# HG changeset patch # User Dylan Holmes # Date 1335739758 18000 # Node ID 299a098a30da61db096f917f218b0c3661fc0557 # Parent 8f3b6dcb9addf1a1eff79540ea3c4a7b7b6d27d6 Saving progress... diff -r 8f3b6dcb9add -r 299a098a30da org/stat-mech.org --- a/org/stat-mech.org Sun Apr 29 02:38:22 2012 -0500 +++ b/org/stat-mech.org Sun Apr 29 17:49:18 2012 -0500 @@ -390,7 +390,7 @@ by specifying any two of the variables arbitrarily, whereupon the third, and all others we may introduce, are determined. Mathematically, this is expressed by the existence of a functional -relationship of the form[fn:: The set of solutions to an equation +relationship of the form[fn:: Edit: The set of solutions to an equation like /f(X,x,t)=/ const. is called a /level set/. Here, Jaynes is saying that the quantities /X/, /x/, and /t/ follow a \ldquo{}functional rule\rdquo{}, so the set of physically allowed combinations of /X/, @@ -633,7 +633,7 @@ so quickly, required a very long time for their first performance, and the essential conclusions of this Section were first arrived at only about 1760---more than 160 years after Galileo's invention of the -thermometer---by Joseph Black, who was Professor of Chemistry at +thermometer---[[http://web.lemoyne.edu/~giunta/blackheat.html][by Joseph Black]], who was Professor of Chemistry at Glasgow University. Black's analysis of calorimetric experiments initiated by G. D. Fahrenheit before 1736 led to the first recognition of the distinction between temperature and heat, and prepared the way @@ -1031,7 +1031,7 @@ Secondly, we have already stressed that the theory being developed must, strictly speaking, be a theory only of equilibrium states, since otherwise we have no operational definition -of temperature . When we integrate over any \ldquo{}path\rdquo{} in the $(V-t)$ +of temperature When we integrate over any \ldquo{}path\rdquo{} in the $(V-t)$ plane, therefore, it must be understood that the path of integration is, strictly speaking, just a /locus of equilibrium states/; nonequilibrium states cannot be represented by points @@ -1067,7 +1067,7 @@ environment can reverse the direction of the process. Recognizing this, we can then say that the paths of integra tion in our equations are to be interpreted physically as -/reversible paths/ . In practice, some systems (such as gases) +/reversible paths/ In practice, some systems (such as gases) come to equilibrium so rapidly that rather fast changes of state (on the time scale of our own perceptions) may be quite good approximations to reversible changes; thus the change of @@ -1100,18 +1100,18 @@ example, a volume of air may be imagined to consist of a number of smaller contiguous volume elements. With this subdivision, the pressure is the same in all subsystems, and is therefore in -tensive; while the volume is additive and therefore extensive . +tensive; while the volume is additive and therefore extensive But we may equally well regard the volume of air as composed of its constituent nitrogen and oxygen subsystems (or we could re gard pure hydrogen as composed of two subsystems, in which the molecules have odd and even rotational quantum numbers -respectively, etc.) . With this kind of subdivision the volume is the +respectively, etc.) With this kind of subdivision the volume is the same in all subsystems, while the pressure is the sum of the partial pressures of its constituents; and it appears that the roles of \ldquo{}intensive\rdquo{} and \ldquo{}extensive\rdquo{} have been interchanged. Note that this ambiguity cannot be removed by requiring that we consider only spatial subdivisions, such that each sub -system has the same local composi tion . For, consider a s tressed +system has the same local composi tion For, consider a s tressed elastic solid, such as a stretched rubber band. If we imagine the rubber band as divided, conceptually, into small subsystems by passing planes through it normal to its axis, then the tension @@ -1151,7 +1151,7 @@ Furthermore, we know that the pressure $P$ is not only the intensive factor of the work; it is also the \ldquo{}potential\rdquo{} which governs mechanical equilibrium (in this case, equilibrium with respect -to exchange of volume) between two systems; i .e., if they are +to exchange of volume) between two systems; ie., if they are separated by a flexible but impermeable membrane, the two systems will exchange volume $dV_1 = -dV_2$ in a direction determined by the pressure difference, until the pressures are equalized. The @@ -1175,7 +1175,7 @@ But we already know that the /temperature/ is the quantity that governs the heat flow (i.e., heat flows from the hotter to -the cooler body until the temperatures are equalized) . So the +the cooler body until the temperatures are equalized) So the intensive factor in $dQ$ must be essentially the temperature. But our temperature scale is at present still arbitrary, and we can hardly expect that such a factorization will be possible for all @@ -1218,7 +1218,7 @@ But we want the entropy $S(U,V)$ to be a definite quantity, like the energy or volume, and so $dS$ must be an exact differential. On an infinitesimal reversible change from one equilibrium state -to another, the first law requires that it satisfy[fn:: The first +to another, the first law requires that it satisfy[fn:: Edit: The first equality comes from our requirement that $dQ = T\,dS$. The second equality comes from the fact that $dU = dQ - dW$ (the first law) and that $dW = PdV$ in the case where the state has two degrees of @@ -1229,7 +1229,7 @@ \end{equation} Thus $(1/T)$ must be an /integrating factor/ which converts $dQ$ into -an exact differential [[fn::A differential $M(x,y)dx + +an exact differential [[fn::Edit: A differential $M(x,y)dx + N(x,y)dy$ is called /exact/ if there is a scalar function $\Phi(x,y)$ such that $M = \frac{\partial \Phi}{\partial x}$ and $N=\frac{\partial \Phi}{\partial y}$. If there is, \Phi is called the @@ -1314,7 +1314,7 @@ In order to fix the relative values of $S$ on different adiabats we need to add the condition, not yet put into the equations, that the integrating factor $w(U,V) = T^{-1}$ is to define a new -temperature scale . In other words, we now ask: out of the +temperature scale In other words, we now ask: out of the infinite number of different integrating factors allowed by the differential equation (1-23), is it possible to find one which is a function only of the empirical temperature $t$? If @@ -1497,7 +1497,7 @@ where f(t) is a function that depends on the particular empirical temperature scale used. But from (1-33) we must then have $f(t) = RT$, where $R$ is a constant, the universal gas constant whose -numerical value (1.986 calories per mole per degree K) , depends +numerical value (1.986 calories per mole per degree K), depends on the size of the units in which we choose to measure the Kelvin temperature $T$. In terms of the Kelvin temperature, the ideal gas equation of state is therefore simply @@ -1554,7 +1554,7 @@ unit could be defined. Thus the old Centigrade and new Celsius scales are the same, within the accuracy with which the Centigrade scale could be realized; so the same notation, ^\circ C, is used -for both . Only in this way can old measurements retain their +for both Only in this way can old measurements retain their value and accuracy, without need of corrections every time a unit is redefined. @@ -1562,7 +1562,7 @@ Exactly the same thing has happened in the definition of the calorie; for a century, beginning with the work of Joule, more and more precise experiments were performed to determine -the mechanical equivalent of heat more and more accurately . But +the mechanical equivalent of heat more and more accurately But eventually mechanical and electrical measurements of energy be came far more reproducible than calorimetric measurements; so recently the calorie was redefined to be 4.1840 Joules, this @@ -1600,167 +1600,261 @@ temperature scale is the same for all substances. -** COMMENT Entropy of an Ideal Boltzmann Gas +** Entropy of an Ideal Boltzmann Gas At the present stage we are far from understanding the physical -meaning of the function $S$ defined by (1-19); but we can investigate its mathematical -form and numerical values. Let us do this for a system con -sisting cf n moles of a substance which obeys the ideal gas +meaning of the function $S$ defined by (1-19); but we can investigate +its mathematical form and numerical values. Let us do this for a +system +consisting of $n$ moles of a substance which obeys the ideal gas equation of state -and for which the heat capacity at constant volume CV is a -constant. The difference in entropy between any two states (1) + +\begin{equation}PV = nRT\end{equation} + +and for which the heat capacity at constant volume +$C_V$ is a constant. The difference in entropy between any two states (1) and (2) is from (1-19), - + +\begin{equation} +S_2 - S_1 = \int_1^2 \frac{dQ}{T} = \int_1^2 +\left[\left(\frac{\partial S}{\partial V}\right)+\left(\frac{\partial S}{\partial T}\right)_V dT\right] +\end{equation} where we integrate over any reversible path connecting the two -states. From the manner in which S was defined, this integral +states. From the manner in which $S$ was defined, this integral must be the same whatever path we choose. Consider, then, a -path consisting of a reversible expansion at constant tempera -ture to a state 3 which has the initial temperature T, and the -.L ' "'1 final volume V2; followed by heating at constant volume to the final temperature T2. Then (1-47) becomes -3 2 I If r85 - on - db — = d — -4 S2 51 J V [aT]v M (1 8) -1 3 -To evaluate the integral over (1 +3) , note that since -dU = T :15 — P dV, the Helmholtz free energy function F E U — TS -has the property dF = --S - P 61V; and of course is an exact -differential since F is a definite state function. The condition -that dF be exact is, analogous to (1-22), +path consisting of a reversible expansion at constant +temperature to a state 3 which has the initial temperature $T_1$, and the +the final volume $V_2$; followed by heating at constant volume to the +final temperature $T_2$. +Then (1-47) becomes + +\begin{equation} +S_2 - S_1 = \int_1^3 \left(\frac{\partial S}{\partial V}\right)_T dV + +\int_3^2 \left(\frac{\partial S}{\partial T}\right)_V dT +\end{equation} + +To evaluate the integral over $(1\rightarrow 3)$, note that since $dU += TdS - PdV$, the Helmholtz free energy function $F \equiv U -TS$ has +the property $dF = -SdT - PdV$; and of course $dF$ is an exact +differential since $F$ is a definite state function. The condition +that $dF$ be exact is, analogous to (1-22), + +\begin{equation} +\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial +P}{\partial T}\right)_V +\end{equation} + which is one of the Maxwell relations, discussed further in -where CV is the molar heat capacity at constant volume. Collec -ting these results, we have -3 -l 3 -1 nR log(V2/V1) + nCV log(T2/Tl) (1-52) -since CV was assumed independent of T. Thus the entropy function +Chapter 2. But [the value of this expression] is determined by the equation of state +(1-46): + +\begin{equation} +\left(\frac{\partial S}{\partial V}\right)_T = \frac{nR}{V} +\end{equation} + +Likewise, along the path $(3\rightarrow 2)$, we have + +\begin{equation} +\left(\frac{\partial S}{\partial T}\right)_V = \frac{n C_V}{T} +\end{equation} + +where $C_V$ is the molar heat capacity at constant volume. +Collecting these results, we have + +\begin{equation} +S_2 - S_1 = \int_1^3 \frac{nR}{V} dV + \int_2^3 \frac{n C_V}{T} dT = +nR\log{(V_2/V_1)} + nC_V \log{(T_2/T_1)} +\end{equation} + +since $C_V$ was assumed independent of $T$. Thus the entropy function must have the form -S(n,V,T) = nR log V + n CV log T + (const.) (l~53) - + +\begin{equation} +S(n,V,T) = nR \log{V} + n C_V \log{T} + (\text{const.}) +\end{equation} From the derivation, the additive constant must be independent of V and T; but it can still depend on n. We indicate this by writing -where f (n) is a function not determined by the definition (1-47). -The form of f (n) is , however, restricted by the condition that -the entropy be an extensive quantity; i .e . , two identical systems -placed together should have twice the entropy of a single system; -Substituting (l—-54) into (1-55), we find that f(n) must satisfy -To solve this, one can differentiate with respect to q and set -q = 1; we then obtain the differential equation -n f ' (n) — f (n) + Rn = 0 (1-57) -which is readily solved; alternatively, just set n = 1 in (1-56) -and replace q by n . By either procedure we find -f (n) = n f (1) — Rn log n . (1-58) -As a check, it is easily verified that this is the solution of -where A E f (l) is still an arbitrary constant, not determined -by the definition (l—l9) , or by the condition (l-55) that S be -extensive. However, A is not without physical meaning; we will -see in the next Section that the vapor pressure of this sub -stance (and more generally, its chemical potential) depends on -A. Later, it will appear that the numerical value of A involves + +\begin{equation} +S(n,V,T) = n\left[R \log{V} + C_V \log{T}\right] + f(n) +\end{equation} + +where $f(n)$ is a function not determined by the definition (1-47). +The form of $f(n)$ is, however, restricted by the condition that +the entropy be an extensive quantity; i.e., two identical systems +placed together should have twice the entropy of a single system; or +more generally, + +\begin{equation} +S(qn, qV, T) = q\cdot S(n,v,T),\qquad 0