15
$\begingroup$

I am am currently studying Newtonian mechanics from a conceptional and axiomatic point of view. Now, if I am not mistaken, one (but surely not all) statement of Newtons second law about nature is, that (almost ?) every motion of a "classical particle" (or "small body") can be described by a second order differential equation on $\mathbb{R}^3$ (or $\mathbb{R}^{3n}$ if one considers a system of $n$ particles). That is, if $I \ni t \mapsto x(t) \in \mathbb{R}^{3n}$ is a motion of $n$ particles in some environment, there is a smooth function $f \colon \mathbb{R}^{3n} \times \mathbb{R}^{3n} \times \mathbb{R} \to \mathbb{R}^{3n}$ (which describes the influence of the environment) such that $\ddot x(t) = f(x(t),\dot x(t), t)$ for all $t \in I$, thereby $I$ is an interval and $\dot x$ denotes the derivative of $x$.

Now my question is, if there is a good, mathematical sound intuition, which kinds motions are not allowed by Newtons second law because of the fact that it is a second order differential equation.

In other words: I want to analyze in detail, what Newtons seconds law tells us about nature. Especially I want to grasp, how the condition to be a second order differential equation gives restrictions to possible conceivable motions of particles. What would be allowed additionally if the equations were of $3$. or higher order?

$\endgroup$
4
  • $\begingroup$ You know that you can convert a higher order differential equation to a lower order one right? Say you have an equation $\partial^3_{ttt}x = f(\ldots)$, just write $y = \dot{x}$, and then your equation becomes $\ddot{y} = f(\dot{y},y,x,t)$ and $\ddot{x} = \dot{y} = g(\cdot{y})$ and voila, you have another second order differential equation... $\endgroup$ Oct 23, 2010 at 12:50
  • $\begingroup$ (ack, the last term should be $g(\dot{y})$.) On the other hand, there is something that you can say about the nature of physics if you assume the corresponding Lagrangian formulation depends only on first derivatives. $\endgroup$ Oct 23, 2010 at 12:52
  • 2
    $\begingroup$ @Willie Wong: Yes, it is clear that one can convert a higher order differential equation to a lower order one, but I don't want to enlarge the space where my ODE lives. For example I want to know if there are typical behaviour which a solution of third order ODE on $\mathbb{R}^3$ can have, but no solution of a second order ODE on $\mathbb{R}^3$, i.e. on the same space. $\endgroup$
    – student
    Oct 23, 2010 at 16:34
  • 2
    $\begingroup$ If you only allow yourself to look at one particle path, any one twice continuously differentiable path is the solution to some second order ODE. You need to look at families of paths, depending on parameterized families of initial condition (infinitely many independent experiments) to see that your paths are the solutions of a second order ODE system, and not a third order one. $\endgroup$
    – Ben McKay
    Mar 26, 2013 at 22:25

9 Answers 9

16
$\begingroup$

Actually, Newton did NOT say that $F = m a$ (i.e., ${d^2 x \over dt^2} = {1\over m} F$) in the Principia. First of all, if he did, no one at that time would have understood what it meant since that was pre-calculus times. What he did say in his 2nd Law was that "Change of motion is proportional to impressed motive force and is in the same direction as the impressed force", i.e., in modern terminology, the instantaneous change of momentum (caused by something like a hammer blow) is equal to the applied impulse. Whenever he used the 2nd Law, he treated a smooth force as the limit of a large number of small impulses. It was only much later that Euler recast the 2nd Law as a 2nd order ODE. This is all discussed in considerable detail in a recent book ``Differential Equations, Mechanics, and Computation'' (that I co-authored with my son Robert). There is a "Web Companion" for the book at the URL http://ode-math.com where you can freely download more than half the book as pdf files. In particular, if you download the first pages of "Chapter 4: Newton's Equations" you will see all of this discussed in considerable detail. One more point: this book was explicitly written to be, as we stress, a conceptual introduction to the subject for someone like yourself who is learning this material for the first time. (See here: http://ode-math.com/NovelFeaturesOfODECM.html)

All of the above is somewhat tangential to your specific question, so let me add that a very major restriction imposed by saying that the laws of motion for say $n$ particles are (equivalent to) 2nd order ODE is that if you know the positions and velocities of the particles at any one instant then their positions at any time in the past and future are uniquely determined by that data. Or as Laplace said in a very famous quote "The current state of Nature is evidently a consequence of what it was in the preceding moment, and if we conceive of an intelligence that at a given moment knows the relations of all things of this Universe, it could then tell the positions, motions and effects of all of these entities at any past or future time. . ." (Of course we now know that the existence of "chaotic behavior" renders that only a very theoretical possibility.)

$\endgroup$
4
  • $\begingroup$ "if you know the positions and velocities of the particles at any one instant then their positions at any time in the past and future are uniquely determined by that data" Are you sure about this? Future evolution is one thing (but even there I don't see why it should follow); but surely not past behaviour? Consider water pouring out of a hole in a bucket (or a large number of small particles, to approximate fluid behaviour). After it's all left the bucket, surely you can't work out how long ago the water ran out, in general? $\endgroup$
    – Zen Harper
    Nov 12, 2010 at 10:32
  • 5
    $\begingroup$ @Zen Harper: I agree that it is highly counter-intuitive that you could discover the past behavior of a fluid from its present state. BUT, if you believe that (1) the fluid is accurately modeled by on the order of $10^3$ Newtonian particles and (2)that the forces between them are Lipschitz continuous functions of their positions and velocities, then it really does follow that if you know the current positions and velocities with ABSOLUTE PRECISION, then you can retrodict. But, in practice, even the tiniest errors in initial data will rapidly magnify and vitiate any pre- (or retro-)dictions. $\endgroup$ Nov 13, 2010 at 6:45
  • 4
    $\begingroup$ This answer doesn't have much to do with the question. $\endgroup$
    – user21349
    Mar 29, 2013 at 0:42
  • $\begingroup$ Interestingly, Newton never says that quantity of motion is a linear function of the velocity either (and hence one cannot conclude that $F = ma$. He does say that it is homogeneous of degree one. If you do mechanics on Finsler manifolds, this actually comes up. $\endgroup$ Aug 8, 2013 at 20:30
6
$\begingroup$

I suspect that the work of Kasner and his students is what you might want. They discuss the differential-geometrical properties of those curves which can arise as the trajectories of second order differential equations. Kasner published a resum\'e of his results in the monograph "Natural families of trajectories" which is still available, perhaps at the Cornell site. Many of the original articles appeared in the Transactions and are also available on internet.

$\endgroup$
1
  • $\begingroup$ That's a very good suggestion. You can find most of his work on JSTOR. $\endgroup$ Aug 8, 2013 at 20:32
4
$\begingroup$

When you play billiard, there are collisions, in which you observe (approximately because of the friction of the carpet, but this can be avoided in more careful experiments) the conservation of mass (the number of balls remain constant), linear momentum (the sum of the velocities) and of energy (the sum of the square of velocities). If the balls have different masses, this has to be pondered by the respective masses. This is confirmed experimentally in system with so many particles that they appear to be continua, even if we have to introduce an internal energy in order to model the fluctations of velocities (the mean of the square is larger than the square of the mean, due to Cauchy-Schwarz). Thus Newtonian Physics is, on a large part, a science of conservation of mass, momentum and energy. Notice that in absence of collision, this conservtion holds true for individual particles, and this means that they move with a constant velocity. The motion of a free particle thus obeys the equation $x''(t)=0$.

In presence of an external field, the total energy is likely to be the sum of the kinetic energy $\frac12m|x'|^2$ that the particle would have in absence of the field, and a term representing the interaction between the particle and the field. Presumably, the total energy the form $E=\frac12m|x'|^2+V(x,x')$, and it is a constant of the motion. Only second order differential equations may be compatible with such first-order conservation laws. At best, an ODE of order $n$ may imply that a first-order expression like $E$ satisfy an ODE of order $n-1$. This implies constancy only if $n=2$.

This explanation is valid also for the PDEs occuring in classical Physics, like Laplace and wave equations. By-products are the inverse square law of the gravitational force (again, Newton) and the Huyghens principle for the propagation of light. Of course, things become much more complex at relativistic and/or quantic regimes. But even then, most equations are second-order because of the presence of frist-order conserved quantities. For instance, the number of particles and the momentum in quantum mechanics are, up to constants, given by $$\int_{{\mathbb R}^3}|\psi|^2dx,\qquad\Re\int_{{\mathbb R}^3}\bar\psi\nabla\psi dx.$$

Needless to say, the first-order conservation laws associated with equations from Physics is the starting point in their mathematical analysis.

$\endgroup$
2
  • $\begingroup$ Huh? The expression you give for quantum mechanical mass is rubbish (or I'm misunderstanding it very much). It simply equates to one (for a normalised wavefunction)! $\endgroup$
    – Noldorin
    Oct 23, 2010 at 15:19
  • $\begingroup$ Ups! That's the number of particles (here, one). I correct it. $\endgroup$ Oct 23, 2010 at 15:42
3
$\begingroup$

The way the question is asked doesn't quite work, because it leads to a trivial answer. The question is "which kinds motions are not allowed by Newton's second law because of the fact that it is a second order differential equation" of the form $\ddot{x}=f(x,\dot{x},t)$. The answer is that essentially any motion is allowed. Give me any twice-differentiable function for $x$, and I can simply define an $f$ for you that depends only on $t$, and equals $\ddot{x}$. Then $x$ is a solution of an equation of the given form.

You can get more interesting possibilities by restricting the form of $f$. For example, if the differential equation is required to have the form $\ddot{x}=f(x,\dot{x})$, with no explicit time dependence, then you can rule out any motion that visits the same point twice, with the same velocity, but doesn't repeat itself. This essentially says that experiments are reproducible, if you set up the same conditions with the particle at the same position and velocity.

Besides the form of $f$ (which things it depends on), the continuity properties of $f$ also make a difference. If $f$ has no explicit time dependence but isn't Lipschitz continuous, then you can get solutions that violate causality. There is a famous example of this type known as Norton’s dome, which you can find out about by googling. The idea is that you can have a motion defined by $x=0$ for $t\le T$, and $x=(t-T)^4$ for $t\ge T$. Regardless of the value of $T$, the function $f(x)$ is the same. In other words, you have a particle that just sits in a certain spot for a long time, and then one day it decides to move. This makes experiments not reproducible. However, there are fundamental reasons why you can't actually create a system with this behavior.

In quantum mechanics, it does matter what order your derivatives are. If you have a Lagrangian that depends on higher derivatives, you get an energy spectrum that's not bounded below. Google "higher derivative theories."

$\endgroup$
2
$\begingroup$

Newton's second law implies basically that the evolution of a mechanical system is completely determined as soon as the particles' initial positions $x(0)$ and velocities $\dot x(0)$ are specified. Since this is a second order equation, the corresponding Cauchy problem $$\begin{cases}\ddot x(t) = f(x(t),\dot x(t), t), &\ t>0 \\\ x(0)=x_0, \\\ \dot x(0)=y_0,\end{cases}\qquad\qquad(*)$$ has a unique local in time solution for any initial data $(x_0,y_0)$ if $f(x,y,t)$ is a Lipschitz function w.r.t. $(x,y)$ (which is almost always the case in applications).

As for possible complexity of motions, there are no obstacles to completely chaotic behaviour of solutions to system $(*)$ whatsoever, provided that $f(x,y,t)$ is a nonlinear function w.r.t $(x,y)$ and any of the following conditions is satisfied:

  • the number of interacting particles $n\geq 3$;
  • the forcing term is nonconservative;
  • the system is essentially nonautonomous (i.e. $\partial_t f(x,y,t)\neq0$ identically).

Just google "chaos in classical mechanics" for numerous examples of innocent looking mechanical systems exhibiting wildly exotic dynamics.

Edit. See also a somewhat related MO question Does every ODE comes from something in physics?.

$\endgroup$
2
$\begingroup$

This is maybe a bit of a simpleton's answer:

Motions of particles are described by second order equations because two particles starting from the same position and with the same velocity follow the same trajectory.

At least no experiment has ever shown evidence of the contrary.

$\endgroup$
1
$\begingroup$

Regarding your question "What would be allowed additionally if the equations were of 3. or higher order?" I would like to mention that the equation you get in classical physics if you consider the effect of an electron's own electromagnetic on itself involves a time derivative of the acceleration, cf. e.g. equation (1) in http://www.philsoc.org/1962Spring/1526transcript.html, and especially remark (III). If I remember it right, more can find in the standard book on classical electrodynamics, Jackson, Classical Electrodynamics.

$\endgroup$
0
$\begingroup$

This is a very deep (or philosophical) question. It seems that in mechanical systems a "tension" (e.g., a plucking of a string) gives rise to an ${\bf acceleration}$ proportional to the "tension", while, e.g., in heat conduction a "tension" (i.e., a temperature gradient) gives rise to a ${\bf velocity}$ proportional to the "tension".

$\endgroup$
0
$\begingroup$

The question is" which kinds motions are not allowed by Newtons second law because of the fact that it is a second order differential equation" and the answer is the "motion of any body in spherical domain around a center of force" like that of an electron around nucleolus in Hydrogen like atoms.Newtons Law being 2nd order differential equation of Position Vector, assumes that the plane of the trajectory of the body instantaneously aligns with the vector joining the center of force and the body ,which is a very strong restriction on the Law of Inertia.Since the Trajectory of micro particles are not necessarily planner curve,the relation between "mover" and the position of the "moved" ie the "equation of motion" should in general contain all 3- unit vectors ( Tangent, Normal & Bi-normal) defining the trajectory at every point which is possible only if the equation of motion is 3rd or higher order which further implies assigning of minimum 2 Inertia to bodies, one being the Newtonian "mass and the other through which the higher order differential equation is accomplished.Newtons Law and all other forms of Law of motion derived from Newtons Law can in no way explain the motion of electron in 3-dimensional region around the center of force not because the trajectory of electron does not exist or the particle behaves as wave(No wave but a complex function to obscure the real position of particle at every instant of time)but just because the equation is 2nd-order.I submitted this answer in March 2013 but find it missing without any valid comment.The fact is that all the Uncertainty Relations of QM follow if the equation of motion is 3rd or higher order

$\endgroup$
1
  • $\begingroup$ If you think an answer of yours disappeared, then perhaps you should bring this up on meta.mathoverflow.net. Maybe there you can tell us what you meant by "without any valid comment." Were there non-valid comments? $\endgroup$ Aug 8, 2013 at 21:00

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.