- #1
unscientific
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Homework Statement
(a) Find ##\dot \phi##.
(b) Find the geodesic equation in ##r##.
(c) Find functions g,f,h.
(d) Comment on the significance of the results.
Homework Equations
The metric components are:
##g_{00} = -c^2##
##g_{11} = \frac{r^2 + \alpha^2 cos^2 \theta}{r^2 +\alpha^2}##
##g_{22} = r^2 + \alpha^2 cos^2 \theta ##
##g_{33} = (r^2+\alpha^2)sin^2 \theta ##
The Attempt at a Solution
Part (a)
The geodesic equation can be written as
##\dot u_\mu = \frac{1}{2} \left( \partial_\mu g_{v\sigma} \right) u^\sigma u^v ##
The 3rd component reads:
[tex]\dot u_3 = \frac{1}{2}\left( \partial_3 g_{v\sigma} \right) u^\sigma u^v [/tex]
Since none of ##g_{ii}## depends on ##\phi##, ##\dot u_3## is simply zero.
Thus, ##\dot u_3 = 0##. Integrating, ##u_3 = J##. But ##J = u_3 = g_{33}u^3##. Therefore
[tex]u^3 = \dot \phi = \frac{u_3}{g_{33}} = \frac{J}{(r^2+\alpha^2)sin^2\theta} [/tex]
Part(b)
Before we start, let's express LHS in terms of ##u^1##. By the product rule, ##\dot u_1 = \dot g_{11} u^1 + g_{11} \dot u^1##. Also, ##\theta = \frac{\pi}{2}## and ##\dot \theta = 0##.
Thus the geodesic equation reads:
[tex]\dot g_{11} u^1 + g_{11} \dot u^1 = \frac{1}{2} \dot r^2 \frac{\partial (g_{11})}{\partial r} + \frac{1}{2} \dot \phi^2 \frac{\partial (g_{33})}{\partial r}[/tex]
[tex]g_{11} \dot u^1 = \frac{1}{2} \dot \phi^2 \frac{\partial (g_{33})}{\partial r} - \frac{1}{2} \dot r^2 \frac{\partial (g_{11})}{\partial r} [/tex]
[tex]\left( \frac{r^2}{r^2 + \alpha^2} \right) \ddot{r} = \frac{1}{2} \frac{J^2}{\left( r^2 + \alpha^2 \right)^2} (2r) - \frac{1}{2} \dot r^2 \frac{2\alpha^2 r}{\left( r^2 + \alpha^2 \right)^2} [/tex]
[tex] r \ddot r = \frac{J^2}{r^2 + \alpha^2} - \dot r^2 \frac{\alpha^2}{r^2 + \alpha^2} [/tex]
Can't see how this would work..