<p>Typical angles can span the range from several degrees to close to <span class="math-container">$90^{\circ}$</span> in some extreme cases. For pulsars with narrow emission cones, the opening angle of the emission cone is approximately
<span class="math-container">$$\rho\approx1.24^{\circ}\left(\frac{r_{\text{em}}}{10\;\text{km}}\right)^{1/2}\left(\frac{P}{\text{s}}\right)^{-1/2}\propto P^{-1/2}$$</span>
with <span class="math-container">$r_{\text{em}}$</span> the emission height and <span class="math-container">$P$</span> the period (Lorimer &amp; Kramer, <em>Handbook of Pulsar Astronomy</em>). 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 <span class="math-container">$P\gg100\;\text{ms}$</span>, which, accordingly, fall in the range <span class="math-container">$1^{\circ}\lesssim\rho\lesssim10^{\circ}$</span>, but fails for millisecond pulsars, some of which have opening angles quite a few times larger.</p>
<p>To give a better visual perspective: The angle of the emission cone, <span class="math-container">$\rho$</span> 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.</p>
<p>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 <span class="math-container">$\rho$</span> from that and the emission height.</p>

CC BY-SA 4.0
2020-12-21T14:38:36.727Z
What do you mean by emission height? Emission of what?
zephyr
CC BY-SA 4.0
2020-12-21T16:20:24.483Z
@zephyr It's the distance at which radio emission takes place.
HDE 226868
CC BY-SA 4.0
2020-12-21T16:39:11.440Z
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.
zephyr
CC BY-SA 4.0
2020-12-21T16:50:28.640Z
@zephyr Good suggestion; done.
HDE 226868
CC BY-SA 4.0
2022-02-07T23:49:24.287Z
@HDE226868 I am also interested in that question (the relative inclination of the poles or beam)
Mark C
CC BY-SA 4.0
2020-12-18T18:46:12.393Z
2020-12-22T03:48:12.693Z
2020-12-22T03:48:12.693Z
HDE 226868
<p>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.</p>
<p>What range of angles might you expect to find for the polar angle of a pulsar's beam cone?</p>

CC BY-SA 4.0
2020-12-26T15:52:09.303Z
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?
HDE 226868
CC BY-SA 4.0
2020-12-31T05:14:31.107Z
I was asking about the the angle of the beam itself.
Jeff Wolski
CC BY-SA 4.0
2020-12-31T05:30:13.020Z
Thanks for clarifying - looks like I understood correctly.
HDE 226868
CC BY-SA 4.0
2022-01-17T18:57:31.420Z
@hde226868 I am also interested in that question (the relative inclination of the poles or beam)
Mark C
<pulsar>
What is a typical polar angle of a pulsar beam?
Jeff Wolski
# What is a typical polar angle of a pulsar beam?

(Asked by: username_0 on 2020-12-18)

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?

**Tags:** pulsar

> **Comments:**
> **username_1**: 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?  
> **username_0**: I was asking about the the angle of the beam itself.  
> **username_1**: Thanks for clarifying - looks like I understood correctly.  
> **username_2**: @hde226868 I am also interested in that question (the relative inclination of the poles or beam)  

---

## 1 answers

### Answer by username_1 on 2020-12-18. Score: 6 | Accepted answer

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.

> **Comments:**
> **username_3**: What do you mean by emission height? Emission of what?  
> **username_1**: @username_3 It's the distance at which radio emission takes place.  
> **username_3**: 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.  
> **username_1**: @username_3 Good suggestion; done.  
> **username_2**: @HDE226868 I am also interested in that question (the relative inclination of the poles or beam)  

---

