Stellar Structure
The fundamental equations governing the internal structure of stars: hydrostatic equilibrium, mass continuity, energy transport, and the virial theorem
Overview
A star is a self-gravitating sphere of plasma in which the inward pull of gravity is balanced by outward pressure gradients. The structure of a star at any instant is determined by four coupled ordinary differential equations — the equations of stellar structure — supplemented by an equation of state and opacity relations. These equations, first systematically formulated by Arthur Eddington in the 1920s, form the foundation of all stellar astrophysics.
In this chapter we derive each equation from first principles, explore the virial theorem and its consequences, derive the Eddington luminosity, and numerically solve the Lane-Emden equation for polytropic stellar models.
1. Hydrostatic Equilibrium
The most fundamental equation in stellar structure expresses the balance between gravity and the pressure gradient. Consider a thin spherical shell at radius \(r\) with thickness \(dr\) inside a spherically symmetric star.
1.1 Force Balance on a Spherical Shell
Let the shell have density \(\rho(r)\), inner radius \(r\), and thickness \(dr\). The mass of the shell element with cross-sectional area \(dA\) is:
$$dm_{\text{shell}} = \rho(r) \, dA \, dr$$
Two forces act on this element in the radial direction:
Gravitational force (inward): The gravitational acceleration at radius \(r\) is due only to the mass \(m(r)\) enclosed within that radius (by the shell theorem). The inward gravitational force on the element is:
$$dF_{\text{grav}} = -\frac{G m(r)}{r^2} \rho(r) \, dA \, dr$$
where \(G\) is Newton's gravitational constant and the minus sign indicates the inward direction.
Pressure force (net): The pressure on the inner face of the shell is \(P(r)\) pushing outward, and on the outer face is \(P(r + dr)\) pushing inward. The net outward pressure force on the element is:
$$dF_{\text{press}} = \left[P(r) - P(r + dr)\right] dA = -\frac{dP}{dr} dr \, dA$$
For a star in mechanical equilibrium (not expanding or contracting), Newton's second law gives \(dF_{\text{grav}} + dF_{\text{press}} = 0\):
$$-\frac{G m(r)}{r^2} \rho(r) \, dA \, dr - \frac{dP}{dr} dr \, dA = 0$$
Dividing through by \(dA \, dr\), we obtain the equation of hydrostatic equilibrium:
Hydrostatic Equilibrium
$$\boxed{\frac{dP}{dr} = -\frac{G m(r) \rho(r)}{r^2}}$$
This equation states that the pressure must decrease outward (since \(dP/dr < 0\)) to support the star against gravitational collapse. It is valid for any self-gravitating fluid in spherical symmetry, from stars to planets to gas clouds.
1.2 Central Pressure Estimate
We can estimate the central pressure by dimensional analysis. Taking \(m(r) \sim M\),\(\rho \sim \bar{\rho} = 3M/(4\pi R^3)\), and integrating from center to surface over a distance \(\sim R\):
$$P_c \sim \frac{G M \bar{\rho}}{R} = \frac{3 G M^2}{4\pi R^4}$$
For the Sun (\(M_\odot = 2 \times 10^{30}\) kg, \(R_\odot = 7 \times 10^8\) m), this gives \(P_c \sim 10^{15}\) Pa, remarkably close to the detailed solar model value of \(2.34 \times 10^{16}\) Pa.
1.3 Beyond Simple Equilibrium
If the star is not in equilibrium, we must include the acceleration term. The full Euler equation in the radial direction becomes:
$$\rho \frac{\partial^2 r}{\partial t^2} = -\frac{dP}{dr} - \frac{G m(r) \rho}{r^2}$$
The dynamical timescale for gravitational collapse (free-fall timescale) is obtained by setting the pressure gradient to zero:
$$t_{\text{ff}} \sim \sqrt{\frac{R^3}{GM}} = \frac{1}{\sqrt{G \bar{\rho}}}$$
For the Sun, \(t_{\text{ff}} \approx 30\) minutes. The fact that the Sun has existed for \(\sim 4.6 \times 10^9\) years shows that hydrostatic equilibrium is maintained to extraordinary precision.
2. Mass Continuity Equation
The second structural equation relates the enclosed mass to the density profile. Consider a thin spherical shell at radius \(r\) with thickness \(dr\).
2.1 Derivation
The mass enclosed within radius \(r\) is:
$$m(r) = \int_0^r 4\pi r'^2 \rho(r') \, dr'$$
Differentiating with respect to \(r\), we immediately obtain the mass continuity equation:
Mass Continuity
$$\boxed{\frac{dm}{dr} = 4\pi r^2 \rho(r)}$$
This is simply a statement that the mass of a thin shell equals its volume times its density. Together with hydrostatic equilibrium, these two equations already constrain the pressure and density profiles given an equation of state \(P = P(\rho, T)\).
2.2 Boundary Conditions
The stellar structure equations require boundary conditions at the center and surface:
At the Center (r = 0)
\(m(0) = 0\) — no enclosed mass
\(L(0) = 0\) — no enclosed luminosity
\(dP/dr|_{r=0} = 0\) — regularity condition
At the Surface (r = R)
\(m(R) = M\) — total stellar mass
\(P(R) \approx 0\) — vanishing surface pressure
\(T(R) = T_{\text{eff}}\) — effective temperature
2.3 Mean Density and the Density Profile
The mean density of a star is:
$$\bar{\rho} = \frac{3M}{4\pi R^3}$$
For the Sun, \(\bar{\rho}_\odot \approx 1410\) kg/m³, but the central density reaches \(\rho_c \approx 1.5 \times 10^5\) kg/m³ — about 100 times the mean value. Real stars show strong density concentration toward the center, with typical ratios \(\rho_c / \bar{\rho} \sim 50{-}100\) for main-sequence stars.
3. Energy Transport
Energy generated in the stellar core must be transported outward to the surface. There are three mechanisms: radiation, convection, and conduction (important only in degenerate matter). Which mechanism dominates depends on the local conditions.
3.1 Radiative Energy Transport
In a radiative zone, energy is carried by photons diffusing outward through the stellar plasma. The mean free path of a photon is \(\ell = 1/(\kappa \rho)\), where \(\kappa\) is the opacity (cross-section per unit mass). In the Sun's interior, \(\ell \sim 1\) cm, so photons undergo an enormous number of scatterings before reaching the surface.
The radiative flux is given by the diffusion approximation:
$$F_{\text{rad}} = -\frac{c}{3\kappa\rho} \frac{d(aT^4)}{dr} = -\frac{4acT^3}{3\kappa\rho} \frac{dT}{dr}$$
where \(a = 4\sigma/c\) is the radiation density constant and \(\sigma\) is the Stefan-Boltzmann constant. The luminosity at radius \(r\) is \(L(r) = 4\pi r^2 F_{\text{rad}}\). Solving for the temperature gradient:
Radiative Temperature Gradient
$$\boxed{\frac{dT}{dr} = -\frac{3\kappa\rho L(r)}{16\pi a c T^3 r^2}}$$
The temperature decreases outward (negative gradient) as expected. Higher opacity or luminosity produces a steeper temperature gradient, which can trigger convective instability.
3.2 Convective Transport and the Schwarzschild Criterion
Convection occurs when the radiative temperature gradient becomes too steep. Consider a blob of gas displaced upward by \(dr\) adiabatically (no heat exchange with surroundings). The blob remains in pressure equilibrium with its surroundings, but its temperature and density may differ.
The blob's temperature changes adiabatically:
$$\left(\frac{dT}{dr}\right)_{\text{ad}} = -\frac{\gamma - 1}{\gamma} \frac{T}{P} \frac{dP}{dr} = -\frac{\gamma - 1}{\gamma} \frac{\mu m_H g}{k_B}$$
where \(\gamma = c_P/c_V\) is the adiabatic index, \(\mu\) is the mean molecular weight, \(m_H\) is the hydrogen mass, and \(g = Gm(r)/r^2\) is the local gravitational acceleration. For an ideal monatomic gas, \(\gamma = 5/3\), giving \(\nabla_{\text{ad}} = (\gamma-1)/\gamma = 2/5\).
The Schwarzschild criterion for convective instability states: convection occurs when the actual (radiative) temperature gradient is steeper than the adiabatic gradient:
Schwarzschild Criterion for Convection
$$\boxed{\left|\frac{dT}{dr}\right|_{\text{rad}} > \left|\frac{dT}{dr}\right|_{\text{ad}} \quad \Longleftrightarrow \quad \nabla_{\text{rad}} > \nabla_{\text{ad}}}$$
Here \(\nabla = d\ln T / d\ln P\). When \(\nabla_{\text{rad}} > \nabla_{\text{ad}}\), a displaced blob is hotter and less dense than its surroundings, so buoyancy drives it further upward — the configuration is unstable to convection.
3.3 Deriving the Adiabatic Gradient
For an ideal gas with equation of state \(P = \rho k_B T / (\mu m_H)\), an adiabatic process satisfies \(P \propto \rho^\gamma\), which combined with the ideal gas law gives \(T \propto P^{(\gamma-1)/\gamma}\). Taking the logarithmic derivative:
$$\frac{d\ln T}{d\ln P}\bigg|_{\text{ad}} = \frac{\gamma - 1}{\gamma} \equiv \nabla_{\text{ad}}$$
Converting to the radial gradient using \(dT/dr = (dT/dP)(dP/dr)\) and hydrostatic equilibrium:
$$\left(\frac{dT}{dr}\right)_{\text{ad}} = \nabla_{\text{ad}} \frac{T}{P} \frac{dP}{dr} = -\nabla_{\text{ad}} \frac{T}{P} \frac{Gm\rho}{r^2}$$
For a fully ionized ideal gas with \(\gamma = 5/3\), \(\nabla_{\text{ad}} = 0.4\). In the Sun, the outer convection zone (\(r > 0.71 R_\odot\)) has \(\nabla_{\text{rad}} > 0.4\)due to the high opacity of partially ionized hydrogen, triggering vigorous convection.
3.4 Energy Generation Equation
The fourth structural equation relates the luminosity gradient to the local energy generation rate \(\varepsilon\) (energy per unit mass per unit time):
Energy Generation
$$\boxed{\frac{dL}{dr} = 4\pi r^2 \rho \varepsilon}$$
The energy generation rate depends on temperature and density through the nuclear reaction rates. For the pp-chain, \(\varepsilon_{\text{pp}} \propto \rho T^4\), while for the CNO cycle, \(\varepsilon_{\text{CNO}} \propto \rho T^{16}\).
4. The Virial Theorem
The virial theorem is one of the most powerful results in stellar astrophysics, connecting the total kinetic (thermal) energy of a star to its gravitational potential energy.
4.1 Derivation from Hydrostatic Equilibrium
Start with the equation of hydrostatic equilibrium and multiply both sides by \(4\pi r^3\):
$$4\pi r^3 \frac{dP}{dr} = -\frac{G m(r)}{r^2} \cdot 4\pi r^3 \rho = -\frac{G m(r)}{r} \cdot 4\pi r^2 \rho$$
Now integrate over the entire star from \(r = 0\) to \(r = R\):
$$\int_0^R 4\pi r^3 \frac{dP}{dr} \, dr = -\int_0^R \frac{G m(r)}{r} 4\pi r^2 \rho \, dr$$
The right-hand side is the gravitational potential energy:
$$\Omega = -\int_0^M \frac{Gm}{r} \, dm$$
For the left-hand side, we integrate by parts. Let \(u = P\) and \(dv = d(4\pi r^3/3)\):
$$\int_0^R 4\pi r^3 \frac{dP}{dr} \, dr = \left[4\pi r^3 P\right]_0^R - 3\int_0^R 4\pi r^2 P \, dr$$
The boundary term vanishes since \(P(R) = 0\) at the surface and \(r^3 P\) is finite at \(r = 0\). This gives:
$$-3\int_0^R 4\pi r^2 P \, dr = \Omega$$
The integral \(\int P \, dV\) is related to the thermal energy. For an ideal gas with \(P = (\gamma - 1) u\) where \(u\) is the internal energy density, we have \(\int P \, dV = (\gamma - 1) K\) where \(K\) is the total thermal energy. For a monatomic ideal gas (\(\gamma = 5/3\)):
Virial Theorem for Self-Gravitating Systems
$$\boxed{2K + \Omega = 0}$$
More generally, for an adiabatic index \(\gamma\):\(3(\gamma - 1) K + \Omega = 0\). For \(\gamma = 5/3\), this reduces to \(2K + \Omega = 0\).
The total energy of the star is:
$$E = K + \Omega = \frac{\Omega}{2} = -K$$
This has a remarkable consequence: when a star radiates energy (\(E\) decreases), \(\Omega\) becomes more negative (the star contracts), but \(K\) increases (the star heats up). Stars have negative heat capacity — losing energy makes them hotter!
4.2 Kelvin-Helmholtz Timescale
Before nuclear reactions were understood, Lord Kelvin and Hermann von Helmholtz proposed that the Sun's luminosity is powered by gravitational contraction. The timescale for this process — the Kelvin-Helmholtz timescale — is the time for the star to radiate away its gravitational potential energy at its current luminosity:
$$t_{\text{KH}} = \frac{|\Omega|}{L} \sim \frac{G M^2}{R L}$$
For the Sun, using \(\Omega \approx -3GM_\odot^2/(5R_\odot)\) for a uniform density sphere:
$$t_{\text{KH},\odot} = \frac{3GM_\odot^2}{5R_\odot L_\odot} \approx \frac{3 \times (6.67 \times 10^{-11})(2 \times 10^{30})^2}{5 \times (7 \times 10^8)(3.8 \times 10^{26})} \approx 1.5 \times 10^7 \text{ years}$$
This is about 15 million years — far shorter than the Sun's actual age of 4.6 billion years. The discrepancy showed that gravitational contraction alone cannot power the Sun, motivating the search for nuclear energy sources.
5. The Eddington Luminosity
Sir Arthur Eddington realized that there is a maximum luminosity a star can sustain in hydrostatic equilibrium, set by the balance between the outward radiation pressure force and the inward gravitational force.
5.1 Radiation Pressure Force
A photon carries momentum \(p = E/c\). When photons diffuse through stellar matter, they transfer momentum to the gas via scattering and absorption. The radiation force per unit volume is:
$$f_{\text{rad}} = \frac{\kappa \rho}{c} F = \frac{\kappa \rho L}{4\pi r^2 c}$$
where \(F = L/(4\pi r^2)\) is the radiative flux and \(\kappa\) is the opacity. The gravitational force per unit volume is:
$$f_{\text{grav}} = \frac{G m(r) \rho}{r^2}$$
5.2 Derivation of the Eddington Limit
Hydrostatic equilibrium requires \(f_{\text{grav}} \geq f_{\text{rad}}\). The maximum luminosity occurs when the radiation force exactly balances gravity at every radius:
$$\frac{\kappa \rho L_{\text{Edd}}}{4\pi r^2 c} = \frac{G m(r) \rho}{r^2}$$
The density \(\rho\) and \(r^2\) cancel from both sides. Setting \(m(r) = M\) (the total mass, appropriate for the outer layers where most of the luminosity has been generated):
Eddington Luminosity
$$\boxed{L_{\text{Edd}} = \frac{4\pi G M c}{\kappa}}$$
For electron scattering opacity \(\kappa_{\text{es}} = 0.2(1 + X)\) cm²/g (where \(X\) is the hydrogen mass fraction), this gives:
$$L_{\text{Edd}} \approx 3.3 \times 10^4 \left(\frac{M}{M_\odot}\right) L_\odot$$
5.3 Implications
The Eddington luminosity has profound consequences:
- Maximum stellar mass: Stars with \(L > L_{\text{Edd}}\) drive powerful winds that strip their outer layers. This limits the most massive stars to \(\sim 150{-}300 M_\odot\).
- Accretion limit: Matter cannot accrete onto compact objects faster than the Eddington rate \(\dot{M}_{\text{Edd}} = L_{\text{Edd}}/(\eta c^2)\) without being blown away by radiation pressure.
- Super-Eddington luminosities: Brief episodes of \(L > L_{\text{Edd}}\)can occur in novae, supernovae, and luminous blue variables, accompanied by dramatic mass loss.
6. Polytropic Models and the Lane-Emden Equation
A polytrope is a stellar model in which the pressure and density are related by a power law:
$$P = K \rho^{1 + 1/n}$$
where \(K\) is a constant and \(n\) is the polytropic index. Despite their simplicity, polytropes capture essential physics:
- \(n = 0\): Uniform density (incompressible fluid) — crude planet model
- \(n = 1.5\): Fully convective star (adiabatic with \(\gamma = 5/3\))
- \(n = 3\): Eddington standard model (radiation-dominated star)
- \(n = 5\): Infinite radius, finite mass — the Plummer model
6.1 Derivation of the Lane-Emden Equation
We introduce the dimensionless variable \(\theta\) defined by \(\rho = \rho_c \theta^n\), where \(\rho_c\) is the central density. Then \(P = K\rho_c^{1+1/n} \theta^{n+1} = P_c \theta^{n+1}\).
Substituting into the hydrostatic equilibrium equation and introducing the dimensionless radius \(\xi = r / \alpha\) where:
$$\alpha = \sqrt{\frac{(n+1) P_c}{4\pi G \rho_c^2}}$$
We combine hydrostatic equilibrium and mass continuity by taking \((1/r^2)(d/dr)\) of the hydrostatic equation and using \(dm/dr = 4\pi r^2 \rho\):
$$\frac{1}{r^2} \frac{d}{dr}\left(\frac{r^2}{\rho}\frac{dP}{dr}\right) = -4\pi G \rho$$
Substituting the polytropic relation and the dimensionless variables yields the Lane-Emden equation:
Lane-Emden Equation
$$\boxed{\frac{1}{\xi^2}\frac{d}{d\xi}\left(\xi^2 \frac{d\theta}{d\xi}\right) = -\theta^n}$$
with boundary conditions \(\theta(0) = 1\) and \(\theta'(0) = 0\). The stellar surface is at \(\xi = \xi_1\) where \(\theta(\xi_1) = 0\).
Analytical solutions exist only for \(n = 0\) (\(\theta = 1 - \xi^2/6\)),\(n = 1\) (\(\theta = \sin\xi / \xi\)), and \(n = 5\)(\(\theta = (1 + \xi^2/3)^{-1/2}\)). For all other values, numerical integration is required.
6.2 Physical Quantities from Lane-Emden Solutions
Once \(\theta(\xi)\) is known, the stellar mass and radius follow:
$$R = \alpha \xi_1, \qquad M = 4\pi \alpha^3 \rho_c \left(-\xi_1^2 \theta'(\xi_1)\right)$$
The ratio of central to mean density is:
$$\frac{\rho_c}{\bar{\rho}} = \frac{-\xi_1}{3\theta'(\xi_1)}$$
7. Numerical Simulation: Lane-Emden Equation and Stellar Profiles
We now solve the Lane-Emden equation numerically for various polytropic indices using a fourth-order Runge-Kutta integrator, and compute the resulting stellar structure profiles (density, pressure, mass, and temperature).
Lane-Emden Equation: Polytropic Stellar Models
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7.1 Recovering Physical Profiles
From the Lane-Emden solution, we can reconstruct the full dimensional stellar profile. The pressure profile follows from \(P(r) = P_c \theta(\xi)^{n+1}\), where:
$$P_c = \frac{4\pi G \alpha^2 \rho_c^2}{n+1} = \frac{G M^2}{4\pi R^4} \frac{(n+1)}{(-\xi_1^2\theta'(\xi_1))^2}$$
The gravitational potential energy for a polytrope of index \(n\) is:
$$\Omega = -\frac{3}{5-n} \frac{GM^2}{R}$$
For \(n = 3\) (Eddington standard model), this gives \(\Omega = -3GM^2/(2R)\), compared to \(-3GM^2/(5R)\) for the uniform density (\(n = 0\)) case. The stronger central concentration of realistic models increases the binding energy.
7.2 Connection to Real Stars
While no real star is exactly a polytrope, polytropic models capture key features:
Low-Mass Stars (\(M \lesssim 0.3 M_\odot\))
Fully convective throughout, well-modeled by \(n = 1.5\) polytrope (adiabatic with \(\gamma = 5/3\)).
The Lane-Emden solution gives \(\xi_1 = 3.654\) and \(\rho_c/\bar{\rho} = 5.99\).
Massive Stars (\(M \gtrsim 10 M_\odot\))
Radiation pressure dominant, approximated by \(n = 3\) polytrope (Eddington standard model).
The Lane-Emden solution gives \(\xi_1 = 6.897\) and \(\rho_c/\bar{\rho} = 54.2\).
Modern stellar structure codes solve the full set of four coupled ODEs with realistic opacity tables (e.g., OPAL), nuclear reaction networks, and equations of state that account for partial ionization, radiation pressure, and electron degeneracy. The polytropic models remain invaluable for building physical intuition and as starting points for numerical iteration.
Chapter Summary
The four equations of stellar structure — hydrostatic equilibrium, mass continuity, energy transport (radiative or convective), and energy generation — together with an equation of state and opacity law, completely determine the structure of a star.
The virial theorem \(2K + \Omega = 0\) provides a powerful global constraint, revealing that stars have negative heat capacity. The Kelvin-Helmholtz timescale\(t_{\text{KH}} \sim GM^2/(RL)\) sets the thermal adjustment timescale.
The Eddington luminosity \(L_{\text{Edd}} = 4\pi GMc/\kappa\) sets the maximum luminosity for hydrostatic equilibrium. The Lane-Emden equation provides analytically tractable models that capture the essential density concentration of real stars.
Video Lectures: Transport in Stellar Interiors
KITP: Transport in Stellar Interiors
Mapping Cosmic Magnetism in the Space Between Stars — Susan E. Clark (Stanford)
Practice Problems
Problem 1:Derive the central pressure of a uniform-density star of mass M and radius R using the equation of hydrostatic equilibrium.
Solution:
1. Hydrostatic equilibrium: $\frac{dP}{dr} = -\frac{G m(r) \rho}{r^2}$.
2. For uniform density $\rho_0 = \frac{3M}{4\pi R^3}$, the enclosed mass is $m(r) = \frac{4}{3}\pi r^3 \rho_0$.
3. Substituting: $\frac{dP}{dr} = -\frac{4\pi G \rho_0^2}{3} r$.
4. Integrate from $r$ to $R$ (where $P(R) = 0$): $P(r) = \frac{2\pi G \rho_0^2}{3}(R^2 - r^2)$.
5. Central pressure ($r = 0$): $P_c = \frac{2\pi G \rho_0^2 R^2}{3} = \frac{3GM^2}{8\pi R^4}$.
6. For the Sun: $P_c \approx 1.3 \times 10^{14}$ Pa (the true value is ~$2.5 \times 10^{16}$ Pa due to central concentration).
Problem 2:Using the virial theorem, estimate the mean temperature of the Sun. Assume it is an ideal gas of ionized hydrogen.
Solution:
1. Virial theorem: $2K + \Omega = 0$, where $K$ is total thermal kinetic energy and $\Omega$ is gravitational potential energy.
2. Gravitational energy of a uniform sphere: $\Omega = -\frac{3GM^2}{5R}$.
3. From the virial theorem: $K = -\frac{\Omega}{2} = \frac{3GM^2}{10R}$.
4. For an ideal gas: $K = \frac{3}{2}Nk_BT$, where $N = M/(\mu m_H)$ with $\mu = 0.5$ (fully ionized H).
5. Setting equal: $\bar{T} = \frac{G M \mu m_H}{5 k_B R}$.
6. Substituting solar values: $\bar{T} \approx \frac{(6.67\times10^{-11})(2\times10^{30})(0.5)(1.67\times10^{-27})}{5(1.38\times10^{-23})(7\times10^8)} \approx 2.3 \times 10^6$ K.
Problem 3:Calculate the Kelvin-Helmholtz timescale for the Sun and explain why gravitational contraction cannot explain the Sun's luminosity over geological time.
Solution:
1. The Kelvin-Helmholtz timescale: $t_{\text{KH}} = \frac{|\Omega|}{L} \approx \frac{GM^2}{RL}$.
2. For the Sun: $G = 6.67\times10^{-11}$, $M_\odot = 2\times10^{30}$ kg, $R_\odot = 7\times10^8$ m, $L_\odot = 3.83\times10^{26}$ W.
3. $t_{\text{KH}} = \frac{(6.67\times10^{-11})(2\times10^{30})^2}{(7\times10^8)(3.83\times10^{26})} = \frac{2.67\times10^{50}}{2.68\times10^{35}}$.
4. $t_{\text{KH}} \approx 10^{15}$ s $\approx 3 \times 10^7$ years $\approx 30$ million years.
5. The Earth is $\sim 4.5 \times 10^9$ years old, which is ~150 times longer than $t_{\text{KH}}$.
6. Therefore, gravitational contraction alone cannot power the Sun over geological timescales. Nuclear fusion (with timescale $t_{\text{nuc}} \sim 10^{10}$ years) is required.
Problem 4:Calculate the Eddington luminosity for a 50 M☉ star with electron-scattering opacity κ = 0.34 cm²/g. Is such a star radiation-pressure dominated?
Solution:
1. Eddington luminosity: $L_{\text{Edd}} = \frac{4\pi G M c}{\kappa}$.
2. Convert: $\kappa = 0.34$ cm$^2$/g $= 0.034$ m$^2$/kg, $M = 50 \times 2\times10^{30} = 10^{32}$ kg.
3. $L_{\text{Edd}} = \frac{4\pi (6.67\times10^{-11})(10^{32})(3\times10^8)}{0.034}$.
4. $L_{\text{Edd}} = \frac{4\pi \times 2.0\times10^{29}}{0.034} = \frac{2.51\times10^{30}}{0.034} \approx 7.4\times10^{31}$ W.
5. In solar units: $L_{\text{Edd}}/L_\odot \approx 7.4\times10^{31}/3.83\times10^{26} \approx 1.9\times10^5$.
6. The mass-luminosity relation gives $L \sim L_\odot(M/M_\odot)^{3.5} \approx 8.8\times10^5 L_\odot$ for 50$M_\odot$, which exceeds $L_{\text{Edd}}$. Such massive stars are indeed radiation-pressure dominated and drive powerful stellar winds.
Problem 5:Show that a star in hydrostatic equilibrium has negative heat capacity, and explain the physical consequences for stellar evolution.
Solution:
1. Total energy: $E = K + \Omega$. By the virial theorem, $2K + \Omega = 0$, so $K = -\Omega/2$.
2. Therefore $E = K + \Omega = -K = \Omega/2 < 0$ (the star is gravitationally bound).
3. Since $K = \frac{3}{2}Nk_B\bar{T}$, the mean temperature is $\bar{T} = \frac{2K}{3Nk_B}$.
4. The heat capacity: $C = \frac{dE}{d\bar{T}} = \frac{d(-K)}{d\bar{T}} = -\frac{3}{2}Nk_B < 0$.
5. Physical consequence: when a star loses energy (radiates), $E$ decreases, so $K$ increases and the star gets hotter.
6. This negative heat capacity drives stellar evolution: nuclear fuel depletion causes contraction, heating, and ignition of heavier fuels in successive shells, leading to the onion-skin structure of evolved massive stars.