Is the Lagrangian a kind of Energy

[Kevin_spencer2]

If we have that Lagrangian is equal to L=T-V and Hamiltonian is just H=T+V my question is if for every lagrangian we could consider is just the energy of a particle with a potential -V(x) instead of V(x) what would happen if V(-x)=-V(x)?

In fact if we substitute velocities by momenta in the Lagrangian the Euler-Lagrange equation for a particle is just (putting -V(x) instead of V(x)) the Hamilton equation $$\dot p =-\frac{dH}{dp}$$.

then it occurred to me if we could use this fact to quantizy the dynamics of a particle using Lagrangians as if they were the pseudo-energy of the system to get energy levels.

 

[dextercioby]

Nope, it doesn't work that way. Lagrangians are not observables, they cannot be measured. There's nothing "pseudo" about them. Their unit is energy, just like for a Hamiltonian, but nothing more. They're just a tool. Very useful in quantum physics, i must say.

Daniel.

 

[Kevin_spencer2]

then, what would we get if we put $$L(p,q)|\Phi >$$ where our $$\Phi$$ is just the wave function of a quantum system

 

[StatMechGuy]

then, what would we get if we put $$L(p,q)|\Phi >$$ where our $$\Phi$$ is just the wave function of a quantum system

I'm not sure why you'd want the eigenstates of the lagrangian. It's certainly observable, but I think that's beyond the point of this thread (being in classical mechanics). I personally have never seen it used, which doesn't mean that it's not of value, it's just that lagrangians enter quantum mechanics in a way that is a little independent of state vectors.

You can measure the lagrangian by just measuring the kinetic and potential energy, certainly. But the physical quantity of most interest is the action, which is just $$S = \int_t L(q, \dot{q},t)$$, and which is the basis for a lot of least-action principles, etc.

 

[Kevin_spencer2]

- the question is without prior knowledge of the system if they give you a quantity $$F(\dot q , q)$$ how could you know if you've got the energy of the whole system or just the Lagrangian?, in fact you could write the minimum-principle in the form $$S=\int_{a}^{b}dt \mathcal H(q,\dot q, t)$$ where H has -V(x) instead of V(x), the question is that it's easier to work with Lagrangians rather than Hamiltonians, and that for example the Lagrangian for GR is just $$(-g)^{1/2}R$$, however the Hamiltonian is just 0, which arises to lots of problems.

 

[StatMechGuy]

The hamiltonian need not equal T + V; it depends on the form of the lagrangian. The lagrangian is also not, generally, T - V -- look to magnetic phenomena to confirm this.

 

[dextercioby]

- the question is without prior knowledge of the system if they give you a quantity $$F(\dot q , q)$$ how could you know if you've got the energy of the whole system or just the Lagrangian?, in fact you could write the minimum-principle in the form $$S=\int_{a}^{b}dt \mathcal H(q,\dot q, t)$$ where H has -V(x) instead of V(x), the question is that it's easier to work with Lagrangians rather than Hamiltonians, and that for example the Lagrangian for GR is just $$(-g)^{1/2}R$$, however the Hamiltonian is just 0, which arises to lots of problems.

The Hamiltonian is not 0 for GR, search for ADM Hamiltonian formulation of general relativity.

Daniel.