Physics 556 Stellar Astrophysics Prof. James Buckley. Lecture 5

Transcription

1 Physics 556 Stellar Astrophysics Prof. James Buckley Lecture 5

2 Thermodyamics

3 Equatio of State of Radiatio The mometum flux ormal to a surface (mometum per uit area per uit time) is the same as the ormal force per uit area or pressure. If oe cosiders a reflectig eclosure cotaiig a isotropic black-body radiatio field, the photo actually trasfers twice its mometum with each collisio, but oly hits the wall for half of the total solid agle. Averagig over agles ad frequecy oe obtais the pressure. p ν = 2 I ν cos 2 θ dω check uits: c half p = 2 I ν dν cos 2 θ dω c half [dye cm 2 Hz 1 ] π/2 cos 2 θ dω =2π cos 2 θ si θ dθ = 2π 3 cos3 θ p = 4π I ν dν 3c half u = dν du ν (Ω) = I ν c dω I ν c = 4π I ν dν = 4π c c B(T ) p = 1 3 u = 1 3 at 4 [erg cm 2 s 1 Hz 1 ]/[cm s 1 ]= [dye cm cm 2 s 1 Hz 1 ]/[cm s 1 ]= π/2 = 2π 3

4 Thermodyamics First law of thermodyamics: Coservatio of eergy, but says othig about the cocept that eergy ca either go ito useful forms (potetial eergy or a sprig, chemical bod, etc) or ito dissipatio ito a radom state from which it is difficult to extract mechaical work. Carot seior oted that there is also a dissipatio of eergy ito radom motios, heat etc. ot useful eergy Carot juior oted that wheever heat eergy falls through a temperature differece, useful work ca be extracted - otherwise this is ot possible. This is basically a first statemet of the secod law of thermodyamics. Rudolf Clausius Clausius described etropy as the trasformatio-cotet or dissipative eergy use of a thermodyamic system or workig body (chemical species) durig a chage of state (e.g., give by state variables such as temperature or pressure). TdS = dq = du + P dv ad total etropy always icreases

5 Distributio Fuctios Defie the particle distributio fuctio f(p) as the umber of particles per uit volume per mometum iterval d 3 p = f(p)d 3 p The mea value of a physical variable x is the give by x = xf(p)d 3 p First we defie Ω to be the umber of microstates correspodig to a thermodyamic macrostate where M etities are partitioe with 1 i the first state, 2 i the secod etc. M! Ω = 1! 2!,... The thermodyamic etropy of M particles is defied to be S M = k l Ω Stirlig s approximatio for >> 1 : l! l + O(l ) S M k M l M M i ( i l i i ) substitutig M = i i S M = km i i M l i M + cost

6 Boltzma Etropy Defiig the probability of fidig a particle i state i: p i i /M S M = km i i l + cost = km p i l p i + cost M M i i the defie the distributio fuctio (or phase space desity) as f i = p i 3 p 3 x l f i = l p i l 3 p 3 x = l p i cost probability per volume of phase space withi a iterval of 3 p of p i ad { withi a iterval of 3 x of x i Etropy per particle s p = S M /M = k i p i l p i = k i f i l f i 3 p 3 x + k p i cost i cost P i p i=cost Now, 3 x could be cosidered a macroscopic volume (if we are ot iterested i the desity i space, just i mometum). I that case 3 x/m =1/ where is the umber desity of particles. Now, takig the limit as 3 p the sum becomes a itegral s = k f l fd 3 p

7 Review of Variatioal Methods Recall the priciple of least actio from mechaics: S L(x, ẋ)dt where L(x, ẋ) =T V ad typically L(x, ẋ) = 1 2 mẋ2 V (x) L L δs = δx + x ẋ δẋ dt = L ẋ δẋ = L t ẋ δx L δx t ẋ b L a t ẋ δx dt = L b ẋ δx = must vaish o edpoits a L δs = x L δxdt= t ẋ = (Euler Lagrage eq.) Here we cosider the Lagragia for a sigle geeralized coordiate x. Costraits o the trajectories (ofte betwee the coordiates) ca be write i the the form g(x, y; t) = L x L t ẋ λ 1(t) g x = L } if t ẋ = mẍ L y L t ẏ λ 2(t) g y = F = ma λ(t) g/ x looks like a force of costrait.

8 Variatioal Methods It is useful to view the problem of fidig a extremum with costraits as equivalet to the problem of fidig the extremum of a ew fuctio, or modified actio. I the case of classical mechaics the modified actio ca be writte S = L(x, ẋ)dt S = [L(x, ẋ) λ(t)g(x)] dt Of course the same method (Lagrage multipliers) is applicable to ay optimizatio problem. Cosider the case where we wat to fid the extremum of a fuctio F (x) where x {x 1,..., x subject to the costraits G 1 (x) =,..., G m (x) = the the miimum ca be foud by simulaeously solvig the costrait equatios ad the equatio F x = λ G 1 1 x λ G m m x where F x F,... ad this is just equivalet to miimizig a ew fuctio K(x) F (x) λ 1 G 1 (x)... λ m G m (x)

9 Maximizig Etropy Startig with this expressio for the the Boltzma etropy s = k f l fd 3 p with costraits o the total umber (volume desity) of particles ad total eergy fd 3 p = Fidig the extremum of etropy without costraits would give: δs = k δ(f l f)d 3 p = fed 3 p = u subject to costraits, we use a modified fuctio with two Lagrage multipliers λ 1 ad λ 2 k δs = δ (f l f) λ 1(f cost) λ 2 (fe cost) d 3 p = or we combie some of the costats to defie two ew lagrage multipliers α ad β allowig us to rewrite the variatioal problem as δ [f l f + α(f cost) + β(fe cost)] d 3 p =

10 Maxwell-Boltzma Distributio δ [f l f + α(f cost) + β(fe cost)] d 3 p = f δff + l fδf + αδf + βeδf d 3 p = [l f +1+α + βe] δf d 3 p = i order for this to be true for a arbitrary δf we have l f +1+α + βe = f = e (1+α) e βe or f = Ce βe If we like, we ca actually view the temperature T as a ew quatity that is iversely proportioal to the Lagrage multiplier for eergy coservatio β, ad we will see that as we work through all of the statistical mechaics this ew quatity (a glorified Lagrage multiplier) is just what we always meat by the thermodyamic temperature! β 1 kt f = Ce E/kT

11 Maxwell-Boltzma Distributio f = Ce E/kT Normalizig : f(p)4πp 2 dp = Usig Feyma s favorite trick or differetiatig uder the itegral 4π Ce p2 /2mkT p 2 dp = 4πC e αp2 dp where α 1/2mkT α so we are left with the problem of solvig the itegral I 2 e x2 dx = 1 2 e x2 dx I 2 = e x2 dx e y2 dy = e (x2 +y 2) dxdy =2π e r2 rdr = π e αp2 dp = 1 α e x2 dx = 1 2 π α 4πC α 1 2 π α = C = π 3/2 α 3/2 = (2πmkT ) 3/2

12 Maxwell-Boltzma Distributio 1 Maxwell Boltzma Distributio : f(p) = 2πmkT 3/2 e p2 /2mkT The more familiar expressio gives this i terms of velocity. f(p) is really d/d 3 p so we ca write f(p)d 3 p = d = f(v)d 3 v f(p)d 3 p = f(p)4πp 2 dp = f(p)4πm 3 v 2 dv = f(v)4πv 2 dv m Maxwell Boltzma Distributio : f(v) = 2πkT 3/2 e mv 2 /2kT

13 Ideal Gas Law (Derived!) da cos θ We ca derive the ideal gas law ad the equatio of the thermal eergy desity of the gas from the Maxwell-Boltma distributio (ad hece from the priciple of maximizatio of etropy). We start by writig a expressio for the chage i mometum of a particle p icidet o a surface (or aother particle) at agle θ p =2mv cos θ θ vdt da P = 4πm 3 df = 2mv cos θ/dt dp/dt per particle m 2πkT P =4π π/2 f(v)dωv 2 dv desity i d 3 v dp = df da P = kt v cos θdadt vol swept out mv 4 f(v)dv cos 2 θ si θdθ P =4π cos3 m 3/2 mv 4 e mv2 /2kT dv 3 2πkT 3/2 5/2 2kT e x2 x 4 dx = kt m e x2 x 4 dx = 8 3 π kt Q.E.D. similarly, we ca show that U = 3 2kT, agai startig with the priciple of maximum etropy ad costraits o the eergy ad umber of particles i a orelativistic, classical system.

14 Equatio of State of Gas For a classical gas (i a star), we start with the familiar ideal gas law PV = kt, or takig ito accout both electros ad uclei i the total umber of degrees of freedom Ideal gas law P = kt = e + uclei,i i kt Avogadro s umber, N ca be combied with atomic mass umber of the uclei A i to give the total umber of uclei of type i per gram = N /A i =( /A i ) atoms/gram i = ρ N X i A uclei,i i i where X i is defied as the mass fractio of uclei of type i, ad ρ is the total mass desity of the gas Similarly, the umber of electros ca be calculated from the umber of uclei ad the ioizatio state of each ucleus (e.g., Z i for fully ioized uclei) combiig the two terms e = ρ i X i N Z i A i = ρ N X i (1 + Z i ) A i i ad fially P gas = ρkt µm u where 1/µ /ρn ad m u is equal to 1/12 the mass of Carbo 12

15 Equatios of State For a ideal gas, there are three traslatioal degrees of freedom, each with 1 2 kt ad U gas = 3 2 NkT Summarizig: For Radiatio: u P r = 1 3 at 4 r = at 4 For a ideal gas: U gas = 3 2 NkT P gas = kt where = ρ i X i N A i (1 + Z i )

16 Specific Heat for a Ideal Gas First law of thermodyamics: dq heat added = du + p dv iteral eergy work TdS = dq = du + P dv Specific Heat at costat volume: C v = C v = δq dt V δq dt = V U T = T V S T = T V 3 2 NkT C v = 3 2 Nk Heat capacity (at costat volume) per particle: c v = 3 2 k

17 Gas Etropy TdS = dq = du + P dv Divide both sides of the equatio by N: S Td = du N N + P dv N For a ideal gas du = 3 2 NkdT = C v dt = c v NdT = N/V, d(1/) = d(v/n)= dv/n S 1 Td = c v dt + Pd N defiig the etopy per particle s p S/N: 1 T ds p = c v dt + Pd substitutig P = kt ad dividig by T ds p = 3 2 k dt T kd T 3/2 = kd l T 3/2 s gas = s p = k l + costat