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:

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

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

  • 4
    $\begingroup$ A similar quantity plays a role in the derivation of the virial theorem. In analytical mechanics you see products of positions and momenta (more specifically, generalized coordinates and generalized momenta) in several places, one being in the computation of action-angle coordinates. $\endgroup$
    – secavara
    May 12, 2022 at 16:55
  • $\begingroup$ Your last derivation only holds if the particle's velocity is constant as it moves between the endpoints. That's true for a free particle, I suppose, but I suspect it doesn't hold in many other circumstances. $\endgroup$ May 16, 2022 at 19:16

2 Answers 2


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:

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

  • $\begingroup$ This is interesting thank you for this derivation. Is there a similar derivation for the Coriolis force? $\endgroup$ May 14, 2022 at 13:25
  • $\begingroup$ I have done the calculation here in an inertial frame, the coriolis force appears in a non inertial frame, right? @bananenheld $\endgroup$ May 14, 2022 at 13:27
  • $\begingroup$ But I mean the equivalent for the coriolis force in comparison to what the centripital force is to the centrifugal force. $\endgroup$ May 14, 2022 at 16:35

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:

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

  • $\begingroup$ Yeah that is pretty interesting observation thank you. Could you think of an interesting effect or experiment that shows this coincidence of values? $\endgroup$ May 12, 2022 at 17:30
  • $\begingroup$ I cannot think of a problem where the angle will be constantly $\theta= \frac{\pi}{4} + n\pi$, if that's what you are asking. However, there are a lot of mechanics problems in which the angle will be momentarily $\frac{\pi}{4} + n\pi$. Such an example could be a bird flying, with you being the origin. $\endgroup$ May 12, 2022 at 20:12

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