# Doubt in Arnold's "Mathematical Methods of Classical Mechanics", Chapter 2

My question is about Arnold's book "Mathematical Methods of Classical Mechanics", chapter 2, section B (pg. 16).

He talks about systems with one degree of freedom, i.e. systems described by $$\ddot x = f(x). \tag{1}$$ I am confused about his description of phase flow with such systems. Here is what he writes:

1. $$1$$ is equiv. to the system of two equations $$\dot x := y,\qquad \dot y=f(x) \tag{2}$$

2. Consider the plane with coordinates $$x$$ and $$y$$, called the phase plane, whose points are called phase points. The RHS of (2) determines a vector field on the phase plane, called the phase velocity vector field.

3. A solution of (2) is a motion $$\varphi:\mathbb{R} \rightarrow \mathbb{R}^2$$ of a phase point in the phase plane, such that the velocity of the moving point at each moment of time equals the phase velocity vector at the location of the phase point at that moment.

4. $$\varphi$$'s image is called phase curve. Thus, phase curve is given by the parametric equations $$x = \varphi(t);\ y = \dot\varphi(t)$$.

I am unable to understand this construction or motivate it entirely. In step (2), when we consider the plane with coordinates $$x$$ and $$y$$, is $$y := \dot x$$, or is $$y$$ another direction we may move in (i.e. we are considering 2-dimensional motion)?

I am especially unsure how this relates to (3) - here, a motion would be $$\varphi:t \mapsto (x(t),y(t))$$, i.e. 2-dimensional motion, so that I am unsure what "a motion of a phase point" means.

Also, how does the RHS of (2) determine a vector field - what is the vector we associate to each point $$(x,y)$$ in the phase plane?

Essentially we transform the second order differential equation on the real line to an equivalent first order differential equation on the plane (i.e. two free variables instead of one). I will describe this process below:

The original problem is to find a real valued function $$\phi$$ so that $$\phi^{\prime \prime}(t)= f( \phi(t))$$ (i would recommend not using $$x$$ for both the function and the variable of space).

Now define a vector field $$F : \mathbb{R}^2 \to \mathbb{R}^2$$ by $$F(y_1,y_2) = (y_2,f(y_1))$$ (this is the "phase velocity vector field"). Then the original problem is equivalent to finding a function $$\varphi : \mathbb{R} \to \mathbb{R}^2$$ that satisfies $$\varphi^\prime (t) = F(\varphi_1(t), \varphi_2(t))$$, where $$\varphi_{1/2}$$ is the first/second component of $$\varphi$$.

This can be seen as follows: By definition of $$F$$ we have $$\varphi_1^\prime (t) = \varphi_2 (t)$$ and $$\varphi_2^\prime(t) = f(\varphi_1(t))$$. Combining the two gives $$\varphi_1^{\prime \prime}(t) = f( \varphi_1(t))$$. So then $$\varphi_1$$ is just the solution to the original problem.

Now back to Arnold: The phase plane is $$\mathbb{R}^2$$. The first variable $$y_1$$ is just $$x$$ (i.e. the spatial variable of the original problem) and the second variable $$y_2$$ is the velocity/momentum variable of the system which is independent of the first variable. The dependence of the two for the actual physical motion only comes in through $$F$$.

Edit: To add some physical motivation: All we need to know to predict the motion of a particle is its momentum and position at a given time. Therefore position and momentum are the core variables of our theory. The importance of momentum is buried in our original formulation, which treats position in a special way and momentum as a secondary variable that only enters through a boundary condition.

This is why we want to transform the problem to the phase plane where momentum and position are treated equally. The phase velocity vector field then makes sure that we have the "right physics" on the phase plane.

The two approaches are special cases of Lagrangian mechanics and Hamiltonian mechanics as remarked in the answer by qmechanic. The equations $$\varphi^\prime (t) = F(\varphi_1(t), \varphi_2(t))$$ are the Hamiltons equations of the system.

• Thank you so much for your in-depth explanation, this makes a lot of sense! From your answer, I understand the role that $F$ plays - namely, flow lines of $F$ yield solutions of $(1)$. However, I wanted to ask if there is a physical interpretation of the way that $F$ is defined. From Arnold, pg. 17, it seems that level curves of $E(x,x')$, where $E$ is the total energy, are flow lines of $F$, but I am unable to see entirely why $F$ is defined this way (aside from the fact that its flow lines correspond to solutions of $F$, which is a mathematical rather than physical motivation). Jun 3, 2023 at 11:08
• @algebroo i edited my answer to include some physical motivation for the switch to the phase plane.
– jd27
Jun 4, 2023 at 6:28
• Thank you so much! Jun 5, 2023 at 7:17

Ref. 1 is using 2 formulations of a 1D system:

1. A Lagrangian formulation with 1 variable $$x$$ (position) satisfying 1 second-order ODE (Lagrange equation).

2. A Hamiltonian formulation with 2 variables $$(x,y)$$ (position, momentum) in a 2D phase space satisfying 2 first-order ODEs (Hamilton's equations).

Ref. 1 is here reinterpreting $$(x,y)$$ as a position and $$(\dot{x},\dot{y})$$ as a velocity in the 2D phase space.

References:

1. V.I. Arnold, Mathematical methods of Classical Mechanics, 2nd eds., 1989; $$\S$$4B p. 16.

After seeing the others' very helpful answers, I arrived at the following conclusion that summarizes my insights from the post:

Given $$\ddot{x} = f(x)$$, say we want $$\varphi:\mathbb{R} \rightarrow \mathbb{R}$$ such that $$x = \varphi(t)$$ solves this ODE. Then, we have the following equivalence:

$$\varphi:\mathbb{R} \rightarrow \mathbb{R}$$ is a solution of the ODE $$\leftrightarrow \vec{\varphi}:\mathbb{R} \rightarrow \mathbb{R}^2, \vec{\varphi}(t) = (\varphi(t),\dot{\varphi}(t))$$ is level curve of $$E(x,\dot{x}) \leftrightarrow \vec{\varphi}$$ is flow line of $$\vec{F}(x,y) = (y,f(x))$$.

$$(1) \implies (2)$$ by law of conservation of energy; $$(2) \implies (3)$$ since vectors perpendicular to the gradient of $$E$$ are precisely parallel to $$\vec{F}$$, and $$(3) \implies (1)$$ directly.