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201902210301 Exercise 6.2.1-6.2.3

Again consider polar coordinates on a flat plane. The transformation equations between polar coordinates r, \theta (the primed coordinate system) and Cartesian coordinates x, y (the unprimed coordinate system) are given by equation 6.15. Consider also the vector \mathbf{v} whose components are v^x=1 and v^y=0.

Lowering an index as given by Equation (6.5):

A_\mu \equiv g_{\mu\nu}A^\nu.

In Cartesian coordinate system the metric tensor is

\mathbf{g}=\begin{bmatrix} g_{xx} & g_{xy} \\ g_{yx} & g_{yy} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

Thus

v_x\equiv g_{x\nu}v^\nu = g_{xx}v^x+g_{xy}v^y = (1)(1) + (0)(0)=1

v_y\equiv g_{y\nu}v^\nu = g_{yx}v^x+g_{yy}v^y = (0)(1) + (1)(0)=0

The definition of covector is given by Equation (6.2):

B'_\nu = \displaystyle{\frac{\partial x^\mu}{\partial x'^\nu}}B_{\mu}

So,

v_r=\displaystyle{\frac{\partial x}{\partial r}}v_x + \displaystyle{\frac{\partial y}{\partial r}}v_y = (\cos\theta )(1) + (\sin\theta )(0) = \cos\theta

v_\theta =\displaystyle{\frac{\partial x}{\partial \theta}}v_x + \displaystyle{\frac{\partial y}{\partial \theta}}v_y = (-r\sin\theta )(1) + (r\cos\theta )(0) = -r\sin\theta

The metric tensor for the polar coordinate basis is given by Equation (5.19):

g_{\mu\nu}\equiv \mathbf{e}_{\mu}\cdot\mathbf{e}_\nu = \begin{bmatrix} \mathbf{e}_{r}\cdot\mathbf{e}_r & \mathbf{e}_{r}\cdot\mathbf{e}_\theta \\ \mathbf{e}_{\theta}\cdot\mathbf{e}_r & \mathbf{e}_{\theta}\cdot\mathbf{e}_\theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & r^2 \\ \end{bmatrix}

Exercise 6.2.3.

One can show that in the polar coordinate system, v^r=\cos\theta and v^\theta =-\sin\theta /r (see Problem P6.1). Show that v'^\mu v'_{\mu}=1. Does this make sense?

Equation (6.5):

A_{\mu} \equiv g_{\mu\nu}A^\nu

Hence

\begin{aligned} v_r & \equiv g_{r\nu}v^\nu \\ \cos\theta & = g_{rr}v^r + g_{r\theta} v^\theta \\ \cos\theta & = (1)(v^r) + (0)(v^\theta ) \\ v^r & = \cos\theta \end{aligned}

\begin{aligned} v_\theta & \equiv g_{\theta\nu}v^\nu \\ -r\sin\theta & = g_{\theta r}v^r + g_{\theta\theta} v^\theta \\ -r\sin\theta & = (0)(\cos\theta ) + (r^2)(v^\theta) \\ v^\theta & = \displaystyle{\frac{-\sin\theta }{r}} \\ \end{aligned}

are checked.

\begin{aligned} v'^\mu v'_{\mu} & = v^r v_r + v^\theta v_{\theta} \\ & = (\cos\theta )(\cos\theta ) + (\displaystyle{\frac{-\sin\theta}{r}})(-r\sin\theta ) \\ & = \cos^2\theta + \sin^2\theta \\ & = 1 \end{aligned}.

Remark. Invariant norm.

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