# Legendre transform

1. Jul 10, 2006

### eljose

Legendre transform....

If we define a function f(r) with r=x,y,z,.... and its Legnedre transform
g(p) with $$p=p_x ,p_y,p_z,......$$ then we would have the equality:

$$Df(r)=(Dg(p))^{-1}$$ (1) where D is a differential operator..the

problem is..what happens when g(p)=0?...(this problem is usually found in several Hamiltonians of relativity) then (1) makes no sense since a 0 matrix would have no inverse..how do you define Legendre transform then...

2. Jul 10, 2006

### HallsofIvy

Staff Emeritus
I don't see any problem with "defining the Legendre transform", just defining its inverse- which does not necessarily exist. What kind of functions, f, and p, give g(p)= 0? Why does that not have an inverse?

3. Jul 10, 2006

### eljose

The problem of Zero Legendre-transform arises for example in Quantum Field theory...let be the "Lagrangian density":

$$L= \sqrt (-g)R$$ where g is the determinant of the "metric" and R is Ricci scalar?...of course if we use a "dot" to indicate derivative respect to time we have that:

$$L=L(g_ab , \dot g_ab,x,y,z)$$ now we definte the "momenta"..

$$\pi _ab =\frac{ \partial L}{\partial \dot g_ab)$$

then the Legendre transform for Quantum gravity is defined by:

$$H=\pi _ab \dot g_ab -L$$ where the metric is given by:

$$g_ab =N(t)dt^{2}-g_ ij dx^{i}dx^{j}$$ i,j=1,2,3 (einstein sum convention)

where N(t) is somehow a "lapse" function with a physical meaning so you get the "constraint"..

$$H=0$$