A New Interpretation of Legendre’s Transformation And

687Mat.-wiss. u.Werkstofftech. 2012, 43, No. 8 DOI 10.1002/mawe.201200937

A new interpretation of Legendres transformation andconsequences

Eine neue Interpretation der Legendre-Transformation und Folgerungen

H.-J. Hoffmann

The Legendre transformation has found widespread application in thermodynamics, Hamilton-Lagrange-mechanics and optics. It attributes the values of the coordinates (x,y (x)) representing

the points of a monotonic piecewise smooth functional curve y (x) the slopes mx (x) dy x dx

and

the intercepts y (mx) of their tangents on the y-axis. Thus, the initial curve y (x) is represented by

the ordered set of all slopes mx (x) dy x dx

of its tangents together with their intercepts y (mx) on

the y-axis. It is shown that the transformed or conjugated function must basically be supple-mented by a homogeneous linear function of the relevant variable. This is usually neglected inthe literature. In addition, a new interpretation of the Legendre transformation is presentedand discussed: For this purpose the derivative mx (x) is considered as the proper initial functionand integrated between x0 and x. This integral is complemented by the integral of x (mx) ((theinverse function of mx (x)) over mx between mx0 = mx (x0) and mx (x), if mx (x) and x (mx) are strictlymonotonic. The sum of both integrals yields the area (xmx x0mx0). Legendres transformationis obtained by reordering the respective terms. The procedure of transformation corresponds tointegration by parts. Some examples and consequences of the properties considered are demon-strated and discussed using the simple model of two-state systems. The general results of thepresent work remove possible internal inconsistencies in thermodynamics.

Keywords: Legendre transformation / thermodynamic potential functions / Massieu-Gibbs functions /third law of thermodynamics / two-state systems /

Die Legendre Transformation wird in der Thermodynamik, der Hamilton-Lagrange-Mechanik undin der Optik hufig angewendet. Dabei werden den Werten der Koordinaten (x,y (x)), die die

Punkte einer monotonen stckweise glatten Kurve y (x) darstellen, die Steigungen mx (x) dy x dx

und die Abschnitte der Tangenten y (mx) auf der y-Achse zugeordnet. Somit wird die ursprngli-

che Kurve durch die geordnete Menge der Steigungenmx (x) dy x dx

zusammen mit den Abschnit-

ten all ihrer Tangenten y (mx) auf der y-Achse dargestellt. Es wird gezeigt, dass die transformierteoder konjugierte Kurve grundstzlich durch eine homogene lineare Funktion der jeweiligen Va-riablen ergnzt werden muss, was in der Literatur gewhnlich vernachlssigt wird. Zustzlichwird eine neue Interpretation der Legendre-Transformation vorgestellt und diskutiert. Hierzuwird die Ableitungmx (x) als die eigentliche Ausgangsfunktion betrachtet und zwischen den Wer-ten x0 und x integriert. Dieses Integral wird durch das Integral von x (mx) (die inverse Funktion vonmx (x)) ber mx zwischen mx0 = mx (x0) und mx = mx (x) ergnzt, wobei vorausgesetzt ist, dass mx (x)und x (mx) streng monoton sind. Die Summe beider Integrale ergibt die Flche (xmx x0mx0). DieLegendre-Transformierte erhlt man durch Umordnung der Terme. Die Prozedur der Transforma-tion entspricht einer partiellen Integration. Es werden einige Beispiele und Folgerungen aus denbetrachteten Eigenschaften an Hand eines einfachen Zwei-Zustnde-Modells gezeigt und disku-tiert. Die generellen Ergebnisse dieser Arbeit beseitigen mgliche innere Widersprche in derThermodynamik.

Schlsselwrter: Legendre-Transformation / thermodynamische Potenzialfunktionen / Massieu-Gibbs-Funktionen / dritter Hauptsatz der Thermodynamik / Zwei-Zustnde-Systeme /

Technische Universitt Berlin, Hardenbergstrae 40, 10623 Berlin

i 2012WILEY-VCH Verlag GmbH & Co. KGaA,Weinheim www.wiley-vch.de/home/muw

Corresponding author: Prof. a.D. Dr. Hans-Juergen Hoffmann, Fachge-biet Glaswerkstoffe, Technische Universitt Berlin, Hardenbergstrae40, 10623 BerlinE-mail: [email protected]

H.-J. Hoffmann Mat.-wiss. u.Werkstofftech. 2012, 43, No. 8

1 Introduction to the application of theLegendre transformation

Legendre's transformation is a powerful tool to switch from onefunction y x of a given variable x to a new function of which theindependent variable mx mx x dy x dx is the derivative ofy x claiming that no information given by y x is lost. It is usedto switch between thermodynamic potential functions [16],between Lagrange and Hamilton functions [7] or in optics [8, 9].To be specific in the following, the transformation of thermody-namic potential functions is considered. The results, however,are not restricted to thermodynamics but are applicable also toother fields, whenever the Legendre transformation is used.The paper was motivated by a new interpretation of the mean-

ing of the Legendre transformation to be presented in detail insections 4 and 5.Afore, sections 2 and 3 remind the fundamental motivation

and standard interpretation of the Legendre transformationtogether with some basic properties. General results are that aconstant function is retained whereas a homogeneous linearfunction yields zero. Thus, functions connected by a Legendretransformation may require special considerations, if a homoge-neous linear contribution appears or is missing, such asS0T / T or T0S / S, e. g., in the energy-like thermodynamicpotential functions (T , temperature; T0, special value of the tem-perature; S, entropy; S0, special value of the entropy). Such linearhomogeneous contributionsmay be due to the entropy of mixing(whichmay increase from S 0 to S0 at a given temperature T0),the residual entropy of frozen-in defects or the creation of (non-equilibrium) defects or structural phase transitions of first order,just to give some examples.

Finally, the properties of the Legendre transformation areillustrated in section 6 for the internal and the free energy of two-state systems as a special example from thermodynamics. Somegeneral consequences of the present findings are summarized insection 7.

2 Conventional introduction of the Legendretransformation

In order to shift from the variable x of the initial function y x tothe conjugated variable mx mx x dy x dx one might assumethat it would be sufficient to replace x by the function x x mx which is the inverse of the derivative of y x . However, this sub-stitution is not helpful to obtain a unique function of mx as canbe seen by the following arguments (see [14]): Starting with thefunction y x and its derivativedy x dx

mx 1

under the proviso that (1) can be solved unambiguously withrespect to x, and inserting the result x x mx as a substitutionfor x into y x we obtain a new function y x mx ysub mx or

y ysub dydx

: 2

Solving (2) with respect to dy=dx yields

dydx

y1sub y 3

where y1sub designates the inverse function of ysub provided that aninverse exists. In this respect it is reminded that an inverse existsonly for intervals in which the function is strictly monotonicincreasing or decreasing.The differential equation (3) does not depend explicitly on x.

Thus, if y x is a solution of (3) also y x const: correspondingto a one-dimensional infinite plurality of solutions. Therefore,simple substitution x x mx does not yield a unique represen-tation of the original function y x in reverse. This ambiguityresults in further difficulties if y is a function of additional varia-bles v;w; :::z. The problem of ambiguity seems being solved ifeach element of the set of coordinates x; y x or each point ofthe original curve is attributed the respective tangent with theslope given as the derivative (1). This is shown in Fig. 1. A pointwith the coordinates x1; y1 x1 is chosen on the curve y x as anexample. The tangent at that point with the slopemx1 is depictedby the dashed line. That straight line intersects the y-axis at yy1which is determined by the shift yy1 y1 calculated from the slopeof the tangentmx1 y1yy1x10 or

mx1 y1 yy1x1 : 4

Eq. (4) yields

yy1 y1 x1mx1 y1 x1 dy x dxxx1

: 5

Thus, an initial point with the coordinates x1; y1 x1 is attrib-uted a tangent characterized by its slope mx1 and its intercept yy1on the y-axis (represented by the pair of numbers mx1;yy1 ). yy1depends on mx1, x1 and y1. The original value x1 of the x-coordi-nate is redundant, since it is buried in the data characterizing thetangent (as x1 in the pair of coordinates of the point of osculationand in the slope of the tangent). Both numbers x1; y1 x1 areused in the simple instruction how to perform the transforma-tion and to obtain both the slope mx1 and the intercept yy1 on they-axis.This procedure is represented in Fig. 1 for just one point of the

curve y x . To make it work for all points on y x in Fig. 1 wedrop in (5) the special index 1, introduce on the right-hand sideformx1 the new variable

mx dy x dx mx x 6

and replace in the second term and in the expression y x1 thevalue x1 by

x x mx : 7Eq. (7) is obtained by solving (6) for x as a function ofmx .Generalizing (5) yields the Legendre transform or conjugated

function

yy mx y x dy x dx x y x mx mxx mx : 8

688


Mat.-wiss. u.Werkstofftech. 2012, 43, No. 8 A new interpretation of Legendres transformation and consequences

The first part of this equation is the coded rule, whereas thesecond part describes the practical performance, as it points atthe essential step how to determine the functional yy mx -depend-ence by inserting x x mx . Since (8) describes the operationwhen performing the Legendre transformation, it may also becalled Legendre's operator. If with the proviso mx 6 const: Eq.(6) cannot be solved explicitly for x one inserts numerical data forx andmx x and determines the transformation numerically.To demonstrate that yy mx is indeed independent of x we ver-

ify that the derivative of (8) with respect to x is zero according to

dyy mx dx

dy x dx

mx dx mx dx mx mx 0: 9

Often dimensions are attributed to the numerical values of thefunctions. As yy mx represents the points of the intercepts of therespective tangents with the y x -axis, the dimension of thetransformed function is necessarily the same as the initial func-tion.

Since the new variable mx dy x dx mx x is obtained as thederivative of the initial function with respect to the initial varia-ble, the pair x;mx is a conjugated pair of variables and the pairy x ;yy mx is called a conjugated pair of Massieu-Gibbs orpotential functions [2].

Applying the Legendre transformation (8) a second time, oneexpects to return to the original variable and function y x . Thisis seen by inserting the appropriate expressions into (8)

Thus, the Legendre transformation is self-inversing. From thecurled bracket in the last line of (10) we see that the conjugatedvariable ofmx in yy mx is x ordyy mx dmx

x : 11

As an alternative to (8) we may define the Legendre transfor-mation or operator by

yy mx y x dy x dx x

y x mx mxx mx f g

mxx mx y x mx 12without loosing information, since yy mx corresponds just tothe image of yy mx mirrored around the mx-axis. Merging bothrepresentations (8) and (12) of the Legendre transformation oneobtains

yy mx y x mx dy x dx x mx

y x mx mxx mx f g : 13

Both transformations are used in the literature, depending onthe special applications. In the following, however, we restrict thediagrams and the discussion to the transformation yy mx yy mx . The results can be applied also to the transformationwith the negative sign by considering the mirror symmetrybetween both representations.

3 Further properties of the Legendretransformation

3.1 Properties of the functions to be transformed

In order that the Legendre transformation works properly y x must not be a constant y a0 const: or a function proportionalto x such as y x a1 x, since in these cases (6) yieldsmx dydx 0 or mx

d a1x dx

a1. In the first case we have noproper variable or its conjugate and in the second case the inversex x mx of the new variable mx x a1 does not exist to beinserted into (8), since the slopemx of a straight line consists of asingle value and does not represent a proper variable either.Referring to a graph, we consider to transform a linear func-

tion y x a0 a1 x in the coordinate system of Fig. 1. Then,all tangents to this linear function possess the same intercepty 0 a0 on the y-axis. The conjugated function shrinks to asingle constant value yy a0 and a conjugated variable is not

defined. Thus, in the interval under consideration necessarilydy x dx

mx 6 const: say d2y x dx2

> 0 ord2y x dx2

< 0, i. e. the func-

tion y x must be truly convex or concave. In Fig. 1 a convex func-tion (seen from below) has been chosen for illustration.

3.2 A possible extension of Legendres transformation

In section 2, the intercept of the y-axis by the tangent has beeninterpreted as the new function yy mx . This procedure andassignment is arbitrary. In principle one can attribute the inter-cept of the tangent on any parallel of the y-axis a new function of

the slope variabledy x dx

mx . A parallel to the y-axis is depic-ted in Fig. 2 for illustration. The intercept of the tangent on thatparallel (which is shifted by a constant xa from the y-axis) can becalculated by analogy to (4) and (5) from

y1 ayy1 mx1 x1 xa 14or

ayy1 y1 mx1x1 mx1xa: 15

689


yy mx dyy mx dmx mx y x mx mxx mx f g d y x mx mxx mx f g

dmxmx

y x mxxf g dy x mx dxdx mx dmx

x mx mx dx mx dmx

mx 10

y x mxxf g mx dx mx dmx x mx mxdx mx dmx

mx y x :


Replacing x1 and mx1 as the variables x and mx the last equa-tion can be generalized to

ayy mx y x mx mxx mxxa yy mx mxxa: 16Here an additional homogeneous linear function mxxa with

xa const: appears. On the other hand, one also realizes howsuch an additional termmxxa is lost in the Legendre transforma-tion by choosing the special value xa 0, i. e. the intercept of thetangent on the y-axis. The same is true if we apply the Legendretransformation a second time. Then, we may add (or subtract,depending on the shift of the parallel to the yy-axis) mxax withmxa const:Thus, any pair of Legendre conjugates are a priori indefinite

with respect to additional functions mxxa or xmxa, which have tobe determined separately according to the problem to be solved.Of course, this includes the possibility that these homogeneouslinear functions may vanish. Both parameters, the absolute shifta0 and the linear homogeneous functions mxax or mxxa, can beused to adjust the potential functions y x or yy mx to experimen-tal observations or theoretical requirements under consideration.

This consequence has not yet been taken into account adequatelyin the literature when introducing the Legendre transformationand dealing with it.

4 A new interpretation of the Legendretransformation

4.1 Motivation

In the standard interpretation of the Legendre transformation[15] the functional curve is represented by the set of its tangentsas shown in the previous sections. This interpretation bearssome difficulties for the beginner, since just one point of eachtangent, namely the point of osculation, coincides with the initialcurve. The other points of each tangent do not represent the ini-tial curve at all, since they do not coincide generally with any ofthe points of the initial curve (Fig. 1). The transformed curve,too, has only one point in common with each tangent, namelythe intercept of the tangent on the y-axis (i. e. for x 0), whicheven may not belong to the interval of definition of y x , if thevariable x possesses a lower limit at x0 > 0, Fig. 1. In addition,the value of the intercept on the y-axis is attributed necessarily anew name as a function of the new variablemx , although the tan-gents are shown in the initial x y-coordinate system.Instead of the intercept of the tangent with the y-axis one may

choose an intercept on a parallel to that axis. This more generalcase with xa 6 0 as described in section 3.2 has not yet been con-sidered in the literature. Thus, the interpretation of the Legendretransformation by a set of tangents is incomplete and may cause

690


Figure 1. Constructing the Legendre transformation of a functiony x . The tangent at the point x1; y1 x1 is defined by the equationy1 x mx1x yy1 with the slopemx1 dy x dx

xx1

obtained from the

derivative of y x at x x1. It intercepts the yaxis atyy1 y1 x1 mx1x1. To obtain the complete Legendre transform thespecial value x1 of the coordinate is replaced by the variable x.

Bild 1. Graphische Darstellung zur Herleitung der Legendre-Transfor-mation von y x . Die Tangente im Kurvenpunkt x1; y1 x1 ist defi-niert durch die Gleichung y1 x mx1x yy1 mit der Steigungmx1 dy x dx

xx1

, die sich aus der Ableitung von y x bei x x1ergibt. Sie schneidet die y-Achse bei yy1 y1 x1 mx1x1. Um dieLegendre-Transformierte vollstndig zu erhalten, wird der spezielleWert x1 der Koordinate durch die Variable x ersetzt.

Figure 2. Same as Fig. 1 but supplemented by the tangent intercept-ing a parallel a shifted from the y-axis by xa. The value of this inter-ception by the tangent is ayy1 y1 x1 mx1xmx1xa.Bild 2. Wie Bild 1, jedoch ergnzt durch den Schnittpunkt der Tan-gente mit der Parallelen a, die um xa gegen die y-Achse verschobenist. Dieser Schnittpunkt der Tangente liegt bei ayy1 y1 x1 mx1xmx1xa.


some confusion. In the following sections 4.2 and 4.3, this tradi-tional approach is avoided by comparing suitable integrals of thevariables entering the pair of conjugated or transformed func-tions.

4.2 A special case

Using the notions and definitions from sections 2 and 3, whendescribing Legendres transformation, the function y x may berepresented generally by the definite integral

y x y x0 Zx

x0

dy x dx

dx Zx

x0

mx x dx: 17

That integral represents the area between the xaxis and thefunction mx x (which only needs being integrable) between thelimits x0 and x. In (17) we have assumed that x0 andmx x0 mx0 are the coordinates of the starting point or lowerboundary of the interval of integration. To simplify argumentsand to demonstrate the essentials of the following procedure, wehave adjusted in Fig. 3 the x and the mxaxis in such a waythat x0 0 and mx x0 mx0 0. Furthermore, y x and yy mx (which will be defined below) are shifted in such a way thaty x0 0 0 and yy mx0 0 0. Then (17) simplifies to

y x Zx

0

dy x dxdx Zx

0

mx x dx: 18

In Fig. 3 A we do not represent y x by its tangents but by theintegral (18) which is the (criss-cross) shaded area below the

curve mx x dy x dx in the interval between x0 0 and thevariable upper boundary x on the xaxis. The function y x represents unambiguously the area as a function of the upperlimit x of the interval of integration.

That area under the curve mx x dy x dx with respect to thexaxis may be complemented by the integral of the same curvewith respect to the mxaxis on the left of Fig. 3 A. To calculatethat integral with respect to the mxcoordinate we must firstsolve mx x dy x dx for x mx as the integrand. x mx is theinverse ofmx x . The corresponding integral reads generallyZmxmx0

x mx dmx yy mx yy mx0 ; 19

where mx mx x and mx0 mx x0 are the values of the inte-grand of (17) at x and x0. (We have extracted a minus sign on theright-hand side of (19) to be consistent with (5).) For the presentchoice of parameters x0 0 and mx x0 mx0 0, however,(19) simplifies to

Zmx0

x mx dmx yy mx : 20

Since the integral is positive for allmx > 0, the function yy mx is negative. The sum of (18) and (20) yields

y x yy mx Zx

0

mx x dx Zmx0

x mx dmx xmx; 21

which is represented for a special set of parameters x and mx inFig. 3 B by the criss-cross area. If we subtract area B from area Awe obtain Fig. 3 C or

yy mx Zmx0

x mx dmx Zx

0

mx x dx xmx

y x xmx: 22

691


Figure 3. Legendre transformation demonstrated for a function which is represented by a definite integral, (A) The function y x Rx0mx x dx to

be transformed is represented by a definite integral with mx x dy x dx , (B) The product mx x x to be subtracted from y x in (A), (C) The proce-dure A-B=C results in the Legendre transform or the conjugated function yy mx

Rmx0x mx dmx represented by a definite integral. Reversing

the sign of the Legendre transform complements y x yielding mx x x.Bild 3. Legendre-Transformation einer Funktion, die durch ein bestimmtes Integral dargestellt wird. (A) Die zu transformierende Funktion

y x Rx0mx x dx ist dargestellt durch ein bestimmtes Integral, wobeimx x dy x dx . (B) Das Produktmx x x, das von y x in (A) abzuziehen ist.

(C) Das Ergebnis der Differenz A-B = C ist die Legendre-Transformierte oder die konjugierte Funktion yy mx Rmx0x mx dmx, die ebenfalls durch

ein bestimmtes Integral dargestellt wird. Nach Umkehrung des Vorzeichens ergnzt die Transformierte die ursprngliche Funktion y x zumx x x.


The result yy mx for the example of Fig. 3 A as shown inFig. 3 C illustrates the meaning of Legendre's transformation: Itis now formally expressed by A B C wherein the symbolsrepresent the relevant areas in the charts of Fig. 3.Eq. (22) coincides perfectly with the Legendre transformation

(8). Thus, the Legendre transform of a function represented bythe integral (18) with the upper boundary x corresponds to thenegative complementary area to fill the rectangle xmx, wheremxis the value of the integrand at the upper boundary.

4.3 The general case

The general case is illustrated in Fig. 4. Summing up the inte-grals (17) and (19) we obtain the area given by xmx x0mx0, i. e.Zx

x0

mx x dx Zmxmx0

x mx dmx xmx x0mx0 23

or

y x y x0 yy mx yy mx0 xmx x0mx0 : 24Each integral (17) and (19) depends on a different variable x or

mx related by the definitions (6) and (7).Eq. (24) can be rearranged to give

yy mx y x xmx yy mx0 y x0 x0mx0 : 25Since the left-hand side of (25) depends on the variables x and

mx mx x (representing the upper boundary) whereas theright-hand side depends on different independent parameters x0andmx0 mx x0 (representing the lower boundary), both sidesmust be a constant. The constant is zero if we define for the right-hand side

yy mx0 x0 y x0 x0mx0 x0 :This yields consistently that the left-hand side of (25) is also

zero or

yy mx y x mx x mx mx y x mx x mx mx 26and

y x yy mx x xmx x : 27If the constant is different from zero, it may shift the functions

y and yy on both sides of (25). Thus, (26) and (27) are exactlyLegendre transformations defined and obtained without the tan-gential interpretation and construction. Eq. (27) shows that theconjugated variable of mx is x . Depending on which integralin (23) is easier to solve, we obtain the complement with a nega-tive sign just by subtracting xmx x0mx0 .

Eqs. (26) and (27) show the perfect correspondence with thedefinition (8) of the Legendre transformation, just the derivationand interpretation are different: The transformed function is nolonger interpreted as the intercept of the yaxis by the tangent tothe initial curve, but as the area needed to complement the initialintegral to yield the rectangle or product xmx of the conjugatedvariables.

5 A tautology: The Legendre transformationcorresponds to integration by parts

The indefinite integralRmx x dx reminds us to apply the rule of

integration by parts, which reads in this particular case

y x Z

mx x dx mx x x Z

x mx dmx : 28

Rearranging (28) we obtain

yy mx Z

x mx dmx Z

mx x dx mx x x

y x mx mxx mx 29which also corresponds exactly to the Legendre transformationEq. (8). Thus, the Legendre transformation results from an inte-gration by parts. In reverse, one may now reinterpret the tech-nique of integration by parts based on the following simple con-siderations: The left-hand side of (28) is the integral to be solved.It is expressed by the area of the rectangle mx x x (first term onthe right-hand side of (28)), from which one has to subtract thearea of the complement of the first integral needed to fill up therectangle (second term of right-hand side of (28)). This comple-ment is expressed by a different integral which may be solvedeasier than the initial one. This is the known motivation to applythemethod of integration by parts.

Of course, inserting mx a1 as the derivative of y x a1xinto (29) one obtains yy mx

Rx mx dmx

Ra1dx a1x

0, which agrees with the statement in section 3.1 that thehomogeneous linear function y x a1x does not yield a conju-gated function with a conjugated variable.On the other hand, onemay use the derivative of (28) to obtain

mx x dx d mx x x x mx dmx: 30This is a special application of the differentiation of a product

692


Figure 4. Areas of integrals to be considered for the transformation, ifthe lower boundary of the integration and the integrand at the lowerboundary are different from zero.

Bild 4. Integralflchen, die bei der Transformation zu bercksichtigensind, wenn die untere Integrationsgrenze und der Integrand bei derunteren Integrationsgrenze nicht null sind.


d u x v x dx

du x dx

v x u x dv x dx

31

or after rearranging and considering x just as an auxiliary varia-ble

u v dv d u v v vdu d uv u v u du 32which can be cast into (30) if we set u x mx x , v x x,du dmx and dv dx. Fig. 5 illustrates the meaning of (32) asan area for finite differencesu andv

uv u vv vv u v v

u vv u v v vv v uuv

vu uvuv 33which yields (32) in the limitv! 0 andu! 0 after rearrang-ing.

6 Discussion and consequences

6.1 An example: Thermodynamic functions of two-statesystems

The present approach is to be applied to two-level or two-state sys-tems, which can be treated conveniently. We consider one mole,NA, of such independent systems with the states of energy e0 0(ground state, index 0) and e1 > 0 (excited state, index 1),Fig. 6. The probabilities to find such a system under thermalequilibrium in the ground or excited state are

p0 T 11 exp e1

kBT

34

and

p1 T exp e1

kBT

1 exp e1kBT

; 35

wherein exp e0kBT

1 and exp e1

kBT

are the Boltzmann

factors with the Boltzmann constant kB and the temperature T .The ratio of Boltzmann factors indicates the ratio of the occupa-tion probabilities of the respective states, whereas the denomina-tor of (34) and (35) represents the sum of all Boltzmann factorsof a given system. Fig. 7 shows the occupation probabilities p0 T and p1 T as a function of the temperature T in units of e1=kB .At low temperature, nearly all systems occupy the ground state,whereas with increasing temperature the occupation probabilityapproaches 0.5.The internal energy of one mole NA of such systems is given

by

U T NA 0p0 T e1p1 T

NAe1exp e1

kBT

1 exp e1kBT

: 36

693


Figure 5. Complementary areas resulting from differences u > 0andv > 0 of the variables u and v.

Bild 5. Sich ergnzende Flchen, die sich aus den Variablen u und vmit den Differenzenu > 0 undv > 0 ergeben.

Figure 6. Energy scheme of a two-state system.

Bild 6. Energieschema eines Zwei-Zustnde-Systems.

Figure 7. Occupation probabilities p0 T and p1 T of the ground andthe excited state as a function of the temperature, T , in units e1=kB (e1is the energy difference between excited and ground state, kB is Boltz-mann's constant).

Bild 7. Besetzungswahrscheinlichkeiten p0 T und p1 T des Grund-und des angeregten Zustands in Abhngigkeit von der Temperatur Tin Einheiten von e1=kB (e1 ist die Energiedifferenz zwischen Grund-und angeregtem Zustand, kB ist die Boltzmann-Konstante).


The internal energy is not a potential function of the variabletemperature T , but of the variables entropy S, volume V ,particle number N and further possible extensive variables tobe specified if necessary, i. e.

dU S;V ;N TdS pdV dN 37with the pressure p and the chemical potential . Consideringthe volume and the particle number, N NA, as constant for thepresent consideration, we have to determineU S or

dU dU S dS

dS TdS with dU S dS

T: 38

Since the entropy is a strictly monotonic increasing functionof the temperature, S T , if the other variables are kept constant,we can consider the temperature as a parameter variable. Thenwe obtain from (38) the well-known general relation

dU S T dT

CV T dU S dSdS T dT

T dS T dT

39

between the molar specific heat capacity, CV T , and the molarentropy capacity,

dS T dT

CV T T

, at constant volume. Inserting

(36) we obtain themolar specific heat capacity

CV T dU T dT NAkBe1

kBT

2 exp e1kBT

1 exp e1kBT

2 40

and the specific molar entropy capacity

dS T dT

CV T T

NAkBT

e1

kBT

2 exp e1kBT

1 exp e1kBT

2 : 41

In Fig. 8 the molar quantities of the internal energy, U T , thespecific heat capacity, CV T , entropy, S T , and specific entropycapacity, dS T =dT CV T =T , are shown as functions of thetemperature, T , in units of e1=kB . U T approaches NAe1=2with increasing temperature, whereas CV T and dS T dT

CV T T

pass through maxima at kBTmax=e1 0:4168 with the valueC Tmax;NA 0:4392NAkB and kBTmax

e1 0:3083 with the value

C T ;NA T

TTmax

1:2335NAk2B

e1, respectively.

Both CV T and dS T dT CV T T

approach zero with increas-

ing temperature: This occurs when the occupation probability ofthe ground and the excited state approach 1/2. Then the molarstorage capacities of internal energy and entropy in the two-statesystems are exhausted. Integrating (41) we obtain the molarentropy as a function of the temperature

S T;NA ZT

0

C T;NA T

dT NAkB n 1 exp e1kBT

e1kBT

exp e1kBT

1 exp e1kBT

42

which approaches with increasing temperaturelimT!1

S T ;NA NAkBn 2 .Eqs. (36) and (42) are parametric representations of the molar

internal energy, U, and entropy, S, with the parameter T. Com-bining U T and S T for the same value of the parameter T weare able to represent the molar thermodynamic potential func-tion U S in Fig. 9. To apply the results of sections 4 and 5 weneed the derivative of U S or the dependence dU S

dS T S ,

which is just the inverse of (42). It is depicted by the solid curve

in Fig. 10 (A). The integral U S RS0T S dS is represented by

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Figure 8. Molar internal energy, U T , entropy, S T , specific heatcapacity, CV T , and specific entropy capacity, dS T =dT CV T =T , atconstant volume as a function of the temperature, T , in units of e1=kBof a two-state system.

Bild 8. Molare innere Energie U T , Entropie S T , spezifische Wrme-kapazitt CV T und spezifische Entropiekapazitt dS T =dT CV T =T bei konstantem Volumen als Funktion der Temperatur T inEinheiten von e1=kB eines Zwei-Zustnde-Systems.

Figure 9. Molar internal energy, U S , at constant volume as a func-tion of the molar entropy, S, in units of NAkB, of a two-state system.

Bild 9. Molare innere Energie U S bei konstantem Volumen als Funk-tion der molaren Entropie S in Einheiten von NAkB eines Zwei-Zu-stnde-Systems.


the (criss-cross) shaded area in Fig. 10 (A) for a special value of S.

If we subtract from U S the product TS dU S dS

S shown in

Fig. 10 (B) by the rectangle), we obtain the complement

RT0S T dT F T known as the molar thermodynamic poten-

tial function free energy with the variable T . It is given as theintegral of S T which is shown in Fig. 10 (C) by the solidcurve. The sequence from Fig. 10 (A) to (C) shows the transitionfrom the molar internal energyU S with the variable entropy,S, to the molar free energy F T with the variable temperature,T dU S

dS, by the Legendre transformation. F T is given in

Fig. 10 (C) by the area between S T and the Taxis as a func-tion of the upper boundary T. The appertaining integral functionF T is represented by the full curve in Fig. 11.

6.2 An example for a homogeneous linear function

The effect of a linear homogeneous function is shown by extend-ing the example of the two-state systems from the previous sec-tion 6.1. The molar entropy of the ensemble of those systems

starts from S 0 0 for T 0 as given in Eq. (42). This is true ifthe number of possible positions or lattice points of the two-statesystems is exactly as large as the number of two-state systems.Then, only one ordering is possible.

If the distribution of the two-state systems over their possiblelattice points is not specified but arbitrarily (this occurs if thenumber of lattice points exceeds the number of two-state sys-tems), one does not know which lattice point is occupied andwhich one is not. Such an ensemble carries depending on themixing ratio a constant amount of entropy of mixing S0 even atT 0 and the total molar entropy as a function of temperature isgiven by the sum

S T S0 S1 T 43where we changed the name of the entropy defined by (42) intoS1 T . Starting from S 0 the entropy is an increasing functionof the mixing ratio. Thus, the value of S0 depends on the mixingratio. It characterizes the individual mixing of the two-statesystems, not the entropy due to the occupation of their internalstates. Furthermore, we assume that each of the two-statesystems carries energy in its internal states, only. Then, theentropy of the mixing does not carry energy. For this case the

695


Figure 10. An example of the Legendre transformation demonstrating the transition from the molar internal energy U S RS0T S dS, which

is represented by a definite integral (see Fig. 3), to the molar free energy F T . (A) The function molar internal energy U S RS0T S dS

to be transformed represented by a definite integral with T S dU S dS

, (B) The product T S S to be subtracted fromU S RS

0T S dS in (A). (C) The result of the Legendre transformation or the conjugated function molar free energy F T RT

0S T dT

represented by a definite integral or the result of the procedure A B = C.

Bild 10. Beispiel fr eine Legendre-Transformation dargestellt an Hand des bergangs von der molaren inneren Energie

U S RS0T S dS, die durch ein bestimmtes Integral dargestellt ist (siehe Fig. 3), zur molaren freien Energie F T . (A) Die zu transformier-

ende Funktion U S RS0T S dS ist durch ein bestimmtes Integral dargestellt, wobei T S dU S =dS. (B) Das Produkt T S S, das von-

U S RS0T S dS in (A) abzuziehen ist. (C) Das Ergebnis fr die Legendre-Transformierte oder konjugierte Funktion molare freie Energie

F T RT0S T dT , die durch ein bestimmtes Integral dargestellt wird, bzw. das Ergebnis der formalen Vorgehensweise A-B = C.


temperature T as a function of S is depicted in Fig. 12 fromwhichone calculates the molar free energy as

F T ZT

0

S T dTF T

ZT

0

S T dT S0T ZT

0

S1 T dT 44

and the molar internal energy

U S ZS

0

T S dS ZS

S0

T S dS 45

The result for the molar free energy (44) is shown in Fig. 13 by

the dashed curve for comparison with RT0S1 T dT (solid curve

from Fig. 11). The absolute difference between the curves is S0T .Differentiating (44) with respect to T , one obtains the correct

molar entropy S dF T dT

S0 S1 T .In agreement with our expectations, we calculate from the

molar free energy the correct molar internal energy using theLegendre transformation as follows

U S F T T dFdT

S0T ZT

0

S1 T dT T S0 S1

TS1 ZT

0

S1 T dT

S S0 T ZT

0

S S0 dT

ZS

S0

T S dS : 46

Starting with the molar internal energy in reverse, one may setthe zero of the entropy at S0 forU S . Then we have the same sit-uation for F T as in Figs. 10 and 11. In fact, we will show thatthe contribution S0T in (44) is just a mathematical term due tothe residual entropy S0 at T 0. Usually entropy added to a sys-tem causes an increase of the temperature. Thus, the entropy is amonotonically increasing function of the temperature. The mix-ing entropy S0, however, is independent of the temperature (seetextbooks on thermodynamics, such as [10]). Here we have thestrange situation that part of the entropy, namely the mixingentropy represented by a constant, does not increase the temper-ature, whereas the other part is a function of the temperature(then, the increase of entropy is necessarily related to an increaseof temperature). The entropy S0 depends on the number ofvacant places for the two-state systems and on the degree of mix-ing. Thus, the molar free energy at any temperature far from

696


Figure 11. Molar free energy, F T , as a function of temperature, T , inunits of e1=kB of a two-state system.

Bild 11.Molare freie Energie F T in Abhngigkeit von der TemperaturT in Einheiten von e1=kB eines Zwei-Zustnde-Systems.

Figure 12. T S dependence of a two-state system in analogy toFig. 10 A with the residual entropy S0 at T 0. While the integralU S RS

0T S dS corresponds to that of Fig. 10, the complementary.

integralRT0S T dT possesses an additional contribution S0T which is a

homogeneous linear function of the temperature. Note that

F T RT0S T dT is the molar free energy at constant volume and.

constant particle number.

Bild 12. T S Abhngigkeit fr ein Zwei-Zustnde-System analog zuFig. 10 A jedoch mit einer Restentropie S0 fr T 0. Whrend dasIntegral U S RS

0T S dS gleich ist wie in Fig. 10 hat nun das komple-

mentre IntegralRT0S T dT einen zustzlichen Beitrag von S0T , einer

homogenen linearen Funktion der Temperatur. Man beachte, dass

F T RT0S T dT die molare freie Energie bei konstantem Volumen

und konstanter Teilchenzahl ist.


T 0 depends just on the entropy S0 at zero temperature. Thisis confusing, since S0 characterizes rather the distribution of thetwo-level systems under consideration as an individual propertyof mixing, not its thermal behaviour. The molar free energy pos-sesses a contribution S0T (the area on the left-hand side of thedotted vertical line at S0 in Fig. 12 which increases linearly withthe temperature T) without a real exchange of energy. It is notrelated to a real increase or decrease of energy with the temper-ature, T . As the contribution S0T does not account for a realexchange of energy in the thermodynamic energy-like potentialfunctions, such a formal or artificial contribution to the freeenergy has to be removed by adding or subtracting a suitablehomogeneous linear function.This is supported by the following equivalent consideration:

The molar energy of the mixing has been assumed being zero,whereas the molar entropy of mixing S0 > 0. As the molarentropy of mixing cannot be removed by thermal conduction, weare not able to remove even part of it by lowering the temperaturefrom say T2 to T1. Only the difference of the free energy

F T2 F T1 RT2T1

S1 T dT can be exchanged, whereas thedifference S0 T2 T1 has no practical influence when usingthe free energy.Thus, including the entropy of mixing S0 into the entropy is

useless or even misleading. Its contribution must be removed byadding to (44) the homogeneous linear function S0T , whichyields the former result without mixing entropy (solid curve inFig. 13). Adding S0T removes the misleading individual (due tothe individual degree of mixing) characteristic from the system.Thus, one cautiously has to correct contributions in the conju-gated potential function due to horizontal parts of T S in theinterval 0 S S0 by taking profit of the additional homogene-ous linear functions.

Even if the molar mixing entropy S0 carries molar mixingenergy E0 (such as for the creation of a constant concentration ofvacancies and interstitials in a crystalline solid) the same consid-erations are valid, since that energy can be exchanged only if theentropy of mixing is removed. The problem described here is notan exception but occurs due to the fact that the mixing entropy isindependent of the temperature. A similar effect occurs if oneconsiders phase transitions of first order, where the entropy alsovaries at constant temperature. The same is true if one considersand includes the entropy of information (which corresponds toentropy of mixing).

As for other variables, such as the volume V p , we considerthat V changes under constant pressure p const: at phase tran-sitions of first order, such as in the melting and in particular in the boiling transition. Then, onemust perform considerationssimilar to those for the entropy above and add or remove addi-tional homogeneous linear functions of the variable volume V orpressure p, depending on the respective potential function.

Confirming the results in section 3.2. linear homogeneouscontributions necessarily appear in the thermodynamic potentialfunctions due to constant contributions of the conjugated variableswhereas the concomitant variables are changing. Such constantcontributions require special considerations.

One may note that the Legendre transformation of (46) fromthe molar free energy F T with the variable T to the molar inter-

nal energy with the variable S makes S0T disappear, as can beconcluded generally from Eq. (16) of section 3.2, and yields

immediately (see also Fig. 12 for the last step)U S RSS0

T S dS:A constant amount of energy, which may be carried with theentropy of mixing S0, has to be included in the molar internalenergy.

Alternatively, transforming (46) with the shifted variableS1 S S0 to the new one T we obtain the useful F T inFigs. 11 and 13. For simplicity, one may presume that theentropy S T 0 0 and omit a possible constant contributionof the entropy at T 0. Then the molar internal energy is writ-ten as U S RS

0T S dS instead of (46) for the given mixed

system. Applying the Legendre transformation to this functionyields F T as shown in Fig. 11 or the solid curve in Fig. 13. Thisexample illustrates how caremust be taken to deal with all contri-butions to the variables and conjugated variables in the thermo-dynamic potential functions. It applies in particular to the resid-ual entropy (such as the mixing entropy) in the limit T 0. Inthis respect it is reminded that the definition of the free energy instatistical thermodynamics (see [4, 10], e. g.)

F T ;V kBTnZ 47ignores discussing a term S0T . (Z is the partition function,which depends on the temperature, T , and the volume, V .) Thus,any entropy of mixing, which is independent as a function oftemperature, is ignored, if one applies the relation

S T dF T dT

to (47). This is justified only if such contributi-

697


Figure 13.Molar free energy, F T RT0S T dT , as a function of tem-

perature, T , in units of e1=kB of a two-state system similar to Fig. 11but with the entropy given by S T S0 S1 T , wherein S0 is theresidual entropy at T 0 (dashed curve). The solid curve representsF T from Fig. 11.Bild 13.Molare freie Energie F T RT

0S T dT in Abhngigkeit von

der Temperatur T in Einheiten von e1=kB eines Zwei-Zustnde-Sys-tems hnlich wie in Fig. 11 aber mit der Entropie S T S0 S1 T ,wobei S0 die Restentropie fr T 0 bedeutet (gestrichelte Kurve). Diedurchgezogene Kurve entspricht F T von Bild 11.


ons do not play a role and do not change during the processes,such as in most cases of mixed isotopes of a given chemical ele-ment.Another example may be taken from [11, 12], where melting of

one-component systems is shown to be a mixing process andglass formation a freezing-in of the mixed units or particles. Theglass contains necessarily residual (mixing) entropy S0 andenthalpy H0 in the limit T 0 (enthalpy, since glass formationoccurs normally under constant pressure p0) below the meltingtemperature.

A mixing effect with a constant mixing entropy and enthalpymust also be taken into account for homogeneous crystallinematerials if the defects are frozen-in upon cooling crystals fromhigh temperature when defects, such as vacancies or interstitials,are thermally induced with a concentration depending stronglyon the temperature. After freezing-in during cooling, however,their residual concentration is constant. In alternative mecha-nisms defects may be produced athermally by different kinds ofradiation, ion implantation, cold working or doping. The mini-mum energy and entropy to produce a constant concentration n0of such mixing units are larger than zero. This causes a temper-ature-independent contribution to the entropy S n0 and to theenergy-like potential functions if the other variables are kept con-stant.

7 Summary

The Legendre transformation is widely used to switch betweenthermodynamic potential functions, between Lagrange andHamilton functions or in optics. Usually that transformation isexplained and visualized by a tangential construction accordingto section 2. In the present article, a new interpretation of theLegendre transformation is demonstrated, if the function to betransformed y x is represented by an integral y x y x0 Rx

x0mx x dx. Adding the complement yy mx yy mx0

Rmxmx0

x mx dmxwe obtain

y x y x0 yy mx x yy mx0 xmx x x0mx0 : 48

This yields the Legendre transformation

yy mx x y x xmx x 49

if we define

yy mx0 y x0 x0mx0 : 50

Usually the potential function y x is adjusted to a specialvalue by adding a constant a0 valid for a given set of variables

(typically: temperature T 298:15K, pressure p 101325Paand the quantity of substance or number of particles or units ofNA 1mole 6:02214179 1023particles). Once the necessaryconstant is added, the conjugated or Legendre transformed func-tion is adjusted automatically, since the constant is retained inthe transformation (see section 3.1).

On the other hand, the variable x (or the conjugated variablemx) may give rise to a homogeneous linear function yad x a1xor yyad mx b1mx as an additional contribution in the respectivepotential function. If such contributions to the potential func-tions do not represent a true effect of the variable considered,they can be removed without difficulties by subtraction of thehomogeneous linear function. Such homogeneous linear termsarise generally, if the variable increases whereas the conjugate isconstant (or vice versa). This occurs at phase transitions of firstorder or due to mixing and if the variable and its conjugate, re-presented by the axis of the coordinate system, do not start simul-taneously at the origin (zero) of both axes.The present considerations are demonstrated for energy-like

Massieu-Gibbs functions as an example. Of course, they can beapplied to all Massieu functions.

8 References

[1] G. Falk, Theoretische Physik auf der Grundlage einer allgemei-nen Dynamik, Bd. II Allgemeine Dynamik, Thermodynamik.Corrected reprint ISBN-3-540-04174-5, Springer, Berlin(1988).

[2] G. Falk, Physik. Zahl und Realitt. ISBN 3-7643-2550-X, Bir-khuser, Basel (1990).

[3] H. B. Callen, Thermodynamics. John Wiley & Sons, Inc.,New York (1960).

[4] G. Adam, O. Hittmair,Wrmetheorie. third ed., ISBN 3-528-23311-7, Vieweg, Braunschweig (1988).

[5] A. Mnster, Chemische Thermodynamik. Verlag Chemie,Weinheim (1969).

[6] E. A. Guggenheim, Thermodynamics. North-Holland,Amsterdam (1967).

[7] V. I. Arnold, Mathematical Methods of Classical Mechanics.ISBN 0-387-96890-3 Springer, New York, N. Y. (1989).

[8] A. V. Gitin,Optics Communications 2008, 281, 3062.[9] A. V. Gitin,Optics Communications 2009, 282, 757.[10] R. Becker, Theorie der Wrme. ISBN 3-540-03559-1,

Springer, Berlin (1961), English translation, RichardBecker: Theory of Heat. second ed., Springer, Berlin,ISBN 0387-03730-6 (1966).

[11] H. J. Hoffmann,Glass Sci. Technol. 2005, 78, 218.[12] H. J. Hoffmann,Mat.-wiss. u. Werkstofftech. 2011, 42, 753.

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