Infinite series of derivatives of position when starting from rest

Question

Suppose you have an object with zero for the value of all the derivatives of position. In order to get the object moving you would need to increase the value of the velocity from zero to some finite value. A change is velocity is acceleration, so the value of the acceleration would have to increase from zero to some value. You would also need to increase the jerk from zero to some value to have a change in acceleration. Is there an infinite series of higher derivatives of position for this to work? Or am I missing something?

Answer 1

I) Disclaimer: In this answer, we will only consider the mathematically idealized classical problem, which is interesting in itself, although admittedly academic and detached from actual physical systems.

II) It is in principle possible that all time derivatives of the position $x(t)$ vanishes at $t=0$, and yet the particle does move away (from where it was at $t=0$).

$\uparrow$ Fig. 1: A plot of position $x$ as a function of time $t$.

Example: Assume that the position is given by the following infinitely many times differentiable function

$$ x(t)~:=~\left\{ \begin{array}{ccl} \exp\left(-\frac{1}{|t|}\right)&\text{for} & t~\neq~ 0, \\ \\ 0 &\text{for} & t~=~ 0. \end{array} \right. $$

Note that the Taylor series for the $C^{\infty}$-function $x:\mathbb{R}\to\mathbb{R}$ in the point $t=0$ is identically equal to zero.$^1$ So the function $x$ and its Taylor series are different for $t\neq 0$. In particular, we conclude that the smooth function $x$ is not a real analytic function.

III) If you like this Phys.SE question you may also enjoy reading about What situations in classical physics are non-deterministic? , Non-uniqueness of solutions in Newtonian mechanics and Norton's dome and its equation.

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$^1$ Sketched proof that $x\in C^{\infty}(\mathbb{R})$: Firstly, obviously, $x$ is $C^{\infty}$ for $t\neq 0$. By induction in $n\in\mathbb{N}_0$, for $t\neq 0$, it is straightforward to deduce that the $n$'th derivative is of the form $$x^{(n)}(t)~=~\frac{P_n(t,|t|)}{Q_n(t,|t|)}x(t), \qquad t\neq 0,$$ for some polynomials $P_n$ and $Q_n\neq 0$. Secondly, it follows that the $(n\!+\!1)$'th derivative at the origin $$x^{(n+1)}(0)~=~\lim_{t\to 0} \frac{x^{(n)}(t)}{t}~=~0$$ exists and is zero "because exponentials beat polynomials". $\Box$

Answer 2

There are smooth functions that have all derivatives equal to zero at a point (in fact, outside a bounded interval) yet are not constant. You can for example find a smooth function $f$ such that $f(t)$ and all its derivatives are $0$ whenever $t \le 0$ and $f(t) = 1$ whenever $t \ge 1$ (and all the derivatives of $f$ are zero when $t \ge 1$). Then if you take the force in Newton's second law to be $$F(t) = mf''(t)$$ the solution to the IVP $$m\ddot{x} = F \qquad x(0) = \dot{x}(0) = 0$$ is clearly a smooth motion starting at $x = 0$ with all derivatives of displacement also zero, and ending at $x = 1$, with all derivatives of displacement also zero.

However for a time-independent force the IVP $$m\ddot{x}(t) = F(x(t)) \qquad x(0) = \dot{x}(0) = 0 $$ when $F(x(0)) = 0$ the solution is a particle at rest. This follows from uniqueness of solutions to ODEs. Physically when $F = 0$ there is an equilibrium position, and a particle at rest at an equilibrium position remains at rest there. Of course this is really just Newton's first law.

Answer 3

Your question is "Is there an infinite series of higher derivatives of position for this to work?"

Answer: No.

Acceleration can jump from zero to something. When it does, its derivative is not defined, so the series of position derivatives stops after the second one.

From the question: "A change is velocity is acceleration, so the value of the acceleration would have to increase from zero to some value."

The first part is true, we need non-zero acceleration. The second part is misleadingly phrased. It can be read as "acceleration has to increase continuously", which is false. In contrast to velocity, acceleration can jump from zero to something.

Philosophically speaking, there is no law that says "in physics, derivatives are always smooth". Some derivatives just jump.

Answer 4

Mathematically, that's how it would work out. If you solved the initial value problem where all time derivatives of displacement were zero, then you would have zero displacement. Don't overthink it :)

Answer 5

Taking the Taylor's series approach at $t=0$ with respect to time, with all the derivatives zero the result would be zero motion.

If motion is present then some higher order derivatives must be an impulse at $t=0$ such that the derivative below that is a step function.

In valvetrain dynamics this is usually done by specifying linear jerk (step snap, impulse crackle, undefined pop).

So at $t=0$ you have the 4th derivative step from zero to $\ddddot{x}>0$ and making jerk linear $\dddot{x} = C_1 t$, acceleration a parabola $\ddot{x} = C_2 t^2$, velocity $\dot{x}=C_3 t^3$ and position $x=C_4 t^4$.

In some critical applications you might specify linear snap increasing the order of the curve by 1 to avoid certain undesirable dynamics of the valve spring.

_NOTE: Snap Crackle and Pop are real terms.