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Landau & Lifshitz write on the first page of chapter 2 of their Mechanics book (p.13)

The number of independent integrals of motion for a closed mechanical system with $s$ degrees of freedom is $2s-1$.

Then they go on

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 $t_0$ in time. Eliminating $t + t_0$ from the $2s$ functions $$q_i=q_i(t+t_0,C_1,C_2, \ldots, C_{2s-1}),\qquad \dot{q}_i=\dot{q}_i(t+t_0,C_1,C_2, \ldots, C_{2s-1}), $$ we can express the $2s-1$ arbitrary constants $C_1,C_2, \ldots, C_{2s-1}$ as functions of $q$ and $\dot{q}$ (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
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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

2 Answers 2

Well I was still confused after reading these answers so I found a book "Cosmology and astrophysics through problems" by T Padmanabhan that finally answered the question in a (fairly) clear way.

Basically for a system with n degrees of freedom, you have $n$ Euler-Lagrange (E-L) equations. Since these equations are second order partial differential equations, to completely solve each equation (and thereby get an "equation of motion" or EOM) you need to specify two initial conditions, or 2 constants, per equation.

Thus for $n$ EOM you have $2n$ constants in total. Now each EOM is going to be written in terms of two variables - one position coordinate and the other a time derivative of position (speed). Call these, respectively, $q_i$ and $\dot {q_i}$ for the $i$th equation ($i$ going from 1 to $n$). Let the two constants per equation be $A_i$ and $B_i$.

Now apparently if the equations of motion ($n$ in total) are integrable, then you can write $q_i$ and $\dot{q_i}$ individually as functions of the initial conditions and time. So you have:

$$ q_i = q_i(t,A_i,B_i)\text{ and }\dot{q_i} = \dot{q_i}(t,A_i,B_i) $$

Then the book states that you can in principle invert these equations such that the constants are written in terms of $q_i$, $\dot{q_i}$ and $t$. That is,

$$ A_i = A_i(q_i,\dot{q_i},t)\text{ and }B_i = B_i(q_i,\dot{q_i},t) $$

So now we have $2n$ equations that each equal a constant or initial condition. But by definition an integral of motion is an equation that only depends on the initial conditions - it doesn't change with time. Since all of these $2n$ equations are equal to an initial condition, then you have $2n$ integrals of motion.

But one of those constants can actually be discarded, which will be $t_0$, the initial time, which is chosen arbitrarily. It is assumed that all integrals of motion are time-independent, so therefore this integral of motion doesn't contribute anything to the overall system - i.e. $t_0$ provides no new information. Thus we can remove one of the constants.

So you have $2n-1$ constants (or integrals of motion) left over. I know that in this proof a lot of assumptions were made, to be honest it doesn't seem very rigorous to me to continuously say "in principle", or to say "there will always be a $t_0$" (that isn't obvious to me either). I wish it were more explicitly proven but this is the best I could find. I wish physics was more like math :(. At least I know there is a clear, rigorous answer even if its difficult to understand.

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Hi Peter, Welcome to Physics.SE! I've added MathJax (Latex) encoding to show the formula; you can view the How-to here if you want to add more or answer other questions! –  Kyle Kanos yesterday

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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