Revisions to What are the points in spherical coordinates?

added 159 characters in body

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edited Oct 19, 2019 at 10:59

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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 $$ \vec{P}'=\vec{P} + d\theta\ \vec{e}_\theta+d\phi\ \vec{e}_\phi+dr\ \vec{e}_r.$$

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. $$\vec{P} =\begin{pmatrix}x \\ y \\ z\end{pmatrix} =\begin{pmatrix}r\sin\theta\cos\phi \\ r\sin\theta\sin\phi \\ r\cos\theta \end{pmatrix} \tag{1}$$ $$\vec{P}' =\begin{pmatrix}x' \\ y' \\ z' \end{pmatrix} =\begin{pmatrix}x+dx \\ y+dy \\ z+dz \end{pmatrix}$$

With this notation you get the 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 $$ \vec{P}'=\vec{P} + d\theta\ \vec{e}_\theta+d\phi\ \vec{e}_\phi+dr\ \vec{e}_r.$$

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 $$ \vec{P}'=\vec{P} + d\theta\ \vec{e}_\theta+d\phi\ \vec{e}_\phi+dr\ \vec{e}_r.$$

changed spherical coordinates to the usual convention (north pole at \theta=0)

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edited Oct 18, 2019 at 17:49

Thomas Fritsch

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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. $$\vec{P} =\begin{pmatrix}x \\ y \\ z\end{pmatrix} =\begin{pmatrix}r\cos\theta\cos\phi \\ r\cos\theta\sin\phi \\ r\sin\theta \end{pmatrix} \tag{1}$$$$\vec{P} =\begin{pmatrix}x \\ y \\ z\end{pmatrix} =\begin{pmatrix}r\sin\theta\cos\phi \\ r\sin\theta\sin\phi \\ r\cos\theta \end{pmatrix} \tag{1}$$ $$\vec{P}' =\begin{pmatrix}x' \\ y' \\ z' \end{pmatrix} =\begin{pmatrix}x+dx \\ y+dy \\ z+dz \end{pmatrix}$$

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

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

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

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. $$\vec{P} =\begin{pmatrix}x \\ y \\ z\end{pmatrix} =\begin{pmatrix}r\cos\theta\cos\phi \\ r\cos\theta\sin\phi \\ r\sin\theta \end{pmatrix} \tag{1}$$ $$\vec{P}' =\begin{pmatrix}x' \\ y' \\ z' \end{pmatrix} =\begin{pmatrix}x+dx \\ y+dy \\ z+dz \end{pmatrix}$$

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

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

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

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. $$\vec{P} =\begin{pmatrix}x \\ y \\ z\end{pmatrix} =\begin{pmatrix}r\sin\theta\cos\phi \\ r\sin\theta\sin\phi \\ r\cos\theta \end{pmatrix} \tag{1}$$ $$\vec{P}' =\begin{pmatrix}x' \\ y' \\ z' \end{pmatrix} =\begin{pmatrix}x+dx \\ y+dy \\ z+dz \end{pmatrix}$$

With this notation you get the 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 $$ \vec{P}'=\vec{P} + d\theta\ \vec{e}_\theta+d\phi\ \vec{e}_\phi+dr\ \vec{e}_r.$$

added 96 characters in body

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edited Oct 18, 2019 at 12:44

Thomas Fritsch

41k
13
75
144

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 don't behave like vector components components with the usual simple rules for vector addition and multiplication. And it would easily leadslead 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. $$\vec{P} =\begin{pmatrix}x \\ y \\ z\end{pmatrix} =\begin{pmatrix}r\cos\theta\cos\phi \\ r\cos\theta\sin\phi \\ r\sin\theta \end{pmatrix} \tag{1}$$ $$\vec{P}' =\begin{pmatrix}x' \\ y' \\ z' \end{pmatrix} =\begin{pmatrix}x+dx \\ y+dy \\ z+dz \end{pmatrix}$$

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

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

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

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