4 ms·
Watch that youtube link in my sibling comment first, but I actually wanted to take a crack at an answer. What's the change in mass, ∆m, as the star cools? We'll
by acidbaseextract 10y ago
Watch that youtube link in my sibling comment first, but I actually wanted to take a crack at an answer. What's the change in mass, ∆m, as the star cools? We'll assume the mass of the neutron star is 1.5 solar masses.
E = mc^2
∆m = ∆E/c^2
So really, what's ∆E? A hyper hand-wavy estimate:
∆E = Q = mc∆T (c being specific heat)
Specific heat by mass is really hard to predict, but by moles it is fairly constant, well within an order of magnitude. So we'll discuss mass in moles.
m = neutron star moles
m = (mass of neutron star / mass of neutron) / Avogadro's #
m = 2.9580163e33 mol
c = 24 J / (mol * K)
∆T = 1e12 K - 1e6 K
→
∆E = 24 * 2.9580163e33 * (1e12 - 1e6) J = 7.0991681e45 J
Substituting that back into the original, as a neutron star cools:
∆m = 7.0991681e45 J / c^2 = 7.89888982e28 kg
Which is like 2.6% of the mass of the original star, so a pretty solid chunk. But that number is pulled out of my ass—I am not a physicist.
But there are even more weird effects going on, due to the warping of gravity the mass of neutron stars can be up to 20% less than you'd expect based on its baryonic (neutron) constituents (questions 4 and 7):
https://www.astro.umd.edu/~miller/teaching/questions/neutron.html https://www.astro.umd.edu/~miller/teaching/questions/neutron...
- m_mueller 10y agoThank you! About your last equation, wouldn't J/m^2/s^2 come out as kg? 7.9E2 would be very low then, no? I wonder how much energy is stored electromagnetically and through nuclear forces though. These things are supposed to have extremely strong EM fields and I imagine every piled up nucleus like a little atomic spring that has been depressed as much as possible. Wouldn't most of the stored energy be in there?