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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: $$F_{12}=-F_{21}$$

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Why are you asking this(curious)? – namehere Nov 29 '12 at 13:21
@namehere During the calculation of the properties of black holes – Neo Nov 29 '12 at 20:42
Which black hole metric are you using? note that Schwarschild, Kerr, and Newmann metrics are stationary(Schwarschild is also static), so they don't represent physical black holes. I'm also not very sure how Newton's third law affects black holes. Note that objects only under the influence of gravitation in General Relativity are in free fall, so that their 'motion' cannot be described by forces, and instead you should look at the geodesic equation. Also, with the metric being generally different at general points, any apparent 'force' that is exerted should NOT obey Newton's Third Law. – namehere Nov 30 '12 at 0:27
@namehere: the Kerr model is closer to a physical black hole than almost any exact model is to any physical body. – Jerry Schirmer Sep 18 '13 at 16:43
up vote 4 down vote accepted

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).

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"Newton's third law will come up false in a variety of electromagnetic applications" i cant immediately recall any effects\phenomena of this sort(unless you count quantum stuff). enlighten me! – namehere Nov 29 '12 at 15:51
@namehere: the simplest case of this is if you have two charged particles of different masses exerting a force on each other, they will radiate asymmetrically, and this will produce different radiation back-reaction forces, so you will observe that the force that the first particle exerts on the second particle will be different than vice versa. Looked at using momentum, you will see that the radiation carries momeuntum, and overall momentum is conserved. – Jerry Schirmer Nov 29 '12 at 16:42
ohhh, thankyou, I couldnt recall that. – namehere Nov 30 '12 at 0:21
Yes, this is an elegant proof. +1 (a short while ago.). – centralcharge Sep 18 '13 at 16:20

Newton's Third Law partially holds in General Relativity.

In General Relativity, spacetime is "curved", and momentum(and hence force) at a point cannot be directly meaningfully corresponded to momentum(and force) at a different point in spacetime. To correspond quantities from different points in spacetime, we need something called "parallel transport" which maps the momentum(or force) space at one point to the momentum(or force) space at another point, given a path. As you can see, the resulting momentum(and force) at the destination is dependent on the path chosen and the particular geometry of the particular spacetime. So forces at a distance cannot really obey things like Newton's Third Law.

However, locally(aka at one spacetime point), Newton's Third Law (sort of) does hold. Basically Newton's Third Law is equivalent to the conservation of momentum. In General Relativiy, this holds in the form of $$\nabla_\beta T^{\alpha\beta} = 0$$ where $T^{\alpha\beta}$ is the energy-momentum tensor and $\nabla_\beta$ denotes the covariant derivative. The energy-momentum tensor which you guessed it, represents the energy and momentum plus some other components irrelevant here at a spacetime point, and covariant derivative is basically a derivative modified slightly say that it works in curved spacetime. So the equation says the derivative of momentum is zero and this can be interpreted as equal to momentum being conserved, which is equal to Newton's Third Law.

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+1 your answer is good too – Neo Nov 29 '12 at 20:43
Yes this answer is also good because of the mention of the sem tensor. I'll +1 this tomorrow, when my daily vote limit is reset. – centralcharge Sep 18 '13 at 16:20

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