Evaluating $\frac{d}{dt}(\boldsymbol{r} \times d\boldsymbol{r})$ - time derivative of vector crossed with infinitesimal - Kepler's second law

Question

I am trying to evaluate this quantity where $\boldsymbol{r}$ is a position vector, and $t$ is time:

$\frac{1}{2}\frac{d}{dt}(\boldsymbol{r} \times d\boldsymbol{r})$

which is the area of a triangle I am trying to calculate. I have made the following attempt which I am unsure about

$\frac{d}{dt}(\boldsymbol{r} \times d\boldsymbol{r}) = (\frac{d\boldsymbol{r}}{dt} \times d\boldsymbol{r}) + (\boldsymbol{r} \times \frac{d\boldsymbol{r}}{dt})$

I understand that $d\boldsymbol{r}$ crossed with itself would equal zero, but is this necessarily true for the time derivative of $\boldsymbol{r}$ crossed with $d\boldsymbol{r}$?

This is related to Kepler's second law, where $\boldsymbol{r} \times d\boldsymbol{r}$ relates to an unknown area which is not necessarily a segment of an ellipse at this point. By my understanding the first equation should equal $\boldsymbol{r} \times \frac{d\boldsymbol{r}}{dt}$ which implies $\frac{d\boldsymbol{r}}{dt} \times d\boldsymbol{r}$ should be zero, but how can I justify this? I am also trying to argue this statement without the use of angular momentum, but please let me know if this is not possible.

I am also unsure if I would label the quantity $\boldsymbol{r} \times d\boldsymbol{r}$ as an area, $\boldsymbol{A}$ or an elemental area, $d\boldsymbol{A}$.

Thank you and apologies if this has made its way to the wrong part of stackexchange.

Answer 1

Why $\frac{d\mathbf{r}}{dt} \times d\mathbf{r}$ equals $0$

First, we must remind ourselves that $d\mathbf{r}$ is a vector representing the change in position over some amount of time $dt$. It is clear that this change in position will be in the direction of the particle's velocity. In a sense, this is what defines the 'instantaneous' velocity: it is the change in position of a body divided by the infinitesimal change in time. This means that $\frac{d\mathbf{r}}{dt}$ and $d\mathbf{r}$ are trivially parallel; therefore, their cross product is $0$.

Angular momentum

To my knowledge, it is necessary to use angular momentum in some form in order to prove Kepler's Second Law. This is because it only holds true for orbits where there is no net torque, i.e., when angular momentum is conserved (in a system with no external torque, $\mathbf{L}$ is a constant vector). $$ \mathbf{L} = \mathbf{r} \times m\frac{d\mathbf{r}}{dt} \quad \Leftrightarrow \quad \frac{\mathbf{L}}{m} = \mathbf{r} \times \frac{d\mathbf{r}}{dt} \tag{I} $$

Using $A$ vs $dA$

As area is a scalar quantity, it makes sense to take the magnitudes of these vectors (although this is not necessary. If you prefer $\mathbf{A}$ and $d\mathbf{A}$, the mathematics below is only different in that the magnitude of vectors would not be taken.

The magnitude of the cross product between two vectors is equal to the area of the parallelogram they form when placed tip-to-tail in a plane. As $dt \to 0$, half of this parallelogram becomes a better approximation for the sector of an ellipse. This means that one would label $\left|\mathbf{r} \times d\mathbf{r}\right|$ as an elemental area, as is approaches $0$ with $dt$. Finding an expression for the (non elemental) area $A$: \begin{align} \frac{dA}{dt} &= \frac{1}{2}\left|\frac{d}{dt}\left(\mathbf{r} \times d\mathbf{r}\right)\right| \\ &= \frac{1}{2}\left|\mathbf{r} \times \frac{d\mathbf{r}}{dt}\right| \\ &= \left|\frac{\mathbf{L}}{2m}\right| \tag{by I} \\ A &= \frac{L(t_1 - t_0)}{2m} \tag{integrating from $t_0$ to $t_1$} \end{align} showing that for a given orbit with constant angular momentum, the area swept in a time interval is solely dependent on the length of that time interval.

This was my first answer, I hope you found it helpful!