The Balancing Act — Hydrostatic Equilibrium

Section 6 of 7

Estimating the Core Temperature

Part 5: Estimating the Core Temperature

Pressure is the macroscopic requirement. To finish answering our guiding question, we connect that required pressure to microscopic particle motion — and therefore to temperature. We run Reading the Math a second time, with one honest difference: there is no new derivative to approximate here. The derivative work was already spent in Part 3 getting $P_c$ ; now we only need two pressure estimates to agree.

① No new derivative — equate the two pressure scales. Part 3 gave the hydrostatic requirement $P_c \sim G M^2/R^4$ . The ideal-gas law gives the thermal pressure the core actually supplies, $P_c \sim \rho\,k_B T_c/(\mu m_p)$ . A star in balance must satisfy both, so set them equal — using the same mean density $\rho \sim M/R^3$ :

$\frac{M}{R^3}\cdot\frac{k_B T_c}{\mu m_p} \;\sim\; \frac{G M^2}{R^4}.$

② Extract the scaling. Cancel one factor of $M$ and multiply by $R^3$ , leaving $k_B T_c/(\mu m_p) \sim G M/R$ . Rearranging for $T_c$ gives the core-temperature scaling.

The left-hand side is the thermal-energy scale per particle; the right-hand side is the gravitational-energy scale per particle. Hydrostatic support requires these to be comparable — so gravity alone predicts a stellar core temperature of order $10^7\,\mathrm{K}$ , no nuclear physics needed.

③ Name the assumption. This pass inherits every row of the Part 3 audit — it is built on top of $P_c \sim G M^2/R^4$ — and adds one of its own:

We assumed	by replacing	What it costs
gas pressure dominates	$P \to P_{\text{gas}}$ , dropping $P_{\text{rad}} = \tfrac{1}{3}aT^4$	fine for the Sun; in massive stars $P_{\text{rad}}/P_{\text{gas}} \propto M^2$ grows and lowers the required $T_c$

The audit only ever grows: each Reading-the-Math pass stacks its assumptions on the ones before. Reading 5 collects the full stack and tests where it finally breaks.

Generated log-log plot of core temperature scale in megakelvin versus stellar mass, showing a hydrostatic scaling curve based on T proportional to M over R, a Sun point, and a dashed comparison for the unrealistic fixed-radius case. — Figure 11Hydrostatic equilibrium sets a fusion-scale temperature for every main-sequence star, but the core temperature rises only modestly with mass because more massive stars are also larger.ASTR 201 (generated)

Worked example: the Sun’s core temperature

Worked Example 2The Sun's core temperature

Problem

Estimate the Sun’s core temperature from $T_c \sim \mu G M_\odot m_p/(k_B R_\odot)$ , using $\mu = 0.6$ , $G = 6.674\times10^{-8}\,\mathrm{cm^3\,g^{-1}\,s^{-2}}$ , $M_\odot = 2.0\times10^{33}\,\mathrm{g}$ , $m_p = 1.67\times10^{-24}\,\mathrm{g}$ , $k_B = 1.38\times10^{-16}\,\mathrm{erg\,K^{-1}}$ , and $R_\odot = 7.0\times10^{10}\,\mathrm{cm}$ .

StepEvaluate the numerator

$(0.6)(6.674\times10^{-8})(2.0\times10^{33})(1.67\times10^{-24}) \approx 1.34\times10^{2}\,\mathrm{cm^3\,s^{-2}}$ .

StepEvaluate the denominator

$(1.38\times10^{-16})(7.0\times10^{10}) \approx 9.66\times10^{-6}\,\mathrm{erg\,K^{-1}\,cm}$ .

Dimensional check

$\dfrac{\mathrm{cm^3\,g^{-1}\,s^{-2}}\cdot\mathrm{g}\cdot\mathrm{g}}{\mathrm{erg\,K^{-1}\,cm}} = \dfrac{\mathrm{g\,cm^2\,s^{-2}}}{\mathrm{erg}}\,\mathrm{K} = \mathrm{K}$ , because $1\,\mathrm{erg} = 1\,\mathrm{g\,cm^2\,s^{-2}}$ . ✓

Result

$T_c \sim (1.34\times10^{2})/(9.66\times10^{-6}) \approx 1.4\times10^{7}\,\mathrm{K} = 14\,\mathrm{MK}$ . Detailed solar models give $\approx 1.5\times10^{7}\,\mathrm{K} = 15\,\mathrm{MK}$ , so this stripped-down estimate is already in the correct solar ballpark — a scaling success, not an exact stellar-structure solution.

Observable

The Sun's measured mass and radius

The Sun’s mass ( $M_\odot = 2.0\times10^{33}\,\mathrm{g}$ ) and radius ( $R_\odot = 7.0\times10^{10}\,\mathrm{cm}$ ) — both measured from binary orbits and angular size plus distance.

Model

Hydrostatic equilibrium plus the ideal-gas picture

Combine hydrostatic equilibrium, gas-pressure dominance, the ideal-gas law, and the mean-density scaling $\rho \sim M/R^3$ .

Inference

A core temperature of order 10⁷ K

$T_c \sim 10^7\,\mathrm{K}$ — about 15 million kelvin, hot enough for nuclear fusion. The temperature needed for fusion is set by gravity.

What this temperature means

A core temperature of $T_c \approx 1.5\times10^7\,\mathrm{K}$ corresponds to an average thermal energy per particle of $\tfrac{3}{2}k_B T_c \approx 3.1\times10^{-9}\,\mathrm{erg} \approx 1.9\,\mathrm{keV}$ , or about $2\,\mathrm{keV}$ (using $1\,\mathrm{keV} = 1.602\times10^{-9}\,\mathrm{erg}$ ). At these temperatures atoms are fully ionized, so the solar core is a plasma of free electrons, protons, and helium nuclei.

But this is still not enough thermal energy to overcome the proton–proton Coulomb barrier classically. That barrier is of order $1\,\mathrm{MeV} = 10^3\,\mathrm{keV}$ , hundreds of times larger than the typical thermal energy. So classical thermal motion alone should not allow fusion — the missing ingredient is quantum tunneling, the subject of Reading 3.

Problem

Using $T_c \propto M/R$ and the main-sequence mass–radius relation $R \propto M^{0.8}$ , how does core temperature scale with mass? Is a $10\,M_\odot$ star’s core hotter or cooler than the Sun’s?

Step	Physics	What it determines
Gravity	$g(r)$	inward pull
Hydrostatic equilibrium	$dP/dr$	pressure gradient
Pressure scaling	$P_c$	central pressure
Ideal gas	$P \leftrightarrow T$	temperature
Virial theorem	$K,\ U$	energy balance