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

Question

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.

Answer 1

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 :

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

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 $\dot q = 0$

Answer 2

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.

Answer 3

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: $$H=-L.$$