# Difference between revisions of "Main sequence"

### Main Sequence

\We understand now how to calculate the luminosity produced by fusion, in addition to understanding how luminous a star needs to be from energy transport. We have

${\displaystyle L_{\rm {fusion}}=\int dM_{r}\epsilon (\rho ,T)\,\!}$

and

${\displaystyle L\propto M^{3}{\rm {\;\;(Thomson)}}\,\!}$
${\displaystyle L\propto M^{5.5}R^{-0.5}{\rm {\;\;(free-free)}}\,\!}$
${\displaystyle L\propto M^{4/7}R^{2}{\rm {\;\;(Convection)}}\,\!}$

When the fusion luminosity is equal to the luminosity that can leak out, then the star is in steady state. This condition, plus hydrostatic equilibrium, give the properties of stars on the main sequence. In other words, given the mass, you can solve for ${\displaystyle L,T_{c},T_{\rm {eff}},R}$. \\We can start with an object that is not fusing, but is radiating energy away. That is, it is undergoing Kelvin Helmholtz contraction. As the object contracts, its central temperature increases, which in turn increases the luminosity due to fusion (once it is hot enough for at least some fusion to occur). This fusion luminosity will eventually increase to the point that it is equal to the luminosity being radiated away, at which point the object ceases contracting. Once this happens, the star ceases contracting, the central temperature stays fixed, the luminosity is fixed, and the star is on the main sequence. Well, almost fixed. The composition of the star will change as it fuses, which will cause very slow changes in these properties even while it is on the main sequence. This phase of hydrogen fusing lasts longer than any other stage of a star’s lifetime. \\For the Sun, we showed that radiative diffusion, Thomson scattering, and the virial theorem imply that ${\displaystyle L\sim L_{\odot }}$, independent of fusion. Then, knowing that fusion of hydrogen in to helium is the energy source, we determined that the central temperature of the Sun is ${\displaystyle T_{c}\sim 10^{7}}$ K. We then used the virial theorem to determine the radius of the star to be ${\displaystyle R\sim 10^{11}}$ cm. The way to think about this is that the central temperature is not set by the virial theorem, but fusion sets the central temperature, and this and the virial theorem set the radius. Once we know the radius and luminosity, we can use the Stefan-Boltzmann equation to determine the effective temperature to be ${\displaystyle T_{\rm {eff}}\sim 6000}$ K. \\What we want to do now is to do this for other stars, and explain the main sequence portion of the Hertzsprung-Russell diagram. In principle, this is easy, as we can just follow the exact same procedure for other stars as we did for the Sun. What complicates this is that for other stars, the opacity, energy generation, energy transport, and dominant pressure may be different. This is because all of these properties depend on density and temperature, both of which depend on the mass of the star. Thus, in order to do this correctly, we need to know what the dominant mechanism for each of these things is for other stars. \\We showed on homework that at the center of the Sun, free-free opacity and Thomson opacity are roughly of equal importance. Thomson opacity is constant with density and temperature, but ${\displaystyle \kappa _{f-f}\propto \rho T^{-7/2}}$. For stars more massive than the Sun, the central temperature is larger and the density is lower, meaning Thomson scattering dominates for stars more massive than the Sun, and free-free dominates for stars less massive than the Sun. \\The energy produced by the CNO cycle is about 1% of the energy produced by the proton-proton chain in the Sun. Since the CNO cycle scales as ${\displaystyle T^{20}}$ whereas the p-p chain scales as ${\displaystyle T^{4}}$, the CNO cycle is dominant for stars somewhat more massive than the Sun, and the proton-proton chain is dominant for stars the Sun’s mass and lower. \\Gas pressure dominantes for all but the most massive stars. Stars less massive than the Sun have convective envelopes and radiative cores. Stars more massive than the Sun have convective cores and radiative envelopes. Knowing all of this, we can look at the main sequence of mass ranges of stars. \\First, we’ll look at stars less massive than the Sun down to about one third of a solar mass. These have free-free opacity, photons transport the energy, proton-proton chain, and are gas pressure dominated. We know that

${\displaystyle L_{\rm {rad}}\propto M^{5.5}R^{-0.5},\,\!}$
${\displaystyle L_{p-p}=\int \epsilon _{p-p}dM_{r}=L_{\rm {rad}},\,\!}$
${\displaystyle L_{p-p}\sim \epsilon _{c}M,\,\!}$
${\displaystyle L_{p-p}\sim \rho T^{4.5}M,\,\!}$
${\displaystyle T_{c}\propto {\frac {M}{R}},\,\!}$
${\displaystyle L_{p-p}\sim {\frac {M^{6.5}}{R^{7.5}}}.\,\!}$

Equating the radiative losses to the energy generation,

${\displaystyle L_{p-p}\sim L_{\rm {rad}},\,\!}$
${\displaystyle {\frac {M^{6.5}}{R^{7.5}}}\propto {\frac {M^{5.5}}{R^{0.5}}},\,\!}$
${\displaystyle R\propto M^{1/7},\,\!}$
${\displaystyle R=R_{\odot }\left({\frac {M}{M_{\odot }}}\right)^{1/7}\,\!}$

The last step is possible because we are able to normalize to the Sun. Likewise for central temperature, we find

${\displaystyle T_{c}\propto {\frac {M}{R}}\propto M^{6/7},\,\!}$
${\displaystyle T_{c}\propto (1.5\times 10^{7}K)\left({\frac {M}{M_{\odot }}}\right)^{6/7}.\,\!}$

The radial change with mass is tiny, so we can just say

${\displaystyle L=L_{\odot }\left({\frac {M}{M_{\odot }}}\right)^{5.5}.\,\!}$

We can also solve for ${\displaystyle R}$ in terms of ${\displaystyle L}$, and find

${\displaystyle R\propto L^{1/40}.\,\!}$

This is tiny, so we can just say that

${\displaystyle T_{\rm {eff}}=5800K\left({\frac {L}{L_{\odot }}}\right)^{1/4}.\,\!}$

This is everything we would want to know for a star that is between roughly 0.3 and 1 solar mass, but outside this mass range, these results do not hold. Let’s do this again then for stars that are more massive than the Sun, ranging from roughly 1–50 solar masses. Here, CNO is dominant, gas pressure is dominant, and Thomson opacity is dominant. We showed in lecture that

${\displaystyle L_{\rm {rad}}=L_{\odot }\left({\frac {M}{M_{\odot }}}\right)^{3}.\,\!}$

The energy generated by fusion is

${\displaystyle L_{\rm {fusion}}=\int \epsilon dM_{r}\sim \epsilon _{c}M,\,\!}$
${\displaystyle \epsilon _{c}\propto \rho T^{20},\,\!}$
${\displaystyle \epsilon _{c}\propto {\frac {M^{21}}{R^{23}}}.\,\!}$
${\displaystyle L_{\rm {fusion}}\propto {\frac {M^{22}}{R^{23}}}.\,\!}$

Equating the radiative luminosity to the fusion luminosity gives

${\displaystyle {\frac {M^{22}}{R^{23}}}\propto M^{3},\,\!}$
${\displaystyle R\propto M^{19/23}\propto M^{0.8},\,\!}$
${\displaystyle R=R_{\odot }\left({\frac {M}{M_{\odot }}}\right)^{0.8}.\,\!}$

Using the virial theorem gives the central temperature dependence of

${\displaystyle T_{c}=(1.5\times 10^{7}K)\left({\frac {M}{M_{\odot }}}\right)^{0.2}.\,\!}$

For stars on the CNO cycle, the temperature only needs to change slightly to keep up with the extra luminosity. Finally, Stefan-Boltzmann gives

${\displaystyle T_{\rm {eff}}=5800K\left({\frac {M}{M_{\odot }}}\right)^{3/8}.\,\!}$

Or, in terms of luminosity, useful for plotting the HR diagram,

${\displaystyle T_{\rm {eff}}=5800K\left({\frac {L}{L_{\odot }}}\right)^{1/8}.\,\!}$

Thus there is a large range of luminosities for stars on the main sequence, despite a relatively small range in effective temperatures. \\We could do this again, this time for stars more massive than 50 solar masses. These only change in that the gas pressure no longer dominates, with radiation pressure taking over. This gives ${\displaystyle L\propto M}$. We will not actually do this. Likewise, we could do fully convective stars, but we’ll leave that for homework. \\Instead, let’s explain why massive stars have convective cores. Convection sets in when there is a negative entropy gradient. A more useful condition in practice is that entropy sets in when

${\displaystyle {\frac {d\ln T}{d\ln P}}>{\frac {\gamma -1}{\gamma }}.\,\!}$

Remember that ${\displaystyle \gamma }$, the adiabatic index, is for the pressure providing particles. Then, for gas pressure dominated stars, we have ${\displaystyle \gamma =5/3}$. We showed that the left hand side can be written

${\displaystyle {\frac {d\ln T}{d\ln P}}={\frac {1}{4}}{\frac {P}{P_{\rm {rad}}}}{\frac {L}{L_{edd}}}{\frac {L_{r}/L}{M_{r}/M}}.\,\!}$

Thus convection sets in if

${\displaystyle {\frac {P}{P_{\rm {rad}}}}{\frac {L}{L_{edd}}}{\frac {L_{r}/L}{M_{r}/M}}>{\frac {8}{5}}.\,\!}$

Rearranging this in terms of the enclosed mass, convection sets in if

${\displaystyle {\frac {M_{r}}{M}}<{\frac {5}{8}}{\frac {P}{P_{\rm {rad}}}}{\frac {L}{L_{edd}}}{\frac {L_{r}}{L}}.\,\!}$

For the CNO cycle, ${\displaystyle L_{r}\approx L}$ even for very small radii. We can get a rough sense of when this is important by looking at the Sun (even though it does not strictly apply for the Sun). We have ${\displaystyle P/P_{\rm {rad}}\sim 3000}$, and ${\displaystyle L/L_{edd}\sim 4\times 10^{-5}}$. This gives convection for

${\displaystyle {\frac {M_{r}}{M}}<0.1.\,\!}$

This is why stars more massive than the Sun have convective cores. The Sun only barely avoids this, because it is dominated by the proton-proton chain, which is slightly less temperature sensitive. The convective core sets in for stars about 1.3 ${\displaystyle M_{\odot }}$ and above.

### Minimum and Maximum Masses of Stars

\Not every ball of gas in hyrdostatic equilibrium is a star. This is reserved for objects that are undergoing, or at some point undergo, fusion of hydrogen in to helium. Stars only exist in between a range of masses. The lower mass limit is set by where degeneracy pressure kicks in. Masses below this limit are called brown dwarfs, and never fuse hydrogen in to helium. The upper limit is less well understood both observationally and theoretically. The upper limit has something to do with being supported by radiation pressure, but the details are unclear. \\So far, we have considered gas in stars to be a classical ideal gas. To understand the minimum mass of stars, we need to consider quantum mechanical effects in a gas. The way to think about this is that the quantum mechanical nature of particles is important if the de Broglie wavelength is of order the separation between particles. The de Broglie wavelength is just the wavelength associated with the thermal momentum,

${\displaystyle \lambda \approx {\frac {h}{p_{th}}}.\,\!}$

When this wavelength is much smaller than the distance between particles, the gas does not care about quantum mechanics. So the condition for quantum mechanics to be important is

${\displaystyle {\frac {h}{p_{th}}}>n^{-1/3}.\,\!}$

Or, writing in terms of the density, and putting the momentum in terms of the temperature,

${\displaystyle n>\left({\frac {mkT}{h^{2}}}\right)^{3/2}.\,\!}$

We will define this density as the quantum number density, ${\displaystyle n_{Q}}$. The particle with the lowest value of ${\displaystyle n_{Q}}$ is the one for which quantum mechanics is important first. Since there is an ${\displaystyle m}$ in the numerator, that means that electrons become degenerate first. The quantum number density does depend on temperature, since the thermal momentum affects the de Broglie wavelength of particles. For electrons, we find

${\displaystyle n_{q}\approx 10^{26}T_{7}^{3/2}{\rm {\;cm^{-3}}}.\,\!}$

At the center of the Sun, the temperature is about ${\displaystyle 10^{7}}$, and the density is ${\displaystyle n\sim n_{Q}}$. For this reason, detailed models of the Sun always treat quantum mechanics when solving for the Sun’s structure. We can see how the importance of this scales for other stars. Assuming a scaling between ${\displaystyle M}$ and ${\displaystyle R}$ of

${\displaystyle R\propto M^{3/4},\,\!}$

we find

${\displaystyle T_{c}\propto M^{1/4}\,\!}$

and

${\displaystyle \rho \propto M^{-5/4}.\,\!}$

Thus, as ${\displaystyle M}$ decreases, the central temperature decreases and the density increases, which means that quantum mechanics becomes increasingly important at lower masses. For increasing ${\displaystyle M}$, the opposite is true, and quantum mechanics is less important for stars more massive than the Sun. \\When ${\displaystyle n>>n_{Q}}$, we consider the gas to be fully degenerate. This is sometimes called the low ${\displaystyle T}$ limit, where low temperature is a very relative term. The number of particles as a function of momentum is

${\displaystyle n(p)={\frac {g/h^{3}}{e^{(E_{p}-\mu )/kT}\pm 1}}.\,\!}$

Here, ${\displaystyle E_{p}}$ is the total energy of the particle (mass plus momentum) and ${\displaystyle \mu }$ is the chemical potential. The chemical potential is defined as

${\displaystyle \mu =\left({\frac {dE}{dN}}\right)_{S,V}.\,\!}$

This is a way of quantifying how changing the number of particles changes the enrgy of a system. The full thermodynamic expression is

${\displaystyle dE=TdS-PdV+\mu dN.\,\!}$

Finally, in the ${\displaystyle n(p)}$ expression, the ${\displaystyle +}$ is for particles which satisfy the Pauli exclusion principle, and the ${\displaystyle -}$ is for Bose-Einstein particles. In the classical limit, the quantum mechanical nature of this expression (${\displaystyle \pm }$) is unimportant. So QM is unimportant when the exponential term is much greater than 1. Or, alternatively, the number density is much less than the quantum mechanical number density. We are actually interested in the other limit, and the fermion expression (${\displaystyle +}$), since electrons, protons, and neutrons are fermions. Then this distribution function is

${\displaystyle n(p)={\frac {g/h^{3}}{e^{(E_{p}-\mu )/kT}+1}}.\,\!}$

Let’s first imagine very low energies. Then ${\displaystyle E_{p}}$ is much smaller than ${\displaystyle \mu }$, and ${\displaystyle kT}$ is much smaller than ${\displaystyle \mu }$. This means the exponential term goes to 0, and the entire expression goes to

${\displaystyle n(p)\approx {\frac {2}{h^{3}}}.\,\!}$

Note: the remainder is missing, as note taker had to leave