Potential from a simple Quadrupole expansion

In summary, the problem at hand involves finding the potential due to a quadrupole moment of a given charge distribution. Two methods are being used to solve the problem, but they are giving different answers. The first method involves using the general expansion for the quadrupole moment and plugging it into the potential formula, while the second method involves using the addition theorem for spherical harmonics to find the general potential for a multipole moment. Both methods should give the same result, but they are giving different answers. The discrepancy may be due to the lack of understanding of the unit vectors and the definition of the vacuum constant.
  • #1
khfrekek92
88
0
Hi everyone! I'm currently working on this problem for which I am getting inconsistencies depending on how I do it. I'm trying to find the potential due to the quadrupole moment of the following distribution:
+q at (0,0,d), -2q at (0,0,0), and +q at (0,0,-2d)

I am doing this using two different methods and they both get different answers:1) Using the general expansion Qij=sum[ql{3ril*rjl-rl^2deltaij)] and plugging into Vquad=1/(8*pi*epsilonor^3)sum[Qij*ni*nj]

This method gives me some constant divided by r^3, with NO angular dependence whatsoever.2) Going back to the basics and using the very general potential by substituting in the addition theorem for spherical harmonics, etc to find the general potential for a general multipole moment: qlm=integral[rho*r^l*Ylm],
and
Vmulti=sum[1/(epsilono(r^(l+1)*(2l+1)*Ylm(theta,phi)*qlm]

Notice how this answer definitely depends on theta for dipole moment and above (when the spherical harmonics introduce cos(theta)'s into them.)

Doing it this method gives me the same constant divided by r^3 that I found earlier, except now it is multiplied by (3cos^2(theta)-1) which comes from exactly the Y20 spherical harmonic. These two methods SHOULD give the same results, but these are radically different... Any ideas?

Thanks!

PS sorry for the lack of Latex, but I figured most people should get the gist of it
 
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  • #2
khfrekek92 said:
This method gives me some constant divided by r^3, with NO angular dependence whatsoever.
Where does the formula for the potential come from? It does not look right.

What is "epsilonor"?
 
  • #3
You can see it derived here:
http://physicspages.com/2012/04/03/quadrupole-moment/

And sorry, epsilonor is just from my lack of latex understanding.. it is really supposed to be the vacuum constant times the magnitude of r:

ε*|r|
 
  • #4
So ni*nj are the components of the vector r? Well, then your potential should depend on the direction of r.
If it does not, please show your work.
 
  • #5
Yes that is correct. However, because all 3 charges are on the z axis, the only non-zero component of the quadrupole moment tensor is Q_zz. Every other one goes to 0.

Then, only one term will be in the potential summation, the Q_zz term. This term corresponds to nz*nz (which are two unit vectors in the z direction) which boils down to 1.
 
  • #6
Those are not the unit vectors! The "unit" here refers to the whole vector r, its z component can be smaller.
 

1. What is the concept of potential from a simple Quadrupole expansion?

The potential from a simple Quadrupole expansion is a mathematical representation of the potential energy between two charged particles in terms of their distance and orientation. It takes into account the quadrupole moment of the particles, which is a measure of their distribution of charge.

2. How is the potential from a simple Quadrupole expansion calculated?

The potential from a simple Quadrupole expansion is calculated using the formula V = kQ/r^2 [1+ (3cos^2θ -1)/2], where V is the potential energy, k is the Coulomb constant, Q is the quadrupole moment, r is the distance between the particles, and θ is the angle between the axis of the particles and the line connecting them.

3. What are the assumptions made in the simple Quadrupole expansion?

The simple Quadrupole expansion assumes that the particles are point charges, the distance between them is much greater than their size, and that the particles are stationary.

4. How is the potential from a simple Quadrupole expansion useful in scientific research?

The potential from a simple Quadrupole expansion is useful in many areas of scientific research, such as in studying molecular interactions, determining the shape and orientation of molecules, and in the analysis of electric fields in the presence of multiple charges.

5. Can the potential from a simple Quadrupole expansion be used to calculate the potential energy between more than two charged particles?

Yes, the potential from a simple Quadrupole expansion can be extended to calculate the potential energy between multiple charged particles by summing up the individual potentials from each pair of particles. This is known as the multipole expansion.

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