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I've found many articles on the geometry of pulsar beams, but I have not been able to find what would be a typical angle for the beam cone. Illustrations I've found seem to look like it might be 1 or 2 degrees.

What range of angles might you expect to find for the polar angle of a pulsar's beam cone?

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  • $\begingroup$ Hi, Jeff. I deleted my answer because I believe I misinterpreted which angle you were looking for - I wrote mine about the angular size of the beam, given what you said about the sizes you had found. But it looks like you were instead talking about the angle between the pulsar's rotational axis and magnetic axis - is that correct? $\endgroup$
    – HDE 226868 ♦
    Dec 26, 2020 at 15:52
  • $\begingroup$ I was asking about the the angle of the beam itself. $\endgroup$
    – Jeff Wolski
    Dec 31, 2020 at 5:14
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    $\begingroup$ Thanks for clarifying - looks like I understood correctly. $\endgroup$
    – HDE 226868 ♦
    Dec 31, 2020 at 5:30
  • $\begingroup$ @hde226868 I am also interested in that question (the relative inclination of the poles or beam) $\endgroup$
    – Mark C
    Jan 17, 2022 at 18:57

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Typical angles can span the range from several degrees to close to $90^{\circ}$ in some extreme cases. For pulsars with narrow emission cones, the opening angle of the emission cone is approximately $$\rho\approx1.24^{\circ}\left(\frac{r_{\text{em}}}{10\;\text{km}}\right)^{1/2}\left(\frac{P}{\text{s}}\right)^{-1/2}\propto P^{-1/2}$$ with $r_{\text{em}}$ the emission height and $P$ the period (Lorimer & Kramer, Handbook of Pulsar Astronomy). The emission height is the distance from the center of the pulsar at which the radio emission occurs; it is typically much larger than the radius of the pulsar. The expression can be derived by making some assumptions about the location of the last open magnetic field lines. The relation works quite well for long-period pulsars with $P\gg100\;\text{ms}$, which, accordingly, fall in the range $1^{\circ}\lesssim\rho\lesssim10^{\circ}$, but fails for millisecond pulsars, some of which have opening angles quite a few times larger.

To give a better visual perspective: The angle of the emission cone, $\rho$ depends on several things. In the classical Julian-Goldreich model, the pulsar is surrounded by plasma, which is forced to corotate with the neutron star by its magnetic field. If we take this to its logical extreme, there must be a radius at which rotating plasma would have to travel faster than the speed of light; this radius defines a surface we refer to as the light cylinder.

Any magnetic field lines which cross the light cylinder's surface do not close. Assuming that the emission cone must fall within the larger cone of open field lines, we can determine the largest allowed angular radius of this cone-within-a-cone if we know the emission height - the point above the pulsar at which radio emission occurs. Not surprisingly, the size of the light cylinder is determined by the rotational period; from the argument above, we can estimate $\rho$ from that and the emission height.

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  • $\begingroup$ What do you mean by emission height? Emission of what? $\endgroup$
    – zephyr
    Dec 21, 2020 at 14:38
  • $\begingroup$ @zephyr It's the distance at which radio emission takes place. $\endgroup$
    – HDE 226868 ♦
    Dec 21, 2020 at 16:20
  • $\begingroup$ Ah, okay. So the height from the surface at which the radio emissions are generated? It may help to expand the answer to include that as it isn't completely clear to me without the source book on hand. $\endgroup$
    – zephyr
    Dec 21, 2020 at 16:39
  • $\begingroup$ @zephyr Good suggestion; done. $\endgroup$
    – HDE 226868 ♦
    Dec 21, 2020 at 16:50
  • $\begingroup$ @HDE226868 I am also interested in that question (the relative inclination of the poles or beam) $\endgroup$
    – Mark C
    Feb 7, 2022 at 23:49

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