What is the meaning of r' in the Multipole Expansion?

[Tony Hau]

Homework Statement

A sphere of radius R, centred at the origin, carries charge density $$\rho(r,\theta) = k\frac{R}{r^{2}}(R-2r)sin\theta,$$
where $k$ is a constant. Find the approximate potential for points on the z axis, far from the sphere.

Relevant Equations

The multipole expansion: $V(r)=\frac{1}{4\pi \epsilon_o}\sum_{l=0}^{\infty}\frac{1}{r^{l+1}}\int(r^{'})^{l} P_l (cos\alpha)\rho(r^{'})d\tau^{’}$

The diagram of the problem should look something like this:

,which is just the normal spherical coordinate.To calculate the potential far away, we use the multipole expansion.

$I_o$ in the expansion is ok, because $(r^{'})^{0} = 1$.

However, I am wondering how I should calculate $I_1$, because $(r^{'})^{1} = r^{'}$; I have to care what $r^{'}$ actually means. I know $r$ is just the normal spherical coordinate $r$.

Can anyone kindly explain? Thanks for your answer in advance because I have learned a great deal from the forum!

 

[Abhishek11235]

$r'$ is the coordinate of charge particles inside sphere.

 

[Tony Hau]

$r'$ is the coordinate of charge particles inside sphere.

Then what is $r$?

 

[DrClaude]

It is called a dummy argument. It still represents the radial coordinate and should disappear after integration (which is a definite integral, don't forget the bounds). It is used so you don't confuse it with the $r$ that is outside the integral.

 

[PeroK]

Then what is $r$?

To add perhaps a bit more detail:

That's the coordinate at which you are evaluating the potential. You have a potential that is a function of $r$, and to evaluate it you must integrate over the region of space where there is non-zero charge density. You can't use $r$ for the dummy integration variable, so it's common to use $r'$ for this.

 

[Tony Hau]

I think $\vec r$ refers to the direction of the radial distance of the potential at a general point from the centre of the coordinate system, whereas $\vec r^{'}$ is the direction from the centre of the coordinate system to the infinitesmal charge; the angle between them is $\alpha$, which is also the variable in the Legendre polynomial $P_{l}(cos\alpha)$.

$\alpha$ is quite different from $\theta$ because $\theta$ is the angle from the z axis to $\vec r$.

 

[etotheipi]

I think $\vec{r}$ refers to the direction of the radial distance of the potential at a general point from the centre of the coordinate system

I don't know what a "direction of the radial distance" means. $\vec{r}$ is the position vector of the point at which you're evaluating the potential.

 

[Tony Hau]

I don't know what a "direction of the radial distance" means. $\vec{r}$ is the position vector of the point at which you're evaluating the potential.

My poor English, that's what I want to say.

 

[vanhees71]

Hint: Check your "relevant equation"!