Would a tachyon be able to escape a black hole?

Question

Or at least escape from a portion of the hole inside the photon horizon?

Answer 1

Your question only really makes sense for a localized tachyon, i.e. one whose wavefunction in position space is constrained to a finite region of space (i.e. has compact support) because that is the only kind that will "fit" inside a horizon of a black hole. And the answer to your question for this kind of tachyon is that it cannot escape a black hole.

The edges of a localized tachyon fulfilling the Klein Gordon equation with an imaginary mass parameter can only spread at a speed less than $c$, even though the plane waves (momentum eigenstates) that make it up have phase velocities greater than $c$. That is, if we plot the region where the wavefunction is nonzero, the boundary of this region can only grow at a speed less than $c$. This is a consequence of the Paley-Wiener theorem, which shows that the Fourier transform of any function with compact support cannot itself have a compact support and must include arbitrarily high frequency components. This is elegantly summarized in John Baez's Article "Do Tachyons Exist?". The theory making use of the Paley-Wiener theorem is summarized in QMechanic's answer to "Do Tachyons Move Faster than Light?.

Since the disturbance of a localized tachyon cannot spread faster than $c$, it therefore cannot escape the inside of a black hole's event horizon. In concluding this, we also need to assume (or at least I will, because I only know classical General Relativity) that the inside of a black hole is exactly as e.g. the Schwarzschild or Kerr solutions (in the Kerr case we need to limit the angular momentum so that there is an event horizon) to the Einstein field equations would describe a black hole: no talk of firewalls or other speculative recent quantum phenomena. We also need to assume it is valid to simply transcribe a solution to the Klein-Gordon (or other suitable wave equation) onto the spacetime inside a black hole. So these ideas will apply for a big black hole, where the spacetime inside the horizon is of very low curvature compared to the scale over which the tachyonic field is nonzero.

Answer 2

Selene Routely's argument conflates group-velocity intuition with the rigorous, microlocal structure of hyperbolic propagation and misapplies the Paley–Wiener theorem to a context where it plays no causal role. In fact the tachyonic Klein–Gordon operator

$$ \Box_g\phi - |m|^2\phi=0 $$

on any globally hyperbolic Lorentzian manifold remains a normally hyperbolic operator whose principal symbol is identical to that of the massless wave operator. By the standard theory of hyperbolic partial differential equations, the domain of dependence property guarantees that any compactly supported Cauchy datum can only influence points within its future light cone at speed $c$. The imaginary mass term alters the dispersion relation

$$ \omega^2 = |\mathbf{k}|^2 - |m|^2 $$

and indeed permits phase and group velocities in excess of $c$, but neither quantity governs the propagation of singularities or the leading edge of support. That role is dictated exclusively by the characteristic set determined by the vanishing of the principal symbol, namely the null cone.

The invocation of the Paley–Wiener theorem is therefore misplaced. That theorem addresses the analytic continuation and decay properties of the spatial Fourier transform of functions with compact Euclidean support. It says nothing about the finite-speed domain-of-dependence that arises from the Lorentzian signature of the d’Alembertian. In curved spacetime one instead relies on the Hadamard parametrix construction and standard energy estimates, which ensure finite propagation speed to prove that singularities propagate strictly on null geodesics. No invocation of entire-function bounds or high-frequency tails is required to see that the field’s wavefront set cannot cross the event horizon.

Moreover, one must distinguish between kinematic superluminality (group velocity exceeding $c$ for certain Fourier modes) and dynamical causality. In relativistic field theory the true “signal velocity” is the front velocity, which in every well-posed hyperbolic system coincides with the maximal speed of the characteristic cone. For the Klein–Gordon operator tachyonic or not that maximal speed is exactly $c$. Consequently nothing constructed from solutions of

$$ (\Box_g - |m|^2)\phi=0 $$

can carry information or support beyond the event horizon once initialized inside it.

Finally, transcribing flat-space momentum eigenstates into a Schwarzschild or Kerr interior neglects the breakdown of global Killing symmetries and the subtleties of horizon crossing. One cannot simply Fourier-decompose into plane waves and expect those modes to maintain their asymptotic dispersion properties deep in curved geometry. A proper treatment uses hyperbolic PDE theory on Lorentzian manifolds, which again enforces that the horizon is a true causal barrier for any wavefront. In sum, the impossibility of tachyonic escape from a black hole follows not from Paley–Wiener but from the immutable hyperbolic character of the relativistic wave equation and the consequent finite front velocity equal to $c$.

Nonetheless, despite tachyons (as noted above) superluminal phase or group velocities, tachyonic fields obey a normally hyperbolic wave equation whose characteristics lie on the light cone. The “front velocity” that governs the leading edge of any compact-support disturbance is exactly c. Since the event horizon is a causal boundary for all signals traveling at or below c, no tachyonic excitation, no matter how one assembles its momentum modes, can carry information or support from inside the horizon to the exterior. In short, tachyons cannot escape a black hole.

Answer 3

Yes. Since faster-than-light travel is required to leave the black hole, and tachyons apparently propagate faster than light, such a thing would be possible. If tachyons existed...

Answer 4

Yes, it would. Tachyons are a hypothetical object that can travel faster than light. They also require infinite energy to slow down as they grow faster the more energy they lose. A tachyon positioned right may as well get stuck in the center for a Planck or two I don’t know but that would speed it up A L O T. Tltr: the closer a tachyon is to the center of a black hole, the faster it is.