Newton's third law and General relativity

Is Newton's third law valid at the General Relativity?

Newton's second law, the force exerted by body 2 on body 1 is: $$F_{12}$$ The force exerted by body 1 on body 2 is: $$F_{21}$$

According to Newton's third law, the force that body 2 exerts on body 1 is equal and opposite to the force that body 1 exerts on body 2: $$F_{12}=-F_{21}$$

Answer

First, let's note that newton's third law is really equivalent to conservation of momentum, by example of object one exerting a force on object two, and vice versa, and these two forces being the only forces in the universe:

$$\begin{align} F_{12} &= -F_{21}\\ m_{2}a_{2} &= -m_{1}a_{1}\\ \int m_{2}a_{2} dt &= -\int m_{1}a_{1} dt\\ m_{2}v_{2f}-m_{2}v_{2i} &= m_{1}v_{1i}-m_{1}v_{1f}\\ m_{1}v_{1f} + m_{2}v_{2f} &= m_{1}v_{1i} + m_{2}v_{2i}\\ \sum p_{f} &= \sum p_{i} \end{align}$$

Now, we know that we are looking for conservation of momentum, rather than just Newton's third law (and conservation of momentum is a more general concept anyway--Newton's third law will come up false in a variety of electromagnetic applications, but conservation of momentum will still be true). How do we get conservation of momentum? Well, the motion of a particle can be found by looking for the minium of something known as the Lagrangian:

$$L = KE - PE$$

It turns out that there is a result called Noether's theorem that says that if the Lagrangian is doesn't change when you modify your variables in a certain way, then the dynamics defined by that Lagrangian will necessarily have a conserved quantity associated with that transformation. It turns out that conservation of momentum arises when the invariance is a translation of the coordinates: $x^{a^\prime} = x^{a} + \delta^{a}$. Now, let's go back to general relativity. Here, the motion of a particle is the one that maximizes the length of:

$$\int ds^{2} =\int g_{ab}{\dot x^{a}}{\dot x^{b}}$$

If the metric tensor $g_{ab}$ has a translation invariance, this motion will necessarily have a conserved momentum associated with it, and will not otherwise. Note: common solutions, like the Schwarzschild solution of GR are NOT translation invariant--that's because the model assumes that the central black hole does not move. A more general solution that included the back-reaction of the test particle's motion WOULD have a conserved momentum (and would end with a moving black hole after some orbiting was completed).