Advanced Thermodynamics   Quiz #1

  1. An ordinary deck of cards contains 52 cards, half of which are red and half of which are black. If we drew an even number of cards from the deck, we'd expect on average to see that same 50% color distribution. But many games draw 5 cards, an uneven number. In a 5 card hand then, we'd expect that the most likely distribution would be 3 cards of one color and 2 of the other. Show how much more likely (by what factor) that is than a hand with one card of one color and the rest of the other.


That ratio is [WRRRBB+WBBBRR] / [WRRRRB+WBBBBR], but since reds and blacks are equally abundant we get the same ratio with WRRRBB / WRRRRB! That is, WRRRBB=WBBBRR and WRRRRB=WBBBBR so we needn't complicate our ratio with redundancies. Now by RRRBB we mean any hand with 3 R and 2 B in it; RBRRB would do as well as BRBRR, etc.

WRRRBB / WRRRRB = [(26×25×24×26×25)/(3! 2!)] / [(26×25×24×23×26)/(4! 1!)] since each time we add a colored card to our hand, there's one less of it to draw from the deck, and it doesn't matter what order the cards of any given color appear in our hand.

When all the canceling is done, WRRRBB / WRRRRB = [25×4!×1!] / [23×3!×2!] = 2.174. So double odds on 4-of-a-color vs 3-of-a-color is a sucker bet.

  • Imagine a nondegenerate ladder of equally-spaced energy levels like that for a harmonic oscillator but with associated wavefunctions which could be occupied by either fermions or bosons. In the case of the fermions, not only would energy have to be conserved but also no two particles could occupy the same wavefunction. A system of bosons would only have to conserve energy.

    The diagram below shows 4 fermions with a system (total) energy of 6 quanta. Fortunately, that's the only distribution possible. It has an associated entropy. Now let the 4 particles be bosons. What's the ratio of boson to fermion distributions? And what's the entropy difference (in units of k) between the boson and fermion possibilities?


    • The "x" occupancies were the ones given for a configuration of 6 quanta shared among 4 oscillators with no two oscillators having the same energy level. It yields a weight W1=4!/(1!1!1!1!)=24 and a set of occupancy probabilities of 25% each for an entropy (of mixing) proportional to -4×(0.25×ln[0.25]) = ln(4) = 1.386.

    The "o" occupancies add the distributions which don't care about multiple occupancies. These plus the original ("x") are available to those bosons.

                                             7 quanta<Pj>
     o       6 quanta.012
      o      5 quanta.036
       oo    4 quanta.072
    x    ooo  3 quanta.119
    x  o   ooooo2 quanta.179
    x o oo ooo oo1 quantum.250
    xoooooooooo o 0 quanta.334
    2441212126446Weights, Wi
    .286.048.143.143.143.071.048.048.071Wi/(Wtot=84)

    From the weights of each configuration, we find their sum, Wtot, and the relative importance (probability) of each configuration, Wi/Wtot. That let's us calculate the probability of occupancy of each state, <Pj> = sum over i configurations Pj;i×Wi/Wtot.

    (Note how the probabilities fall off as we climb the ladder, but note also that they don't fall off with constant ratio! We're nowhere near the large number limit here.)

    And the corresponding entropy (of mixing) will be proportional to the sum over -Pj×ln(Pj) or 1.636. So the difference in entropy between bosons and fermions in this example is proportional to 1.636-1.386=0.250 rounded. We should scale that by 4 to 1.001 to account for 4 oscillators rather than one, but we've not covered that scaling yet.

    If, instead, you'd used the simpler S=klnW, the entropy difference would have been proportional to ln(84/24)=1.253 more than the factor of 4 accounted for by the number of oscillators! The difference is that our P×ln(P) isn't the Boltzmann one; our system is too small. So it's derivation using the large number approximation doesn't apply in this case. The best choice for entropy difference would be the S=klnW one, since that doesn't use the large number approximation.

    But any of those answers will be deemed correct since we haven't looked into the scaling of these formulas with number of systems.
    Last modified 30 September 1997.