Where Light Can Escape A Black Hole

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Maybe you've read somewhere that particles can escape a black hole via quantum tunneling, Hawking radiation or pair production; however, according to classical physics, there is no way light can escape--or is there?

Light escapes a gravitational potential if it has sufficient escape velocity. In terms of energy, a photon's kinetic energy must exceed the gravitational potential energy. A black hole's event horizon, whose distance from the center is the Schwarzschild radius (r), is the place where kinetic energy equals potential energy.

Theoretically, if a photon passes through the event horizon toward the black hole's center, it can't escape. You will note that the second equation above assumes a photon has mass (bar-m). This should be converted to a mass equivalent: fh/c^2 (f=frequency; h=Planck's constant; c=light speed; G=Newton's constant; m=black hole mass). We can now derive the Schwarzschild radius:

Houston, we have a problem! The second equation above assumes that a photon's kinetic energy is (1/2)fh. Electromagnetism has kinetic energy (photons) and potential energy (which involves charge). If photons are kinetic energy, then it only makes sense to use a photon's total energy (fh) when deriving the radius (r).

If (1/2)fh equals the gravitational potential energy, then fh must be greater than the gravitational potential energy--and the photon should be able to escape.

The math above indeed shows that fh is greater than the gravitational potential. There is another consideration: does the photon have more than zero energy? The last equation above shows that fh would equal (1/2)fh if fh is zero. When a photon enters a gravitational field it gains energy or frequency (f). When it tries to leave, it loses energy or frequency (f). The equations below demonstrate this (d=distance the photon travels; g=gravitational acceleration):

The total frequency (ft) of the photon is the sum of the initial frequency (fo) plus or minus the change in frequency (triangle f). How much energy the photon gains or loses depends on its angle of entry or exit relative to the unit normal line (n). If the photon is parallel to the unit normal (n), cos(0) equals 1, the photon will lose a maximum amount of energy when exiting a black hole. If the photon leaves at an angle, it will lose less energy.

So, if a photon enters the event horizon it will gain energy. When it tries to leave, it will lose energy. If it has more than zero energy, will it escape? Let's compare the escape velocity (c) to the gravitational equivalent. As you recall, we derived this inequality:

Let's multiply both sides by C^2/fh, then simplify:

Since c is greater than .707c, the photon will escape. In fact it will escape from any point that is outside the radius of Gm/c^2--one half the Schwarzschild radius.