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All formulations of classical mechanics that I am aware of end up describing particle motion in terms of second-order differential equations in terms of the position vectors $\mathbf{r}_i(t)$. Of course you can always turn them into first-order equations by adding some new variables $\mathbf{u}_i(t) = \mathbf{\dot{r}}_i(t)$ representing velocity, but that does not count because the state space size is increased.

I know this is obviously a consequence of Newton's 1st and 2nd laws. But I would like to know how much of this is empirical, how much of this is a consequence of convenient definitions, and how much of this is a consequence of more fundamental properties of nature.

I have seen this question where the answers mostly address (to my satisfaction) the reason why the order is not greater than 2, but I am not very satisfied with the reasons why the equations could not be order 1 in principle. I don't consider this a duplicate, I am specifically looking for reasons why first-order equations of motion would not work, and this is my question. I added my own thoughts about what I believe the answer might be below.

The way I see it, this causes position and velocity to be fundamentally different from higher time derivatives, in the sense that one can freely specify an initial position and velocity, and then all the behavior of the particle's motion can be predicted. But is there an intuitive reason why initial velocity itself should not be determined by the particle's initial position, say? Of course, Newton's first law states that objects on which no forces are acting will move at constant velocity in inertial reference frames, but I don't think this fully answers the question, as someone might conceive that we defined forces and inertial reference frames in such a way that this property holds (so then you would ask "why are these definitions the right ones?").

My guess is that perhaps this is a consequence of Lorentz invariance in Special Relativity (because in this setting, space and time are fundamentally related by the constant speed of light). But if that were the case, this would mean that pre-relativistic classical mechanics would not have good reason to assume this. The idea of Galilean and/or Lorentz invariance already introduces a special place for velocity in physics, as a free parameter of the group of symmetries of spacetime.

Ultimately, there is this seemingly arbitrary appearance of the number $2$ (instead of $1$, say) which I can't completely justify, and I don't know whether there are better reasons other than it being an empirical fact or convenient definition.

Imagine the following thought experiment: you are in empty space, and you have a ball standing still in front of you. When you push the ball, it starts moving. Why is it natural to expect the ball to continue moving, instead of immediately stopping? Suppose you had not learned about Newton's laws...

PhysicsStudent
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  • The equations of motion are not generically of 2nd order. Consider this motion equation: $$p=p_0+vt+\frac12at^2+\frac16 jt^3+\frac1{24}st^4+\frac1{120}ct^5+\cdots$$ $p$ is position, $v$ speed, $a$ acceleration, $j$ jerk, $s$ snap, $c$ crackle etc. Often in our World we have constant acceleration, though, such as the effect gravity. So, all higher-than-2nd-order terms are zero. But not necessarily and not always. It is a good question, why constant acceleration is such an abundant phenomenon, but the motion equations do not inherently reflect that - we just often only see the reduced versions. – Steeven Mar 04 '19 at 08:10

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OP is describing Aristotelian mechanics (AM) $$ m\dot{q}^i~=~f^i(q,t).$$

  1. Symmetry objections are e.g. that AM has no time-reversal symmetry.

  2. Dynamical objections are e.g. that AM is dissipative, has no closed orbits if the force is conservative gradient field $f_i(q)=-\frac{\partial V(q)}{\partial q^i}$, and has no conventional stationary action principle.

Qmechanic
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If you push a ball in empty space, you observe it’s subsequent uniform motion regardless of whether you are aware of Newton’s equations of motion or not.

More vividly, when a kid kicks a ball, it’s trajectory will depend on the velocity of the foot, independently of the kid’s knowledge of the school’s physics curriculum.

It is just an empirical fact that motion depends on both position and velocity.

That should answer your titular question. Now the question of why is it the case that motion depends not just on position but on velocity too is a different question, which I cannot answer without being speculative. It just looks like it would be too boring perhaps?

Andrea
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When you push the ball, it starts moving. Why is it natural to expect the ball to continue moving, instead of immediately stopping?

You could intuitively argue conservation. During the time when you're pushing the ball, you're exerting a force(push) on it. And that $\mbox{force}\times\mbox{distance_moved_while_pushing}$ is the $\mbox{work}\equiv\mbox{energy}$ you imparted to the ball when you pushed it (i.e., $\mbox{work}=\mbox{force}\times\mbox{distance}$, as usual).

So that work energy can't just "disappear" after you stop pushing. It's conserved as the $\frac12mv^2$ kinetic energy of the ball. So if the ball just stopped moving, all by itself, then that would violate conservation. (Of course, if you don't already intuitively believe conservation, we're back where we started:)

(By the way, re your not-really-relevant relativity remarks, $\vec v$ would be measured in the $\mbox{cm}$ frame where both you and the ball were stationary before the push. And then, to be completely accurate, we'd have to account for kinetic energy due to both your "backward" speed and the ball's "forward" speed after the push.)