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  3. 23. consider a 2d spacetime whose metric is: ds^2

23. Consider a 2D spacetime whose metric is: ds^2 = −dt^2 + [f(q)]^2 dq^2 (3) where...

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Physics

23. Consider a 2D spacetime whose metric is: ds^2 = −dt^2 +[f(q)]^2 dq^2 (3) where f(q) is any function of the spatialcoordinate q. a) Show that the t component of the geodesic equationimplies that dt dτ is contant for a geodesic. b) The q component ofthe geodesic equation is hard to integrate, but show that requiringu · u = −1 implies that f dq dτ is a constant for a geodesic. c)Argue then that the trajectory q(t) of a free particle in thisspacetime is such that dq dt = constant f . d) Imagine that wetransform to a new coordinate system with coordinates t and x wherex = F(q) and F(q) is the antiderivative of f(q). Show that themetric in the new coordinate system is the metric for flatspacetime, so the spacetime described by equation 3 is simply flatspacetime in disguise. We know geodesics in flat spacetime obey dxdτ = (f dq) dτ = constant, so the result in part b) is notsurprising.

Answer & Explanation Solved by verified expert
3.7 Ratings (652 Votes)
a For the metricThe Lagrangian isorwhere dot represents derivative with respect to tauThe geodesicis just the Euler    See Answer
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