The great photon disappearence act!

[Red Rule]

As I understand it a photon:
-always travels at c.
-has no rest-mass but it has some mass when it has energy. (in accordance to m=e/c^2)
-it's frequency is also related to it's energy.
but I'm far from an expert in this field.

Now if one were to take a light source and a gravitational source.
Let's take a lightbulb and a black hole.
(as a black hole is a point mass and a lightbulb emits light in all directions-we'd also need an extension cord for the lightbulb)
if said lightbulb were to fall towards the black hole, it's no stretch of the imagination to say that there would atleast 1 photon whose velocity vector is direcly opposite to the gravitational vector.
Would it then continue to move away from the black hole, losing energy as it did and then dissapear(as it would lose all it's properties)? or would it lose no energy at all and escape the black hole?

The first scenario seems to violate conservation of energy (unless the field absorbs it).
The second one, to me, seems plausible but as I said I'm certainly no expert in this field so I suspect I might be wrong as it leads to a possible violation of the definition of a black hole.

So, assuming I am wrong (as I probablly am), where did I go wrong? are my assumptions wrong or am I missing something important?

[Robert'); DROP TABLE *;]

In the event that a photon is inside the event horizon of the black hole, something rather weird happens: the photon hits the black hole, and contributes its energy to it. If the lightbulb is not inside the event horizon, then the photon escapes with some non-zero energy. (The rest of which is converted into gravitational potential energy.)

[Red Rule]

I am intriguied, but after a little reading I fear that further questioning will result in me having a terrible headache.
So many things to study, so little time.

[quantropy]

if said lightbulb were to fall towards the black hole, it's no stretch of the imagination to say that there would atleast 1 photon whose velocity vector is direcly opposite to the gravitational vector.

Once the lightbulb has crossed the event horizon this is no longer true, all paths lead to the singularity, just as all paths for us lead into the future.

[Roĝer]

or would it lose no energy at all and escape the black hole?

A photon that is emitted outside the event horizon in the right direction will escape, but it will lose energy, in the form of being redshifted. So your nice blue lightbulb will appear to become red as it approaches the black hole.

[Levi]

So what happens to photons that are emitted away from the black hole inside the event horizon?

[Xanthir]

There is no "away from the black hole" inside the event horizon. All paths lead toward the singularity.

[Antimony-120]

Or to put it another way, they might be going "away" at c, but space is moving inwards at greater than c, so ultimatly they move inwards anyways. They just move inwards as slowly as possible.

[quantropy]

It's rather difficult to visualize, but spacetime is still 4 dimensional within the event horizon, but the radial direction now becomes the time dimension. That means there must be another space dimension orthogonal to the two spherical angle dimensions. That's presumably the direction the photon travels in, but it's no longer a radial direction - I would think that it has the same status as the spherical angle dimensions.

[MHD]

Inside a black hole...

Well, what most people don't realise is this: Inside an event horizon there is no such thing as "outwards."

Once you have fallen inside, every direction is towards the singularity.

You have to travel backwards in time to get out.

[Quizatzhaderac]

or would it lose no energy at all and escape the black hole?

A photon that is emitted outside the event horizon in the right direction will escape, but it will lose energy, in the form of being redshifted. So your nice blue lightbulb will appear to become red as it approaches the black hole.

Also important to not is that from your perspective the light-bulb will take longer to reach the event horizon then it would under Newtonian mechanics. So while you see a lower energy of light you see it for longer.

[WarDaft]

Inside a black hole...

Well, what most people don't realise is this: Inside an event horizon there is no such thing as "outwards."

Once you have fallen inside, every direction is towards the singularity.

You have to travel backwards in time to get out.

But you can't observe yourself to have fallen inside the event horizon - it seems to shrink away beneath you, and there will always (well, until you die) in fact be a direction you can point to that to you appears to be 'out' even after you are inside the horizon according to a distant observer.


... random thought, what would happen if the Schwarzschild radius of the observable universe became larger than the observable universe?

[Tass]

... random thought, what would happen if the Schwarzschild radius of the observable universe became larger than the observable universe?

You'd be in a closed universe, doomed to collapse to a singularity in finite time.

[Roĝer]

... random thought, what would happen if the Schwarzschild radius of the observable universe became larger than the observable universe?

You'd be in a closed universe, doomed to collapse to a singularity in finite time.

But the calculation for the Schwarzschild radius doesn't include dark energy/cosmological constant, which at such low densities becomes very relevant.

[Tass]

... random thought, what would happen if the Schwarzschild radius of the observable universe became larger than the observable universe?

You'd be in a closed universe, doomed to collapse to a singularity in finite time.

But the calculation for the Schwarzschild radius doesn't include dark energy/cosmological constant, which at such low densities becomes very relevant.

Sure.