# Constructing Lagrangian from the Hamiltonian

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 $L$.

• The Legendre transform is involutive Commented Jun 20, 2015 at 17:28
• @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). Commented Feb 12, 2022 at 19:11
• The answers below explain how the Legendre transform works. Commented Feb 13, 2022 at 9:22
• @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" Commented Feb 13, 2022 at 14:42

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.

• "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. Commented Feb 12, 2022 at 19:06
• @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. Commented Feb 13, 2022 at 0:45
• 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 :) Commented Feb 13, 2022 at 14:40

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.

• 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. Commented Jun 20, 2015 at 17:59
• I have removed the second part because now the strategy of answering that is obvious
– SRS
Commented Jun 21, 2015 at 3:53
• The problem is that H is written jn terms of q and its derivatives, not in terms of p and q. Commented Feb 12, 2022 at 23:43

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 $L$.

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