Can Photons Escape the Gravitational Pull of Black Holes?

[JPC]

hey
just wondering

how come photons can't get out of black holes , because the gravity equation is :

GMm / d²

But , a photon mass = 0 , so the gravity equation with a photon should be equal to 0, so it should be able to get out


Or, is it that when gravity becomes greater , the length of time becomes greater, or something like that

or that the photon will get out of the black hole , but in a very very very long time


and , i just need the general idea with a few equations if possible, I am only in grade 11

 

[marlon]

hey
just wondering

how come photons can't get out of black holes , because the gravity equation is :

GMm / d²

But , a photon mass = 0 , so the gravity equation with a photon should be equal to 0, so it should be able to get out


Or, is it that when gravity becomes greater , the length of time becomes greater, or something like that

or that the photon will get out of the black hole , but in a very very very long time


and , i just need the general idea with a few equations if possible, I am only in grade 11

The formula you quoted comes from classical physics. Black holes are described by general relativity and photons are described by quantum theory. So, you are using a formula that is NOT valid for the stuff you want to describe. This is why what you ask won't work.

regards
marlon

 

[JPC]

Oh ok

but is there any simple general idea explanation to the fact that photons can't escape black holes ?

 

[Hurkyl]

It doesn't really make sense to speak of "photon" and "black hole" in the same breath.

QFT (quantum field theory) is what tells us about photons, but it doesn't talk about black holes. GR (general relativity) is what tells us about black holes, but it doesn't talk about photons. Attempts to merge QFT and GR have not been successful, so one cannot reliably talk about any hypothetical situation involving both.


But, I think you don't really need to speak of photons to ask your question: you can just ask your question with any hypothetical particle traveling at c.

Honestly, I don't think you will get any understanding out of any answer you hear, unless you make some effort to learn the mathematics behind it -- in this case, differential geometry. Worse, you might misunderstand the answer precisely because you do not know the mathematics behind it.

But, I'll give it a try anyways: to an observer sitting on the event horizon of a black hole, it would appear as if the event horizon were traveling at c. A particle inside the black hole could never catch up with the event horizon to make its escape!


You've probably seen some cartography in school, right? You've learned about Mercator projections, equal-area projections, and maybe other ways of drawing a map of the Earth. You know that, to represent the two-dimensional surface of the Earth on a flat two-dimensional sheet of paper, some distortions must be made. You can ensure that the map looks right near certain points, but you cannot make the map look right everywhere simultaneously.

The same phenomenon happens in GR -- we cannot "map" the universe without introducing distortions. Schwarzschild coordinates are a useful "map" for a distant observer studying a black hole, but it does not look "locally correct" for an observer at the event horizon.

In a similar vein, these two descriptions are equivalent for someone standing on the Earth:
(1) Gravity causes things to accelerate downwards at 9.8 m/s²
(2) There is no gravity, but the surface of the Earth is accelerating upwards at 9.8 m/s²

(and, in some senses, (2) is the more appropriate description)

 

[DaveC426913]

Well, Hurkyl's answer is probably better than mine, but I'll post mine in the name of brevity.

Gravity does not bend the path of photons. Gravity bends the space that the photons travel through. The photons travel in a straight line as always - it's the line that's bent!

Near a black hole, space is bent so drastically that the paths the photons follow bend right back into the black hole. So, it's not that the photons have any trouble at all getting where they're going - it's just that "where they're going" isn't outside the BH!

 

[anantchowdhary]

a photons invariant mass is 0 as it cannot exist at rest.But at c ,a photon has energy and therefore relativistic mass

 

[lightarrow]

hey
just wondering

how come photons can't get out of black holes , because the gravity equation is :

GMm / d²

But , a photon mass = 0 , so the gravity equation with a photon should be equal to 0, so it should be able to get out
Or, is it that when gravity becomes greater , the length of time becomes greater, or something like that
or that the photon will get out of the black hole , but in a very very very long time
and , i just need the general idea with a few equations if possible, I am only in grade 11

I think your question is wrong because, even in classical mechanics, you don't consider the fact that a = F/m so, from F = -GMm/d^2, we have: a = -GM/d^2 and it doesn't depend on the mass m of the object.
Infact, even from classical mechanics light is bent from a massive object (but the bent is half of the real one; the other half was explained by Einstein with General Relativity).

 

[Hurkyl]

Infact, even from classical mechanics light is bent from a massive object (but the bent is half of the real one; the other half was explained by Einstein with General Relativity).

Not so; in classical mechanics, light is a (classical) electromagnetic wave. In particular, light is not a massless particle. Gravity plays no role in Maxwell's equations, thus no bending of light.

 

[Chris Hillman]

This is a FAQ

Hi, JPC,

how come photons can't get out of black holes

Just wanted to point out that the underlying question here is a FAQ; see http://www.math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html
I'd recommend the popular book by Wald or the one by Geroch which are cited on this page (originally by myself): http://www.math.ucr.edu/home/baez/RelWWW/reading.html#pop . Happy reading!

BTW, I agree with Hurkyl that it is probably a bad idea to talk glibly of "photons" (or "gravitons") in the context of gtr. Much better to talk of a "laser pulse".

Not so; in classical mechanics, light is a (classical) electromagnetic wave. In particular, light is not a massless particle. Gravity plays no role in Maxwell's equations, thus no bending of light.

To avoid possible confusion, I hasten to qualify this: Maxwell's field equations are conformally invariant, and so are some classical relativistic theories of gravitation, such as Nordstrom's theory. There is no bending of light in such theories. OTH, Maxwell's theory of EM can be incorporated into gtr (and similar theories of gravitatation) by letting the stress-energy of the EM field act as a source term in Einstein's field equation, and by taking a suitable curved spacetime generalization of Maxwell's field equations. Although this takes a bit of explaining, it is not hard to find exact solutions in gtr which model both null and non-null EM fields (these are called "electrovacuum solutions" if there is no matter present; in particular, no electrically charged matter). See for example "Exact Electrovac Solutions from a Symmetry Ansatz" at http://www.math.ucr.edu/home/baez/RelWWW/group.html

a photons invariant mass is 0 as it cannot exist at rest.But at c ,a photon has energy and therefore relativistic mass

I agree with anyone who decries "relativistic mass"; this is terribly misleading terminology and should be avoided. Say rather than a laser pulse possesses kinetic energy (the so-called "relativistic mass" is basically just the kinetic energy) and momentum. See Taylor and Wheeler, Spacetime Physics for more about this.

 

[JPC]

DaveC426913 , i saw your image about why photons can't escape

But, how can a loop be greated with bending spacetime , because to my thinking it would mean that a path made of a line made of many : paths between '2 points of the paper (that are not next to each other) ' is created

because if u have a straight line , and u want it to make a loop , at some point when u curve it the two ends have to meet and connect to each other//////
and , lightarrow, i meant the classical physics equation for the Force of the gravity interaction between the two objects, not the g value

 

[DaveC426913]

Don't take the diagram too literally. Remember that those photons were surely emitted from infalling matter, they're not simply orbiting.

The point is simply that photons will follow a "straight line through spacetime" - though that straight line will not have an end outside the BH.

 

[JPC]

But if there's no 'end'
it would mean the bended line they follow goes on for infinity in a limited volume

 

[DaveC426913]

But if there's no 'end'
it would mean the bended line they follow goes on for infinity in a limited volume

Why is that a problem? The photon's universe has shrunk to a few hundred km in diameter. It crosses its universe and comes back to where it started from, over and over.

(This is an ideal situation, with a few sparse photons around an otherwise empty black hole. What really happens is that photons are absorbed and emitted by all the infalling matter, and in fact have quite short trajectories. )

 

[jnorman]

another way to look at this is to recognize that within the event horizon of a black hole, the gravity is so great that the escape velocity exceeds the speed of light.

 

[quantum123]

Laplace actually predicted a black hole also using Newton's laws.