# Entropy: Disorder or energy dispersal?

The first definition of entropy given by Clausius is I believe this $$S=Q/T$$ It is as I understand a common fact to understand entropy and maybe often teach it as a measure of disorder through the statistical definition of Boltzmann or Gibbs( depending on the ensemble) $$S=k\lnΩ$$ My question depending entropy, after some searching (look at A MODERN VIEW OF ENTROPY by Frank L. LAMBERT ) is this:

Is the physical meaning of entropy to be understood only in statistical terms as disorder because of the change in the statistical weights $Ω,$ or by looking to the thermodynamics as well, move to a definition of entropy as energy dispersal? In other words, conceive the physical meaning of entropy as a dispersal of the energy inside (or maybe at some points outwards) the system under consideration, where dispersal stands for a more wide allocation through the interior parts of the system( classical or quantum mechanical).

• Can i post a part from some papers or articles i have read so my question can become more specific? Apr 17 '15 at 9:33
• This are to links for further information on the approach mentioned in the question-this is for energy as energy dispersal khimiya.org/volume15/Entropia.pdf entropysite.oxy.edu/jung.pdf Apr 17 '15 at 9:45
• Is there a problem to this question? Apr 18 '15 at 8:54
• Jan 7 '18 at 19:04
• Related 'Does the fact that there are two different mathematical definitions of entropy imply there are two different kinds of entropy?' physics.stackexchange.com/questions/519293/… May 3 '20 at 22:57

S=Q/T and S=klnΩ do not contradict each other. To understand this, first you need to understand S=Q/T does not describe the absolute entropy. It describe the entropy change. For example, when 1J is extracted from a block of stone at 300K, the stone's entropy is reduced by S = Q/T = 1J/300K. (Strictly speaking, we should use dS = dQ/T, as the temperature of the stone may change. To get the total change of S, we need to integrate over the entire energy extraction process: $$\Delta$$S = $${\int} dQ/T$$ ).
Now, microscopically, when you extract 1J from the stone, it cools down a little bit (lower temperature). As temperature changes the phonon (lattice vibration in the stone) distribution (a Bose-Einstein distribution since phonons are bosons), the available microscopic states for the lattice vibration (Ω) becomes less, and the statistical equation $$S=k ln Ω$$ will give you the same result of reduced entropy (or strictly, $$\Delta$$S = $${\int} dQ/T = k \ln Ω_{after} - k \ln Ω_{before}$$ ).