Scaling Factors of Curvilinear Coordinates
Buildup
In the curvilinear coordinate system, the scale factor is an element that multiplies each component so that they have dimensions of length. For instance, the polar coordinate system is represented by $(r,\theta)$, where the distance the coordinates move as $\theta$ changes is the length of the arc, which is $l=r\theta$. Here, things like $r$ are called scale factors. Let’s say the variable of an arbitrary coordinate system is $(q_{1},q_{2},q_{3})$. Then, using the scale factor, the infinitesimal length, area, and volume in any coordinate system can be expressed as follows.
$$ \begin{align*} d\mathbf{r} &=h_{1}dq_{1}\hat{\mathbf{q}_{1}}+h_{2}dq_{2}\hat{\mathbf{q}_{2}}+h_{3}dq_{3}\hat{\mathbf{q}_{3}} \\ ds^{2} &=(h_{1}dq_{1})^{2}+(h_{2}dq_{2})^{2}+(h_{3}dq_{3})^{2} \\ dV &= h_{1}h_{2}h_{3}dq_{1}dq_{2}dq_{3} \end{align*} $$
At this time, $h_{i}=\sqrt{g_{ii}}=\sqrt{\frac{ \partial \mathbf{r}}{ \partial q_{i} }\cdot \frac{ \partial \mathbf{r}}{ \partial q_{i} }}$ is. The scale factor, infinitesimal length, area, and volume in each coordinate system are as follows.
Formulas
Polar coordinate system:
$$ h_{1}=1,\quad h_{2}=r $$
$$ \begin{align*} d\mathbf{r} &=dr\hat{\mathbf{r}}+rd\theta \hat{\boldsymbol{\theta}} \\ \\ ds^{2}&=dr^{2}+r^{2}d\theta^{2} \\ \\ dV&= rdrd\theta \end{align*} $$
Cylindrical coordinate system:
$$ h_{1}=1, \quad h_{2}=\rho,\quad h_{3}=1 $$
$$ \begin{align*} d\mathbf{r}&=d\rho \hat{\mathbf{\rho}}+\rho d\phi \hat{\boldsymbol{\phi}}+dz\hat{\mathbf {z}} \\ \\ ds^{2}&=d\rho ^{2}+\rho ^{2}d\phi^{2}+dz^{2} \\ \\ dV&= \rho d\rho d\phi dz \end{align*} $$
Spherical coordinate system:
$$ h_{1}=1,\quad h_{2}=r,\quad h_{3}=r\sin\theta $$
$$ \begin{align*} d\mathbf{r}&=dr \hat{\mathbf{r}}+r d\theta \hat{\boldsymbol{\theta}}+r\sin\theta d\phi\hat {\mathbf{\boldsymbol{\phi}}} \\ \\ ds^{2}&=dr ^{2}+r ^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \\ \\ dV&= r^{2}\sin\theta dr d \theta d \phi \end{align*} $$
Proof
The proof for the polar coordinate system is written in as much detail as possible, and the proofs for the others are written briefly.
Polar coordinate system
$q_{1}=r$, $q_{2}=\theta$, and the following holds true.
$$ \mathbf{r}=x\hat{\mathbf{x}}+y\hat{\mathbf{y}}=r\cos \theta \hat{\mathbf{x}} + r \sin \theta \hat{\mathbf{y}} $$
Therefore, it is as follows.
$$ \begin{align*} h_{1} &= \sqrt{\frac{ \partial \mathbf{r}}{ \partial q_{1}} \cdot \frac{ \partial \mathbf{r}}{ \partial q_{1} }} =\sqrt{\frac{ \partial \mathbf{r}}{ \partial r} \cdot \frac{ \partial \mathbf{r}}{ \partial r }} \\ &= \sqrt{\frac{ \partial (r\cos \theta \hat{\mathbf{x}} + r \sin \theta \hat{\mathbf{y}})}{ \partial r} \cdot \frac{ \partial (r\cos \theta \hat{\mathbf{x}} + r \sin \theta \hat{\mathbf{y}})}{ \partial r }} \\ &= \sqrt{(\cos \theta \hat{\mathbf{x}} + \sin \theta \hat{\mathbf{y}})\cdot(\cos \theta \hat{\mathbf{x}} + \sin \theta \hat{\mathbf{y}})} \\ &= \sqrt{\cos ^{2}\theta + \sin^{2}\theta} \\ &=1 \\ \\ h_{2}&=\sqrt{\frac{ \partial \mathbf{r}}{ \partial q_{1}} \cdot \frac{ \partial \mathbf{r}}{ \partial q_{1} }} =\sqrt{\frac{ \partial \mathbf{r}}{ \partial \theta} \cdot \frac{ \partial \mathbf{r}}{ \partial \theta }} \\ &= \sqrt{\frac{ \partial (r\cos \theta \hat{\mathbf{x}} + r \sin \theta \hat{\mathbf{y}})}{ \partial \theta} \cdot \frac{ \partial (r\cos \theta \hat{\mathbf{x}} + r \sin \theta \hat{\mathbf{y}})}{ \partial \theta }} \\ &= \sqrt{(-r\sin\theta \hat{\mathbf{x}} + r\cos \theta \hat{\mathbf{y}})\cdot(-r\sin\theta \hat{\mathbf{x}} + r\cos \theta \hat{\mathbf{y}})} \\ &= \sqrt{r^{2}\sin ^{2}\theta + r^{2}\cos^{2}\theta} \\ &=\sqrt{r^{2}} \\ &=r \end{align*} $$
Hence, the following formula is obtained.
$$ \begin{align*} d\mathbf{r}&=h_{1}dq_{1}\hat{\mathbf{q}_{1}} + h_{2}dq_{2}\hat{\mathbf{q}_{2}} \\ &=dr\hat{\mathbf{r}}+rd\theta \hat{\boldsymbol{\theta}} \\ \\ ds^{2}&=(h_{1}dq_{1})^{2}+(h_{2}dq_{2})^{2} \\ &=dr^{2}+r^{2}d\theta^{2} \\ \\ dV&=h_{1}h_{2}dq_{1}dq_{2} \\ &= rdrd\theta \end{align*} $$
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Cylindrical coordinate system
$q_{1}=\rho$, $q_{2}=\phi$, $q_{3}=z$, and the following holds true.
$$ \mathbf{r}=\rho\cos \phi \hat{\mathbf{x}} + \rho \sin \phi \hat{\mathbf{y}} +z\hat{\mathbf{z}} $$
Therefore, $h_{1}$, $h_{2}$ are derived similarly to the polar coordinate system.
$$ h_{1}=1,\quad h_{2}=r $$
Calculating $h_{3}$ results in the following.
$$ \begin{align*} h_{3} &=\sqrt{\frac{ \partial \mathbf{r}}{ \partial z}\cdot \frac{ \partial \mathbf{r}}{ \partial z }} \\ &= \sqrt{(\hat{\mathbf{z}})\cdot(\hat{\mathbf{z}})} \\ &=1 \end{align*} $$
Therefore,
$$ \begin{align*} d\mathbf{r}&=d\rho \hat{\mathbf{\rho}}+\rho d\phi \hat{\boldsymbol{\phi}}+dz\hat{\mathbf{z}} \\ \\ ds^{2}&=d\rho ^{2}+\rho ^{2}d\phi^{2}+dz^{2} \\ \\ dV&= \rho d\rho d\phi dz \end{align*} $$
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Spherical coordinate system
$q_{1}=r$, $q_{2}=\theta$, $q_{3}=\phi$, and the following holds true.
$$ \mathbf{r}=r\sin\theta\cos\phi \hat{\mathbf{x}} + r \sin \theta\sin \phi \hat{\mathbf{y}} +r\cos\theta\hat{\mathbf{z}} $$
Therefore, it is as follows.
$$ \begin{align*} h_{1} &=\sqrt{\frac{ \partial \mathbf{r}}{ \partial r}\cdot \frac{ \partial \mathbf{r}}{ \partial r }} \\ &= \sqrt{\sin^2{\theta}\cos^{2}\phi +\sin^{2}\theta\sin^{2}\phi+\cos^{2}\theta} \\ &=\sqrt{\sin^{2}\theta+\cos^{2}\theta} \\ &=1 \\ \\ h_{2} &=\sqrt{\frac{ \partial \mathbf{r}}{ \partial \theta}\cdot \frac{ \partial \mathbf{r}}{ \partial \theta }} \\ &= \sqrt{r^{2}\cos^{2}\theta \cos^{2}\phi + r^{2}\cos^{2}\theta\sin ^{2}\phi+r^{2}\sin^{2}\theta} \\ &=\sqrt{r^{2}\cos^{2}\theta+r^{2}\sin^{2}\theta} \\ &=r \\ \\ h_{3}&=\sqrt{\frac{ \partial \mathbf{r}}{ \partial \phi}\cdot \frac{ \partial \mathbf{r}}{ \partial \phi }} \\ &=\sqrt{r^{2}\sin^{2}\theta \sin ^{2}\phi+r^{2}\sin\theta^{2}\cos^{2}\phi} \\ &= \sqrt{r^{2}\sin ^{2}\theta} \\ &=r\sin\theta \end{align*} $$
Hence, it is as follows.
$$ \begin{align*} d\mathbf{r}&=dr \hat{\mathbf{r}}+r d\theta \hat{\boldsymbol{\theta}}+r\sin\theta d\phi\hat{\mathbf{\boldsymbol{\phi}}} \\ \\ ds^{2}&=dr ^{2}+r ^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \\ \\ dV&= r^{2}\sin\theta dr d \theta d \phi \end{align*} $$
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