Statistical Mechanics Solutions: R K Pathria Book

[rgshankar76]

I wish to know about web sites or other resourses from where i can get solutions for all the end of the chapter problems for the book on statistical Mechanics by R K Pathria

 

[dukemaster]

there is' t...

:(

 

[Count Iblis]

Why not post the problems here? Most statistical mechanics textbook problems are trivial...

 

[dukemaster]

i 'm coursing statistical mechanics and i need to do all of the excersices of pathria ...

i have do some of then , chapter 1,2 now.. .. but i can't understand the 1.1 and the 2.1

http://img228.imageshack.us/img228/3023/dibujojd7.png



well ...i found some kind of solution but i don't know it's really good ... www.mtholyoke.edu/~mktrias/physics/pathria_1.1.pdf[/URL]


i hope any help ... thanks

 

[Count Iblis]

In the last line of the PDF file, you got the expression proportional to:

W(x) = [x(E-x)]^M

with M = 3N/2 and x = E1

You should then proceed, not by saying: "this is a binomial distribution etc. etc." but by expanding around the maximum. The maximum is at x = E/2. Let's put x = E/2 + y, and call

W(x) = W(E/2 + y) = P(y)

and expand:

Log[P(y)] = M [Log(E/2 + y) + Log(E/2 - y)] =

M[2 Log(E/2) + Log(1 + 2y/E) + Log(1 - 2y/E)] =

Let's expand in powers of 2y/E


M [2 Log(E/2) - 4 y^2/E^2 + ...]

So, for small y we have:

P(y) = const. Exp(-4 y^2/E^2)

If you are careful and keep the constant terms, the Gaussian should automatically be correctly normalized.

 

[dukemaster]

ok .. thanks a lot.


and if you know any of the exercise 2.1 let me know...

thans .

 

[ALBERTO666]

Hi dukemaster !

I don't write english well, my language is spanish... but that's not important here, the important is the physic. The page that you checked isn't bad, but Margaret Trias solved the problem for the ideal gas case. I think that the book want the solution in the general case. I'll tell you the solution of the fisrt part, you can do the last part knowing the first.

Supose that the microestates number is function of E1 and E2 like: O(E1,E2) , where O is Omega (in spanish). The logarithm Ln(O(E1,E2)) decreases more quickly than the fuction O(E1,E2), then, we'll take an expansion in Taylor series about [E1] (averge value), this is:

Ln(O(E1,E2)) =Ln(O([E1],E2))+(E1-[E1])(d_1)Ln(O([E1],E2))+))+(E1-[E1])^2(d_2)Ln(O([E1],E2))+...

where (d_1) is the first derivate with respect E1 and (d_2) the second derivate with respect the same.

Check that the first term in the expansion is constant, the secod is zero (that's the equilibrium condition) and the third is differente to zero, then:

Ln(O(E1,E2)) =C+(E1-[E1])^2 (d_2)Ln(O([E1],E))

taking the exponetial in both sides:

O(E1,E2)) =Aexp{(E1-[E1])^2 (d_2)Ln(O([E1],E))}

that's is the Gaussian in the parameter E1. If you take the case of ideal classical gas O(E)=cte E^(3N/2), you'll find the solution of b).

Greetings to all!

I hope it will serve my comment.

 

[dukemaster]

thanks ALBERTO666 !

o como decimos en chile " Muchas Gracias" jejeje .. I'm Chilean and my english is't too good jeje ...

i saw your answer and i repeat ... thanks.

to do this excercises i need to expand my mind .. and thing more than usual..jeje. well ... Bye