1. Nuclear Astrophysics 1 Lecture 2 Thurs. Oct. 27, 2011 Prof. Shawn Bishop, Office 2013, Ex. 12437 www.nucastro.ph.tum.de

2. THERMODYNAMIC PROPERTIES: EQUATIONS OF STATE 2

3. On the Road to Pressure Integral Momentum transfer imparted to surface dS: 3

4. Flux Volume Volume of parallelepiped Particles within this volume, moving in direction of p, can strike the bottom surface dS in a unit time 4

5. Pressure Integral Number of all particles with momentum between p and p + dp,  and  + d ,  and  + d  striking dS will be: Number per unit vol with momentum p Flux Volume Fraction of solid angle subtended by dS XX Total hitting dS 5

6. Homework: Ideal gas: Use And its derivatives, to show the pressure integral gives in the well known result: P = nkT, where n is the number of particles per unit volume. 6

7. Electron in a box: Length L Schroedinger Wave Equation: Fill the box with N electrons. Each level can take 2 electrons (Pauli). Electrons fill box from lowest energy level up to a maximum energy level we call Only positive integers n x,y,z are valid. States in the positive octant (1/8) of sphere in n x, n y, n z coordinates contribute. Volume of this sphere:, is the radius of sphere at the Fermi Level with energy. N can now be related to the number of levels 7

8. Energy of Electron Gas in Box The pressure of this electron gas is then given by (and ) Where is the electron number concentration. Pressure NOT dependent on temperature, but only on the electron volume concentration! Degenerate electron gas does NOT behave like an ideal gas. 8

9. Relativistic Degenerate Electron Gas 9 We previously had on page 7 for kinetic energy of electron in orbital “n”: Let us equate this to and get: For the pressure integral, we need We also had, from page 7, for the total number of electrons in the box: These two equations give us p as a function of N, by substituting for n: Solve for N, and take dN/dp. Result is:

10. 10 For relativistic electrons, and the pressure integral becomes: From previous page, we have all we need to get the Fermi momentum, Finally, we have the relativistic degenerate electron pressure:

11. THE PATH TO RADIATION PRESSURE 11

12. Photon Gas in Box: Length L Maxell’s Equation for the Electric field: With E y and E z given by similar expressions by cyclic shifting of the cosine in the right (  ) direction Substitution of these back into Maxell’s Eqn yields: Mode Energy: 12

13. 13 Total energy of the photon gas: Each mode will have some average number of photons in it. Call this number. We need to eventually find a formula for Photon pressure: Thermodynamic Partition Function Z for a mode (review your thermodynamics) Energy in a mode:

14. 14 Total Energy: Remember from page 12, that n is a triplet of integers: n x, n y, n z. We replace the sum by an integral over dn x, dn y, dn z, and change to spherical coordinates: And we recall from page 12 that 2 polarizations Change variable

15. 15 Pressure (at last!) was found, on page 13, to be U/3V:

16. Hydrostatic Equilibrium 16 The gravitational body forces integrated over the volume V must be balanced by the pressure acting on the total surface area A. Gauss’ Law (Divergence Theorem): Use, where v is a constant vector (has no (x,y,z) dependence) and also use the following identity: Now Div. Thm: But v was chosen to be a constant (non-zero) field. So, term in ( ) is zero. Use it on RHS in force balance equation (*). = 0 (*)

17. 17 Force balance: (From last identity on previous page) This result must hold throughout the entire volume V, and since volume shape is arbitrary, we therefore have: In spherical geometry, using, where M is the enclosed mass inside radius r, we finally have:

18. Summary of Results so far. 18

19. Lower Bound for Central Pressure 19 Enclosed Mass within radius r: Derivative form: We can now write: Consider the function: At the star’s surface and, so Consider now: Always negative Therefore,

20. 20 The function is monotonically decreasing from the star’s center to the surface. Therefore, At the center and (convince yourself of this with homework problem below) Homework: Prove by two applications of L’Hosptial’s Rule to Finally: For our Sun: For any star: