You can't use eq
\delta U=T\delta S-P\delta V
Second law of thermodynamics is defined by eq
d U=TdS-Pd V
You can't simply change derivatives with variations. Give reference for that.
Also is interesting that Landau then says
\frac{\partial^2 U}{\partial S^2}(\delta S)^2+2\frac{\partial^2 U}{\partial S\partial V}\delta S\delta V+\frac{\partial^2 U}{\partial V^2}(\delta V)^2>0
If such an inequality holds for arbitrary \delta S and \delta V then
\frac{\partial^2 U}{\partial...
Yes you're right. Landau says in his book. In thermodynamics equlibrium G is minimum. Thus for any small deviation from equilibrium the change of the quontity G musy be positive. So
\delta U -T\delta S+P\delta V>0
In other words the minimum work which must be done from bring this part of...
If in some thermodynamics system preasure P and temperature T are constant then Gibbs potential has minimum.
G=U-TS+PV
Variation of G is
\delta G=\delta U-T \delta S+P \delta V
U=U(S,V)
\delta U=(\frac{\partial U}{\partial S})_V\delta S+(\frac{\partial U}{\partial V})_S\delta...