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”.


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 $r_0(\alpha)$. enter image description here

  • $\begingroup$ 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 $\endgroup$
    – safesphere
    Apr 1, 2023 at 3:03
  • $\begingroup$ 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. $\endgroup$
    – JanG
    Apr 1, 2023 at 8:48
  • $\begingroup$ inis.iaea.org/search/search.aspx?orig_q=RN:32008938 $\endgroup$
    – safesphere
    Apr 1, 2023 at 13:02
  • $\begingroup$ 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. $\endgroup$
    – JanG
    Apr 1, 2023 at 14:50
  • $\begingroup$ The luxon ($\delta_1=0$) with energy zero has stable circular orbit on expanding outwards horizon. Can luxon have zero energy? $\endgroup$
    – JanG
    Apr 2, 2023 at 7:59

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – JanG
    Apr 3, 2023 at 18:24
  • $\begingroup$ Correct should be $r=1/\sqrt{K}$. $\endgroup$
    – JanG
    Apr 4, 2023 at 6:11
  • $\begingroup$ 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 $\endgroup$
    – Dale
    Apr 4, 2023 at 10:43
  • $\begingroup$ My knowledge about tachyons is rudimentary. However, I still look for the physical interpretation of mathematical results above. $\endgroup$
    – JanG
    Apr 4, 2023 at 11:22
  • $\begingroup$ I don’t think there is a physical interpretation. You can draw transcendent tachyon lines in any hypersurface. $\endgroup$
    – Dale
    Apr 4, 2023 at 11:32

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