Energy of an electric quadrupole in an Electric Field

[karanmohan]

Hello, I have tried to look for related threads but could not find any. Please excuse me if this is a repetition. I was curious about the energy of an electric quadrupole moment in an electro-magnetic field.

Basically, i am trying to follow first order perturbation theory and derive the matrix element of an atom interacting with a field in Bohm's Quantum Theory book - upon expansion, we get terms that "look" like an electric dipole, magnetic dipole and so on.

For example, the energy of an electric dipole in an electric field is E dotted with d (dipole moment). Likewise, that of a magnetic dipole would mu (magnetic moment) dotted with B. I am looking for a similar expression for the quadrupole.

Any help, as well as references are greatly appreciated. Thanks again!

 

[Naty1]

I can't answer your question directly, but this may help:

http://en.wikipedia.org/wiki/Electric_Quadrupole_Radiation

 

[karanmohan]

hmm...thanks for the link friend, but it doesn't give me enough detail unfortunately :(

 

[Meir Achuz]

$$U_Q = -\frac{1}{3}\bf[Q]:[{\nabla} E]$$.

 

[karanmohan]

thanks a lot clem, that really helps a lot...do you have reference by any chance?

 

[Meir Achuz]

Section 2.4 of Franklin's "Classical Electromagnetism" discusses electric quadrupoles.

 

[JJfortherear]

What does the operation : signify? I assume it yields a scalar here, since the expression is for energy, but how is it defined? Thank you

 

[SiggyYo]

What does the operation : signify? I assume it yields a scalar here, since the expression is for energy, but how is it defined? Thank you

What clem said about Franklin's "Classical Electromagnetism" are wise words in this context :P

Section 2.4 of Franklin's "Classical Electromagnetism" discusses electric quadrupoles.

 

[Meir Achuz]

$$U_Q = -\frac{1}{3}\bf[Q]:[{\nabla} E]$$.

That is a misprint.
The correct equation is
$$U_Q = -\frac{1}{3}\bf[Q]:[{\nabla}\nabla E]$$
$$=-\frac{1}{3}[(\bf[Q]\cdot\nabla)\cdot\nabla] E$$.