# Integrals of Motion

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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Elaborate on what? The dynamics is second-order so there will always be 2N parameters (I don't like that Landau calls them integrals of motion since they are not what is understood by that term in modern physics, i.e. dynamically conserved quantities) for the system with N degrees of freedom. Picking initial time eliminates one of the parameters and leaves you with 2N -1. –  Marek Aug 22 '11 at 5:40
Why people often say "Landau writes", "Landau calls"? It is Landau and Lifshitz who prepared this physics course. Moreover, there is no single sentence that Landau wrote in his famous course. All "paperwork" was due to Lifshitz, Landau was the inspirer, the scientific adviser and the editor of the course. Sometimes people even joking like: "In the physics course by L&L there is no single word of Landau and no single thought of Lifshitz" (c) –  Physicsworks Aug 22 '11 at 9:21
@Physicsworks: so in view of Landau's supposedly horrible case of writer's block in his later life, should we say "Landau calls" and "Lifshitz writes"? :-) –  Willie Wong Aug 22 '11 at 16:27
@Willie Wong: yeah :) BTW: Since early 30's Lifshtiz even wrote scientific papers for Landau (I heard this from one of Landau's coworkers). –  Physicsworks Aug 22 '11 at 18:19
@Marek: I am an beginning undergraduate so 'naturally' am not familiar with the mathematical formalism (or maybe it is just because am a dumb ass). Whatever be the reason, I am thinking of the 2n parameters as the n co-ordinates and n velocities(varying in time). I agree with the fact that the origin of time can be chosen arbitrarily. However the following argument escapes me. I have tried to made sense of it, but have little belief in my ideas. So, HELP ME OUT!!! –  Sourav Aug 23 '11 at 1:04
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 $$x=C_1(t-t_0).$$