Physics with Calculus/Thermodynamics/Entropy

Definition
Entropy is a measure of how organised a system is. A system with low entropy is one with a organised, or unlikely configuration. For example: if you have a glass of water with average temperature 20 degrees, a low entropy state would be the top half 30 degrees and bottom half 10 degrees, whereas the highest entropy state is the entire glass 20 degrees.

Mathematically
To quantify this, we find the total number of microstates(Ω); all the possible states the system could be in. For a coin: Ω = 2 (heads / tails), 2 coins: Ω = 4 (HH/HT/TH/TT), N coins: Ω = 2N. If you have two systems with Ω1 and Ω2 microstates, Ωtotal = Ω1Ω2.

To apply this to a system of particles, use a volume V with N atoms. Classically, atoms can have an infinite number of states, making Ω infinite. But due to results from quantum physics, Ω is in fact finite. Given a total energy E and N atoms with mass m:

$$ \Omega = \frac{c}{N!}\left(\frac{V(2mE)^{3/2}}{h^3}\right)^{N}$$

c is a constant that depends on geometry.

h is Planck's constant.

A few notes here:


 * 1) It is possible to derive this from a pseudo-quantum argument, which says that there are only finite positions in the box equal to the volume divided by h3 and number of possible momenta is the surface area of a hypersphere. But since these are both aphysical and irrelevant, I might as well just state the formula(learn quantum mechanics!).
 * 2) I put a constant in the formula, because we are mainly interested in how the number of states grow rather than the exact number.
 * 3) The N! is only for indistinguishable particles. This is another quantum effect that means nature can't tell two atoms of the same kind apart.
 * 4) The numbers that statistical mechanics works with are on the order of N = 1027, so Ω is generally very, very big.

Entropy is defined as $$S = k_b \ln(\Omega)$$, where kb is Boltzmann's constant. So:

$$ S = k_b \ln\left(\frac{c}{N!}\left(\frac{V(2mE)^{3/2}}{h^3}\right)^{N}\right) = k_b (\ln c - \ln N! + N(\ln V + 3/2 \ln 2mE - 3 \ln h)) $$

Note: the units are fuddled in the last bit, it won't matter though. Stirling's approximation says for large N: $$ N! \sim N \ln(N) - N $$. So:

$$ S = N k_b(\ln(V)+\frac{3}{2}\ln(E) + D) $$

Since the important part is the change, I put the constants into D(assuming particle conservation). Also, $$E = \frac{1}{2} fNk_bT$$, so:

$$\Delta S = N k_b(\ln\frac{V_2}{V_1}+3/2 \ln\frac{T_2}{T_1})$$

$$\Delta S = N k_b\ln\frac{V_2}{V_1}$$ at constant temperature.

$$\Delta S = 3/2 N k_b \ln\frac{T_2}{T_1}$$ at constant volume.

It is also useful to define temperature in terms of entropy:

$$T = \frac{1} = \frac{2E}{3Nk_b}$$