Scaling Factors of Curvilinear Coordinates
📂Mathematical Physics 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 , θ ) (r,\theta) ( r , θ ) θ \theta θ l = r θ l=r\theta l = r θ r r r ( q 1 , q 2 , q 3 ) (q_{1},q_{2},q_{3}) ( q 1 , q 2 , q 3 )
d r = h 1 d q 1 q 1 ^ + h 2 d q 2 q 2 ^ + h 3 d q 3 q 3 ^ d s 2 = ( h 1 d q 1 ) 2 + ( h 2 d q 2 ) 2 + ( h 3 d q 3 ) 2 d V = h 1 h 2 h 3 d q 1 d q 2 d q 3
\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*}
d r d s 2 d V = h 1 d q 1 q 1 ^ + h 2 d q 2 q 2 ^ + h 3 d q 3 q 3 ^ = ( h 1 d q 1 ) 2 + ( h 2 d q 2 ) 2 + ( h 3 d q 3 ) 2 = h 1 h 2 h 3 d q 1 d q 2 d q 3
At this time, h i = g i i = ∂ r ∂ q i ⋅ ∂ r ∂ q i h_{i}=\sqrt{g_{ii}}=\sqrt{\frac{ \partial \mathbf{r}}{ \partial q_{i} }\cdot \frac{ \partial \mathbf{r}}{ \partial q_{i} }} h i = g ii = ∂ q i ∂ r ⋅ ∂ q i ∂ r
Polar coordinate system:
h 1 = 1 , h 2 = r
h_{1}=1,\quad h_{2}=r
h 1 = 1 , h 2 = r
d r = d r r ^ + r d θ θ ^ d s 2 = d r 2 + r 2 d θ 2 d V = r d r d θ
\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*}
d r d s 2 d V = d r r ^ + r d θ θ ^ = d r 2 + r 2 d θ 2 = r d r d θ
Cylindrical coordinate system:
h 1 = 1 , h 2 = ρ , h 3 = 1
h_{1}=1, \quad h_{2}=\rho,\quad h_{3}=1
h 1 = 1 , h 2 = ρ , h 3 = 1
d r = d ρ ρ ^ + ρ d ϕ ϕ ^ + d z z ^ d s 2 = d ρ 2 + ρ 2 d ϕ 2 + d z 2 d V = ρ d ρ d ϕ d z
\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*}
d r d s 2 d V = d ρ ρ ^ + ρ d ϕ ϕ ^ + d z z ^ = d ρ 2 + ρ 2 d ϕ 2 + d z 2 = ρ d ρ d ϕ d z
Spherical coordinate system:
h 1 = 1 , h 2 = r , h 3 = r sin θ
h_{1}=1,\quad h_{2}=r,\quad h_{3}=r\sin\theta
h 1 = 1 , h 2 = r , h 3 = r sin θ
d r = d r r ^ + r d θ θ ^ + r sin θ d ϕ ϕ ^ d s 2 = d r 2 + r 2 d θ 2 + r 2 sin 2 θ d ϕ 2 d V = r 2 sin θ d r d θ d ϕ
\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*}
d r d s 2 d V = d r r ^ + r d θ θ ^ + r sin θ d ϕ ϕ ^ = d r 2 + r 2 d θ 2 + r 2 sin 2 θ d ϕ 2 = r 2 sin θ d r d θ d ϕ
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_{1}=r q 1 = r q 2 = θ q_{2}=\theta q 2 = θ
r = x x ^ + y y ^ = r cos θ x ^ + r sin θ y ^
\mathbf{r}=x\hat{\mathbf{x}}+y\hat{\mathbf{y}}=r\cos \theta \hat{\mathbf{x}} + r \sin \theta \hat{\mathbf{y}}
r = x x ^ + y y ^ = r cos θ x ^ + r sin θ y ^
Therefore, it is as follows.
h 1 = ∂ r ∂ q 1 ⋅ ∂ r ∂ q 1 = ∂ r ∂ r ⋅ ∂ r ∂ r = ∂ ( r cos θ x ^ + r sin θ y ^ ) ∂ r ⋅ ∂ ( r cos θ x ^ + r sin θ y ^ ) ∂ r = ( cos θ x ^ + sin θ y ^ ) ⋅ ( cos θ x ^ + sin θ y ^ ) = cos 2 θ + sin 2 θ = 1 h 2 = ∂ r ∂ q 1 ⋅ ∂ r ∂ q 1 = ∂ r ∂ θ ⋅ ∂ r ∂ θ = ∂ ( r cos θ x ^ + r sin θ y ^ ) ∂ θ ⋅ ∂ ( r cos θ x ^ + r sin θ y ^ ) ∂ θ = ( − r sin θ x ^ + r cos θ y ^ ) ⋅ ( − r sin θ x ^ + r cos θ y ^ ) = r 2 sin 2 θ + r 2 cos 2 θ = r 2 = r
\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*}
h 1 h 2 = ∂ q 1 ∂ r ⋅ ∂ q 1 ∂ r = ∂ r ∂ r ⋅ ∂ r ∂ r = ∂ r ∂ ( r cos θ x ^ + r sin θ y ^ ) ⋅ ∂ r ∂ ( r cos θ x ^ + r sin θ y ^ ) = ( cos θ x ^ + sin θ y ^ ) ⋅ ( cos θ x ^ + sin θ y ^ ) = cos 2 θ + sin 2 θ = 1 = ∂ q 1 ∂ r ⋅ ∂ q 1 ∂ r = ∂ θ ∂ r ⋅ ∂ θ ∂ r = ∂ θ ∂ ( r cos θ x ^ + r sin θ y ^ ) ⋅ ∂ θ ∂ ( r cos θ x ^ + r sin θ y ^ ) = ( − r sin θ x ^ + r cos θ y ^ ) ⋅ ( − r sin θ x ^ + r cos θ y ^ ) = r 2 sin 2 θ + r 2 cos 2 θ = r 2 = r
Hence, the following formula is obtained.
d r = h 1 d q 1 q 1 ^ + h 2 d q 2 q 2 ^ = d r r ^ + r d θ θ ^ d s 2 = ( h 1 d q 1 ) 2 + ( h 2 d q 2 ) 2 = d r 2 + r 2 d θ 2 d V = h 1 h 2 d q 1 d q 2 = r d r d θ
\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*}
d r d s 2 d V = h 1 d q 1 q 1 ^ + h 2 d q 2 q 2 ^ = d r r ^ + r d θ θ ^ = ( h 1 d q 1 ) 2 + ( h 2 d q 2 ) 2 = d r 2 + r 2 d θ 2 = h 1 h 2 d q 1 d q 2 = r d r d θ
■
Cylindrical coordinate system q 1 = ρ q_{1}=\rho q 1 = ρ q 2 = ϕ q_{2}=\phi q 2 = ϕ q 3 = z q_{3}=z q 3 = z
r = ρ cos ϕ x ^ + ρ sin ϕ y ^ + z z ^
\mathbf{r}=\rho\cos \phi \hat{\mathbf{x}} + \rho \sin \phi \hat{\mathbf{y}} +z\hat{\mathbf{z}}
r = ρ cos ϕ x ^ + ρ sin ϕ y ^ + z z ^
Therefore, h 1 h_{1} h 1 h 2 h_{2} h 2
h 1 = 1 , h 2 = r
h_{1}=1,\quad h_{2}=r
h 1 = 1 , h 2 = r
Calculating h 3 h_{3} h 3
h 3 = ∂ r ∂ z ⋅ ∂ r ∂ z = ( z ^ ) ⋅ ( z ^ ) = 1
\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*}
h 3 = ∂ z ∂ r ⋅ ∂ z ∂ r = ( z ^ ) ⋅ ( z ^ ) = 1
Therefore,
d r = d ρ ρ ^ + ρ d ϕ ϕ ^ + d z z ^ d s 2 = d ρ 2 + ρ 2 d ϕ 2 + d z 2 d V = ρ d ρ d ϕ d z
\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*}
d r d s 2 d V = d ρ ρ ^ + ρ d ϕ ϕ ^ + d z z ^ = d ρ 2 + ρ 2 d ϕ 2 + d z 2 = ρ d ρ d ϕ d z
■
Spherical coordinate system
q 1 = r q_{1}=r q 1 = r q 2 = θ q_{2}=\theta q 2 = θ q 3 = ϕ q_{3}=\phi q 3 = ϕ
r = r sin θ cos ϕ x ^ + r sin θ sin ϕ y ^ + r cos θ z ^
\mathbf{r}=r\sin\theta\cos\phi \hat{\mathbf{x}} + r \sin \theta\sin \phi \hat{\mathbf{y}} +r\cos\theta\hat{\mathbf{z}}
r = r sin θ cos ϕ x ^ + r sin θ sin ϕ y ^ + r cos θ z ^
Therefore, it is as follows.
h 1 = ∂ r ∂ r ⋅ ∂ r ∂ r = sin 2 θ cos 2 ϕ + sin 2 θ sin 2 ϕ + cos 2 θ = sin 2 θ + cos 2 θ = 1 h 2 = ∂ r ∂ θ ⋅ ∂ r ∂ θ = r 2 cos 2 θ cos 2 ϕ + r 2 cos 2 θ sin 2 ϕ + r 2 sin 2 θ = r 2 cos 2 θ + r 2 sin 2 θ = r h 3 = ∂ r ∂ ϕ ⋅ ∂ r ∂ ϕ = r 2 sin 2 θ sin 2 ϕ + r 2 sin θ 2 cos 2 ϕ = r 2 sin 2 θ = r sin θ
\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*}
h 1 h 2 h 3 = ∂ r ∂ r ⋅ ∂ r ∂ r = sin 2 θ cos 2 ϕ + sin 2 θ sin 2 ϕ + cos 2 θ = sin 2 θ + cos 2 θ = 1 = ∂ θ ∂ r ⋅ ∂ θ ∂ r = r 2 cos 2 θ cos 2 ϕ + r 2 cos 2 θ sin 2 ϕ + r 2 sin 2 θ = r 2 cos 2 θ + r 2 sin 2 θ = r = ∂ ϕ ∂ r ⋅ ∂ ϕ ∂ r = r 2 sin 2 θ sin 2 ϕ + r 2 sin θ 2 cos 2 ϕ = r 2 sin 2 θ = r sin θ
Hence, it is as follows.
d r = d r r ^ + r d θ θ ^ + r sin θ d ϕ ϕ ^ d s 2 = d r 2 + r 2 d θ 2 + r 2 sin 2 θ d ϕ 2 d V = r 2 sin θ d r d θ d ϕ
\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*}
d r d s 2 d V = d r r ^ + r d θ θ ^ + r sin θ d ϕ ϕ ^ = d r 2 + r 2 d θ 2 + r 2 sin 2 θ d ϕ 2 = r 2 sin θ d r d θ d ϕ
■
