# Potential from a Quadrupole using Legendre polynml's

1. Mar 21, 2010

### Vapor88 1. Problem Statement:
There are charges of q placed at distance +a and -a from the origin on the z-axis. There is a charge at the origin of -2q. Express the potential of this point-like linear quadrupole in Legendre polynomials

The distance between origin and point is r, the distance between (0,0,a) and r is $$l$$ and the distance between (0,0,-a) is $$l'$$

3. The attempt at a solution

The potential can be expressed as
$$(q/(4\pi r)*(1/l + 1/l' + 2/r)$$

Using law of cosines

$$l = \sqrt{a^2 + r^2 - 2ar cos(\theta)}$$
$$l' = \sqrt{a^2 + r^2 + 2ar cos(\theta)}$$

Solving for 1/l and 1/l' gives
$$1/l = \frac{1}{r} (1 + a^2/r^2 - a/r cos (\theta))^{-1/2}$$
$$1/l' = \frac{1}{r} (1 + a^2/r^2 + a/r cos (\theta))^{-1/2}$$

that's supposed to be raised to the -0.5, but i don't know how to make it look right in latex

Now

$$V = ( \frac{q}{4 \pi \epsilon_{o}} \frac{1}{r} [(1+( \frac{a}{r})^2 - \frac{a}{r} cos (\theta))^{-1/2} + (1+ (\frac{a}{r})^2 + \frac{a}{r} cos (\theta))^{-1/2} - 2]$$

We had an example in class of a dipole that used the expansion

(1+x)^n = 1 + nx + ...

And only used the first two terms of it. Do I need to do the same thing to this, or can I go straight to the Legendre polynomials. I'm not sure where to go from here.

Also... If you take a^2/r^2 to go to zero, then the potential goes to zero as well. So I can't do that.

2. 3. Mar 22, 2010

### Vapor88 I just worked further into the problem, tried using that expansion the same way, don't know if I can really do that though.

But I took the a^2/r^2 + a/r cos (theta) = x

In the end, I did get something over r^3, which makes sense because it's a quadrupole, but I missed the point of the problem completely, using Legendre polynomials.

4. Mar 22, 2010

### vela Staff Emeritus
The idea is to go straight to the Legendre polynomials using the well-known result

$$\frac{1}{\sqrt{r^2+a^2-2ar\cos\theta}}=\sum_{l=0}^\infty \frac{a^l}{r^{l+1}}P_l(\cos\theta)$$

when a<r.

5. Mar 22, 2010

### Vapor88 Ok, so I got a bit further in the problem, he wants us to use the first three terms of binomial expansion, I only did the first two to get my results above, but they did not need Legendre polynomials. Though I do not know why my answer is incorrect. If I use the first three terms of the binomial expansion for both of the 1/(...)^1/2, then simplify, I end up with the following

$$\frac{q}{4 \pi \epsilon_{o} r} * [ - \frac{a^2}{r^2} + \frac{3a^4}{4r^4} + \frac{3a^2}{4r^2} cos^2( \theta)]$$

Now, my misunderstanding is in the Legendre polynomials. I know the definitions of Po P1 P2, etc. but I don't get how to apply them. I see that there is a cos^2 term, so I know that that's related to P2
$$P_{2} = \frac {(3x^{2}-1)}{2}$$
$$P_{1} = x$$
$$P_{0} = 1$$

/edit... P2 is supposed to be (3x^2-1)/2

And I imagine that in those terms, I have x = cos (theta)

I'm just very sketchy on the details of Legendre polynomials. Everyone in my class had my professor for quantum, I haven't taken quantum yet, so he really did not spend much time explaining it. As I understand it, the first two terms in my equation get Po applied to them, but I don't know how to do this.

6. Mar 22, 2010

### vela Staff Emeritus
You made a mistake in your algebra somewhere. The cos^2 term shouldn't have a 4 in the denominator.

7. Mar 22, 2010

### Vapor88 For
$$[1 + \frac{a^2}{r^2}- \frac{a}{r} cos( \theta)]^{-1/2}$$

I expand to get

$$[1 - \frac{1}{2} ( \frac{a^2}{r^2} - \frac{a}{r} cos( \theta)) + \frac{- \frac{1}{2} (- \frac{1}{2} - 1)}{2!} ( \frac{a^4}{r^4} - \frac{2a^3}{r^3} cos( \theta) + \frac{a^2}{r^2} cos^2( \theta))]$$

Which gets simplified to

$$[1 - \frac{1}{2} ( \frac{a^2}{r^2} - \frac{a}{r} cos( \theta)) + \frac{3}{8} ( \frac{a^4}{r^4} - \frac{2a^3}{r^3} cos( \theta) + \frac{a^2}{r^2} cos^2( \theta))]$$

For the other term

$$[1 + \frac{a^2}{r^2}+ \frac{a}{r} cos( \theta)]^{-1/2}$$

When expanded, everything stays the same except all terms are positive

$$[1 - \frac{1}{2} ( \frac{a^2}{r^2} + \frac{a}{r} cos( \theta)) + \frac{3}{8} ( \frac{a^4}{r^4} + \frac{2a^3}{r^3} cos( \theta) + \frac{a^2}{r^2} cos^2( \theta))]$$

then, when you add em together you get

$$V = \frac{q}{4 \pi \epsilon_{o} r} * [ - \frac{a^2}{r^2} + \frac{3a^4}{4r^4} + \frac{3a^2}{4r^2} cos^2( \theta)]$$

...er there's still a 4 in the denominator... hrm

Oh! and the 1's disappeared because of the -2 seen in the first post

But yeah... I keep reworking it and get 3/8 + 3/8 = 3/4 for that last term

Last edited: Mar 22, 2010
8. Mar 22, 2010

### vela Staff Emeritus
Oh, you dropped the two way in the beginning. You wrote

$$1/l = [1 + \frac{a^2}{r^2} - \frac{a}{r} \cos(\theta)]^{-1/2}$$

but it should be

$$1/l = [1 + \frac{a^2}{r^2} - 2 \frac{a}{r} \cos(\theta)]^{-1/2}$$

9. Mar 22, 2010

### Vapor88 Ok, so I believe this is correct, however, I'm still stumped as far as legendre polynomials go. I have no idea how to implement them.

$$V = \frac{q}{4 \pi \epsilon_{o} r} * [ - \frac{a^2}{r^2} + \frac{3a^4}{4r^4} + \frac{3a^2}{4r^2} cos^2( \theta)]$$

Given this, I can see that it sort of looks like P2, but would that make it -1+1+3/2cos^2-1/2

This doesn't seem right at all.

10. Mar 22, 2010

### vela Staff Emeritus
How is this different from what you wrote before? You still have the four in the denominator of the cos^2 term that doesn't belong there.

11. Mar 22, 2010

### Vapor88 It's not, but I keep reworking it and get that. here is the work for how I got there

$$[1 - \frac{1}{2} ( \frac{a^2}{r^2} - \frac{a}{r} cos( \theta)) + \frac{3}{8} ( \frac{a^4}{r^4} - \frac{2a^3}{r^3} cos( \theta) + \frac{a^2}{r^2} cos^2( \theta))]$$

Then you add that to

$$[1 - \frac{1}{2} ( \frac{a^2}{r^2} + \frac{a}{r} cos( \theta)) + \frac{3}{8} ( \frac{a^4}{r^4} + \frac{2a^3}{r^3} cos( \theta) + \frac{a^2}{r^2} cos^2( \theta))]$$

You can see what cancels out and what gets carried through. If those two equations are correct, then the cos^2 term's coefficient is 3/4

Working back further...

For the expansion:
The 3/8 comes from n(n-1)/2! where n = -1/2

The x^2 in the expansion's third term becomes this
$$\frac{a^4}{r^4} + \frac {2a^3}{r^3} cos ( \theta) \frac{a^2}{r^2} cos^2( \theta)$$

The 1's in front get canceled by the 2 in the first post. All of the cos(theta) terms cancel. The 3/8 gets distributed to the cos^2 term in both of them, then when I add them together, I get 3/4. What coefficient am I supposed to be getting, then? If I am making an algebra mistake, I am still not seeing it. Can you please point it out, I've just reworked it three times, and gotten the same thing.

12. Mar 22, 2010

### vela Staff Emeritus

13. Mar 22, 2010

### Vapor88 *facepalm*

thanks!

14. Mar 22, 2010

### Vapor88 Alrighty...

After banging my head on the desk, I've got the following.

$$V = \frac{q}{4 \pi r \epsilon_{o}} [- \frac{a^2}{r^2} + \frac{3a^4}{4r^4} + \frac{3a^2}{2r^2} cos^2( \theta)]$$

Now... to implement those Legendre polynomials.

15. Mar 22, 2010

### Vapor88 Ok, here's what I know of Legendre plynml's from my notes

They are integrals such that

$$\int P_{l} (x) P_{m} (x)dx = 0$$

When l =/= m

When l = m, that integral is equal to
$$\frac{2}{2l+1}$$

For this problem, I want to take x to be cos(theta)? Does that turn the first two terms into terms that can be handled with Po and the third with P2?

16. Mar 22, 2010

### gabbagabbahey What is $P_2(\cos\theta)$? How about $P_0(\cos\theta)$?

17. Mar 22, 2010

### Vapor88 $$P_{0} = 1$$

$$P_{1} = cos ( \theta)$$

$$P_{2} = \frac{3cos^2 ( \theta) - 1}{2}$$

I'm fairly confident that this is correct.

Are these my $$P_{l}$$ values? And would a^2/r^2 be Pm?

Does this mean that I'll have three integrals, all evaluated from -1 to 1, such that

$$\int P_{l} (cos \theta) P_{m} (cos \theta)d(cos \theta) = 0$$

Last edited: Mar 22, 2010
18. Mar 22, 2010

### gabbagabbahey Right. Now look at your expression for the potential..

$$V = \frac{q}{4 \pi r \epsilon_{o}} [- \frac{a^2}{r^2} + \frac{3a^4}{4r^4} + \frac{3a^2}{2r^2} cos^2( \theta)]$$

The first thing you should do is drop the $\frac{a^4}{r^4}$. In your Taylor expansion, you are assuming that $a$ is very small compared to the distance from the origin at which you are measuring the potential. (If you don't make this assumption, you must keep all of the terms in the Taylor expansion) So, certainly $\frac{a^4}{r^4}$ should be much much smaller than $\frac{a^2}{r^2}$, right?

After that, you should notice that

$$- \frac{a^2}{r^2} + \frac{3a^2}{2r^2} cos^2( \theta)=\frac{a^2}{r^2}\left(\frac{3\cos^2\theta-1}{2}\right)- \frac{a^2}{2r^2}=\frac{a^2}{r^2}P_2(\cos\theta)+\frac{a^2}{2r^2}P_0(\cos\theta)=\frac{a^2}{r^2}\left(P_2(\cos\theta)+\frac{1}{2}P_0(\cos\theta)\right)$$

What exactly are you planning on integrating?

Edit: You seem to have made an error in your Taylor expansion, to second order in $\frac{a}{r}$, I get

$$V\approx\frac{q}{4 \pi r \epsilon_0}\left[- \frac{a^2}{r^2} + \frac{3a^2}{r^2} cos^2( \theta)\right]=\frac{q}{2 \pi \epsilon_0}\left(\frac{a^2}{r^3}\right)P_2(\cos\theta)$$

Last edited: Mar 22, 2010
19. Mar 22, 2010

### Vapor88 What am I planning on integrating? lol... I'm not really sure, I'm very, very shaky on these legendre polynomials, they're completely new to me, but everyone else in my class is familiar with them.

Anyways, I see from there, when plugged into the potential equation, I get an r^-3, which is expected in quadrupoles. Is further work necessary?

20. Mar 22, 2010

### Vapor88 I used the binomial expansion, not the Taylor expansion... They are different, right?

21. Mar 22, 2010

### gabbagabbahey $$a^2+r^2\pm2ar\cos\theta$$ is a trinomial, not a binomial. You should use a Taylor expansion instead.