I am aware that closed form solution for 2-body problem does not exist. What I am looking for is a numerical solution. I was thinking on how to start to approach this very problem. I am aware of lagrangian formulation of general relativity where we cook up a lagrangian of different interactions to form the action integral. Pure gravitational action which would be something like

$$A=-m\int ds$$

Now since $ ds^2=g_{\mu\nu}dx^\mu dx^\nu$, if we let the metric tensor $g_{\mu \nu}$ to be a non-flat one (say) schwarzchild or kerr then minimizing the action would lead us to the geodesic equation for a particle in that spacetime.

This formulation literally means that to incorporate the effects of gravity in any situation just instead of flat metric use a non-flat metric. Which means for a 2-body problem what I need is a metric that is specifically calculated for such a situation. So how to even begin to solve such a problem?

Computer simulations of relativistic bodies under mutual interaction with each other have been created. They must have formulated $n$-body interaction in some numerical way. So how do they do it for a general case?

Edit: I looked up there is a metric for 2-body named "Curzon-Chazy metric" but it's not for a general case.


1 Answer 1


Very good question. The truth is that the full-blown-fair-and-square two body problem is fairly complicated. The thing is that you have to be solving simultaneously a system of differential equations which incorporate (i) Einstein's filed equations for the space-time metric generated by the world lines of both bodies, coupled with (ii) the geodesic equations of each body in the said metric. If $x_1(\tau)$ and $x_2(\tau)$ are the sought geodesic wold-lines of body 1 and 2 with respect to proper time $\tau$, the equations should be something like this

$$d\tau = \sqrt{g_{\mu \nu}(x)\, dx^{\mu}\, dx^{\nu}} \,\,\,\, \text{(the definition of proper time)}$$

$$T^{\mu \nu}[x_1, x_2](x) = m_1\, \frac{dx_1^{\mu}}{d\tau} \, \frac{dx_1^{\nu}}{d\tau}\, \delta\big(x - x_1(\tau)\big) + m_2\, \frac{dx_2^{\mu}}{d\tau} \, \frac{dx_2^{\nu}}{d\tau}\, \delta\big(x - x_2(\tau)\big) \,\,\,\, \text{(the energy momentum tensor generated by particle 1 and 2)} $$ $$R_{\mu \nu}[\,g\,](x) - \frac{1}{2}\,\, R[\,g\,](x) \,\, g_{\mu\nu}(x) = \frac{8\pi G}{c^4} \, T_{\mu \nu}[x_1,x_2](x) \,\,\,\, \text{(Einstein field equations for the gravitational field generated by the two bodies)} $$ $$\frac{d^2x_1^{\lambda}}{d\tau^2} + \Gamma^{\lambda}_{\mu \nu}[\,g\,](x_1)\,\frac{dx_1^{\mu}}{d\tau} \, \frac{dx_1^{\nu}}{d\tau} =0$$ $$\frac{d^2x_2^{\lambda}}{d\tau^2} + \Gamma^{\lambda}_{\mu \nu}[\,g\,](x_2)\,\frac{dx_2^{\mu}}{d\tau} \, \frac{dx_2^{\nu}}{d\tau} =0$$ $$\,\,\,\, \text{(equations of motion for each of the two bodies)} $$ The square brackets mean that the given quantity is obtained from the quantities in the square brackets. The function $\delta$ is Dirac's delta function. It might be more appropriate to parametrize the world-lines in terms of the time coordinate of the coordinates $x$, i.e. to write $\tau$ in terms of the coordinate time $x^0$. I think you can see how intricately coupled all these equations are.

I should mention here that the system above is written in its full form with respect to arbitrary coordinates (observer). However, you can simplify the equations drastically, if you pick coordinates attached to one of the mass-points, say point 2. In other words, you are solving the problem from the point of you of an observer traveling with body 2. In such coordinates, the trajectory $x_2(\tau)$ is a straight time-like line and the metric tensor along $x_2(\tau)$ is the Minkowski flat metric. But away from the straight line trajectory of $x_2$, the metric is not Minkowskian. These coordinates are called Fermi (free falling) coordinates. Then, you have to solve the equations $$R_{\mu \nu}[\,g\,](x) - \frac{1}{2}\,\, R[\,g\,](x) \,\, g_{\mu\nu}(x) = \frac{8\pi G}{c^4} \, T_{\mu \nu}[x_1](x) \,\,\,\, $$ $$\frac{d^2x_1^{\lambda}}{d\tau^2} + \Gamma^{\lambda}_{\mu \nu}[\,g\,](x_1)\,\frac{dx_1^{\mu}}{d\tau} \, \frac{dx_1^{\nu}}{d\tau} =0$$ where the energy momentum tensor $$T^{\mu \nu}[x_1](x) = m_1\, \frac{dx_1^{\mu}}{d\tau} \, \frac{dx_1^{\nu}}{d\tau}\, \delta\big(x - x_1(\tau)\big) + m_2\, \frac{dx_2^{\mu}}{d\tau} \, \frac{dx_2^{\nu}}{d\tau}\, \delta\big(x - x_2(\tau)\big) \,\,\,\,$$ has a second term $$m_2\, \frac{dx_2^{\mu}}{d\tau} \, \frac{dx_2^{\nu}}{d\tau}\, \delta\big(x - x_2(\tau)\big) = 0 \,\,\,\, \text{ when $\mu \neq 0$ or $\nu \neq 0$ }$$ and $$m_2\, \frac{dx_2^{\mu}}{d\tau} \, \frac{dx_2^{\nu}}{d\tau}\, \delta\big(x - x_2(\tau)\big) = m_2 \, \delta\big(x - (x^0,0,0,0)\big) \,\,\,\, \text{ when $\mu = \nu = 0$ }$$

  • $\begingroup$ Sorry for late comment. I did not understand your equations completely but I got a good idea about the problem, I guess. So I'll take my answer as: **From Einstein's field equation what we usually do is to put in the metric and find the energy and mass flow and distribution but instead if we do the reverse process of finding the metric given the mass-energy distribution through the equations then since energy and mass could be arbitrary distributed we basically solve the n-body problem ** I got the idea on how to tackle n-body problem in general relativity. Thanks!! $\endgroup$ Jun 25, 2018 at 13:23

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