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PHYS 633: Introduction to Stellar Astrophysics Spring Semester 2006 Rich Townsend ([email protected])

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PHYS 633: Introduction to Stellar Astrophysics Spring Semester 2006 Rich Townsend (Rhdt@Bartol.Udel.Edu) PHYS 633: Introduction to Stellar Astrophysics Spring Semester 2006 Rich Townsend ([email protected]) The Equation of State The equation of state for stellar matter is used to relate one of the thermo- dynamic state variables (pressure P , temperature T or density ρ) to the other two. It can be constructed by combining the individual equations of state for the three classes of particle making up stellar material: ions (and, at low tem- peratures, atoms), free electrons, and photons. Specifically, we can write the total pressure P as P = Pion + Pe + Prad, (1) where Pion, Pe and Prad – the partial pressures of ions, electrons and radiation, respectively – each depend in general on the temperature T , density ρ and chemical composition Xi. The reason we can break down P in this way comes from the fact that the three differing classes of particle interact — either with themselves, or with other classes of particle — only at short ranges; there are no long-range forces present. So, how can we calculate the partial pressures appearing in the above ex- pression? Let’s first focus on the ions. To a high degree of accuracy, these can be treated as an ideal gas; while there are some situations at very high densities where the ion gas can behave strangely (e.g., crystallization or neutronization), these situations are never encountered during ‘normal’ stellar evolution. Hence, we write down the equation of state for the ion gas as Pion = nionkT, (2) where nion is the number density of the ions (i.e., the total number of ions per unit volume). To relate nion to the mass density ρ, recall from our treatment of the chemical composition that the mass fraction Xi of element i is defined by n m X = i i , (3) i ρ where ni is the number of nuclei of element i per unit volume, and mi is the corresponding mass of each nucleus. Rearranging, we have Xiρ ni = , (4) mi which can also be interpreted as the number density of ions of element i, since there is one nucleus per ion. Hence, writing the total ion number density as X X Xiρ n = n = , (5) ion i m i i i 1 we have the equation of state for the ion gas as X Xi P = k ρT. (6) ion m i i 1 Let’s introduce µi as the molecular weight of the ions of element i, so that mi µi = (7) mu (we don’t worry about the mass of any electrons still bound to the ion). Then, the equation of state can be expressed as X Xi P = R ρT ; (8) ion g µ i i as usual, Rg ≡ k/mu is the gas constant per atomic mass unit. We can also introduce a mean ionic molecular weight !−1 X Xi µ = , (9) ion µ i i to obtain Rg Pion = ρT. (10) µion With the ions dealt with, let’s now examine the electrons. As we will discuss further on, there are cases that arise, even during normal stellar evolution, where the electrons cannot be treated as an ideal gas. The departures from ideal behavior are a consequence of degeneracy, and typically occur in the deeper parts of low-mass stars, where densities become very large. For the moment, however, let’s consider cases where the density is not so high, and we can treat the electrons as an ideal gas. Then, the equation of state governing them is Pe = nekT, (11) where ne is the electron number density. To calculate ne, we need information regarding the ionization state of the gas, again a topic that we will examine in detail further on. For the moment, let’s consider the two limiting cases of ionization. If the stellar material is so cold that none of the atoms are ionized, then there are no free electrons; in this limit, ne is zero, and likewise Pe. In the opposite limit, where all atoms are completely ionized, then element i will contribute Zi free electrons per nucleus, where Zi is the element’s atomic number. Thus, we have X ne = Zini, (12) i 1Recall, when an astronomer speaks of the molecular weight of something, they are referring to the mass in units of the atomic mass unit mu. 2 which by use of eqn. (4) can be written X ZiXi n = ρ. (13) e m i i Substituting this back into the equation of state, we find that Rg Pe = ρT, (14) µe where we have introduced the mean electronic molecular weight as !−1 X ZiXi µ = . (15) e µ i i Note that this is not the mass (in mu) per electron in the gas; it is the mass of a given sample divided by the number of electrons in the sample, so it can be thought of as the amount of ionic mass ‘belonging’ to each electron. It’s often useful to combine the individual equations of state for ions and electrons, to obtain a single equation describing the total gas pressure Pgas. Assuming as we do above that the electrons can be treated as an ideal gas, then we write the gas equation of state as R P = P + P = g ρT, (16) gas ion e µ where µ is the mean molecular weight of ions and electrons together. In the neutral (non-ionized) limit, µ is just equal to the mean ionic molecular weight µion given by eqn (9). In the opposite, fully ionized limit, we have " #−1 X Xi(1 + Zi) µ = , (17) µ i i which includes contributions from ions and electrons. To demonstrate how this expression can be applied, let’s consider an ionized gas comprising hydrogen, helium and metals, with mass fractions X, Y and Z2. These elements have individual molecular weights µi = 1, 4 and ∼ 2Zi, and we therefore find that 3 1 −1 µ = 2X + Y + Z , (18) 4 2 where we have approximated (1 + Zi)/Zi ≈ 1. This expression is a well-known result that crops up quite frequently. Another useful expression in the limit of full ionization is the mean electronic molecular weight, which from eqn. (15) is 1 1 −1 µ = X + Y + Z . (19) e 2 2 2 Don’t confuse the metal mass fraction Z with the atomic number Zi of element i. 3 Using the closure relation X + Y + Z = 1, this may also be written as 2 µ = . (20) e 1 + X Now let’s look at the contribution to the total pressure P from the radiation pressure Prad. Deep in the interior, the radiation field is very close to that of a black body, and Prad therefore depends only on the temperature, a P = T 4. (21) rad 3 Combining this with the equation of state for the gas (16), the full equation of state becomes R a P = P + P = g ρT + T 4. (22) gas rad µ 3 Often, this is just written in the form 1 R P = g ρT, (23) β µ where P β ≡ gas (24) P is the ratio between gas and total pressure. The limit β → 1 corresponds to zero radiation pressure, and β → 0 to zero gas pressure. 4.
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