Chapter 2 Stellar Structure Equations 1 Mass conservation equation

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2 Momentum conservation and the equation of hydrostatic equilibrium The next equation we introduce comes from the conservation of momentum; or more precisely, in LTE the net force acting on any microscopically large but macroscopically small region is zero. Inside a star in LTE, the forces involve are gravity and pressure gradient due to gas or photon. Specifically, dP dr = −Gm(r)ρ(r) r 2 . (3) We denote the central pressure P (0) by P c . And clearly, P (R) can be approximated by 0. Eq. (3) can be rewritten as dP dm = − Gm 4πr 4 (m) . (4) Hence, Since r < R inside the star, we have P (R) − P (0) = < − ∫ M 0 ∫ M 0 dP dm dm Gm 4πR 4 dm = − GM 2 8πR 4 . (5) P c > GM 2 8πR 4 (6) in any star. In particular, this inequality tells us that the central pressure of our Sun is at least 4.4 × 10 14 dyn/cm 2 ≈ 4 × 10 8 atm. Multiplying Eq. (4) by the volume V (r) = 4πr 3 /3, we have ∫ P (R) ∫ M P (0) V (r)dP = − 1 3 2 0 Gm r dm . (7)
Note that the R.H.S. above is Ω/3 where Ω is the gravitational potential energy of the entire star. Using integration by parts, the L.H.S. above equals − ∫ V (R) P dV = − ∫ M P/ρ dm. Hence, we arrive at 0 0 Ω = −3 ∫ M 0 P ρ dm . (8) This equation is one form of the virial theorem (sometimes known as the global form of the virial theorem). We can use the non-relativistic classical ideal gas to obtain the usual form of virial theorem. The kinetic energy density of an ideal gas ε is simply given by ε = N 3 kT and the pressure of an ideal gas is P= N kT= 2 ε. Substituting V 2 V 3 this relation into above equation we get ∫ M ∫ P V (R) ∫ V (R) 3 0 ρ dm = 3 P dV = 2 εdV = 2U = −Ω (9) 0 Thus, virial theorem tells us that 2U + Ω = 0 where U is the thermal energy of a star making up of non-relativistic classical ideal gas. Consequently, the total energy of the star is U + Ω < 0. In other words, the star is in a bound state. One consequence of the virial theorem is that upon gravitational contraction, the star becomes hotter, more tightly bound and have to radiate some energy to space. In contrast, if the star is made up of extremely relativistic particles, the pressure equals one-third the energy density. Hence, in this case, virial theorem implies that U + Ω = 0. In other words, a star making up of extremely relativistic particles can be in LTE only when its total energy is 0. So, such a star is not bounded. Another consequence of the virial theorem concerns the conversation law. Consider a star making up of non-relativistic classical ideal gas. Suppose further that the timescale concerned is small enough that the total energy of the star is roughly a constant. Since 2U + Ω = 0, the total energy of the star can be expressed as a function of U (or Ω) alone. Thus, the thermal energy and the gravitational potential energy of the star are also conserved. In turns out that this consequence of the virial theorem is useful to understand several properties of late stage stellar evolution. 3 0
Page 1: Chapter 2 Stellar Structure Equatio
Page 5 and 6: ( d 2 P dr 2 )c = − ( dρ Gm + ρ
Page 7 and 8: as adiabatic expansion. With this i
Page 9 and 10: By substituting θ and ξ in the La
Page 11 and 12: ighter component: M = 8.30 × 10 33
Page 13 and 14: • The nuclear burning timescale:
Page 15 and 16: Note that we have assumed that the
Page 17 and 18: where Z s = ∑ i [ ] −(Es,i −
Page 19 and 20: The ion pressure is given by where
Page 21: Note that the first fraction in Eq.

Chapter 2 Stellar Structure Equations 1 Mass conservation equation

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