Since the equation of mechanics are of second order in time, we know that for $N$ degrees of freedom we have to specify $2N$ initial conditions. One of them is the initial time $t_0$ and the rest of them, $2N-1$ are initial positions and velocity. Any function of these initial condition is a constant of motion, by definition. Also, there should be exactly $2N-1$ algebraically independent constants of motion.

On the other hand, Noether's procedure gives us integrals of motion as a result of variational symmetries of the action. These integrals of motion are also conserved but they are not always $2N-1$ in number. In consequence, we classify the system by their integrability.

So, what is the difference between the constant of motion and integral of motion? Why do non-integrable systems have less integrals of motion when they should always have $2N-1$ constants of motion?

  • 2
    $\begingroup$ If you like this question you may also enjoy reading this and this Phys.SE post. $\endgroup$
    – Qmechanic
    Mar 4, 2013 at 21:47

1 Answer 1


1) A constant of motion $f(z,t)$ is a (globally defined, smooth) function $f:M\times [t_i,t_f] \to \mathbb{R}$ of the dynamical variables $z\in M$ and time $t\in[t_i,t_f]$, such that the map $$[t_i,t_f]~\ni ~t~~\mapsto~~f(\gamma(t),t)~\in~ \mathbb{R}$$ doesn't depend on time for every solution curve $z=\gamma(t)$ to the equations of motion of the system.

An integral of motion/first integral is a constant of motion $f(z)$ that doesn't depend explicitly on time.

2) In the following let us for simplicity restrict to the case where the system is a finite-dimensional autonomous$^1$ Hamiltonian system with Hamiltonian $H:M \to \mathbb{R}$ on a $2N$-dimensional symplectic manifold $(M,\omega)$.

Such system is called (Liouville/completely) integrable if there exist $N$ functionally independent$^2$, Poisson-commuting, globally defined functions $I_1, \ldots, I_N: M\to \mathbb{R}$, so that the Hamiltonian $H$ is a function of $I_1, \ldots, I_N$, only.

Such integrable system is called maximally superintegrable if there additionally exist $N-1$ globally defined integrals of motion $I_{N+1}, \ldots, I_{2N-1}: M\to \mathbb{R}$, so that the combined set $(I_{1}, \ldots, I_{2N-1})$ is functionally independent.

It follows from Caratheodory-Jacobi-Lie theorem that every finite-dimensional autonomous Hamiltonian system on a symplectic manifold $(M,\omega)$ is locally maximally superintegrable in sufficiently small local neighborhoods around any point of $M$ (apart from critical points of the Hamiltonian).

The main point is that (global) integrability is rare, while local integrability is generic.


$^1$ An autonomous Hamiltonian system means that neither the Hamiltonian $H$ nor the symplectic two-form $\omega$ depend explicitly on time $t$.

$^2$ Outside differential geometry $N$ functions $I_1, \ldots, I_N$ are called functionally independent if $$\forall F:~~ \left[z\mapsto F(I_1(z), \ldots, I_N(z)) \text{ is the zero-function} \right]~~\Rightarrow~~ F \text{ is the zero-function}.$$ However within differential geometry, which is the conventional framework for dynamical systems, $N$ functions $I_1, \ldots, I_N$ are called functionally independent if $\mathrm{d}I_1\wedge \ldots\wedge \mathrm{d}I_N\neq 0$ is nowhere vanishing. Equivalently, the rectangular matrix $$\left(\frac{\partial I_k}{\partial z^K}\right)_{1\leq k\leq N, 1\leq K\leq 2N}$$ has maximal rank in all points $z$. If only $\mathrm{d}I_1\wedge \ldots\wedge \mathrm{d}I_N\neq 0$ holds a.e., then one should strictly speaking strip the symplectic manifold $M$ of these singular orbits.

  • $\begingroup$ Here we implicitly assume that the Hamiltonian $H$ itself is functionally independent function. In particular, we exclude the case where $H$ is identically zero. $\endgroup$
    – Qmechanic
    Jun 2, 2018 at 15:46
  • $\begingroup$ 1. If I understand correctly, the set of integrals of motion or the first integrals is a subset of the constants of motion which depend only on the phase space variables $(q,p)$ and does not depend on time explicitly. 2. They are constants whether or not equations of motions are satisfied. Is this correct? $\endgroup$ Feb 17, 2020 at 7:28
  • $\begingroup$ @mithusengupta123: Thanks for the feedback. 1. Correct. 2. Incorrect. They are not necessarily constant if the EOMs are not satisfied. $\endgroup$
    – Qmechanic
    Feb 17, 2020 at 7:53
  • $\begingroup$ Can you write their difference in simple English? I couldn't follow the mathematical definition very well. Thanks in advance. $\endgroup$ Feb 17, 2020 at 9:36

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