Different definitions for effective potential in static spherically symmetric spacetimes – which is right? In the paper Existence and stability of circular orbits in general static and spherically symmetric spacetimes authors define the effective potential as
$$ V\equiv \frac{1}{g_{rr} g_{tt}}~[E^2-g_{tt}~(1+\frac{L^2}{r^2})] \tag{1}$$ whereas in Boundary Orbits: 1 Static Spacetimes authors define it as $$V \equiv g_{tt}~(1+\frac{L^2}{r^2 }) \tag{2}$$ and in Theoretical Search for Gravitational Bound States of Tachyons $$V \equiv \frac{E^2}{g_{tt}}-~(1+\frac{L^2}{r^2 }). \tag{3}$$ Which definition is correct, or rather, are all three definitions admissible?
 A: The short answer is yes, all above definitions are admissible!
To explain it I would like to quote professor Tiberiu Harko, who kindly answered my question in private communication as follows:
"The potential in static general relativity is rather arbitrary. As opposed to general relativity, the potential is an effective quantity, which does not have a direct physical meaning, like in Newtonian gravity. If you can write the equation of motion for $r$ in the form $\frac{\it{1}}{\it{2}}~\dot{r}^2+something=constant$, you can define something as an effective potential, by analogy with classical mechanics, $V\equiv something$. This would be a kind of "standard" definition. But, if for mathematical or other reason, it is more convenient to define the effective quantity in another way, I don't think any problem in this. The effective potential in general relativity is mostly a mathematical tool. However, the definition may be important, because some physical
quantities, like the marginally stable circular orbits, are obtained from it. But even in this case, once you impose the condition $\dot{r}=0$, you can handle the problem in various ways."
