Question

Is there a way to derive a Lagrangian that's true to the second order?

Answer 1

Well, I don't know this... if you get the answer, do tell me....

Answer 2

context? what does our potential energy field look like?

Answer 3

There is no context, just curious since when deriving the Euler-Lagrange equation it just seems to only be true up to the first order, so I wondered if there was a way to derive to be true to the second or even higher orders.

Answer 4

Which part of the Euler-Lagrange equation is only true to first order? Do you mean in terms of only considering the first variation? Higher variations aren't needed for simply finding an extremum of a functional, although they have natural importance in classifying the extremum (compare with second derivative test; second variation can hopefully tell us it's a maximum or minimum). Do you mean in the fact the Lagrangian only involves velocities and not higher derivatives?

Answer 5

Yeah, I am asking why it only ends up with first derivatives and nothing higher, although your comment about the "second variation" to determine if I'm at a local min or max seems useful as well. In fact, I suppose is there some kind of Hessian of second variations we can use?

Answer 6

yes: http://liberzon.csl.illinois.edu/teaching/cvoc/node57.html

Answer 7

and the reason for the lack of higher derivatives has to do with physics itself -- consider Newton's law characterizes dynamics in terms of accelerations \(\ddot q\)

Answer 8

https://en.wikipedia.org/wiki/Ostrogradsky_instability

Answer 9

Ah these are both great, thanks. I have a few questions to go with this,

What are degenerate Lagrangians, is this related at all to degenerate orbitals in QM?

What is conjugate momenta? Is this somehow something relating the position and momentum of particles by something like a Fourier transform or is this something to do with \(L^2\) integrable spaces?

Answer 10

degeneracy of the Lagrangian tells us we have some redundancy because there seem to be more degrees of freedom than there actually are, it seems. it's not directly related to degenerate orbitals, no, but the word degenerate has similar meaning in both

Answer 11

conjugate momenta has to do with the fact that in the Hamiltonian formulation we have canonical coordinates \(q_i,p_j\) where \(p_j\) are the corresponding 'momenta'

Answer 12

https://en.wikipedia.org/wiki/Hamiltonian_mechanics#As_a_reformulation_of_Lagrangian_mechanics

Answer 13

Can you possibly give me the gist of what a Symplectic manifold is? I'm familiar with manifolds in the mathematical sense if we start say from this definition of a topological space: https://proofwiki.org/wiki/Definition:Topological_Space .