2
$\begingroup$

Let's use the spherical coordinates so that $\vec P=(r, \theta, \phi)$. In this context i've read that it's possible to write $$\vec P'=\vec P + d\theta\ \vec e_\theta+d\phi\ \vec e_\phi+dr\ \vec e_r$$ but i don't understand the meaning of this since it looks like $$(r', \theta', \phi')=(r, \theta, \phi)+d\theta\ \vec e_\theta+d\phi\ \vec e_\phi+dr\ \vec e_r$$ that doesn't sound to have any sense. Probably i'm misunderstanding something in the notation, can somebody clarify the question?

Note that $ \vec e_\theta,\vec e_\phi,\vec e_r$ aren't the versos but $\vec e_\theta=\frac {\partial \vec P} {\partial \theta}$ and so on

$\endgroup$
  • 1
    $\begingroup$ The notation is weird. Maybe $\vec{dP'}= \ldots $ would make more sense $\endgroup$ – ja72 Oct 19 at 17:28
6
$\begingroup$

It would be misleading to write $$\vec P=(r, \theta, \phi)$$ or $$\vec P=\begin{pmatrix}r \\ \theta \\ \phi \end{pmatrix}$$ because the curvilinear coordinates $r,\theta,\phi$ don't behave like vector components with the usual simple rules for vector addition and multiplication. And it would easily lead to nonsensical conclusions (as you already noticed).

It is better to imagine $\vec{P}$ with cartesian components $x,y,z$ because these truly behave like vector components. Using the definition of spherical coordinates you have: $$\vec{P} =\begin{pmatrix}x \\ y \\ z\end{pmatrix} =\begin{pmatrix}r\sin\theta\cos\phi \\ r\sin\theta\sin\phi \\ r\cos\theta \end{pmatrix} \tag{1}$$ $$\vec{P}' =\begin{pmatrix}x' \\ y' \\ z' \end{pmatrix} =\begin{pmatrix}x+dx \\ y+dy \\ z+dz \end{pmatrix}$$

With this notation you get the basis vectors $\vec e_\theta,\vec e_\phi,\vec e_r$ in a straight-forward way by differentiating (1).

$$\vec{e}_r = \frac{\partial\vec P}{\partial r} =\begin{pmatrix}\sin\theta\cos\phi \\ \sin\theta\sin\phi \\ \cos\theta \end{pmatrix}$$ $$\vec{e}_\theta = \frac{\partial\vec P}{\partial \theta} =\begin{pmatrix}r\cos\theta\cos\phi \\ r\cos\theta\sin\phi \\ -r\sin\theta \end{pmatrix}$$ $$\vec{e}_\phi = \frac{\partial\vec P}{\partial \phi} =\begin{pmatrix}-r\sin\theta\sin\phi \\ r\sin\theta\cos\phi \\ 0 \end{pmatrix}$$

It is meant with the vectors above that you have equation $$ \vec{P}'=\vec{P} + d\theta\ \vec{e}_\theta+d\phi\ \vec{e}_\phi+dr\ \vec{e}_r.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.