How Is Mutual Energy Calculated Between Non-Parallel Dipoles?

[MathematicalPhysicist]

i need to show that the mutual energy between two dioples p1 and p2 (not necessarily parallel to each other) is $$U=-\frac{p_1\cdot p_2}{|r|^3}-3\frac{(p_1\cdot r)(p_2\cdot r)}{|r|^5}$$
where r is the vector from p_1 to p_2. (the p's are moments of diople).

i tried using this equation: $$U=\int dV \rho_2 * \phi_1$$
and also this :$$\phi=\frac{p\cdot r}{|r|^3}$$
(phi is the potential and rho is the density).
$$\rho_2=-1/4\pi\nabla^2\phi_2=-1/4\pi[\frac{1}{|r|^2}@/@r(r^2@\phi/@r)+\frac{1}{|r|^2*sin(\theta)}@/@\theta(sin(\theta)@\phi/@\theta)]$$
where @ stands for peratial derivative, but i didnt get to the desired answer.
any pointers?

 

[Pythagorean]

What about

$$U=-{m\cdot B}$$

where m is the vector for one of the dipoles in field B from the other dipole.

 

[Pythagorean]

you'll have to dig the equation for the field (B) of dipole, too, I reckon.

 

[MathematicalPhysicist]

by B, you mean the magnetic field, well i haven't learned it yet (i mean we haven't touched it in class as of yet, i myself read it form purcell), i pretty much sure i don't need here to use B, perhaps something else?

 

[Pythagorean]

So these are electric dipoles then? I guess p usually denotes electric dipole and m is for magnetic dipole.

 

[Pythagorean]

anyway, same equation, just:

U = -p.E

instead of -m.B

and dig up the E field for an electric dipole

 

[MathematicalPhysicist]

well for E i found already, shouldn't i prove that it equals p.E? or in other words how to derive it?

 

[Pythagorean]

well for E i found already, shouldn't i prove that it equals p.E? or in other words how to derive it?

If you want to find a derivation for U = -p.E, you might also remember that potential Energy is

the integral of F.dl

and the Force from a dipole is

F = (p.grad)E