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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.

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    $\begingroup$ See here. $\endgroup$
    – march
    Commented May 15, 2022 at 4:02

2 Answers 2

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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.

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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.

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