# Escape from black hole for different observers

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)!

So,

## 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.

• 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. – user140561 Apr 9 '17 at 1:34
• 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. – user140561 Apr 9 '17 at 12:13

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