Separation of variables in spherical coordinates (electrostatics)

[milkism]

Homework Statement

Need help with setting up the first part of the problem.

Relevant Equations

See solution.

Problem:

Solution:
When I looked at an example problem, they started writing the potential in terms of the Legendre polynomials.
The example problem:

This is what I did:
$$V_0 \alpha P_2 (\cos(\theta)) \Rightarrow \frac{\alpha 3 \cos ^2 (\theta)}{2} - \frac{\alpha}{2} \Rightarrow \frac{\alpha 3}{2} - \frac{\alpha}{2}= 1 \Rightarrow \alpha = 1$$
Hence,
$$V(\theta) = V_0 P_2 (\cos(\theta)).$$
Have I done right?

 

[milkism]

Homework Statement: Need help with setting up the first part of the problem.
Relevant Equations: See solution.

Problem:
View attachment 325294

Solution:
When I looked at an example problem, they started writing the potential in terms of the Legendre polynomials.
The example problem:
View attachment 325295
This is what I did:
$$V_0 \alpha P_2 (\cos(\theta)) \Rightarrow \frac{\alpha 3 \cos ^2 (\theta)}{2} - \frac{\alpha}{2} \Rightarrow \frac{\alpha 3}{2} - \frac{\alpha}{2}= 1 \Rightarrow \alpha = 1$$
Hence,
$$V(\theta) = V_0 P_2 (\cos(\theta)).$$
Have I done right?

Also with radial dependance do they just mean calculate A_l and B_l?

 

[PhDeezNutz]

Yes you are supposed to find $A_\ell$ and $B_\ell$ and plug them into the generic formula

$V \left(r, \theta \right) = \sum_{\ell = 0}^{\infty} \left( A_\ell r^\ell + \frac{B_\ell}{r^{\ell + 1}}\right) P_\ell \left( \cos \theta\right)$

On physical grounds $A_\ell = 0$ outside and $B_\ell = 0$ inside because we cant have the potential blowing up at infinity or at the origin.

 

[pasmith]

You may want to start with $$\left.\begin{array}{r} P_2(\cos \theta) = \frac{1}{2}(3 \cos^2 \theta - 1) \\ P_0(\cos \theta) = 1 \end{array}\right\} \quad \Rightarrow\quad \cos^2 \theta = \frac23P_2(\cos \theta) + \frac13P_0(\cos \theta)$$

 

[vela]

Hence,
$$V(\theta) = V_0 P_2 (\cos(\theta)).$$
Have I done right?

No. Write out the righthand side as a polynomial in $\cos\theta$, and you'll see it doesn't equal the given $V(\theta)$.

In the example problem, you have the potential $V = k \cos 3\theta= k(4\cos \theta^3-3\cos\theta)$, which is a third-order polynomial of $\cos\theta$. That means it's an element of the vector space consisting of polynomials of degree 3 or fewer.

The first four Legendre polynomials are a basis for this vector space, so any element of the vector space can be expressed as a linear combination of the Legendre polynomials. In other words, you can say
$$4k\cos^3\theta - 3k\cos\theta = c_3 P_3(\cos\theta) + c_2 P_2(\cos\theta) + c_1 P_1(\cos\theta) + c_0 P_0(\cos\theta).$$ The problem is then finding the appropriate values for the $c_i$'s.

A straightforward way to do that is to simply expand out the righthand side and then equating coefficients from the two sides of the equation. You'll get a system of four linear equations which is simple to solve. Griffiths used a useful shortcut to deduce $c_0 = c_2 = 0$, so he ended up with a system of two equations and two unknowns.

In your problem, you have $V=V_0 \cos^2\theta$. It's an element of the vector space consisting of polynomials of degree 2 or fewer, and the first three Legendre polynomials will span that space. So the problem is to find $c_0$, $c_1$, and $c_2$ so that
$$V_0 \cos^2\theta = c_2 P_2(\cos\theta) + c_1 P_1(\cos\theta) + c_0 P_0(\cos\theta).$$ Expand the righthand side and then equate coefficients to get the equations you need to solve.

 

[milkism]

Thanks you all, first time doing such problem and I completely understand it now!

 

[milkism]

Also for finding the electrostatic dipole moment of the system can we just use the surface charge density on the surface to calculate it?

 

[milkism]

Also for finding the electrostatic dipole moment of the system can we just use the surface charge density on the surface to calculate it?

I get zero for electrostatic dipole moment of the system, don't know if it's right.
Using this formula:

I get $$\frac{V_0 \epsilon_0}{R} \left( 5\cos ^2 (\theta) - \frac{4}{3} \right)$$.
Then for electrostatic dipole moment in spherical coordinates:

Computing the integral I get zero.

 

[PhDeezNutz]

Having solved this very problem myself, for the outside I do not get a 1/r^2 term for the potential. So I’d have to agree with you about your dipole moment.

 

[milkism]

Having solved this very problem myself, for the outside I do not get a 1/r^2 term for the potential. So I’d have to agree with you about your dipole moment.

Is it even possible for there to not be a dipole moment?

 

[PhDeezNutz]

Absolutely. The next higher up moment is the quadrupole which in the simplest illustration is two back to back opposing dipoles and therefore the dipole moments cancel out.

The Quadrupole is a 2nd rank (symmetric) tensor and is not so easy to visualize. This moment is scarcely dealt with in Griffiths however there is an exercise pertaining to it.

 

[milkism]

Absolutely. The next higher up moment is the quadrupole which in the simplest illustration is two back to back opposing dipoles and therefore the dipole moments cancel out.

The Quadrupole is a 2nd rank (symmetric) tensor and is not so easy to visualize. This moment is scarcely dealt with in Griffiths however there is an exercise pertaining to it.

My next problem is to calculate the mono-di and quadrapole of 4 charges haha, luckily no octopole problem.

 

[milkism]

Absolutely. The next higher up moment is the quadrupole which in the simplest illustration is two back to back opposing dipoles and therefore the dipole moments cancel out.

The Quadrupole is a 2nd rank (symmetric) tensor and is not so easy to visualize. This moment is scarcely dealt with in Griffiths however there is an exercise pertaining to it.

I think the dipole moment is zero because the angular dependant potential is symmetric (even), just like how symmetric molecules have no dipole moment

 

[PhDeezNutz]

That makes sense to me.

When you calculate the dipole moment of the 4 charges I bet it’s going to be zero.

Edit: your instructor knows how to choose good problems. It’s nice how one leads into the other with emphasis. I’m a fan.