Apply the Legendre Transformation to the Entropy S as a function of E

AI Thread Summary
The discussion focuses on applying the Legendre transformation to the entropy function S as a function of energy E, volume V, and particle number N. Participants emphasize the importance of expressing the differential dS in terms of the independent parameters (E, V, N) and then transforming it to the parameters (T, V, N). The transformation involves expressing dE in terms of dT, dV, and dN to derive a new potential, referred to as free energy F. The Sackur-Tetrode equation is mentioned as a goal of this process. Overall, the conversation highlights the steps necessary to successfully complete the task using differential forms and Legendre transformations.
GravityX
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Homework Statement
Apply Legendre Transformation to Entropy
Relevant Equations
##g(m)=f(x(m))-m*x(m)## and ##x(m)=(f')^-1(m)##
Hi,

Unfortunately I am not getting anywhere with task three, I don't know exactly what to show

Bildschirmfoto 2022-11-28 um 16.30.54.png

Shall I now show that from ##S(T,V,N)## using Legendre I then get ##S(E,V,N)## and thus obtain the Sackur-Tetrode equation?
 
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You should rather think in terms of differentials. The "natural independent parameters" for ##S## are ##(E,V,N)##. Now you want to get another potential with the "natural independent parameters" ##(T,V,N)##. So first write down the differential ##\mathrm{d} S## in terms of ##(E,V,N)## and then think about, how to Legendre transform to a new potential with the other set of independent parameters.
 
Thanks vanhees71 for your help 👍

I have now represented ##ds## as follows

$$ds=\frac{\partial S}{\partial E}dE+\frac{\partial S}{\partial V}dV+\frac{\partial S}{\partial N}dN$$

$$ds=\frac{1}{T}dE+\frac{P}{T}dV-\frac{\mu }{T}dN$$

Now I would just have to get rid of the ##dE## or rather I would have to express ##dE## with the help of ##dT##, ##dV##, ##dN##, right?
 
Write it in the form of ##\mathrm{d} E=...## then find a new potential, ##F## ("free energy"), such that instead of a differnetial with ##\mathrm{d} s##, ##\mathrm{d} V## and ##\mathrm{d}N## you get one with ##\mathrm{d} T##, ##\mathrm{d}V## and ##\mathrm{d}V##. Note that
$$\mathrm{d}(Ts)=s \mathrm{d} T + T \mathrm{d} s!$$
 
Thanks for your help vanhees71 👍 I think I got it now 😀
 
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