201902210253 Exercise 6.1.1-6.1.3

Consider polar coordinates on a flat plane. The transformation equations between the polar coordinates $r$ , $\theta$ (the primed coordinate system) and cartesian coordinates $x$ , $y$ (the unprimed coordinate system) are

Equation (6.15a):

$x=r\cos\theta$ , $y=r\sin\theta$

Equation (6.15b):

$r=\sqrt{x^2+y^2}$ , $\theta =\tan^{-1}\bigg( \displaystyle{\frac{y}{x}} \bigg)$

Consider also the scalar function $\varPsi =bxy=br^2\cos\theta\sin\theta$ .

Calculate the four transformation partials $\displaystyle{\frac{\partial x^\mu}{\partial x'^\nu}}$ for the transformations given above:

$\displaystyle{\frac{\partial x}{\partial r}}=\cos\theta$ , $\displaystyle{\frac{\partial x}{\partial \theta}}=-r\sin\theta$ , $\displaystyle{\frac{\partial y}{\partial r}}=\sin\theta$ , $\displaystyle{\frac{\partial y}{\partial \theta}}=r\cos\theta$ .

It can be shown that the gradient of $\varPsi$ in cartesian coordinate system is

$\partial_x\varPsi =by$ and $\partial_y\varPsi =bx$

and that of $\varPsi$ in polar coordinate system is

$\partial_r\varPsi =2br\cos\theta\sin\theta$ and $\partial_\theta\varPsi =br^2[\cos^2\theta -\sin^2\theta]$ .

To make practice of the covector transformation rule:

Equation (6.3):

$\partial_\mu\varPsi \equiv \displaystyle{\frac{\partial \varPsi}{\partial x^\mu}}$

Equation (6.4):

$\partial'_\nu \equiv \displaystyle{\frac{\partial \varPsi}{\partial x'^\nu}} = \displaystyle{\frac{\partial x^\mu}{\partial x'^\nu}}\displaystyle{\frac{\partial \varPsi}{\partial x^\mu}} = \displaystyle{\frac{\partial x^\mu}{\partial x'^\nu}}\partial_\mu \varPsi$

Thus,

$\begin{aligned} &\quad\enspace \displaystyle{\frac{\partial x}{\partial r}}\partial_x\varPsi + \displaystyle{\frac{\partial y}{\partial r}}\partial_y\varPsi \\ & = (\cos\theta )(by) + (\sin\theta )(bx) \\ & = (\cos\theta )(br\sin\theta ) + (\sin\theta )(br\cos\theta) \\ & = 2br\cos\theta\sin\theta \\ & = \partial_r \varPsi \end{aligned}$

and

$\begin{aligned} &\quad\enspace \displaystyle{\frac{\partial x}{\partial \theta}}\partial_x\varPsi + \displaystyle{\frac{\partial y}{\partial \theta}}\partial_y\varPsi\\ & = (-r\sin\theta )(by) + (r\cos\theta )(bx) \\ & = (-r\sin\theta )(br\sin\theta ) + (r\cos\theta )(br\cos\theta ) \\ & = br^2[\cos^2\theta -\sin^2\theta ] \\ & = \partial_\theta \varPsi \end{aligned}$

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