Finding an elegant proof of rectilinear motion

Question

Say you've got a body with position r(t), subject to a force collinear with r(t) but otherwise not specified.

So r(t) × r''(t) = 0 for all t (with x being the cross product).

Since r(t) x r''(t) = (r(t) x r'(t))',

we've got r(t) x r'(t) = c with c a constant.

When c is not the zero vector, this is a nice proof that the body is moving in a plane, since at any time its position vector is orthogonal to c. The way I understand it, the elegance comes from the fact that we're using a conserved quantity to derive the answer, without having to use any integration or infinitesimals.

Now when c = 0, the speed and position are also collinear at any given point in time.

We've got r(t + dt) = r(t) + r'(t) dt, so r(t + dt) and r(t) are collinear, and the body's motion is linear.

I've been trying to find a proof for this degenerate case that's similar to that of the general case, using a conserved quantity, but no luck so far. The natural analogue would be to use a primitive of r(t) x r'(t), but I haven't found it, either by myself or looking online.

Is this primitive known? Or would there be another similar way of proving that the motion is linear?

Answer 1

The reason you're not able to prove it is that it's just not true.

It just goes to show that "proving" things with an informal intuition about infinitesimals is risky.

For $\mathbf{c} = \mathbf{0}$ the motion is not only not rectilinear, it's not even necessarily in a single plane!

Let's define a helper function: $$u(t) = t^3(1-t)^3$$ We have: $$u(0) = u'(0) = u''(0) = u(1) = u'(1) = u''(1) = 0$$

Let $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$ be three orthogonal vectors. Define:

$$\mathbf{r}(t) = \begin{cases} u(t)\cdot\mathbf{e}_1 \text{ for } 0 \le t \le 1 \\ u(t-1)\cdot\mathbf{e}_2 \text{ for } 1 \le t \le 2 \\ u(t-2)\cdot\mathbf{e}_3 \text{ for } 2 \le t \le 3 \end{cases}$$

Easy to check that $\mathbf{r}(t) \times \mathbf{r}'(t) = \mathbf{0}$ and $\mathbf{r}(t) \times \mathbf{r}''(t) = \mathbf{0}$ but the motion is not in a single plane.

Now let's add an extra assumption that $\mathbf{r} \ne \mathbf{0}$.

In that case, the conserved quantity is the direction of $\mathbf{r}$: $$\mathbf{d}(t) = \frac{\mathbf{r}(t)}{\lVert\mathbf{r}(t)\rVert}$$

Let's compute the derivative: $$ \begin{align} \mathbf{d}' &= {\lVert\mathbf{r}\rVert}^{-2}\left(\mathbf{r}'\lVert\mathbf{r}\rVert - \mathbf{r}\cdot\frac {\mathbf{r}\cdot\mathbf{r}'}{\lVert\mathbf{r}\rVert}\right) \\ &= {\lVert\mathbf{r}\rVert}^{-3}(\mathbf{r}'(\mathbf{r}\cdot\mathbf{r}) - \mathbf{r}(\mathbf{r}\cdot\mathbf{r}')) \\ &= {\lVert\mathbf{r}\rVert}^{-3}(\mathbf{r}\times(\mathbf{r}'\times\mathbf{r})) \\ &= {\lVert\mathbf{r}\rVert}^{-3}(\mathbf{r}\times \mathbf{0}) \\ &= \mathbf{0} \end{align} $$

I used the "triple product" formula $\mathbf{a}\times(\mathbf{b}\times \mathbf{c}) = \mathbf{b}(\mathbf{c}\cdot\mathbf{a}) - \mathbf{c}(\mathbf{a}\cdot\mathbf{b})$.

So in this case indeed the direction doesn't change and hence the motion is rectilinear.