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What happens when you increase entropy in a system? Does it increase the instability? Is disorder the same as entropy?

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closed as unclear what you're asking by Kyle Kanos, Jim, John Rennie, Brandon Enright, Qmechanic Apr 23 '14 at 17:04

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

Please explain your notion of instability, and how you intend to increase the entropy – Lord_Gestalter Apr 22 '14 at 7:54
Could you elaborate on the context and system? – JamalS Apr 22 '14 at 8:57
I'm not an expert in this field which is why i'm posting this question, but when i say instability i mean chaos and all the dynamics associated with chaos, i.e. unstable, bifurcations, perturbations, words that are similar and have fancier spelling, etc. One person told me if you increase entropy, the system becomes more stable and i know entropy has an equilibrium state, but i don't see that makes sense since entropy is related to increasing the temperature of a system, which creates more collisions. I don't know what the difference is between a QM or GR context. – user136128 Apr 23 '14 at 5:48

Is disorder the same as entropy?

In statistical physics, the definition of entropy varies and also depends on the particular ensemble chosen, but roughly speaking, is a measure of the number of microstates of a system. A particular definition is given by the expression,

$$S=-k_B \log \Omega$$

where $\Omega$ is the number of microstates, and $k_B$ is Boltzmann's constant. Alternatively, the entropy can be computed from the partition function of a system which is a preferred approach in some cases. If we relate disorder roughly to the number of ways to arrange a system, then we can say that entropy and disorder, although not equal, are roughly the same type of measure.

What happens when you increase the entropy of a system?

The answer depends entirely on the system described. Therefore we present the simplest example, that of an ideal gas of $N$ indistinguishable particles. Assuming they do not interact, the Hamiltonian for each:


The partition function, $Z_0$, of the ideal gas is given by an integral over phase space,

$$Z_0=\frac{1}{(2\pi\hbar)^3}\int \! \mathrm{d}^3 q \mathrm{d}^3 p \, \, e^{-\beta H(q,p)} = V \left( \frac{mk_B T}{2\pi\hbar^2}\right)^{3/2}$$

where we have used the standard formula for a Gaussian integral in $p$, substituted $\beta=1/k_BT$ and assumed the integral over $\mathrm{d}^3 x $ yields the volume of the system. For $N$ particles, we raise the partition function to the power $N$, and divide by $N!$ due to the indistinguishability of the particles,

$$Z_N = \frac{Z^N_0}{N!}$$

To compute the entropy, given the partition function, we employ the relation, $$S= k_B \frac{\partial }{\partial T}\left( T \log Z_N \right)= Nk_B \left( \log \left( \frac{V}{N\lambda^3}\right) +\frac{5}{2}\right) $$

where we have used Stirling's formula for the approximation of the factorial, and defined the thermal de Broglie wavelength,

$$ \lambda = \sqrt{\frac{2\pi\hbar^2}{mk_BT}}$$

To answer your question for this simple case, we see that if we increase the entropy of the system, this must be accomplished by, for example: increasing the volume of the system, the temperature, or selecting particles of a higher mass, as $S \propto T, m$. Generally speaking, modifying the entropy will therefore requite the modification of the parameters describing the system.

Regarding instability, the answer largely depends, once again, on the context. We could be discussing anything from the instability of a quantum mechanical system to a solution in general relativity.

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