Light, black holes and gravity

[JSK333]

Gravity is "The tendency of a mass of matter toward a center of attraction; esp., the tendency of a body toward the center of the earth; terrestrial gravitation" from most physics dictionaries.

So, here's a basic question.

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

Solomon

 

[mathman]

Although photons have zero REST mass, they have energy. Gravity acts on energy just as it does on mass. The famous 1919 eclipse observation, which was a dramatic confirmation of general relativity, was based on the bending of starlight due to the sun's gravity. In simple terms, the escape velocity from a black hole is greater than the speed of light, therefore light can't escape.

 

[maximus]

in Newtonian theory of gravity, one would suppose that this effect would never take place becuase as you said photons are massless. i don't know what mathman meant by no-rest mass becuase photons are never at rest, but he was right: they have energy. in the spec. theory of relitivity, energy, like its equivalent in mass, are affected by the curvature of spacetime. since a black hole is a pretty heavy distortion of spacetime, it is impossible for light to escape, thus making it black.

 

[garytse86]

or you could say that light is still traveling in a straight line in four dimensional space, but since a black hole distorts space time around it, it curves space-time in three dimensional space. Light will always follow a straight line geodesic in 4-dimensional space.

 

[marcus]

Originally posted by JSK333
Gravity is "The tendency of a mass of matter toward a center of attraction; esp., the tendency of a body toward the center of the earth; terrestrial gravitation" from most physics dictionaries.

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.

 

[marcus]

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.

 

[ranyart]

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 black hole is? I've 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?

 

[marcus]

Originally posted by ranyart
Could it be that the understanding you are relaying is that you have no comprehension of what a black hole is? I've 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_grav = 1/sqrt(1 - R_s/R)

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_s = 2GM/c²

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

 

[ShawnD]

Planck's formula E = hf tells us the energy for a photon of light at a given frequency and Einstein's formula E =mc^2 relates energy to mass and speed. If we turn the quations into one we get hf = mc^2, then we can move it around and get m = hf/c^2 as the mass of a photon. h (Planck's constant) is a constant, f (frequency) is always > 0 for any photon and c (speed of light) is a constant (for vacuums in this case).

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

 

[neutroncount]

Originally posted by ShawnD
Planck's formula E = hf tells us the energy for a photon of light at a given frequency and Einstein's formula E =mc^2 relates energy to mass and speed. If we turn the quations into one we get hf = mc^2, then we can move it around and get m = hf/c^2 as the mass of a photon. h (Planck's constant) is a constant, f (frequency) is always > 0 for any photon and c (speed of light) is a constant (for vacuums in this case).

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 :)

 

[maximus]

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?

 

[Andersen]

I'm not sure i understand your question, but photons/light do/does have momentum.

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

Joe

 

[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.