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White holes, if they exist (unlikely but still fun to think about), are regions of space that can't be entered.

They are time-reversals of black holes (GR is time-symmetric). However, a time reversal of gravitational attraction is still gravitational attraction. This leaves us with a paradox: a region of spacetime that is both attractive and un-enterable.

In light of this conflict, what would happen to an observer falling toward a white hole from rest at a large distance?

Kevin Kostlan
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    A general piece of advice would be to learn how to interpret Penrose diagrams. Personally, I can never make sense out of any of this kind of stuff without drawing a Penrose diagram and staring at it. I have a couple of free online books that introduce Penrose diagrams. With almost no math: http://www.lightandmatter.com/poets/ . With more math: http://www.lightandmatter.com/genrel/ . Look in either book's index under "Penrose diagram." –  Nov 12 '19 at 20:28
  • Ben’s advise above would make sense, if anyone could draw a diagram for an attractive white hole (which has not become a black hole), but no one can: https://physics.stackexchange.com/questions/740527 – safesphere Nov 18 '23 at 21:59

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Short answer: Fall of an observer toward white hole would convert this white hole into a black hole.

Longer answer. Let us consider a nonrotating white hole that is not currently radiating. The metric outside of this white hole would be the Schwarzschild solution, right until the moment that white hole “explodes”, after which there would be a cloud of “explosion debris” (which could be anything: an expanding cloud of gas, a cloud that would later collapse into a black hole, some radiation and a white hole of smaller mass, a pair of white holes etc.). An inertial observer falling toward this white hole would be moving along a geodesic of Schwarzschild metric. And since by the clock of a far away stationary observer the falling observer would never cross the surface of antihorizon $r=2M$ (in units $G=c=1$), this falling observer would remain outside the white hole until the moment when white hole explodes. After that the falling observer would either be incinerated by the explosion or (if she survived) would continue moving in around explosion debris.

The above description is only possible if we completely disregard the mass of the falling observer. If this observer instead has a mass $m$ (which we consider very small in comparison with the mass $M$ of the white hole), then when the falling observer has radial coordinate $r=2(M+m)$ an event horizon would form enveloping both observer and the white hole. The falling observer and the white hole merge, creating a black hole with mass $M+m$. Timescale for emergence of event horizon is essentially time (by outside clock) needed for an observer to reach $r=2(M+m)$ since the moment of crossing some characteristic surface outside the white hole, say a photon sphere $r=3M$: $$ \tau \approx 2 M \,\ln \frac{M}{m}. $$

This conversion of a white hole into black would happen for any form of mass–energy approaching the antihorizon, no matter how small it is. For example, a typical CMB photon at $\approx 3\,\text{K}$ would convert a white hole with the mass of $10^6\,M_\odot$ in a matter of days after crossing the photon sphere. So a white hole of stellar mass or larger cannot survive for any significant periods of time in the present universe: white holes are unstable against conversion into black holes.

This instability was discovered by Eardley in the 70s:

This white to black conversion scenario has a loophole: white hole could be continuously radiating. Then particles falling toward white hole could be deflected (radiation pressure) and white hole antihorizon could be receding fast enough that no mass–energy would approach it close enough to convert it into the black hole.

A.V.S.
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  • Interesting to consider https://www.sciencedirect.com/science/article/pii/S0550321316301274, which claims that event horizons are prevented because of Hawking radiation. – Kevin Kostlan Nov 13 '19 at 13:31
  • "when the falling observer has radial coordinate $r=2(M+m)$ an event horizon would form enveloping both observer and the white hole" - Wouldn't this process be gradual rather than abrupt? As the falling observer is approaching the horizon, the horizon is expanding toward the observer until they asymptotically meet at $r=2(M+m)$. Then, in any external coordinates, the falling observer will never cross the expanded horizon due to the infinite time dilation. If so, in which frame do you see the horizon "enveloping both observer and the white hole"? Thanks for your insight! – safesphere Dec 07 '19 at 22:53
  • Great answer but limited to one type of white holes defined as “a compact, long-delayed region of the big bang”. With the paper behind a paywall, the physical meaning of this is unclear. It these regions are compact and delayed, they are not FLRW-expanding and should become black holes on their own without anything falling to them. Even if somehow they become Oppenheimer-Snyder white holes, the logic of the “enveloping horizon” is flawed, so at least based on the abstract and description the linked paper doesn’t seem credible. – safesphere Nov 18 '23 at 19:48
  • To expand on why the “enveloping horizon” logic is wrong, when an object $m$ (let it be a small black hole for example) approaches a black hole $M$, the spacetime between them is curved stronger. Since the position of the horizon depends on the curvature, the horizon of $M$ expands toward $m$ until it “meets” $m$ at $r=2(M+m)$. No “enveloping horizon” forms outside of $m$. Here is a mathematically correct numerical gravity video: https://m.youtube.com/watch?v=Y1M-AbWIlVQ&feature=youtu.be - Finally, since the external metrics are the same, $M$ can be a white hole with exactly the same result. – safesphere Nov 18 '23 at 20:32
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In general, there is no gravitational attraction. GR defines no such concept - at least not in general. You have curved spacetime on which test particles move on timelike geodesics ("straight lines in 4D spacetime that go from past to future") and GR will tell you how is the spacetime curved. Sometimes, gravitation can be even repulsive as is the case near a black hole singularity of charged black hole, where there is repulsion proportional to charge of the black hole squared that acts on the neutral particles also (so we cannot interpret the repulsion as due to electromagnetic force and must admit, that repulsion is due to gravitational interaction)

Moreover. The point of white holes - at least those i came across studying simple black holes - is that they live in the past. You can think about them as the singularities from which universe started. Therefore, it makes perfect sense you cannot fall into them, since for that you would need to go back in time. But if i am wrong in this case and there are different white holes that do not offer themselves for this interpretation, i would be happy if (the reader) could let me know in the comment.

Umaxo
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  • (1) “there is no gravitational attraction. GR defines no such concept” - It is defined as the gravitational acceleration in the coordinates of a remote observer. The direction of the acceleration defines attraction vs. repulsion. (2) “white holes - at least those i came across studying simple black holes - is that they live in the past” - This is only one specific type based on the maximally extended solution, which is unphysical. You also can take a regular Schwarzschild solution and reverse it in time to obtain another type of white holes that can exist in the present and future. – safesphere Nov 18 '23 at 17:18