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201903030417 Exercise 8.5.2

Eq. (8.41):

0=\displaystyle{\frac{\mathrm{d}^2\theta}{\mathrm{d}s^2}}-\sin\theta\cos\theta\bigg( \displaystyle{\frac{\mathrm{d}\phi}{\mathrm{d}s}} \bigg)^2

Eq. (8.42):

0=\displaystyle{\frac{\mathrm{d}}{\mathrm{d}s}}\bigg(\sin^2\theta \displaystyle{\frac{\mathrm{d}\phi}{\mathrm{d}s}}\bigg)

Integrating on Eq. (8.42) gives Eq. (8.43):

\sin^2\theta \displaystyle{\frac{\mathrm{d}\phi}{\mathrm{d}s}}=\textrm{Const.}\equiv \displaystyle{\frac{c}{R}}\Rightarrow \displaystyle{\frac{\mathrm{d}\phi}{\mathrm{d}s}=\frac{R}{\sin^2\theta}}.

The definition of pathlength along the curve is given by Eq. (8.16):

g_{\mu\nu}\displaystyle{\frac{\mathrm{d}x^\mu}{\mathrm{d}s}} \displaystyle{\frac{\mathrm{d}x^\nu}{\mathrm{d}s}} = + \bigg( \displaystyle{\frac{\mathrm{d}s}{\mathrm{d}s}} \bigg) = +1.

Previously to Exercise 8.5.2., the metric tensor is given by Eq. (8.40):

g_{\mu\nu}=\begin{bmatrix} R^2 & 0 \\ 0 & R^2\sin^2\theta \end{bmatrix}.

Eq. (8.45) begins:

\begin{aligned} 1 & = R^2\bigg( \displaystyle{\frac{\mathrm{d}\theta}{\mathrm{d}s }} \bigg)^2 +R^2\sin^2\theta \bigg( \displaystyle{\frac{\mathrm{d}\phi}{\mathrm{d}s}} \bigg)^2 \\ 1 & = R^2\bigg( \displaystyle{\frac{\mathrm{d}\theta}{\mathrm{d}s }} \bigg)^2 +R^2\sin^2\theta \bigg( \displaystyle{\frac{c}{R\sin^2\theta}} \bigg)^2 \\ 1 & = R^2\bigg( \displaystyle{\frac{\mathrm{d}\theta}{\mathrm{d}s}} \bigg)^2 + \displaystyle{\frac{c^2}{\sin^2\theta}} \\ \bigg( \displaystyle{\frac{\mathrm{d}\theta}{\mathrm{d}s}} \bigg)^2 & = \displaystyle{\frac{1}{R^2}\bigg( 1-\frac{c^2}{\sin^2\theta} \bigg)} \\ \displaystyle{\frac{\mathrm{d}\theta}{\mathrm{d}s}} & = \pm \displaystyle{\frac{1}{R}\sqrt{1-\frac{c^2}{\sin^2\theta}}} \\ & = \pm \displaystyle{\frac{1}{R\sin\theta}}\sqrt{\sin^2\theta -c^2} \end{aligned}

and this is Eq. (8.46).

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