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Why is the Hamiltonian constructed from the Lagrangian?

  1. Feb 28, 2013 #1
    I understand how to use Hamiltonian mechanics, but I never understood why you construct the Hamilitonian by first constructing the Langrangian, and then performing a Legendre transform on it.

    Why can't you just construct the Hamiltonian directly? Does it have to do with the generalized coordinate [itex] p_i [/itex] being more difficult to think about than the coordinate [itex] \dot{q_i} [/itex]? This seems like the only reason to start with the Lagrangian first, if your goal is to build the Hamiltonian.
     
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  3. Feb 28, 2013 #2

    Simon Bridge

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    I understand that you can construct the Hamiltonian directly - it's pretty much what you do in quantum wave mechanics.

    The reason it tends not to be taught that way in classical mechanics is that the Lagrangian is usually easier to build (and use.)
     
  4. Feb 28, 2013 #3

    Mute

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    The Lagrangian is just "T-V": kinetic energy minus potential. For many systems of interest, it turns out that the Hamiltonian is T+V; however, that not need be the case. There are systems for which the Hamiltonian is not simply T+V, and I would guess that one may not be able to write down the Hamiltonian without first going through the Lagrangian.
     
  5. Mar 1, 2013 #4

    stevendaryl

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    I think that's exactly right. Velocities are easy-to-understand quantities, and momenta are a little more complicated. The relationship between velocity and momentum is not simple, in general. For example, in the presence of an electromagnetic field corresponding to vector potential [itex]A^\mu[/itex], the momentum is not [itex]m \dfrac{dx^\mu}{d \tau}[/itex], but is [itex]m \dfrac{dx^\mu}{d \tau} - e A^\mu[/itex]. That relationship comes out automatically using a simple Lagrangian, but it would be hard to guess if you started with a Hamiltonian.

    On the other hand, there is a formalism that's equivalent to Lagrangian that uses the Hamiltonian. Instead of extremizing the action [itex]\int L(q,\dot{q}) dt[/itex], one extremizes the action: [itex]\int (H(q,p) - p \dot{q}) dt[/itex]

    This approach puts coordinates and momenta on sort of the same footing.
     
  6. Mar 1, 2013 #5
    Interesting responses. Thanks, I see now the advantage!
     
  7. Mar 1, 2013 #6

    WannabeNewton

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    What you will find if you learn classical mechanics from the viewpoint of symplectic geometry is that the Hamiltonian is relatively easier to start with. The cotangent bundle of the configuration space has a "god - given" symplectic 2 - form that relates the differential of the Hamiltonian to the corresponding Hamiltonian vector field via the symplectic form. If you wanted to do the same thing relating to Lagrangians; however, you would have to do some more work on your part.
     
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