Is the ultimate reason that the neutron star loses angular momentum that the magnetic-dipole radiation is not spherically symmetric and photons have momentum?

Just for the loss of angular momentum, particles emitted from the neutron star surface are not necessary, right?

  • $\begingroup$ I'm voting to close this question as off-topic because you haven't made the connection to astronomy clear. Better on Physics. $\endgroup$
    – James K
    Commented Dec 22, 2016 at 19:26

1 Answer 1


Radiation carries a momentum per unit volume given by $$ \vec{p} = \epsilon_0 \vec{E} \times \vec{B}$$

It therefore carries an angular momentum of $$ \vec{L} = \vec{r}\times \vec{p} = \epsilon_0 \vec{r} \times (\vec{E}\times \vec{B})$$

For plane-polarised transverse electromagnetic waves propagating in the $\hat{r}$ direction, then $\vec{L}=0$. However, if the axis of a magnetic dipole is inclined to the rotation axis then the radiation is not a single pure transverse wave; the dipole moment is the sum of two oscillating magnetic dipole moments that are perpendicular and oscillate $\pi/2$ out of phase. These produce two transverse waves, with arbitrary amplitudes, perpendicular polarisations and a $\pi/2$ phase shift between them - a.k.a. elliptically polarised light. Elliptically polarised light carries angular momentum.

  • $\begingroup$ Magnetic-dipole radiation is not spherically symmetric, then a neutron star can decelerate. But the main reason is elliptically polarised light carries angular momentum? $\endgroup$ Commented Dec 23, 2016 at 12:02
  • $\begingroup$ Indeed dipolar fields are not spherically symmetric but that does not necessarily carry away angular momentum. The elliptical polarisation induced by a rotating dipole moment is I think crucial. $\endgroup$
    – ProfRob
    Commented Dec 23, 2016 at 12:44
  • $\begingroup$ I am not sure if you are right... $\endgroup$ Commented Jan 2, 2017 at 4:45
  • $\begingroup$ About what @questionhang ? $\endgroup$
    – ProfRob
    Commented Jan 2, 2017 at 8:00
  • $\begingroup$ I am not sure the elliptical polarization is the main reason. $\endgroup$ Commented Jan 2, 2017 at 11:38

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