In between two singularities, does a particle experience gravity?

If a particle is exactly at the midpoint between two singularities, will its net experience of the gravitational force (ignoring the rest of the universe, natch) be zero? I guess it ought to be assumed to be a point particle: I get the feeling that hadrons' internal structures aren't completely uniform and so their quarks would experience slightly different gravitational pulls as they are not quite at the exact same point. Or something like that, blah blah. Anyway, let's go with an electron for argument's sake.

Does the curvature of spacetime at the midpoint return to zero because their effects cancel? Does that mean that, in between two black holes, a region is formed within the event horizons for which the singularities are naked due to the opposing gravitational fields canceling out?

And similarly, with the "ringularity" of a Kerr black hole, is there a singularity that is naked to anything within that disc of spacetime? Though at least that region itself would be contained beyond and hidden by the event horizon, I suppose.

It sounds fun - to observe a naked singularity you just have to be bold enough to pass between two black holes that are sufficiently close.

  • $\begingroup$ For a particle on a free falling trajectory only the first order of gravity vanishes, the second order, which is what causes "spaghettification" is non-zero, so there will be an effective tugging force between the two singularities and a pressure term from the direction perpendicular to their connecting line. No need to go to general relativity, though, the very same thing already happens in Newtonian gravity. It's a real problem for those folks who are trying to generate microgravity environments, e.g. on the ISS. $\endgroup$
    – CuriousOne
    Dec 22 '15 at 4:29
  • 2
    $\begingroup$ What do zero gravitational-force points have to do with naked singularities? $\endgroup$ Dec 22 '15 at 4:39
  • $\begingroup$ Because naked singularities are not hidden by an event horizon. The event horizon is produced by the overwhelming gravitational force warping spacetime so even photons can't escape. If the net force is zero, then spacetime wouldn't be warped at that point. Or that's my logic, and I was trying to find out if it were right. $\endgroup$
    – Peter Zed
    Dec 22 '15 at 18:48
  • $\begingroup$ I'm not familiar with much higher physics, CuriousOne. So spaghettification will occur even without a net force? And it occurs in the direction of the pressure term (perpendicular to the connecting line)? Is it possible to simplify what happens, even if the analogy isn't comprehensive and wholly accurate? $\endgroup$
    – Peter Zed
    Dec 22 '15 at 18:52
  • $\begingroup$ Ah, gotcha. (Note, put an 'at-symbol' @ ahead of a person's name to 'mention' them --- sending them a notification. i.e. @CuriousOne). Thanks for clarifying. $\endgroup$ Dec 22 '15 at 20:26

From your comment:

The event horizon is produced by the overwhelming gravitational force warping spacetime so even photons can't escape. If the net force is zero, then spacetime wouldn't be warped at that point.

The Event Horizon is the surface within which nothing can escape. It doesn't matter if there are zero-force points within the horizon, because even if there is no force at that point there is still too-much-force-to-escape all around it. So unfortunately, this is not a way to produce a naked singularity (or perhaps, fortunately?!)

Note that an observer which falls into a BH will observe the singularity without a problem --- they'll soon fall into it, as-well. A singularity is only 'naked' if there is no event horizon surrounding it at all.

  • $\begingroup$ It's not just at zero-force points within the horizon. If you bring the two close enough- say, so that they're just sitting at the edge of each other's event horizons- then there would be a whole region of "low enough"-force points where the gravitational pulls would partially cancel. If the physics is that simple to formulate, which I suspect it isn't (for example that "first order" gravitation business is not something I even grasp). $\endgroup$
    – Peter Zed
    Dec 23 '15 at 0:29

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