|MadSci Network: Astronomy|
In general, the material of which the forming star is composed undergoes a large decrease in entropy during the formation process, and the rest of the universe (since total entropy must never decrease) endures an even larger increase. Star formation is not understood in detail; it is in fact one of the outstanding mysteries of contemporary astrophysics. Thus we must content ourselves with a somewhat schematic answer. Nevertheless, the general picture is clear.
The material of the new star comprises a cloud of gas, held together by gravity, and cooling by radiation to the exterior. As a result of this loss of heat by radiation, the gas pressure in the interior would drop, except that gravity keeps squeezing the cloud into a smaller and smaller space. Thus the pressure cannot drop: it rises, while the volume decreases. Furthermore, as the volume decreases, the force of gravity, driven by its 1/R² dependence on distance R, steadily increases. As a result, the material in the star is compressed and heated; and the more heat it loses, the hotter it gets.
Let's look at the entropy effects of this process from two different points of view, the thermodynamic and the statistical mechanical.
while the entropy gain of the rest of the universe is
By conservation of energy, the heat dQ lost by the cloud (neglecting any other energy coupling effects, like pressure, magnetic fields, etc) must equal the heat gained by radiation in the rest of the universe; and since T2 >> T1 (typical numbers might be T2 = 1000 K in a newly forming star and T1 = 25 K in a cloud),
This is, of course, simply another example of the entropy change which occurs when a hot body loses energy to a cooler one. The finite difference in temperature is the tip-off that the process is irreversible, and that the total entropy must be decreasing.
For a simple monatomic gas, the co-ordinates of the i-th particle would be (x,y,z)i, and its momentum would be (px,py,pz)i. The co-ordinate factor in the volume would be V, the product of the allowed range of (x,y,z)i, and the momentum factor would be the volume of a sphere in momentum space of radius p, the mean magnitude of the momentum. The value of p is related to the kinetic energy by K.E. = p²/2µ, where µ is the mean molecular weight; and hence to the temperature T. Thus we get for the momentum factor in the phase space volume Vp:
for each particle. Taking then the product of V and Vp, all to the N-th power to account for all particles, and the logarithm of this volume, finally, to get the entropy, we obtain:
where k is Boltzmann's constant.
For something more complex than a monatomic gas, the 3/2 factor multiplying ln(T) turns out to be 1/(gamma-1), where gamma is the thermodynamic ratio of specific heats, but this is peripheral to our main concern. What we are interested in here is that in its collapse, the proto-star changes from a size of ~1018 cm in its space dimension to something like 1011 cm, so the coordinate volume factor V decreases by an enormous factor of roughly 1021 per particle. The momentum factor Vp in the phase space volume increases by a factor of perhaps 403/2 per particle (based on T changing from 25 K to 1000 K, our previous example), or a few hundred at most. So the volume factor wins out, and the total entropy of the material of the star decreases by a large factor, as we concluded from thermodynamics.
Ref: See The Feynman Lectures, Vol I, Chapter 44, for a more detailed discussion of the formula above for S(V,T).
Try the links in the MadSci Library for more information on Astronomy.