The common idea is that gravity can't be shielded against. I highly doubt that.

Let us create a sphere: $$s: r^2 = x^2+y^2+z^2$$ which is centred on a point $M$.

Now we start putting tiny black holes all over this sphere and increase their number to infinity. We now created a bowl of black holes, which is empty in the middle. Now any outside information cannot pass the bowl. Therefore any outside gravitational force cannot apply to something in the middle. Of course, this bowl is collapsing, but since we can choose $r$ this does not matter (ugh, does not matter?, it is all about matter). Assuming we would put a body in the middle, this body would not feel any force but its own weight, when r is only big enough.

Could this idea, theoretically, shield against gravity? I consider that changes in the structure of the sphere should make no impact at all...

I think the abstraction of this question is: What will happen to gravitational information when stopped by a black hole.

EDIT: This experiment does not contradict Newton's third law since also the inside of the sphere cannot apply force to the outside.

It does not matter, that the sphere can be stretched as long as we presume that it cannot be torn up. If we choose $r$ big enough, we will feel this stretching since it can only be stretched with the speed of light. If we would now imagine that we would have the heaviest thing possible outside the sphere, which would without the sphere apply massive force on us, we have shielded (for at least some time) against its gravity.

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    $\begingroup$ How do you keep the shell of black holes from collapsing? $\endgroup$
    – zeta-band
    Commented Apr 26, 2019 at 22:56
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    $\begingroup$ You do not need to. They will collapse, but you choose r so that this takes a long enough time. $\endgroup$
    – TVSuchty
    Commented Apr 26, 2019 at 23:15
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    $\begingroup$ How do you keep the black holes in a sphere if they are asymmetrically perturbed? $\endgroup$
    – ProfRob
    Commented Apr 26, 2019 at 23:21
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    $\begingroup$ How do you keep the black holes in a sphere if they are asymmetrically perturbed? Just presume they are equally distributed about the (w)hole sphere, or just assume small changes in time. $\endgroup$
    – TVSuchty
    Commented Apr 26, 2019 at 23:52
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    $\begingroup$ Yes, but it is not the same (information). Imagine you create a sphere of really tiny black holes and outside of this: The heaviest thing you can possibly imagine. Maybe the sphere will be stretched (not destroyed), but it cannot be stretched faster than with lightspeed. If r now is big enough this almost leaves us with no gravity change in the centre and you (in the middle of the sphere) will still not feel the heaviest thing, because you are surrounded by a singularity. $\endgroup$
    – TVSuchty
    Commented Apr 27, 2019 at 7:17

1 Answer 1


Now any outside information cannot pass the bowl

You are assuming gravity is caused by a type of information that moves, as is the case for photons in the EM force.

It is not, gravity is the configuration of space itself. Adding black holes reconfigures space, but it does not make it disappear. The mass-energy outside the sphere still has an effect on the end-result configuration.

It's easy to confuse the two cases because the changes to that configuration travel at the speed of light, so it seems like "something" is moving, but it's simply not the same at the bottom.

  • $\begingroup$ How does space change when it hits a wall of singularity? $\endgroup$
    – TVSuchty
    Commented Apr 27, 2019 at 15:41
  • $\begingroup$ And why would then gravitational information travel with light speed at all? $\endgroup$
    – TVSuchty
    Commented Apr 27, 2019 at 15:50
  • $\begingroup$ Space changes at the singularity the same way it does anywhere. I'm sure you've seen the rubber-sheet analogy, the sheet still exists on either side of the masses you put on it. It would be the same in this case. $\endgroup$ Commented Apr 27, 2019 at 16:04
  • $\begingroup$ Unfortunately, I am a high-school student... $\endgroup$
    – TVSuchty
    Commented Apr 27, 2019 at 16:38
  • $\begingroup$ That's NOT unfortunate, trust me. Wait until your eye's start going and you'll know what I mean. But here is what I was talking about: youtube.com/watch?v=MTY1Kje0yLg - gravity is a continuum, its not an "exchange" of any sort. At least not in the main model we use, GR. $\endgroup$ Commented Apr 27, 2019 at 17:07

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