In Landau-Lifshitz Classical mechanics textbook, it is said that there are generally $2s-1$ integrals of motion where $s$ is the number of degrees of freedom. Why is that? I couldn't find anywhere an answer.

  • 7
    $\begingroup$ How about the paragraph right after that sentence in Landau and Lifshitz? Or this page $\endgroup$
    – peek-a-boo
    Jan 7, 2021 at 12:15
  • $\begingroup$ @Qmechanic I did not find satisfying the answers already present, so I voted to reopen the qustion as I have another answer. $\endgroup$ Jan 7, 2021 at 14:31
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/55861/2451 , physics.stackexchange.com/q/8626/2451 , physics.stackexchange.com/q/44576/2451 and links therein. $\endgroup$
    – Qmechanic
    Jan 7, 2021 at 14:36
  • $\begingroup$ None is satisfying to me. The answer I like is elementary. Consider the vector field $Z\neq 0$ tangent to the solutions of motion in $\mathbb{R}\times TQ$. It is easy to prove that, in a neightborhood of every state, there is a coordinate system such that one coordinate say $x^1$ is tangent to $Z$. In this chart $x^2,\ldots, x^{2n+1}$ are constants during the evolution and thus are $2n$ (parametrically time-dependent) constant of motion (not $2n-1$) $\endgroup$ Jan 7, 2021 at 15:09
  • $\begingroup$ This answer is completely independent of the chosen formulation. All other answers are technically related to some precise approach, but the question here is completely general. One can replace $\mathbb{R} \times TQ$ for $\mathbb{R} \times M$ where $M$ is the space of the states of a given dynamical system and the possible motions are first order equations on $\mathbb{R} \times M$ (reducing to the first order also equations of second order just by doubling the number of variables as is the case in Lagrangian or Hamiltonian Mechanics). $\endgroup$ Jan 7, 2021 at 15:16

2 Answers 2


Every autonomous dynamical system arising from mechanics can be described as a first-order ODE on a $2n$-dimensional manifold $M$ (the spacetime of kinetic states of Lagrangian mechanics with local coordinates $q,\dot{q}$ or the space of phases of Hamiltonian mechanics with coordinates $q,p$). Every state $s\in M$ determines a unique solution of the ODE describing the motion with that initial condition at $t=0$. As a consequence, the motions are the integral lines of a vector field $Z$ defined on $M$.

In other words, the equations are actually of the form $$\frac{ds}{dt} = Z(s(t))\:.$$ For instance, in Hamiltonian formulation, the right-hand side is the Hamiltonian vector field generated by a given Hamiltonian. $$\frac{dq^k}{dt} = \frac{\partial H}{\partial p_k}\:,\quad \frac{dp_k}{dt} = -\frac{\partial H}{\partial q^k}\:, \quad k=1,\ldots, n$$ so that $$Z = \sum_{k=1}^n\frac{\partial H}{\partial p_k} \frac{\partial}{\partial q^k}- \sum_{k=1}^n\frac{\partial H}{\partial q^k} \frac{\partial}{\partial p_k}\:.$$

In the Lagrangian case, the existence of $Z$ relies on the well-known fact that the equation of Euler-Lagrange can be written in normal form: $$Z = \sum_{k=1}^n A^k \frac{\partial}{\partial \dot{q}^k} + \sum_{k=1}^n\dot{q}^k\frac{\partial}{\partial q^k}$$ in order that $$\frac{dq^k}{dt}= \dot{q}^k\:, \quad \frac{d\dot{q}^k}{dt}= A^k(q,\dot{q})\:,\quad k=1,\ldots, n $$ are the E.L. equations, for some Lagrangian $L=L(q,\dot{q})$, written in normal form.

Unless $Z(s_0)=0$, there is a neighborhood $U$ of every $s_0\in M$ such that $Z(s)\neq 0$ for $s\in U$. As is well-known from differential geometry (I am assuming that $M$ and the vector field $Z$ are at least $C^1$) and as is easy to prove, it is always possible to equip a smaller neighborhood $V\subset U$ of $s_0$ with a system of coordinates $t,x^2,\ldots, x^{2n}$ constructed like this. $t$ is the parameter of the integral curves of $Z$ computed starting from a $2n-1$ submanifold transverse to $Z$ where $t=0$ and the remaining coordinates are coordinates on that submanifold.

So let us denote by $y^1,\ldots, y^n$ the constructed local coordinates around $s_0$.

In these coordinates, the solution of the motion with initial conditions $y^1_0,\cdots, y^{2n}_0$ at time $0$ have a trivial expression: $$y^1(t) = y^1_0+t\:,\quad y^{k}(t)= y_0^k\:, \quad k=2,\ldots, 2n.$$ In other words, the coordinates $y^k$ remain constant along the motion if $k=2,\ldots, 2n$.

These constants of motion are also functionally independent as they are coordinates of a local chart on $M$.

It should be clear that we cannot have more than $2n-1$ functionally independent integrals of motion. If they existed they would define a coordinate system around each state. So, when their values are given, then the state is fixed and no evolution is possible. Around a state where $Z\neq 0$ this is not possible.

ADDENDUM. I prove here the existence of the said coordinate system in the autonomous case.

Let us suppose that $Z(s_0) \neq 0$. Choose a local chart in $M$ around $s_0$ equipped with coordinates $x^1,\ldots, x^{2n}$. Exploiting a suitable bijective linear transformation, we can always change the coordinate system in order that, exactly at $s=s_0$, $$Z(s_0) = c \frac{\partial}{\partial x^1}|_{s_0}\:, \quad c>0\:.$$ By continuity, the component of $Z$ along $x^1$ cannot vanish in a neighborhood of $s_0$. Therefore, the $2n-1$ submanifold $C \subset M$ locally described by $x^1=0$ is transverse to $Z$ around $s_0$. I stress that $C$ is covered by local coordinates $x^2,\ldots, x^{2n}$ as a consequence.

Since $Z$ is $C^1$, it admits a $C^1$ local flow $$\phi^{(Z)}: A \to M$$ where $A \subset \mathbb{R}\times M$ is an open set containing the points $(0,s)$. By definition of flow, the jointly $C^1$ map $$A\ni (t,s) \mapsto \phi_t(s)\in M$$ is the solution of the differential equation $$\frac{dx}{dt}= Z(x(t))$$ with initial condition $s$ and evaluated tat the value $t$ of the parameter: $$\phi_t(s) = x(t; s)\:.$$

Let us consider the $C^1$ map $$I \times C \ni (t,s)\mapsto \phi_t(s) \in M$$ where $I \subset (-\infty,+\infty)$ is a sufficiently small interval containing the origin such that the right-hand side is well defined when $s\in C$. As before $C$ is the $2n-1$ dimensional submanifold $C$ transverse to $Z$. If necessary, we take $C$ sufficiently narrowed around $s_0$. Notice that $I$ exists because $A$ is open.

It is easy to prove that, working in local coordinates $x^1,\ldots, x^{2n}$, the Jacobian matrix of the above map has non zero determinant exactly at $(0,s_0)$.

(That is because the Jacobian matrix evaluated at that point has the first row made of the components of $Z$ in coordinates $x^1,\ldots, x^{2n}$ and the remaining rows made of the rows of the identity matrix $(2n-1)\times (2n-1)$. The determinant is therefore the first component of $Z$ at $s_0$, which does not vanish by hypothesis.)

Applying the theorem of the inverse function, we have that, in a neighborhood of $(0,s_0)$, the local coordinates $(t,x^{2}, \ldots, x^{2n})$ define an admissible $C^1$ local chart on $M$.

The found coordinates around $s_0$ denoted by $y^1,\ldots, y^n$ are such that the solution of the motion with initial conditions $y^1_0,\cdots, y^{2n}_0$ at time $0$ have a trivial expression: $$y^1(t) = y^1_0+t\:,\quad y^{k}(t)= y_0^k\:, \quad k=2,\ldots, 2n.$$ As said, the coordinates $y^k$ remain constant along the motion if $k=2,\ldots, 2n$. As said, we have $2n-1$ functions defined on $M$ around $s_0$ which are constants of motion. These functions are also functionally independent (their Jacobian matrix has maximal rank), just because they are local coordinates.


In Landau's mechanics, one can read in chap 6 about the energy:

Landau page 17

What is it you do not understand?

  • $\begingroup$ We do not understand why it can ALWAYS be taken as an additive constant in time. An explicit proof would be appreciated ;) $\endgroup$
    – Cheng
    Sep 23, 2022 at 23:58
  • $\begingroup$ Check this answer: physics.stackexchange.com/q/13832 $\endgroup$
    – Shaktyai
    Sep 24, 2022 at 0:22
  • 1
    $\begingroup$ The subtle point, which would deserve an explanation here, is why one may write the constants $C_k$ as (smooth!) functions of the coordinates $q$ and $\dot{q}$. This fact is (obviously) true and I proved it in my answer. $\endgroup$ Sep 25, 2022 at 8:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.