Landau writes in chapter 2 of his Mechanics book "The number of independent integrals of motion for a closed mechanical system with s degrees of freedom is 2s-1". Then he goes on to show how -- "Since the equations of motion for a closed system do not involve time explicitly, the choice of the origin of time is entirely arbitrary, and one of the arbitrary constants in the solution of the equations can always be taken as an additive constant t0 in time. Eliminating t + t0 from the 2s functions, we can express the 2s-1 arbitrary constants as functions of q and q’ (generalized co-ordinates and velocities) and these functions will be the integrals of the motions." Could someone elaborate?
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I think the source of your confusion is mathematics, not physics. It is important here (and L&L did mention this) that the system of differential equations is autonomous. If this is the case, than along with solution $q_i=q_i(t,C_1,\dots,C_{2s}),$ it has a solution $q_i=q_i(t-t_0,C_1,\dots,C_{2s}).$ Because the former is the general solution, the later must reduced to it, that is, it should be $$q_i(t-t_0,C_1,\dots,C_{2s})=q_i(t,C_1'(C_1,\dots,C_{2s},t_0),\dots,C_{2s}'(C_1,\dots,C_{2s},t_0)).$$ Putting $C_{2s}=0$, yields $$q_i(t-t_0,C_1,\dots,C_{2s-1})=q_i(t,C_1'(C_1,\dots,C_{2s-1},t_0),\dots,C_{2s}'(C_1,\dots,C_{2s-1},t_0)).$$ For example, in the case of a free 1D motion $x=C_1t+C_2$. Making the shift in time one has: $$x=C_1t+(C_2-C_1t_0)$$ (expression in brackets is $C_2'$). And letting $C_2=0$ we find $$x=C_1(t-t_0).$$ |
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:-)– Willie Wong Aug 22 '11 at 16:27