Potential of a charged ring in terms of Legendre polynomials

[patric44]

Homework Statement

find the potential of a charged ring in terms of Legendre polynomials

Relevant Equations

dV = kdq/|r-r'|, k is set =1

hi guys
I am trying to calculate the the potential at any point P due to a charged ring with a radius = a, but my answer didn't match the one on the textbook, I tried by using
$$
V = \int\frac{\lambda ad\phi}{|\vec{r}-\vec{r'}|}
$$
by evaluating the integral and expanding denominator in terms of Legendre i got the following answer
$$
V = 2\pi\lambda\sum_{n=0}^{\infty}P_{n}(cos\theta)\left(\frac{r}{a}\right)^{n}\qquad,for\;\;a>r
$$
but the book answer was
$$
V = 2\pi\lambda\sum_{n=0}^{\infty}\frac{(-1)^{n}(2n)!}{2^{2n}(n!)^{2}}P_{2n}(cos\theta)\left(\frac{r}{a}\right)^{2n}\qquad,for\;\;a>r
$$
what I am doing wrong

 

[ergospherical]

Can you write out your working? I think you should only get even-numbered polynomials.

 

[patric44]

Can you write out your working? I think you should only get even-numbered polynomials.

I expanded the inverse of $|\vect{r}-\vect{r^'}|$ as follows
$$
\left|\vec{r}-\vec{r^{'}}\right|^{-1}=\left(r^2+a^2-2ra\cos\theta\right)^{\frac{-1}{2}}=\frac{1}{a}\left(1-2\frac{r}{a}\cos\theta+\frac{r^2}{a^2}\right)^{\frac{-1}{2}},
$$
comparing with the generating function of the Legendre polynomials, I got
$$
V=2\pi\lambda\left(1-2\frac{r}{a}\cos\theta+\frac{r^2}{a^2}\right)^{\frac{-1}{2}}=2\pi\lambda\sum_{n=0}^{\infty}P_n\left(\cos\theta\right)\left(\frac{r}{a}\right)^{n}.
$$

 

[vela]

You incorrectly assumed the angle between $\vec r$ and $\vec r'$ is $\theta$.

 

[patric44]

You incorrectly assumed the angle between $\vec r$ and $\vec r'$ is $\theta$.

Isn't it just $\theta$!

 

[ergospherical]

The vectors are not necessarily at the same azimuthal angle $\phi$.

 

[patric44]

The vectors are not necessarily at the same azimuthal angle $\phi$.

How should I take care of this in the expansion of $|\vec{r}-\vec{r^{'}}|^{-1}$?

 

[ergospherical]

Write $\mathbf{r}$ and $\mathbf{r}’$ as Cartesian vectors and evaluate $\mathbf{r} \cdot \mathbf{r}’$ that way.

 

[PhDeezNutz]

The vectors are not necessarily at the same azimuthal angle $\phi$.

I think we should find a way to dispense with the azimuthal angle entirely since this is an axis-symmetric potential.

 

[ergospherical]

I think we should find a way to dispense with the azimuthal angle entirely since this is an axis-symmetric potential.

If $\mathbf{r}$ is at azimuthal angle $\phi$ and $\mathbf{r}'$ is at azimuthal angle $\phi'$, then the expansion of $|\mathbf{r} - \mathbf{r}'|^{-1}$ contains terms in $\cos^k{(\phi - \phi')}$. This azimuthal dependence vanishes upon integrating over $\phi'$ from $0$ to $2\pi$.

 

[PhDeezNutz]

OP

The general form of an axis symmetric potential in the near region is

$\sum_{n=0}^{\infty} A_n r^n P_n \left( \cos \theta \right)$

Treating the on-axis potential in the near region $\theta = 0$ makes all of the Legendre Polynomials equal to 1 ; $P_n \left( \cos 0 \right) = P_n \left( 1 \right) = 1$ for all n.

So the on-axis potential is

$\sum_{n=0}^{\infty} A_n r^n$

Once you find the on-axis Taylor Expansion you just slap on the relevant Legendre Polynomials

On axis $r = z$

$\Phi \left( z \right) = \frac{1}{4 \pi \epsilon_0} \int \frac{\lambda}{\sqrt{R^2 + z^2}} R \, d \phi' = \frac{\lambda}{2 \epsilon_0} \frac{R}{\sqrt{R^2 + z^2}}$

Invoke the smallness parameter $r \ll R$ and taylor expand

Hint: $\frac{1}{\sqrt{1+x^2}} = 1 - \frac{x^2}{2} + \frac{3x^4}{8} - \frac{5x^6}{16} + \frac{35x^8}{128}$

This hint matches up with the coefficients in your desired answer.

I wouldn't worry about expanding the whole thing, matching the first few terms is enough to draw a conclusion IMO.

Edit: I was more or less copying and pasting one of my posts from another thread. So just chanage $R$ to $a$

 

[kuruman]

In post #1 you state "I am trying to calculate the the potential at any point P $\dots$"

Doesn't that include the region $r>a$?

 

[Dr Transport]

If memory serves me correctly, there is a trick to finding it easily for all \vec{r}

 

[vanhees71]

It's easier to think in terms of the Green's function,
$$V(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 G(|\vec{x}-\vec{x}'|) \rho(\vec{x}').$$
Of course
$$G(\vec{x}-\vec{x}')=\frac{1}{4 \pi |\vec{x}-\vec{x}'|} \; \Rightarrow \; \Delta_x G(\vec{x}-\vec{x}')=\delta^{(3)}(\vec{x}-\vec{x}').$$
The multipole expansion is then derived from the multipole expansion of the Green's function,
$$G(\vec{x}-\vec{x}') = \sum_{l=0}^{\infty} \sum_{m=-l}^l \frac{4 \pi}{2l+1} \frac{r_{<}^l}{r_{>}^{l+1}} \mathrm{Y}_{lm}^*(\vartheta',\varphi') \mathrm{Y}_{lm}(\vartheta,\varphi).$$
Here $r_<=\text{min}(r,r')$ and $r_{>}=\text{max}(r,r')$, and I used the spherical harmonics normalized to 1, i.e.,
$$\int_0^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi \sin \vartheta \mathrm{Y}_{l'm'}^*(\vartheta,\varphi) \mathrm{Y}_{lm}(\vartheta,\varphi)=\delta_{ll'} \delta_{mm'}.$$
The relation to the Legendre polynomials for axisymmetric problems (rotational symmetry around the $z$ axis) is
$$\mathrm{Y}_{l0}(\vartheta,\varphi)=\sqrt{\frac{2l+1}{4 \pi}} P_l(\cos \vartheta).$$
Of course all $\mathrm{Y}_{l0}$ do not depend on $\varphi$, because $\mathrm{Y}_{lm}(\vartheta,\varphi) \propto \exp(\mathrm{i} m \varphi)$. The orthogonality relation thus is
$$\int_{-1}^1 \mathrm{d} u P_{l'}(u) P_{l}(u)=\frac{2}{2l+1} \delta_{ll'}.$$

 

[PhDeezNutz]

Isn't it just $\theta$!View attachment 294210

Theta is the angle from the z-axis.

 

[kuruman]

How meaningful is this multipole expansion for $(r_{<},~r_{>})\approx a$ when $\vartheta = \frac{\pi}{2}$?

 

[vanhees71]

I'm not sure, whether there's a misunderstanding concerning my convention. In my notation $\vartheta$ is the polar and $\varphi$ the azimuthal angle, i.e., the relation between Cartesian and spherical coordinates of the position vector is
$$\vec{x}=\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} r \cos \varphi \sin \vartheta \\ r \sin \varphi \sin \vartheta \\ r \cos \vartheta \end{pmatrix}, \quad r \in \mathbb{R}_{>0}, \quad \vartheta \in (0,\pi), \varphi \in [0,2 \pi).$$
The spherical coordinates are singular along the $x_3$-axis, i.e., for $\vartheta \in \{0,\pi \}$.

So the idea is to just use the expansion of the Green's function in terms of spherical harmonics
$$\frac{1}{4 \pi |\vec{x}-\vec{x}'|} = \sum_{\ell=0}^{\infty} \sum_{m=-\ell}^{\ell} \frac{4 \pi}{2 \ell+1} \frac{r_>^{\ell}}{r^{\ell+1}} \mathrm{Y}_{\ell m}^*(\vartheta',\varphi') \mathrm{Y}_{\ell m}(\vartheta,\varphi).$$
Then the potential is given by (for $r<a$, i.e., $r_{>}=a$ and $r_{<}=r$, \quad $\vartheta'=\pi/2$)
$$\Phi(\vec{x})=4 \pi \int_0^{2 \pi} \mathrm{d} \varphi' \frac{a \lambda}{4 \pi|\vec{x}-\vec{x}'|} = 4 \pi \lambda \int_0^{2 \pi} \mathrm{d} \varphi' \sum_{\ell=0}^{\infty} \sum_{m=-\ell}^{m=\ell} \left (\frac{r}{a} \right)^{\ell} \mathrm{Y}_{\ell m}^*(\pi/2,\varphi') \mathrm{Y}(\vartheta,\varphi).$$
The integral makes all contributions with $m \neq 0$ vanish. Further
$$\text{Y}_{\ell 0}(\vartheta'=\pi/2)=\sqrt{\frac{2 \ell+1}{4 \pi}}
\mathrm{P}_{\ell}(0)=\begin{cases} \sqrt{\frac{2 \ell+1}{4 \pi}}
\frac{(-1)^{\ell/2} \ell!}{4^{\ell/2} (\ell/2)!^2} & \text{for} \quad
\ell \quad \text{even} \\ 0 & \text{for} \quad \ell \quad \text{odd}. \end{cases}$$
Plugging all this into the integral you should get the result of the textbook.

For $r>a$ the only difference is that instead of $(r/a)^{\ell}$ you have the factor $(a/r)^{\ell+1}$ in front.