Why does a particle initially at rest at origin with acceleration as square of its $x$ coordinate ever move?

Question

Consider a particle initially at rest at origin, with acceleration, $a$, such that $ a(x)=x^2$.

Since the particle is at origin, initial acceleration would be 0. It's also at rest initially. Its $x$-coordinate should stay 0 till it acquires a velocity, to acquire a velocity it needs to be accelerated, to accelerate it needs to have a non-zero $x$-coordinate.

This should imply that the particle would forever be at rest and at origin, but the initial equation can be solved to get $x$-coordinate as a function of time such that $ x(t)=\cfrac{6}{t^2}$.

What mistake am I making here?

Answer 1

You found a solution to the ODE that arises from Newton's Second Law ($\ddot{x} = x^2$). It's just not the solution that's consistent with the initial conditions you asked for.

In general, there are a wide variety of possible motions that are consistent with a given equation of motion. For a very simple example, the equation $\ddot{x} = - g = - 9.8 \text{ m/s}^2$ has a bunch of possible solutions (ignoring units): $$ x(t) = 1 - 4.9 t^2 \\ x(t) = -4 + 5 t - 4.9 t^2 \\ x(t) = 10^6 - \pi t - 4.9 t^2 $$ The particle "chooses" which of these motions it will follow according to the initial conditions it has. In the case where $\ddot{x} = - g$, you have to specify $x_0 = x(0)$ and $v_0 = \dot{x}(0)$ in order to specify the motion — as you recall from the first-year kinematic equation $x(t) = x_0 + v_0 t - \frac{1}{2} g t^2$. And what's more, you don't have to specify these values at $t = 0$—any "initial time" will do the trick just as well.

So in your case, one of the possible solutions to the ODE $\ddot{x} = x^2$ is $x(t) = 6/t^2$. But this solution is not consistent with the initial conditions you are assuming ($x_0 = v_0 = 0$); so the particle will not follow this motion. Instead, it will execute the solution $x(t) = 0$ (for all $t$), which—as you can verify—both satisfies $\ddot{x} = x^2$ at all times and has $x(0) = \dot{x}(0) = 0$.

If, in contrast, you had demanded the initial conditions $x(1) = 6$ and $\dot{x}(1) = -12$, then the solution $x(t) = 6/t^2$ would in fact be the correct solution to the equations of motion, consistent with the initial conditions.

Answer 2

I do not think that the particle would ever start moving if $\ddot{x} = x^2$ is truly its equation of motion. You could imagine this as the particle being trapped at exactly the top of some potential well (e.g. a ball exactly at the top of a hill). You would see movement if the particle were slightly left or right of x=0 (e.g. x=0.00001). This is clear if we look at the potential (ball on a hill) picture too, the particle is clearly in an unstable equilibrium.

Your solution for the differential equation $$\ddot{x}=x^2\tag{1}$$ is not well defined at $t=0$ as $\frac{6}{0^2}$ is problematic! If you require the position at $t=0$ to be $0$, you must apply this as a boundary condition when solving the differential equation (1). However, the solution to (1) is not easy, I believe it is the Weierstrass Functions which is probably not useful for a dynamics problem. I believe that your solution ($x=6/t^2$) is actually the first term of the expansion of this function. While you do correctly recover $a(x) = x^2$ from this solution, I'm not sure if it has any physical meaning at $t=0$. Also, the particle would only ever have $x=0$ at $t=\infty$ according to this solution!

Answer 3

This is a fine example of how mathematical models are not physics. The equation implies that there's a positive force on the particle wherever it is, unless it's exactly at the origin. But nothing is exact in real physics. "Particles" are mathematical objects that model physical objects, but no physical object has all of the properties of a mathematical particle. It is not possible to place a physical object exactly anywhere, or even to identify where such a point (another non-physical mathematical object) is.

So, if you do this experiment, the particle will either stay put or move. If it stays put, and you look closely enough, it will turn out that there's some additional force holding it it place. If it moves, you'll attribute that to some combination of extra force and/or imprecise initial placement if you examine the details.