I gave this a shot as well as since your value for E(T) → ∞ as T → ∞, while I would think the system should cap out at εN.
I get a different value for S(E), reasoning:
If E/ε is 1, there are N microstates, since 1 of N positions is at energy ε.
If E/ε is 2, there are N(N-1) microstates.
etc. etc, giving for E/ε = x that there are N!/(N-x)!
so S = ln [N!/(N-x)!] = ln(N!) - ln((N-x)!) = NlnN - (N-x)ln(N-x)
S(E) = N ln N - (N—E/ε) ln (N—E/ε)
Can you explain how you got your equation for the entropy?
Going on I get E(T) = ε(N—e^(ε/T − 1) )
This also looks wrong, as although E → ∞ as T → ∞, it also doesn’t cap at exactly εN, and E → -∞ for T→ 0...
I’m expecting the answer to look something like: E(T) = εN(1 - e^(-ε/T))/2 which ranges from 0 to εN/2, which seems sensible.
EDIT: Nevermind, the answer was posted while I was writing this. I’d still like to know how you got your S(E) though.
S(E) is the log of the number of states in phase space that are consistent with energy E. Having energy E means that E/ε particles are excited, so we get (N choose E/ε) states. Now take the log :)
I gave this a shot as well as since your value for E(T) → ∞ as T → ∞, while I would think the system should cap out at εN.
I get a different value for S(E), reasoning:
If E/ε is 1, there are N microstates, since 1 of N positions is at energy ε. If E/ε is 2, there are N(N-1) microstates. etc. etc, giving for E/ε = x that there are N!/(N-x)!
so S = ln [N!/(N-x)!] = ln(N!) - ln((N-x)!) = NlnN - (N-x)ln(N-x)
S(E) = N ln N - (N—E/ε) ln (N—E/ε)
Can you explain how you got your equation for the entropy?
Going on I get E(T) = ε(N—e^(ε/T − 1) )
This also looks wrong, as although E → ∞ as T → ∞, it also doesn’t cap at exactly εN, and E → -∞ for T→ 0...
I’m expecting the answer to look something like: E(T) = εN(1 - e^(-ε/T))/2 which ranges from 0 to εN/2, which seems sensible.
EDIT: Nevermind, the answer was posted while I was writing this. I’d still like to know how you got your S(E) though.
S(E) is the log of the number of states in phase space that are consistent with energy E. Having energy E means that E/ε particles are excited, so we get (N choose E/ε) states. Now take the log :)