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Does gravity pass through the event horizon to the other side affecting objects on the other side?

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or does it stop gravity from passing through it?

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AI says it does pass through one moment and then the other moment it says it cannot pass through.

Does the gravity stack between all the objects?

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Or is the gravity adopted by the black hole?

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    $\begingroup$ It’s hard to see a black hole on a black background. $\endgroup$
    – Ghoster
    Commented 2 days ago
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    $\begingroup$ The planet feels the gravity of the black hole and the star behind it, but describing the star’s gravity as “passing through” the black hole is incorrect. $\endgroup$
    – Ghoster
    Commented 2 days ago
  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – ACuriousMind ♦
    Commented 23 hours ago
  • $\begingroup$ 1) The black has no infinite mass. In most cases, its mass is roughly like of a larger star. 2) You have got a lot of answers, I think neither of them answers really your question. That is because your real question is, how the nonlinearity of the GR affects the gravity of the planet, compared to the naive superposition of the gravity of the two masses (star + BH). 3) Calculating this analytically is probably far above the skills and time of all of us, numerical calculation would be possible but still many work. Thus, no one knows the answer. I think, probably there is some masking. $\endgroup$
    – peterh
    Commented 16 hours ago

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In the best theory of gravity that we have now (general relativity) gravity isn't a force that "goes through" things but rather an effect of the curvature of spacetime. Both the star and the black hole contribute to that curvature so spacetime near the planet is affected by both.

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  • $\begingroup$ Yes, but how? Will it be stronger as the newtonian or weaker? My super-intuitive insight is that it will be likely weaker, so there is some masking. $\endgroup$
    – peterh
    Commented 14 hours ago
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    $\begingroup$ Far from the black hole it'll be basically same as the Newtonian case. Close to it it'll be quite different, but that's true of black holes in general. GR is nonlinear, so my guess is that the overall "pull" would be stronger than Newton would have predicted, but you'd have to solve the equations (or more likely do a numerical simulation) to tell for sure. $\endgroup$
    – Eric Smith
    Commented 10 hours ago
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$c$ is not so much the speed of light, it is the universal speed limit of causality. That includes gravity (= spacetime curvature). The event horizon is the place where light cannot propagate outwards anymore, and the same is true for any causal influence. Nothing behind the event horizon can ever influence the outside.

As such, gravity cannot "go through a black hole's event horizon to affect objects on the other side". Everything that we observe about a black hole is the warped spacetime around it. An object near a black hole influences an object on the other side via the spacetime around the black hole, not through the black hole itself.

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There is some helpful discussion on a NASA hosted site that answers a related question about how gravitons could escape a black hole.

particles don't escape from "inside" a black hole, because the extreme curvature doesn't allow any "escape trajectories" ... Now, signals (i.e. photons) involve time changing fields. Indeed no such signal can come out of the horizon. However, the electrostatic field of a charge does not convey any signal; it can therefore escape from the black hole. In technical parlance it is a space-like object, and those can cut across horizons. ... one could substitute gravitons for photons. the ones that correspond to the longitudinal part are those providing the static Newtonian (far away) fields of black holes (and all other objects). The ones producing gravitational radiation require "shaking" the gravitational lines of force and those come from outside the horizon. (Drs. Louis Barbier and Demos Kazanas)

So if you are thinking of gravity as being mediated by gravitons, then incoming gravitons will follow the same rules as incoming light. Anything directed at the blackhole will be absorbed behind the event horizon. The gravitational field outside the blackhole can be affected by the blackhole outside of the event horizon and it is the action of the blackhole on that external field which generates outward moving gravitons.

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    $\begingroup$ So if you are thinking of gravity as being mediated by gravitons, then incoming gravitons will follow the same rules as incoming light. Virtual particles and real particles are not at all similar. You said that the virtual gravitons mediating gravity are space-like. But the real photons of light are light-like, not space-like. They don’t “follow the same rules”. $\endgroup$
    – Ghoster
    Commented 2 days ago
  • $\begingroup$ @Ghoster I don't believe I said anything about virtual partcles. $\endgroup$
    – Freedom
    Commented 2 days ago
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    $\begingroup$ @Ghoster the discussion is about fields being spacelike, the particles are not. The point in the nasa article is that fields themselves are spacelike objects as the do not convey signal themselves, only their excitations do $\endgroup$
    – Freedom
    Commented 2 days ago
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    $\begingroup$ I don’t think it even makes sense to refer to a field as space-like since a field exists at all points in spacetime, with pairs of those points having all three types of spacetime interval. Can you point me to a mathematical definition of a space-like field? $\endgroup$
    – Ghoster
    Commented 2 days ago
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    $\begingroup$ I also don’t think it’s helpful to explain gravity in terms of gravitons, given that gravitons have not been detected and there is no accepted theory of quantum gravity. The generally-accepted theory of gravity is General Relativity. $\endgroup$
    – Ghoster
    Commented 2 days ago
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I imagine it similar to the earth and the moon. A satellite will experience variations in the earth's pull, based on the position of the moon.

Now while the "inside" of the event horizon can be considered "infinite gravity", the position of the event horizon will be influenced (slightly) by the presence of nearby objects (since the slope of the gravity well will be changed slightly).

So in my imagination, a nearby star would not move the back hole, but it would change the position of the event horizon at both the closest and the furthest points relative to the star.

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I'm not an expert but I believe that both fields will be superposed:

  1. If we ever assume that there is something like gravitons (virtual or real) despite having no quantum gravity theory then my best guess would be that a) real gravitons wouldn't be able to get out, but they likely don't matter to your question because you are asking about the strength of the fields and not about the symmetry principles between fields and particles as per the standard model b) virtual gravitons (mediating the gravitational pull) should get out. If they didn't then we wouldn't be able to see any gravitational pull from objects with masses of billions of suns. Maybe we are completely wrong about black holes, but you asked the question, so let's assume that they really do exist:)

This brings me to the point where we simply leave the unknowns of the quantum gravity and look at the spacetime. Several posts like by Eric Smith describe what should happen in this case. They definitely will superpose. If linearly or nonlinearly I don't know but the planet will experience a stronger pull than just by either the star or black hole.

There is BTW a nice video on PBS spacetime in YouTube where the concept of gravitational pull from inside the event horizon to the outside event horizon gets explained. I just don't know the title, it was years back since I watched it...

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