B L = 0 in the equation for effective potential energy?

sqljunkey
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Hi,

What would happen if I set L in this equation to zero? I can have an L that is zero right?
 
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Sure. That would be zero angular momentum. What do you expect to happen?

(EDIT: Which equation, exactly? There are many on that page.)
 
L can be zero. From context, L is just the angular momentum, and L=0 corresponds to a radial infall. This is the GR forum, so I'd guess you are most likely interested in the GR case, though it's possible you are interested in the Newtonian case as well.
 
Start with a thorough study of the Kepler problem in Newtonian physics. There (almost) everything can be solved in analytical form with standard elementary functions. After that it's easier to understand the general-relativsitic treatment (test particle in a Schwarzschild spacetime).
 
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From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
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