R, dr and d²r and curvilinear coordinates

Jhenrique

Hellow everybody!

If $d\vec{r}$ can be written in terms of curvilinear coordinates as $d\vec{r} = h_1 dq_1 \hat{q_1} + h_2 dq_2 \hat{q_2} + h_2 dq_2 \hat{q_2}$ so, how is the vectors $d^2\vec{r}$ and $\vec{r}$ in terms of curvilinear coordinates?

Thanks!

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HallsofIvy

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1. The vector $\vec{r}$ does NOT lie in the tangent plane of a surface and so cannot be written in such a way.

2. I don't know what you mean by "$d^2\vec{r}$". The second derivative? Second derivatives are NOT vectors, they are second order tensors.

Jhenrique

2. I don't know what you mean by "$d^2\vec{r}$". The second derivative? Second derivatives are NOT vectors, they are second order tensors.
In polar coordinates $\vec{r} = r \hat{r}$

Aplying d of derivative, we have: $d\vec{r} = dr \hat{r} + r d\theta \hat{\theta}$

Aplying d again, we have: $d^2\vec{r} = (d^2r - d\theta^2) \hat{r} + (2 d\theta dr + r d^2 \theta)\hat{\theta}$

But, I'd like of see this result/operation in curvilinear coordinates, just this.

joshmccraney

in spherical coordinates:

$$\vec{r}=r\hat{e_r}$$

understand $r=f(r)$ and $\hat{e_r}=f(\theta , \phi )$

it is well understood in multivariable calculus that, given $z=f(x,y)$ we have $dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$ (this is intuitive as well).

now by the product rule we have $$d\vec{r}=dr\hat{e_r}+r d\hat{e_r}$$

thus it seems we have only to compute $d \hat{e_r}$. by the above theorem, $d \hat{e_r}=\frac{\partial \hat{e_r}}{\partial \theta}d\theta+\frac{\partial \hat{e_r}}{\partial \phi}d\phi$. a geometric illustration works best from here, but ill outline the procedure.

by definition of a partial derivative, we have $\frac{\partial \hat{e_r}}{\partial \theta}=\lim_{\Delta \theta \to 0} \frac{\hat{e_r}(\theta +\Delta \theta, \phi)-\hat{e_r}(\theta , \phi)}{\Delta \theta}$. recognize (by drawing a picture perhaps) that $\hat{e_r}(\theta +\Delta \theta, \phi)-\hat{e_r}(\theta , \phi)= \sin(\phi) \hat{e_\theta}\Delta \theta$. after considering the limit we arrive at simply $d \hat{e_r}=\sin(\phi) \hat{e_\theta}d\theta+\frac{\partial \hat{e_r}}{\partial \phi}d\phi$

the $\phi$ term is conducted in a similar fashion. if you're lost in the geometry, sketch it out (it makes much more sense that way). hope this helps!

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