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:

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

Answer 3

EDIT: After some research, your intuition is correct. Momentum times distance (also some dot product or projection of it into a given vector) represents the concept of "action," which is deeply related to Newtonian and Lagrangian mechanics. It often shows up in formulations of the "Principle of Least Action":

$\delta\int mvds=0$

Another form of this is the ubiquitous Euler Lagrange Equations:

$\frac{d}{dt}\frac{\partial L}{\partial \dot q_k}-\frac{\partial L}{\partial q_k}=0$

From what I've learned about the history of it all, many 17-18th century contemporaries were debating about which properties truly described motion. At the time, it was yet to be decided. Was it momentum, energy (i.e. Vis Viva), Force, Action or some unknown quantity that described motion most succinctly? Newton, Leibniz, Lagrange, Euler and Maupuertis were trying to get at the essence of mechanical motion. It turns out that the principle of least action unifies energy methods (D' Alembert's Principle of Virtual Work) and Newtonian methods through the Euler-Lagrange Equation. Though I don't have the best intuition for this, the principle shows that the EoM's derived by F=ma, and the EoM's derived by Energy methods are necessarily the same.

Reference:

[1] https://en.wikipedia.org/wiki/Action_(physics) [2] https://youtu.be/Q10_srZ-pbs (Great video by Veritasium, articulating the concept much more aptly)

ORIGINAL RESPONSE 24MAR25: Yes, fyi I believe that's more of a graduate level question. It's implicated in the calculus of variations that appear when deriving Hamilton's and Lagrange's Equations. (You could think of these as forms of energy analysis). I'm still building my intuition for what it means physically -- and working on putting it into more plain language. Not quite there yet, however at a more technical level it's a part of the integrated sum of the variation in kinetic and potential energy over a given time interval for single & multi-variable systems.

$\delta T+\delta W =\sum_{i=1}^Nm_i \frac{d}{dt}(\dot{r_i} \cdot \delta{r_i})$

$\int_{t_1}^{t_2}(\delta T+\delta W )dt=\int_{t_1}^{t_2}\sum_{i=1}^Nm_i \frac{d}{dt}(\dot{r_i}\cdot \delta{r_i})dt={...}$

$...=\int\sum_{i=1}^N m_i d(\dot r_i\cdot \delta r)=\sum_{i=1}^N m_i d(\dot r_i\cdot \delta r)|_{t_1}^{t_2}$

At base, this means that the amount of energy in a system is conserved over a given time interval if $\delta r_i=f(x,y,t)=f(t_a)=0$ for any given time $t_a$. Basically there would be no bifurcations or variation of position in time (i.e. position would be an explicit function of time).

Citation:

(1) Section 4.8 Hamilton's Principles (you can find the following book in the public domain online)

Baruh, H. Analytical Dynamics. New York, NY: McGraw-Hill, 1998. ISBN: 9780073659770

(2) I've been going to this youtube channel while I get a handle on analytical dynamics.

Dr. Shane Ross's intro to hamiltonian systems: https://www.youtube.com/watch?v=pB-aleLeKL0&t=300s

(3) Dr. Ross's profile handle: https://www.youtube.com/user/RossDynamicsLab