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I am a mathematician. I have a (somewhat long term) goal of understanding some of the physical insights that have influenced my area of research. To this end I read Arnold's Mathematical methods in classical mechanics a while ago, but something I didn't understand has been bugging me ever since.

In the first chapter Arnold defines a motion of $n$ particles in $\mathbb{R}^3$ as a map $\mathbf{x}:\mathbb{R} \rightarrow \mathbb{R}^N$ for $N=3n$. The first chapter is then about what types of motion are allowed. In section 2D Arnold makes the following observation:

According to Newton's principle of determinacy all motions of a system are uniquely determined by their initial positions ($\mathbf{x}(t_0) \in \mathbb{R}^N$) and initial velocities ($\mathbf{\dot{x}}(t_0) \in \mathbb{R}^N$).

This seems important and I expect that this would have an impact on what types of functions $\mathbf{x}$ could be. He goes on:

In particular, the initial positions and velocities determine the acceleration. In other words, there is a function $\mathbf{F} : \mathbb{R}^N \times \mathbb{R}^N \times \mathbb{R} \rightarrow \mathbb{R}^N$ such that $$ \mathbf{\ddot{x}} = \mathbf{F}(\mathbf{x},\mathbf{\dot{x}},t). $$

So the implication of Newton's determinacy principle is that the acceleration obeys a second order differential equation. This seems completely vacuous to me. Any function $\mathbf{x}$ obeys a second order differential equation (as long as it is twice differentiable).

Could someone please explain to me what Arnold is saying here. I feel like I am missing something important.

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    $\begingroup$ Comment to the question (v1): The function $\bf F$ is part of the (already) given input, while the function $\bf x$ is the sought-for output, not the other way around. In other words, $\bf F$ is not arbitrary. It is fixed. For fixed $\bf F$, one can in principle say a lot about possible $\bf x$. It obeys a particular given 2nd-order ODE, not an arbitrary 2nd-order ODE. $\endgroup$ – Qmechanic Apr 13 '14 at 20:54
  • $\begingroup$ Yes absolutely, thanks. I didn't make that clear enough. We do not know what $\mathbf{x}$ is and the point here is that we can use Newton's principle of determinacy (and Galileo's principle of relativity) to make some qualitative statements about what $\mathbf{x}$ is. $\endgroup$ – sedn Apr 13 '14 at 21:30
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    $\begingroup$ I think Arnold is saying that one qualitative statement we can make about $\mathbf{x}$ is that it obeys a second order ODE. To me this seems like a completely vacuous thing to say. My question is, why is it not vacuous, or how have I misinterpreted what Arnold is saying. $\endgroup$ – sedn Apr 13 '14 at 21:33
  • $\begingroup$ Any single-valued function obeys a second-order differential equation. The point of determinacy here, I think, is that $x$ is given by a single-valued function. $\endgroup$ – Scott Lawrence Apr 16 '14 at 4:53
  • $\begingroup$ I also find that the second quotation is confusing or just wrong (with all due respect to the author). In physics, Newton's second law rather says that the acceleration is proportional to a given force F. Here, one talks at first about "initial" position and speed, but then gives an ODE at any time. However, I'm not an expert but the equation is not arbitrary or "vacuous" at all, since a general second order ODE should be $F(\mathbf{\ddot{x}},\mathbf{\dot{x}}, \mathbf{x}, t) = 0$? $\endgroup$ – Noix07 May 10 '15 at 8:39
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The trajectories are uniquely determined means that the theorem of existence and uniqueness applies (so, the differential equation has to be sufficiently regular).

Newton's principle states more: the system is fully determined by the position and the speed, that is, by $2n$ constants, where $n$ is the dimension of the space. As you have $n$ equations (one per spacial coordinate), they are completely determined if and only if they are of second order.

The statement is not that the function $x$ obeys a second order differential equation, it says that the dynamics are directed by a second order DE.

Edit:

In other words, The key is that there is one set of ODE for any possible initial condition. You can construct a first order ODE for a given trajectory, but it will be useless if you change the initial conditions.

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  • $\begingroup$ Ok thanks, I think this is getting at what I am not understanding. What would be the mathematically precise way of saying your last paragraph? $\endgroup$ – sedn Apr 16 '14 at 9:54
  • $\begingroup$ For a given system, there is a set of $n$ second order ODE whose solution is the evolution of the system. $\endgroup$ – Davidmh Apr 16 '14 at 10:00
  • $\begingroup$ The rest of Newton's derivation is to make that statement constructive. $\endgroup$ – Davidmh Apr 16 '14 at 10:00
  • $\begingroup$ I'm not sure I see the difference between "there is a set of n second order ODE whose solution is the evolution of the system" and "the system obeys a second order DE". $\endgroup$ – sedn Apr 16 '14 at 10:14
  • $\begingroup$ The key is that there is one set of ODE for any possible initial condition. You can construct a first order ODE for a given trajectory, but it will be useless if you change the initial conditions. $\endgroup$ – Davidmh Apr 16 '14 at 10:17
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A very simple motivation for writing $\ddot{\bf x}(t) = {\bf F}({\bf x}, \dot{\bf x}, t)$, which might shed some light, is the following. We are given ${\bf x}(t)$ and $\dot{\bf x}(t)$ and we desire to calculate ${\bf x}(t + \delta t)$ and $\dot{\bf x}(t + \delta t)$. Now, ${\bf x}(t + \delta t) = {\bf x}(t) + \dot{\bf x}(t) \delta t$, which we may calculate. But $\dot{\bf x}(t + \delta t) = \dot{\bf x}(t) + \ddot{\bf x}(t) \delta t$, which in order to calculate we require to know $\ddot{\bf x}(t)$. By Newton's law of determinacy, the motion is calculable given ${\bf x}(t)$ and $\dot{\bf x}(t)$. Thus there must be a function ${\bf F} : \mathbb{R}^N \times \mathbb{R}^N \times \mathbb{R} \rightarrow \mathbb{R}^N$ such that $\ddot{\bf x}(t) = {\bf F}({\bf x}, \dot{\bf x}, t)$. What this function is constitutes a law of nature.

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  • $\begingroup$ May I inquire as to the reason for the down-vote? $\endgroup$ – P. Plowman Nov 6 '15 at 16:27
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What I understand from the qualitative statement is that of all possible laws that acceleration could "obey," it actually obeys a 2nd ODE! It is irrelevant, whether or not, other functions also obey the same law.

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