Perturbing the equilibrium of Module 3: the wave equation, Newton's isothermal sound speed against Laplace's adiabatic one, the acoustic cutoff of a stratified atmosphere, and the solar large frequency separation, which rules out an isothermal Sun with the tabulated model alone and shows that a perfect gas with $\Gamma_1 = 5/3$ and a linear temperature profile falls short in the outer $1.7$ per cent of the radius.
Module 3 set the velocity to zero and found the structure that the pressure gradient must have to hold a gas up against gravity. This module asks what happens when that equilibrium is disturbed a little. The answer is sound, and the method that finds it, linear perturbation theory, is the method that every instability in Modules 5 to 7 reuses.
The first calculation of the speed of sound from mechanics was Newton's. He treated the air as a gas whose pressure is proportional to its density at fixed temperature, and his result, in modern notation, is
$$ c_{\rm N} \;=\; \sqrt{\frac{P}{\rho}} \;=\; \sqrt{\frac{k_BT}{\mu\,m_u}}, \tag{1.1}$$with $P$ the pressure, $\rho$ the mass density, $k_B$ Boltzmann's constant, $T$ the temperature, $\mu$ the mean molecular weight and $m_u$ the atomic mass unit, as in Module 3, equation (3.2). For dry air at $0\ ^\circ$C, (1.1) gives $280.02\ \mathrm{m\,s^{-1}}$. The measured speed is $331.45\ \mathrm{m\,s^{-1}}$. Newton's value is $0.8448$ of it, low by $15.5$ per cent. The error is not in the measurement and not in the arithmetic. It is in the relation assumed between pressure and density, the closure, and §3.1 finds it.
The Sun supplies the astrophysical half of the question. Its surface moves up and down with a period near five minutes. The motion is the superposition of many standing sound waves trapped inside the star, and their frequencies are measured precisely enough that the spacing below carries an uncertainty of $0.1\ \mu$Hz. Near the frequency of maximum power, $\nu_{\max} = 3090\ \mu$Hz, modes of the same angular degree and consecutive radial order are spaced by the large separation
$$ \Delta\nu_{\odot} \;=\; 135.1\pm0.1\ \mu\mathrm{Hz}. \tag{1.2}$$§5 shows that this spacing is set by the time sound takes to cross the star. That time depends on the sound speed at every depth, and the sound speed depends on the closure. So (1.2) tests the closure inside the Sun, as the $331.45\ \mathrm{m\,s^{-1}}$ tests it in air.
The module follows the three-part shape of this book. §2 and §3 derive the terrestrial result and check Newton and Laplace against the measured speed of sound in air. §4 and §5 change the regime: gravity stratifies the gas, which introduces a cutoff frequency tied to the scale height of Module 3, and a star is a closed cavity, which quantises the frequencies. §6 and §7 take the result to the Sun. The isothermal closure is refuted by a bound computed from the tabulated model alone. The adiabatic closure is consistent with the measurement in the interior. In the outer $1.7$ per cent of the radius, where the model table stops, a perfect gas with $\Gamma_1 = 5/3$ falls short of what the measurement requires.
The equations to be perturbed are those of an inviscid, adiabatic fluid. Module 2 derived them: continuity (6.3), the Euler equation (7.3), and the adiabatic energy equation in (8.4). With the Lagrangian derivative $D/Dt \equiv \partial/\partial t + \mathbf{u}\cdot\nabla$ of Module 2, (6.1), they read
$$ \frac{D\rho}{Dt} \;=\; -\rho\,\nabla\cdot\mathbf{u}, \tag{2.1}$$ $$ \rho\,\frac{D\mathbf{u}}{Dt} \;=\; -\nabla P + \rho\,\mathbf{g}, \tag{2.2}$$in which $\mathbf{u}$ is the bulk velocity and $\mathbf{g}$ the gravitational acceleration. Module 2 wrote the adiabatic condition for a gas with a fixed ratio of specific heats. A gas whose ionisation changes as it is compressed needs a more general coefficient.
Equation (2.4) follows from (2.3) by the chain rule along a parcel's path, since $s$ is constant along that path. Module 2's condition $D(P/\rho^\gamma)/Dt = 0$ is (2.4) with $\Gamma_1=\gamma$ constant. The last sentence of Definition 1 is quoted, not derived; the thermodynamics of an ionising gas is not developed in this book, and the statement is used in §7.3 only as the candidate explanation of a measured shortfall.
Equations (2.1), (2.2) and (2.4) are nonlinear, because products such as $\rho\,(\mathbf{u}\cdot\nabla)\mathbf{u}$ appear. Close to an equilibrium the products of small quantities are smaller still, and dropping them gives linear equations.
The dropped term $(\mathbf{u}_1\cdot\nabla)\mathbf{u}_1$ is the one that matters most. For a wave of speed $c$ and wavelength $\lambda$, $\partial\mathbf{u}_1/\partial t$ is of order $|\mathbf{u}_1|c/\lambda$ and $(\mathbf{u}_1\cdot\nabla)\mathbf{u}_1$ is of order $|\mathbf{u}_1|^2/\lambda$. Their ratio is $|\mathbf{u}_1|/c$, the Mach number of the fluid motion. Linear theory is therefore the theory of motions whose speed is small compared with the sound speed. When the ratio approaches one, the wave steepens into a shock, which is the subject of Module 8.
The quantity $c$ defined in (2.8) has the dimensions of a speed. §3 shows that it is the speed of sound.
Take the simplest equilibrium: no gravity, so that hydrostatic balance gives $\nabla P_0 = \mathbf{0}$, and a uniform density $\rho_0$. Then $P_0$, $\rho_0$ and $c$ are constants.
This book writes a perturbation as proportional to $\exp[i(\mathbf{k}\cdot\mathbf{x}-\omega t)]$ and every dispersion relation in the form $\omega^2 = \ldots$. With that convention a real $\omega$ is an oscillation and $\omega^2\lt 0$ is exponential growth. Module 5 will write the Jeans relation in the same form, where the sign of the right-hand side decides stability.
The proof used only one property of $c^2$: that it is the constant ratio $P_1/\rho_1$ in the wave. Which ratio that is depends on whether the gas exchanges heat during an oscillation. That is the whole of the disagreement between Newton and Laplace.
If heat flows fast enough to keep each parcel at the equilibrium temperature, the compression is isothermal. Then $P = \rho k_BT/(\mu m_u)$ at fixed $T$ gives $P_1/\rho_1 = P_0/\rho_0$, and $c$ is Newton's $c_{\rm N}$ of (1.1). If no heat flows during a period, the compression is adiabatic, (2.4) holds, and $c^2 = \Gamma_1P_0/\rho_0$. For an ideal gas this is Laplace's speed of sound,
$$ c_{\rm L} \;=\; \sqrt{\frac{\gamma P}{\rho}} \;=\; \sqrt{\frac{\gamma\,k_BT}{\mu\,m_u}} \;=\; \sqrt{\gamma}\;c_{\rm N}. \tag{3.3}$$Which limit applies is decided by comparing two times. Heat diffuses with a thermal diffusivity $\chi$, in $\mathrm{cm^2\,s^{-1}}$, and crosses a distance $\ell$ in a time of order $\ell^2/\chi$. The temperature contrast in a wave is between a compression and the neighbouring rarefaction, whose separation is set by the wavelength; take $\ell = 1/|\mathbf{k}|$. The oscillation lasts a time of order $1/\omega$. The compression is adiabatic when heat cannot cross $\ell$ within that time:
$$ \frac{1/\omega}{\ell^2/\chi} \;=\; \frac{\omega\chi\,|\mathbf{k}|^2}{\omega^2} \;=\; \frac{\omega\,\chi}{c^2} \;\ll\; 1, \tag{3.4}$$where the middle step used (3.2). Criterion (3.4) is an order-of-magnitude estimate. It is not proved here, because the linearised energy equation with conduction is not solved in this module.
For dry air at $0\ ^\circ$C, with $\chi = 0.187\ \mathrm{cm^2\,s^{-1}}$ and $c = 331.32\ \mathrm{m\,s^{-1}}$, the ratio in (3.4) is $1.07\times10^{-6}$ at $1$ kHz and $2.14\times10^{-5}$ at $20$ kHz, the upper limit of hearing. It reaches one only at $9.3\times10^{8}$ Hz, where the wavelength is $355$ nm, $5.2$ mean free paths of air. At that wavelength the continuum description of Module 1 is itself failing (Problem C1). Across the whole range in which air behaves as a fluid, sound in air is adiabatic.
The thermal diffusivity is computed from handbook values for dry air at $0\ ^\circ$C: thermal conductivity $0.0243\ \mathrm{W\,m^{-1}\,K^{-1}}$, density $1.293\ \mathrm{kg\,m^{-3}}$, specific heat $1005\ \mathrm{J\,kg^{-1}\,K^{-1}}$. They are inputs, not checked against a source here. The conclusion depends only on the order of magnitude of $\chi$, and would survive an error of a factor of $10^3$ at every audible frequency (the ratio would stay below $0.022$). The mean free path of $68$ nm is the room-air value used in Module 1.The two predictions can now be compared with a measurement. For dry air, $\mu = 28.9647$ as in Module 3 and $\gamma = 7/5$.
| Closure | Sound speed at $273.15$ K | Ratio to measured $331.45\ \mathrm{m\,s^{-1}}$ |
|---|---|---|
| Isothermal, Newton (1.1) | $280.02\ \mathrm{m\,s^{-1}}$ | $0.8448$ |
| Adiabatic, Laplace (3.3), $\gamma=7/5$ | $331.32\ \mathrm{m\,s^{-1}}$ | $0.9996$ |
The residual of Laplace's value is $0.13\ \mathrm{m\,s^{-1}}$, or $4\times10^{-4}$. It is thirteen times the quoted measurement uncertainty, so it is real. This module does not attribute it. Two inputs that could contribute are the ideal-gas law, which real air departs from slightly at one atmosphere, and the composition; the measurement was made at $300$ ppm of carbon dioxide, while $\mu = 28.9647$ is the standard dry-air value. Neither is computed here. Newton's residual, $51.43\ \mathrm{m\,s^{-1}}$, is $395$ times Laplace's, $0.1303\ \mathrm{m\,s^{-1}}$ before rounding.
The measured value, $(331.45\pm0.01)\ \mathrm{m\,s^{-1}}$ for dry air at $273.15$ K, one atmosphere, zero frequency and $300$ ppm of carbon dioxide, is from Smith & Harlow (1963). It was read as quoted in section 2.3 of Gavioso et al. (2025); the 1963 paper was not read directly. The value is trusted as a measurement: Gavioso et al. list it among experimental determinations, not among calculations.A star is not uniform. Gravity stratifies it, and Module 3 derived the simplest stratified state: the isothermal plane-parallel layer, whose pressure and density both fall as $e^{-z/H}$ with height $z$, where $H = k_BT/(\mu m_u g)$ is the pressure scale height of Module 3, Proposition 3. This section perturbs that state. The result is a new frequency, below which sound cannot propagate upward.
Restrict attention to vertical motion, $\mathbf{u}_1 = u(z,t)\,\hat{\mathbf{z}}$, with all perturbations depending on $z$ and $t$ only. Waves with a horizontal wavenumber add a second restoring force, buoyancy, and are treated in Module 6.
Three consequences follow, and each is read directly from (4.4) or from the substitution.
A cutoff. For real $k$, (4.4) gives $\omega^2\ge\omega_{\rm ac}^2$. A wave with $\omega\lt\omega_{\rm ac}$ must have $k^2 = (\omega^2-\omega_{\rm ac}^2)/c^2 \lt 0$, so $k = \pm i\kappa$ with
$$ \kappa \;=\; \frac{\sqrt{\omega_{\rm ac}^2 - \omega^2}}{c}. \tag{4.5}$$Such a wave does not propagate; $\psi$ varies as $e^{\pm\kappa z}$, and the physical solution in a semi-infinite atmosphere is the one that decays upward. The wave is evanescent. In cyclic frequency the cutoff is
$$ \nu_{\rm ac} \;=\; \frac{\omega_{\rm ac}}{2\pi} \;=\; \frac{c}{4\pi H} \;=\; \frac{\Gamma_1\,g}{4\pi c}, \tag{4.6}$$where the last form uses $H = c^2/(\Gamma_1g)$ from (4.1). This is the relation that Module 3's closing table promised: the cutoff frequency contains that module's scale height.
Growth of the amplitude with height. Since $\rho_0\propto e^{-z/H}$, the factor $e^{z/2H}$ in $u = e^{z/2H}\psi$ equals $[\rho_0(0)/\rho_0(z)]^{1/2}$. A propagating wave, with $|\psi|$ constant, has a velocity amplitude that grows as $\rho_0^{-1/2}$. The kinetic energy density $\tfrac12\rho_0u^2$ is proportional to $|\psi|^2$ and stays constant. Over $1000$ km of the solar atmosphere, with $H = 143.0$ km, the velocity grows by a factor of $33.0$ (Problem D1). Linear theory holds only while $|u|\ll c$, so a wave that propagates far enough upward must steepen into a shock, the subject of Module 8.
Energy in an evanescent wave. For $\omega\lt\omega_{\rm ac}$ the energy density $\tfrac12\rho_0u^2\propto|\psi|^2\propto e^{-2\kappa z}$ decays upward. The velocity amplitude varies as $e^{(1/2H-\kappa)z}$, and by (4.5) $\kappa\lt\omega_{\rm ac}/c = 1/2H$, so the velocity of an evanescent wave still grows with height, more slowly than that of a propagating one. "Evanescent" describes the energy, not the velocity (Problem C2).
Fig. 1 draws (4.4) and (4.5) for the solar photosphere. The dashed line is the uniform gas of Proposition 2. Stratification lifts the propagating branch above it and ends it at $\nu_{\rm ac}$ when $k=0$. Below $\nu_{\rm ac}$ the evanescent branch runs from $\kappa=0$ to $\kappa = 1/2H$ at zero frequency.
Evaluate (4.6) at the photosphere. The surface gravity from the IAU nominal $GM_\odot$ and the radius of the tabulated solar model of §6 is $g_\odot = 2.7398\times10^{4}\ \mathrm{cm\,s^{-2}}$. The surface composition of that model, $X = 0.7583$ and $Y = 0.22905$ in its last row, with $Z = 1 - X - Y = 0.01265$, gives for a neutral gas, with $16$ taken as the mass per particle of the heavy elements,
$$ \mu \;=\; \frac{1}{X + Y/4 + Z/16} \;=\; 1.2250. \tag{4.7}$$At $T = T_{\rm eff} = 5772$ K with $\Gamma_1 = 5/3$, the sound speed is $c = 8.081\ \mathrm{km\,s^{-1}}$, the scale height is $H = 143.0$ km, and
$$ \nu_{\rm ac} \;=\; \frac{c}{4\pi H} \;=\; 4497\ \mu\mathrm{Hz}. \tag{4.8}$$Module 3 evaluated the photospheric scale height with a round input $\mu = 1.3$ and found $134.6$ km, with $g = 2.742\times10^{4}\ \mathrm{cm\,s^{-2}}$ from the IAU radius; this module's table-radius $g$ gives $134.7$ km. The value $1.2250$ used here is computed from the model's composition. With $\mu = 1.3$ the cutoff would be $4633\ \mu$Hz; by (4.6) at fixed $T$ and $g$, $\nu_{\rm ac}\propto\mu^{1/2}$.
The measured cutoff of the Sun is "around $5000\ \mu$Hz" (Jiménez, García & Pallé 2011), varying by $100$ to $150\ \mu$Hz over the activity cycle. The prediction (4.8) is $0.8994$ of it. By (4.6), $\nu_{\rm ac}\propto T^{-1/2}$ at fixed $g$, $\mu$ and $\Gamma_1$, so matching $5000\ \mu$Hz requires $T = 4669$ K, which is $0.809\,T_{\rm eff}$. If the $150\ \mu$Hz maximum-to-minimum variation is centred on $5000\ \mu$Hz, the required temperature runs from $4532$ K at $5075\ \mu$Hz to $4812$ K at $4925\ \mu$Hz, that is from $0.785$ to $0.834\,T_{\rm eff}$.
This comparison is a consistency tie, not a check. The solar atmosphere is not isothermal, so no single temperature is the right one to put into (4.6); $T_{\rm eff}$ is a convenient reference, not a prediction of where the cutoff is set. The measured value carries no formal uncertainty in the source. What the comparison shows is that an isothermal atmosphere with $\Gamma_1 = 5/3$ and the neutral $\mu$ of (4.7) reproduces the measured cutoff at $4669$ K, $0.809\,T_{\rm eff}$, and at $0.785$–$0.834\,T_{\rm eff}$ across the activity variation. Within that model the gas must be cooler than $T_{\rm eff}$. The real atmosphere is not isothermal, so (4.6) is not the exact cutoff there, and the comparison does not locate the gas that sets it. The source's own derivation of $5000\ \mu$Hz is described there and not reproduced here.The ratio of the measured cutoff to $\nu_{\max}$ is $5000/3090 = 1.618$. The five-minute modes, whose period at $\nu_{\max}$ is $5.39$ minutes, lie below the cutoff. Upward they are evanescent. The atmosphere reflects them, and they are trapped inside the star. That trapping is what makes the Sun a resonant cavity, and §5 finds the frequencies of the cavity.
A wave trapped between two reflecting boundaries interferes with its own reflection, and only frequencies at which the reflections reinforce survive. The spacing of those frequencies is set by the time sound takes to cross the cavity.
The surface condition $P_1 = 0$ is an idealisation of a free surface. A different reflecting condition changes the phase of the reflection. In the uniform column of Proposition 5, $u=0$ instead of $P_1=0$ at one end gives $\nu_n = (n+\tfrac12)/(2\tau)$: every frequency shifts by $1/(4\tau)$ and the spacing is unchanged. In the sphere the rigid condition gives $\tan kR = kR$, whose roots approach $(n+\tfrac12)\pi$ only as $n$ grows, so there the spacing is unchanged only asymptotically. In a real star $c$ varies with radius. The next proposition shows, in one dimension, that the spacing is still $1/(2\tau)$ at high frequency, with $\tau$ the integral (5.1).
The bound (5.3a) shows the rate: the departure from $n/(2\tau)$ falls as $1/\nu_n$, which is the form of the second-order term in (5.4). A sphere adds two things that the column lacks: the centre, where the radial equation is singular, and gravity, which makes the upper reflection occur at the cutoff of §4 rather than at a rigid end. Tassoul (1980) carried out the asymptotic analysis for a star. For modes of radial order $n$ and angular degree $\ell$ it gives
$$ \nu_{n,\ell} \;\simeq\; \left(n + \frac{\ell}{2} + \varepsilon\right)\Delta\nu_{\rm as}, \qquad \Delta\nu_{\rm as} \;=\; \left(2\int_0^R\frac{dr}{c}\right)^{-1} \;=\; \frac{1}{2\tau}, \tag{5.4}$$with a phase constant $\varepsilon$, plus a second-order term in $1/\nu$. The boundaries change $\varepsilon$ and the second-order term, as the free-surface condition did in Proposition 4. The spacing stays $1/(2\tau)$, the result that Propositions 4 and 5 prove in the smaller settings.
Relation (5.4) is quoted, not derived for a sphere. It is stated in this form, with $\Delta\nu_{\rm as} = (2\int_0^R dr/c)^{-1}$, as equations (1) and (7) of Mosser et al. (2013), who attribute it to equations 65–74 of Tassoul (1980). It is trusted as standard asymptotic theory of stellar oscillations, valid for high radial order. The Sun's modes near $\nu_{\max}$ have radial orders near $n = \nu_{\max}/\Delta\nu \approx 23$.A measured spacing is not exactly $\Delta\nu_{\rm as}$. The value (1.2) is an average of $\nu_{n+1}-\nu_n$ over the modes near $\nu_{\max}$, where the second-order term of Tassoul's expansion still bends the spacing. Mosser et al. (2013) fit that bending across many stars and give
$$ \Delta\nu_{\rm as} \;=\; (1+\zeta)\,\Delta\nu_{\rm obs}, \tag{5.5}$$with a solar value $\zeta_\odot\simeq 2.6$ per cent. The asymptotic spacing of the Sun is therefore larger than the measured one, $1.026\times135.1 = 138.61\ \mu$Hz. Both values are carried through §7, and each conclusion there is stated for both.
The correction $\zeta_\odot\simeq2.6$ per cent is stated in section 4.2 of Mosser et al. (2013) without the step that produces it. Their main-sequence fit (equation 20), $\zeta = 0.57/n_{\max}$ with coefficient $0.57\pm0.02$ and $n_{\max} = \nu_{\max}/\Delta\nu_{\rm obs}$, gives $2.49$ per cent with the values of (1.2); the $0.1$ percentage-point difference changes no conclusion below. They give no separate uncertainty for the solar value; its trust is that of an empirical fit, and the checks below are built so that their conclusions do not depend on whether $\zeta$ is applied.The comparison object is the standard solar model BS2005-AGS,OP of Bahcall, Serenelli & Basu (2005), used in Module 3 and stored as data/bs05_agsop.dat. Its $1284$ rows give $P$ and $\rho$ from $r = 0.00161\,R$ to $r = 0.98308\,R$, with $R = 6.9598\times10^{10}$ cm. They do not give $\Gamma_1$, so the sound speed must be computed from an assumed closure, and that is what makes the table a test of closures.
Integrating (5.1) over the table, from its first to its last row, with the trapezoid rule and adding $r_0/c_0 = 2.2$ s for the innermost $0.00161\,R$, gives
$$ \tau_{\rm tab} \;=\; \int_0^{0.98308R}\frac{dr}{c} \;=\; \begin{cases} 2963.2\ \mathrm{s}, & c^2 = \tfrac{5}{3}P/\rho,\\[2pt] 3825.4\ \mathrm{s}, & c^2 = P/\rho. \end{cases} \tag{6.1}$$Simpson's rule on the same rows gives $2963.15$ s against the trapezoid's $2963.16$ s for $\Gamma_1 = 5/3$, and halving the rows, with the last row kept, gives $2963.20$ s, so the quadrature error is below $0.1$ s. The ratio of the two values is $3825.4/2963.2 = 1.2910 = \sqrt{5/3}$, as it must be, since the two sound speeds differ by that constant factor at every row.
The table stops $11{,}776$ km below the surface, $1.69$ per cent of the radius. The outer layer is omitted, and the sound speed there is lowest, so it contributes a crossing time out of proportion to its thickness. Fig. 2 shows this on a logarithmic depth axis. What is known about the layer without a model is only its sign: every part of it adds a positive time $dr/c$ to $\tau$. The whole of §7 rests on that sign.
The model table is trusted as in Module 3: as a faithful copy of BS2005-AGS,OP, whose surface helium matches the paper, and not as a description of the Sun. Its $P$ and $\rho$ are the model's; the sound speed computed here is not the model's own sound speed, because the model's $\Gamma_1$ is not tabulated. The radius used is the table's own constant.The measured spacing (1.2) implies an acoustic radius $\tau = 1/(2\Delta\nu)$. With $\Delta\nu_{\rm obs} = 135.1\pm0.1\ \mu$Hz it is $3701.0\pm2.7$ s. With the asymptotic value $138.61\ \mu$Hz of (5.5) it is $3607.2$ s. Each check compares a closure with these two totals.
An isothermal closure makes $c^2 = P/\rho$ at every depth. By (6.1), the tabulated interior alone then takes $3825.4$ s to cross. The untabulated layer adds a positive time, so the whole star takes longer, and
$$ \Delta\nu_{\rm iso} \;\le\; \frac{1}{2\times3825.4\ \mathrm{s}} \;=\; 130.70\ \mu\mathrm{Hz}. \tag{7.1}$$The conclusion does not depend on the second-order correction. Mosser et al.'s correction moves the comparison value away from the bound. A correction of the opposite sign would have to exceed $3.25$ per cent to save the isothermal closure, and the measured statistical uncertainty is $0.07$ per cent. The strength of the check comes from its being a bound. A best estimate for the layer would need a model of the layer; an upper limit needs only the fact that crossing times add.
The check refutes the isothermal closure given the model's $P$ and $\rho$. It does not show that a different structure could not have an isothermal sound speed and the measured $\Delta\nu$. What it shows is that the Sun's measured spacing and the standard model's stratification together exclude Newton's closure, as the $331.45\ \mathrm{m\,s^{-1}}$ excluded it in air.
With $c^2 = \tfrac{5}{3}P/\rho$ the table takes $2963.2$ s, and the corresponding bound $\Delta\nu\le168.74\ \mu$Hz lies above both measured and asymptotic values, so the bound does not refute the adiabatic closure. The outer layer must then supply the remainder:
$$ \tau_{\rm layer} \;=\; \tau - \tau_{\rm tab} \;=\; \begin{cases} 3607.2 - 2963.2 \;=\; 644.0\ \mathrm{s}, & \Delta\nu_{\rm as},\\[2pt] 3701.0 - 2963.2 \;=\; 737.8\ \mathrm{s}, & \Delta\nu_{\rm obs}. \end{cases} \tag{7.2}$$A positive remainder is a weak test. It would be satisfied by any closure whose interior crossing time is shorter than the total, and it says nothing about whether $644$ s is the right amount. §7.3 asks whether a simple gas can supply that amount.
The adiabatic assumption is also an assumption about heat. By (3.4) it requires $\omega\chi/c^2\ll1$, with $\chi$ now the radiative diffusivity of the stellar gas. That criterion is not evaluated for the Sun in this module. It is expected to hold deep in the star, where photons diffuse slowly, and to fail near the photosphere, where they escape; neither statement is computed here, and the size of the failure is a subject of Module 13.
To ask whether a simple gas supplies $644$ s, the layer needs a structure. Take the layer between the table's last row, at depth $D = 11{,}776$ km and $T_{\rm top} = 81{,}200$ K, and the photosphere at $T_{\rm eff} = 5772$ K, and assume $T$ linear in the depth $z$ below the surface. The choice is motivated by Module 3: a polytropic layer with uniform $\mu$ and $g$ has $T$ linear in depth, by Proposition 4 of that module. With a uniform $\Gamma_1$ and $\mu$, $c^2 = \Gamma_1k_BT/(\mu m_u)$, and the crossing time of the layer is
$$ \tau_{\rm layer} \;=\; \int_0^{D}\frac{dz}{c} \;=\; \frac{2\left(\sqrt{T_{\rm top}} - \sqrt{T_{\rm eff}}\right)}{\sigma\,\sqrt{\Gamma_1k_B/(\mu m_u)}}, \qquad \sigma \;=\; \frac{T_{\rm top}-T_{\rm eff}}{D}, \tag{7.3}$$which follows from $dz = dT/\sigma$ and $\int T^{-1/2}dT = 2T^{1/2}$.
The slope of the linear profile is $\sigma = 6.41\ \mathrm{K\,km^{-1}}$. An $n = 3/2$ layer of fully ionised gas, from Module 3's (4.2), would have $7.78\ \mathrm{K\,km^{-1}}$ and reach $97{,}361$ K at depth $D$; a neutral one would have $16.15\ \mathrm{K\,km^{-1}}$ and reach $195{,}907$ K. The tabulated $81{,}200$ K is closer to the fully ionised adiabat, and the linear profile is not far from what a polytrope of that composition would give. It remains an assumption. The real layer includes a thin superadiabatic region under the photosphere where the gradient is steeper (a standard solar-model result, not computed here), and nothing here shows that the linear profile over- or under-estimates the crossing time.
| Gas in the layer, $\Gamma_1 = 5/3$ | $\mu$ | $\tau_{\rm layer}$ | Fraction of $644.0$ s | Fraction of $737.8$ s | $\Delta\nu$ of the whole star |
|---|---|---|---|---|---|
| fully ionised, surface composition | $0.5901$ | $425.8$ s | $0.661$ | $0.577$ | $147.54\ \mu$Hz |
| $\mu_{\rm eff}$ of the table's last row | $0.6585$ | $449.8$ s | $0.698$ | $0.610$ | $146.50\ \mu$Hz |
| neutral, (4.7) | $1.2250$ | $613.5$ s | $0.953$ | $0.832$ | $139.79\ \mu$Hz |
A fully ionised gas has $\mu = 1/(2X + 3Y/4 + Z/2)$ (Module 3, (3.2)), which is $0.5901$ for the surface composition. Here $\mu_{\rm eff} = \rho k_BT/(P_{\rm gas}m_u)$ is computed from the table's own $P$, $\rho$ and $T$, with the radiation pressure $a_{\rm rad}T^4/3$ subtracted from $P$. The crossing time scales as $\mu^{1/2}$, so the neutral value $\mu = 1.2250$ is the most favourable of the three. It is also not a possible value for the whole layer: hydrogen is not neutral at $81{,}200$ K, and the table's own last row gives $\mu_{\rm eff} = 0.6585$.
State what this check does and does not show. It does not measure $\Gamma_1$: the required values come from the linear-$T$ assumption, and a cooler profile would lower the requirement. It shows that a gas of fixed composition, fixed ionisation and $\Gamma_1 = 5/3$, with this profile, cannot fill the gap for any $\mu$ between fully ionised and neutral. Closing it needs a lower sound speed in the layer: a $\Gamma_1$ below $5/3$, a $\mu$ that changes with depth, a cooler profile, or a combination. The first two are what Definition 1 says a partly ionised gas does, and the table's $\mu_{\rm eff}$ rises outward below the layer, consistent with the second.
| $r/R$ | $T$ (K) | $\mu_{\rm eff}$ from the table | $\mu$, fully ionised | Excess |
|---|---|---|---|---|
| $0.8493$ | $9.504\times10^{5}$ | $0.6012$ | $0.5901$ | $+1.9\%$ |
| $0.9499$ | $2.690\times10^{5}$ | $0.6145$ | $0.5901$ | $+4.1\%$ |
| $0.9698$ | $1.524\times10^{5}$ | $0.6290$ | $0.5901$ | $+6.6\%$ |
| $0.9800$ | $9.740\times10^{4}$ | $0.6491$ | $0.5901$ | $+10.0\%$ |
| $0.9831$ | $8.120\times10^{4}$ | $0.6585$ | $0.5901$ | $+11.6\%$ |
The model's effective molecular weight rises above the fully ionised value outward, by $11.6$ per cent at the last row. A rising $\mu$ means fewer free particles per unit mass, which is what recombination of helium and hydrogen produces. The table does not state the mechanism, and this module attributes the rise to ionisation as the standard interpretation, not as a computed result. The inference also assumes the table's gas pressure is ideal; a non-ideal correction that lowers $P$ at fixed $\rho$ and $T$ would raise $\mu_{\rm eff}$ in the same way, and this module does not separate the two. The hydrogen and helium ionisation zones lie in the untabulated layer and below it, and they lower both $\Gamma_1$ and the sound speed there.
Fig. 2 summarises the three solar checks. Only the isothermal closure is excluded outright, by a bound. The adiabatic interior is consistent, and a perfect gas in the outer layer is not enough to reach the measurement: in the right panel all three perfect-gas points sit to the right of the band.
The equations of this module are the linearised equations (2.6)–(2.8), applied to a uniform gas, to an isothermal atmosphere, and through the asymptotic relation (5.4) to a star. The table names each idealisation and where it is repaid.
| Idealisation made here | What it costs | Where it is repaid |
|---|---|---|
| Linear: $|\mathbf{u}_1|\ll c$ | Excludes steepening. A wave propagating upward above the cutoff grows as $\rho_0^{-1/2}$, by a factor of $33.0$ over $1000$ km of the solar atmosphere, and leaves the linear regime | Module 8 (shocks) |
| Adiabatic perturbations | Confirmed in air by Check 1. In the Sun, fails near the photosphere where radiation carries heat out of a compression within a period; this damps the modes and shifts their frequencies | Module 13 (radiation hydrodynamics) |
| Vertical waves only in §4 | Omits buoyancy, which acts on waves with a horizontal wavenumber and supports a second family of oscillations, gravity waves | Module 6 (convection), through the buoyancy frequency |
| Gravity held fixed | Omits the perturbation of gravity by $\rho_1$. Expected to be small for the high-order modes near $\nu_{\max}$ (not computed here); decisive for long wavelengths, where it drives collapse | Module 5 (Jeans instability) |
| Isothermal atmosphere in §4.1 | The cutoff depends on $T^{-1/2}$, and the real atmosphere has no single temperature. The measured $5000\ \mu$Hz needs $4669$ K | Not repaid in this module |
| Asymptotic relation (5.4), quoted for a sphere | Proved here only for a uniform sphere and a one-dimensional column. The second-order term moves the measured spacing by $\zeta_\odot\simeq2.6$ per cent | Not repaid; Tassoul (1980) and Mosser et al. (2013) |
| $\Gamma_1$ assumed, not computed | Check 4 shows $5/3$ is not enough in the outer layer but cannot measure the $\Gamma_1$ there. A solar model tabulated to the surface with its own $\Gamma_1$ would turn Check 3 into a direct test of $\Delta\nu$; Model S of Christensen-Dalsgaard et al. (1996) is one candidate, not examined here | Not repaid in this module |
| No magnetic field, no rotation | Omits magnetic tension and pressure as restoring forces, and the rotational splitting of mode frequencies | Module 12 (magnetohydrodynamics, Alfvén waves) |
One limitation belongs to the checks. Checks 2 to 4 combine a measured spacing with a model's $P$ and $\rho$. They test closures given that structure. Module 3's Check 3 showed that the same model misplaces the base of the convection zone by $0.0147\,R$, so its stratification is not exact. The isothermal refutation is $3.25$ per cent in $\Delta\nu$ against a bound, and a structural error in the model would have to shift the tabulated crossing time by at least that much in the right direction to reverse it. That has not been computed here.
Every number quoted in this module is produced by figs/m04_numbers.py or is arithmetic on its printed values, every problem answer by figs/m04_problems_check.py, and every figure by figs/m04_build_figs.py. Checks 2 and 3 need only the following, run from afd/figs.
"""Checks 2 and 3 from the tabulated model BS2005-AGS,OP.
tau = int dr/c over the table; Delta nu = 1/(2 tau). Run from afd/figs.
"""
import m04_numbers as M
ssm = M.load_ssm()
tau_iso = M.acoustic_radius_table(ssm, 1.0)
tau_ad = M.acoustic_radius_table(ssm, 5/3)
dnu_as = (1 + M.ZETA_SUN)*M.DNU_MEAS
print(f"isothermal bound = {1e6/(2*tau_iso):.2f} microHz")
print(f"layer needs {1/(2*dnu_as) - tau_ad:.1f} s (asymptotic), "
f"{1/(2*M.DNU_MEAS) - tau_ad:.1f} s (measured)")
assert 1/(2*tau_iso) < M.DNU_MEAS < dnu_as
Running it prints an isothermal bound of $130.70\ \mu$Hz and a layer requirement of $644.0$ s (asymptotic) and $737.8$ s (measured). The assertion fails if the isothermal bound ever reaches the measured spacing.
Solutions are given for every problem. Each numerical answer is verified by figs/m04_problems_check.py.
| Symbol | Meaning | Units (CGS) | First used |
|---|---|---|---|
| $P$, $\rho$, $T$ | pressure, mass density, temperature | $\mathrm{dyn\,cm^{-2}}$, $\mathrm{g\,cm^{-3}}$, K | (1.1) |
| $\mu$, $m_u$, $k_B$ | mean molecular weight, atomic mass unit, Boltzmann's constant | —, g, $\mathrm{erg\,K^{-1}}$ | (1.1) |
| $c_{\rm N}$, $c_{\rm L}$ | Newton's isothermal and Laplace's adiabatic sound speeds | $\mathrm{cm\,s^{-1}}$ | (1.1), (3.3) |
| $\nu_{\max}$ | frequency of maximum oscillation power | Hz | §1 |
| $\Delta\nu$, $\Delta\nu_{\rm obs}$, $\Delta\nu_{\rm as}$ | large separation; measured near $\nu_{\max}$; asymptotic | Hz | (1.2), (5.4) |
| $\mathbf{u}$, $\mathbf{g}$ | bulk velocity, gravitational acceleration | $\mathrm{cm\,s^{-1}}$, $\mathrm{cm\,s^{-2}}$ | (2.1), (2.2) |
| $\Gamma_1$, $\gamma$ | first adiabatic exponent; ratio of specific heats | — | Definition 1 |
| subscripts $0$, $1$ | equilibrium value; first-order perturbation | — | Definition 2 |
| $c$ | adiabatic sound speed $(\Gamma_1P_0/\rho_0)^{1/2}$ | $\mathrm{cm\,s^{-1}}$ | (2.8) |
| $\mathbf{k}$, $k$, $\omega$, $\nu$ | wavevector, vertical wavenumber, angular and cyclic frequency | $\mathrm{cm^{-1}}$, $\mathrm{s^{-1}}$, Hz | (3.2) |
| $\chi$ | thermal diffusivity | $\mathrm{cm^2\,s^{-1}}$ | (3.4) |
| $H$ | pressure scale height $k_BT/(\mu m_ug)$ | cm | (4.1) |
| $\psi$ | scaled vertical velocity $e^{-z/2H}u$ | $\mathrm{cm\,s^{-1}}$ | (4.3) |
| $\omega_{\rm ac}$, $\nu_{\rm ac}$ | acoustic cutoff $c/2H$, and $c/(4\pi H)$ | $\mathrm{s^{-1}}$, Hz | (4.3), (4.6) |
| $\kappa$ | decay rate of an evanescent wave | $\mathrm{cm^{-1}}$ | (4.5) |
| $X$, $Y$, $Z$ | mass fractions of hydrogen, helium, heavier elements | — | (4.7) |
| $\tau$, $s$ | acoustic radius; acoustic coordinate | s | (5.1), Proposition 5 |
| $Z$ (impedance), $V$, $a$ | acoustic impedance $\rho_0c$; potential and coefficient of (5.3) | $\mathrm{g\,cm^{-2}\,s^{-1}}$; $\mathrm{s^{-2}}$; $\mathrm{s^{-1}}$ | (5.3) |
| $n$, $\ell$, $\varepsilon$ | radial order, angular degree, phase constant | — | (5.2), (5.4) |
| $\zeta$ | fractional correction from measured to asymptotic $\Delta\nu$ | — | (5.5) |
| $D$, $\sigma$ | depth of the untabulated layer; its assumed temperature gradient | cm; $\mathrm{K\,cm^{-1}}$ | (7.3) |
| $\mu_{\rm eff}$ | molecular weight inferred from the table's $P$, $\rho$, $T$ | — | §7.3 |
The symbol $Z$ is used both for the heavy-element mass fraction in (4.7) and for the impedance in Proposition 5; the two never appear in the same equation.
Every published value this module compares against, with what was taken from it.
data/bs05_agsop.dat, as in Module 3.Module 4 of Astrophysical Fluid Dynamics. Previous: Module 3 — Hydrostatic Equilibrium