1. I was reading Landau & Lifschitz's book on Mechanics, and came across this sentence on p.19:

"There are no other additive integrals of the motion. Thus every closed system has seven such integrals: energy, three components of momentum, and three components of angular momentum".

However, no proof is given for this statement. Why is it true?

2. I find the statement somewhat counter-intuitive; it says at the beginning of the second chapter that for any mechanical system with $s$ degrees of freedom, there are at most $2s-1$ integrals of motion.

But the above statement would seem to imply that a system with three degrees of freedom has at least $2s+1$ integrals of motion. Why is this not a contradiction?

3. Finally, these integrals of motion correspond neatly to homogeneity of time (energy), homogeneity of space (momentum), and isotropy of space (angular momentum).

From this perspective it also makes sense why energy is "one-dimensional", since there is only one time dimension, and why momentum and angular momentum are "three-dimensional", since space has three dimensions.

However, why do the only additive integrals of motion correspond to these properties? What is special about them which guarantees that they have additive integrals of motion and that no other property can?

Even if you don't know the answer to all of these questions, I would really appreciate any help or insight you could give me. I was really enjoying this book until I thought of this question and now I am hopelessly confused.

  • $\begingroup$ Thank you for your help, although I don't really understand what you are saying. I wasn't trying to generalize the equation, I was just pointing out that for $s=3$ seven is strictly greater than $2s-1=5$. So in this case, wouldn't any system with three or fewer degrees of freedom be "superintegrable"? That seems surprising to me. I know that no other properties of spacetime are known, but why would additive integrals of motion correspond ONLY to known properties of spacetime? This is what I don't understand. $\endgroup$ May 4, 2016 at 20:19
  • $\begingroup$ Wait so why is the system a "black box" from the outside? This sounds like a really cool example, but I don't think I understand what you are talking about. I get at least the proportionality of energy to magnitude of momentum squared by the kinetic energy for a free particle. Also is what you are saying that for $s\le3$ there are no contradictions because in that case the additive integrals of motion can't be independent? So that the statement in the book only really applies for four or more dimensions of freedom, for which there is no problem/subtleties to consider? $\endgroup$ May 5, 2016 at 0:57
  • $\begingroup$ Also if you put your comments as an answer I will upvote them; I find them to be very helpful. We want two particles so we can connect them with a spring and thus reduce the degrees of freedom while somehow leaving the number of integrals of motion unaffected? Am I understanding this correctly? $\endgroup$ May 5, 2016 at 0:58
  • $\begingroup$ @CosmasZachos Sorry, one more question: is it possible to prove any of this using Poisson brackets? I.e. using either $\partial f /\partial t + [H,f]=0$ (42.3 in L&L) if and only if $f$ is an integral of motion (where $H$ is the Hamiltonian), or $[f,g]=constant$ (42.15 in L&L) for any two integrals of motion $f,g$ (Poisson' theorem)? Or do Poisson brackets have nothing to say about in particular additive integrals of motion? $\endgroup$ May 5, 2016 at 7:56

2 Answers 2


OK, as per your request…. My sense is you want to learn everything about integrability from here, and combine issues which confuses them, instead of separating them…. How about you supplement L&L with Arnold’s book?

  1. The seven additive integrals of L&L are the additive conservation laws of the isolated center of mass system, and standard center of mass conservation theorems dictate that they are fixed in the absence of external forces and torques, and so work in/outputs, by Newton’s action-reaction laws: you sum all energies, or momenta, or angular momenta, and their sums,since the system is “closed” are preserved (just as in a black hole!). But… they need not be independent, as, e.g. for one free particle in the box, $E\propto \vec{P}^2$, i.e., & is not an absolute lower bound on the number of conserved integrals. (and for one free particle, J being 0/meaningless, reduces the indep conserved integrals to 3.) To compare with 2., I’ll use the much simpler isolated system of 2 particles in 2d, so the rotation group is one dimensional, and the conserved additive integrals, instead of 7 are now just 4: E, $\vec{P}$, and J.

  2. This is a broad abstract statement for an upper bound to the number of independent integrals of phase-space motion, not necessarily additive . In a 2 s-dimensional phase space, each independent conserved integral specifies an independent hypersurface on which trajectories lie, and specify the phase-space point must run on their intersection. The most restrictive case is 2 s - 1 hypersurfaces, whose common intersection is a line, the trajectory of a (multidimensional) phase space point; one more constraint and the line is intersected to a point, so the point does not move in time! Systems with this maximal number of constraints are called maximally superintegrable, like the Kepler problem, or most baby freshman physics problems. As an overkill aside, all these problems are described much more symmetrically by the equivalent Nambu mechanics picture: the classical part expresses some of it in PB language. For invariants in involution, see this.

    • So, now, consider two particles connected by a spring, in 2d, starting with the k=0 limit, so free particles. $$L=\frac{1}{2}M\left(\dot{X}^2+\dot{Y}^2\right)+\frac{1}{2}m\left(\dot{x}^2+ \dot{y}^2 \right)-\frac{1}{2}k\left(r-d\right)^2= \frac{1}{2}M\left(\dot{X}^2+\dot{Y}^2\right)+\frac{1}{2}m\left(\dot{r}^2+ r^2\dot{\theta}^2 \right)-\frac{1}{2}k\left(r-d\right)^2.$$ for capitals being the center of mass coordinates, $r=\sqrt{x^2+y^2}$, and θ the angle between the relative coordinates. Now the equations of motion for θ are already integrated to constant $r^2\dot{\theta}\equiv J$, so we can drop its kinetic term in favor of a term $mJ^2/(2r^2)$ relocated into the potential part of the lagrangian.

    • Let us count overall conserved quantities, first for k=0, the free case: in Cartesian coords, we have 2 comps of momentum for 2 particles, so 4 in total; plus J and each of the two energies E and ε for the capital and lower case variables ? Not quite, since E is not independent of the c.m. momenta, and ε of the internal ones. The independent integrals appear to be 5. However, the external, additive ones, are E+ε, J, and $\vec{P}$, so fewer than the independent ones. Turning on the interaction (spring, nonvanishing k) destroys the conservation of the two components of the internal momentum, but $\epsilon_x$, $\epsilon_y$ and J are still preserved (2×2-1 for x,y, oscillators being maximally superintegrable) and the independent integrals overall are 5, yet again, forestalling your paradox.

    • Finally a word on your Poisson theorem question. Being totally schematic and cavalier about factors, you can see that, given the invariants $\epsilon_x, \epsilon_y, J$ of this double oscillator, $\{ \epsilon_x, J\} \sim K\equiv p_x p_y +xy$, also easy to confirm to be time independent, as per the Jacobi identity. Is there a 4th invariant? It can't be: we saw above maximal superintegrability only allows for 3. But note, fixing signs, factors, etc, that $\epsilon_x \epsilon_y=J^2+K^2$, so one of the four is dependent on the other three, nonlinearly. Phew!....


There are no contradiction at all and your questions can be answered at once.

An isolated system with $s$ degrees of freedom has $2s-1$ integrals of motion since the solutions for the coordinates $q_i$ involve $2s-1$ constants (determined by the initial conditions). The $2s$th can be solved in terms of $t-t_0$ where $t_0$ can be arbitrarily chosen. However not all of these integrals of motion are additive. By additive integral of motion is meant one that is the sum of correspondent integral of motion of the isolated subsystems.

These additive integral of motion, normally called conserved quantities, are the ones originated from some continuous symmetry of the system and can be computed via Noether's Theorem. However, unless I missed some subtlety there are 10 conserved quantities: energy E (associated to time translation), momentum $\vec P$ (associated to three space translation), angular momentum $\vec L$ (associated to three rotations) and the vector $t\vec P-M\vec R_{cm}$ (associated to three Galileo's boosts). You can read more in section three of this reference.

  • $\begingroup$ Hi thanks for your help! Can you elaborate on the second part of your answer? Are you saying that Noether' theorem is what shows that the additive integrals of motion can correspond only to symmetries of the system (e.g. homogeneity of space or time, isotropy of space)? Also why are there 10 conserved quantities associated with time? Three space translation - this is momentum? Three rotations - this is angular momentum? What are Galileo's boosts? I have not heard that term before. Thank you again so much for your help! $\endgroup$ May 5, 2016 at 0:53
  • $\begingroup$ Please consider taking another look at the answer. I tried to express myself better. $\endgroup$
    – Diracology
    May 5, 2016 at 1:08
  • $\begingroup$ That's an amazing reference; thank you so much again! $\endgroup$ May 5, 2016 at 2:37
  • $\begingroup$ Sorry, one more question: is it possible to prove any of this using Poisson brackets? I.e. using either $\partial f /\partial t + [H,f]=0$ (42.3 in L&L) if and only if $f$ is an integral of motion (where $H$ is the Hamiltonian), or $[f,g]=constant$ (42.15 in L&L) for any two integrals of motion $f,g$ (Poisson' theorem)? Or do Poisson brackets have nothing to say about in particular additive integrals of motion? $\endgroup$ May 5, 2016 at 7:56

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