Infinite gravitational potential question
For this question I use Newtonian gravity only. Relativistic gravitational/kinematic effects are ignored.
It is known that the gravity surrounding a point particle of mass $M$ can be described by gravitational potential $\Phi$ where $$\Phi=-\frac{GM}{r}$$ $r$ is simply the distance from the particle. The gravitational acceleration $\mathbf{a}$ of a test particle (negligible mass) is defined as $$\mathbf{a}=-\nabla\Phi$$ Because Newtonian gravity is linear in potential, the potential of any arbitrary distribution of multiple point masses is the sum of the potential of all the masses.
I came across a question that I can't seem to answer it myself when trying to analyze two-body problem using the concept of gravitational potential. Assume that there are two point particles $A$ and $B$ with mass $M$ at points $(-a,0,0)$ and $(a,0,0)$ in Cartesian coordinates, initially at rest relative to each other. $\Phi$ in this case is time dependent because both particles will move towards each other due to mutual gravitational interaction. 483
However let consider initial situation where $\Phi$ is $$\Phi=-\frac{GM}{\sqrt{(x+a)^2+y^2+z^2}}-\frac{GM}{\sqrt{(x-a)^2+y^2+z^2}}$$ and the acceleration at a point is $-\nabla\Phi$, as expected. Acceleration becomes ill-defined at points $(-a,0,0)$ and $(a,0,0)$ because it contains the term $0/0$. I think this is not a problem if you interpret this as the acceleration of a test particle, because it is infinitely close to either $A$ or $B$ and therefore experiences infinite acceleration. But, if I need to calculate the acceleration of $A$ itself, obviously the wrong answer is obtained because you can calculate the acceleration using inverse square law and it is finite. One way around this problem is to consider only the potential due to $B$ $$\Phi=-\frac{GM}{\sqrt{(x-a)^2+y^2+z^2}}$$ This gives answer in agreement with inverse-square law. But it raises the conceptual question: Why can we ignore the self-potential of the particle being accelerated? (This is not an issue for massless test particle)
Also, how does this work in nonlinear system, such as general relativity where gravitational field also gravitates and produces even more gravity? I don't think you can just ignore the potential because of self-coupling of gravity. 766