Multipole Expansion when r<r'

In summary, the conversation discusses finding the multipole expansion for a situation where the distance from the observation point is less than the distance from the source point. The solution is obtained through a Taylor expansion, with a small mistake in the coefficient of r^l. The correct coefficient should be 1/r'^{l+1} instead of -1/r'^{l+1}.
  • #1
Vuldoraq
272
1

Homework Statement



Find the multipole expansion for the case when r<r', where r is the distance from the origin to the observation point and r' is the distance from the origin to the source point.


Homework Equations



[tex]\frac{1}{\sqrt{r^{2}+r'^{2}-2rr'\cos(\theta)}}=\sum_{l=0}^{\infty}(A_{l}r^l+B_{l}r^{-l-1})[/tex]

The Attempt at a Solution



To solve this I considered the case where [tex]\cos{\theta}[/tex] is equal to one and obtained,

[tex]\frac{1}{r-r'}=\sum_{l=o}^{\infty}A_{l}r^{l}[/tex]

The Bl terms must be zero since otherwise the sum would blow up as r goes to zero.

Now I have Taylor expanded the 1/(r-r') to get,

[tex]\frac{1}{r-r'}=\frac{1}{r'}\frac{1}{\frac{r}{r'}-1}=\frac{-1}{r'}*(1+\frac{r}{r'}+\frac{r^{2}}{r'^{2}}+...)[/tex]

Which means,

[tex]\frac{1}{r-r'}=-\sum_{l=0}^{\infty}\frac{r^{l}}{r'^{l+1}}[/tex]

Which means,

[tex]-A_{l}=\frac{1}{r'^{l+1}}[/tex]

Now this is the right answer except for the minus sign. Please could someone help me out in finding where I have gone wrong? I am geussing it is in my Taylor expansion, which is as follows,

[tex]\frac{1}{(-1)+x}=-[1+x+x^{2}+x^{3}+...][/tex]

My thanks for any help. Also Sorry about the Latex, it won't generate my first equation, even though I can detect no error in the code.
 
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  • #2


Hi there,

It seems like you have made a small mistake in your Taylor expansion. The correct expansion for 1/(r-r') is:

1/(r-r') = 1/(r'-r) = 1/r' * 1/(1-r/r') = 1/r' * (1 + r/r' + (r/r')^2 + ...)

So the coefficient of r^l should be 1/r'^{l+1}, not -1/r'^{l+1} as you have it. This should fix the issue with the minus sign.

Also, just a small note, in your first equation, you have "\cos(\theta)" but it should be "\cos{\theta}" to generate the proper LaTeX.

I hope this helps! Keep up the good work.
 
  • #3


Dear student,

Your attempt at solving this problem is on the right track. However, there are a few issues with your solution that I will address.

First, in your Taylor expansion, you have missed the negative sign in front of the series. It should be:

\frac{1}{(-1+x)}=-[1+x+x^{2}+x^{3}+...]

This is because when you substitute (-1) for x, you get:

\frac{1}{(-1)+(-1)}=-[1+(-1)+(-1)^{2}+(-1)^{3}+...]

Which simplifies to:

\frac{1}{-2}=-[1-1+1-1+...]

And from here, you can see that the series is indeed negative.

Next, in your solution, you have assumed that \cos{\theta} is equal to one. This is not always the case for r<r'. In fact, for r<r', \cos{\theta} could be any value between -1 and 1. Therefore, you cannot assume that \cos{\theta} is equal to one.

To fix this, you can use the fact that for r<r', \cos{\theta}=\frac{r'}{r}. This is because in this case, the angle between r and r' is obtuse, and the cosine of an obtuse angle is negative. Therefore, you can rewrite your original equation as:

\frac{1}{\sqrt{r^{2}+r'^{2}-2rr'\cos(\theta)}}=\frac{1}{\sqrt{r^{2}+r'^{2}-2rr'\frac{r'}{r}}}=\frac{1}{r'}\frac{1}{\sqrt{\frac{r^{2}}{r'^{2}}+1-2\frac{r}{r'}}}

Now, you can use your Taylor expansion for \frac{1}{(-1+x)} to expand the last term in this equation. I will leave it to you to complete the solution using this approach.

I hope this helps. Good luck with your studies!

Sincerely,
 

1. What is meant by "Multipole Expansion when r<r'"?

Multipole Expansion when r<r' is a mathematical technique used in physics to approximate the potential or field of a distribution of charges or masses when the distance from the point of interest is smaller than the characteristic size of the distribution.

2. How is the multipole expansion calculated when r<r'?

The multipole expansion when r<r' is calculated by expressing the potential or field in terms of a series of terms, each representing a different type of multipole moment. The higher order moments are typically smaller and can be neglected when r<r'.

3. What are the advantages of using multipole expansion when r<r'?

The main advantage of using multipole expansion when r<r' is that it allows for a more efficient and accurate calculation of the potential or field. It is especially useful when dealing with systems that have a large number of particles or charges.

4. Are there any limitations to using multipole expansion when r<r'?

Yes, there are some limitations to using multipole expansion when r<r'. It is only applicable for systems where the distance from the point of interest is smaller than the characteristic size of the distribution. It also assumes that the particles or charges are evenly distributed, and the potential or field is smooth and continuous.

5. Can multipole expansion be used for any type of potential or field?

No, multipole expansion is typically used for systems with a gravitational or electrostatic potential or field, where the potential or field decays with distance. It is not applicable for systems with strong or rapidly varying potentials or fields, such as those in nuclear physics or quantum mechanics.

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