# Time derivative of unit velocity vector?

Let's say I have some parametric curve describing the evolution of a particle $$\mathbf{r}(t)$$. The velocity is $$\mathbf{v}(t) = d\mathbf{r}/dt$$ of course. I am trying to understand what the expression for $$\frac{d}{dt}\frac{\mathbf{v}}{\Vert\mathbf{v}\Vert}$$ is. I think it should have something to do with the perpendicular acceleration to the particle's tangent vector, but I can't seem to get any meaningful expression that would tell me this. Helping hand would be appreciated here.

• See here. Commented May 15, 2022 at 4:02

As the magnitude of a unit vector cannot change, $$\dfrac {d\hat v}{dt}$$ is related to the rate of change of the direction of the velocity.
The following is valid for any unit vector $$q=Q/|Q|$$, where $$Q$$ is a function $$Q: \mathbb{R}\rightarrow \mathbb{R}^D$$. Therefore, $$q(s)$$ is a curve in $$\mathbb{R}^D$$ such that $$|q(s)|=1$$ for all values of the real parameter $$s$$. Given that $$|Q| =\sqrt{Q\cdot Q}$$, direct calculation gives:
$$\frac{dq}{ds}= \frac{d }{ds} \frac{Q}{\sqrt{Q\cdot Q}} = \frac{1}{|Q|}\perp \frac{dQ}{ds} \, ,$$
where $$\perp$$ is the operator that projects any vector on the $$(D-1)$$-dimensional subspace orthogonal to $$q$$. In index notation (using Einstein notation):
$$\frac{dq_i}{ds}= \frac{1}{|Q|}\perp_{ij} \frac{dQ_j}{ds}$$
where $$\perp_{ij}=\delta_{ij}-q_i q_j$$. This is consistent with the idea that the derivative must be orthogonal to $$q$$, which comes from the fact $$d|q|/ds=0$$. A totally analogous reasoning is used in special relativity to show that the 4-velocity and the 4-acceleration are "orthogonal" or that (as pointed out in the comments) a centripetal acceleration does not change the speed.