I've become confused over the use of spherical coordinates when working with dipole moments. It would probably be best o use an example to show where I'm confused.

If we have a pure dipole, with a dipole moment pointing angle $ \theta $ from the vertical axis (which we'll call $x$), it's dipole moment is given by the vector $ \vec{p} = pcos( \theta) \hat{x} + psin( \theta) \hat{y} + 0 \hat{z}$ in Cartesian coordinates, but my book claims that it is given by (in spherical coordinates) $ \vec{p} = pcos( \theta) \hat{r} + psin( \theta) \hat{\theta}$ .

How is that possible? If you want to change from Cartesian to spherical, you use the conversation equations, which equal $\vec{p} = p \hat{r} + \theta \hat{ \theta} + 0 \hat{ \phi}$. The rest of the problem asks to find the torque on this dipole moment due to a conductor a distance away (method of images).

I get the process, but the coordinates are really causing problems for me.

I'm not asking for help solving the problem, I'm mostly confused with the coordinates, I think my book may have made a mistake.


2 Answers 2


The dipole is presumably, not at the origin! Although they have vaguely double used $\theta$; this is how the situation should look:

dipole moment

  • $\begingroup$ I see, thanks for the graph, it finally clicked, I always get the units jumbled in my head. $\endgroup$
    – Astrum
    Jul 24, 2013 at 7:13
  • $\begingroup$ @Astrum I'm glad I could help. $\endgroup$
    – Ali
    Jul 24, 2013 at 7:15
  • $\begingroup$ The dipole could be at the origin as well. This will not change the result. $\endgroup$
    – Trimok
    Jul 24, 2013 at 10:25
  • $\begingroup$ @Trimok it definitely will. At the origin $\hat r$ and $\hat \theta$ are not well defined! $\endgroup$
    – Ali
    Jul 24, 2013 at 10:31
  • $\begingroup$ @Ali : $r \hat r$ is not necessarily the position of the origin of the dipole. it could be an (external) new choice of coordinates, independant of the position of the origin of the dipole. $\endgroup$
    – Trimok
    Jul 24, 2013 at 11:03

There's nothing wrong or right in choosing Cartesian over Spherical coordinates or vice-versa. Its just that in some cases, the math in one can be easier to handle over the other.

Maybe you are getting confused in the $\theta$ chosen for both notations(they are different).

The $\theta$ you chose for the Cartesian notation, is the angle $\vec p$ makes with $\hat x$.

The $\theta$ they have chosen is the angle $\vec p$ makes with $\hat r$. Maybe that's why your conversion equation doesn't work.

More on Spherical coordinate notation here.


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