We want to investigate the Dynamical Stability, Thermal Stability and Eddington Luminosity limits of a star. And likewise how this affects the minimum and maximum mass of a star. And finally what processes are responsible for fueling the star and how this affects the energy production.
To understand dynamical stability we need to check its adiabatic index if it is close to \(\frac{4}{3}\) Here is a demonstration of why the adiabatic index close to 4/3 is bad.
For something to be in adiabatic condition it must satisfy the following: \[PV^{\gamma} = \text{constant}\] \[d(PV^{\gamma}) = 0\] \[P dV^{\gamma} + V^{\gamma} dP = 0\] \[P dV^{\gamma} = -V^{\gamma} dP\] \[P \gamma V^{\gamma-1}dV = -V^{\gamma} dP\] \[PdV = -\frac{V}{\gamma}dP\]
We can also check ideal gas law \[PV = nkT \sim U\] \[dPV = d U\] \[VdP + PdV = d U\] \[VdP -\frac{V}{\gamma}dP = d U\] \[(1 -\frac{1}{\gamma})dP = \frac{d U}{V}\] \[(\frac{\gamma -1}{\gamma})dP = \frac{d U}{V}\] \[P = \frac{\gamma}{\gamma -1}\frac{ U}{V}\] We see that for the virials theroem says \[\Omega = -3 \bar P V\] \[\Omega = -3 \frac{\gamma}{\gamma -1}\frac{ U}{V} V\] \[\Omega = -3 \frac{\gamma}{\gamma -1}U\] \[U + \Omega = U - 3 \frac{\gamma}{\gamma -1}U\] \[(1 - 3 \frac{\gamma}{\gamma -1})U = 0 \Rightarrow \gamma = \frac{4}{3}\] if \(\gamma = \frac{4}{3}\) then we get zero!
A photon gas is a collection of photons in thermal equilibrium, behaving as a statistical system with specific thermodynamic properties. We can think of this in relation to radiation pressure within stars. We can see that the photon energy density: \[ u \propto aT^4\] Since photons are relativistic we see that equation of state has no dependence on density. \[P = 1/3 U/V = \frac{1}{3} u\] \[P = \frac{1}{3} aT^4\] \[P = \frac{1}{3} aT^4\] This is blackbody radiation.
We see that for there to be gas pressure to dominate over the radiation pressure or else the adiabatic index will be close to 4/3. Let us see how this is the case.
If radiation dominates then, we are relativistic. This means \[P \sim -1/3 U/V \] \[U = PV/(\gamma -1)\] \[3 = 1/(\gamma -1)\Rightarrow \gamma = 4/3\] We know this is bad!
The dividing line here is:
\[\boxed{P_{rad} = \frac{1}{3} aT^4 \sim nkT \Rightarrow T \propto n^{1/3}}\]
We see that if we are gas pressure dominate then we are non-relativistic. This means \[P_{gas} \sim \rho k T \sim M/R^3 \cdot M/R \Rightarrow M/R^4 \]
This means anything that is relativistic will be unstable INCLUDING things within the classical regime that is below the relativistic limit for the gas but have radiation pressure dominating. This in turn means it sets a maximum mass for a star! \[\frac{P_{rad}}{P_{gas}} \sim \frac{1/3 aT^4}{M/R^4} \sim \frac{ (M/R)^4}{M/R^4} \sim M^2\] Depending on models this is approximately \(M \sim 100 M_{\odot}\)
Thermal instability is when temperature is not coupled to pressure in the equation of state. This is the line of degeneracy. This means that when it heats up, the pressure does not increase to expand the material and cool it down. Furthermore we will see that the luminosity of the nuclear reaction is highly coupled to the temperature and thus when it heats up luminosity increases increasing radiation pressure dramatically which pushes it into dynamical instability.
We want to see how bright can something shine before it blows away the particles that are graviationally bound to it. We balance \[F_{rad} = F_{g}\] \[F_{g} = \frac{GMm_p}{r^2}\] On the other hand the radiation force is given by \[F_{rad} = \frac{L \sigma_T}{4\pi r^2 c}\] Which means we can equate the two: \[\frac{L \sigma_T}{4\pi r^2 c} = \frac{GMm_p}{r^2}\] \[L = \frac{GMm_p c}{\sigma_T}\]
Nuclear fusion is the process of two or more atomic nuclei joining together to form a single heavier nucleus. However to be able to successfully fuse two nuclei together we need to overcome the coulomb barrier. This is not very easy to do.
We consider the distance for strong force interaction is through the heisenberg uncertainty principle. \[\Delta t \Delta E \geq h\] \[\Delta t = \frac{h}{E}\] \[\Delta t = \frac{h}{mc^2}\] And then we can find the distance by \[d \sim c \Delta t = \frac{h}{mc}\sim 10^{-13}cm\] We see that we need to get close enough to this potential well \[E \sim \frac{Z_eZ_p}{d} \sim 1MeV \sim kT \Rightarrow T \sim 10^{10}K\] At this temperature this will completely destroy the nuclei as the binding energy of the proton is at relativisitic and is unstable.
We see then that quantum tunneling is the process of a particle passing through a potential barrier without having enough energy to do so. And is what brings nuclear fusion to happen.
We can calculate the tunneling probability by solving for the wave function of the particle in the potential barrier. \[P = \exp{\sqrt{E_g/E}}\] Where \(E_g = [\pi Z_eZ_p]^2 mc^2\) is gamow energy peak.
We see this compared with the energy tail of the maxwellian distribution which produces a window for which fusion happens. This is because the tail is \[e^{-E/kT} \cdot e^{\sqrt{E_g/E}}\] This prdouces a peak at \[E_0 = (\sqrt{E_g}kT)^{2/3}\]
We can we can then calculate the energy rate production as a function of temperature. \[\epsilon \propto \rho T^4\] while the CNO fusion \[\epsilon \propto \rho T^{18}\]!
Let us talk more in detail about the P-P chain vs the CNO cycle:
P-P Chain: pp chain: This is a direct fusion process. Protons collide directly with other protons to eventually form Helium. It relies on the weak interaction specifically the conversion of a proton to a neutron: and produces a neutrino as a by product. pp chain (Low Barrier): The fusion involves protons (\(Z=1\)) colliding with protons (\(Z=1\)). The electrostatic repulsion (Coulomb barrier) is relatively low, allowing tunneling to occur at lower temperatures. It has a low temperature sensitivity (\(\epsilon_{pp} \propto T^4\)). It is like a "tortoise"—slow and steady, dominating at lower temperatures.
CNO Cycle: CNO cycle: This is a more complex fusion process. It involves carbon, nitrogen, and oxygen nuclei (with \(Z=6,7,8\)) undergoing a series of reactions to eventually form Helium. It relies on the strong interaction and the conversion of a proton to a neutron: and produces a neutrino as a by product. CNO cycle (High Barrier): The fusion involves carbon, nitrogen, and oxygen nuclei (with \(Z=6,7,8\)) undergoing a series of reactions to eventually form Helium. The electrostatic repulsion (Coulomb barrier) is relatively high, requiring higher temperatures for tunneling to occur. It has a high temperature sensitivity (\(\epsilon_{CNO} \propto T^{18}\)). It is like a "cheetah"—fast and furious, dominating at higher temperatures. The CNO cycle is more efficient at higher temperatures, and it is more sensitive to temperature changes.
The crossover point where the CNO cycle becomes more efficient than the pp chain is approximately \(1.5 \times 10^7 K\)
The upper bound of the mass of a star is set by the dominance of radiation pressure over gas pressure which turns the adiabatic index closer to 4/3. This means the photon gas is the dominat pressure component over the non-relativistic gas. Photon gas is dynamically unstable. The upper limit is about 100 \(M_{\odot}\).
The lower bound is set by the temperature of which nuclear fusion can occur. We can find the maxium temperature and make sure it surpasses the temperature needed to ignite fusion.
We know that the pressure at the core is a mix of degenerate and normal matter \[\boxed{P = kN_rn_e^{5/3}+n_ikT} \] We know that the electrons go degenerate first before the ions. We can replace each with densities. \[P = kN_r(\frac{\rho}{\mu_em_e})^{5/3}+(\frac{\rho}{\mu_im_i})kT \] We remember we need this to be hydrostatic equilibrium. So we have: \[\frac{-Gm}{r^2} = \frac{1}{\rho}\frac{dP}{dr}\] \[\frac{-Gm}{r^2} rho = P/r\] \[\frac{-Gm}{r} rho = P\] \[\rho^{1/3} = m^{1/3}/r\] \[-Gm \rho^{1/3}/m^{1/3} rho = P\] \[\boxed{-Gm^{2/3} \rho^{4/3}= P}\] \[kN_r(\frac{\rho}{\mu_em_e})^{5/3}+(\frac{\rho}{\mu_im_i})kT = -Gm^{2/3} \rho^{4/3}\] We simplify and swallow the constants \[A \rho^{1/3} + B \rho^{2/3} = kT\] We can then find the maximum temperature by solving this quadratic like equation. The solution also gives rise to the temperature dependence of \(T \propto M^{4/3}\). The mass needs to be \(M > \# T^{3/4}\) where the constant comes from properly solving the quadratic equation. \[\boxed{M > \# T_{ignite}^{3/4}}\]