Exploring Lagrangians that Cannot be Transformed with Legendre

[irycio]

Hi!
Our TA told us, that it may be not always possible to change lagrangian into hamiltonian using Legendre transformation. As far as I'm concerned the only such possibility is that we can not substitute velocity (dx/dt) with momenta and location(s). And so, we've been tryging to come up with an idea of such terms in lagrangian that would not allow us to calculate the velocity explicitly. Now, of course, one can think of terms like (\frac{dx}{dt})^5+\alpha (\frac{dx}{dt})^4+... and so on, but they do not seem very physical to me ;).
And hence my question-do you guys know any examples of real-life :D lagrangians that wouldn't be subject to Legendre transformation?

Cheers and Mery Christmas :D

 

[lalbatros]

Hello irycio,

I found on wikipedia an obvious condition necessary to make the L to H transformation possible:

see there: en.wikipedia.org/wiki/Legendre_transformation#Hamilton-Lagrange_mechanics

Would there be some real-world examples where this condition is violated?

Michel

 

[lalbatros]

Just found this: http://www.worldscibooks.com/etextbook/7689/7689_chap01.pdf

You can read:

"... gauge theories are certainly the most important examples of singular Lagrangian systems ..."

I am not so familiar with this language.
I can only guess that a gauge theory involves some gauge invariance and that Maxwell's theory is the only example I know.
If this is the case, the Maxwell's theory should have a singular Lagrangian.

I remember that the quantification of Maxwell's theory is usually built on the Hamiltonian of the field.
I am almost sure that, at some point in the derivation, a specific gauge was chosen.
I can even remember that the Coulomb gauge was preffered, but I don't remember why.

Could you see why the Maxwell's theory lagrangian is singular?

Michel

My guess is that "the jacobian" will indeed be singular (0), but that most of "minor determinants" will not be,
and that chosing a specific gauge amounts to get rid a the useless degress of freedom.

 

[torquil]

Could you see why the Maxwell's theory lagrangian is singular?

Consider the Lagrangian F^2 of pure Maxwell theory. F doesn't contain the time derivative of A^0, the time-component of the 4-vector potential.

If you haven't chosen a gauge, the generalized coordinates q_i in this case are the four components of A.

Since L doesn't depend on dA^0/dt, there is a whole row and a whole column in the matrix of second derivatives of L given above, that vanishes. One when i=0 and one when j=0. Thus the determinant vanishes.

 

[lalbatros]

Thanks torquil.
Could you explain how chosing the Coulomb gauge (or the Lorenz gauge) removes this singularity?
Thanks again

 

[torquil]

Sorry, I don't have the time to write it up here, but I can recommend section 8.2 in Weinberg book "The quantum theory of fields, Vol.1".

 

[lalbatros]

Thanks torquil !
The three volumes are waiting on my shelves since a few years.
I will have a look.
Michel