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!

A Generalized Coordinates and Porn

  1. Jun 30, 2016 #1
    Yes, that is a serious title for the thread.

    Could someone please define GENERALIZED COORDINATES?

    In other words (and with a thread title like that, I damn well better be sure there are other words )
    1. I understand variational methods, Lagrange, Hamilton, (and all that).
    2. I understand the pendulum and the distinction between x/y and r/theta
    3. I understand how generalized velocities can depend on generalized coordinates and so on.
    4. I understand how they represent the minimum variables needed to describe a system...

    OK. But could someone provide a clear, concise definition of the word "generalized?" What makes x/y Cartesian and r/theta "generalized?" When does one have the right to attach the modifier "generalized" to a coordinate system describing a mechanical (or otherwise) system?

    What is a generalized coordinate?
    (I know it when I see it -- like porn -- but I can't define it.)
     
  2. jcsd
  3. Jun 30, 2016 #2
    Generalized coordinates are local coordinates on configuration manifold
     
  4. Jun 30, 2016 #3
    Wow... that was good... thanks!

    May I ask for one more thing?

    It turns out the in classical mechanics, the kinetic energy is not just a function of the generalized velocities. It is also a function of the generalized coordinate.

    (As you must well know, KE = 0.5 * m * v * v. But when the coordinates are generalized, the coordinate also appears in the KE.)

    In the context of your previous answer, could you demonstrate why this happens?
     
  5. Jun 30, 2016 #4
    the kinetic energy is Riemann metric on configuration manifold (precisely speaking, quadric part of the kinetic energy)

    Assume we have a particle of mass ##m## moving on a plane. The kinetic energy is ##T=m(\dot x^2+\dot y^2)/2##; now express the kinetic energy in terms of polar coordinates: ##x=r\cos\phi,\quad y=r\sin\phi##;
     
  6. Jun 30, 2016 #5
    And, thank you once again!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Generalized Coordinates and Porn
Loading...