2019
DOI: 10.1007/s10773-019-04242-0
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Can Quantum Particles Cross a Horizon?

Abstract: The prevalent opinion that infalling objects can freely cross a black hole horizon is based on the assumptions that the horizon region is governed by classical General Relativity and by specific singular coordinate transformations it is possible to remove divergences in the geodesic equations. However, the coordinate transformations usually used to demonstrate the geodesic completeness are of class C 0 , while the standard causality theory requires that the metric tensor to be at least C 1 . Introduction of C … Show more

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Cited by 6 publications
(4 citation statements)
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“…This means that particle equations have not horizon crossing plane wave solutions and we need to set adequate boundary conditions on the Schwarzschild sphere. In this case one obtains the exponential solutions with the complex phases that correspond to the absorptions and reflections of particles by the BH horizon [37,38]. In this model the event horizon surrounds an impenetrable sphere of the ultra-dense matter, reflected from which particles could obtain energy from the gravitational field and imitate some burst-type signals, like GRBs, FRBs, ultra-high energy cosmic rays, or the LIGO events [46].…”
Section: Discussionmentioning
confidence: 99%
See 2 more Smart Citations
“…This means that particle equations have not horizon crossing plane wave solutions and we need to set adequate boundary conditions on the Schwarzschild sphere. In this case one obtains the exponential solutions with the complex phases that correspond to the absorptions and reflections of particles by the BH horizon [37,38]. In this model the event horizon surrounds an impenetrable sphere of the ultra-dense matter, reflected from which particles could obtain energy from the gravitational field and imitate some burst-type signals, like GRBs, FRBs, ultra-high energy cosmic rays, or the LIGO events [46].…”
Section: Discussionmentioning
confidence: 99%
“…Both lagrangians give the same equations of motion if they take a constant value for the solution, i.e. two formalisms become equivalent only if we fix the condition (37). But, for the Schwarzschild case (1) this condition is satisfied only if r = r S .…”
Section: Schwarzschild Geodesicsmentioning
confidence: 99%
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“…Indeed, the three from six non-zero independent components of the mixed Riemann tensor for Schwarzschild metric blow up at the horizon. So, we consider a model where at the BH horizon, the spacetime is not Minkowskian (see, for example [91][92][93][94]). In summary, the extension of geodesics across the Schwarzschild horizon by singular diffeomorphisms presents difficulties, even at the classical GR level.…”
Section: Quantum Vacuum At a Horizonmentioning
confidence: 99%