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B Question about Hamiltonian and kinetic energy

  1. Nov 30, 2016 #1
    basically, as far as I know we can derive 1/2mv2 from

    for wave equation we use Hamiltonian H=P2/2m+V where P and V are both operators
    However, I wonder how we can say that P2/2m is the term for kinetic energy because
    ∫F⋅ds=∫(dp/dt)⋅ds=1/2mv2 is saying that knowing F and path s, we can determine (p2)/2m. and given with an appropriate boundary condition, vice versa.

    Then, we can say that P2/2m is a functional F[p,s]=∫(dp/dt)⋅ds in which we have to determine p and s separately?

    Then KE is dependent on both s and p, and this seems a bit ambiguous to me because in Hamiltonian KE is dependent on momentum and potential energy is dependent on position. Dimension-wise, it would agree with energy but I wonder how we took (p2)/2m from ∫F⋅ds=∫(dp/dt)⋅ds=1/2mv2
    Last edited: Nov 30, 2016
  2. jcsd
  3. Dec 6, 2016 #2

    Simon Bridge

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    Look at the Hamiltonian mechanics description from classical mechanics.
    It is often glossed over in favour of Lagrangian mechanics - but many courses will give you a motivation for the Schrodinger equation from there.

    Also recall that ##\frac{1}{2m}\hat p^2## is not the kinetic energy - it is the kinetic energy operator.
  4. Dec 6, 2016 #3

    Thank you. I just understood this way. Although I learned that the definition of energy is ∫F⋅ds=1/2mv2, it is just an equation that explains the physical quantity energy w.r.t an observable "position". Equivalent expression w.r.t another observable momentum is ∫dp/dt⋅ds=∫dp⋅ds/dt and this is the same expression but written in terms of different observable.

    So yeah, I think thats it.
  5. Dec 6, 2016 #4

    Simon Bridge

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    That's a puzzling way to claim "understanding" there. Lets see if I can show you what I mean:

    The equation you are looking at is this:
    ##\qquad\qquad \int_C\vec F\cdot d\vec s = \frac{1}{2}mv^2##
    ... the LHS is called "work done by force F"... and it is specifically the amount of work needed to increase the speed from rest to ##v## - when nothing else is going on. So this is a long way from defining kinetic energy. What if the initial speed is not zero? Then the RHS has to be ##\frac{1}{2}m(v^2-u^2)##, using the usual suvat notation, right?

    If the object is subject only to a potential-energy ##V(\vec r)## then##\vec F = -\vec\nabla V## ... but ##\vec F## may be any combination of forces. It is ossible for force F to be needed to lift a mass m against gravity at constant speed ... since that constant speed can be anything, it does not always follow that ##v^2=\frac{2}{m}\int_c\vec F\cdot d\vec s## ... which is implied in what you wrote down.

    If you take the Newtonian definition of the net force, then ##\vec F = \frac{d}{dt}\vec p## ... and we should probably write the kinetic energy as ##p^2/2m## and include the possibility to a change in potential energy as well. So:

    $$\int_C \left(\frac{d}{dt}\vec p\right)\cdot d\vec s = \frac{1}{2m}|\vec p_f-\vec p_i|^2 + V_f - V_i$$ ... so the work done by the net force is equal to the change in total mechanical energy.

    Is the energy stored as rotation kinetic too? What about heat? Should these be included in the definition for kinetic energy too?
    All that is just in the classical regime.

    What I noticed from your original post is that you seemed to expect to understand a broad theory in terms of a subset of that theory you feel you do understand ... in post #3 you appear to be saying that your understanding of that subset is based on a very narrow set of situations within that subset that you feel comfortable with.

    As I have tried to explain to you: this is not how "understanding" works.
  6. Dec 19, 2016 #5

    Thanks for your reply. It helped a lot really :)
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