Quadrupole Potential: Help Derive and Decipher This Beast!

1. May 25, 2009

UAR

$$(\textbf{q}\bullet\nabla)(\textbf{p}\bullet\nabla)\frac{1}{r} = -(\textbf{p}\bullet\textbf{q})\frac{1}{r^{3}} + 3(\textbf{p}\bullet\textbf{r})(\textbf{q}\bullet\textbf{r})\frac{1}{r^{5}}$$

r is distance between field point and dipole source, $$\textbf{p}$$ is dipole moment, and I believe $$\textbf{q}$$ may be quadrupole moment tensor (what is that anyways?),

How is the above equation derived ? and exactly how is it related to the more physically and mathematically lucid dipole potential below:

$$(\textbf{p}\bullet\nabla)\frac{1}{r} = -(\textbf{p}\bullet\textbf{r})\frac{1}{r^{3}}$$

Last edited: May 25, 2009
2. May 25, 2009

tiny-tim

Welcome to PF!

Hi UAR! Welcome to PF!

(use \cdot instead of \bullet )
It's derived from that equation simply by using it twice, first with p· and then with q·

(and because (q·)(p·r) = p·q )

3. May 25, 2009

UAR

Thanks Tiny-tim!

However, while you are still online, excuse my slowness: why is:

$$(\textbf{q}\cdot\nabla)(\textbf{p}\cdot\textbf{r})=\textbf{p}\cdot\textbf{q}$$ ?

4. May 25, 2009

tiny-tim

Because p.r = xpx + ypy + zpz,

so (q·∇)(p.r) = … ?

5. May 25, 2009

UAR

Aaah! Thanks! I was hesitant to do that due to a (poor notation)-induced irrational fear that $$p_{x'}$$ was a function of $$x$$. But now I see it is not, since $$\nabla$$ is w.r.t field point $$\textbf{r}$$, while $$\textbf{p}$$ depends only on source pts $$\textbf{r'}$$.

One more question: What is $$\textbf{q}$$ ?

By the way, you are truly a good mentor. Thanks for your help and keep up the great work!!!

6. May 26, 2009

tiny-tim

No idea … it could be anything, and the equation would still work

7. May 26, 2009

UAR

Hi Tiny Tim,

Anyone else care to help physically and mathematically elucidate $$\textbf{q}$$ in the quadrupole potential equation above. I read somewhere that it is called (or is related to ?) the "quadrupole moment tensor" (what exactly is that by the way?).