Simple Legendre Transform

1. Simple Legendre Transform

Consider the relation y = 2x 2 +5

(12) We call y as the basis function, y = y (0) , i.e.,

y (0) = 2x 2 + 5. (13) We can use a series of points to draw the curve ABCD (through point geometry) as shown in

Figure 3 . The slope Figure 3 . The slope

(x)

Slope ξξξ ξ=12 at x=3

y (0)

CS

Slope ξ=4 at x=1 ξξξ

y (1) =3

3 Slope ξ =8 at x=2 ξξξ

F Slope ξξξ ξ =12 at x=3

Figure 3: Illustration of Legendre transform.

y (0) ′(x) = ∂y / ∂x = ξ = 4x, i.e., (14)

y 2 = (1/8) ξ + 5. (15) Can we draw the same curve y (0) vs x in the y (0) - x plane which we drew with Eq. (13), but

just by using Eq. (15)? Given slope, we cannot place y (0) in y (0) -x plane. Equation (15) is a differential equation, and by replacing x with the slope information regarding the value of x is lost. From Eq. (15),

y (0) = f( ξ), (16) which is a relation that does not provide the same information as the expression

y (0) = f(x). (17) Therefore, we require a different function y (1) ( ξ) that describes the curve y (0) (x). In the

context of Eq. (14) ξ = 4at x = 1, ξ = 8 at x = 2, and so on. A tangent drawn at point A (x =1) intersects the y (0) axis at point E (y (0) = 3) ( Figure 3 ) and one at point B (x=2) intersects the

y (0) axis at point F (y = –3). The intercepts will be denoted as y (1) . The locus tangent to all the lines is the

curve y (0) . Thus, we can construct the curve y (x).

0 1 2 3 Hence, a curve may be drawn either along points (point (0) y

5 7 13 23 geometry, Eq. (12)) or with a series of slopes or lines

0 4 8 12 (line geometry). Line geometry requires a functional

5 3 –3 –13 relation between the intercept y ( ξ) and ξ to draw the

y (1)

Table 1: Values of x, y (0) , ξ and curve as y (0) (x) while point geometry requires a func-

y (1) for the basis function y (0) = 2x y (1) for the basis function y (0) = 2x

(18) Replacing y (0) in Eq. (18) using Eqs. (15) and (14), we obtain the relation

y (1) ( ξ) = 5 – ξ 2 /4. (19) Equations (18) and (19) provide the same information as does Eq. (12). The functional relation

y (1) ( ξ) is known as the first Legendre transform of y (0) with respect to x. For the above exam- ple (cf. Eqs. (12)–(19)), Table 1 presents values of x, y (0) , ξ and y (1) for the basis function (y (0)

=2x 2 + 5). It is apparent from Eq. (18) or (19) that the first Legendre transform slope of ξ is an independent variable and y (1) is a dependent variable. In the context of the basis function x is the independent variable and y (0) is the dependent variable. Differentiating Eq.(18)

dy (1) = dy – dx ξ – x dξ y . (20) From Eq. (17),

dy (0) = ξ dx, i.e., (21) Using the result in Eq.(20)

dy (1) = ξ dx – dx ξ – x dξ = –x dξ, or x = – ∂y (1) / ∂ξ. (22)

a. Relevance to Thermodynamics We cannot measure the entropy directly, so that it must be derived from other meas- urements. We can use the relation U = U(S) and measure isometric temperatures, which repre- sent the slope

V . Thereafter, the function A = y ( ξ) can be generated to construct the U(S) curve.

ξ = (∂U/∂S) (1)

a. Example 1 The following measurements are made of u (1) and T = ( ∂u/∂s) v .

–16.2 –32.0 Can you plot u (0) vs s ; It is known for an incompressible liquid that

y (0) =u (0) = 273 c (exp (s/c) –1), (A) where the specific heat c = 4.184 kJ kg –1 K –1 . Also apply the first Legendre transform

for Eq. (A) to obtain an expression for u (1) (T). Solution

At specified values of (0) ξ the intercepts u are known. Hence, the u (s) curve can be constructed from the above data.

b) Using the first Legendre transform, u (0) =u (1) –( ∂u/∂s) v,s =u (1) – Ts.

(B) Differentiating Eq. (A)

( 0 ∂u / ∂s) v = 273 exp(s/c) = T, or s/c = ln (T/273). (C) Using Eq. (C) in (A) ( 0 ∂u / ∂s) v = 273 exp(s/c) = T, or s/c = ln (T/273). (C) Using Eq. (C) in (A)