0
$\begingroup$

What information does the quadrupole provide?

I've seen many definitions on the internet, but I don't understand the relation between them.

  • What is the relation between knowing that the quadrupole moment is between four charges and that it is a measurement of how much the nucleus charge distribution deviates from spherical symmetry?
  • Why does a dipole come from cancelling two monopole fields and a quadrupole comes from cancelling two dipole fields?
  • Does a tripole field exist?
$\endgroup$
0
2
$\begingroup$

Moments measure how charge (or any other quantity of interest) is distributed in space. Instead of specifying a continuous charge density function $\rho(x,y,z)$, you can specify discrete numbers: its monopole moment (i.e., total charge)

$$q=\int \rho\,dV,$$

its dipole moments

$$p_i=\int \rho x_i\,dV,$$

its quadrupole moments

$$Q_{ij}=\int \rho x_ix_j\,dV,$$

its octupole moments

$$O_{ijk}=\int \rho x_ix_jx_k\,dV,$$

etc. The complete set of moments contains the same amount of info, but in a form that can be more useful. In many cases, only the first few moments are important, because the effects (field strengths, forces, energies, torques, etc.) of higher-order moments drop off more quickly with distance.

Do not focus on the two-point-charges, four-point-charges, etc. cases. Moments are much more general than that.

From the definition of the quadrupole moment, you can see that spherical symmetry implies $Q_{11}=Q_{22}=Q_{33}$ and the off-diagonal elements are zero. Departures from this indicate a non-spherical distribution.

The moments I have shown are Cartesian moments, which are the simplest to understand. You can also define moments using spherical harmonics, and these are more elegant because you don’t get a proliferation of indices.

$\endgroup$
3
  • $\begingroup$ How are the off-diagonal elements zero? $\endgroup$
    – amyah
    Mar 31 '20 at 20:57
  • $\begingroup$ Express $x$, $y$, and $z$ in terms of $r$, $\theta$, and $\phi$. Assume $\rho$ is only a function of $r$, and do the angular integrals. $\endgroup$
    – G. Smith
    Mar 31 '20 at 21:25
  • $\begingroup$ Ok got it thank you $\endgroup$
    – amyah
    Mar 31 '20 at 21:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.