Finding the Hamiltonian if I'm given the Lagrangian

[fluidistic]

Homework Statement

Determine the Hamiltonian corresponding to the an-harmonic oscillator having the Lagrangian $L(x,\dot x )=\frac{\dot x ^2}{2}-\frac{\omega ^2 x^2}{2}-\alpha x^3 + \beta x \dot x ^2$.

Homework Equations

$H(q,p,t)=\sum p_i \dot q _i -L$.
$p _i=\frac{\partial L}{\partial \dot q _i}$.

The Attempt at a Solution

Using the equations in 2), I get $p_x= \dot x +2 \beta x \dot x$.
So that $p_x ^2= \dot x ^2 (1+4 \beta x +4 \beta ^2 x^2)$.
Now if I can write the following function in function of only p, q and t then it would be the Hamiltonian, but I couldn't do it.
Here's the function: $\dot x^2 \left ( \frac{1}{2} +\beta x \right ) +\omega ^2 \frac{x^2}{2}+\alpha x^3$.
I see absolutely no way to get rid of the x's terms (or q in this case).
Any help is appreciated. Thanks!

 

[Pengwuino]

I don't see what the issue is here. You already know how to exchange all $\dot{x}$ for momenta based off your first calculation. So just do that.

 

[mathfeel]

I see absolutely no way to get rid of the x's terms (or q in this case).
Any help is appreciated. Thanks!

Why do you want to get rid of $x$? The term you want to get rid of are $\dot{x}$, with the help of the definition of momenta.

But you are on the right track.

 

[fluidistic]

Whoops guys you are right. This is so new to me that I don't have almost any intuition in this stuff yet.
I get $H(p,q,t)= p_x ^2 \frac{\left (\frac{1}{2}+\beta x \right ) }{1+4 \beta x +4 \beta ^2 x^2}+\omega ^2 \frac{x}{2}+ \alpha x^3$. I hope it's right and it's the answer the exercise is looking for.
By the way so far it seems like that the Hamiltonians I've seen so far depend on $p _i ^2$ rather than $p_i$. Is there any example of a system whose Hamiltonian depends on $p_i$ rather than $p_i ^2$?

 

[jambaugh]

... Is there any example of a system whose Hamiltonian depends on $p_i$ rather than $p_i ^2$?

The problem there is that to carry out the Legendre transformation from $L(q,\dot{q})$ to [itex]H(q,p)[itex], the mapping must be invertible. You need to be able to express [itex]\dot{q} = F(q,p)[itex]where $p \equiv \partial L / \partial \dot{q}$.<br /> <br /> In particular the matrix of elements $M_{ij} = \frac{\partial^2 }{\partial \dot{q}_i\partial \dot{q}_j}L$ must be invertible.<br /> <br /> Given it is symmetric it can be diagonalized via a linear transformation on the q's. To be invertible none of the diagonal terms can be zero.<br /> <br /> From there I think you can get to a definite <b>No</b> to your question.<br /> <br /> Now you can get only mixed quadratic factors. For $L = \dot{x}\dot{y}$ you'll get $H = p_x p_y$.[/itex][/itex][/itex][/itex]

 

[fluidistic]

The problem there is that to carry out the Legendre transformation from $L(q,\dot{q})$ to [itex]H(q,p)[itex], the mapping must be invertible. You need to be able to express [itex]\dot{q} = F(q,p)[itex]where $p \equiv \partial L / \partial \dot{q}$.<br /> <br /> In particular the matrix of elements $M_{ij} = \frac{\partial^2 }{\partial \dot{q}_i\partial \dot{q}_j}L$ must be invertible.<br /> <br /> Given it is symmetric it can be diagonalized via a linear transformation on the q's. To be invertible none of the diagonal terms can be zero.<br /> <br /> From there I think you can get to a definite <b>No</b> to your question.<br /> <br /> Now you can get only mixed quadratic factors. For $L = \dot{x}\dot{y}$ you'll get $H = p_x p_y$.[/itex][/itex][/itex][/itex]

[itex][itex][itex][itex] Thank you very much. I must admit it's over my head for now; I hope I'll be able to fully understand this after taking the mathematical methods used in physics course.<br /> But the answer is good to know for me, at least I can know I obtained an error when I get that the Hamiltonian of a system depends on p_i alone.<br /> By the way, it seems what you said concern only systems where the energy is not dissipated, i.e. the Lagrangian and Hamiltonian do not depend explicitly on time. Would the conclusion be the same if $L(q,\dot q ,t)$ and $H(p,q,t)$ rather than just $L(q, \dot q)$ and $H(p,q)$? I'm guessing that it's an obvious "yes", but I just want to be 100% sure. <br /> Thanks.[/itex][/itex][/itex][/itex]

 

[mathfeel]

By the way so far it seems like that the Hamiltonians I've seen so far depend on $p _i ^2$ rather than $p_i$. Is there any example of a system whose Hamiltonian depends on $p_i$ rather than $p_i ^2$?

Yes, acoustic phonon (quantum of lattice vibration) in solid has low energy effective Hamiltonian $H = v |p|$, where $v$ is the speed of sound in the solid.

You can't really have $H \propto p$ because then you can keep have larger negative momentum and having lower and lower energy, which is strange. Well not completely, but that's a whole different discussion involving special relativity.