Time derivatives of the unit vectors in cylindrical and spherical

Question

In cylindrical and spherical coordinates, the position vectors are given by $\mathbf{r}=\rho \widehat{\boldsymbol{\rho}}+z \hat{\mathbf{k}}$ and $\mathbf{r}=r \hat{\mathbf{r}}$, next to next, and their derivatives with respect to time are $$ \dot{\mathbf{r}}=\dot{\rho} \hat{\boldsymbol{\rho}}+\rho \dot{\varphi} \widehat{\boldsymbol{\varphi}}+\dot{z} \hat{\mathbf{k}} $$ $$ \dot{\mathbf{r}}=\dot{r} \hat{\mathbf{r}}+r \dot{\theta} \widehat{\boldsymbol{\theta}}+r \sin \theta \dot{\varphi} \widehat{\boldsymbol{\varphi}} $$ I wonder what would be the time derivatives of the unit vectors of the basis themselves. Is there any straightforward way to deduce their expressions?

Answer 1

It doesn't make a lot of sense to talk about the "time derivatives of the unit vectors of the basis themselves", because the unit vectors themselves are constant with respect to time. Instead, in curvilinear coordinates, the basis vectors are vector fields, with different values from point to point in space. The unexpected terms that arise in the expressions you've written are because the unit vectors are not constant with respect to space, and any trajectory that moves through space will see these unit vectors vary because of their motion through space.

To make this more concrete, think about $\hat{r}$ as a vector field: $\hat{r}(r,\theta,\phi)$. If we then have a particle that has a trajectory given by $\mathbf{r}(t) = r(t) \; \hat{r}(r(t),\theta(t),\phi(t))$, then we can derive the velocity vector as follows: \begin{align*} \frac{d\mathbf{r}}{dt} &= \frac{dr}{dt} \hat{r} + r \frac{ d \hat{r}}{dt} \\ &= \frac{dr}{dt} \hat{r} + r \left( \frac{\partial \hat{r}}{\partial r} \frac{dr}{dt} + \frac{\partial \hat{r}}{\partial \theta} \frac{d\theta}{dt} + \frac{\partial \hat{r}}{\partial \phi} \frac{d\phi}{dt} \right) \end{align*} where we have used the product rule and the multi-variable chain rule.

If you carefully calculate $\frac{\partial \hat{r}}{\partial \theta}$ and $\frac{\partial \hat{r}}{\partial \phi}$, you can show that they're equal to $\hat{\theta}$ and $\sin \theta \hat{\phi}$ respectively, and so the expression you have for spherical coordinates is obtained. The key point to note here, though, is that the $\hat{r}$ vector field does not depend inherently on $t$; it's only a function of your position in space.