When a particle falls through the event horizon, it cannot escape the black hole anymore, as doing so would need a speed greater than c.

Is the above statement valid for all observers? For an asymptotic observer it sure is true, as the coordinate speed of a photon past the event horizon is (in natural units)

$$v = \left(1 - \frac{2M}{r}\right)\tag{1}$$

where $r$ is the distance from the black hole, and $M$ its mass.

Now, if the photon were to move from some $R$ such that $0< R < 2M$ to $r = 2M$, it would take infinite time to do so.

For a particle that is not a photon, the same holds for an asymptotic observer. However, in the particle frame, calculating the proper time for the particle to move from $0 < R < 2M$ to $r = 2M$, we get$†$

$$\Delta\tau = \frac{4M}{3}\left[1 - \left(\frac{R}{2M}\right)^{3/2}\right]\tag2$$ which is finite. Yet, its future light cone is contained within $r \leq 2M$ (even in Kruskal-Szekeres coordinates)!


can the particle escape the black hole in its own frame?

$\dagger$Consider the outgoing orbit equation: $$\left(\frac{\mathrm{d}r}{\mathrm{d}\tau}\right)^2 = E^2 - \left(1 - \frac{2M}{r}\right)\left(1 + \frac{L^2}{r^2}\right)$$ where $E$ is the energy per unit mass and $L$ is the angular momentum per unit mass. For a radial approach, $\mathrm{d}\phi = 0$, thus $L = 0$. If $E = 1$(fall from rest at infinity), we have $$\mathrm{d}\tau = \frac{\mathrm{d}r}{\left(\frac{2M}{r}\right)^{1/2}}$$ integrating gives $(2)$.

Note: the above is a modification of an infalling particle calculation taken from Schutz (2008).

$\ddagger$In the diagram linked, any future light cone for a particle that crossed the $r = 1$ line is entirely contained inside the horizon, as the future light cone is not distorted and is inside the $\mathrm{II}$ region.

  • $\begingroup$ Also, note that this question is NOT a duplicate of Why is a black hole black? nor of Irreversibility of Hawking radiation emission and Noether's theorem, as this question does not involve reversibility and is asking specifically about the escape of a particle in its own frame, not in the frame of a distant observer. $\endgroup$
    – user140561
    Apr 9, 2017 at 1:34
  • $\begingroup$ For context: the above question grew out from a comment discussion(see the comments in Ari's answer) in this question, where it is discussed the possibility of a particle escaping the black hole in its own frame. $\endgroup$
    – user140561
    Apr 9, 2017 at 12:13

1 Answer 1


No. Since the statement "escape from the black hole" can be rephrased in term of physical quantities like proper length, there is no dependence from the reference frame. If it's true in one reference frame, it must be true in every other one.

On top of that, the black hole is a GLOBAL concept, referring to the causal structure of the spacetime. Remember that given a spacetime M the definition is:

$$Black \,Hole \equiv \left[ M - J^-(\mathscr{I}^+) \right]$$

In words, the complementary of the causal past of future null infinity. The causal structure doesn't care about reference frames.


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