https://www.researchgate.net/post/What_is_the_correct_energy_in_a_VASP_calculation
Entropy: A concept that is not a physical quantity
https://www.researchgate.net/publication/230554936_Entropy_A_concept_that_is_not_a_physical_quantity
Dear All,
What is the difference between TOTEN energy in OUTCAR and E0 energy in OSZICAR ?
Why OSZICAR E0 number of iterations are not equal to TOTEN iterations. For example if any system is converging in 20 iteration which is given in OSZICAR then OUTCAR will show TOTEN energy steps more than 20. Why is more iteration in OUTCAR compare to OSZICAR.
When we use command like "grep E0 OSZICAR" if any systen is converging in 20 iteration then it will show 20 E0 values among then last one 20th is converged one.
When we do same for "grep TOTEN OUTCAR" then it shows more than 20 TOTEN energies roughly 5-6 times E0 energies in OSZICAR. Why so ?
Thank You
Shilendra
Calculus is not "take for granted".
ΔQ/T can not be turned into dQ/T. That is, the so-called "entropy " doesn't exist at all.
It is well known that calculus has a definition.
Any theory should follow the same principle of calculus; thermodynamics, of course, is no exception, for there's no other calculus at all, this is common sense.
Based on the definition of calculus, we know:
to the definite integral ∫T f(T)dQ, only when Q=F(T), ∫T f(T)dQ=∫T f(T)dF(T) is meaningful.
As long as Q is not a single-valued function of T, namely, Q=F( T, X, …), then,
∫T f(T)dQ=∫T f(T)dF(T, X, …) is meaningless.
1) Now, on the one hand, we all know that Q is not a single-valued function of T, this alone is enough to determine that the definite integral ∫T f(T)dQ=∫T 1/TdQ is meaningless.
2) On the other hand, In fact, Q=f(P, V, T), then
∫T 1/TdQ = ∫T 1/Tdf(T, V, P)= ∫T dF(T, V, P) is certainly meaningless. ( in ∫T , T is subscript ).
We know that dQ/T is used for the definite integral ∫T 1/TdQ, while ∫T 1/TdQ is meaningless, so, ΔQ/T can not be turned into dQ/T at all.
that is, the so-called "entropy " doesn't exist at all.
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