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!

 

[Jhenrique]

UP!?

 

[Jhenrique]

up... pliz!

 

[HallsofIvy]

You can get banned from this board for "bumping" threads.

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.

 

[member 428835]

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!