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Unformatted text preview: Looking at Multinomial Distributions... N! f ( n1 ) = Can we plot f(n j ) vs n j? n1 !(N - n1 !) ln [ f (n )] is a monotonic function max [ f (n )] = max [ ln f (n )] (for small N, see HW #1) For large N, it is hardly possible, so... use ln f (n ) instead gln ( n1 ) f = 0 will yield the max of f (n ) 1 n1
F or very large values of n, we can use the Stirling approximation ln N ! N x ln N - N N! ln [ f (n1 )] = ln = ln ( N !) - ln ( n1 !) - ln ( ( N - n1 ) !) n 1 !(N - n1 !) ln [ f (n1 )] = N g ( N ) - N - n1 g ( n1 ) + n1 - ( N - n1 ) g ( N - n1 ) + ( N - n1 ) ln ln ln g ln ( n1 ) l n g ( N ) - n1 gn ( n1 ) - ( N - n1 ) gn ( N - n1 ) f N l l = ln n1 n1 1 1 = - ln(n1 ) - n1 + 1 ( N - n1 ) + ( N - n1 ) ln n1 ( N - n1 ) = - ln(n1 ) + ln ( N - n1 ) g ln ( n1 ) f = 0 n1 1 for N - n1 =1 n1 N n = this is the n1 for which f (n1 ) maximizes 2 for the shape of the distribution, we can always use a Taylor expansion around the max of the function ln f + n1 ln f (n1 ) = ln f n1 n1 ( 1 n1 - n1 + 2 ) 2 ln f n12 n1 (n -n )
1 2 1 +K N 1 2 1 + ln f (n1 ) ln + 0 + ( - 1) f 2 2 N N-N 2 ( ) 2 n - N 1 2 2 N N ln f (n1 ) g ln - 1 - f N n 2 g 2 g 2 g N f (n1 ) g f e 2 2 N - 1- n N 2 2 shape of the distribution N f (n1 ) f e g 2 - 2 n1 - N 2 N ( ) 2 this is a gaussian distribution
f(n1) 1 with standard deviatio n = N 4 to =1/4 N1/2 for N j (10 ) 23 N/2 n1 it is a gaussian, centered around 1023 w ith a f(n ) 1 23 11.5 wid th j 10 = 0.25 10 4 1023 0.25 1011.5 0.5
1 N/2 n1 Without the constraint, df ( x,1, x2 ,g= xr )== g j So far, we look at a function with a single variable n1. gf x g j dx j max f min 0 finds the max of f ( x,1, x2 , xr ) Since there are no constraints on x j , dx j can have any value gf x g j = 0 for every dx j max f min
1 2 3 How to maximize a function of multiple variables, when the variables are not independent? (n n n etc) i.e maximize the Multinomial distribution with f ( x,1,x 2 ,K xr ) with gx
j j =N Since there is a constraint, =0 max f min n = const.
j gf the dx j are not independent x g j ? Lagrange Undetermined Multipliers co nstrain t: g( x,1, x2 ,g xr ) = 0 and
j dg ( x,1, x2 ,C= xr )== g Cg dx j gx j 0 finds the max of g( x,1, x2 , xr ) gf g dx - j = 0 undetermined multiplier gj gx xj j for a particular term with j = m, dxm is not independent, gf x g g m gmax f
min and we define g gg x g max m f
min g this term is therefore cancelled in the g ( ) dx
j j for all other terms, dx j are independet f g - =0 xj xj for j m For j = 1,2 k, we have k equations of the form f g - =0 xj xj When several constraints are present, g1( x1, x2 ,K xk ) g2 ( x1, x2 ,K xk ) K gn ( x1, x2 ,K xk ) f g g g - 1 - 2 - K - n =0 xj xj xj xj Some "old" Thermo, to remember our roots... Extensive properties depend on the system size (V, S, mass, etc) Intensive properties do not depend on syst. Size (T, p, etc) Isolated systems evolve SPONTANEOUSLY toward simple "terminal states" (equilibrium). E1 E2 1st Law: Internal energy is extensive E1+ E2 =E E is conserved ( changes are the result of doing something to the system (dE = dQ+dw) X = extensive mechan. properties. Change in X involves work Adiabatic walls control exchange of heat Equilibrium is macroscopically characterized by (E,X) C a small number of control variables Equilibrium is reached (kept) by constraints. What terminal state will be reached if a constraint is removed or changed? Entropy: Extensive property, S(E,X) 2nd Law: If state B can be adiabatically reached from state A, then SBCSA A process which can be exactly retraced by infinitesimal changes in its control variablesCC reversible arbitrarily slowly moving from one equilibrium state to another C trajectory within the manifold of equilibrium states Natural processes do not proceed through the manifold of equilibrium states C cannot be retraced C irreversible If ACB is adiabatic SBCSA and if it is also reversible, then BCA is also adiabatic C SACSB C SA = SB for adiabatically reversible processes If ACB is irreversible, BCA cannot be reached adiabatically then SB>SA S C0 for adiabatic process S(E,X) C S S dS = dE + d X g X E X E S S S g S = dE + dX1 + dX 2 + K + dX n g X E X1 X2 Xn E E E for a reversible process, the system is always @ equilibrium (dS = 0), and, forceext = forcesyst dQ = dQrev dE = dQrev + f d X f S S S 0 = dQrev + f d X + d X g X E E X X E dQ = 0 f S S Adiabatic and Reversible = - f g E X E X which also holds for reversible non-adiabatic processes, because S, E and X are functions of state for an adiabatic process, Postulate: Entropy is a monotonically increasing f n of E a S > 0 g X E intensive extensive E from here, we can define T g 0 g X S S g S and therefore dS = dE + f X becomes - d g X E E X 1 1 dS = dE - f dX T T E = E (S, X ) dE = T dS + f d X ...
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This note was uploaded on 05/29/2011 for the course CHM 6461 taught by Professor Bowers during the Spring '08 term at University of Florida.
- Spring '08