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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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    $\begingroup$ 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. $\endgroup$
    – Marek
    Aug 22, 2011 at 5:40
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    $\begingroup$ 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) $\endgroup$ Aug 22, 2011 at 9:21
  • $\begingroup$ Concerning constants of motion vs. integrals of motion, see this Phys.SE post. $\endgroup$
    – Qmechanic
    Mar 1, 2015 at 21:57

4 Answers 4

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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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  • $\begingroup$ Why put $C_{2s} = 0$? $\endgroup$
    – Cheng
    Sep 14, 2022 at 3:06
  • $\begingroup$ It is autonomous only for closed systems right? Or at least when the Lagrangian is time independent? $\endgroup$
    – Cheng
    Sep 14, 2022 at 4:26
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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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    $\begingroup$ 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! $\endgroup$
    – Kyle Kanos
    Mar 1, 2015 at 21:37
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One more easy way to understand it is as follows. Motion of a particle can be described as a trajectory in 2n dimensional phase space (composed of n positions and n conjugate momenta). Now a function of 2n variable of the form f1(q1,q1,..qn)=c will give a 2n-1 dimensional hypersurface. So in order to describe a 1-d trajectory you will need 2n-1 such equations i.e f1(q1,q1,..qn)=c1,f2(q1,q1,..qn)=c2...etc. And hence there are 2n-1 functions of the dynamical variables which are constants of motion. Hope that helps.

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I am not a native speaker of English and still a high school student. So the explanations below may contain some errors or unclear parts.

​ When I am reading the Mechanics, I am also very confused about this paragraph. But, luckily, I finally found one explanation with clear and step-by-step derivations. The original explanation was posted in a Chinese Q&A community by an already deleted account. So, here, I will present this idea in English, in the hope to help other beginners just like me.

​ Firstly, from the Lagrangian Equation, we will have a system of simultaneous equation: $$ \left\{ \begin{array}{ll} \frac{\partial L}{\partial q_1}-\frac{\text{d}}{\text{d}t}\frac{\partial L}{\partial \dot q_{1}} = 0\\\cdot\\\cdot\\\cdot\\ \frac{\partial L}{\partial q_s}-\frac{\text{d}}{\text{d}t}\frac{\partial L}{\partial \dot q_{s}} = 0 \end{array} \right. $$ ​ Then, we can write the left side as $g_i(q;\dot q ;\ddot q)$, this is because Lagrangian of a closed system only depends explicitly on the generalised coordinates and velocity.

​ [Here, the $q$ and $\dot q$ without superscript or subscript just represent the collection of $q_i$ or $\dot q_i$, $i.e.$ $g_i(q;\dot q ;\ddot q)$ $ = $ $g_i(q_1,...q_s,$$\dot q_1,...\dot q_s,$$\ddot q_1,...,\ddot q_s)$ ]

​ As a result, we can rewrite the system as $$ \left\{ \begin{array}{ll} g_1(q;\dot q ;\ddot q) =0 \\ \cdot \\ \cdot \\ \cdot \\ g_s(q;\dot q ;\ddot q) =0 \end{array}\right. $$ ​ Then we can treat $q$ and $\dot q$ as constants, and rewrite the system as a linear system of $\ddot q$.

​ [I think, mostly, we can separate the accelerations from each other as all the accelerations come from the $\frac{\text{d}}{\text{d}t}\frac{\partial L}{\partial \dot q}$ terms. ] $$ \left\{ \begin{array}{ll} R_{11}\ddot q_1 + R_{12}\ddot q_2 + ...+ R_{1s}\ddot q_s = K_1 \\ \cdot \\ \cdot \\ \cdot \\ R_{s1}\ddot q_1 + R_{s2}\ddot q_2 + ...+ R_{ss}\ddot q_s = K_s \end{array}\right. $$ ​ [$R_{ij}$ and $K_i$ are just some functions about $q$ and $\dot q$]

​ By some eliminations, we should be able to get $$ \left\{ \begin{array}{ll} \ddot q_1 = Z_1 (q; \dot q) \\ \cdot \\ \cdot \\ \cdot \\ \ddot q_s = Z_s (q; \dot q) \end{array}\right. $$ ​ [$Z_i$ are just some functions depends on $q$ and $\dot q$ ]

​ Now, before we manipulate the equations further more, we need to recall some mathematical tricks for solving the second-order differential equations. For example, in the simple harmonic motion, we have $$ a = -mx^2 $$ we will not integrate it directly with time, instead, we integrate both sides with $\text{d}x$ as $$ a = \frac{\text{d}v}{\text{d}x} \frac{\text{d}x}{\text{d}t} = v\frac{\text{d}v}{\text{d}x} $$ ​ Back to our question, we can apply the similar tricks. $$ \ddot q_i =\frac{\text{d}\dot q_i}{\text{d} q_1} \frac{\text{d}q_1}{\text{d}t} =\dot q_1\frac{\text{d}\dot q_i}{\text{d}q_1} $$ ​ Thus, we can have $$ \left\{ \begin{array}{ll} \dot q_1\frac{\text{d}\dot q_1}{\text{d}q_1} = Z_1 (q; \dot q) \\ \dot q_1\frac{\text{d}\dot q_2}{\text{d}q_1} = Z_2 (q; \dot q) \cdot \\ \cdot \\ \cdot \\ \dot q_1\frac{\text{d}\dot q_s}{\text{d}q_1}= Z_s (q; \dot q) \end{array}\right. \Rightarrow \left\{ \begin{array}{ll} {\text{d}\dot q_1} = \frac{Z_1 (q; \dot q)}{\dot q_1}{\text{d}q_1} \\ {\text{d}\dot q_2} = \frac{Z_2 (q; \dot q)}{\dot q_1}{\text{d}q_1}\\ \cdot \\ \cdot \\ \cdot \\ {\text{d}\dot q_s} = \frac{Z_s (q; \dot q)}{\dot q_1}{\text{d}q_1} \end{array}\right. $$ ​ Then, we integrate both side, $$ \left\{ \begin{array}{ll} \dot q_1 + A_1= M_1(q;\dot q) \\ \dot q_2 + A_2= M_2(q;\dot q)\\ \cdot \\ \cdot \\ \cdot \\ \dot q_s + A_s= M_s(q;\dot q) \end{array}\right. $$ ​ [$A_i$ are some arbitrary constants which, we will see, belong to the 2s-1 constants]

​ Now, similar to the argument above, we treat $q$ as constants. Here, I didn't find it very intuitive to assume that all $\dot q_i$ can be separated from each other. But for the sake of the argument, we just accept that we can solve the system into $$ \left\{ \begin{array}{ll} \dot q_1 = N_1(q;A) \\ \dot q_2 = N_2(q;A)\\ \cdot \\ \cdot \\ \cdot \\ \dot q_s = N_s(q;A) \end{array}\right. $$ [A represents the collection of $A_i$]

​ Then we do our math trick again, but, notice, we cannot apply it for the first equation $$ \left\{ \begin{array}{ll} \dot q_1 = N_1(q;A) \\ \text dq_2 \dot q_1 = N_2(q;A)\text d q_1\\ \cdot \\ \cdot \\ \cdot \\ \text dq_s \dot q_1 = N_s(q;A)\text d q_1 \end{array}\right. \Rightarrow \left\{ \begin{array}{ll} \dot q_1 = N_1(q;A) \\ \text dq_2 = \frac{N_2(q;A)}{ N_1(q;A)}\text d q_1\\ \cdot \\ \cdot \\ \cdot \\ \text dq_s = \frac{N_s(q;A)}{ N_1(q;A)}\text d q_1 \end{array}\right. $$ ​ Then we can solve all the equations except the first one. $$ \left\{ \begin{array}{ll} q_2 +B_2 = P_2(q;A)\\ \cdot \\ \cdot \\ \cdot \\ q_s +B_s = P_s(q;A) \end{array}\right. $$ ​ [$B_i$ are $s-1$ arbitrary constants which belong to the $2s-1$ constants at the end]

​ Again, by treating $q_1$ as a constant and solving the system, we can get $$ \left\{ \begin{array}{ll} q_2 = Q_2(q_1;A;B)\\\cdot \\ \cdot \\\cdot \\q_s = Q_s(q_1;A;B)\end{array}\right. $$ ​ Then how about the first equation regarding $\dot q_1$ ?

​ Let us try another approach for the equation. $$ \begin{align} & \, \,\,\,\,\,\,\, \frac{\text d q_1}{\text dt} = N_1(q,A)\\ &\Rightarrow t - t_0 = F(q,A) \end{align} $$ Then adding this new equation to the system above, $$ \left\{ \begin{array}{ll} F(q,A) = t-t_0\\ q_2 = Q_2(q_1;A;B)\\\cdot \\ \cdot \\\cdot \\ q_s = Q_s(q_1;A;B)\end{array}\right. $$ ​ By assuming the system is solvable, we should get our final result, all the coordinate can be represented by $$ q_i = C_i(t-t_0; A;B) $$ ​ Don't forget, A is the collection of $s$ arbitrary constants and B is the collection of $s-1$ arbitrary constants. Hence, in total, we can have only $2s-1$ arbitrary constants here.

​ In a nutshell, I think the reduction of 1 arbitrary constant here is just simply because we do no need to take time into considerations and we can find a nice relationship of $\text d t= f(q)\text d q_1 $.

​ Hope you find it helpful!

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  • $\begingroup$ Your mentioned the trick $$\ddot q_i =\frac{\text{d}\dot q_i}{\text{d} q_1} \frac{\text{d}q_1}{\text{d}t} =\dot q_1\frac{\text{d}\dot q_i}{\text{d}q_1}$$ However, $$\ddot q_i = \frac{d}{dt} \frac{dq}{dt} = \Sigma_ j \frac{\partial \dot q_i}{\partial q_j} \frac{d q_j}{dt} $$ Are you suggesting that we invert $q_1 (t)$ and express $q_j(t) = q_j(q_1)$? Then $\dot q_i (q_1, q_2, ..., q_s) = \dot q_i(q_1)$. $\endgroup$
    – Cheng
    Sep 14, 2022 at 3:58
  • $\begingroup$ @Cheng, yep, I did not realise the detail here until I saw your comment. I think you are correct; I am essentially changing the variable here. $\endgroup$
    – Abl grp
    Oct 31, 2022 at 11:35

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