Problem Set 4 Solutions 3.20 MIT Dr. Anton Van Der Ven Fall 2002

Transcription

1 Problem Set 4 Solutions 3.20 MIT Dr. Anton Van Der Ven Fall 2002 Problem -22 For this problem we need the formula given in class (Mcuarie -33) for the energy states of a particle in an three-dimensional infinite well, namely Now we can make a table nxn yn z h 2 8ma 2 n x 2 n y 2 n z 2, n x,n y,n z,2... n x n y n z n 2 x n 2 2 y n z Degeneracy Problem -29

2 Start with the differential form for E de TdS pdv E T S T T p We can use a Maxwell relation on S T from FT,V S T p T V So, E T p T T V p Problem -4 Show that x x 2 x 2 x 2 x x 2 x 2 2x x x 2 x 2 2x x x 2 x 2 2 x x x 2 x 2 x 2

3 Problem -43 Here we have to plot the Gaussian px 2 to see what happens as 0 x x 2 exp 2 2 for several values of As 0, the function becomes sharper and sharper (remember the area under the curve is contrained to be, as we will see in problem -44a). Thus, the Gaussian approaches a delta function. Problem -44 Gaussian distribution is px 2 exp x x (a) show pxdx 2 exp x x dx Let

4 u x x 2, then du dx 2 So we now have expu2 du And using standard integral tables or the math software, it can be show that this integral is equal to.

5 (b) n th central moment for n 0,,2, and 3 For n 0 x x 0 see part a For n x x x x pxdx x x 2 x x 2 exp 2 2 dx Let u x x 2, then du dx 2 Then x x u expu 2 du 0 u expu 2 du I u expu 2 du 0 II For I let v u and dv du. Then I 0 u expu 2 du 0 v expv 2 dv II Thus, u expu 2 du 0 For n 2 x x 2 x x pxdx x x 2 2 x x 2 exp 2 2 dx Let

6 u x x 2, then du dx 2 x x u 2 expu 2 du 22 u 2 expu 2 du Now lets take a crack at this integral... u 2 expu 2 du We have to do this by parts. Remembering how to do that... yx y x yx y x yx yx For our case, let y 2uexpu 2 du and x u 2 y expu 2 and x 2

7 So, u 2 expu 2 du u expu expu Which leaves us with, x x 2 2 For n 3 x x 3 x x 3 pxdx x x 3 2 x x 2 exp 2 2 dx As before, let u x x 2, then du dx 2 Then, x x u 3 expu 2 Since e u2 is a symmetric function around the origin and u 3 is an antisymmetric function around the origin, the integral of the product of the two functions is zero. x x 3 0 (c) limpx lim x x 2 exp 2 2 x x, the delta function The delta function is defined as

8 x a xdx a and xdx So lets see if this works for our case lim 0 2 exp x a2 2 2 xdx Let u xa 2 and du dx 2 lim expu 2 2 a 0 Since 0 expu 2 a a expu 2 a Problem -49 Maximize WN,N 2,...N m N! m N! under the contraints that N N and E N. Since the natural log is a monotonic function, we can maximize lnw. M lnwn,n 2,...N m ln N! m N! Using Sterling approximation, this can be written as

9 NlnN N N lnn N But the maximization must be constrained therefore we intoduce Lagrange multipliers M NlnN N N lnn N N N E N N Remembering N N and taking the derivative of M with respect to N M N lnn E 0 lnn E lnn E, with N exp expe Problem -50 We want to show that the maximum of WN,N 2,...N m N! occurs for N N 2 N s N s Maximize m N! M lnw NlnN N N lnn N subect to the constraint N N. So we must use Lagrange multipliers: M NlnN N N lnn N N N

10 M lnn 0 N lnn N exp But now we must determine s s N exp Sexp N exp N s So N N s

11 Problem -5 Here we use Lagrange multipliers again with the constraint P. N M P lnp N P Maximize M P lnp 0 P exp Determine N P Nexp exp N Thus P N Problem 2- From stat mech we get E p N, p And from thermo we have

12 E T p N,T T p To show why cont T lets see what happens if we let T where is a constant E N, T p T p or E T p N,T T p but from a Maxwell relation we know p T S N,T and we can write E T S N,T de TdS PdV N,T p This statement violates the second law of thermodynamics because it implies that TdS ds T

13 Problem 2-2 Given n n! n!n n! is n n? n is the value of n that maximizes n. From problem -50, we know that n n. 2 n is given by n n n 0 n n 0 n n! n!nn! n! n!nn! 2-2- We also know that (given in the problem): n y x n x n n! n n 0!n n! If we take the derivative of y with respect to x we get n y n x n n x n n! n!n n! n 0 If we let x : n n2 n n n! n!n n! n 0 Furthermore,

14 n 2 n n 0 n! n!n n! So if we put this all together back in (2-2-) we get n n n 0 n n 0 n n! n!nn! n! n!nn! n2n 2 n n 2 n n n Problem 2-5 Show that S k P lnp We have the following three relations already and S ln T P exp exp E E kln Starting with S we can write S T kln E exp E 2 kln S k E exp E kln S k ln exp E exp E kln

15 S k ln exp E exp E kln P Since P S k P ln exp S k P ln exp E E kln P k P ln S k P ln exp E S k P lnp Problem 2-8 E E expe E expe ln E ln Problem 2-0 First derive E 2 ln T starting from the result of problem 2-8, namely E ln

16 From the chain rule E ln T T We know T so 2, thus E 2 ln T We can check this by taking 2 ln T. 2 ln T 2 T 2 T exp E 2 T exp E E exp E 2 The 2 cancels and we are left with 2 ln T E exp E definition of E So we have shown again that indeed E 2 ln T. Now for the relation involving p p p exp E E exp E exp E N,T p ln N,T

17 Problem 2- Starting with F ln (Remember F A S F T V,N S ln kln Eqn T V,N p F T,N p ln T,N Eqn F E TS E F TS E ln T ln kln T V,N E 2 ln T V,N Eqn. 2-3 Problem 2-3 E V. We can start For a particle confined to a cube of length a we are asked to show p 2 3 with the equation for the energy states of a particle in an three-dimensional infinite well, namely E h 2 8ma 2 n x 2 n y 2 n z 2, n x,n y,n z,2... Remembering that a 3 V or a V 3 we can write E in terms of V E h2 V 2 3 8m n x 2 n y 2 n z 2 p E 2 3 h2 V 5 3 8m n x 2 n 2 y n 2 z 2 3 V h 2 V 2 3 8m n x 2 n 2 y n 2 z E So we have

18 p 2 3 E V and taking the ensemble average gives us p 2 3 E V Problem 2-4,T N! 2m h 2 3N 2 V N For p, p ln ln N! 3N 2 ln p N V T.N ln 2m h 2 NlnV Now for E E 2 ln T 2 3N 2 E 3N 2 2mk h 2 2m h 2 Ideal gas equation of state is obtained when ft N ln lnft NlnV p ln T.N lnft T,N NlnV p N V

19 Problem 2-5 We are given as exp h 2 exp h 3N exp U o substituting in h k exp 2T exp T 3N exp U o To find c v we need to use the following two relations E 2 c v ln T E T Using your favorite math package (Maple), we can get E as E Uo e 2 3Nke Uo 2Nke Uo k 9Nke 2 kuo e T 2 2U o e Uo 4U o e Uo k 2U o e 2 kuo and c v as c v E T 3 T 2 kn 2 e T e T 2 Now we can take the lim T c v andwefindthat

20 lim T 3 T 2 k 2 e T N e T 2 3Nk