1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Physical interpretation of one of Hamilton's equations

  1. Oct 11, 2012 #1
    1. The problem statement, all variables and given/known data

    I've attached a picture from a passage of my book (Liboff, Quantum Mechanics) with which I am having difficulty. Specifically, equation 1.25 claims to possess a certripetal force factor (in the text underneith) and a moment arm factor. I see both of these terms present. However, shouldn't the centripetal force factor reduce to mv2/r? Also, i assume the moment arm is included because torque is formally (clasically) r x F. However, here θ denotes the polar angle, and so rcos(θ) is the z-component of r. As a result, I dont see how this expression amounts to the torque. Can someone clarify this for me?


    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Oct 11, 2012 #2
    Okay, I may have made some progress already:

    For everything below, phi = σ.

    In this case, mv2/r is equal to m(rσ)2/r = mrσ2. Now, r x F = |r| |F| sin(ζ), where ζ is the angle between r and F. But this must mean here that sin(θ)cos(θ) = sin(ζ), and this is (perhaps?) what I don't see.

    **Upon second thought, this centripetal force arises from motion in the phi direction at constant r and theta, so the cross product is simply the product of the magnitudes r and F (since the angle between these two vectors is always 90 degrees). Thus, disregard my comments above.
     
    Last edited: Oct 11, 2012
  4. Oct 13, 2012 #3
    Still looking for help!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Physical interpretation of one of Hamilton's equations
Loading...