Physics Made Easy

Astrophysics II

Fundamental principle I: stars are self-gravitating bodies in dynamical equilibrium due to a balance of gravity and internal pressure forces.

Equation of hydrostatic equilibrium: consider a small volume element at a distance r from the centre- cross section hydrostatic-eqm-star.jpgδS, length δr,

(P_{r+\delta r}-P_r)\delta S+\frac{GM_r}{r^2}(\rho_r\delta_S\delta_r)=0



Equation of distribution of mass

M_{r+\delta r}-M_r=(\frac{dM_r}{dr})\delta r=4\pi r^2\rho_r\delta r

\frac{dM_r}{dr}=4\pi r^2\rho_r

Dimensional analysis: use this to estimate the central pressure. Consider a point at r=\frac{R_s}{2} and approximate- \frac{dP}{dr} \sim \frac{P_c}{R_s}, \rho_r \sim \frac{3M_s}{4\pi R_s^3}(\bar{\rho}), M_r \sim \frac{M_s}{2}

Dynamical timescale: the dynamical timescale is the time it would take the star to collapse completely if pressure forces were negligible.

Equation of motion: (\rho \delta S \delta r)\ddot{r}=-(\frac{GM_r}{r^2})(\rho \delta S \delta r)

Inward displacement: s=\frac{1}{2}gt^2=\frac{1}{2}\frac{GM_r}{r^2}t^2

Put s~Rs, r~Rs, Mr~ Ms to estimate.

Virial theorem:

Start with \frac{dP_r}{dr}=\frac{-gM_r\rho_r}{r^2}

$latex 4\pi r^3dP_r=-(\frac{GM_r}{r})4\pi r^2\rho_rdr

Integrate 4\pi[r^3 P_r]^{r=R_s}_{r=0}-3\int_0^{R_s}P_r4\pi r^2dr=-\int_0^{R_s}\frac{GM_r}{r}4\pi r^2 \rho_r dr

We can cancel the first term as it is zero at both limits, leaving-

\int_0^{R_s}3P_r4\pi r^2dr=\int_0^{R_s}\frac{GM_r}{r}4\pi r^2 \rho_r dr

\int_0^{R_s}3P_r4\pi r^2dr=\int_0^{R_s}\frac{GM_r}{r}dM_r

\int_0^{R_s}3P_r4\pi r^2dr=-\Omega (total gravitational energy) (1)

Thermal energy per unit volume: u=\frac{1}{2}nfkT (f represents degrees of freedom).

\gamma=\frac{f+2}{f}, \frac{f}{2}=\frac{1}{\gamma-1}, n=\frac{\rho}{\mu m_H}

Substitute all this in to get u=\frac{1}{\gamma-1}\frac{\rho kT}{\mu m_H}

Now recall the Ideal gas law P=\frac{\rho kT}{\mu m_H}

-> u=\frac{1}{\gamma-1}P

and total thermal energy is U=\int_0^{R_s}\frac{1}{\gamma-1}P_r4\pi r^2 dr

Now back to (1):

3(\gamma-1)\int_0^{R_s}\frac{1}{\gamma-1}P_r 4\pi r^2dr=-\Omega




For a fully ionised, ideal gas \gamma=\frac{5}{3}

-> 2U+\Omega=0

Total energy E=U+\Omega=-U=\frac{\Omega}{2}


Now for a quick summary to remind you of the steps to take if you ever need to derive this

  1. Start with equation of hydrostatic equilibrium
  2. Multiply by 4πr3 and integrate over the radius of the star
  3. Substitute for gravitational energy
  4. Consider thermal energy

Fundamental principle II- negative ‘heat capacity’: for the above case, if total energy decreases, thermal energy increases and the star heats up.

Nuclear burning is self regulatory: if nuclear burning and thus total energy decrease, the core contracts and heats up; this causes nuclear burning to then increase again, as it is highly temperature dependent. Conversely, if there is an increase in nuclear burning, the core will expand and cool, decreasing it again.

Unstable stars: if γ=4/3, then E=0, i.e. the star has zero binding energy. In this case, the star is easily disrupted, causing rapid mass loss.

Periods when nuclear burning is not active: during star formation, energy is lost from the surface, so the proto-star contracts and heats up until hydrogen burning is ignited. When hydrogen fuel is exhausted, the same thing happens, and if the star is massive enough, fusion of heavier elements can occur.

Fundamental Principle III: since stars lose energy by radiation, stars supported by thermal pressure require an energy source to avoid collapse.

Thermal timescale: (aka Kelvin-Helmholtz timescale)

t_{th}\sim \frac{GM_s^2}{2R_sL_s} ~ 15 million years, i.e. too short to provide energy for a stellar lifetime. Since thermal energy~gravitational energy, it is clear that for stars there must be another mechanism- this is where nuclear fusion comes in.

Nuclear timescale: t_n=\eta \frac{M_c}{M}\frac{Mc^2}{L}

where η is the efficiency- for H->He fusion, \eta =(\frac{4M_H-M_{He}}{4M_H})=7\times 10^{-3}

and c is the mass of the core.

Working this out gives us the timescale we would expect for a star.

Energy loss at a stellar surface is compensated for by energy release from nuclear reactions in the stellar interior.

L_s=\int_0^{R_s}\epsilon_r \rho_r 4\pi r^2 dr

where εr is the nuclear energy released per unit mass per second.

\frac{dL_r}{dr}=4\pi\epsilon_r\rho_r r^2 for any elementary shell.

Energy transport

  1. Conduction: negligible except in degenerate matter.
  2. Radiation: interior consists of X-ray photons which undergo a random walk over ~5×103 yr and are degraded to optical frequencies.
  3. Consider a spherical shell, area A=4πr2, radius r, thickness dr

    Radiation pressure P_{rad}=\frac{1}{3}aT^4 momentum flux

    Rate of deposition of momentum in r->r+dr \frac{-dP_{rad}}{dr}dr4\pi r^2 (i)

    Opacity, κ, is a measure of absorption. \frac{dI}{I}=-\kappa\rho dx

    Solution I=I+0e^{-\tau}

    Where we define optical depth \tau \equiv \int\kappa\rho dx

    \frac{1}{\kappa}{\rho} is the mean free path.

    If τ>>1, material is optically thick

    If τ<<1, material is optically thin

    Rate of momentum absorption in shell \frac{L_r\kappa\rho}{c }dr, which is equal to equation (i).

    So putting it all together, our final equation is L_r=-4\pi r^2 \frac{4ac}{3\kappa\rho}T^3\frac{dT}{dr}

  4. Convection: Consider a bubble rising with initial pressure and density P_1, \rho_1, rising by an amount dr. star-convection.jpgAmbient pressure and density are \bar{P}_1, \bar{\rho_1}. The bubble expands adiabatically, i.e. P_2=P_1(\frac{\rho_2}{\rho_1})^{\gamma}

Assuming the bubble remains in pressure equilibrium with the ambient medium, then P_2=\bar{P}_2=\bar{P(r+dr)}\simeq P_1+\frac{dP}{dr}dr


\rho_2\simeq\rho_1(1+\frac{1}{P_1}\frac{dP}{dr}dr) ^{\frac{1}{\gamma}}

\rho_2\simeq\rho_1(1+\frac{1}{P_1\gamma}\frac{dP}{dr}dr) (binomial expansion).

We have convective instability when the bubble keeps rising, i.e. if \rho_2-\bar{\rho}_2<0


\bar{rho}_2\simeq \rho_1+\frac{d\rho_1}{dr}dr

Now substitute the expressions we worked out for ρ2 and \bar{\rho}_2 into our condition for instability.

-> Instability when \rho_(1+\frac{1}{P\gamma}\frac{dP}{dr}dr) > \rho+\frac{d\rho}{dr}dr (dropping the subscripts).

\frac{\rho}{\gamma P}\frac{dP}{dr}>\frac{d\rho}{dr}

Alternatively, we can express the instability condition as

actual temperature gradient $latex > adiabatic (critical) temperature gradient

Fundamental principle summary: Before we forge ahead, let’s just check that we’re not too lost on everything covered above.

· Stars are self-gravitating bodies in dynamic equilibrium due to a balance of gravity and internal pressure forces (hydrostatic equilibrium).

· Stars lose energy by radiation from the surface. Stars supported by thermal pressure require an energy source to avoid collapse, .e.g. nuclear and gravitational energy (3(γ-1)U+Ω=0).

· Temperature structure is largely determined by the mechanism by which energy is transported from the core to the surface; conduction, convection and radiation.

· The central temperature is determined by the characteristic temperature for the appropriate nuclear fusion reactions (107 K for hydrogen, 108 K for helium).

· Normal stars have a negative ‘heat capacity’ (virial theorem), i.e. they heat up when their total energy decreases.

· In a non-degenerate core, nuclear burning is self-regulatory (negative feedback system).

· The global structure of a star is determined by the simultaneous satisfaction of these principles, and the local properties of a star are determined by the global structure (so it all boils down to making sure everything works as we’ve said it does). Mathematically, this requires the simultaneous solution of a set of coupled, non-linear differential equations with boundary conditions, but before you hide under your covers at the prospect of this, that isn’t something we’re even going to attempt here.

Equations of stellar structure summary aka the ones to remember-

1.      \frac{dP_r}{dr}=-\frac{GM_r\rho_r}{r^2}

2.      \frac{dM_r}{dr}=4\pi r^2\rho_r

3.      \frac{dL_r}{dr}=4\pi r^2\rho_r(\epsilon_r-T\frac{dS}{dt}

4.      \frac{dT_r}{dr}=\frac{-3\kappa_rL_r\rho_r}{16\pi acr^2T_r^3}


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