# Crab pulsar radiation

Regarding the radiation from the Crab pulsar and nebula, some reading leads me to the following conclusions.

1. The rotation energy of the neutron star is the source of the non-thermal electromagnetic radiation from the neutron star and the nebula. The rotation also supplies the energy for the electrons in the pulsar wind. The magnetic field of the neutron star acts as an intermediary.
2. Most of the non-thermal electromagnetic energy is emitted by the neutron star as magnetic dipole radiation at the spin frequency of the pulsar (Crab: $\lambda=10^7\text{ m}$).
3. A small fraction of the non-thermal electromagnetic energy is emitted by the surrounding nebula as radio waves ($\lambda\ll10^7\text{m}$), visible light, röntgen and gamma radiation. This fraction may be curvature or synchrotron radiation or result from inverse compton scattering.
4. The surface of the neutron star produces blackbody radiation at a temperature of about 2 MK. The source of this energy is thermal.
5. A young pulsar like the Crab also looses a lot of thermal energy by neutino emission (URCA process, neutrino bremsstrahlung).

Are these conclusions more or less correct?

• Partly self-contradictory. If the rotational KE is the source of the power and goes mostly into long wavelength radio waves, then these must dominate the energy output? It's also very hard to absorb very long wavelength radio waves. – Rob Jeffries May 16 '15 at 8:08
• 10km is not $10^{7}$ m. The rest seems more-or-less ok. Neutron stars of course cool with time. Neutrinos only dominate the cooling for $10^4$ years. Beg your pardon about the absorption of the radio waves. Yes, 30Hz is well below the plasma frequency of the nebula, so they do get absorbed. – Rob Jeffries May 16 '15 at 20:09
• Hopefully the refrased conclusions above aren't self-contradictory anymore. I will ask the professor who wrote that the $10^7$ m wavelength radiation "is absorbed by the gas in the Crab Nebula" – gamma1954 May 16 '15 at 20:16

1. Most of the non-thermal electromagnetic energy is emitted by the neutron star as magnetic dipole radiation at the spin frequency of the pulsar (Crab: λ=107 m).

No, this is not correct. The vast majority of electromagnetic radiation is produced only after the Poyting dominated cold relativistic wind passes through the termination shock. Here is where the plasma is shock-heated and enters the nebula bubble.

Even the jets (and their radiative emission) are directly caused by this post shock flow. Almost all of the radition produced by the Crab is in one way or the other a result of shock heating via the termination shock. All of the main radiative features of the nebula can be modeled with astounding accuracy via the post shock flow alone.

I'd say your other points are about right (obviously omitting lots of intricate details). I hope this helps.

Edit. If you would like to read a paper that covers some aspect of what I have said above, see MNRAS paper [Observations of "wisps" in magnetohydrodynamic simulations of the Crab Nebula] (http://arxiv.org/pdf/0907.3647v3.pdf).

• This is interesting and I'd like to understand it. Most (elementary) literature claims that the spin down of the pulsar can be equated with the power lost in magnetic dipole radiation and that it is this radiation that is ultimately the power source for the nebula. Are you saying that the dominant way that energy is directly transferred into the nebula from the pulsar is different and that the amount of non-thermal radiation is not dominated by magnetic dipole emission? Is that true of just the nebula (true because the dipole emission is absorbed I guess), but what about from the pulsar? – Rob Jeffries Jun 8 '15 at 15:38
• At the stellar surface, the pulsar’s huge magnetic fields and rapid rotation induce enormous electric fields within the magnetosphere, these consequently tear particles from the stellar surface and accelerate them to high energies. Plasma then fills the magnetosphere and the extreme magnetic field present is sufficient to cause the plasma to rigidly co-rotate. However, this co-rotation must cease somewhere near the light cylinder, and the particles flow along the opened magnetic field lines, carrying away energy in the form of an ultrarelativistic magnetized wind. – MoonKnight Jun 8 '15 at 19:10
• OK, I get this. But how does the kinetic energy flux of these particles compare with the power emitted by magnetic dipole radiation? As I say, most texts say that $\partial_t (0.5 I\omega^2)$ equals the magnetic dipole radiation power. And this number is pretty big. – Rob Jeffries Jun 8 '15 at 19:14
• The spin-down power of the pulsar is spent in the generation of magnetic field in the wind zone. Outside of the rigid co-rotating plasma the wind is cold and highly collomated, in almost all models, the relativistic wind leaving the magnetosphere must be Poynting flux dominated. However, somewhere between the light cylinder and the termination shock the wind changes from Poynting flux dominated to particle flux dominated - this leads us to the termination shock where the radiation mechanisms kick in as the flow is decelerated and shock heated. As for your question... – MoonKnight Jun 8 '15 at 19:19
• Any radiative energy produced inside the magnetosphere will be small compared to the energy carried off in the wind. How do I now this, well [without dusting off the pencil] if you look at the overall luminosity of the nebula coupled with it's expansion rate you can infer the energy required to inflate such a nebula (if you are happy with what I have said so far). This luminosity alone will be a vast proportion of the spin-down power of the star, hence we can say that the energy carried away in the ultrarelativistic wind dominates all other mechanisms of energy release... – MoonKnight Jun 8 '15 at 19:26