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- Shapiro & Teukolsky describe a simple derivation by Landau of the maximum mass limit for white dwarfs and neutron stars. They say that the result is the same for both types of object. But is that correct?
In Section 3.4 of Shapiro & Teukolsky (1983), a simple derivation, due to Landau, of the maximum mass limit for white dwarfs and neutron stars is given. I will briefly describe it here and then pose my question.
The basic method is to derive an expression for the total energy (excluding rest mass energy--this is not mentioned in Shapiro & Teukolsky, but I take it that it is assumed to be constant and so drops out of the variational method being used) of the object, and then ask under what conditions that energy will have a minimum, indicating a stable equilibrium. The energy is the sum of two terms: a positive term coming from the Fermi energy of the fermions that are producing the degeneracy pressure, and a negative term coming from the gravitational potential energy.
The object is assumed to contain ##N## fermions that produce degeneracy pressure, and some number of baryons of mass ##m_B## that produce the mass of the object. For an object of radius ##R##, the number density of fermions is ##n = N / R^3##, and the momentum of a fermion is proportional to ##\hbar n^{1/3}##. So the Fermi energy per fermion is the momentum times ##c##, or
$$
E_F \approx \frac{\hbar c N^{1/3}}{R}
$$
The gravitational potential energy per fermion is given by
$$
E_G \approx - \frac{G M m_B}{R}
$$
where ##M = N m_B##. The total energy is then ##E = E_F + E_G##, and an argument is then given (which I am not questioning) that there will only be a minimum of ##E## if ##N## is less than a critical value, which denotes the maximum mass.
My question is about the formula ##M = N m_B##. For a neutron star, that is obviously true, since the neutrons play both roles: they provide the degeneracy pressure and they provide the mass. For a white dwarf, however, it does not seem correct: ##N## is the number of electrons, but ##M## comes from the baryons, and the number of baryons will be larger than the number of electrons (since not all of the baryons are protons). So it seems like the formula for white dwarfs should be ##M = (N / f) m_B##, where ##f < 1## is the fraction of baryons that are protons (so ##N / f## is the total number of baryons).
I don't see this mentioned anywhere in Shapiro & Teukolsky, so I'm wondering if I've missed something, or if they just left it out, presumably because they were only giving a rough order of magnitude argument and weren't concerned about numerical factors of order unity. The cube root of the factor ##f## would be what would appear in the final formula for the maximum mass [Edit--this turns out not to be correct, it should be f^2, see post #14], so the effect on the result would not be large. But it would mean that the results, even at this rough order of magnitude level, would be different for white dwarfs and neutron stars; as far as I can tell from the description in Shapiro & Teukolsky, Landau's original 1932 paper in which he published his result said they were the same to this level of approximation.
The basic method is to derive an expression for the total energy (excluding rest mass energy--this is not mentioned in Shapiro & Teukolsky, but I take it that it is assumed to be constant and so drops out of the variational method being used) of the object, and then ask under what conditions that energy will have a minimum, indicating a stable equilibrium. The energy is the sum of two terms: a positive term coming from the Fermi energy of the fermions that are producing the degeneracy pressure, and a negative term coming from the gravitational potential energy.
The object is assumed to contain ##N## fermions that produce degeneracy pressure, and some number of baryons of mass ##m_B## that produce the mass of the object. For an object of radius ##R##, the number density of fermions is ##n = N / R^3##, and the momentum of a fermion is proportional to ##\hbar n^{1/3}##. So the Fermi energy per fermion is the momentum times ##c##, or
$$
E_F \approx \frac{\hbar c N^{1/3}}{R}
$$
The gravitational potential energy per fermion is given by
$$
E_G \approx - \frac{G M m_B}{R}
$$
where ##M = N m_B##. The total energy is then ##E = E_F + E_G##, and an argument is then given (which I am not questioning) that there will only be a minimum of ##E## if ##N## is less than a critical value, which denotes the maximum mass.
My question is about the formula ##M = N m_B##. For a neutron star, that is obviously true, since the neutrons play both roles: they provide the degeneracy pressure and they provide the mass. For a white dwarf, however, it does not seem correct: ##N## is the number of electrons, but ##M## comes from the baryons, and the number of baryons will be larger than the number of electrons (since not all of the baryons are protons). So it seems like the formula for white dwarfs should be ##M = (N / f) m_B##, where ##f < 1## is the fraction of baryons that are protons (so ##N / f## is the total number of baryons).
I don't see this mentioned anywhere in Shapiro & Teukolsky, so I'm wondering if I've missed something, or if they just left it out, presumably because they were only giving a rough order of magnitude argument and weren't concerned about numerical factors of order unity. The cube root of the factor ##f## would be what would appear in the final formula for the maximum mass [Edit--this turns out not to be correct, it should be f^2, see post #14], so the effect on the result would not be large. But it would mean that the results, even at this rough order of magnitude level, would be different for white dwarfs and neutron stars; as far as I can tell from the description in Shapiro & Teukolsky, Landau's original 1932 paper in which he published his result said they were the same to this level of approximation.
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