Physics 576 Stellar Astrophysics Prof. James Buckley. Lecture 13 Thermodynamics of QM particles

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1 Physics 576 Stellar Astrophysics Prof. James Buckley Lecture 13 Thermodynamics of QM particles

2 Reading/Homework Assignment Read chapter 3 in Rose. Midterm Exam, April 5 (take home) Final Project, May 4 (end of reading period)

3 Virial Theorem We can combine the equations of hydrostatic equilibrium and mass continuity to obtain dp (r) dr dm(r) dr = dp ρgm(r) dm = r 2 4πr 2 ρ 4πr 3 dp = GM(r) r Integrating over the (spherically symmetric) star dm and defining V (r) = 4 3 πr3 as the volume contained within radius r 3 Ps P c V (r)dp = Ms o GM(r) r dm Integrating by parts, i.e. using d(p V ) = P dv + V dp we have 3[P V ] s c 3 Vs V c P dv = Ms 0 GM(r) r dm where P s = 0 (really it is just small compared with P c ), V c = 0 and V s = 4 3 πr3 Vs Ms GM Virial Theorem : 3 P dv 0 0 r dm thermal energy Ω, gravitational potential energy =0

4 Virial Theorem Say that we have a bound (or bounded) system of particles, we can show that in equilibrium, there is a simple relationship between the kinetic and potential energy. We define a new quantity called the virial G dg dt = dp i dt r i i P i F i r i p i r i i=1 + i p i dr i dt P i m ivi 2=2T For a bound system, we can not have G increasing without bound, so dg/dt, on average, must be zero or G would increase without bound and particles would escape the system. So, if we take the time average over both sides of the equation we have dg =0= F i r i +2T dt i

5 Virial Theorem ow let s consider the first term on the RHS, i F i r i. We start by realizing that in a closed system, the force on any particle is just the sum of the forces exerted by every other particle (excluding this one) or F k = = F k r k = k=1 k=1 j=1,j<k j=1,j=k F jk r k + F jk k=1 j=1,j=k F jk r k From ewton s third law of motion, for every action there is an equal and opposite reaction F jk = F kj k=1 j=1,j>k F jk r k Substitute this into the second term on the RHS, then interchange j and k labels F k r k = F jk (r k r j ) k=1 k=1 j=1,j<k

6 Virial Theorem For problems where the force can be written as the gradient of a potential (as for gravity, the only case we are really interested in) we can write the force on particle k due to particle j in terms of the distance between the two particles r jk r k r j F jk = rk V = dv dr rk r j r jk F k r k = k=1 k=1 j=1,j<k F jk (r k r j )= k=1 j<k dv dr (r k r j ) 2 r jk = k=1 j<k dv dr r jk For astrophysics we are typically only interested in 1/r 2 gravitational forces, but just for fun let s generalize the result to any power-law force V (r) =ar n F k r k = k=1 k=1 j<k dv dr r jk = k=1 j<k nar n 1 jk r jk = nv (r jk )=nv tot k=1 j<k

7 Virial Theorem dg dt =0= F i r i +2T i and substituting for the first term on the RHS for the case of a gravitational potential we obtain V = 2T (Virial Theorem for Gravity)

8 Virial Theorem 3 Vs 0 P dv = E g On the left hand side, we are just summing up P (r) V for each little shell of the star (of volume V ). ow, let s relate this to the thermal energy for an ideal gas: P gas V = kt U gas = 3 2 kt so P gas V = 2 3 U gas For an ideal gas, the virial theorem states 3 Vs 0 P dv =3 2 3 U gas = E g E g V = 2U gas 2T So we see that for a gas (or more rigorously, a nonrelativistic gas) the virial theorem as expressed above reduces to the standard form from classical mechanics.

9 Quantum Statistics We can proceed in a similar fashion to describe the distribution function for quantum mechanical particles. In this case, we need to take into account two important differences: QM particles are indistinguishable QM particles come in two varieties, bosons and fermions In either case the procedure is to: 1. Calculate the number of microstates Ω corresponding to a given macrostate (i.e., a given distribution of particles in momentum) be doing some combinatorics. 2. Maximize the entropy ln Ω subject to constraints to find the distribution function

10 Quantum Statistics de Broglie said if a wave can be like a particle, then a particle can be like a wave Photons : E = ω, p= E c = ω c Light also behaves like a plane wave: E(r,t)=E o e i(k r ωt) or more generally as a wave packet E(r,t)= n c n e i(k n r ω n t) and the probability of finding a photon is P E(r,t) 2 ow, we would like (as a lawyer like de Broglie) for particles to have equal rights, and to behave as wave packets: ψ(r,t)= c n e i(k n x ω n t) n and, using the relationships for light as a guide, we have ω E, k p ψ(r,t)= n c n e i p r i E t

11 Quantum Mechanics E(r,t)= i c i e i(k i r ωt) We see that a spatial derivative / x or more generally has the effect of bringing down a factor of ik and a time derivative of bringing down a factor of iω ik t iω and if we didn t already have a wave equation, we could build one by starting with k 2 (ω/c) 2 = 0 then substituting the operators k i, ω iω Well, for particles, no one had a wave equation so some smart fellows (Schrödinger, Dirac, etc.) decided to reverse engineer some equations in the same way. They noticed that starting with de Broglie s wave function, spatial and time derivatives had the effect of bringing down the momentum and energy ψ(r,t)= i c i e i p r i E t i p t ie but to mystify everyone, they turned this around and said we start with the Hamiltonian and write moment and energy as operators p i E i t

12 on-relativistic Quantum Statistics Then substituting into the classical hamiltonian: H = p2 2m + V (r) =E one obtains Schrödinger s equation 2 2m 2 ψ(r,t)+v (r)ψ(r,t)=i t ψ(r,t) then multiplying Schrödinger s equation by ψ and then multiplying the complex conjugate of S.E. by ψ and taking the difference one obtains a nice equation for conservation of probability density of the form P (r,t) t + S =0 where the probability density is P (r,t) ψ ψ and the particle current is S(r,t)= 2im [ψ ψ (ψ )ψ] now, of course, that was not confusing enough and people made up commutation relations like [p, x] = i which gan be seen to follow trivially from the association of momentum with the spatial derivative operator when applied to a wavefunction ψ

13 QM Density of States To derive the density of states for a nonrelativistic, quantum system start with Schrödinger s equation: 2 2m 2 ψ(r,t)+v (r)ψ(r,t)=i t ψ(r,t) for a cubic region with sides of length L and V = 0 inside, the solutions of the S.E. have the form of standing waves: ψ n (x, y, z, t) =C sin k x x sin k y y sin k z ze ie nt/ where the wavefunction vanishes at the boundaries if k x = n x π/l, etc. giving ψ n (x, y, z, t) =C sin n xπx L sin n yπy L sin n zπz L e ie nt/ with n x, n y, n z positive integers to avoid over-counting states of the same energy substituting into S.E. E= h2 8mL 2 (n2 x + n 2 y + n 2 z) E = 2 2m (k2 x + ky 2 + kz)= 2 2 k 2 2m ormalizing the wave function so the total probability ψ ψ inside the box is one, we get C = 8L 3/2

14 Classical Density of States for Photons Consider standing electromagnetic waves in a cubic cavity of dimensions L L L E = cos(k x x) cos(k y y) cos(k z z) cos(ωt)ê x z da =4πk 2 dk k!!!!!""# k L k y =2π/L da = 4 k dk dk " 2 y to satisfy the wave equation k 2 = k 2 x + k 2 y + k 2 z = ω2 c 2 k x = n xπ L,k y = n yπ L,k z = n zπ L where n x,n y,n z are odd integers d =2 4πk2 dk (2π/L) 3 ρ(ν) = d dν dν = 4πν2 dν c 3 Using E = hν the density of states becomes ρ(e) = 4π h 3 c 3 E2 de ote that in this case we had a linear relationship between E, ν, k and n, while for QM particles we have a quadratic relationship between n and E

15 R Quantum Density of States For nonrelativistic particles, we can define the wave vector k according to the de Broglie prescription: z da =4πk 2 dk da = 4 k dk " 2 dk k y k p = 1 2mE d = 4πk2 dk (2π/L) 3 = 4π 2mE (2π/L) 3 2 2m 1 2 E 1/2 de x k!!!!!""# k y = 2π L L d de = 4πL3 h 3 m 2mEdE Assuming that the nonrelativistic particles are spin 1/2 fermions, we need to multiply by 2 for the number of spin states and we can write the density of states g(e)de (the number of quantum states per unit volume in the energy interval (E,E + de) as g(e)de = 1 d L 3 de = 8π h 3 m 2mEdE = 2d3 p h 3