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Mathematical Methods of Physics: For Undergraduate

Sunil Karna

Section 1.7 Examples C

Example 1.7.1.

Prove that a cylindrical coordinate system is orthogonal.
Solution.
The position vector of any point P in a cylindrical coordinate is
\begin{equation*} \vec{r}= x\hat{i}+y\hat{j}+z\hat{k}= \rho\cos\phi\hat{i}+\rho\sin\phi\hat{j}+z\hat{k} \end{equation*}
Now,
\begin{equation*} \frac{\partial \vec{r}}{\partial \rho}, \frac{\partial \vec{r}}{\partial \phi}, \frac{\partial \vec{r}}{\partial z} \end{equation*}
are the tangent vectors along the curves \(\rho, \phi,z,\text{,}\) respectively.
\begin{equation*} \therefore \quad \frac{\partial \vec{r}}{\partial \rho} = \cos\phi\hat{i}+\sin\phi\hat{j}, \end{equation*}
\begin{equation*} \frac{\partial \vec{r}}{\partial \phi} = -\rho\sin\phi\hat{i}+\rho\cos\phi\hat{j}, \end{equation*}
\begin{equation*} \text{and} \quad \frac{\partial \vec{r}}{\partial z} = \hat{k}. \end{equation*}
The unit vectors along the tangents are
\begin{equation*} \hat{e}_{1}=\hat{e}_{\rho}=\frac{\frac{\partial \vec{r}}{\partial \rho}}{\mid\frac{\partial \vec{r}}{\partial \rho}\mid} \end{equation*}
\begin{equation} = \frac{\cos\phi\hat{i}+\sin\phi\hat{j}}{\sqrt{\cos^{2}\phi + \sin^{2}\phi}} = \cos\phi\hat{i}+\sin\phi\hat{j},\tag{1.7.1} \end{equation}
\begin{equation*} \hat{e}_{2}=\hat{e}_{\phi}=\frac{\frac{\partial \vec{r}}{\partial \phi}}{\mid\frac{\partial \vec{r}}{\partial \phi}\mid} \end{equation*}
\begin{equation} = \frac{-\rho\sin\phi\hat{i}+\rho\cos\phi\hat{j}}{\sqrt{\rho^{2}\sin^{2}\phi + \rho^{2}\cos^{2}\phi}} = -\sin\phi\hat{i}+\cos\phi\hat{j},\tag{1.7.2} \end{equation}
\begin{equation} \hat{e}_{3}=\hat{e}_{z}=\frac{\frac{\partial \vec{r}}{\partial z}}{\mid\frac{\partial \vec{r}}{\partial z}\mid} = \hat{k}\tag{1.7.3} \end{equation}
Then,
\begin{equation*} \hat{e}_{1}\cdot\hat{e}_{2} = (\cos\phi\hat{i}+\sin\phi\hat{j})\cdot(-\sin\phi\hat{i}+\cos\phi\hat{j}) = 0, \end{equation*}
\begin{equation*} \hat{e}_{1}\cdot\hat{e}_{3} = (\cos\phi\hat{i}+\sin\phi\hat{j})\cdot (\hat{k})=0, \end{equation*}
and
\begin{equation*} \hat{e}_{2}\cdot\hat{e}_{3} = (-\sin\phi\hat{i}+\cos\phi\hat{j})\cdot (\hat{k}) = 0. \end{equation*}
Hence \(\hat{e_{1}}, \hat{e_{2}}\text{,}\) and \(\hat{e_{3}}\) are mutually perpendicular and the coordinate system is orthogonal.

Example 1.7.2.

Express the velocity \(v\) and acceleration \(a\) of a particle in cylindrical coordinates.
Solution.
The position vector of any point in a cylindrical coordinate is defined as
\begin{equation*} \vec{r}= x\hat{i}+y\hat{j}+z\hat{k}= \rho\cos\phi\hat{i}+\rho\sin\phi\hat{j}+z\hat{k} \end{equation*}
\begin{equation*} =\rho\cos\phi(\cos\phi\hat{e}_{\rho}-\sin\phi\hat{e}_{\phi})+\rho\sin\phi(\sin\phi\hat{e}_{\rho}+\cos\phi\hat{e}_{\phi})+z\hat{e}_{z} \end{equation*}
\begin{equation*} =\rho\hat{e_{\rho}} +z \hat{e_{z}} \end{equation*}
from solving the entities (1.7.1)–(1.7.3)
\begin{equation*} \left[\hat{e}_{\rho}= \cos\phi\hat{i}+\sin\phi\hat{j}, \hat{e}_{\phi} = -\sin\phi\hat{i}+\cos\phi\hat{j}, \hat{e}_{z} = \hat{k}\right] \end{equation*}
We get -
\begin{equation} \hat{i}= \cos\phi\hat{e}_{\rho}-\sin\phi\hat{e}_{\phi},\tag{1.7.4} \end{equation}
\begin{equation} \hat{j}= \sin\phi\hat{e_{\rho}}+\cos\phi\hat{e_{\phi}}, \tag{1.7.5} \end{equation}
\begin{equation} \hat{k}=\hat{e_{z}}\tag{1.7.6} \end{equation}
and,
\begin{equation*} \vec{v}= \frac{d \vec{r}}{d t} = \frac{d\rho}{d t}\hat{e}_{\rho} +\rho\frac{d\hat{e}_{\rho}}{d t}+ \frac{d z}{d t}\hat{e}_{z} +z\frac{d\hat{e}_{z}}{d t} \end{equation*}
\begin{equation*} = \dot{\rho} \hat{e_{\rho}}+\rho \hat{e_{\phi}}\dot{\phi} + \dot{z}\hat{e_{z}} +0 \end{equation*}
Again differentiating \(v\) w.r.to ’t’, we get -
\begin{equation*} \vec{a}= \frac{d^{2} \vec{r}}{d t^{2}} = \frac{d}{dt}(\dot{\rho} \hat{e}_{\rho}+\rho \hat{e}_{\phi}\dot{\phi} + \dot{z}\hat{e}_{z}) \end{equation*}
\begin{equation*} =\dot{\rho} \frac{d \hat{e}_{\rho}}{dt}+\ddot{\rho}\hat{e}_{\rho} + \rho\dot{\phi}\frac{d \hat{e}_{\phi}}{dt} + \rho\ddot{\phi} \hat{e}_{\phi} + \dot{\rho} \dot{\phi}\hat{e}_{\phi} + \ddot{z}\hat{e}_{z} \end{equation*}
\begin{equation*} =\dot{\rho}\dot{\phi}\hat{e}_{\phi} + \ddot{\rho}\hat{e}_{\rho}+\rho\dot{\phi}(-\dot{\phi}\hat{e}_{\rho})+\rho\ddot{\phi} \hat{e}_{\phi}+\dot{\rho} \dot{\phi}\hat{e}_{\phi} + \ddot{z}\hat{e}_{z} \end{equation*}
\begin{equation*} =(\ddot{\rho}-\rho\dot{\phi}^{2})\hat{e}_{\rho} +(\rho\ddot{\phi}+2\dot{\rho}\dot{\phi})\hat{e}_{\phi} + \ddot{z}\hat{e}_{z} \end{equation*}
 1 
\begin{equation*} \frac{\,d\hat{e}_{\rho}}{\,dt}= \frac{\,d\hat{e}_{\rho}}{\,d\phi}\cdot\frac{\,d\phi}{\,dt} = \dot{\phi}\hat{e}_{\phi} \end{equation*}
after differentiating \(\hat{e}_{\rho}\) w.r.to \('\phi'\text{;}\) also
\begin{equation*} \frac{\,d\hat{e}_{\phi}}{\,dt}= \frac{\,d\hat{e}_{\phi}}{\,d\phi}\cdot\frac{\,d\phi}{\,dt} = -\dot{\phi}\hat{e}_{\rho} \end{equation*}
after differentiating \(\hat{e_{\phi}}\) w.r.to \('\phi'\text{.}\)

Example 1.7.3.

Find an expression for \(\,ds^{2}\) in orthogonal curvilinear system.
Solution.
We know that the position vector
\begin{equation*} \vec{r} = \vec{r}(u,v,w) \end{equation*}
and
\begin{equation*} \,ds^{2} = \vec{\,dr}\cdot\vec{\,dr} =\left(\frac{\partial \vec{r}}{\partial u} \,du + \frac{\partial \vec{r}}{\partial v} \,dv +\frac{\partial \vec{r}}{\partial w} \,dw\right)^{2} \end{equation*}
\begin{equation*} =\left(\frac{\partial \vec{r}}{\partial u}\right)^{2} \,du^{2} +\left(\frac{\partial \vec{r}}{\partial v}\right)^{2} \,dv^{2} + \left(\frac{\partial \vec{r}}{\partial w}\right)^{2} \,dw^{2} \end{equation*}
\begin{equation*} + 2 \left(\frac{\partial \vec{r}}{\partial u}\right)\cdot \left(\frac{\partial \vec{r}}{\partial v}\right)dudv \end{equation*}
\begin{equation*} + 2 \left(\frac{\partial \vec{r}}{\partial v}\right)\cdot \left(\frac{\partial \vec{r}}{\partial w}\right)\,dv\,dw + 2 \left(\frac{\partial \vec{r}}{\partial w}\right)\cdot \left(\frac{\partial \vec{r}}{\partial u}\right)\,dw\,du \end{equation*}
Since, \(\frac{\partial \vec{r}}{\partial u}\text{,}\) \(\frac{\partial \vec{r}}{\partial v}\text{,}\) and \(\frac{\partial \vec{r}}{\partial w} \) are the tangent vectors along the curves \(u,v,w\text{,}\) we have -
\begin{equation*} \left(\frac{\partial \vec{r}}{\partial u}\right)\cdot \left(\frac{\partial \vec{r}}{\partial v}\right)= \left(\frac{\partial \vec{r}}{\partial v}\right)\cdot \left(\frac{\partial \vec{r}}{\partial w}\right)= \left(\frac{\partial \vec{r}}{\partial w}\right)\cdot \left(\frac{\partial \vec{r}}{\partial u}\right)=0 \end{equation*}
which are the conditions of orthogonality.
\begin{equation*} \therefore \quad \,ds^{2}=\left(\frac{\partial \vec{r}}{\partial u}\right)^{2} \,du^{2}+ \left(\frac{\partial \vec{r}}{\partial v}\right)^{2} \,dv^{2}+\left(\frac{\partial \vec{r}}{\partial w}\right)^{2} \,dw^{2}. \end{equation*}

Example 1.7.4.

A force is described by \(\vec{F}=\frac{y}{x^{2}+y^{2}}\hat{i} +\frac{x}{x^{2}+y^{2}}\hat{j}\text{.}\) Find \(\vec{F}\) in a cylindrical coordinates.
Solution.
Use eqns. (1.7.4)–(1.7.6)
\begin{equation*} \hat{i}= \cos\phi\hat{e_{\rho}}-\sin\phi\hat{e_{\phi}}, \quad \hat{j}= \sin\phi\hat{e_{\rho}}+\cos\phi\hat{e_{\phi}}, \quad \hat{k}=\hat{e_{z}} \end{equation*}
to find
\begin{equation*} \vec{F} =\frac{\hat{e_{\phi}}}{\rho}. \end{equation*}

Example 1.7.5.

Show that \(\vec{A}= -\phi\frac{\sin\theta}{r} \hat{e}_{\theta}\) is a solution of \(\nabla \times \vec{A}=-\frac{1}{r^{2}}\hat{e}_{r}\) in spherical coordinates.
Solution.
we have -
\begin{equation*} \nabla\times\vec{A}=\frac{1}{r^{2}\sin\theta}\begin{vmatrix} \hat{e}_{r} & r\hat{e}_{\theta} & r\sin\theta\hat{e}_{\phi}\\ \frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \phi}\\ 0 & -\frac{r\phi\sin\theta}{r} & 0 \end{vmatrix} \end{equation*}
\begin{equation*} = \frac{1}{r^{2}\sin\theta}[\hat{e}_{r}(-\sin\theta)-0+0] = - \frac{1}{r^{2}}\hat{e}_{r} \end{equation*}

Example 1.7.6.

Find the velocity and acceleration of a moving particle in a spherical coordinates.
Solution.
We have -
\begin{equation*} \vec{r}=r\hat{e}^{r}+\theta\hat{e}_{\theta}+\phi\hat{e}_{\phi} \end{equation*}
Where,
\begin{equation*} \hat{e}_{r} = \frac{\frac{\partial \vec{r}}{\partial r}}{\mid\frac{\partial \vec{r}}{\partial r}\mid} = \sin\theta\cos\phi\hat{i}+\sin\theta\sin\phi\hat{j}+\cos\theta\hat{k}; \end{equation*}
\begin{equation*} \hat{e}_{\theta} = \cos\theta\cos\phi\hat{i}+\cos\theta\sin\phi\hat{j}-\sin\theta\hat{k}; \end{equation*}
\begin{equation*} \text{and}\quad \hat{e}_{\phi}=-\sin\phi\hat{i}+\cos\phi\hat{j}. \end{equation*}
Now,
\begin{equation*} \frac{\partial \vec{r}}{\partial t}=\dot{r}\hat{e}_{r}+r\frac{\partial\hat{e}_{r}}{\partial t} +\dot{\theta}\hat{e}_{\theta}+\theta\frac{\partial\hat{e}_{\theta}}{\partial t} +\dot{\phi}\hat{e}_{\phi}+\phi\frac{\partial\hat{e}_{\phi}}{\partial t} \end{equation*}
where,
\begin{equation*} \frac{\partial\hat{e}_{r}}{\partial t} = \dot{\theta}\phi\frac{\partial\hat{e}_{r}}{\partial\theta} + \theta\dot{\phi}\frac{\partial\hat{e}_{r}}{\partial\phi}; \end{equation*}
\begin{equation*} \frac{\partial\hat{e}_{r}}{\partial\theta} = \dot{\theta}\phi\hat{e}_{\theta}; \end{equation*}
\begin{equation*} \text{and} \quad \frac{\partial\hat{e}_{r}}{\partial\phi} = \theta\sin\theta\dot{\phi}\hat{e}_{\phi} \end{equation*}
Similarly, other components can be found to obtain the velocity and acceleration.
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