Hamiltonian vs. Lagrangian

  • Thread starter CMBR
  • Start date
  • #1
13
0

Homework Statement


So I just learned how to derive the equation of motion under the Lagrangian formulation which involves finding the euler-lagrange equation when setting the change in action to zero, chain rule, integration by parts etc.. Then I learnt how to find the equations of motion under Hamiltonian formulation, you take the legendre transformation of the Lagrangian, then take partial derivative of the hamiltonian w.r.t momentum & general coordinates to find the Hamiltons equations.

I feel my fundamental understanding is just not there, is there no concept of "minimum action" when deriving equations of motion under the hamiltonian formulation? Is there no such thing as dS=∫dt (H-H0)=0 as there is dS=∫dt (L-L0)=0 in lagrangian mechanics?


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
13,025
6,911
This might help your understanding:

http://en.wikipedia.org/wiki/Hamiltonian_mechanics

Basically Hamiltonians describe the total energy of the system vs Lagrangian's which describe the difference between kinetic and potential energy to define the action which systems will tend to minimize as they change state ( aka least action or stationary action).

In contrast, the Hamiltonian can interpreted as follows:

Basic physical interpretation

A simple interpretation of the Hamilton mechanics comes from its application on a one-dimensional system consisting of one particle of mass m under no external forces applied. The Hamiltonian represents the total energy of the system, which is the sum of kinetic and potential energy, traditionally denoted T and V, respectively. Here q is the coordinate and p is the momentum, mv. Then

\mathcal{H} = T + V , \quad T = \frac{p^2}{2m} , \quad V = V(q).

Note that T is a function of p alone, while V is a function of q alone.

In this example, the time-derivative of the momentum p equals the Newtonian force, and so the first Hamilton equation means that the force equals the negative gradient of potential energy. The time-derivative of q is the velocity, and so the second Hamilton equation means that the particle’s velocity equals the derivative of its kinetic energy with respect to its momentum.

I recall it was sometimes easier to find solutions using Hamiltonian mechanics first order DE vs the Lagranigian 2nd order DE but don't quote me on this. (too many years ago)

Hamilton's equations consist of 2n first-order differential equations, while Lagrange's equations consist of n second-order equations. However, Hamilton's equations usually don't reduce the difficulty of finding explicit solutions. They still offer some advantages, since important theoretical results can be derived because coordinates and momenta are independent variables with nearly symmetric roles.

Hamilton's equations have another advantage over Lagrange's equations: if a system has a symmetry, such that a coordinate does not occur in the Hamiltonian, the corresponding momentum is conserved, and that coordinate can be ignored in the other equations of the set. Effectively, this reduces the problem from n coordinates to (n-1) coordinates. In the Lagrangian framework, of course the result that the corresponding momentum is conserved still follows immediately, but all the generalized velocities still occur in the Lagrangian - we still have to solve a system of equations in n coordinates.[2]
 

Related Threads on Hamiltonian vs. Lagrangian

  • Last Post
Replies
13
Views
15K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
0
Views
6K
  • Last Post
Replies
3
Views
4K
Replies
3
Views
517
Replies
6
Views
471
Replies
2
Views
1K
Replies
2
Views
4K
Replies
1
Views
471
Top