Minimal mass of a neutron star.

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SUMMARY

The minimal mass of a neutron star, derived using the semi-empirical mass formula, is approximately 0.05 solar masses (Msun), equating to 9 x 1028 kg. The calculation involves parameters such as the binding energy, gravitational constant (G ≈ 6.674 x 10-11 m3 kg-1 s-2), and neutron mass (mn ≈ 1.67 x 10-27 kg). The formula assumes a uniform mass distribution, which is not accurate for neutron stars, highlighting the limitations of the semi-empirical mass formula in this context.

PREREQUISITES
  • Understanding of the semi-empirical mass formula
  • Knowledge of gravitational binding energy concepts
  • Familiarity with neutron stability and interactions
  • Basic grasp of astrophysical mass calculations
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  • Research the implications of gravitational binding energy in neutron stars
  • Study the stability of nucleons under strong interactions
  • Explore advanced astrophysical models for neutron star formation
  • Investigate the limitations of the semi-empirical mass formula in various contexts
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Astrophysicists, theoretical physicists, and students studying stellar evolution and neutron star properties will benefit from this discussion.

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What is the minimal mass of a neutron star, if the semi empirical mass formula is used?
 
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If you mean the formula for the mass of a nucleus, you probably know that it doesn't apply where there are no protons.
 
clem said:
If you mean the formula for the mass of a nucleus, you probably know that it doesn't apply where there are no protons.

I thought the coulomb term would cancel and you would have to add a 0.6GM^2 / R and some how the binding energy would be positive.
 
Why isn't the answer "one neutron"?
 
Vanadium 50 said:
Why isn't the answer "one neutron"?

:biggrin: :approve:
 
It's a straightforward substitution. When you discard negligible factors, you get simply this
E_{binding} = (a_V-a_A) A - a_s A^{2/3} + 0.6 G m_n^2 A^2 / (r_0 A^{1/3}) = 0.6 G m_n^2 A^{5/3} / r_0 - (a_A-a_V) A

A_{min} = (5/3 (a_A-a_V) r_0 / (G m_n^2))^{1.5}

a_A-a_V \approx 8 MeV
G \approx 6.674 * 10^{-11} m^3 kg^{-1} s^{-2}
r_0 \approx 1.25 * 10^{-15} m
m_n \approx 940 MeV \approx 1.67*10^{-27} kg

and we may need to insert a few powers of c to get the dimensions right. one way to do it is this

A_{min} \approx (5/3 * 8/940 * (3*10^8 m/s)^2 * (1.25 * 10^{-15} m / (6.674 * 10^{-11} m^3 kg^{-1} s^{-2} * 1.67*10^{-27} kg))^{1.5} = (5/3 * 8/940 * 9 * 1.25 / (6.674*1.67) 10^{39})^{1.5} \approx 5.4 * 10^{55}

M_{min} = 9*10^{28} kg = 0.05 M_{sun}

and of course we should be happy to get the right order of magnitude, considering how crude our model is ... for example, the formula for gravitational binding energy assumes uniform mass distribution, which is not the case in a neutron star. And applicability of the semi-empirical mass formula to neutron star is a very big and rather unfounded assumption.
 
Vanadium 50 said:
Why isn't the answer "one neutron"?
Because "one neutron" is unstable ?
 
humanino said:
Because "one neutron" is unstable ?

an ordinary star is also unstable
 
malawi_glenn said:
an ordinary star is also unstable
A neutron is [huge]VERY[/huge] stable for strong interaction, but unstable for time scales relevant to stars, and the gravitational interaction.
 
  • #10
The thing Vanadium is pointing out is perhaps that a neutron star is an object of nucleons which is held together by gravity, thus one should not imagine a neutron star as a huge 'nuclei'
 
  • #11
malawi_glenn said:
The thing Vanadium is pointing out is perhaps that a neutron star is an object of nucleons which is held together by gravity, thus one should not imagine a neutron star as a huge 'nuclei'
Yes, and I think we all agree on that.
 
  • #12
humanino said:
Yes, and I think we all agree on that.

well the OP seemed to be confused :-)
 

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