Prove that $g(t) \times \frac{d}{dt} g(t) = 0$

Question

If the vectorial function $r = g(t)$, with values in $\mathbb R^3$ and where $t\in\mathbb R$, is a solution of the differential equation $\frac{d^2}{dt^2} r(t) = t^2 r(t)$, such that $g(0) = 0$, prove that $g(t) \times \frac{d}{dt} g(t) = 0$ for every $t$.

I have tried resolving the differential equation but I don't get anywhere. How can I do this?

Thanks.

Answer 1

Let us use $\mathbf{f}(t)$ to denote the cross product. We are regarding it as a generic function of $t$, not necessarily constant. I use dot notation to indicate time deriative. $$\mathbf{f}(t) = \mathbf{g}(t)\times \dot{\mathbf{g}}(t)$$ We can differentiate $\mathbf{f}$ by applying the following rule $$\frac{d}{dt}\left(\mathbf{a}\times \mathbf{b}\right) = \dot{\mathbf{a}}\times \mathbf{b} + \mathbf{a}\times \dot{\mathbf{b}}$$ from which we get $$\frac{d}{dt}\mathbf{f}(t) = \dot{\mathbf{g}}(t)\times \dot{\mathbf{g}}(t) + \mathbf{g}(t)\times \ddot{\mathbf{g}}(t)$$ The first term in the above sum is zero, since a vector crossed with itself is always zero. The second term we can rewrite using the differential equation as $$\mathbf{g}(t)\times \ddot{\mathbf{g}}(t) = \mathbf{g}(t) \times t^2\mathbf{g}(t) = t^2\left(\mathbf{g}(t)\times \mathbf{g}(t)\right)$$ Again, the above is zero since we have a vector crossed with itself. This tells us that $\mathbf{f}(t)$ is a constant with respect to time. $$\mathbf{f}(t) = \mathbf{g}(t)\times \dot{\mathbf{g}}(t) = \mathbf{C}$$ To determine the constant $\mathbf{C}$, we note that the above equation holds for all time, and in particular for $t=0$. But then $\mathbf{g}(0) = \mathbf{0}$ so that $$\mathbf{C} = \mathbf{g}(0)\times \dot{\mathbf{g}}(0) = \mathbf{0}\times\dot{\mathbf{g}}(0) = \mathbf{0}$$ It follows that $$\mathbf{g}(t)\times \dot{\mathbf{g}}(t) = \mathbf{0}$$ as desired.

Notice that we have in fact shown a much more general result. If $\mathbf{g}$ satisfies $$\ddot{\mathbf{g}}(t) = h(t)\mathbf{g}(t)$$ for any scalar function $h$, then the result continues to hold (although the constant may not be zero). Your particular case is $h(t) = t^2$.