Electric potential of a charged ring

[MMS]

Homework Statement

Find the electric potential of a ring of radius R that is charged uniformly with a linear charge density λ.

Homework Equations

The Attempt at a Solution

I wasn't sure which section to post this in but I finally landed here. It isn't really a problem that I'm troubled solving. Rather, it's something in the solution given I'm troubled understanding.
Here's the full solution: http://docdro.id/lhpIgyS

First off, I believe they messed up a little substituting Y_lm (not Y*_lm) as the exponent is exp(-i*m*phi) rather than exp(i*m*phi).
However, What's really bugging me is what they did after picking phi=0 and using the delta in (8).
I'm really not sure what went there. It seems as if they used some sort of orthogonality expression, I'm assuming that of spherical harmonics. If they did do so, I can't see what manipulations they did to get 2*pi*delta_m0.
It hints that m'=0 but there is no m' here. Also, assuming there was, they picked it to be 0 because of the azimuthal symmetry?
Moreover, if they did use the orthogonality expression of spherical harmonics, why is there no delta_ll'?

Thanks in advance.

 

[TSny]

First off, I believe they messed up a little substituting Y_lm (not Y*_lm) as the exponent is exp(-i*m*phi) rather than exp(i*m*phi).

Yes, you are right. But their mistake won't affect the result.

However, What's really bugging me is what they did after picking phi=0 and using the delta in (8).
...I can't see what manipulations they did to get 2*pi*delta_m0.
It hints that m'=0 but there is no m' here. Also, assuming there was, they picked it to be 0 because of the azimuthal symmetry?

Note that they picked $\phi = 0$. You should be able to use the azimuthal symmetry to see why you can do this.

Once you let $\phi = 0$ you are left with just the integral $\int_{0}^{2\pi} e^{-im\phi'} d\phi'$. What is the result of carrying out this integral? Consider the case where $m = 0$ as well as $m \neq 0$.

 

[MMS]

Note that they picked $\phi = 0$. You should be able to use the azimuthal symmetry to see why you can do this.

Once you let $\phi = 0$ you are left with just the integral $\int_{0}^{2\pi} e^{-im\phi'} d\phi'$. What is the result of carrying out this integral? Consider the case where $m = 0$ as well as $m \neq 0$.

God I hate doing physics late at night.

For m=0 it's instantly 2*pi since it's an integral over 1.
For any m≠0 the integral would be that of sine\cosine from 0 to 2*pi which clearly gives 0.
easier way of writing this would be 2*pi*delta_m0.

Thank you TSny for bringing this to my attention. It was starting to piss me off a little. :P

 

[MMS]

Also, if I may, I want to make sure I understand something in this question or in similar problems so I'd be happy if you could verify the following:

Picking phi=0 here is merely because of the symmetry of the problem that suggests that the final answer won't be dependent of phi.
However, it does not suggest that m=0 necessarily. It simply means that I can pick whichever angle phi I desire to observe the potential of the ring from. For instance, if it were simpler, we could've picked phi=pi or something else.

 

[TSny]

Yes, that's right. For this problem it isn't necessary to choose a value for $\phi$. The factor $e^{im\phi}$ can be pulled out of the integral since it doesn't depend on $\phi'$. So, the integral over $\phi'$ will still force $m$ to be zero. Then, with $m = 0$, $e^{im\phi} = 1$, independent of $\phi$.

 

[MMS]

Yes, that's right. For this problem it isn't necessary to choose a value for $\phi$. The factor $e^{im\phi}$ can be pulled out of the integral since it doesn't depend on $\phi'$. So, the integral over $\phi'$ will still force $m$ to be zero. Then, with $m = 0$, $e^{im\phi} = 1$, independent of $\phi$.

Got it. Thanks for the help TSny!