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).

## Best Answer

Newton's third law comes in two versions:

The

weakNewton's third law says that mutual forces of action and reaction areequalandoppositebetween two particles $i$ and $j$ at position $\vec{r}_i$ and $\vec{r}_j$, $$ \vec{F}_{ij}+\vec{F}_{ji}~=~\vec{0}.\tag{1}$$The

strongNewton's third law says besides eq. $(1)$ that the forces are alsocollinear,$$\vec{F}_{ij} ~\parallel ~\vec{r}_{ij},\tag{2}$$ i.e. parallel to the difference in positions $$\vec{r}_{ij}~:=~\vec{r}_j-\vec{r}_i,\tag{3}$$ or equivalently, act on the same line, as OP writes.The weak Newton's third law is Newton's original formulation. The strong Newton's third law is a non-trivial additional assumption, cf. e.g. this Phys.SE post. If the strong Newton's third law holds, it often facilitates arguments. E.g. when applying D'Alembert's principle or Lagrangian mechanics to a rigid body, one can then ignore all the internal forces, which is an enormous simplification, cf. this Phys.SE post.