# Minimal mass of a neutron star.

1. Mar 5, 2009

### hhhmortal

What is the minimal mass of a neutron star, if the semi empirical mass formula is used?

2. Mar 5, 2009

### clem

If you mean the formula for the mass of a nucleus, you probably know that it doesn't apply where there are no protons.

3. Mar 5, 2009

### hhhmortal

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.

4. Mar 5, 2009

Staff Emeritus
Why isn't the answer "one neutron"?

5. Mar 5, 2009

6. Mar 5, 2009

### hamster143

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.

7. Mar 5, 2009

### humanino

Because "one neutron" is unstable ?

8. Mar 6, 2009

### malawi_glenn

an ordinary star is also unstable

9. Mar 6, 2009

### humanino

A neutron is [huge]VERY[/huge] stable for strong interaction, but unstable for time scales relevant to stars, and the gravitational interaction.

10. Mar 6, 2009

### malawi_glenn

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. Mar 6, 2009

### humanino

Yes, and I think we all agree on that.

12. Mar 6, 2009

### malawi_glenn

well the OP seemed to be confused :-)