# Hamiltonian form: inverting the general momenta

I am studying analytical mechanics and am confused with the part about Hamiltonian form.

My textbook develops the whole theory of Legendre transforms, in order to invert the formula $$\mathbf{p}=\text{grad}_{\mathbf{\dot{q}}} L(\mathbf{q},\mathbf{\dot{q}},t)$$

Yet, when you use Legendre transforms to find the Hamiltionian from the Lagrangian, you always have to explicitly write the generalised velocities in terms of the generalised momenta yourself, so what exactly do they mean by the Legendre formula somehow doing this for you?

The exact quotation is:

The general problem of converting Lagrange's equations into Hamiltonian form hinges on the inversion of the equations that define $\mathbf{p}$, so as to express $\mathbf{\dot{q}}$ in terms of $\mathbf{q},\mathbf{p},t$. This is precisely what Legendre transforms do!

• What is the exact quotation in your book that says so? – Victor Pira Jan 18 '16 at 8:39
• @VictorPira I added it to my post. – Michael Angelo Jan 18 '16 at 8:47
• I believe that statement is a bit misleading, in my mind, the Legendre transform is a way to transform a system dependent on $A$ and $B$ to instead be dependent on $A$ and $C$, where $B=B(A,C)$. It's used extensively in thermodynamics, where you use Legendre transforms to go between, energy, enthalpy, and free energy in various forms. – Mikael Fremling Jan 18 '16 at 8:54

In my opinion, the Mikael Fremling's comment is precisely it. The trick is not just to "relabel" lagrangian variables in terms of $p$. The meaning is "to change the formalism", not just "substitute variables".

I agree that the quoted book formulation might be a bit confusing, but seen from this point of view it does make sense.

• How can you be sure that the general momenta can always be written in terms of generalised coordinates and -velocity? – Michael Angelo Jan 18 '16 at 9:16