Bubbling up some ideas

Last night, one thought plopped one after another in my mind, and one of them was about adding local space anomalies which reverse the force applied through gravity. They're literally static bubbles : They are spherical, don't move relative to the world nor grow or shrink over any meaningful time. Their size ranges from as little as 20cm to 30m in radius, although most are around 1 to 4m in radius.

Because I lack complex physics knowledge and because I want simple things, the way I imagine they work is to simply add a coefficient to the Newtonian gravity equation. This coefficient ranges from 1 (outside the bubble, normal gravity) to -1 (at the bubble's center, reversed gravity). There is a smooth, linear gradation at the borders which roughly goes to 0 after 15-20cm inside the bubble and -1 at double the distance, so ~40cm. The borders' size are not relative to the bubble's own size.

2d cut of the bubble's force field picture

A rough 2D cut of the bubble's force field, where white is reversed-gravity and black is normal gravity. Grey is anything in-between.

So what can you expect from this?

The basics :

  • Objects outside fall normally (quite obviously).
  • Objects fully inside "fall" reverse to local gravity. They go "up" towards the sky and not the ground.

For objects "left" on the borders (ie. dropped without giving force) :

  • They fall towards the ground at lower forces (and so speed) in the first 20cm inside the bubble
  • At roughly 20cm they don't receive any force.
  • Between 20cm and 40cm roughly they fall reverse to local gravity at lower forces.
  • For bubbles less than 40cm in radius you never reach full reverse gravity, and for the really few ones at 20 cm it's just at most a no-gravity zone.


The movement of anything solid inside the bubble is quite easy to figure out (I could actually make a simulation of it like I did there) : Things inside move "up", things outside move "down". The weirdest thing that could happen is at the borders, where -as far as my thinking can go- objects left not exactly on the bubble's top will tend to spin along the round surface, as the forces are not uniformly spread on the volume. Think like you slide a cylinder between your fingers, sort of.

That is, everything above is conceptualized in a vacuum environment, as I happen to also have vacuum in my fluid dynamics lessons. If you left a tennis ball here or there, it would probably be affected by air currents too and move another way. That is, if there are any airflows o_x.

So I wish to know what would happen to air inside and in direct proximity to these bubbles? Would air currents be created spontaneously, and if so, what direction would they take and what speed would they roughly have?

In other words, should I expect some fresh cool winds or stormy, very strong ones, if at all? Would they go upward then downward around the bubble, or maybe something else? Would this affect only large sized-bubbles? This is the kind of thing I'm currently wondering about.

I'm aiming this for a classic Earth-like world. I'd like first and foremost to know what would happen in windless environments, as it'd give a much better understanding of the system. But if by chance you happen to have a glimpse of what would happen in a real context with some wind, It'd be interesting too :).

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    $\begingroup$ Conceptually related: worldbuilding.stackexchange.com/questions/206934/… $\endgroup$
    – Qami
    Commented Sep 29, 2021 at 18:37
  • $\begingroup$ worldbuilding.stackexchange.com/q/178900/62241 the modeling of orbits in regions of reversed gravity might help to understand how air might arrange itself $\endgroup$
    – BMF
    Commented Oct 1, 2021 at 17:17
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    $\begingroup$ @BMF By how I upvoted this, I think I actually looked at this question when writing the other one about cyclic orbits ^^. It's just that I feel like solid objects don't exactly react the same way as fluids. I guess I need more of that knowledge to improve my work-air-flow :p. $\endgroup$ Commented Oct 1, 2021 at 21:47

4 Answers 4


The main effect with regard to air currents in and around the reverse gravity bubbles would be due to the difference in pressure gradients inside and outside the bubbles.

At sea-level there is an approximate pressure gradient of -10 Pa/m (i.e. pressure drops as you rise in altitude), and inside the bubble (ignoring the transition zone between normal and reverse gravity) this gradient would be reversed (i.e. atmospheric pressure inside teh bubble would increase slightly as you rise in altitude). For a 4 m diameter bubble, the upper part of the bubble would have an excess pressure of ~40 Pa compared to the air outside the bubble, and at the bottom surface of the bubble there would be a similar pressure deficit. As a result, air would be forced in through the bottom and would be expelled at the top (a wonderful example of perpetual motion that often accompanies anti-gravity models).

40 Pa is not a lot (about 0.04% of atmospheric pressure), but it would be sufficient for quite significant 'convection' currents to develop. Approximating each of the transition zones at the top and bottom of the 4 m diameter bubble as a vent of diameter 2 m, 1 m in length (to allow for the transition from +ve to -ve gravity), with a pressure difference of 40 Pa, then greatly abusing an airflow model beyond anything it was reasonably designed to calculate, suggests that this would introduce an upward oriented 'draught' with wind velocity of the order of 10-20 m/s (about 35-70 kmph). That is a fairly stiff breeze.

  • $\begingroup$ To build on this, I think you'd end up getting ring-shaped vortices around each bubble, with a "downdraft" arrayed outside the equator of the bubble. (Think like a horizontal mini-tornado, bent around and connected to itself to make a ring.) The downward flux outside the bubble would need to match the upward flux inside the bubble, but I would imagine that it would be spread throughout a greater cross-sectional area and therefore be slower/weaker. $\endgroup$
    – Qami
    Commented Sep 30, 2021 at 14:40
  • $\begingroup$ It took me some time, I wanted to check things and in overall get better informed. All answers at the time of this comment are really interesting and helped me out of this bubble of incertitude, but I must admit your answer is the one I understood the most and gave more practical measures I can relate to. Hence my seal of acceptation :). $\endgroup$ Commented Oct 6, 2021 at 20:01

If you have a force field active in the volume, whose effect depends on the mass present in that volume, you can have buoyancy forces by changing the relative density of the matter in that volume.

The buoyancy forces will then produce currents, with effects proportional to their magnitude.

Nothing much different than the old story on hot and less dense air going up and being replaced by cold and more dense air, with the warning label that wherever "up" is is determined by the local field.

Again, the magnitude of the buoyancy forces will determine the strength of the currents. The bigger the difference in density, the bigger the induced currents.


It's easier to intuitively visualize the behavior if you switch to a potential diagram. The potential of a point is defined as "what energy do I need to bring a particle with one unit of (mass, charge, whatever interacts with the field) from infinity to the point of interest" (in math formalism, force integrated over distance from infinity to the position)

For you gravitational bubble with a radial distribution, it looks something like this:

grav bubble potential

Now, release some gas around it.

  • For the behavior outside the gravity inversion boundary - If the kinetic energy of a gas particle is lower than necessary to go to infinity, it will be trapped in the valley below zero and will form the atmosphere of your bubble. If it is higher, it will escape to infinity and you'll lose it.

  • For the behavior inside the gravity inversion boundary - any particle, irrespective of its energy, will never reach the center of your bubble, but it can get infinitely close as its kinetic energy increases. Such a particle will get reflected by the center but, all in all, you won't get anything spectacular (you can thing of it as a tiny gas amount trying to fight against an ever increasing "pressure" towards the center)

Tactile, trying to grab it, it will be like squishing a ball of nothing that becomes tougher the harder you squish (not unlike a compressed rubber, except that the resistance to deformation increases faster than linear). If follows that the ball will levitate on top of a table (when placed into the gravitational field of a planet) at whatever distance the repulsive potential energy inside the bubble equals the attractive potential energy of the planet acting on the bubble as a whole.

Visually, classic Newton interpretation, gravity does not influence light, so you'll see a "nothing to see there". Einstein GR, gravity is a curvature of space, you'll get a divergent gravitational lens in your hand.

  • $\begingroup$ Thanks for pointing out details on how it would "feel" and look like. That wasn't parts I have given much thoughts as I thought mainly in non-relativistic manners (I've never learned it properly). After having looked at radial distribution, I think I intuitively understand what it means, but I have been quickly (read instantly x) ) overwhelmed by the formulas and diagram. I wish I just knew what each axis mean x_x... $\endgroup$ Commented Oct 1, 2021 at 11:59
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    $\begingroup$ @Tortliena x axis - radius. y-axis the gravitational potential. The quickest way to "feel" what that mean is to think of it as a profile of hills-valleys in the constant gravity. As for the formulas, it's the attractive part up to a radius of 1 ("goes down" in a valley proportional with -1/R) then switches to repulsive inside ("goes up" the hill) proportional with 1/R. The -2 needs to be added to make the plot continuous. some basics in grav potential $\endgroup$ Commented Oct 1, 2021 at 21:13

Oscillation of air on the boundary

opening: "Their size ranges from as little as 20cm to 30m in radius, although most are around 1 to 4m in radius."

I don't think it would relevantly affect its outside, at this size. When the field is modulated in some way, air could move (oscillate) around the boundary and generate a sound. Of course, in a vacuum, that sound would not be audible.

In a small scale situation, as presented in the opening, enough reverse gravity inside the sphere would cause material and gasses inside to move outward, slowly, quickly, explosively, depending on the strength of the field. It will leave a vacuum inside, or low pressure situation. This would attract gasses again, causing the oscillation. But this traveling back is not caused by gravity. Outside the object, any gravitational effect would be near zero, as a result of the small size and weight of the object. Escape velocity from this object would be zero. The only reason gases will tend to move inward is pressure difference. Air flows back, until it is forced out again.

A much larger scale field of this kind could form a hollow sphere

When this field would be half the size of a solar system, the mass inside would become relevant, and matter outside the field would be drawn toward it, or orbit around it. Meanwhile, all matter inside would emigrate outward. The sun will be torn apart. Oscillation at the boundary will happen, but eventually, lots of matter would get concentrated into a certain region at the boundary. When parameters are tuned accurately, the field could form a natural Dyson Sphere.


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