Origin of coordinates in multipole expansion

[mondo]

I am reading Griffiths chapter 3.4.3 on origin of coordinates in multipole expansion (can be found online here https://peppyhare.github.io/r/notes/griffiths/ch3-4/) And I got stuck at this:

For the figure 3.22: the dipole moment $p = qd\hat{y}$ and has a corresponding dipole term in the potential fomulae. The monopole potential $$\frac{q}{4\pi\epsilon_0 r}$$ is not quite correct for this configuration. rather the exact potential is $$\frac{q}{4\pi\epsilon_0 \gamma}$$ (I used gamma for r distance from q to point marked with a dot on 3.22). Then author also says: " The multipole expansion is, remember, a series in inverse powers of r (the distance to the origin), and when we expand 1/γ1/γ, we get all powers, not just the first."

I don't get:
1. Why the monopole is a not correct for configuration shown on 3.22 - after all we have a single charge q
2. Why he uses the r and not r in his exact potential formula?
3. Author also says "The multipole expansion is, remember, a series in inverse powers of r (the distance to the origin)" but in his calculation he uses a distance r which is definitely not in a reference to the origin. What is wrong here?

 

[BvU]

I am reading Griffiths chapter 3.4.3 on origin of coordinates in multipole expansion (can be found online here https://peppyhare.github.io/r/notes/griffiths/ch3-4/) And I got stuck at this:



View attachment 353021

For the figure 3.22: the dipole moment $p = qd\hat{y}$ and has a corresponding dipole term in the potential fomulae. The monopole potential $$\frac{q}{4\pi\epsilon_0 r}$$ is not quite correct for this configuration. rather the exact potential is $$\frac{q}{4\pi\epsilon_0 \gamma}$$ (I used gamma for r distance from q to point marked with a dot on 3.22). Then author also says: " The multipole expansion is, remember, a series in inverse powers of r (the distance to the origin), and when we expand 1/γ1/γ, we get all powers, not just the first."

I don't get:
1. Why the monopole is a not correct for configuration shown on 3.32 - after all we have a single charge q
2. Why he uses the r and not r in his exact potential formula?
3. Author also says "The multipole expansion is, remember, a series in inverse powers of r (the distance to the origin)" but in his calculation he uses a distance r which is definitely not in a reference to the origin. What is wrong here?

Hi,

1. The author develops the exact potential $$\frac{q}{4\pi\epsilon_0 \gamma}$$ for a single charge that is not located at the origin but at $\bf r'$ in powers of $1\over r$ ( 3.91 through 3.96). The unavoidable consequence of developing $1\over \gamma$ in terms of $1\over r$ is the emergence of higher-order terms.

2. Equation 3.96 is the exact expression for $V({\bf r})$ in terms of $1\over r$ that is suitable for comparing the exact expression for $V({\bf r})$ of a general charge distribution with the exact expression for for $V({\bf r})$ of a single charge that is not located at the origin.

3. Distance $r$ is with respect to the origin. I don't understand why you claim it isn't. Can you explain in more detail what you mean ?

$\$

 

[mondo]

Hi, thanks for an answer. I am not sure I get your reationale:

The unavoidable consequence of developing 1γ in terms of 1r is the emergence of higher-order terms.

Why do we even need $$\gamma$$ in the first place? Why can't we reply on r alone?

3. Distance r is with respect to the origin. I don't understand why you claim it isn't. Can you explain in more detail what you mean ?

If you look at 3.22 you can see two distances: r - a disatnce of a "dot/point" from the origin and r which here in math equations we call $$\gamma$$, that is a distance from charge q on the y axis to this dot point at distance r from the origin. Now, I don't understand why do we need this this second distance $$\gamma$$. How is that bringing multipole expansion that r can not?

 

[BvU]

If you look at 3.22 you can see two distances

You mean 3.32$\qquad$ Right ?

To express $V({\bf r})$ as a power series in $1\over r\ \$ (with $r = |{\bf r}| \ )$, you really need $\gamma$

(the

in fig 3.32)

$\$

 

[mondo]

Sorry for a late response.

You mean 3.32 Right ?

Yes, my mistake.

To express V(r) as a power series in 1r (with r=|r| ), you really need γ

(the

in fig 3.32)

Yes but here we are talking about a monopole contribution of a single charge q acting on the origin. Should it rather be $$\frac{1}{4\pi\epsilon_0} \frac{q}{d}$$ so the distance is "d" ?

Also, at the end of page 154, in subsection 3.4.2 author says "For a point charge at the origin, Vmon is the exact potential.." but with a single charge at the origin we get the term $$\frac{1}{r} -> \infty$$ since r is close to 0, right?

 

[BvU]

Yes but here we are talking about a monopole contribution of a single charge q acting on the origin

No we are not.
(The link now doesn't point to pages anymore, very inconvenient !

)

We are working on developing an experssion for $V(\vec r)$ everywhere in terms of multipole contributions.

I don't know what to add without repeating, I'm afraid...

$\$

 

[mondo]

I think I was able to answer some of my questions:
1. In the previous post I was asking why the monopole term does not go to infinity with r -> 0. The answer is: I was looking at wrong r. When the charge is at the origin both r and r become the same distance.
2. In the very first post I was asking why the monopole potential is not exact for the configuration on figure 3.32. The answer is, when the charge is not at the origin we have other terms contributions as well (dipole, quadrupole etc) In general equation 3.96.

However, I don't understand this:

"If the total charge is 0, the dominant term on the potential will be the dipole." How is that possible? If the total charge is 0 the entire potential must be 0 as we always, in every term, multiply by charge.

 

[BvU]

"If the total charge is 0, the dominant term on the potential will be the dipole." How is that possible? If the total charge is 0 the entire potential must be 0 as we always, in every term, multiply by charge

Certainly not !

$\$

 

[mondo]

Certainly not !

Ok, can you please explain how can we have a non zero potential when total charge of the configuration is 0?

 

[renormalize]

Ok, can you please explain how can we have a non zero potential when total charge of the configuration is 0?

Because the total potential $V$ of two opposite charges $q\,,-q$ that aren't located at the same point is nonzero:

(http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dipole.html)

 

[mondo]

Ok I think I got it - the total charge may be 0 but the dipole may still exist because it is an arrangement of equal and opposite charges i.e -q and q (total = 0)