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Why in the irradiament mulipoles of Lienard-Wiechert's potential we say that electric quadrupole give a contribute of the same order of the magnetic dipole? How can we see it from their equations? And is there a physical reason for this? Sorry for my trivial question.

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Try to give a little more context. Right now, to answer your question the person has to know off the top of his/her head what's the quadrupole contribution to the Lienard-Wiechert's potential. ;-) –  Malabarba Dec 25 '10 at 20:31
Are you reading Griffiths or Jackson? Off the top of my head I'd say its because there is no magnetic monopole, so the magnetic dipole is the first magnetic term in the multipole expansion. "Sorry for my trivial question" - please do not apologize. You're learning physics. Nothing is trivial. And if you think it is you might be better of in finance ;) –  user346 Dec 25 '10 at 21:26
I did apologize because sometimes I had seen some questions with comments: "this question is a little trivial for this site", and I know that here there are people that know physics better of me and will think this question is trivial :-) In the question, I meant about order of magnitude: in the irradiation all multipoles fields behave like 1/r, so how con I see that E quadrupole contribute is smaller than E dipole contribute and M dipole is like E quadrupole? –  Boy Simone Dec 26 '10 at 12:21
This site has people that know physics better than you and this world has people that know physics better than them. Don't worry. Everybody is playing catchup :-) –  user346 Dec 27 '10 at 3:21

2 Answers 2

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a good question. The discrepancy between the counting for electric and magnetic multipoles appears because in the potentials, the electric field is encoded in the potential $\phi$ while the magnetic field is encoded in the vector potential $A$.

However, to find the electric field $E$, you take the gradient of $\phi$ combined with the time derivative of $A$ (both with a minus sign). There are no spatial derivatives in the second term so it is the dominant one, in a $1/r$ expansion. To obtain the magnetic field $B$, you need to take the curl of $A$ - which means its spatial derivatives. There is no way to get a term to $B$ from $\phi$ and $A$ that would contain no spatial derivatives.

From a relativity viewpoint, spatial and temporal derivatives are of the same order. However, things are different for the analysis of static objects. It's only the spatial derivatives that produce an extra power of $1/r$.

So the electric field $E$ is of the same order as the vector potential $A$ (the time derivative doesn't add any $1/r$) while the magnetic field $B$ has an extra $1/r$ relatively to the vector potential $A$. Quadrupoles have one extra $1/r$ relatively to the dipoles - the standard expansion in $1/r$. So to go from the electric dipoles to the next order, you either add the $1/r$ by switching from the electric to magnetic fields; or you get the extra $1/r$ by going from electric dipoles to electric quadrupoles. At any rate, the counting of powers is shifted by one between the electric and magnetic multipoles.

Best wishes Lubos

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thank you for the good answer!! It's what I looked for:-) –  Boy Simone Jan 14 '11 at 18:23

The usual derivation of the radiation multipoles involves writing the potentials as an integral over the source distribution, and then doing a Taylor expansion in the source coordinate. In Griffiths's notation, we're evaluating A( r ) by performing an integral over source coordinates r', and the Taylor expansion is in powers of the small quantity r' / r. Electric dipole shows up at first order in this Taylor expansion, and both magnetic dipole and electric quadrupole show up at second order.

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