lag section 6_3
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Draft section of Phil's notes (dated 6.6.15 in the header) on the Boltzmann problem. It solves the two constraints for N1 and N2, checks that df/dNi vanishes at the solution, and shows the second derivative is negative, so the solution is a maximum of f = ln Ω. A Taylor expansion then estimates the peak width; for M near 10^20 particles the fractional half-width is about 10^-10.
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Curvature Calculation for f PhL 6.6.15
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6.3 More details of the Boltzmann problem solution
Start with
f = lnΩ = Σi [(1+lngi)Ni - NilnNi ] = Σi Ni [1 + ln(gi/Ni)] . (6.2.3)
The constraints ΣiNi= M and ΣiNiεi= U can be regarded as two equation in the two unknowns N1 and N2 which are easily solved to obtain,
N1 = + (ε2- ε1)-1[ ε2M - U + Σn=3m (εi- ε2)Ni] = N1(N3,...Nm)
N2 = - (ε2- ε1)-1[ ε1M - U + Σn=3m (εi - ε1)Ni] = N2(N3,...Nm) . (6.3.1)
Notice that, for i = 3,4...m,
= + (ε2- ε1)-1 [ (εi- ε2) ] =
= - (ε2- ε1)-1 [ (εi- ε1) ] = - . (6.3.2)
Now think of f as function of N3,,,,Nm ,
f(N1(N3,...Nm), N2(N3,...Nm), N3, .....Nm) . (6.3.3)
Compute the total derivative df/dNi for i = 3,4...n:
= f1 + f2 + fi where fi ≡
= f1 - f2 + fi (6.3.2)
= ln(g1/N1) - ln(g2/N2) + ln(gi/Ni) (6.2.6) (6.3.4)
Installing the solution values ln(Ni/gi) = λ1 + λ2εi shown in (6.2.8) this becomes
= (λ1 + λ2ε1) - (λ1 + λ2ε2) + (λ1 + λ2εi)
= λ1 [ - + 1] + λ2 [ ε1 - ε2 + εi ]
= λ1 [ - + ] + λ2 [ ε1 - ε2 + εi ]
= λ1 [0] + λ2[0] = 0 (6.3.5)
This is the expected result since the solution is supposed to be a critical point of f given in (6.3.3).
Our main interest is in the second derivative d2f/dNi2 . If this comes out negative at the solution point, we know we have a maximum. Applying d/dNj to the second line in (6.3.5),
= - +
= [ f11 - f12 + f1i] fij ≡
- [ f21 - f22 + f2i]
+ [ fi1 - fi2 + fii] i = 3,4....m (6.3.6)
Recall now that
f = lnΩ = Σi [(1+lngi)Ni - NilnNi ] (6.2.3)
fi = lngi - lnNi (6.2.6)
fij = - δij(1/Ni) (6.3.7)
Therefore (6.3.6) simplifies to
= f11 [ ]2 + f22 [ ]2 + fii
= – { [ ]2 + [ ]2 + } (6.3.8)
This says that the curvature of f in all independent directions Ni for i = 3,4...m is negative and this is true for all vectors N = {Ni}. This is of course then true for the solution vector N. But for the solution vector N we also have df/dNi = 0 from (6.3.5) and therefore the solution is a maximum of f and therefore of Ω.
Finally, we may estimate the width of the peak of f and then of Ω. The energy fractions in (6.3.8) are on the ballpark order of unity and the Ni are on the order of M, the total number of particles. If we do Taylor expansion of f(N) in the i direction only (i = 3,4...m) about the solution N, then
f(Ni + dNi) ≈ f(Ni) + (df/dNi)dNi + (1/2)(d2f/d2Ni) (dNi)2 + ,,,
= f(Ni) + 0 * dNi - (1/2) { [ ]2 + [ ]2 + } (dNi)2
~ f(Ni) + 0 - (1/2) { } (dNi)2 ~ f(Ni) - (dNi)2
so
df = f(Ni + dNi) - f(Ni) ~ - (dNi)2 . (6.3.9)
Therefore,
|df| ~ (dNi)2 dNi ~
~ (6.3.10)
This gives a rough estimate of the fractional distance one must move in the Ni direction to cause a reduction |df| in f from the peak value of f.
Suppose we are interested in finding dNi such that the Ω(N) drops to half its peak value. Then
Ωhalf/Ωpeak = 1/2 ln Ωhalf - ln Ωpeak = 1/2
fhalf - fpeak = ln(1/2) df = fpeak - fhalf = ln(2) = 0.7 (6.3.11)
Then (6.3.10) says
~ = ~ (6.3.12)
For a macroscopic box of gas particles, one might have M = 1020 and then ~ 10-10 . Since this is the fractional half-width of the peak in Ω, the peak is extremely sharp. This then is why, in an ensemble of macroscopic particle systems, all systems will have almost the same N value.
6.4 A numerical example