# Does the mass point move?

There is a question regarding basic physical understanding. Assume you have a mass point (or just a ball if you like) that is constrained on a line. You know that at $t=0$ its position is $0$, i.e., $x(t=0)=0$, same for its velocity, i.e., $\dot{x}(t=0)=0$, its acceleration, $\ddot{x}(t=0)=0$, its rate of change of acceleration, $\dddot{x}(t=0)=0$, and so on. Mathematically, for the trajectory of the mass point one has

$$\left. \frac{d^{n}x}{dt^n}\right|_{t=0} = 0 \textrm{ for } n \in \mathbb{N}_0\mbox{.}$$

My physical intuition is that the mass point is not going to move because at the initial time it had no velocity, acceleration, rate of change of acceleration, and so on. But the mass point not moving means that $x(t) \equiv 0$ since its initial position is also zero. However, it could be that the trajectory of the mass point is given by $x(t) = \exp(-1/t^2)$. This function, together with all its derivatives, is $0$ at $t=0$ but is not equivalent to zero. I know that this function is just not analytical at $t=0$. My question is about the physical understanding: How could it be that at a certain moment of time the mass point has neither velocity, nor acceleration, nor rate of change of acceleration, nor anything else but still moves?

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This needs a lot of rewording, but it's a great question, conceptually. – Sparr Apr 8 '13 at 21:58
For an example of indeterminism in Newtonian classical physics, see also this Phys.SE post. – Qmechanic Apr 8 '13 at 22:16
There is no reason I can't assign $\ddot{x} = \Theta(t - 1)$ for $\Theta$ the Heavyside step function, and see that the mass moves just not now. – dmckee Apr 9 '13 at 1:56
Nice question +1. However it should be noted that (in framework of classical mechanics) only specification of the initial conditions is not enough to find out time evolution of an entity. One should also specify the force. – user10001 Apr 9 '13 at 20:11
BTW there is a practical side to this question, when it comes to designing camshaft lobes for valvetrains and other dynamic systems. – ja72 Apr 9 '13 at 20:49

Ok so the way I see it, let's strip away the physics first. Then we're left with the math. So the question you're posing is, given only $\frac{d^nx(t)}{dt^n}|_{t=c} \forall n$, can we reconstruct the function $x(t)$ uniquely in some domain $(a,b)$ s.t. $c \in (a,b)$?

Well the answer is no. Even if you impose a $C^\infty$ condition on $x(t)$, the example you gave (with $x(0) = 0$) shows that it is not possible. Well, a necessary and sufficient condition requires complex analysis: all holomorphic functions in $(a,b)$ can be reconstructed just based on knowledge of all its derivatives at one point. Note that holomorphicity is equivalent to analyticity.

Thus, we need the extra a priori condition that our function is holomorphic in $(a,b)$, in order to determine it uniquely.

Now let's turn on the physics and try to interpret the question. Here I'm just offering my opinions/speculating, so feel free to correct me or post comments.

I'll first like to point out that this is a purely kinematical question, and not a dynamical one. In other words, this question makes no reference to Newton's laws at all. So this question is actually somewhat different from the Norton's dome paradox, in which the question there is whether Newton's law $F=ma$ should be deterministic or not, since the equation of motion was derived specifically from $F = ma$.

Now why do we think the Norton dome paradox is a problem? Because somehow we inherently believe that Newton's laws should be deterministic - given an initial condition that evolves due to Newton's laws, we should get a unique flow in phase space. But obviously this is only guaranteed if the force equation obeys certain conditions like Lipschitz continuity. The resolution to this problem is either that we arbitrarily impose a Lipschitz continuity condition on the force, or that we say that Newton's theory is not completely correct. Since Norton's dome gives us an actual scenario in which the force is not Lipschitz-continuous, we are forced to take the second view, and the better theory that supersedes Newtonian theory is of course, QM.

Now for this question, it doesn't actually say anything about Newtonian classical mechanics, and so it cannot give any 'insight' about the non-determinism of Newtonian mechanics, unlike the Norton dome problem. While it might seem un-intuitive that given the position, velocity, acceleration, jerk, etc. of a particle at one in time, we do not know its position at all times, I would like to argue that this is purely a failure in our understanding of mathematics, as opposed to a shortcoming of Newtonian mechanics.

Thus the point I'm trying to make is that kinematics is not the same as dynamics, although the two are intimately intertwined. This might be a bold (but true) statement, but kinematics is not physics - it is just math littered with physical terms: Position, velocity, acceleration can be replaced with function, first derivative of function, second derivative of function; 'distance traveled in $t$ seconds' is just 'integral of $x'(t)$' etc. So whatever paradoxes we have about kinematics ultimately stem from misunderstandings about mathematics.

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This is a very good answer, and shows why sound mathematics and technical mathematical details must be paid attention... – daaxix Apr 11 '13 at 7:33

In my eyes it is not possible. Your functions doesn't agree with the initial conditions and is therefore invalid.

If you approximate you solution using a taylor series around zero, you will see that this leads to: $x(t) = 0 \forall t \in \mathbb{R}$

This should prove that no other function, whose taylor series is convergent, could be a solution. But it doesn't prove if there are solutions which don't meet this condition.

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My function agrees with the initial conditions. Simply note that $$\lim_{t\rightarrow 0}\frac{d^n}{dt^n}\exp(-1/t^2) = 0 \forall t \mbox{.}$$ – Physicist Apr 9 '13 at 7:43
I see that the Taylor series of $\exp(-1/t^2)$ is identical to zero. That is why I wrote that this function is not analytical at $t=0$. However, how does this answer my question? 'This should prove that no other function, whose taylor series is convergent, could be a solution.' Why? What do you mean by 'solution'? I have not posed any differential equation. – Physicist Apr 9 '13 at 7:58
I based the taylorseries not on the function $exp(-1/t^2)$, but on the condition that the solution must fullfil that every derivat of the function is zero. Therefore every term of the taylor series is zero. One of your conditions is: $x(t=0)=0$ this doesn't agree with oyur – Stein Apr 9 '13 at 17:22
@ Stein : $\exp(-1/t^2)$ for $t \neq 0$ and $0$ for $t = 0$ is a classic example of a $C^\infty$ function whose Taylor series at $0$ is identically $0$, i.e. all $d^n x(t)/dt^n|_{t=0} = 0$. (And hence its Taylor series at $0$ is trivially convergent). – nervxxx Apr 10 '13 at 7:35

This function, together with all its derivatives, is 0 at t=0

First, a technical observation. The exponent isn't defined for t = 0. What you're thinking of is:

$x(t) = \begin{cases}\exp(-1/t^2)&\text{if }t\ne0,\\ 0&\text{if }t=0,\end{cases}$

How could it be that at a certain moment of time the mass point has neither velocity, nor acceleration, nor rate of change of acceleration, nor anything else but still moves?

The same could be asked about the velocity, acceleration, jerk, etc. Since all derivatives are zero at this point, this is a super stationary point, i.e., the function and all its derivatives are stationary there.

The question then is how do any of the derivatives change since $x^{(n)}(0 + dt) = 0$?

Evidently, we have here a function and its derivatives with the property of being non-zero only for a finite (non-infinitesimal) displacement in time from zero, i.e: $x^{(n)}(0 + \Delta t) <> 0$.

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1. I couldn't understand the last symbol [<>] before 0. What is it ? 2. Suppose there is a function of which all the derivatives are stationary in this interval f(x_0) ~ f(x_0+ndx) but for all other it is nonzero. Then what will happen ? – Arka Karmakar Apr 22 at 10:56

This might not be a very good answer, but anyways:

In principle nothing stops you from considering a motion $x(t) = \exp(-1/t^2)$, you can even find a (time-dependent) force that gives that motion:

$$F(t) = m \ddot x(t)$$

namely

$$F(t) = m e^{-1/t^2} (4-6 t^2))/t^6, \text{ for }t>0$$

This a perfectly reasonable, if somewhat unrealistic force and with your initial conditions, you will get the desired motion. It is unrealistic because physically it would not be possible to excatly generate any kind of time dependent force and even more so such a 'flat' increase.

I think the confusion arose by overlooking that one also has to specify the (potentially time-dependent) forces to determine the motion of a mechanical system.

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Perhaps it's unrealistic, but I don't see why that should be any objection. I think the paradox can be simplified to "why does the object move if every derivative of position is zero?" – Pricklebush Tickletush Apr 9 '13 at 23:11
orbifold, as my answer indicates, your answer leads to the question "but how does the force change from zero to non-zero when all of the time derivatives of the force vanish when t=0?" – Alfred Centauri Apr 9 '13 at 23:46
Alfred Centauri, yes but it is equally impossible to exactly generate any kind of force, so to me this is as unrealistic as a sudden violent onset of a force at t=0, modelled by a step function. Realistically you will only be able to approximate it by forces that have some non zero time derivative at t=0. – orbifold Apr 10 '13 at 10:23

A potential needed for a particle to have the equation of motion of $x = e^{1/2t^2}$ is easily found by calculating the kinetic energy of the particle at any given time as a function of the position. (For simplicity, assume mass = 1.)

$$U(x) = -\frac{1}{2} \dot{x}^2 = -\frac{1}{2t^6}e^{1/t^2} = 4x^2 ln^3(x)$$

A particle that is moving in the negative x direction with a total energy of zero will have that equation of motion.

Another more conceptual system that would have an equation of motion with all derivatives zero at time zero would be a golf ball from the time of first contact by the golf club. Since, at that instant, the club hasn't started to deform the ball, the ball is still stationary. As time goes on, the club will deform the ball so the ball will start moving.

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