Is linear momentum times distance any meaningful quantity? $\vec{r} \cdot \vec{p}$ or $pr = mvr$ comparing to $\vec{r} \times \vec{p}$

Question

Angular momentum is$$ L=\vec{r} \times\vec{p}$$

I was wondering if the dot product has any meaning: $$ ?= \vec{r} \cdot \vec{p}$$

Does it mean anything? It could also be rewritten like $ rp$ or $\Delta x p$. Other ways to write it is:

$$ \int p dx $$ $$ \vec{r} \cdot \vec{p}$$ $$pr = mvr$$

Is it useful in anyway? Or maybe in a certain construction, with centre of mass etc.

EDIT (15/5): I found another way to find something relating these two quantities $p,x$ with the action:

$$\mathcal{S} =\int^{t_{1}}_{t_{0}}Ldt = \int^{t_{1}}_{t_{0}}\frac{1}{2}m\dot{x}^{2}dt $$

$$\begin{split} \mathcal{S} &=\int^{t_{1}}_{t_{0}}\frac{1}{2}m\dot{x}^{2}dt \\ &=\int^{x_{1}}_{x_{0}}\frac{1}{2}m(\frac{dx}{dt})^{2}\frac{dt}{dx}dx \\&=\int^{x_{1}}_{x_{0}}\frac{1}{2}mvdx \\&=\frac{1}{2}mv(x_{1}-x_{0}) \\&= \boxed{\frac{1}{2} p \Delta x} \end{split}$$

Answer 1

You could use it to derive the centripetal acceleration equation.

Suppose we have a particle moving around in a circle and a position vector to it from the center of the circle. Consider the quantity:

$$ Q = r \cdot p$$

$Q$ is equal to zero because $r$ and $p$ are perpendicular.

Differentiating both side:

$$ 0 = ( \frac{dr}{dt}) \cdot p + r \cdot \frac{dp}{dt}$$

We have,

$$ \frac{dr}{dt} = v$$ and,

$$ \frac{dp}{dt} = F$$

This leads us to:

$$ 0 = v \cdot p + r \cdot F$$

Now $p=mv$ and $ r \cdot F = |r| F_{centripetal}$:

$$ 0 = m|v|^2 + |r| F_{centripetal}$$

Rearranging, we find:

$$ F_{centripetal} = - \frac{m|v|^2}{r}$$

Answer 2

Unfortunately, I do not know if $\vec{r} \cdot \vec{p}$ has any physical significance; at least, I have never heard of it.

However, it is interesting to note that for $\theta= \frac{\pi}{4} + n\pi$, where $n \in\mathbb{Z}$, the dot product between position $\vec{r}$ and linear momentum $\vec{p}$ coincides with the magnitude of angular momentum:

$$\vec{r} \cdot \vec{p} = |\vec{r}||\vec{p}|\cos{(\pi/4)}=|\vec{r}||\vec{p}|\sin{(\pi/4)}=|\vec{r} \times \vec{p}|=|\vec{L}|$$