So, as the title says, why is the Hamiltonian the Legendre transform of the Lagrangian?

I know that from quantum mechanics, one can start with the Hamiltonian defined as the generator of time evolution, create the Hamiltonian path integral, and then integrate out the momenta and identify the remaining "object" as the Legendre transformation of the Hamiltonian. So if the Lagrangian is defined as the correct object to out in a the path integral (ie a path integral that only integrates over the different paths of the coordinate of the particle), the above gives the relationship between the Hamiltonian and the Lagrangian.

However, from a purely classical point of view, when one usually starts with the Lagrangian as defined to give the correct equations of motion (or by T-V), I have trouble understanding why the Hamiltonian is defined as the legendre transform. I know that the Legendre transform and this definition of the canonical momenta switches from Euler Lagrange equations to the Hamilton Jacobi equations and that these happen to be two convenient ways to parametrize the dynamics. But is there a better motivation why one performs the Legendre transform on the Lagrangian?

  • 1
    $\begingroup$ In classical mechanics, the Legendre transform allows us to have $2n$ first-order differentials on $2n$ independent variables via Hamiltonian, rather than second-order differentials via Lagrangian. $\endgroup$
    – Kyle Kanos
    Sep 25, 2014 at 13:18

1 Answer 1


Here is one line of motivation:

  1. On one hand, in the Lagrangian formalism, the Lagrangian energy function $$\tag{1} h(q,v,t)~:=~v^i \frac{\partial L(q,v,t)}{\partial v^i}- L(q,v,t)$$ is defined as the Noether charge for time translations.

  2. Noether's theorem states that if the Lagrangian is invariant under time translations, which implies that $$\frac{\partial L(q,v,t)}{\partial t}~=~0,$$ then the energy (1) is conserved $$\frac{d h(q,v,t)}{d t}~\approx~ 0$$ on-shell.

  3. On the other hand, the Hamiltonian function is defined as the Legendre transform $$\tag{2} H(q,p,t)~:=~\sup_v (v^i p_i -L(q,v,t)).$$

  4. Although the Lagrangian energy function (1) and the Hamiltonian function (2) are two different functions which depend on different variables, the two functions do take the same values when the velocities $v^i$ and momenta $p_i$ are paired via the Legendre transformation rules. In this manner, the Hamiltonian gets an interpretation as energy.

For more information, see also e.g. this and this Phys.SE posts.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.