Before studying the effects of convection in red giants, it is important to understand why convection sets in as a subgiant cools. A packet of gas in a star will rise if it is less dense than its surroundings. As long as it finds itself in a cooler and more dense environment, it will continue to rise. Conversely, a packet of gas in a star will fall (due to gravity) if it is more dense than its surroundings. As long as it finds itself in a hotter and less dense environment, it will continue to fall. Complex convection currents can therefore be set up within a star, and their action depends on the variation in temperature gradient from place to place. We examine the condition for convection below.
The ideal gas law (Equation 1.11) may be rearranged as
4.7 The red-giant phase: shell-hydrogen burning
so, if we take natural logarithms of this equation, we obtain logeρ = loge , m k - + logeP − logeT.
Differentiating this equation with respect to P gives d logeρ dP = d loge(m/k) dP + d logeP dP − d logeT dP .
Using the chain rule and setting the first right-hand term to zero (as m and k are constant) gives d logeρ dρ dρ dP = d logeP dP − d logeT dT dT dP.
Now we note that d logex/dx = 1/x, so multiplying through by dP gives
dρ ρ = dP P − dT T .
Considering small finite increments (rather than infinitesimals) gives Δρ ρ = ΔP P − ΔT T . (4.9)
This equation describes how the temperature, pressure and density of a small packet of gas will adjust, relative to its initial properties, when subject to change.
In the following argument we will refer to changes in a packet of rising gas by δ, e.g. δρ, whereas changes in the surroundings will be referred to by Δ, e.g. Δρ.
Now consider a packet of gas at a certain height in a stellar envelope, whose density ρ, temperature T and pressure P match those of its surroundings. If the packet of gas is displaced upwards to a slightly different height where the density, temperature and pressure are ρ + Δρ, T + ΔT and P + ΔP , in general its properties will no longer match those of its surroundings. (Remember that Δρ, ΔT and ΔP can be positive or negative.) Let us suppose the density, temperature and pressure of the packet of gas are now ρ + δρ, T + δT and
P + δP respectively. We can further suppose that:
• the pressure in the packet will rapidly change to match that of the surroundings
where the pressure is P + ΔP , so δP = ΔP ;
• the packet will expand or contract adiabatically, i.e. without input or extraction
of energy, in order to achieve this matched pressure.
In an adiabatic process, the pressure P , density ρ and volume V of an ideal gas obey the relationship
P Vγ= constant or P ∝ ργ, (4.10) where γ (the adiabatic index) of a classical ideal gas is related to the number of degrees of freedom s of the particles in the gas (see box below) by
γ = 1 + s/2
If each gas particle has just the three translational degrees of freedom, then s = 3 and γ = 5/3. However, if the number of degrees of freedom is larger, then γ is smaller. As s becomes large, so γ approaches 1. This situation will occur in a molecular gas where the particles have additional degrees of freedom due to rotation or vibration. It will also be the case if heat can be absorbed by ionizing atoms.
Degrees of freedom
The total energy of a gas particle is said to arise from s degrees of freedom when there are s independent terms in the expression for that energy, each term arising from the square of a component of translational, vibrational or angular velocity. An ideal, monatomic gas has just three degrees of freedom, corresponding to independent velocities in each of the x-, y- and
z-directions. However, molecular ideal gases have a few additional degrees
of freedom associated with molecular rotation and vibration, and non-ideal gases have even more degrees of freedom associated with ionization. Figure 4.9 shows that a diatomic molecule has three forms of energy: translational, rotational and vibrational. These give rise, respectively, to three degrees of freedom due to velocities in three independent directions, two degrees of freedom due to there being two independent axes about which the molecule can rotate (a rolling motion about the axis joining the atoms is not counted), and two degrees of freedom due to the kinetic and potential energies of vibration, giving a total of seven degrees of freedom.
vz (a) translation ω1 ω2 vx vy (b) rotation v (c) vibration: EVIB= EK+ EP
Figure 4.9 Three forms of energy of a diatomic molecule.
Now, let the constant of proportionality in P ∝ ργbe C, so P = Cργ.
Differentiating with respect to density gives dP /dρ = Cγργ−1, i.e. (1/γ) dP =
Cργ−1dρ. Dividing through by P gives (1/γ) dP /P = Cργ−1dρ/P . On the
right-hand side, we then use the fact that P = Cργ to obtain Cργ−1dρ/Cργ=
ρ−1dρ = dρ/ρ. Substituting this into the previous expression therefore gives
1 γ dP P = dρ ρ .
4.7 The red-giant phase: shell-hydrogen burning
permissible to replace the infinitesimals by finite increments: dP → δP and dρ → δρ. Doing so gives: 1 γ δP P = δρ ρ . (4.12)
Now, convection will be possible if the packet of gas is buoyant and
continues to rise, that is if its density is less than that of its surroundings, i.e. (ρ + δρ) < (ρ + Δρ) or simply δρ < Δρ. So substituting Equation 4.12 into Equation 4.9, we have the condition for convection as
1 γ δP P < ΔP P − ΔT T ,
where δP is the pressure change in the packet and ΔP and ΔT are the pressure and temperature changes in the surrounding gas. As noted above, we can assume that δP = ΔP, since the pressure of the packet will respond rapidly to its new environment, and rewrite this condition as
ΔT T < (γ − 1) γ ΔP P . (4.13)
So the critical temperature gradient for convection to occur is dT dr < (γ − 1) γ T P dP dr. (4.14)
Note that since r is measured outwards in the star, both dT /dr and dP /dr are negative (T and P decrease outwards). Therefore, the statement that dT /dr is less than another negative number actually means that the temperature gradient is
steeper than the pressure gradient term. Taking absolute magnitudes of the
gradients, we get ))
))dTdr)))) > (γ − 1)γ TP ))))dPdr)))), (4.15) i.e. the inequality switches from < to >. You may use either form, but you must be unambiguous; merely saying that the temperature gradient must be less than the pressure gradient, without adding that both are negative, could be misleading.
Exercise 4.6 Consider the critical temperature gradient for convection, and evaluate the coefficient (γ − 1)/γ for two values of s: (a) s = 3, and (b) s → ∞. In case (b), under what circumstances is the gas unstable to convection (assuming
dT /dr < 0)? ■
The critical temperature gradient for convection can be written in several different ways. Since dx/x = d logex, we can also write the critical condition for
convection in terms of the logarithmic temperature and pressure gradients d logeT dr < (γ − 1) γ d logeP dr .
One can even go a step further, and make use of the chain rule to remove all reference to the radius r, and consider the change in log (temperature) with log (pressure):
d logeT
d logeP > γ − 1
γ . (4.16)
(Note the change of sense of the inequality from < to > because we have divided both sides by a negative quantity, d logeP/dr.)