- #1

JSK333

So, here's a basic question.

Why cannot/does not light escape from a black hole (gravity well), if photons are inherently massless?

Solomon

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter JSK333
- Start date

- #1

JSK333

So, here's a basic question.

Why cannot/does not light escape from a black hole (gravity well), if photons are inherently massless?

Solomon

- #2

mathman

Science Advisor

- 7,932

- 484

- #3

- 477

- 4

- #4

- 308

- 0

- #5

marcus

Science Advisor

Gold Member

Dearly Missed

- 24,738

- 788

Originally posted by JSK333

So, here's a basic question.

Why cannot/does not light escape from a black hole (gravity well), if photons are inherently massless?

Solomon

I just started a thread in Astronomy forum about the effect of gravity on light. It partially overlaps stuff here altho there are several other issues raised here. You might want to check it out.

- #6

marcus

Science Advisor

Gold Member

Dearly Missed

- 24,738

- 788

Originally posted by JSK333

So, here's a basic question.

Why cannot/does not light escape from a black hole (gravity well), if photons are inherently massless?

Solomon

several people in this thread (mathman, gary etc.) have already pointed out that

the dictionary def is too restricted

it talks only of the effect of gravity on matter

but gravity is modeled by spacetime curvature, which effects everything, not just matter

the geometry of spacetime must effect everything in spacetime.

it cannot single out matter and limit its effects to matter.

light must travel along the geodesics (which may be curved by concentrations of energy), as I think several people already said.

This question about light escaping from a black hole is interesting.

Seen by an observer who is well clear of the hole, for all practical purposes at "infinity" or many hole-radiuses away, the light from near the event horizon of the hole is way way redshifted.

Its wavelength can get so long that it is no longer detectable--no more quantum energy. Infinitely redshifted light dies.

The "event horizon" of a BH, which for a lot of people plays the role of the spherical "surface" around the hole is pretty much defined as the place from which light cannot get clear out to infinity because in the course of doing that it undergoes an infinite redshift---and ceases to exist.

- #7

- 370

- 0

Originally posted by marcus

several people in this thread (mathman, gary etc.) have already pointed out that

the dictionary def is too restricted

it talks only of the effect of gravity on matter

but gravity is modeled by spacetime curvature, which effects everything, not just matter

the geometry of spacetime must effect everything in spacetime.

it cannot single out matter and limit its effects to matter.

light must travel along the geodesics (which may be curved by concentrations of energy), as I think several people already said.

This question about light escaping from a black hole is interesting.

Seen by an observer who is well clear of the hole, for all practical purposes at "infinity" or many hole-radiuses away, the light from near the event horizon of the hole is way way redshifted.

Its wavelength can get so long that it is no longer detectable--no more quantum energy. Infinitely redshifted light dies.

The "event horizon" of a BH, which for a lot of people plays the role of the spherical "surface" around the hole is pretty much defined as the place from which light cannot get clear out to infinity because in the course of doing that it undergoes an infinite redshift---and ceases to exist.

Could it be that the understanding you are relaying is that you have no comprehension of what a blackhole is? Ive seen your post in astronomy, and a number of your postings are blatantly wrong, but I guess your just like the rest of us learning.

Blackholes ain't so Black! has L Booda been transformed into another alias?..has anyone seen or heard from him recently?

- #8

marcus

Science Advisor

Gold Member

Dearly Missed

- 24,738

- 788

Originally posted by ranyart

Could it be that the understanding you are relaying is that you have no comprehension of what a blackhole is? Ive seen your post in astronomy, and a number of your postings are blatantly wrong, but I guess your just like the rest of us learning.

Blackholes ain't so Black! has L Booda been transformed into another alias?..has anyone seen or heard from him recently?

Don't know Booda.

My post above and the corresponding one "effect of gravity on light" I just posted at "Astronomy" forum are mainly expositions of the gravitational redshift formula

1+ z

If you disagree with this please let me know. It is certainly a textbook formula, and exact to the best of my knowledge.

If you would like a reference for it a good one would be F. Shu "The Physical Universe" a great textbook in general astronomy. Probably also plenty of web references available if you google "gravitational redshift".

Among other things, you can use this formula to prove that light can't escape the event horizon (not talking about Hawking radiation from near the EH).

R

Of course you already *know* light cannot escape the EH, but the redshift formula provides a neat quick way to prove it.

- #9

ShawnD

Science Advisor

- 668

- 1

Now we see that light truly does have a calculatable mass :D

- #10

neutroncount

Originally posted by ShawnD

Now we see that light truly does have a calculatable mass :D

Wrong

http://www.physlink.com/Education/AskExperts/ae180.cfm

Photons don't have REST mass, but they do have energy equating to mass. So what your math shows is the mass equivalent of a photon at a certain energy. Photons DO have momentum though and can transfer it giving the appearance of mass.

If this is what you were implying, then retract my entire statement :)

Last edited by a moderator:

- #11

- 477

- 4

Originally posted by neutroncount

Photons don't have REST mass, but they do have energy equating to mass. So what your math shows is the mass equivalent of a photon at a certain energy. Photons DO have momentum though and can transfer it giving the appearance of mass.

i know you weren't talkning to me, but see if this is a correct logical line of thought. the photon has no rest mass, or it would not be able to accelerate to c, but it does have energy, and thus graivity is a factor. does a photon therefore have no inertial value while accelertating to c? basically is it only mass (not its equivelent in energy) that can be affected by inertia? or the energy inherent in a virtual paricles movement does not add to its energy value (therefore its mass) in the same way that it would for a particle with rest mass?

- #12

- 13

- 0

p=E/c (from the definition of the energy-momentum tensor in SR)

Joe

- #13

neutroncount

Originally posted by maximus

i know you weren't talkning to me, but see if this is a correct logical line of thought. the photon has no rest mass, or it would not be able to accelerate to c, but it does have energy, and thus graivity is a factor. does a photon therefore have no inertial value while accelertating to c? basically is it only mass (not its equivelent in energy) that can be affected by inertia? or the energy inherent in a virtual paricles movement does not add to its energy value (therefore its mass) in the same way that it would for a particle with rest mass?

But here's the deal. Photons don't accelerate. They ARE at c, no if, ands, or buts. The creation of a phonton requires it to alreay be moving at c upon it's creation. A photon can neither slow down or speed up. Yes, I know about different mediums but at every point in the medium the propagation of photons is at c. It's the interaction that causes a slowdown.

Share: