Astronomy 112: The Physics of Stars Class 6 Notes: Internal Energy and Radiative Transfer In the last class we used the kinetic theory of gasses to understand the pressure of stellar material. The kinetic view is essential to generalizing the concept of pressure to the environ- ments found in stars, where gas can be relativistic, degenerate, or both. The goal of today’s class is to extend that kinetic picture by thinking about pressure in stars in terms of the associated energy content of the gas. This will let us understand how the energy flow in stars interacts with gas pressure, a crucial step toward building stellar models. I. Pressure and Energy A. The Relationship Between Pressure and Internal Energy The fundamental object we dealt with in the last class was the distribution of particle mometa, dn(p)/dp, which we calculated from the Boltzmann distribu- tion. Given this, we could compute the pressure. However, this distribution also corresponds to a specific energy content, since for particles that don’t have inter- nal energy states, internal energy is just the kinetic energy of particle motions. Given a distribution of particle momenta dn(p)/dp in some volume of space, the corresponding density of energy within that volume of space is Z ∞ dn(p) e = (p) dp, 0 dp where (p) is the energy of a particle with momentum p. It is often more conve- nient to think about the energy per unit mass than the energy per unit volume. The energy per unit mass is just e divided by the density: 1 Z ∞ dn(p) u = (p) dp. ρ 0 dp The kinetic energy of a particle with momentum p and rest mass m is s 2 2 p (p) = mc 1 + − 1 . m2c2 This formula applies regardless of p. In the limit p mc (i.e. the non-relativistic case), we can Taylor expand the square root term to 1 + p2/(2m2c2), and we recover the usual kinetic energy: (p) = p2/(2m). In the limit p mc (the ultra- relativistic case), we can drop the plus 1 and the minus 1, and we get (p) = pc. Plugging the non-relativistic, non-degenerate values for (p) and dn(p)/dp into the integral for u and evaluating gives ∞ 2 ! 1 Z 4πn 2 p 2 −p /(2mkB T ) u = 3/2 p e dp ρ 0 (2πmkBT ) 2m 1 ∞ 2π Z 2 4 −p /(2mkB T ) = 2 3/2 p e dp m (2πmkBT ) 0 Z ∞ 4 4 −q2 = kBT q e dq mπ1/2 0 3 = k T 2m B 3 P = , 2 ρ √ where the integral over q evaluates to 3 π/8. This is the same as the classic result that an ideal gas has an energy per particle of (3/2)kBT . Since the pressure and energies are simply additive, it is clear that the result u = (3/2)(P/ρ) applies even when there are multiple species present. Applying the same procedure in the relativistic, non-degenerate limit gives 1 Z ∞ c 3 n u = p2e−pc/kB T (pc) dp ρ 0 kBT 2 c4 Z ∞ 3 −pc/kB T = 3 p e dp 2m(kBT ) 0 3 = k T m B P = 3 . ρ It is straightforward to show that this result applies to radiation too, by plugging = hν for the energy and the Planck distribution for dn(ν)/dν. Note that this implies that the volume energy density of a thermal radiation field is 4 erad = aT = 3Prad. For the non-relativistic, degenerate limit we have a step-function distribution that has a constant value dn(p)/dp = (8π/h3)p2 out to some maximum momentum 3 1/3 p0 = [3h n/(8π)] , so the energy is 2 ! 1 Z p0 8π p u = p2 dp ρ 0 h3 2m 4π = p5 5mρh3 0 3 2/3 3h2n5/3 = π 40ρm 3 P = , 2 ρ exactly as in the non-dengerate case. 2 Finally, for the relativistic degenerate case we have 1 Z p0 8π u = p2 (pc) dp ρ 0 h3 3 1/3 3 hcn4/3 = π 8 ρ P = 3 ρ B. Adiabatic Processes and the Adiabatic Index Part of the reason that internal energies are interesting to compute is because of the problem of adiabatic processes. An adiabatic process is one in which the gas is not able to exchange heat with its environment or extract it from internal sources (like nuclear burning), so any work it does must be balanced by a change in its internal energy. The classic example of this is a gas that is sealed in an insulated box, which is then compressed or allowed to expand. In many circumstances we can think of most of the gas in a star (that outside the region where nuclear burning takes place) as adiabatic. It can exchange energy with its environment via radiation, but, as we have previously shown, the radiation time is long compared to the dynamical time. Thus any process that takes place on timescale shorter than a Kelvin-Helmholtz timescale can be thought of as adiabatic. To understand how an adiabatic gas behaves, we use the first law of thermody- namics, which we derived a few classes back: du d 1! ∂F + P = q − = 0, dt dt ρ ∂m where we have set the right-hand side to zero under the assumption that the gas is adiabatic, so it does not exchange heat with its environment and does not generate heat by nuclear fusion. We have just shown that for many types of gas u = φP/ρ, where φ is a constant that depends on the type of gas. If we make this substitution in the first law of thermodynamics, then we get d 1! 1 d d 1! d 1! 1 d 0 = φP + φ P + P = (φ + 1)P + φ P dt ρ ρ dt dt ρ dt ρ ρ dt This implies that dP φ + 1 d 1! = − ρP dt φ dt ρ φ + 1! P dρ = φ ρ dt dP φ + 1! dρ = P φ ρ ln P = γa ln ρ + ln Ka γa P = Kaρ , 3 where γa = (φ + 1)/φ and Ka is a constant. Thus we have shown that, for an adiabatic gas, the pressure and density are related by a powerlaw. The constant of integration Ka is called the adiabatic constant, and it is deter- mined by the entropy of the gas. The exponent γa is called the adiabatic index, and it is a function solely of the type of gas: all monatomic ideal gasses have γa = (3/2 + 1)/(3/2) = 5/3, whether they are degenerate or not. All relativistic gasses have γa = 4/3, whether they are degenerate or not. The adiabatic index is a very useful quantity for know for a gas, because it de- scribes how strongly that gas resists being compressed – it specifies how rapidly the pressure rises in response to an increase in density. The larger the value of γa, the harder it is to compress a gas. In a few weeks, we will see that the value of the adiabatic index for material in a star has profound consequences for the star’s structure. For most stars γa is close to 5/3 because the gas within them is non-relativistic. However, as the gas becomes more relativistic, γa approaches 4/3, and resistance to compression drops. When that happens the star is not long for this world. A similar effect is responsible for star formation. Under some circumstances, radiative effects cause interstellar gas clouds to act as if they had γa = 1, which means very weak resistance to compression. The result is that these clouds collapse, which is how new stars form. C. The Adiabatic Index for Partially Ionized Gas The adiabatic index is fairly straightforward for something like a pure degener- ate or non-degenerate gas, but the idea can be generalized to considerably more complex gasses. One case that is of particular interest is a partially ionized gas. This will be the situation in the outer layers of a star, where the temperature falls from the mean temperature, where the gas is fully ionized, to the surface temperature, where it is fully neutral. This case is tricky because the number of free gas particles itself becomes a function of temperature, and because the potential energy associated with ionization and recombination becomes an extra energy source or sink for the gas. Consider a gas of pure hydrogen within which the number density of neutral atoms is n0 and the number densities of free protons and electrons are np = ne. The number density of all atoms regardless of their ionization state is n = ne + n0. We define the ionization fraction as n x = e , n i.e. x is the fraction of all the electrons present that are free, or, equivalently, the fraction of all the protons present that do not have attached electrons. The pressure in the gas depends on how many free particles there are: P = nekBT + npkBT + n0kBT = (1 + x)nkBT = (1 + x)RρT Thus the pressure at fixed temperature is higher if the gas is more ionized, because there are more free particles. 4 The number densities of free electrons and protons are determined by the Saha equation, which we encountered in the first class. As a reminder, the Saha equa- tion is that the number density of ions at ionization states i and i + 1 are related by n 2Z 2πm k T !3/2 i+1 i+1 e B −χ/kB T = 2 e , ni neZi h where χ is the ionization potential and Zi and Zi+1 are the partition functions of the two states.