Can gravity from a massive object outside our cosmological event horizon have any effect on Earth?

Question

The gravity of a very massive luminous celestial object traveling at the speed of light away from us (at the Hubble Sphere) would have an Extremely small, but non-zero gravitation effect on Earth (acceleration). It would be negligible compared to all other gravity, dark energy, etc in our observable universe, but the point of question is zero versus non-zero. Assume the mass is large enough we don't have issues with Heisenberg's uncertainty principle.

If this very massive celestial object was much farther from Earth and traveling away from us much faster such that it was outside our cosmological event horizon (so we would never see any light from it), would the effects of its gravity on earth from it be actually zero, or just an even smaller effect than above?

Benrg, thank you for your point and excellent answer -- wish I were allowed to award 2 correct answers. Yes, the Hubble sphere is not a horizon since gravity fields and light emitted from an object moving at speed c are constantly moving into space with lower escape velocity (oversimplified perhaps). It was used in my question to describe a position far away, but well within the Past Light Cone. To bring this question full circle, my understanding is a massive object outside the Past Light Cone, but inside the Event horizon would have zero effect on Earth, until the specific point in time, t, when the expanding Past Light Cone just barely encompasses the massive object. At which point the full effect of gravity from that massive object (however small) would be felt all at once.

Answer 1

Nothing on the other side of our cosmological event horizon can affect us gravitationally (or in any other way) by the definition of event horizon.

The idea that virtual particles somehow change this is widespread, but wrong. There is no violation of light-cone causality in quantum field theory, nor in quantum gravity as far as anyone knows. Any faster-than-light virtual particles that may seem to exist in perturbation theory (Feynman diagrams) make no contribution to the final result. If they did contribute in any experimentally verifiable way, you could use that to build a faster-than-light radio.

Virtual particles are not responsible for gravitational or electromagnetic fields escaping from black holes. The fields don't escape. The field at a point outside a black hole is due to matter on the past light cone of that point, much of which is on the verge of crossing the event horizon but none of which actually has. Nothing inside the event horizon affects what happens outside, by definition.

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Whether anything outside the observable universe could affect us is a slightly different question. The answer is technically yes, because the observable universe is defined by cutting off our past light cone at a certain time in the past, such as the last scattering time or the end of inflation. There are good reasons to think that our past light cone if extended farther back must encompass a much larger area. Anything in that larger area could influence us (gravitationally or otherwise) while being outside of the observable universe as it's normally defined. But any "interesting" (not just homogeneous and isotropic) influence from outside would show up as an inhomogeneity in the cosmic microwave background, which isn't observed.

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You also mentioned the Hubble sphere. The Hubble sphere is defined as a sphere of radius $c/H$ where $H$ is the Hubble parameter; in the current era that radius is about 14 billion light years. This is not the radius of the observable universe nor the distance to the cosmological horizon. In most cases the Hubble sphere has no physical significance.

Answer 2

At present gravity has not been definitively quantized, and on cosmic scales it is the theory of Genera relativity that holds, and gravitational waves travel with the velocity of light at maximum. So if it is outside the cosmic event horizon which is moving faster than light , the masses out there cannot be detected the same way that the light from outside the horizon is undetectable. In this article common misconceptions are addressed.

If gravity is quantized, and gravitons make up the gravitational waves the way photons make up light , one would be able to write Feynman diagrams for gravity, where virtual particles could exist. The existence of virtual photons is speculated to explain the electric field of charged black holes, for example.

In a covariant gauge, one might find that the dominant contribution to the field comes from photons near the light cone, but as far as we know nobody has done such a calculation.

So even for quantum electromagnetism it is just a speculation.

Considering that even if virtual gravitons do exist, the coupling constant of gravitation is so small that probably no measurable effect would be seen.

Answer 3

The idea (propagated by popular writings) that an instantaneous static gravitational (or electric) field is in conflict with information going faster than light is wrong and shows a misunderstanding of quantum field theory (QFT).

Virtual gravitons (inside the condensate of gravitons that constitutes the instantaneous gravitational field) can travel faster than the speed of light. They can travel at any speed. Faster than light paths for gravitons are canceled only for real gravitons (which are not involved in the instantaneous field), to make them move exactly at the speed of light So you should be able to feel the presence of the big mass on Earth, even if it's behind the horizon. Real photons though can't travel faster than the speed of light, so the mass will not be visible. Gravitational waves (real gravitons) emerging from the mass can't reach us either.

This is not in contradiction with the impossibility for information to travel faster than the speed of light. If a mass moves away from us, the gravitational field will instantaneously diminish for us. No information is transmitted in this case. Only if we could make the mass instantaneously appear and disappear transformation could travel faster than light. This, obviously, is not the case, so information can't travel FTL. If you let the mass move in a way to contain information, the information is transmitted by real gravitons or gravitational waves (or photons, in the electromagnetic case).

So when a mass moves behind whatever event horizon this has an instantaneous influence on all of the spacetime surrounding the mass, including spacetime, say, 40 billion lightyears away from it. But real photons or gravitational waves (both produced by the mass) are not visible anymore. That is, the events that produced them are not visible anymore. So an event horizon is still there.

What if an electron-positron pair is produced? In that case, the static field due to both is present everywhere in the universe at the moment the pair is created. You can't transmit information in this way. That is, by letting pairs appear in such a way that a distant observer can perceive the information you have put in the appearance of the pairs. You can't, in fact, put information in the creation of pairs. You can only send information with the help of real photons (or gravitons), by letting the charges move in an orderly way. Pairs are created in a probabilistic, random way (compare this with the inability to send information FTL by means of the instantaneous, "spooky" action at a distance).

There are physicists who don't believe that there is an instantaneous static field. The static field is in fact quite complicated in QFT and far from resolved. An experiment must decide. Say by creating, again, an electron-positron pair and looking all around the pair at the same time to see how the field accompanying their creation looks like. Needless to say that such an experiment is hard to realize.

Answer 4

A large mass A close to (but outside) the observable universe (OV) horizon can certainly influence the motion of an object B just inside the OV. Photons and gravitons from B can certainly influence Earth (if Earth's survival lasts long enough). Therefore, it seems obvious that A can also influence Earth (if Earth's survival lasts long enough).