High School L = 0 in the equation for effective potential energy?

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Setting L to zero in the effective potential energy equation indicates zero angular momentum, leading to a scenario of radial infall. This situation can be analyzed in both Newtonian and general relativity contexts. In Newtonian physics, the Kepler problem provides a foundational understanding, allowing for analytical solutions. Understanding this framework simplifies the transition to general relativity, particularly in the context of a test particle in Schwarzschild spacetime. A thorough study of these concepts is essential for grasping the implications of L=0.
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).
 
In an inertial frame of reference (IFR), there are two fixed points, A and B, which share an entangled state $$ \frac{1}{\sqrt{2}}(|0>_A|1>_B+|1>_A|0>_B) $$ At point A, a measurement is made. The state then collapses to $$ |a>_A|b>_B, \{a,b\}=\{0,1\} $$ We assume that A has the state ##|a>_A## and B has ##|b>_B## simultaneously, i.e., when their synchronized clocks both read time T However, in other inertial frames, due to the relativity of simultaneity, the moment when B has ##|b>_B##...

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