# How is a Hamiltonian constructed from a Lagrangian with a Legendre transform

many textbooks tell me that Hamiltonians are constructed from Lagrangians like $$L=L(q,\dot{q})$$ with a Legendre transformation to obtain the Hamiltonian as $$H=\dot{q}\frac{\partial L}{\partial \dot{q}}-L$$ but none of the textbooks explain how this is done.

My specific problem is that I have Lagrangians that do not depend on $\dot{q}$ and therefore should have $\frac{\partial L}{\partial \dot{q}}=0$, hence $H=-L$. But my impression from the clues I have is that it is not that simple.

Let's say the Lagrangian is $$L(q)=\ln(q)-(2q-10)\lambda$$ Now as far as I know the Legendre transformation should give a function $f^*(p)=\sup(pq-L(q))$ (this implies $p=\frac{\partial L}{\partial q}$) which is obtained by substituting the stationary point $q_s$ of $\sup(pq-L(q))$ into $pq-L(q)$ thus getting $f^*(p)=pq_s-L(q_s)$ (for instance wikipedia's Legendre Transformation page explains this). Doing this for the example above: $$\frac{\partial (pq-L(q))}{\partial q}=\frac{\partial (pq-\ln(q)+(2q-10)\lambda)}{\partial q}=p-\frac{1}{q}+2\lambda$$ must be 0 for a stationary point, thus $q_s=1/(p+2\lambda)$. And hence the transformation should be $$f^*(p)=p\frac{1}{p+2\lambda}-\ln(\frac{1}{p+2\lambda})+(2\frac{1}{p+2\lambda}-10)\lambda$$

and this should be the Hamiltonian.

But this equation does obviously have nothing to do with the textbook Hamiltonian. Rather, an answer to another question has in a similar case treated $L(q)$ as being dependent on $\dot{q}$ implicitly (Writing $\dot{q}$ in terms of $p$ in the Hamiltonian formulation ... answer by Qmechanic). It mentions using the Dirac-Bergmann method for obtaining the Legendre transform.

Trying something along the lines of this other question the above example seems to give $$p=\frac{\partial L}{\partial \dot{q}}=0$$ and $$p \approx 0$$ (an equality modulo constraint as the answer to the question linked above says). And then $H=\dot{q}p-L$.

The difference seems to be that the Legendre transform is done with respect to two different variables, $q$ and $\dot{q}$ - but it was my understanding that it had to be done with respect to all variables the Lagrangian depends on. So how does the $qp_q$ term vanish if we have only the $\dot{q}p_{\dot{q}}$ term left?

Thanks.

edit: changed ln to \ln as Plane Waves suggested, and sup to \sup. And yes, sup is the supremum over all q, as Vibert said, sorry for forgetting to mention that.

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Where did you get that Lagrangian? It has no actual dynamics... – webb Jun 11 '13 at 23:00
It is a nonlinear programming Lagrangian similar to those you can see for instance here en.wikipedia.org/wiki/Lagrange_multiplier#Examples - basically it can be used for maximizing a target function $\ln(q)$ under the constraint $2q-10 \leq 0$ (the solution is obviously $q=5$). No it has no actual dynamics but a corresponding Hamiltonian must exist nevertheless and there must be a general way to obtain it no matter if the Lagrangian is actually dynamic or not. – 0range Jun 12 '13 at 2:04

## 3 Answers

The fact that $p = \large \frac{\partial L}{\partial \dot{q}} = 0$ introduces a problem in the equivalence between Lagrangian and Hamiltonian representations.

The idea is that the Hamiltonian representation plus the constraint $p = 0$ is equivalent to the Lagrangian representation

The Lagrangian $L$ is a function of $q$ and $\dot q$, that is $L(q, \dot q)$

If we work with the Lagrangian, we will apply the Euler-Lagrange equations which are :

$$\frac{\partial L}{\partial q} = \frac{d}{dt} (\frac{\partial L}{\partial \dot{q}})$$

Because $\large \frac{\partial L}{\partial \dot{q}} = 0$, the equation is simply $\large \frac{\partial L}{\partial q} = 0$, that is $\frac{1}{q} - 2\lambda = 0$, so $q = \frac{1}{2 \lambda}$

Now try to work with the Hamiltonian.

The Hamiltonian $H$ is a function of $q$ and $p$, that is $H(q, p)$

The link between the two is the Legendre transformation :

$$H=\dot{q}\frac{\partial L}{\partial \dot{q}}-L$$

Because your Lagrangian does not depends of $\dot q$, then $p = \frac{\partial L}{\partial \dot{q}} = 0$, and so :

$$H(q, p) = - L(q, \dot q) = - \ln(q) + (2q-10)\lambda$$

From this hamiltonian, you get the equations of movement :

$$\dot q = \frac{\partial H}{\partial p} ~,~\dot p = - \frac{\partial H}{\partial q}$$ So we have :

$$\dot q = 0~,~\dot p = \frac{1}{q} - 2\lambda \tag{1}$$

From this, we cannot recover the equation obtained from Euler-Lagrange equations, we have to add the constraint $p = 0$.

If $p = 0$, it means that $\dot p = 0$, and so :

$$q = \frac{1}{2 \lambda}\tag{2}$$

This is coherent with the fact that $\dot q = 0$

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Trimok Why we have to add constraints with Hamiltonian representation to be equivalent the Lagrangian representation? What is/are physically or mathematically reason/s? Thank you very much for the answers. – user30750 Oct 8 '13 at 15:56
The OP Lagrangian here was very very special, because it depends only on $q$, so we have $p=0$. Now, passing to the hamiltonian $H(p,q)$, we have to keep this information $(p=0)$, if we want to recover the usual Euler-Lagrange equations $(2)$ (see also the beginning of the answer), from the hamiltonian relations $(1)$. I suppose a more rigorous answer would be based on the Legendre transformation – Trimok Oct 8 '13 at 16:19

What is "sup"? Please write "ln" with a backslash $\ln$. Anyway if $L$ is independent of a specific variable, then the canonically conjugate variable is conserved, which means that the energy function (the Hamiltonian) cannot explicitly depend on it. In your case $p_q$ should be conserved so your first intuition is correct: $$H=-L.$$

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'sup' means supremum (over all possible values of $q$). – Vibert Jun 11 '13 at 21:39

Functions like yours are often referred to as "Lagrangians" in economic textbooks and such, but in the context of physics a Lagrangian is a functional, not just a function, and implies the concept of action, which in turn implies a dynamic situation. So you should probably avoid calling it a Lagrangian, at least when in earshot of physicists. Let's call your function $f(q)$ instead for now.

In the Legendre transformation that leads to the Hamiltonian, the argument of the Lagrangian is $\dot{q}$ and the argument of the Hamiltonian (i.e. the Legendre transformation) is $\frac{\partial L}{\partial \dot{q}}$ (or $p$)--which I think is called the canonical momentum conjugate to $q$. Both arguments are dynamic in nature.

In the case of a function (not functional) like yours, there is nothing dynamic happening, so better not to try to extrapolate from the derivation of the Hamiltonian. The Legendre transformation (call it $h$) is simply

$$h \left( m \right)=mq-f(q)$$

Where $m=\frac{df}{dq}$ and $q=q(m)$

This paper might help understand the Legendre transformation better.

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The action is a functional, but it's perfectly fine to call the Lagrangian a function; actually, in case of first-order mechanics it should be a 1-form $L=\mathcal L\mathrm dt$ on the jet bundle $J^1Q$ of some configuration bundle $Q\to\mathbb R$, but if we don't care about generalizations, we can get away with considering the Lagrangian a function $\mathrm TM\times\mathbb R\to\mathbb R$ for $Q=M\times\mathbb R$ – Christoph Jun 12 '13 at 9:21