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R, dr and d²r and curvilinear coordinates

  1. Jan 19, 2014 #1
    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!
     
  2. jcsd
  3. Feb 14, 2014 #2
    UP!!!???
     
  4. Feb 19, 2014 #3
    up.... pliz!
     
  5. Feb 20, 2014 #4

    HallsofIvy

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    Science Advisor

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

    1. The vector [itex]\vec{r}[/itex] 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 "[itex]d^2\vec{r}[/itex]". The second derivative? Second derivatives are NOT vectors, they are second order tensors.
     
  6. Feb 20, 2014 #5
    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.
     
  7. Feb 21, 2014 #6

    joshmccraney

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    Gold Member

    in spherical coordinates:

    [tex]\vec{r}=r\hat{e_r}[/tex]

    understand [itex]r=f(r)[/itex] and [itex]\hat{e_r}=f(\theta , \phi )[/itex]

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

    now by the product rule we have [tex]d\vec{r}=dr\hat{e_r}+r d\hat{e_r}[/tex]

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

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

    the [itex]\phi[/itex] 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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