25
$\begingroup$

Given the Lagrangian $L$ for a system, we can construct the Hamiltonian $H$ using the definition $H=\sum\limits_{i}p_i\dot{q}_i-L$ where $p_i=\frac{\partial L}{\partial \dot{q}_i}$. Therefore, to determine, $p_i$ we need to know $L$.

Now suppose, we are given the Hamiltonian $H$. Can we then reconstruct the Lagrangian $L$? Certainly the relation $L=\sum\limits_{i}p_i\dot{q}_i-H$ is not helpful because there is no prescription how to determine $p_i$ without knowing $L$.

$\endgroup$
4
  • 3
    $\begingroup$ The Legendre transform is involutive $\endgroup$
    – Phoenix87
    Commented Jun 20, 2015 at 17:28
  • $\begingroup$ @Phoenix87 that is clear, but the problem here is that it seems that the OP has H written in terms of q and its time derivative, not of p and q: it not clear how to invert Legendre because OP has no way to know H(q,p). $\endgroup$
    – Quillo
    Commented Feb 12, 2022 at 19:11
  • $\begingroup$ The answers below explain how the Legendre transform works. $\endgroup$
    – Phoenix87
    Commented Feb 13, 2022 at 9:22
  • $\begingroup$ @Phoenix87 the problem is: if you have H(q, dot(q)), and not H(q,p), how to reconstruct L? Certainly they explain how Legendre works, but the interesting part of the question is still unanswered: "Certainly the relation L=... is not helpful because there is no prescription how to determine p without knowing L" $\endgroup$
    – Quillo
    Commented Feb 13, 2022 at 14:42

3 Answers 3

14
$\begingroup$

Yes, there exists a Legendre transformation from $g(p)$ to $f(x)$: $$ f(x)=p(x)x-g(p(x)) $$ with $x=dg/dp$. Here the notation $p(x)$ means $p$ written in terms of $x$. In your case, the Hamiltonian is a function of $p$ and you are transforming it to a function of $\dot{q}$, so you must use Hamilton's equation to get the velocity: $$ \dot{q}_i=\frac{\partial H}{\partial p_i} $$ which you then solve for $p$ (so that it's a function of $\dot{q}$, e.g. $p=h(\dot{q})$). You then have your Lagrangian as $$ L(q,\dot{q})=\dot{q}_ih(\dot{q}_i)-H(q,h(\dot{q})) $$

For the (relativistic) Hamiltonian1, $$ H(q,p)=\sqrt{p^2c^2+m^2c^4}+V(q) $$ the momentum should be $$ p(\dot{q})=\frac{m\dot{q}}{\sqrt{1-\dot{q}^2/c^2}} $$ which was computed using $\dot{q}=\partial H/\partial p$ & then inverting to get $p$ in terms of $\dot{q}$. You should verify that this is correct (but it does look right to the relativistic momentum, $p=\gamma mv$). Then you can just do the substitution and get your Lagrangian.


1. This particular Hamiltonian was included in version 2 of this question, but was since removed; as it still provides an example of the $H\to L$ transform, I kept it in.

$\endgroup$
3
  • $\begingroup$ "Certainly the relation L=∑pq'−H is not helpful because there is no prescription how to determine pi without knowing L." This means that we are given H in terms of q and its derivatives, so it is a "pre-hamiltonian": we have no direct access to p. $\endgroup$
    – Quillo
    Commented Feb 12, 2022 at 19:06
  • $\begingroup$ @Quillo OP's own post begs to differ, though it was later edited out. The fact that it was accepted as the answer suggests they were satisfied with the response. $\endgroup$
    – Kyle Kanos
    Commented Feb 13, 2022 at 0:45
  • $\begingroup$ The quote is present also in the previous version... then OK, there is also the relativistic H bit. But to me the interesting part of the question is that one. Certainly the OP can accept everything they want. I am commenting to hope that the question gets a bit of activity to discuss also this point, maybe I will consider a bounty :) $\endgroup$
    – Quillo
    Commented Feb 13, 2022 at 14:40
12
$\begingroup$

First of all, the hamiltonian contains the coordinates $q_i$ and their momenta $p_i$. You have to calculate the velocities $\dot{q}_i$. For that, you'll need the Hamilton-Jacobi equations $$\dot{q}_i = \frac{\partial H}{\partial p_i}$$The Legendre transform, as noted in the comments, is involutive, so the lagrangian is just the Legendre transform of the hamiltonian $$L = \sum_i p_i \dot{q}_i - H$$ where you have to express everywhere the momenta in terms of the velocities.

Worked-out example: harmonic oscillator. The well-known hamiltonian is $$H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 q^2$$ From the Hamilton-Jacobi we get (unsurprisingly) that $$\dot{q} = \frac{\partial H}{\partial p} = \frac{p}{m}$$ And plug it in the Legendre transform $$L = \dot{q}p - H = \dot{q}(\dot{q} m) - \frac{(\dot{q}m)^2}{2m} - \frac{1}{2}m\omega^2 q^2 = \frac{1}{2}m \dot{q}^2 - \frac{1}{2}m\omega^2 q^2$$Which is indeed the lagrangian for the harmonic oscillator.

$\endgroup$
3
  • $\begingroup$ Note: this answers the first version of the question, wich didn't mention the relativistic hamiltonian. As Kyle addressed this point before me, I won't expand my answer. $\endgroup$
    – Bosoneando
    Commented Jun 20, 2015 at 17:59
  • 1
    $\begingroup$ I have removed the second part because now the strategy of answering that is obvious $\endgroup$
    – SRS
    Commented Jun 21, 2015 at 3:53
  • $\begingroup$ The problem is that H is written jn terms of q and its derivatives, not in terms of p and q. $\endgroup$
    – Quillo
    Commented Feb 12, 2022 at 23:43
8
$\begingroup$

Let us suppress explicit time dependence $t$ from the notation in the following. Hamilton's eqs. are the Euler-Lagrange (EL) eqs. for the so-called Hamiltonian Lagrangian

$$\tag{1} L_H(q,\dot{q},p)~:=~ p_i\dot{q}^i-H(q,p).$$

In other words, the solutions to Hamilton's eqs. are stationary points for the Hamiltonian action

$$\tag{2} S_H[q,p]~:=~\int \! dt~L_H(q,\dot{q},p). $$

Next define the Lagrangian as

$$ \tag{3} L(q,\dot{q})~:=~\sup_p L_H(q,\dot{q},p). $$

Formula (3) is the succinct answer to OP's question about how to construct the Lagrangian from the Hamiltonian.

The Legendre transformation (3) is often referred to as integrating out$^1$ the momentum variables $p_i$. Then the action (2) becomes

$$ \tag{4} S[q]~:=~\int \! dt~L(q,\dot{q}). $$

The stationary points of the action (4) are given by the EL eqs. for $L$.

--

$^1$ If we go beyond classical mechanics, and consider the phase space path integral formulation, then "integrating out the momentum" is exactly what is happening.

$\endgroup$

Your Answer

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