# Centrifugal force in the two body problem?

In the two body problem, the Effective radial potential energy in general relativity is given by

$$V(r)=-\frac{G M m}{r}+\frac{L^{2}}{2\mu r^{2}}-\frac{G(M+m)L^{2}}{c^{2}\mu r^{3}}$$ where the second term is the centrifugal potential energy.

1. Firstly, I am a little bit confused about the second term because from my study of classical physics the centrifugal term only arises if we solve the equations in a rotated frame reference however the second term seem to appear even without that assumption. For example when one finds the potential energy in a schwartzchild metric the second term seems to appear too. Am I missing something here?
2. Secondly, In a rotated frame of reference two "fictitious" forces appear the centrifugal force $$F_{cent}=-m\,\Omega\times(\Omega\times r)$$ and the Coriolis force $$F_{Cor}=-m\,\Omega\times\frac{dr}{dt}$$ Is $F_{cent}$ equal to the derivative of the second term in $V(r)$ and if so then how because they look very different to me?
• The equation of motion for the 2-body problem is usually given in the frame with the centre of mass at rest. How did you reach the conclusion that it is rotating? Jun 5, 2016 at 10:00

1. The centrifugal potential is nothing but the angular part of the kinetic energy of the (reduced mass) particle. To see this, write the Lagrangian in polar coordinates, $$L=T-V=\frac{\mu}{2}\left(\dot r^2+r^2\dot\phi^2\right)-V(r).$$ Since $\phi$ is a cyclic coordinate its conjugated momentum (the angular momentum) is conserved $$p_\phi=\frac{\partial L}{\partial \dot\phi}=\mu r^2\dot\phi=\mbox{constant}.$$ This relation eliminates $\dot\phi$ from the Lagrangian,
$$L=\frac{\mu}{2}\dot r^2+\frac{p_\phi^2}{2\mu r^2}-V(r).$$ Note that although the second term on the RHS is a kinetic term, it does not explicitly depend on velocities so it may be reinterpreted as a potential term, $$L=\frac{\mu}{2}\dot r^2-V(r)_{\mathrm{ef}},$$ where the effective potential $V(r)_{\mathrm{ef}}$ is the sum of the so-called centrifugal potential and the interaction potential, $$V(r)_{\mathrm{ef}}=\frac{p_\phi^2}{2\mu r^2}-V(r).$$ That is a general procedure, whenever one has a cyclic variable one can write its associated kinetic terms without explicit velocities and thus interpret them as potential terms.
2. The equation of motion for the radial coordinate is $$\mu\ddot r=-\frac{\partial V(r)_{\mathrm{ef}}}{\partial r}=\frac{p_\phi^2}{\mu r^3}-\frac{\partial V(r)}{\partial r}.$$ The second term on the RHS is an repulsive term and it equals the centrifugal force, $$\frac{p_\phi^2}{\mu r^3}=\mu r \dot\phi^2=|-\mu\vec\Omega\times(\vec\Omega\times\vec r)|,$$ since $\vec\Omega=\dot\phi \vec e_k$