I White dwarf collapses into a neutron star and Energy is released

AI Thread Summary
The discussion focuses on estimating the energy released during the collapse of a white dwarf into a neutron star, specifically using the Chandrasekhar mass. Participants suggest calculating the gravitational potential energy of both the white dwarf and neutron star, assuming uniform density for both. The energy released in the collapse is estimated to be around 10^53 ergs. Additionally, there is a conversation about the temperature and luminosity of the resulting neutron star, with one participant noting an incorrect estimate of 17 billion solar luminosities per square meter. The approximation of constant density is deemed valid due to the degeneracy pressure supporting both types of stars.
donielix
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Hello everyone.
I am trying to solve a problem whose statement reads as follows:

Neutron stars have radii of ##\sim 10## km. If we assume that before the collapse the nucleus of the white dwarf precursor has a mass equal to that of Chandrasekhar, estimate the energy generated in the collapse.

I've tried to solving it by considering a ##n=\dfrac{3}{2}## polytrope, but no more information is provided and i don't know how to relate mass to the radii with no knowledge of central density (necessary for apply polytropic relations). Any ideas? Thank you and sorry for my bad english
 
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It says "estimate". I would just assume that both stars have a uniform density and calculate the total gravitational potential energy in both cases. You are given the radius of the neutron star and the mass, and you can look up the radius of a white dwarf of Chandrasekhar mass.
 
But how i can relate Mass to Radio, can i use the relationship ##R\sim M^\dfrac{1-n}{3-n}## and then with that initial radii calculate the difference in potential energy between 2 radius?
And why can we assume that potential energy is the only source of energy?
Thanks for you response
 
donielix said:
But how i can relate Mass to Radio, can i use the relationship ##R\sim M^\dfrac{1-n}{3-n}## and then with that initial radii calculate the difference in potential energy between 2 radius?
And why can we assume that potential energy is the only source of energy?
Thanks for you response

Problems like this are intended to test not just your ability to do calculations, but your grasp of the physical processes. The problem is somewhat ambiguous, so you could always ask for clarification. But I would just assume a white dwarf with a mass of 1.4 Msun and a radius of about 10,000 km collapses to a neutron star with a mass of 1.4 Msun and a radius of 10 km. If you assume both the initial white dwarf and the final neutron star have uniform densities, you can calculate the gravitational potential energy (i.e. the binding energy) in both cases, and estimate how much energy is released. What other energy sources did you have in mind? Fusion energy? You could estimate the energy of other sources as well and see how they compare to the gravitational energy released.
 
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Likes stefan r
Ok phyzguy thank you very much for your answer.
Finally I've considered the mass-radio relationship that I previously mencioned and assumed that mass is constant during process.
The result obtained is in order about ##\sim 10^{53}\ \text{erg}## of energy released in the collapse
 
I also got about 10^53 ergs.
 
Great! Thank you for your help
 
As a bonus, estimate temperature of the resulting NS and its luminosity.

Last time I did it, I've got something like 17 billion solar luminosities... from each square meter of NS surface.
 
The approximation of constant density for both stars is valid because they are both supported by degeneracy pressure - the white dwarf by degenerate electrons, and the neutron star by degenerate neutrons...
 
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nikkkom said:
As a bonus, estimate temperature of the resulting NS and its luminosity.

Last time I did it, I've got something like 17 billion solar luminosities... from each square meter of NS surface.
That wouldn't be correct though, stars typically have a more lower surface temperature than interior temperature-- and for exactly that reason.
 
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