How is the Legendre transform defined in cases where g(p) equals 0?

[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...

 

[HallsofIvy]

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?

 

[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