The negative mass Schwarzschild metric has no event horizon.
Why isnt there a particular radius in which spactime flows outwards at the speed of light? This would imply a region of the solution for which exterior particles could not cross.
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asked Nov 1, 2023 at 21:08
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$\begingroup$ The horizon at 2GM/c² would be at a negative radius $\endgroup$– YukterezCommented Nov 2, 2023 at 1:51
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$\begingroup$ “particular radius in which spactime flows outwards at the speed of light” This is characterization of positive mass Schwarzchild metric in outgoing Gullstrand–Painleve coordinates that covers both outside and white hole regions of maximally extended Schwarzchild. $\endgroup$– A.V.S.Commented Nov 2, 2023 at 4:29
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$\begingroup$ @Yukterez What is the meaning of a negative radius in the solution? Is r even allowed to take negative values? $\endgroup$– Independent PhysicsCommented Nov 6, 2023 at 0:13
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$\begingroup$ Manuel asked: "Is r even allowed to take negative values?" - negative r is probably more allowed than negative M, at least in the maximally extended spacetimes the negative r is the extension, while the negative M is said to violate the energy condition. So negative r and M might both be unphysical, but since you asked we're speaking hypothetically. $\endgroup$– YukterezCommented Nov 6, 2023 at 5:29
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2 Answers
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With negative $\rm M$ in Schwarzschild spacetime the lightlike geodesics in static coordinates are in natural units $\rm dr/dt=\pm (r+2)/r$.
In the $\rm \{t,r\}$ spacetime diagram below are the in- (blue) and outgoing (red) photon geodesics for a negative mass singularity at $\rm r=0$ (for the calculation click on the image):
As you can see they can escape from and travel toward $\rm r=0$, and due to the inversed gravitational time dilation they do so with $\rm |dr/dt|>c$ (with positive mass the shapirodelayed coordinate velocity would be slower than $\rm c$, while relative to local observers the photons always have $\rm v=c$).
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$\begingroup$ Why isnt there a region in which spacetime flows outwards at the speed of light (similar to white hole horizon) by which light cannot enter? $\endgroup$ Commented Nov 11, 2023 at 15:59
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$\begingroup$ @Manuel - The white hole is always in the past, therefore nothing can flow there since nothing can flow to the past. If you fly towards where the white hole was you'll end up inside a black hole, not a white one, but you'll cross the horizon in finite proper time, see here and here: "White holes attract matter like any other mass. The white hole event horizon in the past becomes a black hole event horizon in the future, so any object falling towards it will eventually reach the black hole horizon". $\endgroup$– YukterezCommented Nov 11, 2023 at 18:42
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$\begingroup$ The white hole metric is the same as the black hole metric, while the negative mass solution is different. In Gullstrand Painlevé coordinates where the gtr term reflects the freefall/escape velocity ±√(2/r) for a black hole it would be ±√(-2/r) for a negative mass, which is imaginary, in contrast to a white/black hole. $\endgroup$– YukterezCommented Nov 11, 2023 at 19:16
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$\begingroup$ @Yukterez Your chart is technically correct, but overlooks an important fact that ingoing geodesics are hard, if possible at all, to follow. In a black hole case, a fairly wide beam of light will all end up inside. A white hole though is infinitely repulsive, so any photon with an infinitesimal deviation from the ingoing geodesic will be deflected away. And even with a zero initial deviation, the diffraction always creates some deviation, so the probability of a photon or anything else to actually hit the negative singularity is zero. $\endgroup$ Commented Nov 17, 2023 at 22:25
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$\begingroup$ @safesphere - That is true, but you can get arbitrary close by making the initial wavelength arbitrary small, or with a particle by making the initial velocity arbitrary close to the speed of light, and you'd have to aim exactly radial, but the original question was about the horizon, which there is none. $\endgroup$– YukterezCommented Nov 17, 2023 at 23:22
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Bc Goo = 1+2Gm/(c^2*r) is never zero. No negative mass is known; anti matter behaves like regular matter in gravity.
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$\begingroup$ What does "Bc Goo" mean? BTW, you can use the MathJax version of LaTeX on this site. $\endgroup$– PM 2RingCommented Nov 6, 2023 at 5:24
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$\begingroup$ The answer is correct though, despite the sign error and the terrible notation. He meant "because $\rm g_{00}=1-2GM/(c^2 r) \neq 0$", since $\rm M<0$ $\endgroup$– YukterezCommented Nov 6, 2023 at 5:40