r}\right)\left({r' \over r} - 2\cos \theta \right) + {3 \over 8} \left({r' \over r}\right)^2\left( \left({r' \over r}\right)^2 - 4 {r' \over r} \cos \theta + 4 \cos^2 \theta \right) - {5 \over 16} \left({r' \over r}\right)^3 \left(\left({r' \over r}\right)^3 - 6 \left({r' \over r}\right)^2 \cos \theta + 12 \left({r' \over r}\right) \cos^2 \theta - 8 \cos^3 \theta\right) + ...
\right]\;</math>
Distribute:
- <math>{1 \over \mathfrak{r}} = {1 \over r} \left[
1
+ \left(
- {1 \over 2} \left({r' \over r}\right)^2
+ \left({r' \over r}\right) \cos \theta
\right)
+ \left(
{3 \over 8} \left({r' \over r}\right)^4
- {3 \over 2} \left({r' \over r}\right)^3 \cos \theta
+ {3 \over 2} \left({r' \over r}\right)^2 \cos^2 \theta
\right)
+ \left(
- {5 \over 16} \left({r' \over r}\right)^6
+ {15 \over 8} \left({r' \over r}\right)^5 \cos \theta
- {15 \over 4} \left({r' \over r}\right)^4 \cos^2 \theta
+ {5 \over 2} \left({r' \over r}\right)^3 \cos^3 \theta
\right)
+ ...
\right]\;</math>
Combine like terms:
- <math>{1 \over \mathfrak{r}} = {1 \over r} \left[
1
+ \left({r' \over r}\right) \cos \theta
+ \left({r' \over r}\right)^2 \left(
{3 \over 2} \cos^2 \theta
- {1 \over 2}
\right)
+ \left({r' \over r}\right)^3 \left(
{5 \over 2} \cos^3 \theta
- {3 \over 2} \cos \theta
\right)
+ ... \right]\;</math>
Conclusion
If you haven't noticed yet, we have a series with Legendre polynomials.
- <math>{1 \over \mathfrak{r}} = {1 \over r} \sum_{n=0}^\infty \left({r' \over r}\right)^n P_n(\cos \theta)\;</math>
Putting this in, and drawing out the r (since it doesn't vary within the integration):
- <math>V(P) = {1 \over 4 \pi \epsilon_0} \sum_{n=0}^\infty {1 \over r^{n+1}} \int (r')^n P_n(\cos \theta) \rho \ d\tau\;</math>
If we expand the series, we get:
- <math>V(P) = {1 \over 4 \pi \epsilon_0} \left[
{1 \over r} \int \rho\ d \tau
+ {1 \over r^2} \int r' \cos \theta \rho\ d \tau + {1 \over r^3} \int (r')^2 \left( {3 \over 2} \cos^2 \theta - {1 \over 2} \right)\rho\ d \tau + ... \right]\;</math>
The first term is the monopole term; it goes 1/r.
The second is the dipole term. 1/r^2
The third is the quadrupole, 1/r^3, etc...
This is an exact equation. We did not approximations.
However, it is really useful AS an approximation scheme.
Let's say the net charge wasn't 0. Then we could probably ignore the rest of the terms and treat the object as a point charge and get close results.
If it has a net charge of 0, then we'd look to see if the charge was more like a dipole. We're looking at the r' cos theta direction-- that's the axis perpendicular to r! If there is a net positive in one direction, and a net negative the other, then we have a dipole.
If that wasn't the case, then we'd look at the quadrupole term. And so on.
We simply use the first term with a non-zero coefficient, and use that as the charge.
The rest of this section is going to be looking at the different terms, and how to use them.
