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Consider a thought experiment where there is a point mass M present in empty universe. Then another point mass m materializes (This is ONLY a thought experiment) at a distance of 1 light year from M. Then m creates a gravitational field around it which take a year to reach M. But since M was already present, it's gravitational field also is already present. Thus m experiences a force of attraction as soon as it materializes. Then there is an action on m but there is no immediate reaction. Thus Newton's third law is broken. What is going wrong in this thought experiment? Or is Newton's third law not obeyed in this case?

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    $\begingroup$ Re, "gravitational field...which take a year to reach M." That is not how gravity is modeled in Newtonian mechanics. You might then conclude that Newtonian mechanics does not accurately model how the Universe actually works on a large scale, and you would be correct in reaching that conclusion. $\endgroup$ Commented Mar 5, 2019 at 16:55

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What is going wrong in this thought experiment?

When analyzing a thought experiment it is important to explicitly identify which laws of physics you are using for the analysis. In this case the two relevant laws would be either Newton’s universal law of gravitation or Einstein’s general relativity.

Newtonian gravity: under Newtonian gravity there is no finite propagation speed for gravity. As soon as the new mass appears it’s gravitational field instantaneously permeates the universe. The force is immediately felt by the original mass and Newton’s third law holds at all times.

General relativity: under GR Newton’s third law needs to be generalized. What holds in GR is not exactly Newton’s third law, but rather the local conservation of momentum. This includes any momentum carried by fields as well as the momentum of any objects. Also, in GR you cannot simply have matter appear, it must be created from energy, and that energy gravitates just as much as the matter. So in this scenario to have the matter appear would require gathering energy, and the energy would gravitate just as the matter, with any discrepancies between the changes in momentum of the two objects being momentum carried in the fields.

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  • $\begingroup$ I know that you can't have matter appear out of nowhere. But I wanted to know what would happen if such a mass came into existence. Thanks for your answer. $\endgroup$ Commented Mar 5, 2019 at 17:28
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    $\begingroup$ It isn’t a problem with Newtonian gravity to have matter just appear. Although there is no physical mechanisms for matter to appear, Newtonian gravity is consistent with the idea. But with GR it is different. Matter just appearing explicitly violates the Einstein Field Equations. So GR is fundamentally incompatible with the “just appearing” scenario. You cannot logically have both matter just appearing and GR, they are contradictory premises. $\endgroup$
    – Dale
    Commented Mar 5, 2019 at 19:27
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As a few comments suggest, due to the conservation of four-momentum, massive particles can't pop into existence without the requisite energy being present in the locality. This requisite energy would have a gravitational influence according to general relativity and thus, your framing of the question makes it look moot given this understanding.

Nevertheless, the spirit of your question can be saved via formulating it in this way: Suppose a mass-energy configuration $A$ undergoes a change in its configuration so that it experiences a different force due to the gravity of $B$ than it did before--but the influence of this change in $A$ will reach $B$ after a finite time and during that time, $B$ will continue experiencing the same force as before. Thus, the third law is violated.

And, this is simply true. The third law is not a law in relativistic physics for a variety of reasons. The most fundamental of them is the fact that the third law appeals to the values of forces at distant places at the same time. Since the notion of simultaneity of distant events is frame dependent in relativistic physics, such a law is destined to be violated in one legitimate frame or the other--and, thus, is not a law. As explained in another answer, no hell breaks loose due to any of this because the more general principle of the conservation of momentum (from which the third law can be derived in Newtonian physics under a few natural assumptions) is nonetheless valid in relativistic physics.

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  • $\begingroup$ “there is no such law as the conservation (local or global) of mass” This is not correct. Locally the four momentum is conserved and hence all components of the four momentum and the norm of the four momentum are all conserved. The mass is the norm of the four momentum, so it is locally conserved by the same law as the conservation of the four momentum. $\endgroup$
    – Dale
    Commented Mar 5, 2019 at 19:32
  • $\begingroup$ “For example, see electron-positron creation from a photon-photon scattering” This is not a counter example since the system of the two photons has mass equal to the mass of the resulting electron-positron system. The mass of a system is usually greater than the sum of the masses of its constituents. In particular, a system composed of a pair of non-parallel photons has mass >0. $\endgroup$
    – Dale
    Commented Mar 5, 2019 at 19:36
  • $\begingroup$ @Dale I have edited my answer--I realize I was mistaken in citing the example of a photon-photon scattering as the system of two photons would have some mass. I replaced the example with that of a pair production where a single photon decays into an electron-positron pair. $\endgroup$
    – user87745
    Commented Mar 5, 2019 at 20:01
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    $\begingroup$ @Dale Ah, I see the point of your comment now. Yes, I realize that the mass of the whole system would be simply conserved as it is the norm of the total four-momentum. The spirit in which I meant to present my argument for the non-conservation of mass was the non-conservation of the sum of the masses of each particle of the system individually. I know that is not the mass of the system as a whole and that's why my sentences are not perfectly correct and are misleading. I have modified the answer. Thanks! :) $\endgroup$
    – user87745
    Commented Mar 5, 2019 at 20:08
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    $\begingroup$ Excellent edits. I upvoted the edited answer! $\endgroup$
    – Dale
    Commented Mar 6, 2019 at 1:01

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