# What is the gravitational field of a hole in an infinite perfect crystal?

Or equivalently and more interestingly: In the early universe when there was uniform H/He gas everywhere, gravitational field was close to 0 everywhere. Every test particle was pulled from all sides equally. Every volume of space had an equal volume of space, with equal mass, on the opposite side of the test particle, canceling its pull. Overdensity attracts test particle as it pulls stronger than the opposite volume of space. Underdensity repels test particle as it pulls less than the opposite volume of space. Like the dipole repeler. Empty container (vacuum) is a permanent underdensity when in primordial gas cloud. If you put overdensity next to underdensity, overdensity attracts underdensity and underdensity repels overdensity. They chase each other forever with runaway acceleration. This can be used for reactionless propulsion. Am I missing something or have I discovered (effective) antigravity?

• This sounds like one of those 19th century permanent magnet based perpetual motion machine schemes. Commented Jul 11 at 8:50
• It might sound like that, but why would it be wrong. The energy is suplied by the rest of the universe, it's not perpetuum mobile. Commented Jul 11 at 9:35
• It's wrong for the same reason why none of the 19th century magnetic perpetual motion machines worked: there is no net energy gain in these potentials. A simple way of saying it is "What goes up, must come down.". Commented Jul 11 at 9:46
• You didn't explain anything. Commented Jul 11 at 10:38
• It was a joke on potential energy. Commented Jul 11 at 15:02

If you put overdensity next to underdensity, overdensity attracts underdensity and underdensity repels overdensity. They chase each other forever with runaway acceleration.

Not true. A volume with greater than average density attracts everything around it - but it is not repelled by a volume with lower than average density. We know, for example, that the near-vacuum of outer space does not repel anything. So, no, you have not discovered antigravity.

• Underdensity repels test particle as it pulls less than the opposite volume of space. Like the dipole repeler. Underdensity doesn't repel, the opposite volume of space attracts test particle away from underdensity. We know, for example, that the great repeler effectively repels us. It's just that these days underdensities need to be larger than BAO scale. Commented Jul 11 at 11:08
• @Alienfromfuture > the opposite volume of space attracts test particle away from underdensity. We can't conclude that, because we do not know distribution of matter around the bubble. If that distribution is centrally symmetric and concentric with the empty bubble, then test particles are attracted to the bubble. Commented Jul 11 at 12:12
• We do know the distribution. It's almost perfectly homogenous H/He gas that gets even more homogenous down the past lightcone. I'm talking hundreds of ky after the big bang. Commented Jul 11 at 12:30
• @Alienfromfuture If it is homogeneous in finite space, then behaviour on the boundary, and position of this boundary matters. If it is homogeneous in infinite space, then gravitational field is not determined by this distribution (can be "anything"). Commented Jul 11 at 12:40
• Yes but the boundary is cosmological horizon, so I get to eat my cake and keep it too. Commented Jul 11 at 13:36

The equations that govern gravitational fields and matter simply don't make sense when you apply them to an "infinite" volume of matter. Mathematically, we have $$\nabla^2 \phi = 4 \pi G \rho$$, where $$\phi$$ is the gravitational potential and $$\rho$$ is the density of matter. Roughly speaking, this says that if there is matter in some region of space ($$\rho \neq 0$$) then the gravitational potential must vary in that region of space ($$\nabla^2 \phi \neq 0$$.) This contradicts your idea that the gravitational field inside the matter would be zero in the absence of a hole.

This is a tricky notion to wrap your head around, because it seems like you should be able to argue that the gravitational field is zero everywhere. But the problem is that our equations for gravitational fields simply don't work for infinitely sized matter configurations; they throw out nonsensical results like "the gravitational field inside an infinite volume of matter cannot be zero." The correct response to this seeming contradiction is to say that we can't use our tools to predict what would happen in this situation. This would be a problem if we regularly encountered infinite volumes of uniformly dense matter in our day-to-day lives; but thankfully this doesn't happen very often.

• Interesting. This is how I see it. The matter configuration (relativistically) is not infinite, because the past lightcone of test particle is finite and quite short in early universe. Clasically this would be like the Earth, gravitational potential varies but is 0 at the center. Again back to relativity, every point in the early universe was like the center of Earth as every point is at the peak of its light cone, so Poisson's equation is not applicable here. By simple symetry argument a hole in a crystal has the same gravitational field as negative mass particle in vacuum. Commented Jul 11 at 13:00
• Edit: Poisson's equation doesn't apply because it's classical. If the boundary of mass distribution is a cosmological horizon you can have 0 gravitational field everywhere by above argument. Commented Jul 12 at 0:06

The previous answers are contesting the semantics of the question. It is true that the question is provocatively (and in some ways misleadingly) phrased. However, the scenario is fully realizable in principle, and its evolution should be completely predictable. Indeed if you specialize to matter domination, just at scales far below the Jeans scale of the matter (so it remains uniform), all of the dynamics are Newtonian and you don't even need relativity.

It seems to me that this should work. The scenario is that there are two objects:

• a positive point mass $$M$$;
• a hollow shell of radius $$R=(\frac{3M}{4\pi\rho})^{1/3}$$, where $$\rho$$ is the cosmological energy density, so it corresponds to a total "missing" mass $$M$$.

The point mass is pulled away from the hollow shell because there is more mass in the other direction. One immediate complication is that the hollow shell will not move, because since it is massless, the (nongravitational) drag from collisions with the background material will hold it in place (relative to the rest frame of the background). However, that is not a serious obstacle: you can simply fix the shell to the point mass through nongravitational means.

Maximum gravitational acceleration is achieved by keeping the point mass right at the edge of the shell, so the distance between the "masses" is $$R$$. In the subhorizon regime ($$R\ll c/H$$), the missing mass gives rise to a gravitational acceleration $$\frac{GM}{R^2}=\left(\frac{4\pi}{3}\right)^{2/3} G M^{1/3}\rho^{2/3}.$$ The net peculiar acceleration would be $$\dot{v}=\left(\frac{4\pi}{3}\right)^{2/3} G M^{1/3}\rho^{2/3}-Hv-f_\mathrm{drag},$$ where $$-Hv$$ is the "Hubble drag" (responsible for $$v\propto 1/a$$ in the absence of peculiar accelerations) and $$f_\mathrm{drag}\sim \rho v^2 R^2/M\sim v^2\rho^{1/3}/M^{1/3}$$ (up to some numerical coefficient) is the ordinary nongravitational drag from the background. Note that for flat cosmologies, $$H\sim (G\rho)^{1/2}$$.

You could fix the cosmology and the numerical coefficients and solve this, but the key observation is that the drag terms all scale with velocity, while the gravitational acceleration does not. So at any fixed time, there will be a nonzero equilibrium velocity.

Of course, there is no free energy or momentum. The contraption is just exchanging energy and momentum with the background. You could do that through nongravitational means as well, if you wanted.

• What about the curvature around the underdensity? Is it the same as for negetive mass in vacuum in the context of Alcubiere drive? Commented Jul 27 at 2:01
• The spacetime in the hollow shell is flat, whereas outside the shell it is curved. I suppose you could say the difference between them is like if you added a negative mass density.
– Sten
Commented Jul 27 at 3:43
• I got there by a simple simetry argument. The problem is that while with negative mass in vacuum conservation of energy and momentum is obvious, with a hole in false vacuum it seems different. I can not understand where does the energy come from. It doesn't cool the environment, the scheme works at 0K. Considering it's reactionless drive with effective antigravitational field I wouldn't be too surprised if the system is not time translation simetrical in the context of Noether's theorem. I'm trying to figure that out now. Commented Jul 28 at 15:39
• It doesn't work at 0K because it would accrete a cold background.
– Sten
Commented Jul 28 at 16:32
• OD would accrete everything around it including UD, UD would blow away everything including OD. It is not true that underdensity wouldn't move, everything falls at the same rate. It still works at 0K. Commented Jul 30 at 12:38

... overdensity attracts underdensity and underdensity repels overdensity ...

Yes, in an empty universe a pair of matched positive and negative masses initially at rest would be self-accelerating indefinitely. And within cosmological perturbation theory gravitational effect of underdensity is akin to that of a negative mass while overdensity behaves like positive mass (if we ignore higher order nonlinear effects).

But this perturbation must be added to the unperturbed gravitational field. And the initial premise that

In the early universe when there was uniform H/He gas everywhere, gravitational field was close to 0 everywhere.

is certainly wrong. Early universe filled with uniform gas is well described by an FLRW metric, a solution of Einstein equations that is curved (i.e. has nonzero gravitational field) even when purely spatial geometry is flat. Even if we use Newtonian gravity to describe such situation, gravitational field would still be nonzero.

Also keep in mind that attraction/repulsion is about acceleration, second derivative of position. In the early universe the gas is expanding in all directions so gravitational attraction of overdensity region may slow that expansion but (depending on its size and magnitude) may not even reverse it. (And the perturbation that includes not only densities but also fine-tunes velocities would be much more contrived and unlikely to spontaneously appear).

Note, that there is a solution of Einstein equations that indeed has zero Newtonian gravitational field while having a constant positive matter density and zero initial velocity everywhere: Einstein's static universe. There the gravitational effect of uniform mass density is balanced by fine-tuned value of cosmological constant. If we perturb this solution by placing a region of overdensity near region of underdensity, then indeed at first such a pair would be self-accelerating initially, until the perturbations spread to other regions so that there would many more regions of over/underdensity besides initial two. But this is just one of the aspects of the instability of Einstein's static universe: small perturbation tend to grow.

• You are correct. However, FLRW metric is closely aproximated by Einstein's static universe if we're talking about meter sized objects over a period of hours, or years or whatever, in the early universe. It is certanly true that perturbations spread as they are very spread today and incredibly unfortunatelly this mode of propulsion is completelly useless today unless your masses are bigger than 200Mly. Commented Jul 11 at 13:08
• FLRW metric is closely aproximated by Einstein's static universe No. ESU has zero Hubble parameter and zero acceleration, while generic FLRW (and certainly the gas phase of our Universe) does not. I added discussion of acceleration vs. velocity to the answer. Commented Jul 11 at 14:10
• Right, but the change in volume goes to zero as change in cosmic time goes to zero. So if we start at 500ky and end in 500ky+1y the universe barely expanded and if the overdensity has 100kg of excess mass it barely influences local expansion. That's why I'm saying ESU is the limit of FLRW. Yet 1y of runaway acceleration is gonna give substantial speed boost. Commented Jul 12 at 0:02

This can be used for reactionless propulsion. Am I missing something or have I discovered (effective) antigravity?

You are forgetting the role of the mind in all of this, which, a unified field theory of mind is essential to the understanding of antigravity and its applications. So no, you do not have a reactionless drive nor a warp drive here.