The spin-down evolution of a simple vacuum pulsar model is well known. Using the power output of radiation emission of a rotating magnetic dipole, we get the following equation (assuming that the rotation kinetic energy loss equals the electromagnetic power emitted by the rotating dipole) : \begin{equation}\tag{1} \frac{d\,}{dt} \Big( \frac{1}{2} \, I \, \omega^2 \Big) = I \, \omega \, \dot{\omega} = -\, \frac{\mu_0 \, \mu^2 \, \omega^4}{6 \pi c^3} \, \sin^2{\alpha}, \end{equation} where $\alpha$ is the tilt (or inclination) angle of the pulsar's magnetic axis, relative to the rotation axis.

I'm interested in a similar equation for the time evolution of the tilt angle : \begin{equation}\tag{2} \dot{\alpha} = \; .?. \end{equation} According to this paper : https://arxiv.org/abs/1603.01487, we should have this equation (see equation (2) from that article) : \begin{equation}\tag{3} I \, \dot{\alpha} = -\, \frac{\mu_0 \, \mu^2 \, \omega^2}{6 \pi c^3} \, \sin{\alpha} \, \cos{\alpha}. \end{equation} How to demonstrate this equation ?

And how to show that the time variation of the angular momentum (relative to the magnetic moment ?) has the following components ? \begin{align}\tag{4} & I \, \frac{d\omega}{d \, t}, && I \, \omega \, \frac{d \, \alpha}{d \, t}. \end{align} I understand only the first component, which is obvious to me, while the second is not.

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    $\begingroup$ Can you access the following links from the SAO/NASA server? Davis & Goldstein 1970, Michel & Goldwire 1970 $\endgroup$ – Michael Seifert Aug 4 '17 at 14:10
  • $\begingroup$ The short answer seems to be "calculate the torque by integrating the Maxwell stress tensor appropriately over the surface of the star." $\endgroup$ – Michael Seifert Aug 4 '17 at 14:24
  • $\begingroup$ Hmm, the calculations are very laborious, and I don't get the tilt torque part (equation (3) above). Only the standard angular part (equation (1)). Paper articles.adsabs.harvard.edu//full/1970ApJ...159L..81D/… states that the equation (3) could also be derived from symmetry and equation (1) (see the first page). I don't get it. $\endgroup$ – Cham Aug 4 '17 at 15:31

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