# Why can't light escape from a classical black hole?

Photons do not have (rest) mass (that's why they can move at speed of "light").

So, my question is how the gravity of classical$^1$ black hole can stop light from escaping?

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$^1$ We ignore quantum mechanical effects, such as, Hawking radiation.

Black holes affect the causal structure of spacetime in such a manner that all future light cones within a black hole lie within the event horizon of it.

Although photons are massless they have energy and have to obey the geometry of a curved spacetime. Since all future lies within the event horizon, photons are trapped inside the black hole.

• Right! Asking why a photon can't escape from a black hole is exactly like asking why a photon can't travel from here to last Thursday: both are travel into the past. – Ted Bunn Apr 12 '11 at 14:17
• I've seen the shifted light cones in books for decades and they still don't answer the question. If I am in Earth's gravitational field, I am accelerating, not moving toward the center of mass. Consider an accelerating spaceship in flat space for simplicity - someone with a special relativity background should expect light in the + and - direction both move at c, although there is an asymptotic event horizon due to long term behavior. Note, this is inconsistent with a linearly 'tilted' light cone. – Alan Rominger Jun 4 '11 at 5:33

Even though photons have no mass, they are still affected by gravity. That's how we can see black holes - by the way they distort the light going near them.

The reason nothing can escape a black hole is because within the event horizon, space is curved to the point where all directions are actually pointing inside.

The escape velocity from within a black hole's event horizon is faster than the speed of light, hence light cannot go at that speed and thus cannot escape.

• ""Even though photons have no mass, they are still affected by gravity. "" Silly statement. By what means would "gravity" affect a photon, if not by mass? Photon have a mass, what they haven't is rest mass! – Georg Apr 12 '11 at 8:56
• For roughly the last fifty years, the vast majority of physicists have used the word "mass" to mean "rest mass." – Ted Bunn Apr 12 '11 at 14:16
• @Georg, gravity affects photons because gravity actually curves space. It's the same reason a feather and bowling ball would fall at the same speed (neglecting air resistance). Photons don't have mass, they have energy. That's a different point and indicates that the photon actually has its own gravitational field. – RightHandedMonkey Dec 11 '15 at 12:25
• Escape velocity is a terrible concept to use to explain this. There is no such thing as escape velocity for light. Escape energy would make more sense. – B T Mar 21 '17 at 1:16

Gravity is the force which bends the very fabric of Space time. During eclipse scientists have seen the light from distant stars which are near the Sun change their path. So it proves that light is affected by Gravity. Now that you know that light gets affected by gravity, you must also be knowing that the gravitational force of a Black-Hole is immense. As anything on earth needs to have a minimum velocity to overcome the gravitational pull of Earth (which is called escape velocity) is something that Man has been able to achieve, so our space ships and rockets reach Space. But the escape velocity required to overcome the gravitational pull of a black hole is greater than the speed of Light. And as we know that nothing travels faster than light, so Black Hole swallows anything and everything that comes near it including light.

• The concept of escape velocity simply doesn't apply to photons. – B T Mar 21 '17 at 1:16
• Okay. But light shows properties of both particle and waves. So shouldn't a particle have to overcome an escape velocity? – MK Singh Aug 26 '17 at 18:20

Here is a different explanation.

Due the equality principle standing on the surface of a planet and accelerating is equal.

Far from massive bodies, fixed proper acceleration leads to a hyperbolic trajectory in the space-time diagram. This hyperbola's asymptote is diagonal (approaching the speed of light.)

If you imagine this hyperbola you can see that if you shoot a beam of light towards the accelerating object beyond a certain distance, it will never reaches it. This is the Rindler-horizon, beyond it no light reaches you. If you accelerate with $a$ the Rindler horizon is at $c^2/a$ behind you.

The black hole's event horizon is analogous with this. If you hover over a black-hole you are in an accelerating reference frame, so Rindler-horizon exists at the event horizon (using the schwarzschild metric).

Rindler-horizon disappears if the observer stops acceleration. The observer near a black hole stops acceleration if it begins a free fall towards the black hole, it's movement becomes inertial so the event horizon should also disappear. But since gravity is not uniform, the event horizon won't disappear, but remains under the observer as it falls in.

The above statement about a photon having no "rest" mass is actually quite contrary to reality. As an object approaches the speed C its mass becomes relatively smaller and smaller, until it reaches the asymptotic light speed and becomes virtually massless, timeless, and spaceless. The more energy the photon has, the smaller the wavelength, higher the frequency, and higher the rest mass. This equation can be given by mv=hf/c. Technically, you yourself (and this entire planet) resemble but a group of photons to some frame of reference (for example, a galaxy going in the opposite direction to ours at a very high speed). I think gravity effects photons because while they are massless (by definition, when it is moving at the speed of light it has no mass) is because they are, even massless, a photonic bundle of energy.

## protected by Qmechanic♦Apr 12 '14 at 8:13

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