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Integrals of the motion

  1. Nov 20, 2011 #1

    Geofleur

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    I can't seem to understand why, if there are s generalized coordinates, there end up being only 2s-1 integrals of the motion.

    The solutions of Lagrange's equation will have 2s constants. Why couldn't one simply solve the 2s equations for the solutions q_i and dq_i/dt for the 2s constants?

    For 1D motion of a free particle, for example, x = C_1*t + C_2 and dx/dt = C_1. Because the particle's velocity is constant, dC_1/dt = 0. Also, C_2 = x - dx/dt * t, so that dC_2/dt = dx/dt - dx/dt = 0. So C_1 and C_2 both appear to be constants of the motion. Integrals and constants of the motion are the same thing aren't they?

    I'm thinking of the paragraph in Landau and Lifgarbagez's Mechanics where they say that that there are only 2s-1 integrals and that one of the solution constants can be taken as "an additive constant t_0 in the time". This statement also confuses me - the constants are determined by initial positions and velocities and end up involving these quantities. In what sense can you choose one of them as t_0?
     
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  3. Nov 20, 2011 #2

    Geofleur

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    OK, I see that I've missed an important qualification: There are only 2s-1 *independent* integrals of the motion. Is it true, in my example above, that because C_2 = x - C_1*t, it is not independent of C_1? If so, I still do not understand what taking one constant as an additive t_0 in time has to do with it.
     
  4. Nov 21, 2011 #3

    Geofleur

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    Since no one has responded so far, what I'm about to say may be of use to someone.

    An integral of the motion must depend only on the generalized positions and velocities, not on time. So in the last post, C_2 = x - C_1*t is not an integral of the motion (though it IS a constant of the motion).

    I interpret L&L to mean the following on page 13 of Mechanics:

    One can choose one's inertial frame (of which there are an infinity, all moving at constant velocities relative to one another) such that one of the constants in the solution of Lagrange's equations will involve a constant t_0 in such a way that all of the q_i and dq_i/dt's become functions of (t+t_0). Then one can solve for t+t_0, which will eliminate one of the equations for the q_i or dq_i/dt's, giving the rest of the 2s-1 q_i, dq_i/dt's in terms of 2s-1 solution constants. These equations can be inverted to solve for the solution constants, and the resulting expressions will be the independent integrals of the motion.

    NOTE: I worked this out in detail for the case of a falling particle under a constant gravitational field. The one resulting integral of the motion turned out to be x - (dx/dt)^2/(2g).
     
    Last edited: Nov 21, 2011
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