I'd like to ask some questions about flipping two coins related to statistical mechanics, e.g. microcanonical distribution, phase space distribution function etc... after I rephrase the coin flipping problem into the language of statistical mechanics.

In probability theory, given the following problem

  • Random experiment: Toss two coins
  • Example of an Outcome: $10 = (Heads, Tails)$
  • Sample space: $S = {11,10,01,00}$, $|S| = 4$
  • Examples of Events: 2 Heads $= 2H = \{11\}$, $|2H| = 1$, $1H = \{10,01\}$, $|1H| = 2$, $0H = \{00\}$, $|0H| = 1$

I'd like to translate this to statistical mechanics as much as possible, in which:

  • A microstate is an element of the sample space (right?), e.g. $10$ or $01$.
  • A macrostate is an event (a subset of the sample space, right?), e.g. $1H = \{10,01\}$.
  • The statistical weight (statistical probability) of a macrostate is the cardinality of the event, e.g. $|1H| = 2$.
  • The equilibrium distribution is the most likely macrostate which is the macrostate with the highest statistical weight which is the event with the highest cardinality, e.g. $1H = \{10,01\}$ since $|1H| = 2$.

we can find the Maxwell-Boltzmann distribution function $n_i$ for this system by extremizing

$$w(n) = "number \ of \ heads \ in \ n" = \tfrac{2!}{n!(2-n)!}= \tfrac{2!}{n_1!n_2!}$$

with respect to $n_i$ given the constraint equation $n_1 + n_2 = 2$, showing $n_1 = e^0 = 1$ maximizes $w$, $w(1) = |1H| = |\{10,01\}| = 2$ is the maximum, thus the entropy $S = \ln(w) = \ln(2!)$ is at it's largest and the system is most disordered.

  1. What is the microcanonical distribution of a random experiment in which you toss 2 coins?
  2. What is the canonical distribution for this experiment?
  3. Can I choose both, or is there an example of when I should use one or the other, related to this example?
  4. How do I find it's phase space distribution function $\rho = \rho(p,q)$?
  5. Are these stupid questions? If they are, why? Is there a way to make it so that I can derive the phase space distribution function? As far as I can see I think I'm supposed to add another Lagrange multiplier as a way to incorporate the energy, then after expressing $n$ in terms of the $\varepsilon _i$'s I can take a derivative of $w$ with respect to $E$ (so long as I replace $\varepsilon _1 = (E - n_2 \varepsilon _2)/n_1$ and $\varepsilon _2 = (E - n_1 \varepsilon _1)/n_2$ in the Maxwell-Boltzmann distribution that I'm plugging back in to $w$, but that seems crazy and seems to make no sense since if $E$ varies then shouldn't $\varepsilon _1$ and $\varepsilon _2$ also vary? This idea assumes they stay fixed...) to get $\frac{dw}{dE} = \int \delta [H - E]dpdq$, but it seems logically flawed as I've described and even if it worked how would you use it to determine $\rho$ inside the integral? I'm not even sure if this set up allows for the micro or just canonical distribution anyway, hence the question...


  • $\begingroup$ There isn't a constraint in the problem of two flipped coins, so I'm pretty sure the entire mechanism of physical statistical mechanics doesn't make sense. Temperature is the Lagrange multiplier you get from the fixed energy constraint, chemical potential is from the fixed particle number constraint. In your case, nothing is constrained. $\endgroup$ – DanielSank Apr 13 '14 at 5:10

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