# Distance times Velocity

Is there a meaningful physical concept of $distance * velocity$?

Came across something analogous in computer science and was wondering if there was any physical analogue.

• In units this can be transformed to [mechanical-power/acceleration] = $m^2/s$
– user78217
Commented Feb 15, 2017 at 16:37

In diffusion equations, the diffusion coefficient typically has a dimensionality of $\mathrm{L^2T^{-1}}$.

For instance, the heat equation is typically written in this form:

$$\frac{\partial u}{\partial t}-\alpha\nabla u=0,$$

where $u$ is temperature and $\alpha$ is the thermal diffusivity. The thermal diffusivity is defined as:

$$\alpha=\frac{k}{\rho c_p},$$

where $k$ is the thermal conductivity of the medium, $\rho$ is the mass density and $c_p$ is the specific heat capacity. The SI unit of thermal diffusivity is $\mathrm{m^2/s}$.

• Although this does have the dimensionality of $distance*time$ you didn't really explain if there's an actual physical analogue. You gave a dimensional analogue, I don't see that as the same thing. Your answer doesn't seem to address distance or time specifically.
– JMac
Commented Feb 15, 2017 at 21:26
• @JMac I sort of see your point (although I think you mean $distance*velocity$). I think there is a "physical analogue" in the sense you mean, but it may seem a little contrived. We also only know that OP wants an analogue to a quantity that is a distance multiplied with a velocity. We don't know what the distance or the velocity represent or why it's meaningful to multiply them. Without more information I think there isn't much point in going further than dimension analysis to check for analogousness.
– jkej
Commented Feb 15, 2017 at 23:12

Angular momentum for a unit mass. Angular momentum is: $$L= r × (mv)$$

This could also be thought of as an "area speed" (in lack of a better word). For instance if you are painting a wall and $d$ is the width of your brush, and $v$ is the speed you are moving the brush with, the area you would cover per unit time would be $d\cdot v$. I wouldn't say that this is a commonly used physical quantity though.

In fluid dynamics, the line integral of velocity field is called "flow" or "circulation."

The heat equation in $$3+1$$ dimensions $$\frac{∂ρ}{∂t} = α ∇^2ρ$$ with $$ρ = ρ(𝐫,t)$$ lifts to a hyperbolic equation-with-constraint in $$4+1$$ dimensions, as $$\left(∇^2 + \frac{∂^2ρ}{∂t∂u}\right)ρ = 0, \quad \frac{∂ρ}{∂u} = -\frac{ρ}{α},$$ with solutions of the form $$ρ(𝐫,t,u) = e^{-u/α} ρ(𝐫,t)$$, where $$ρ(𝐫,t) = ρ(𝐫,t,0)$$.

The coordinate $$u$$ has the dimension of length-squared/time.

This may be seen, more explicitly, to be the wave-equation-with-constraint. Under the substitution: $$t_± = t ± \frac{u}{V^2}, \quad \hat{ρ}\left(𝐫,t_+,t_-\right) = ρ(𝐫,t,u),$$ where $$V$$ is an arbitrarily-chosen non-zero speed, we have $$\frac{∂ρ}{∂t} = \left(\frac{∂}{∂t_+} + \frac{∂}{∂t_-}\right)\hat{ρ}, \quad \frac{∂ρ}{∂u} = \frac{1}{V^2}\left(\frac{∂}{∂t_+} - \frac{∂}{∂t_-}\right)\hat{ρ},$$ resulting in the equations: $$\left(V^2∇^2 + \left(\frac{∂}{∂t_+}\right)^2 - \left(\frac{∂}{∂t_-}\right)^2\right)\hat{ρ} = 0, \quad \frac{∂\hat{ρ}}{∂t_+} - \frac{∂\hat{ρ}}{∂t_-} = -V^2\frac{\hat{ρ}}{α},$$ with general solutions of the form $$\hat{ρ}\left(𝐫,t_+,t_-\right) = e^{V^2\left(t_- - t_+\right)/(2α)} ρ\left(𝐫,\frac{t_+ + t_-}{2}\right), \quad ρ(𝐫,t) = \hat{ρ}\left(𝐫,t,t\right).$$