Generalized Legendre Transform

2. Generalized Legendre Transform

Consider the generalized expression for a basis function involving more than one variable, i.e.,

y (0) =y (0) (x 1 ,x 2 ,x 3 , ..., x n ) so that y (1) =y (0) – ξ 1 x 1 .

The function y (1) =y (1) ( ξ 1 ) for prescribed x

2 , ..., x n , can also be used to describe Y . The first Legendre transform with respect to x 1 is expressed in the form

(24) If x 1 ,x 3 , ..., x n are held constant, the first Legendre transform with respect to x 2

x 1 =y (1) ( ξ 1 ,x 2 , ..., x n )

(25) Since

x 2 =Y (x 1 , ξ 2 ,x 3 , ..., x n ).

(26) the second Legendre transform is

2 )x 2 =y – ξ 2 x 2 , i.e.,

(29) Generalizing Eq. (29) to the m–th Legendre transform (where m<n),

y (0) ( ξ

1 , ξ 2 ,x 3 ,x 4 ...) = y (x 1 ,x 2 ,x 3 ..) – ξ 1 x 1 – ξ 2 x 2 .

y (m) =y (0) – Σ i=1,m ξ i x i , and (30) the n–th Legendre transform

(31) In the context of the second Legendre transform,

y (0) =y – Σ

(n)

i=1,n ξ i x i .

dy (2) = dy (0) –d ξ 1 dx 1 – ξ 1 dx 1 –d ξ 2 x 2 – ξ 2 dx 2 =

1 dx 1 + ξ 2 dx 2 + ξ 3 dx ξ 3 +.... – d ξ 1 x 1 – ξ 1 dx 1 –d ξ 2 x 2 – ξ 2 dx 2 , i.e., dy (2) = ξ

(32) for m–th Legendre transform

3 dx 3 + ξ 4 dx 4 +…–d ξ 1 x 1 –d ξ 2 x 2 ,

(33) and the n–th Legendre transform

dy (m) = Σ j=m+j,n ξ j dx j – Σ j=1,m d ξ j x j , m < n,

dy (n) = –Σ j=1,n d ξ j x j .

If, in the context of Eq. (32) ξ 2 ,x 3 ,x 4 , …, x n are held constant, then (dy (2) ) ξ2,x3,x4... = –d ξ 1 x 1 , i.e.,

(36) Similarly,

( ∂y (2) / ∂ξ 1 ) ξ 2 , xx 3 , 4 ,... =–x 1 .

( (1) ∂y / ∂ξ

1 ) xxx 2 , 3 , 4 ,... = –x 1 .

Likewise,

(dy (2) ) ξξ 1 , 2 , xx 4 , 5 ,... = ξ 3 dx 3 , i.e.,

ξ 3 ∂y / ∂x 3 ) ξ 1 , xx 2 , 4 ,... =( ∂y / ∂x 3 ) xxx 1 , 2 , 4 ,... , and (38) ( (2) /

∂y (1) ξξ 1 , 2 , xx 3 , 5 ,... = ξ 4 =( ∂y / ∂x 4 ) ξ 1 , xxx 2 , 3 , 5 ,... =( ∂y / ∂x 4 ) xxxx 1 , 2 , 3 , 5 ,... . (39) If the variables in the basis function are extensive, then the Euler equation must be

∂x (0)

satisfied so that

(40) Since, y (n) =y (0) – Σ i=1,n ξ i x i ,y (n) = 0, and Σ i=1,n x i d ξ i = 0,

i=1,n x i ∂y / ∂x i = Σ i=1,n ξ i x i =y .

(41) which is known as the Gibbs–Duhem relation

b. Example 2 Consider a basis function involving five variables. Obtain the first and third Legendre transforms.

Solution The basis function is y (0) (x

1 ,x 2 ,....x 5 ), i.e.,

(A)

y (3) =y (0) – ξ 1 x 1 – ξ 2 x 2 – ξ 3 x 3 (cf. Eq. 29),

(B) dy (3) = –( ξ 4 dx 4 + ξ 5 dx 5 ) – (x 1 d ξ 1 +x 2 d ξ 2 +x 3 d ξ 3 ) (cf. Eq. 32),

(F) Employing Eqs. (C)–(E),

1 ) ξξ 2 , 3 , xx 4 , 5 =( ∂y / ∂ξ 1 ) ξ 2 , xxx 3 , 4 , 5 =( ∂y / ∂ξ 1 ) xxxx 2 , 3 , 4 , 5 = –x 1 , where

y (1) = f( ξ 1 ,x 2 ,x 3 ,...), i.e., ∂y (1) / ∂ξ 1 =–x 1 . Differentiating with respect to x 2 ,

2 =y 2 = ξ 2 , and . ( ∂/∂ξ 2 )( ∂y ) = 1.

More generally,

= ξ k , k > m, and ∂/∂ξ l (y k )= ∂ξ k / ∂ξ l .

(m)

For instance,

More generally,

/ ∂ξ k = ... = ∂y /d ξ k = –x k , k < m, and m ≠ n. Employing Eqs. (C)–(F),

∂y ( (0) / ∂x 5 ) xxxx 1 , 2 , 3 , 4 =– ξ 5 ,

∂y ( (0) / ∂x 4 ) xxxx 1 , 2 , 3 , 5 =– ξ 4 ,

=( ∂y (0) ∂x 5 ξξ 1 , 2 , xx 3 , 4 ∂y ∂x 5 ξ 1 , xxx 2 , 3 , 4 ∂y / ∂x 5 ) xxxx 1 , 2 , 3 , 4 =– ξ 5 ,

ξξ 1 , 2 , xx 3 , 5 =( ∂y / ∂x 4 ) ξ 1 , xxx 2 , 3 , 5 =( ∂y / ∂x 4 ) xxxx 1 , 2 , 3 , 5 =– ξ 4 ,

3 ) ξξ 1 , 2 , xx 4 , 5 =( ∂y / ∂x 3 ) ξ 1 , xxx 2 , 4 , 5 =( ∂y / ∂x 3 ) xxxx 1 , 2 , 4 , 5 =– ξ 3 ,

( (0) ∂y / ∂x

5 ) ξ 1 , xxx 2 , 3 , 4 =( ∂y / ∂x 5 ) xxxx 1 , 2 , 3 , 4 =– ξ 5 , ( ∂y (1) / ∂x 4 ) ξ 1 , xxx 2 , 3 , 5 =( ∂y (0) / ∂x 4 ) xxxx 1 , 2 , 3 , 5 =– ξ 4 ,

( ∂y (1) / ∂x 3 ) ξ 1 , xxx 2 , 4 , 5 =( ∂y (0) / ∂x 3 xxxx 1 , 2 , 4 , 5 =– ξ 3 ,

( ∂y (1) / ∂x )

2 ξ 1 , xxx 3 , 4 , 5 =( ∂y / ∂x 2 ) xxxx 1 , 3 , 4 , 5 =– ξ 2 ,

( ∂y (0) / ∂x 5 ) xxxx 1 , 2 , 3 , 4 =– ξ 5 , ( ∂y (0) / ∂x 4 ) xxxx 1 , 2 , 3 , 5 =– ξ 4 ,

( ∂y (0) / ∂x 3 ) xxxx 1 , 2 , 4 , 5 =– ξ 3 , ( ∂y (0) / ∂x 2 ) xxxx 1 , 3 , 4 , 5 =– ξ 2 , and ( (0) ∂y / ∂x 1 ) xxxx 2 , 3 , 4 , 5 =– ξ 1 .

In general, ∂y (m) / ∂x k = ∂y (m–1) / ∂x k … ∂y (0) / ∂x k =– ξ k , m < k, and m ≠ n, and

11 = ∂ y / ∂x 1 ,y 1k =( ∂/∂x 1 )( ∂y / ∂x k ).