Why the astrophysical Black Hole Candidates are not rotating black holes

TL;DR

This paper challenges the conventional view that astrophysical black hole candidates are rotating Kerr black holes, showing that under certain metric properties, such objects cannot have positive mass if they are rotating, suggesting they may be non-rotating or not black holes.

Contribution

The paper presents a novel theoretical argument based on metric properties that questions the rotation and mass conditions of black hole candidates.

Findings

Rotating black hole candidates with positive mass cannot exist under the specified metric property.

Non-rotating black holes (Schwarzschild) with positive mass are possible if rotation is zero.

Recent observations align with the idea that these candidates may not be true black holes.

Abstract

It is believed that the basic component of the central engine of quasars, micro-quasars, and energetic Gamma Ray Bursts are the rotating or the Kerr Black Holes (BH)[1]. But by using a generic property[2-4] of the metric components of a stationary axisymmertic rotating metric in its standard form}, namely, g_phi \phi = sin^2 theta g_theta theta, where phi is the azimuth angle and theta is the polar angle measured from the axis of symmetry, we have found the unexpected and surprising result that (i) in order to have a mass of a Kerr BH m ge 0, it is necessary that its rotation parameter a=0 and if one insists for an a ge 0, one must have m le 0! Thus if the suspected Black Hole candidates with m >0 are really rotating they cannot be BHs at all which is in agreement with some detailed analysis of recent observations[5-8]. However, if it is assumed that such objects are strictly non-rotating, they could be non-rotating Schwarzschild BHs (a=0) with m ge 0 if we ignore the physical difficulties associated with the existence of such objects. This result calls for new theoretical efforts to understand a vast range of astrophysical phenomenon. If one derives the Kerr Metric in a straightforward manner by using the Backlund transformation, it is seen that a=m sin phi. This relationship confirms that a=m=0 for Kerr BHs.