Why light can't escape a Black Hole's gravity?

[morgsboi]

 

Only matter is affected by gravity. And if light is made up of matter, why won't it destroy anything it goes through?

 

Would this answer be correct:

 

Black holes warp space-time so much that any particle that crosses the event horizon finds that simply moving forward in time also moves it toward the singularity at the center of the black hole. The only way anything could escape the black hole would be if it could travel faster than light. You can view an event horizon as being defined to be the boundary between an escape velocity that is lower than c and an escape velocity that has to be greater than c.

 

 

Another question that I can't answer is: Would light bounce between the event horizon and the mass? Because Professor Stephen Hawking says that not even a black hole is truly black.

So even in the slightest, light should reflect off the black hole but only to get sucked in again as it can't escape the event horizon.

[questionposter]

Light is not matter, it has no mass, it's primary composition is just energy, and energy is another form of matter and vice versa. Both are effected by the fabric of space and they both effect the fabric of space. Light can't escape from a black hole simply because the gravity of a black hole is too powerful, it creates a gravitational well so steep not even light has enough speed to escape it.

 

As far as scientists can tell, nothing comes back from the event horizon except perhaps whatever the result is of black hole evaporation. Once light passes the event horizon, it doesn't come out. Black holes don't emit optical photons, so they are the color of black.

[Klaynos]

A quick comment, don't have time for more, sorry. But Energy is a property of stuff, not stuff itself, so saying photos composition is just energy is meaningless.

[granpa]

light may or may not be affected by gravity but it is certainly affected by gravitational time dilation.

 

time stops completely at the event horizon.

[IM Egdall]

Why light does not escape a black hole -- Recall gravitational red-shift: Say a beam of light in zero gravity has a certain frequency. This same beam of light will have a lower frequency when it is in a gravitational field. In other words, its frequency is stretched towards the red end of the spectrum in the presence of mass/energy. And this so- called gravitational red-shift is relative; the light's frequency is as seen as lower by an observer far away (in zero gravity).

 

 

So what about a black hole? This ultimate source of gravity produces the ultimate red-shift. Light's frequency inside the event horizon is stretched to zero.

 

Edited November 12, 2011 by IM Egdall

[morgsboi]

light may or may not be affected by gravity but it is certainly affected by gravitational time dilation.

 

time stops completely at the event horizon.

 

You say time stops, I don't quite understand how time itself stops. Otherwise the black hole wouldn't be doing anything within the event horizon.

 

Light is not matter, it has no mass, it's primary composition is just energy, and energy is another form of matter and vice versa. Both are effected by the fabric of space and they both effect the fabric of space. Light can't escape from a black hole simply because the gravity of a black hole is too powerful, it creates a gravitational well so steep not even light has enough speed to escape it.

 

As far as scientists can tell, nothing comes back from the event horizon except perhaps whatever the result is of black hole evaporation. Once light passes the event horizon, it doesn't come out. Black holes don't emit optical photons, so they are the color of black.

But how can even the most extreme gravity affect something that has no mass?

And according to Professor Stephen Hawking, even a black hole is not truly black so it must emit something.

[IM Egdall]

You say time stops, I don't quite understand how time itself stops. Otherwise the black hole wouldn't be doing anything within the event horizon.

 

 

 

Its relative. Time stops at the event horizon of a black hole as seen from far away. For an obserer passing through the event horizon, his/her time is running normally.

[Mystery111]

The simple reason can be explained in a simple way.

 

You require energy to leave the earths atsmosphere. This is because gravity is pulling your object (the mass of the earth) from your located origin. What if that mass became so dense that the radial force required to leave your origin became that of the speed of light?

 

Simply, a particle of light cannot escape something when the escape velocity exceeds the speed of light.

[granpa]

What if that mass became so dense that the radial force required to leave your origin became that of the speed of light?

 

Simply, a particle of light cannot escape something when the escape velocity exceeds the speed of light.

 

teh force of gravity at the event horizon is not infinite.

Edited November 14, 2011 by granpa

[Mystery111]

teh force of gravity at the event horizon is not infinite.

 

I nev'r said it was... is this a reply to someone else?

 

People need to qoute. It is hard to keep track

[granpa]

whether you said it or not people could get that impression from what you said.

 

I thought it was obvious who I was responding to

Edited November 14, 2011 by granpa

[Mystery111]

Then explain more clearer please. I responded to his question in two or three lines and I was much more clearer. How did my reply warrent yours. Please explain, before I get my crystal ball out

[imatfaal]

It is clear that it is not necessary for gravity to be infinite in order for the escape velocity to be greater than the speed of light (and I don't recall seeing mystery implying that it was)

 

[math] v_{escape} = \sqrt{\frac{2GM}{r}}[/math]

 

it is quite simple to sub in the speed of light for escape velocity - re-arrange and get the Schwarzchild radius; that radius for a mass from which the escape velocity is c

 

[math] r_{schwarzchild} = \frac{2GM}{c^2} [/math]

[granpa]

pardon me but I believe that your first equation is the equation for nonrelativistic escape velocity

 

obviously if the velocity is near the speed of light then you would need to use a relativistic equation.

Edited November 14, 2011 by granpa

[Mystery111]

It is clear that it is not necessary for gravity to be infinite in order for the escape velocity to be greater than the speed of light (and I don't recall seeing mystery implying that it was)

 

[math] v_{escape} = \sqrt{\frac{2GM}{r}}[/math]

 

it is quite simple to sub in the speed of light for escape velocity - re-arrange and get the Schwarzchild radius; that radius for a mass from which the escape velocity is c

 

[math] r_{schwarzchild} = \frac{2GM}{c^2} [/math]

Agreed.

 

It doesn't matter if it is non-relativistic. Nothing in a black hole can be infinite apart from controversially the singular region of spacetime.

[granpa]

you can use the wrong equation if you want to but I dont see how you can prove anything with it.

 

we know that the actual escape velocity at the event horizon would be less than c due to relativistic effects.

Edited November 14, 2011 by granpa

[Mystery111]

you can use the wrong equation if you want to but I dont see how you can prove anything with it.

 

we know that the actual escape velocity at the event horizon would be less than c due to relativistic effects.

 

It don't need to prove anything. From your original statement, I know fine well this is not what is implied. Please, read what you said again, in response to mine. We will see who is pulling things from the aether.

[imatfaal]

you can use the wrong equation if you want to but I dont see how you can prove anything with it.

 

we know that the actual escape velocity at the event horizon would be less than c due to relativistic effects.

 

Feel free to provide the correct one - it might also be nice for you to show how the correct solution pops out.

 

The Schwarzschild radius can be calculated using the equation for escape speed:

v_esc = (2GM/R)^1/2

For photons, or objects with no mass, we can substitute c (the speed of light) for V_esc and find the Schwarzschild radius, R, to be

R = 2GM/c²

http://imagine.gsfc....black_holes.htm - so it is a common explanationand which might be lacking in subtlety but it is pretty good.

Edited November 14, 2011 by imatfaal

[granpa]

It don't need to prove anything. From your original statement, I know fine well this is not what is implied. Please, read what you said again, in response to mine. We will see who is pulling things from the aether.

 

you were trying to prove that an object cant escape from the event horizon.

 

Feel free to provide the correct one - it might also be nice for you to show how the correct solution pops out.

 

the correct solution is, as I said before, that gravitational time dilation causes time to stop at the event horizon.

[Mystery111]

you were trying to prove that an object cant escape from the event horizon.

 

 

 

the correct solution is, as I said before, that gravitational time dilation causes time to stop at the event horizon.

 

give it a velocity larger than ''c'' and it will. Otherwise, it won't.

[imatfaal]

 

the correct solution is, as I said before, that gravitational time dilation causes time to stop at the event horizon.

That is no solution. Firstly the time slows massively when viewed by those in an external accelerated frame and not for those in free fall along side at the same gravitational potential. Secondly you have your coordinate system wrong if you come up with infinite time dilation - near a blackhole's event horizon you should be using Kruskal-Szekeres coordinates, or Eddington-Finkelstein coordinates - these avoid the mathematical singularites that schwarzchild coordinate systems throw up.

[Mystery111]

That is no solution. Firstly the time slows massively when viewed by those in an external accelerated frame and not for those in free fall along side at the same gravitational potential. Secondly you have your coordinate system wrong if you come up with infinite time dilation - near a blackhole's event horizon you should be using Kruskal-Szekeres coordinates, or Eddington-Finkelstein coordinates - these avoid the mathematical singularites that schwarzchild coordinate systems throw up.

 

You don't need to explain yourself. Bottom line is, he is wrong.

[granpa]

for those that actually want to learn

here is an explanation showing how general relativity is derived from special relativity.

 

 

math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

 

The Relativistic Rocket

 

Below the rocket, something strange is happening...

 

everything in the universe is falling "below" the rocket, but never receding any farther than a distance of -c2/a as measured by you. It all piles up just short of this distance, asymptoting to a plane called a horizon. You see this horizon actually form as the rocket accelerates, because there comes a time when no signal emitted from "below" the horizon can ever reach you. Everything falls toward that plane, and as it does so it begins to redden, due to the increasing red shift of its light, because you are accelerating. Finally it fades out of visibility. In fact, as anything gets closer to the horizon, it ages more and more slowly; time comes to a complete halt there.

Edited November 14, 2011 by granpa

[Mystery111]

I will later. Doctor rocket is not the be all and end all.

[morgsboi]

The simple reason can be explained in a simple way.

 

You require energy to leave the earths atsmosphere. This is because gravity is pulling your object (the mass of the earth) from your located origin. What if that mass became so dense that the radial force required to leave your origin became that of the speed of light?

 

Simply, a particle of light cannot escape something when the escape velocity exceeds the speed of light.

 

So if nothing is faster than light, then why would physics allow a force that pulls stronger than light can travel?