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Not sure how to say this simpler, so we'll try shorter.

We have a big space gun. It makes gravity not be there where it shoots.

It shoots at a black hole.

Does black hole go boom?

And for the tricky question, we'll get crazy and define the space gun a little more rigidly.

So, treat the space gun like it creates a cylinder (that's one of those circular things with height) of totally nullified gravity. That cylinder has a radius much smaller than the event horizon of the black hole. It is aimed straight at the center of the black hole.

Will that still cause the black hole to go boom?

Also, we don't care about the fate of the space gun, or even where it is. Also, we can treat the cylinder as though it has effectively infinite height, unless it really matters.

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closed as unclear what you're asking by Aify, Mark Olson, Cadence, John, L.Dutch Jul 8 '18 at 3:43

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  • $\begingroup$ temporarily flatten their local space-time, effectively canceling gravity? $\endgroup$ – SomeGuy Jul 7 '18 at 18:54
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    $\begingroup$ I am somewhat scientifically literate, as i had lectures on relativity and quantum mechanics, but this feels like it is just physics words lumped together to form grammatically correct, but ultimately nonsensical sentences. Please clarify how these concepts are supposed to work as they are not anything i have ever heard of nor seem to be real concepts at all. $\endgroup$ – ArtificialSoul Jul 7 '18 at 19:22
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    $\begingroup$ @SomeGuy The problem is that you can't actually flatten space time because space time isn't actually a rubber sheet. The only way to get rid of that phenomenon in a local area (scientifically speaking) is to remove the object in the area. I like to think of Gravity as a byproduct of mass interacting with spacetime - you can't "create" or "remove" gravity; it's just there. $\endgroup$ – Aify Jul 7 '18 at 20:00
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    $\begingroup$ Ah, so you are asking for inverse gravity. Like undoing the warp mass creates. If you had an equal object within the other object but with negative mass, you would cancel the gravity of the first. Negative mass does not exist. Gravity, unlike electromagnetism, does not work in two directions. It's just mass warping spacetime towards it. And there are no other ways of creating gravity, but matter (or dark matter). Neither could be used to achieve anything like the cannon you described. $\endgroup$ – ArtificialSoul Jul 7 '18 at 20:19
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    $\begingroup$ This is a valid question (although not very well formulated). It's not tagged as "science-based", so anything internally logical is acceptable. $\endgroup$ – Alexander Jul 8 '18 at 1:57
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EDIT:  The question was edited and it shifted the context. I'm leaving my original answer for fun and giggles (everything below the line), but here's the new answer:

Gravity causes pressure

Gravity doesn't change mass (I'm ignoring things like enough mass leads to enough weight that things fuse and begin to burn... too complex for the example) — but it does change weight. As you add mass the available gravity increases which increases weight. The consequence is the creation of pressure.

However, it's unlikely (though we don't actually know from empirical evidene) that gravity is uniform from the center to the surface. The shell theorem mathematically proves that inside a hollow shell there is no gravity, the distant but more massive portions of the shell countering the gravitic pull of the mass behind you at any point in the shell.

Thus, as you descend toward the center, you only have the gravity of the mass below you to contend with until you get to the center where there is no mass below you and there is no gravity.

Which, curiously, suggests that the center of a supermassive black hole has no gravity, but that's an attribute for another question.

But pressure increases... odd that

Miners have long known that pressure increases as you descend into the earth. At first that seems counter-intuitive to what I just described since, theoretically, only the mass below you is affecting you.

But it's affecting the shell above you, too. And as that shell gets thicker, its weight changes. This is the cool part. In the beginning (methinks), gravity is growing smaller very slowly, but weight is increasing very rapidly, and thus so is pressure.

But, as you get near the center, the pressure toward the center must decrease because gravity (toward the center) has reduced so much that eventually there is no pressure at all.

Toward the center.

Now all the pressure is toward the surface. Using our tried and true "circular horse" method of modeling, it suggests that a homogenous mass will have a boundary (today, because I don't know if anyone has theorized this before — and I could be completely wrong — and they probably have so let me know so I can give credit where credit's due — but today, let's call it the "JBH critical pressure boundary." I'll reveal my true name when I publish....) where the pressure is at its maximum. The pressure from that infinitely thin shell within the volume of the sphere drops to zero as you descend toward the center or as you ascend away from the center to the surface (remember, homogenious mass... on a planet the outer zero-point would be somwhere above the "surface" in or near the surface of the atmosphere, yeah... non linear... cool!).

So, "boom!" has some cool effects

I'm lousy at art, even though a picture here would be worth a thousand words. But, let's say SomeGuy pulls the trigger on that bad boy weapon of his except rather than his cylinder, all gravity in the universe is instantly reduced to zero (this is important for the example, I'll explain in a moment).

You'd get a boom! A really cool boom! Because all mass below the "JBH critical pressure boundary" (TM until proven otherwise!) springs toward the center of the sphere until the pressure reaches a maximum against the nuclear bonds of the material of the mass and then rebounds away from the center.

The mass above the "JBH Critical Pressure Boundary" (look, caps, I'm making it official, remember you heard about it first (maybe) here folks!) simply springs away from the center. In both cases we're dealing with a spring effect. That mass begins oscillating as it compresses (for example) in the direction away from the centr, forcing the mass "above" (away from the center) to spring forward at increased speed.

So, in keeping with our spherical horse simplicity, your explosion has a whole series of "shock waves" comprised of the mass of the black hole. A really big shell of rapidly expanding matter followed by a small shell. I'll leave it to mathematicians to explain how far apart they are. Just expect to be slapped hard a whole bunch of times.

And they stay that way forever, or until you turn the gravity back on

And this goes on and on and on... Heck we're affecting the entire universe (here's that "explain in a moment" moment), so every glob of mass is doing the same thing. It continues to redistribute until shockwaves hit one another and then the biggest game of billiards the universe has ever seen begins. But that might not be interesting.

But, what if we can only affect a limited area?

Let's assume the weapon wasn't quite that godlike, but affected an area, say, twice the diameter of the event horizon. All this mass expands to that point and then begins to re-experience gravity. So it begins to re-coalesce in directions along the plane of the expanding shells, but not toward the center (Shell Theorem). It's very likely that the mass wasn't homegeneous, meaning there's variations of density within the shell, which means new globs are forming.

At this point, let's turn the weapon off.

Would the black hole reform? That would need a ton of math that's beyond my abilities to prove. The force of gravity (which isn't that strong "locally" for each glob) must overcome the velocity (kinetic energy) of the expanding particles. If the expansion was slow enough, it would reform ... eventually.

If fast enough it never reforms. It would take the galaxy time to realize that because from its perspective (a massively large ring while the expanding shell still looks like a small point) all that gravity is still there. I suspect that as the expanding shell passes star systems that it then raises massive havoc with those systems, which are suddenly inside the shell (Shell Theorem) and their trajectories are massively modified.

But they also receive a massive infusion of, well, mass. And that's cool becaues the shockwaves may be traveling fast enough to do this quickly, either pummeling planets into dust or layering mass just on one side causing serious rotational wobble. Heh heh heh...

After a while all stars will be affected as the shell grows larger and larger and must be dealt with, not as an infinetly small point at the center, but an ever larger plane that's much closer but much weaker.

It's catastrophic, kinda, but it wouldn't destroy the galaxy per-se because in all this time the mass has been forming, shifting, trading alliances...

Let the math prove me right or wrong, but I suspect given enough time a new supermassive black hole would form, but it wouldn't be in the same place. Some of the mass escapes the galaxy altogether, meaning all the dynamics will change at least some. Maybe a few of the outer stars (like Earth... Aaaahhhhhh!) spin off as rogue systems into intergalactic space. The time-lapse photography would be cool to watch.

Finally, the cylindrical "beam" of the OP's weapon...

In the condition that gravity can reassert itself so quickly due to the spatially small area effect of the weapon, there would be a boom, a whomping big boom, but reaserted gravity would begin to recoalesce the mass almost instantly. The black hole would take forever and a day to completely reform, but from the perspective of the galaxy, all that mass is still at the center, so it wouldn't notice a thing. The minor fluctuations in gravity caused by the actual shift in density during the explosive and recoalescing process might cause some of the nearby stellar systems to burp a bit, but those near the rim would be unlikely to feel anything at all (only detectable via instrumentation).

At least that's what I think'll happen...


This is the original answer to the original question, left only for historical purposes.

Newton's Third Law — for every action there is an equal and opposite reaction.

Which is a fancy way of saying if something's coming at you at 100 Newtons/s, the only way to bring it to a stop is to apply 100 Newtons/s in the opposite direction.

The answer, therefore, is no. It's not plausible. The force representing the black hole's gravity must go somewhere, and where it's going to go is the weapon. Thanks to gravity being more comprehensible as something that pulls you toward it, you'd "pull" against the black hole and the black hole would "pull" back.

This is an issue because gravity is a force that can actually push. It's not like lights where most of the energy is (or can be thought of as) heat. That's why lasers of tremendous melting capacity don't have a recoil.

But gravity recoils.

HOWEVER

I'm a huge advocate of not getting too buried in the details. If what your weapon does is disrupt gravity or, perhaps more realistically (kinda), disrupts the nuclear bonds of the atoms comprising the black hole... then maybe you have an explanation that would make sense! Especially if you declare it to be independent of gravity so that whole "the black hole will just suck your beam in and consume it like everything else" problem.

I'd like to point out, though, that any civilization with the technology to destroy a black hole... especially one that can do so with a single weapon emplacement (the potential size is staggering)... is basically godlike. In such case, you don't really need to describe the how's. Just the why's.

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  • $\begingroup$ Disrupting the the nuclear bonds in a black hole would do nothing. It is held together by gravity. You need to lower its effective mass to get rid of it. Either by actually somehow getting rid of the mass (e.g. Hawking Radiation), or by getting rid of the warp in space time caused by the mass (probably impossible). $\endgroup$ – SomeGuy Jul 8 '18 at 5:43
  • $\begingroup$ Also, I think you might need to revise your answer based on my hopefully much clearer wording of my question. $\endgroup$ – SomeGuy Jul 8 '18 at 5:45
  • $\begingroup$ Rewritten to account for changes in question. Bam! $\endgroup$ – JBH Jul 8 '18 at 16:49
  • $\begingroup$ This is definitely a very cool image you have invoked. $\endgroup$ – SomeGuy Jul 8 '18 at 18:34
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Canceling Out the Gravity of a Black Hole

Part 1 - Gravity Waves

Well. They'll be big.

Depending one the size of the black hole, you could instantly generate a gravity tsunami.

Imagine the front end of your ship accelerating at 50g while the back end is standing still. That would be bad for you ship

Part 2 - time

The thing about space-time is that it is both space and time.

Your ship is either going to experience a "groundhog day" or will magically be flung into the future.

If time isn't really one-directional, parts of your ship will go to the future while the other parts will be moved to the past. Again - this is bad.

Part 3 - escaping matter

(Handwaving this physics.. .er.. all of it.) The forces that are attempting to repel the sub-atomic particles are so strong that the instant nullification of gravity causes them to accelerate past the speed of light.

Your ship will get pulverized by these particles before anyone sees the light which is generated by Cherenkov Radiation caused by the particles slowing down to under c.

Part 4 - 2 > 1

Since your gun is not powerful enough to encompass the entire black hole, you do nothing more than succeed in splitting it in half.

Instead of 1 gravity well, you now have to deal with 2. They two soon join and release enough energy to simulate a mini big bang. (I believe this was mentioned on "how the universe works")

This would not be a good thing for your ship to experience.

Final Thoughts

From "The Tick"

Gravity is a harsh mistress

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  • $\begingroup$ I think you may need to reformulate your answer based on my revised question. Hopefully it's far more clear now. $\endgroup$ – SomeGuy Jul 8 '18 at 5:44
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Think of gravity as a product of mass locally curving space. I am picturing this antigravity gun as somehow compelling space to uncurve.

A black hole is held together by its gravity. If that disappears then if nothing else the constituent atoms will, if nothing else, drift apart.

But black holes and nearly every other astronomic body are spinning to some degree - the constituent parts have angular momentum. If they were kids on a merry-go-round, then gravity is their hands holding onto the bars. If they let go they are flung off. So to the parts of your black hole. Once gravity disappears they will be flung away from the black hole according to whatever angular momentum they have.

Your black hole will expand into a disc shaped cloud. Once the gravity comes back on, it will recondense into the black hole.

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  • $\begingroup$ Would the contents of the black hole actually be released as matter? $\endgroup$ – SomeGuy Jul 8 '18 at 5:44
  • $\begingroup$ The matter in a black hole is going to be weird stuff. I do not know what happens to matter like that when it is released from compression. I was wondering that in this dwarf black star question. worldbuilding.stackexchange.com/questions/117297/… it comes crashing back in after the ray is turned up, it will get hot because of the release of gravitational potential energy and so on expanding I suppose it must get cold. Black hole matter is hard enough to think about without mysterious antigravity. $\endgroup$ – Willk Jul 8 '18 at 13:03

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