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Equation of motion

  1. Aug 29, 2009 #1
    I came across this from a book saying that:
    If all the co-ordinates and velocities are simultaneously specified, it is known from experience that the state of the system is completely determined and that its subsequent motion can, in principle, be calculated. Mathematically, this means that, if all the co-ordinates q and velocities dq/dt are given at some instant, the accelerations d[tex]^{2}[/tex]q/dt[tex]^{2}[/tex] at that instant are uniquely defined.

    May i know what is meant by the last sentence?
  2. jcsd
  3. Aug 29, 2009 #2


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    This is an example of the application of Lagrangian physics. In Lagrangian physics, you can specify the Lagrangian of the system (the difference between the kinetic and potential energies) as a function of position, velocity and time. The integral of the Lagrangian over time is called the action. The equations of motion for a classical system are ones which make the action stationary. This can be solved in a variety of ways.

    The associated wiki: http://en.wikipedia.org/wiki/Action_(physics [Broken])

    So yes, the statement is true but there is a bit of work that you have to do to get the equations of motion out of it. I'm sure you could probably do it also via a different means, maybe using Hamiltonian physics but I am not very familiar with working with Hamiltonians (as horrible as that fact may be) outside of quantum physics.
    Last edited by a moderator: May 4, 2017
  4. Aug 29, 2009 #3
    It is a somewhat wrong statement. The accelerations are also determined with forces that should be given too. The external and inter-particle forces depend on coordinates and velocities, so if the force dependencies are known, the future of the system is calculated from the initial data and the given forces. It is just like for one-particle system: you specify the initial data and a force to calculate the time dependence of your variables.
  5. Aug 29, 2009 #4
    hmm...but i think the book should be more or less accurate...i quote this from the book "Mechanics" by L.D.Landau...i do have the same opinion as yours when i am reading this...But considering Born2bwire's i think the statement is alright, since the Lagrangian is T-U, in which the potential energy also being taken care of (i.e. if U depend on the position then it's being taken into account) =). Thanks everyone =)
    Last edited: Aug 29, 2009
  6. Aug 29, 2009 #5
    This is correct as long as the accelerations are independent of time. If every parameter is independent of time, then the motion is uniquely defined.
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