ideal gas application

Ideal Gas Application PhL 10.15.16 How would my general analysis on Boltzmann be applied to the Maxwell distribution business. I will first try to "roll my own" and see where I get, later I will look at Livesey. 1. Plan A Box of simple point atoms (monatomic gas, no energy in rotation or vibration). Energy of a particle is (1/2)mv2 and we are 3D. Energy levels are ε(vi) = (1/2)mvi2. Box contains M atoms and has N energy, so M = ΣiNi still applies. Imagine that the vi are discrete somehow. Ni is the number of particles having energy near ε(vi) . How would you write this in math? I think the first step is to make v-space be a lattice of tiny cubes each of volume V. Then consider a thin shell in v-space of thickness dv. The number of states in this volume is then gi = [4πv2dv]/V ≡ g(v) (dv/V) g(v) ≡ 4πv2 where now g(v) is a state density function in v space. I want to think of dv as a small but finite quantity, and the same for V. What about Ni ? This is the number of particles with v lying in the thin shell of volume [4πv2dv]. So you could invent a density function N(v) such that Ni = N(v) [4πv2dv] Note for later use that gi/Ni = g(v) (dv/V)/ N(v) [4πv2dv] = What then would you say about Ω(N) = .... = Πi=1m ? This becomes somehow a product of an infinite number of factors. Each factor corresponds to a "shell" in v-space. I know a few infinite product formulas from GR7 page 14. Now consider . The numerator is g1N = [ g(v) (dv/V)] N(v)[4πv2dv] N1! = (N(v) [4πv2dv])! Again, I am thinking of dv and V as finite, so these are finite numbers. The product for Ω(N) has a factor for v = 0, and v = dv and v = 2dv and so on. So write v = n(dv) Then the product is Πn=1∞ where we ignore the n = 0 term perhaps. So we then have Ω(N) = Πn=1∞ { [ g(v) (dv/V)] N(v)[4πv2dv] / (N(v) [4πv2dv])! = Πn=1∞ { [ g(ndv) (dv/V)] N(ndv)[4π(ndv)*ndv(dv)] / (N(ndv) [4π(ndv)2dv])! = Πn=1∞ Maybe I can just write it like so and they not use it, it is just some huge mess which hopefully converges because the Ni taper off at higher energies, which means in the higher factors. Let's look at the form of one of these factors where: g1N = [ g(v) (dv/V)] N(v)[4πv2dv] N1! = (N(v) [4πv2dv])! As N(v) → 0 for large v, we see that g1N → [ g(v) (dv/V)]0 = 1 if g(v) is reasonable. And we see that N1! → (0)! = 1. Thus if you look at the far right end of the infinite product of factors, you see a product of numbers that each approach 1, so it is then at least conceivable that this limit approaches a finite number. OK, what would I do next in following Lagrange doc? I assume Ni large which is at least valid for occupied energy levels. This would give f = Σi Ni [1 + ln(gi/Ni)] where now we have a sum instead of a product. Installing we get f = Σv N(v) [4πv2dv] [1 + ln()] = Σn=1∞ N(ndv) [4π(ndv)2dv] [1 + ln()] The limit here is not so obvious, but again just take this as what it is. Now just jump to our major result which is Ni = A gi e-βε which we can translate to read N(v) [4πv2dv] = A [4πv2dv]/V exp(-βmv2/2) or N(v) = (A /V) exp(-βmv2/2) Remember that my N(v) is a shell radial density, and thus differs from N(v). Now integrate to get M = !Syntax Error, IN(v) [4πv2dv] = (A /V) 4π !Syntax Error, Iexp(-βmv2/2)v2dv Evaluate the integral with α = βm/2 α = βm/2 !Syntax Error, Iexp(-βmv2/2)v2dv = !Syntax Error, Iexp(-αv2)v2dv According to Schaum page 98 item 15.76 the integral is M = (A /V) 4π * (/4) [βm/2]-3/2 = (A /V) π3/2 [2/(βm)]3/2 = (A /V) [2π/(βm)]3/2 This lets you solve for A to be A = MV [2π/(βm)]-3/2 = MV [βm/(2π)]3/2 and then we seem to have our solution N(v) = (A /V) exp(-βmv2/2) = M [βm/(2π)]3/2 exp(-βmv2/2) and the V's have cancelled! Livesey does not state his final result. Remember that Ni = N(v) [4πv2dv] = number of particles in a thin shell at v with thickness dv so we are finding that Ni = [4πv2dv] M [βm/(2π)]3/2 exp(-βmv2/2) = 4πM [βm/(2π)]3/2 v2 exp(-βmv2/2) dv = 4πM [βm/(2)]3/2 π-3/2 v2 exp(-βmv2/2) dv = 4M π-1/2 [βm/(2)]3/2 v2 exp(-βmv2/2) dv and this then is the traditional Maxwell velocity curve which has a smooth peak. 2. Let's look now at various in-house authors Livesey page 16 gets results similar to me, but he does not quite state the final result with the constant evaluated. He uses symbol f(v) I think in the sense that I use Ni/dv above, and he gets f(v) = C v2exp(-βmv2/2). Zemansky does his derivation starting page 147 and ends up with his final result on page 159 which is this, changing w→v and β = 1/(kT) : dNv/dv = ( 4M/ )(mβ/2)3/2 v2 exp(-βmv2/2) This agrees exactly with my Plan A result, hurray! Big Reif: Page 267 gives this speed distribution F(v) = 4πn (mβ/2π)3/2 v2 exp(-βmv2/2) where I think his n is the particle density in volume space. Now π (1/π)3/2 = π π-3/2 = π-1/2 so he agrees with Zemansky apart from the definition of F(v). 3. Where is the super strong peak I keep talking about in Lagrange doc? A box of gas has the Maxwell speed distribution which is a smooth peak. Based on the above discussion, you can regard it as a plot of Ni which solves the gas problem. This set of many Ni makes the vector N I talk about. Now take an ensemble of 1 billion boxes of gas, and each has a solution vector N which means each one has a speed curve. Fact: The speed curve and its peak are NOT the curve I am talking about with my billion systems comment. Each of the billion boxes of bas has a curve which is its N, and the curves of all the systems are very close. The speed curve IS the stationary point solution vector N. Now you can look at a particular component Ni of this vector and ask about its probability distribution. If this is a value My existing curvature formula does now work very well. Maybe I can back up to my results above which reads, f = Σv N(v) [4πv2dv] [1 + ln()] // which is lnΩ = !Syntax Error, Idv N(v) [4πv2] [1 + ln()] But I know that, g(v) ≡ 4πv2 N(v) = M [βm/(2π)]3/2 exp(-βmv2/2) = C exp(-βmv2/2) C = M [βm/(2π)]3/2 = = Then f = !Syntax Error, Idv [4πv2] C exp(-βmv2/2) [ 1 - ln [ CV exp(-βmv2/2) ] which is just some number. So how do you talk about a derivative like ? Something is wrong here.