# What is the reasoning behind the idea that light cannot escape from a black hole? [duplicate]

According to the definition, light cannot escape from a black hole.

How did scientists deduce that light cannot escape from a black hole?

• Because (as you noted)... it is the definition of a black hole. Not a lot of deduction there. Commented Feb 26, 2015 at 18:21
• Duplicate of Why is a black hole black Commented Feb 26, 2015 at 19:55
• @JohnRennie Dang. I was hoping nobody found a duplicate. I put a lot of love into my answer. Sigh, que sera sera
– Jim
Commented Feb 26, 2015 at 19:59
• I don't think that's the same. A black hole could be black because it is so absorbent of light. Commented Feb 26, 2015 at 20:09
• @Jiminion: Anything that absorbs light or any other form of energy will heat up and start emitting black body radiation. A black hole can only be black if nothing can get out. Commented Feb 27, 2015 at 6:21

Scientists asked the question "How does a body of arbitrary mass affect spacetime around it?" To answer this question, they took Einstein's General Relativity and applied it to the description of a spherically symmetric spacetime (meaning you can rotate any way you like and it looks the same) centered on a body of arbitrary mass, $M$.

I'll spare you the dirty details of this calculation, but what they found it that spacetime can be interpreted (loosely) as "falling" towards a gravitating body. They also found that the direction of an object's motion through time for any path that led away from the gravitating body was given by the sign (positive or negative) of $$1-\frac{2GM}{c^2R}$$ where $G$ is Newton's Gravitational constant, $c$ is the speed of light, and $R$ is the radius of the body. What does this mean? If the sign is positive, that means that paths away from the body move forward in time (meaning we can follow them, since we only move forward). If the sign is negative, paths leading away from the body go backward in time, which means the only way to move away from the body is to go back in time; alternatively, it means any path you choose to take would lead towards the body. If the above expression comes out to be zero, then in GR that means that it is a path light would take. This means that only light could follow paths leading away from the body if the expression gives $0$ and not even light could move away if it is negative.

This is how we came to deduce the existence of a black hole. If $1-\frac{2GM}{c^2R}<0$, nothing can escape and if it equals zero, only light can escape. So they said, that means any body of mass $M$ that has a radius $R\le\frac{2GM}{c^2}$ would have a gravitational pull so strong that not even light could enter it and escape afterwards. Because this means that no light could ever come from such a body (and therefore it would look black), and because things fall into it but don't come back out (like a hole), they dubbed it a Black Hole

• So, basically the original discovery of black holes was completely theoretical [though they did find a lot of black holes later] Commented May 29, 2017 at 5:06
• @PhyEnthusiast yup, they only had the math at first that said black holes should exist. It wasn't until much later that we found real evidence that told us what we thought was reality
– Jim
Commented May 30, 2017 at 11:49

When the escape velocity of a body of mass is faster than the speed of light, you have a black hole.

This is because gravity affects light the same way it affects matter.

The speed of light is invariant, so it is not possible to accelerate it above $c$. From Newton's law we know that the escape velocity $v$ from any given mass is $v_{esc}=\sqrt{\frac{2 G M}{r}}$ so if the mass $M$ is high enough and the distance $r$ to the mass small enough $v>c$. Since nothing can have more than $c$ even light cannot escape from there. The border where $v$ approaches $c$ is called the Schwarzschildradius.

• It's convenient that the speed of light escape velocity for a black hole and Newtonian mechanics coincide this time. But it is just coincidence. Newtonian gravity is not valid near the event horizon of a black hole nor was this equation used to determine the Schwarzschild radius. You'll note that this Newtonian formulation says that you only need the magnitude of the velocity to be $c$ to escape. GR goes further to say that the direction must be radially outwards as well, else there is no escape
– Jim
Commented Feb 26, 2015 at 19:40
• I did not know that yet, thanks for completing my answer! Commented Feb 26, 2015 at 19:42
• @JimdalftheGrey just a coincidence? Are you joking? Commented Feb 26, 2015 at 20:01
• @SeñorO I'm not joking that explaining this through using Newtonian mechanics and gravity is not correct and that it may coincide but does not mean it should be treated as the proper way to approach the situation
– Jim
Commented Feb 26, 2015 at 20:03
• @SeñorO No, I'm sorry, Jim is right. A Newtonian black hole would be qualitatively different from a GR one. In particular, you could escape from inside a Newtonian black hole without ever going faster than light, just as you can escape from Earth without ever reaching escape velocity. And light emitted from inside a Newtonian black hole would leave, go a long distance away, and eventually turn around and come back.
– user10851
Commented Feb 26, 2015 at 21:39