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Generalized momentum and Hamiltonian over a non inertial reference frame

  1. Jun 29, 2012 #1
    Hi there. I need help to work this out.

    A particle with mass m is studied over a rotating reference frame, which rotates along the OZ axis with angular velocity [tex]\dot\phi=\omega[/tex], directed along OZ. It is possible to prove that the potential (due to inertial forces) can be written as:
    [tex]V=\omega \cdot L-\frac{1}{2}m(\omega\times r)^2[/tex]
    L denotes the angular momentum round the origin O. Determine:
    a) The generalized moment taking as generalized coordinates the cartesian coordinates (X,Y,Z) taken over the rotating system.
    b) The generalized moment taking as generalized coordinates the cylindrical coordinates [tex](\rho,\phi,Z)[/tex] taken over the rotating system.
    c) Use the corresponding Legendre transformation, assuming there are no additional forces to find the Hamiltonian. Demonstrate that the Hamiltonian is:
    [tex]H=H_0-\omega \cdot L[/tex]
    Where H0 is the hamiltonian for a free particle.

    Excuse my english :P

    I don't know how to start. I've tried making a transform from x', y',z' inertial coordinates, using a rotation. Lets say:
    [tex]x'=X \cos\phi-Ysin\phi[/tex]

    Should I just use this transformation to get the kinetic energy and then just set L=T-V?

    Thanks for your help :)
  2. jcsd
  3. Jul 3, 2012 #2
    Last edited: Jul 3, 2012
  4. Jul 3, 2012 #3
    Alright, it was easier than what I thought. The fictitious potential contained all the terms that appear considering the velocity over the non inertial reference frame. So all what I had to do was [tex]T=\frac{m}{2}(\dot x^2+\dot y^2)[/tex]

    I had to use the transformation to realize about it, but I think that what I did is quiet correct. Anyway, I couldn't completely verify the equality in c), I neither did b). On c), I get something that looks pretty much like what it gives, but I've probably made some algebra mistake somewhere, and I get an extra term [tex]H=H_0-\omega \cdot L+\frac{m \omega^2}{2}(x^2+y^2)[/tex]


    From the transformation I got:
    [tex]\dot x^2+\dot y^2=\dot x'^2+\dot y'^2-2\omega \dot x'^2y'+2\omega \dot y'x'+\omega^2(x'^2+y'^2)[/tex]

    Thats the square of the velocity for a particle moving on the rotating frame with a speed [tex]\dot x+\dot y[/tex] with respect to the rotating frame, measured from the inertial reference frame.
    Last edited: Jul 3, 2012
  5. Jul 15, 2012 #4
    I finally got what I was looking for, but I'm not sure why. I had to use in the first place the lagrangian obtained using the fictitious potential. From this lagrangian I've obtained the generalized momentums with respect to the rotating frame. That cofuesed me a little bit, because I had moments with respect to both reference frames, I wasn't sure to which corresponded the ones that appeared in the fictitious potential, but now I'm pretty sure those correspond to the moments taken in the inertial reference frame. Once I got the generalized moments, using the lagrangian with the fictitious potential, I had to construct the Hamiltonian, using those moments, but when considering the lagrangian I just had to consider the lagrangian with out the fictitious potential. I think that information is already given in the generalized moments. When I was constructing the hamiltonian that I have previously posted, I was considering the Lagrangian as the one with the fictitious potential included. But the result is obtained by considering the lagrangian with out that potential, with the kinetic energy just as the sum of the components:
    [tex]\dot x+\dot y[/tex]

    That is: [tex]H=p_x \dot x+p_y \dot y-\frac{m}{2}(\dot x^2+\dot y^2)[/tex]
    Where the momentums are obtained from the Lagrangian with the fictitious potential.

    Assuming there is no other potential energy. In that way I obtain the result given by the exercise. I realized about it because just in that way I don't get the extra terms, but I didn't get the more profound reasoning on why it must be done this way, on a deeper physical sense. In the first place I thought of using the fictitious potential in the lagrangian for this hamiltonian, but it's like some combination of things, which confuses me a bit.
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