Is Bertrand's Theorem Consistent with GR? Bertrand's Theorem - Please observe the gif of orbits with different exponents in the denominator. (I couldn't get the image upload feature to work for this)
Note, the orbit with a 1.9 exponent precesses in the reverse direction of the orbit, and the orbit with a 2.1 exponent precesses in the same direction as the orbit.
Mercury's orbit precesses in the same direction as the orbit. source
Which should mean any force equation which is consistent with Bertrand's Theorem should produce a lesser force/acceleration than Newton's inverse square law for Mercury.
But the force equation approximations of GR produce a force/acceleration greater than Newton's inverse square law as demonstrated in my previous question here.
So per the title, is Bertrand's Theorem consistent with GR or am I missing something?
 A: There are several issue here.

*

*Bertrand's result applies to forces described by a scalar potential. GR isn't a theory of that kind (it's a tensor field), so the theorem doesn't apply.


*If Wikipedia is to be believed, Bertrand's theorem is only about existence or nonexistence of precessing orbits; it says nothing about the direction of the precession. That said, Asaph Hall's article, which is online here, also attributes to Bertrand the formula $π/\sqrt{n+3}$ for the angle between perihelion and aphelion in the case of small eccentricity, where $n$ is the exponent ($-2$ in the Newtonian case). That's where Hall got the exponent of $-2.00000016$.


*When you change the exponent in the force law, in order for the units to work out, you also have to change the gravitational constant. Because it has different units, it's incomparable to the usual $G$. It can't be equal, or larger or smaller. There will always be some crossover radius outside of which the force with the larger negative exponent is weaker, and inside of which it's stronger. So an exponent like $-2.00000016$ is actually consistent with the idea that GR is stronger than Newtonian gravity at short distances (which is supported by the fact that the force goes to infinity, in some sense, at $r=r_s>0$).


*But GR doesn't behave like a $r^n$ force law with $n\ne-2$. Nor should you expect it to: it's almost meaningless that you can fit Mercury's precession by varying $n$, because you're varying one parameter and fitting one data point. To justify that change to Newtonian's theory over all the other changes you could have made, you'd have to show that it fits a lot of other data also – which it doesn't, of course.


*It's a bit iffy in any case to compare Newtonian and GR predictions at the same $r$ because it's not entirely clear what "the same $r$" means; one of the theories has no fixed background that can be used to unambiguously measure distances.
