If we assumed that the big bang hypothesis is true, then the universe is in cooling and expanding phase, and this is justified by the star spectrum analysis. What about the entropy variation of universe? I think that It is decreasing according to Boltzmann theory.
Now, if we consider that the universe is an isolated system (we don't know and we can't define what can be the external medium to universe), its entropy is increasing (there is no exchange of entropy and the universe evolution is not reversible) because of entropy generation
Dear sir
I think in the beginning the entropy of universe is zero then it increases.
OK. But it is not evident to define the entropy at the beginning. We should be careful about one thing: what is the beginning? What was the situation before the beginning? Here we know only the Planck walls of time and space (10^-43s for time and 10^-34m for space). Furthermore, the beginning is it described by classical Physics or quantum physics when all universe matter was embedded in a space of the dimension of an egg. What is the dimension of the thermal wave length by comparison to the inter-atomic distances? U = 0 (all matter is in static state) yields to T = 0K and to S = 0 will be not true in the quantum description because of wave function recovery.
The positive mass-energy of the Universe was created at the Big Bang and very shortly thereafter at the expense of its negative gravitational energy, so the total energy of the Universe is ALWAYS zero. The entropy S = k ln OMEGA of the initial state of the Universe (zero mass, zero gravitational energy) was zero, because there is only one way to have nothing (initial OMEGA = 1). Thereafter the entropy S of the Universe increases, but not as fast as its maximum possible entropy Smax, so an entropy gap Smax - S is available to drive processes.
Entropy: A concept that is not a physical quantity
https://www.researchgate.net/publication/230554936_Entropy_A_concept_that_is_not_a_physical_quantity
Greetings,
What do you mean exactly by "A concept that is not a physical quantity"?
According to the second Principe, the differential of entropy is the sum of two terms: one related to the heat exchange at constant temperature and the second is related to the entropy generation, it means to the irreversibility of the thermodynamic transformation. How it is not physical then? It is Physico-chemical in many cases if you want. If you think that it is not directly measurable, I tell recall you that many and many physical concepts are calculated indirectly and can never be measurable.
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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