# Is black hole event horizon a transcendent tachyon?

In static spherically symmetric spacetime with metric $$ds^2=e^{2\nu}c^2dt^2-e^{2\lambda} dr^2-r^2 (d\theta^2+\sin{\theta}~d\phi^2)\equiv \delta_1~c^2 d\tau^2 \tag{1}$$ the geodesic equation in equatorial plane ($$\theta=\pi/2$$) reads $${e}^{2\nu+2\lambda}~\frac{\dot{r}^2}{c^2}=k^2-{e}^{2\nu}~(\delta_{1}+\frac{l^2}{c^2 r^2})\equiv k^2-V(r) \tag{2}$$ where the constants of motion are defined as $$k\equiv {e}^{2\nu}~\dot{t},~~~l\equiv r^2~\dot{\phi}~~~.$$ Correspondingly to space-, null- and timelike geodesics $$\delta_{1}$$ can take values $$~-1,0,1$$ . For circular geodesic it applies $$\dot{r}=\ddot{r}=0$$. Hence, a geodesic of curvature radius $$r_{0}$$ has constants $$k^2=\delta_{1}\frac{e^{3\nu}_{0}}{e^{\nu}_{0}-r_{0}~\partial_{r}{e^{\nu}_{0}}},~~~l^2=\delta_{1} c^2 \frac{r_{0}^3~\partial_{r}{e^{\nu}_{0}}}{e^{\nu}_{0}-r_{0}~\partial_{r}{e^{\nu}_{0}}}~~~. \tag{3}$$ For the special geodesic defined through condition $$e^\nu(r_0)~=~0$$ they result in $$k^2=0,~~~l^2=-\delta_{1}~ c^2 r_0^2~~~. \tag{4}$$ The second equation seems to indicate that such a geodesic can be followed only by spacelike particle ($$\delta_1=-1$$) with zero rest energy ($$k^2=0$$) and non-vanishing momentum ($$l^2\geq 0$$) or timelike particle $$(\delta_1=1)$$ with imaginary angular momentum ($$l^2\leq 0$$).

Assuming the first case the effective Potential reads $$V=e^{2\nu}~(-1+\frac{r_0^2}{r^2}) \tag{5}$$ and satisfy $$V(r_0)=\frac{dV}{dr}(r_0)=\frac{d^2V}{dr^2}(r_0)=0. \tag{6}$$ Thus, that special geodesic represents the innermost circular stable orbit (ICSO).

However, there only known particles in General Relativity satisfying theses conditions are transcendent tachyons. While at the same time the equation $$e^{\nu}(r_0)=0$$ defines event horizon of the curvature radius $$r_0$$, the question arises if one could interpret the event horizon as a transcendent tachyon.

Although such interpretation would contradict the prevailing understanding of event horizon as an imaginary 2-sphere it would have some other merits. Particles colliding with transcendent tachyon reflect without energy change but with charge switch, see Fig.5 in “Particles beyond the light barrier”. That has some resemblance to so called "firewall" theory. Another interesting point is that transcendent tachyon geodesic defined by equation (4) is stable. The corresponding effective potential has minimum there and the whole process of black hole formation could be understood as spontaneous symmetry breaking phenomenon, see Fig. 4 in “Tachyons and Solitons in Spontaneous Symmetry Breaking in the Frame of Field Theory”.

Appendix

I wanted to examine parametric stability of perfect fluid (sphere) spacetime using geodesic equation and effective potential $$V(r)$$. The parameter is $$\alpha\equiv r_S/R$$. For the congruence of radial geodesics $$V(r)=e^{2\nu}$$. In case of Schwarzschild constant density star the potential for $$\alpha<8/9$$ (green line) has a regular ($$\sim r^2$$) minimum at the center that for $$\alpha=8/9$$ (red line) degenerates ($$\sim r^4$$) and transforms to local maximum for $$\alpha>8/9$$ (blue line). Quasi-statically, the black hole formation process looks like classical "pitchfork" catastrophe. The former central minimum moves outwards until $$\alpha=1$$ is reached. In terms of radial geodesics the local minima ($$r_0$$) are stable and the local maximum unstable. I wanted to see the corresponding stability of circular orbits starting at different $$r_0$$ (minimums) and their possible physical interpretation. After all, the minimums at $$r_0(\alpha)$$ represent moving event horizon of curvature radius $$r_0(\alpha)$$.

• Studies suggest that we may observe only the real part of tachyon transcendent mass, which is suppressed comparing to the rest mass of a corresponding bradyon. So transcendent tachyons are observed as luxons and the horizon is observed as null +1 Commented Apr 1, 2023 at 3:03
• Interesting, can you name some literature about it? I have supplemented my question by two references. I still try to understand the black hole formation process looking for parametric stability of static solutions.
– JanG
Commented Apr 1, 2023 at 8:48
• inis.iaea.org/search/search.aspx?orig_q=RN:32008938 Commented Apr 1, 2023 at 13:02
• Looks very reasonable. I will try to get pdf. By the way, I have just improved my argumentation by timelike particle with imaginary mass as the second possible explanation.
– JanG
Commented Apr 1, 2023 at 14:50
• The luxon ($\delta_1=0$) with energy zero has stable circular orbit on expanding outwards horizon. Can luxon have zero energy?
– JanG
Commented Apr 2, 2023 at 7:59

I think that the big point you are missing with this analysis is that the event horizon is a 3 dimensional submanifold or hypersurface, not a 1D worldline (geodesic or otherwise). The fact that a hypersurface contains some kind of geodesic does not make that hypersurface that kind of geodesic.

As a directly relevant counterexample, consider a simple electromagnetic plane wave. This is the prototypical null surface. It should be immediately obvious that a plane wave also contains “transcendent tachyon” geodesics. Nevertheless, the hypersurface is a null surface and it is physically formed by lightlike particles.

In fact, since tachyons in principle can accelerate, it is possible to describe any hypersurface as being composed of transcendental tachyons.

• I appreciate your answer. First, I am using geodesics to examine parametric stability of spacetime of perfect fluid in equilibrium. Second, I regard the hyperspace of constant $t$ as spheres bundle (not foliation!) with fiber $S^2$ and base $R$ labeled by $r\equiv 1/K$ where K is Gaussian curvature of $S^2$. Thrid, a transcendent tachyon does not move but rests (it is everywhere on event horizon sphere), similar to Higgs bosons - at least in my understanding of transcendent tachyon.
– JanG
Commented Apr 3, 2023 at 18:24
• Correct should be $r=1/\sqrt{K}$.
– JanG
Commented Apr 4, 2023 at 6:11
• Ok, but none of that changes my answer in the least. The reasoning given in the OP does not support the claim that the EH is a transcendent tachyon
– Dale
Commented Apr 4, 2023 at 10:43
• My knowledge about tachyons is rudimentary. However, I still look for the physical interpretation of mathematical results above.
– JanG
Commented Apr 4, 2023 at 11:22
• I don’t think there is a physical interpretation. You can draw transcendent tachyon lines in any hypersurface.
– Dale
Commented Apr 4, 2023 at 11:32

The real problems in this topic arise when you want to reconcile the events with quantum mechanics (qm). But to understand and appreciate the problems it is best to first ignore qm and learn about the details of gravity and black holes using only classical (i.e. non quantum mechanical) physics - as you are doing now. Indeed, one finds that black holes with matter present are unstable, because hatter will fall in while nothing can come out. Unless you add thermodynamics: entropy can only increase! But then you’ll find that qm is inevitable ...

• You're absolutely right about this point. I have even quoted you on Physics on this subject. What I'm trying to show is that in classical GR, the appearance of what we call a singularity does not mean the breakdown of Einstein's theory. We can study this situation using a simple mathematical model of a static perfect fluid sphere. In this case, Einstein's field equations reduce to a parametric linear differential equation.
– JanG
Commented Jun 20 at 14:12
• Using the effective potential, we can show that the so-called gravitational collapse can be seen as a quasi-static process of spontaneous symmetry breaking. If the matter at the center is subjected to an unlimited pressure, quantum mechanics comes into play, otherwise the pressure would be infinite. This was also the view of Einstein, who believed that spacetime singularities could only occur in matter, and that the theory of matter, quantum mechanics, would prevent them.
– JanG
Commented Jun 20 at 14:14
• The transient Oppenheimer-Synder model of gravitational collapse suffers from the simplification of matter as pressureless dust. In this case, the fluid particles follow geodesics with proper time as an affine parameter along them. Thus, the assumption determines the result obtained. For fluids with pressure, however, the situation is quite different. A static consideration of gravitational collapse is there possible, as the theory of gravastar proves.
– JanG
Commented Jun 20 at 16:10
• What would be the result of the instability, Gerard? Commented 2 days ago