4 ms·

Regarding the city-size: you are right that "city" is a whole range of sizes but so is the size of a neutron star, they are not all of the same size. Don't read

by eis 4y ago

Regarding the city-size: you are right that "city" is a whole range of sizes but so is the size of a neutron star, they are not all of the same size. Don't read too much into it, they just want to give a very rough idea. You can easily calculate how much mass it is losing via E=mc^2. If something emits neutrons then it is losing energy and therefore mass. There are no energyless particles. Everything has a certain energy and by that mass. Some particles have no rest-mass but that is something different. Fun fact: you could create a black hole from beams of light.

paulrouget 4y ago
I don’t think you can just use E=mc2 in that case. The rotation of the star is so high that you have to consider relativistic angular momentum effect in the equation: https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relati...
- eis 4y ago
  While you are correct that the neutron star has angular momentum and then you'd use the E^2 = (pc)^2 + m0^2c^4 for it, the question was if it is losing mass by emitting neutrons and how much. For that, the momentum energy of the Neutron star can be ignored. What does though play a role is the momentum of the particles that get shot into space and due to the escape velocity of these stars being so high it would still make sense to use this formula instead of E = mc^2. BTW these stars emit a lot more than neutrons just in case the GP thought due to their name that that's what they emit mostly. Most of the emission of neutron stars is electromagnetic radiation (EMR) aka photons. Actually neutron stars have two ways to lose energy: slowing down their rotational speed ("spin down") and losing mass. Which one of these contributes more to the EMR varies. For magnetars for example, only about 1% of the radiation is thought to be powered by rotation and for those we could actually use E = mc^2 to calculate how much rest mass the star is losing due to its radiation. That's my layman understanding at least :)
pdonis 4y ago
> If something emits neutrons Pulsars don't emit neutrons, they emit radio waves. They are neutron stars, so they're made of neutrons, but they don't emit any.

by eis 4y ago

paulrouget 4y ago

I don’t think you can just use E=mc2 in that case. The rotation of the star is so high that you have to consider relativistic angular momentum effect in the equation: https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relati...

eis 4y ago
While you are correct that the neutron star has angular momentum and then you'd use the E^2 = (pc)^2 + m0^2c^4 for it, the question was if it is losing mass by emitting neutrons and how much. For that, the momentum energy of the Neutron star can be ignored. What does though play a role is the momentum of the particles that get shot into space and due to the escape velocity of these stars being so high it would still make sense to use this formula instead of E = mc^2. BTW these stars emit a lot more than neutrons just in case the GP thought due to their name that that's what they emit mostly. Most of the emission of neutron stars is electromagnetic radiation (EMR) aka photons. Actually neutron stars have two ways to lose energy: slowing down their rotational speed ("spin down") and losing mass. Which one of these contributes more to the EMR varies. For magnetars for example, only about 1% of the radiation is thought to be powered by rotation and for those we could actually use E = mc^2 to calculate how much rest mass the star is losing due to its radiation. That's my layman understanding at least :)

pdonis 4y ago

> If something emits neutrons Pulsars don't emit neutrons, they emit radio waves. They are neutron stars, so they're made of neutrons, but they don't emit any.