How Do You Derive Temperature from Entropy?

jorgen
Messages
12
Reaction score
0
Hi all,

I have determined the entropy previously which gives the following results

S = k*(N Log N - n Log n -(N-n)*Log(N-n)

I have furthermore deduced the relation between state n and the energy as

n = \frac{E+N*\epsilon}{2*\epsilon}

I know that the temperature is

\frac{\partial S}{\partial E}=\frac{1}{T}

I am not quite sure how to proceed - I could replace small n with an expression containing the energy but what about capital N - should I isolate that from the relation and insert that into the entropy expression and then differentiate with respect to the energy - any hints or commands appreciated thanks in advance.

Best
J
 
Physics news on Phys.org
Replacing 'n' in your expression for the entropy seems like a good start to me. The next question is "which variables are held fixed when you compute the partial derivative of S with respect to E"? ...Surely if 'N' is fixed during this differentiation, the problem is simplified.:wink:
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top