Deriving Lagrangian mechanics from Hamiltonian mechanics

Question

I have just finished reading Susskind's Theoretical Minimum. I think I got it well, even found some mistakes in few equations The book starts from Newtonian Mechanics, then derives Lagrangian formalism and then Hamiltonian one, which I guess is very standard way of the subject introduction. But it kept me pondering, if we start with Hamilton's formulation, then if we are able to get Euler-Lagrange equations. I will state that in questions.

1. If we have any abstract system defined via Hamiltonian mechanics, is there a proof that there exists equivalent reformulation of the system in Lagrangian mechanics?
2. If such a proof exists, is it constructive? That is, if there is a canonical/algorithmic way of constructing the Lagrangian from a given Hamiltonian?
3. Can you outline the proof, show counterexample (or point to such proof/counterexample)? In case of the proof, I am especially interested in how to derive Euler-Lagrange equation from Hamilton's Equations.
4. Bonus question: What if we would made Hamilton's Equations more symmetric, by removing the minus sign? Does it have a deep fundamental meaning on abstract level, or does it rather comes from the fact that we want to keep the convention(is it really only a convention?) that in Newtonian Mechanics we want force to act in a direction that makes the related potential smaller?

Thank you for your answers ;)

Answer 1

I will stick to the one-dimensional case, i.e. a system with a unique Lagrangian coordinate $q=q^1$. The procedure I am about to discuss easily extends to the case of many variables.

If $H(t,q,p)$ is smooth and such that the function at $t$ and $q$ fixed $$s = \frac{\partial H(t,q,p)}{\partial p}\tag{1}$$ can be smoothly inverted, obtaining $$p= p(s,q,t),\tag{2}$$ then $$L(t,q,s) := sp(s,q,t) - H(t,q,p(s,q,t))\tag{3}$$ defines a Lagrangian function. The $t$-parametrized curves which — in the variables $q,p$ — solve the Hamilton equations generated by $H$ one-to-one correspond — in the variables $q,\dot{q}:= s$ — to the curves which solve the Euler Lagrange equations generated by $L$ above.

PROOF. A lengthy computation starting from $(3)$ produces the following three identities. $$\frac{\partial L(t,q,s)}{\partial t}= - \frac{\partial H(t,q,p)}{\partial t}\Biggr|_{p=p(s,q,p)}\:,\tag{I1}$$ $$\frac{\partial L(t,q,s)}{\partial q}= - \frac{\partial H(t,q,p)}{\partial q}\Biggr|_{p=p(s,q,p)}\:,\tag{I2}$$ $$\frac{\partial L(t,q,s)}{\partial s} = p(s,q,p)\:.\tag{I3}$$ Now consider a curve $t\mapsto (q(t),p(t))$ which solves Hamilton's equations and its corresponding curve in the Hamiltonian setting. $t\mapsto (q(t),s(t)):= \left(q(t), \frac{\partial H(t,q,p)}{\partial p}\Biggr|_{(t,q(t),s(t))}\right)$. The Hamilton equations are $$\frac{\mathrm{d}p}{\mathrm{d}t}= - \frac{\partial H(t,q,p)}{\partial q}\Biggr|_{(t,q(t),p(t))}\:, \quad \frac{\mathrm{d}q}{\mathrm{d}t}= - \frac{\partial H(t,q,p)}{\partial p}\Biggr|_{(t,q(t),p(t))}\:.$$

The former Hamilton equation, using $(\text{I}3)$ on the left-hand side and $(\text{I}2)$ on the right-hand side, can be rephrased to $$\frac{\mathrm{d}}{\mathrm{d}t}\Biggr|_{(t,q(t),s(t))} \frac{\partial L}{\partial s}= \frac{\partial L}{\partial q}\Biggr|_{(t,q(t),s(t))}\:.$$ The latter Hamilton equation, according to $(1)$, can be rephrased to $$s(t) = \frac{\mathrm{d}q}{\mathrm{d}t}\:.$$ If we change the name of the variable $s$ to $\dot{q}$, we have the standard EL equations: $$\frac{\mathrm{d} }{\mathrm{d}t}\Biggr|_{(t,q(t),\dot{q}(t))} \frac{\partial L}{\partial \dot{q}}= \frac{\partial L}{\partial q}\Biggr|_{(t,q(t),s(t))}\:.$$ $$\dot{q}(t) = \frac{\mathrm{d}q}{\mathrm{d}t}\:.$$

Notice that all the procedure is reversible so that every curve that satisfies EL's equations solves Hamilton's equations when passing to Hamiltonian variables.

REMARK. The crucial point here is the invertibility of the transformation $(1)$ (at fixed $q$ and $p$). That is equivalent to the invertibility of the Legendre transformation for the Lagrangian associated with $H$. If one starts from a singular Hamiltonian, i.e., such that $(1)$ cannot be inverted like this $$H(q,p) = kp+ hq\:,$$ then all the above procedures cannot be used. That is the same as for singular Lagrangians, which do not admit Hamiltonian formulation.