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Compact Objects

Universite Paris Saclay

What are compact objects? How are they formed?

Compact objects are the final remnants of stellar evolution, born when normal stars exhaust their nuclear fuel and can no longer support themselves against gravitational collapse. They represent the ultimate fate of stars and are laboratories for extreme physics: high densities, strong gravity, and exotic states of matter.

Stellar evolution in a nutshell

The fate of a star is determined primarily by its initial mass. The diagram below summarizes the main evolutionary paths:

What are they? Properties of compact objects

Compactness parameter

A useful quantity to characterize compact objects is the compactness:

C=GMRc2C = \frac{GM}{R c^2}

This dimensionless parameter measures how close an object is to forming an event horizon (where C=0.5C = 0.5 corresponds to the Schwarzschild radius Rs=2GM/c2R_s = 2GM/c^2).

ObjectTypical Radius (km)Typical Mass (MM_\odot)Compactness CCDensity ρ\rho (g/cm³)
Sun7×1057 \times 10^51106\sim 10^{-6}1\sim 1
White Dwarf104\sim 10^40.5–1.4104\sim 10^{-4}106\sim 10^6
Neutron Star10\sim 101.2–2.20.1\sim 0.1–0.31014\sim 10^{14}1015
Black HoleRs=2GM/c2R_s = 2GM/c^23\gtrsim 30.5Singularity

White Dwarfs

White dwarfs are the remnants of low- and medium-mass stars. They are supported against gravitational collapse by electron degeneracy pressure, a quantum mechanical effect arising from the Pauli exclusion principle.

Neutron Stars

Neutron stars are the remnants of core-collapse supernovae from massive stars. When the core collapses beyond white dwarf densities, electrons are captured by protons (p+en+νep + e^- \to n + \nu_e), producing a star composed primarily of neutrons. They are supported by neutron degeneracy pressure and nuclear repulsion at supranuclear densities.

Black Holes

When the remnant core exceeds the maximum mass for a neutron star, nothing can prevent complete gravitational collapse. A black hole forms, characterized by an event horizon — a surface from which nothing, not even light, can escape.

How do we measure their properties?

Determining the radius

Determining the mass

Stellar structure equations

The internal structure of a star (or compact object) in hydrostatic equilibrium is governed by two fundamental equations.

This equation simply states that the mass enclosed within radius rr increases with rr according to the density profile.

This equation balances the inward force of gravity against the outward pressure gradient. For compact objects, the pressure PP is not thermal but comes from degeneracy or nuclear forces.

The Chandrasekhar limit

The Chandrasekhar limit is the maximum mass of a white dwarf supported by electron degeneracy pressure. We derive it under simplifying assumptions using polytropic models and the Lane–Emden equation.

Proof 1 (Derivation of the Chandrasekhar limit)

Assumptions:

  • The white dwarf is spherical and in hydrostatic equilibrium.

  • The equation of state is that of a completely degenerate electron gas.

  • The star is composed of fully ionized matter with mean molecular weight per electron μe\mu_e (for helium, μe=2\mu_e = 2; for carbon‑oxygen, also μe=2\mu_e = 2).

  • General relativistic effects are neglected.

1. Hydrostatic equilibrium and mass continuity

For a spherically symmetric star, the mass enclosed within radius rr is

dm(r)dr=4πr2ρ(r),m(0)=0.\frac{dm(r)}{dr} = 4\pi r^2 \rho(r), \qquad m(0)=0.

Hydrostatic equilibrium balances pressure against gravity:

dP(r)dr=Gm(r)ρ(r)r2.\frac{dP(r)}{dr} = -\frac{G m(r) \rho(r)}{r^2}.

Differentiating the second equation and using the first eliminates m(r)m(r):

ddr(r2ρdPdr)=4πGr2ρ.(1)\frac{d}{dr}\left( \frac{r^2}{\rho} \frac{dP}{dr} \right) = -4\pi G r^2 \rho. \tag{1}

2. Polytropic equation of state

For many equations of state it is convenient to adopt a polytropic form

P=Kρ1+1n,P = K \rho^{1 + \frac{1}{n}},

where KK is a constant and nn is the polytropic index.
For white dwarfs:

  • Non‑relativistic degenerate electrons: Pρ5/3P \propto \rho^{5/3}n=3/2n = 3/2.

  • Ultra‑relativistic degenerate electrons: Pρ4/3P \propto \rho^{4/3}n=3n = 3.

The Chandrasekhar limit corresponds to the ultra‑relativistic case, so we focus on n=3n = 3.

3. Dimensionless variables and the Lane–Emden equation

Introduce the dimensionless quantities

ρ(r)=ρcθn(r),r=αξ,\rho(r) = \rho_c \theta^n(r), \qquad r = \alpha \xi,

with ρc\rho_c the central density and α\alpha a length scale to be chosen. Substituting P=Kρc1+1/nθn+1P = K \rho_c^{1+1/n} \theta^{n+1} into Eq. (1) gives:

1ξ2ddξ(ξ2dθdξ)=θn,\frac{1}{\xi^2} \frac{d}{d\xi}\left( \xi^2 \frac{d\theta}{d\xi} \right) = -\theta^n,

provided we set

α=[(n+1)K4πGρc1n1]1/2.\alpha = \left[ \frac{(n+1)K}{4\pi G} \rho_c^{\frac{1}{n}-1} \right]^{1/2}.

Equation (2) is the Lane–Emden equation. The boundary conditions are θ(0)=1\theta(0)=1, θ(0)=0\theta'(0)=0 (regularity at the centre). The surface of the star is at the first zero ξ1\xi_1 where θ(ξ1)=0\theta(\xi_1)=0.

4. Mass of the polytrope

The total mass is

M=0R4πr2ρdr=4πρcα30ξ1ξ2θndξ.M = \int_0^R 4\pi r^2 \rho \, dr = 4\pi \rho_c \alpha^3 \int_0^{\xi_1} \xi^2 \theta^n \, d\xi.

Using the Lane–Emden equation, the integral can be evaluated:

0ξ1ξ2θndξ=ξ12θ(ξ1).\int_0^{\xi_1} \xi^2 \theta^n \, d\xi = -\xi_1^2 \theta'(\xi_1).

Hence

M=4πρcα3[ξ12θ(ξ1)].(3)M = 4\pi \rho_c \alpha^3 \left[ -\xi_1^2 \theta'(\xi_1) \right]. \tag{3}

For n=3n=3, the Lane–Emden equation has been solved numerically; the results are

ξ16.89685,ξ12θ(ξ1)2.01824.\xi_1 \approx 6.89685, \qquad -\xi_1^2 \theta'(\xi_1) \approx 2.01824.

5. Electron degeneracy pressure in the ultra‑relativistic limit

For a completely degenerate electron gas, the pressure depends on the electron number density nen_e. In the ultra‑relativistic limit (pFmecp_F \gg m_e c),

P=c12π2(3π2ne)4/3.P = \frac{\hbar c}{12\pi^2} (3\pi^2 n_e)^{4/3}.

The electron density is related to the mass density by ne=ρμemun_e = \dfrac{\rho}{\mu_e m_u}, where mumpm_u \approx m_p is the atomic mass unit and μe=A/Z\mu_e = A/Z (for fully ionized matter). For typical white dwarf compositions (carbon‑oxygen), Z/A=1/2Z/A = 1/2, so μe=2\mu_e = 2.

Thus

P=c12π2(3π2μemp)4/3ρ4/3.P = \frac{\hbar c}{12\pi^2} \left( \frac{3\pi^2}{\mu_e m_p} \right)^{4/3} \rho^{4/3}.

Comparing with the polytropic form P=Kρ1+1/nP = K \rho^{1+1/n} for n=3n=3, we identify

K=c12π2(3π2μemp)4/3.K = \frac{\hbar c}{12\pi^2} \left( \frac{3\pi^2}{\mu_e m_p} \right)^{4/3}.

6. Eliminating ρc\rho_c and α\alpha from the mass formula

From the definition of α\alpha for n=3n=3:

α2=4K4πGρc2/3=KπGρc2/3.\alpha^2 = \frac{4K}{4\pi G} \rho_c^{-2/3} = \frac{K}{\pi G} \rho_c^{-2/3}.

Hence α3=(KπG)3/2ρc1\alpha^3 = \left( \dfrac{K}{\pi G} \right)^{3/2} \rho_c^{-1}.

Substitute into Eq. (3):

M=4πρc(KπG)3/2ρc1[ξ12θ(ξ1)]=4π(KπG)3/2[ξ12θ(ξ1)].M = 4\pi \rho_c \left( \frac{K}{\pi G} \right)^{3/2} \rho_c^{-1} \left[ -\xi_1^2 \theta'(\xi_1) \right] = 4\pi \left( \frac{K}{\pi G} \right)^{3/2} \left[ -\xi_1^2 \theta'(\xi_1) \right].

Notice that ρc\rho_c has cancelled! This means that for n=3n=3 the total mass is independent of the central density – a unique mass is obtained.

Insert the numerical factor and KK:

MCh=4π(1πGc12π2(3π2μemp)4/3)3/2×2.01824.M_{\text{Ch}} = 4\pi \left( \frac{1}{\pi G} \frac{\hbar c}{12\pi^2} \left( \frac{3\pi^2}{\mu_e m_p} \right)^{4/3} \right)^{3/2} \times 2.01824.

7. Simplifying the constants

First, combine the powers:

(c12π2)3/2(3π2μemp)2=(c)3/2(12π2)3/29π4μe2mp2.\left( \frac{\hbar c}{12\pi^2} \right)^{3/2} \left( \frac{3\pi^2}{\mu_e m_p} \right)^2 = \frac{(\hbar c)^{3/2}}{(12\pi^2)^{3/2}} \frac{9\pi^4}{\mu_e^2 m_p^2}.

Also 4π×(1πG)3/2=4π×π3/2G3/2=4π1/2G3/24\pi \times \left( \dfrac{1}{\pi G} \right)^{3/2} = 4\pi \times \pi^{-3/2} G^{-3/2} = 4\pi^{-1/2} G^{-3/2}.

Thus

MCh=2.01824×4π1/2G3/2(c)3/2(12π2)3/29π4μe2mp2.M_{\text{Ch}} = 2.01824 \times 4\pi^{-1/2} G^{-3/2} \frac{(\hbar c)^{3/2}}{(12\pi^2)^{3/2}} \frac{9\pi^4}{\mu_e^2 m_p^2}.

Now combine the pure numbers. A careful evaluation (using π3.1416\pi \approx 3.1416) gives

MCh=3.098μe2(c)3/2G3/2mp2.M_{\text{Ch}} = \frac{3.098}{\mu_e^2} \frac{(\hbar c)^{3/2}}{G^{3/2} m_p^2}.

Inserting the physical constants in SI units:

  • =1.055×1034\hbar = 1.055 \times 10^{-34} J·s,

  • c=2.998×108c = 2.998 \times 10^8 m/s,

  • G=6.674×1011G = 6.674 \times 10^{-11} m³/kg·s²,

  • mp=1.673×1027m_p = 1.673 \times 10^{-27} kg,

  • M=1.989×1030M_\odot = 1.989 \times 10^{30} kg.

One finds

MCh=1.457(2μe)2M.M_{\text{Ch}} = 1.457 \left( \frac{2}{\mu_e} \right)^2 M_\odot.

For μe=2\mu_e = 2, this yields

MCh1.44M.\boxed{ M_{\text{Ch}} \approx 1.44 M_\odot }.

Beyond white dwarfs: neutron stars and the TOV limit

For neutron stars, the situation is more complex due to:

The maximum mass for a neutron star (the TOV limit) is typically 2.22.5M2.5 M_\odot, with the exact value providing crucial constraints on nuclear physics.

Summary and key takeaways

These extreme objects continue to challenge our understanding of physics and provide unique laboratories for testing general relativity, nuclear physics, and quantum mechanics in regimes inaccessible on Earth.