# Geometry / dynamics analogues

In 3-space geometry we have curvatures when a point is proceeding along a curved arc. Similarly when particle motion occurs with respect to time we have accelerations.

Is there a one to one correspondence or combination relation among curvatures/ tangent rotations on the one hand and accelerations/velocities on the other?

EDIT1:

I hope to be able to see a table of dynamic/geometric analogues.

e.g., for plane curves

Centripetal acceleration ~ normal curvature Coreolis acceleration ~ ? tangential acceleration ~ ? Angular momentum ~ ? ~ geodesic torsion ?

Most importantly a query remains with me longtime:

What are the characteristic accelerations or velocities of a particle moving along a curved asymptotic line of a ( negative curved ) surface?

EDIT2

added tags: curves and differential geometry

• Incidentally, if my answer does address what you were asking, it's probably a good idea to change the tags to "curves" and "differential-geometry". :) – Andrew D. Hwang Feb 27 '15 at 16:38

$\newcommand{\Reals}{\mathbf{R}}\DeclareMathOperator{\Tgt}{\mathbf{T}}\DeclareMathOperator{\Nml}{\mathbf{N}}$Your question is qualitative and open-ended, but perhaps the following observations get at what you're asking:

Let $I$ be an open interval of real numbers, and suppose $\gamma:I \to \Reals^{n}$ is a twice continuously-differentiable curve with non-vanishing velocity. For each $t$ in $I$, the velocity $\gamma'(t)$ is non-zero, so we may write $$\gamma'(t) = \|\gamma'(t)\|\, \frac{\gamma'(t)}{\|\gamma'(t)\|} = v(t) \Tgt(t),$$ with $v(t) = \|\gamma'(t)\|$ the speed at $t$ and $\Tgt(t)$ the unit tangent vector.

By the product rule, the acceleration decomposes as $$\gamma''(t) = v'(t) \Tgt(t) + v(t) \Tgt'(t). \tag{1}$$ Remarkably, this is the decomposition of the acceleration $\gamma''(t)$ into parallel and orthogonal components with respect to $\Tgt(t)$, i.e., into tangential and normal components: Since $\Tgt(t) \cdot \Tgt(t) = 1$ for all $t$, we have $2\Tgt(t) \cdot \Tgt'(t) = 0$ for all $t$.

If $\Tgt'(t) \neq 0$, we may write $$\Tgt'(t) = \|\Tgt'(t)\| \frac{\Tgt'(t)}{\|\Tgt'(t)\|} = \frac{\kappa(t)}{v(t)} \Nml(t),$$ with $\kappa(t)/v(t) = \|\Tgt'(t)\|$. Qualitatively, (1) says acceleration can be resolved into a term $v'(t) \Tgt(t)$ measuring how the speed changes at each instant, and a term $v(t) \Tgt'(t) = \kappa(t) \Nml(t)$ measuring how the direction changes at each instant.

The function $\kappa(t)$, the "rate of rotation of $\Tgt(t)$, normalized according to the speed $v(t)$", has a pleasant geometric interpretation as the reciprocal radius of curvature of $\gamma$ at $t_{0}$. Let $s:I \to \Reals$ be the arc length measured along the curve starting at $\gamma(t_{0})$, i.e., $$s(t) = \int_{t_{0}}^{t} \|\gamma'(\tau)\|\, d\tau,\quad \text{so that s'(t) = v(t) and s(t_{0}) = 0.}$$ It's straightforward to check that the curve $$C_{t_{0}}(t) = \left[\gamma(t_{0}) + \frac{1}{\kappa(t_{0})} \Nml(t_{0})\right] - \frac{1}{\kappa(t_{0})} \cos \bigl(\kappa(t_{0}) s(t)\bigr) \Nml(t_{0}) + \frac{1}{\kappa(t_{0})} \sin \bigl(\kappa(t_{0}) s(t)\bigr) \Tgt(t_{0}),$$ which parametrizes the osculating circle, of radius $r(t_{0}) = 1/\kappa(t_{0})$ lying in the plane through $\gamma(t_{0})$ and spanned by $\Tgt(t_{0})$ and $\Nml(t_{0})$, agrees with $\gamma$ to second order at $t_{0}$, in the sense that $$C_{t_{0}}(t_{0}) = \gamma(t_{0}),\qquad C_{t_{0}}'(t_{0}) = \gamma'(t_{0}), \qquad C_{t_{0}}''(t_{0}) = \gamma''(t_{0}).$$

In words, if $\gamma'(t_{0})$ and $\gamma''(t_{0})$ are non-zero, then so far as velocity and acceleration (at that instant) are concerned, a particle may as well be moving along a circle of radius $1/\kappa(t_{0})$.

Particularly, if a particle moves at unit speed, i.e., $\gamma$ is "parametrized by arc length", then $\Tgt'(t) = \kappa(t) \Nml(t)$, and the rate of rotation of $\Tgt$ at $t_{0}$ is precisely the reciprocal radius of the osculating circle to $\gamma$ at $t_{0}$.

As user86418 pointed out, your question is too vast. Here is a table with analogous concepts in mechanics and geometry that I can remember now. You can see more details about each one by a simple search on google.

## Geometry | Mechanics

tangent vector$\quad\quad$ velocity vector

curvature of a curve$\quad\quad$ acceleration intensity

geodesic curvature$\quad\quad$ tangential acceleration intensity

normal curvature$\quad\quad$ normal acceleration intensity

geodesics$\quad\quad$ path taken by a free particle

Clairaut's relation$\quad\quad$ angular momentum