Light near a black hole's event horizon

[beecee]

I saw a discussion of black holes of different sizes having differing gravity forces at the event horizons, but if the horizon is established by C, and C is constant, shouldn't the gravity at the surfaces be equal?

 

 

As most have said, the escape velocity of the EH of any BH is "c".

If any photon is emitted just at this side of the EH of any BH, it will be forced to arc back and be swallowed by the BH, unless that photon is emitted directly radially away.

In that case the photon will always hover just above the EH, never quite getting away but also never being swallowed by the BH.

What you may have seen is a discussion of the "criticality" of the spacetime curvature near the EH and inside a BH.. You or I could in effect cross the EH of the SMBH at the center of our galaxy without any undue immediate effects.....But approach and cross the EH of a stellar size BH and one would quickly be spaghettified and ripped asunder possibly even before crossing the EH. This is due to the tidal gravitational effect.

Of course as one approached the singularity the spaghettification and being ripped apart will certainly take place in any BH no matter the size.

[robinpike]

When light is outside of a black hole's event horizon and is moving away from the event horizon - I understand that the light is red shifted, but can someone just confirm that the light still moves at the speed of light at all times?

 

I was reading another post on black holes, and it talked about light 'hovering' near the event horizon and wondered what that meant.

[MigL]

Light does not 'hover', accelerate or decelerate.

It is constrained to move at c at all times because it is a massless particle.

[swansont]

I was reading another post on black holes, and it talked about light 'hovering' near the event horizon and wondered what that meant.

 

 

Objects might be described that way. Time dilation near an event horizon will be large, so regardless of what the object experiences, a remote observer would see slowed motion.

 

As MigL says, light travels at c in any local frame. However, a remote observer (not in the local frame) will observe time dilation, known as Shapiro delay. That remote observer will also observe the trajectory of the light to be curved (the local observer sees local space as flat). AFAIK the effects compensate for each other, much as in special relativity with length contraction and time dilation.

[MigL]

You will never see a photon 'hover'.

It will be red-shifted to a near infinite wavelength, but will always move at c.

 

You can't use a mathematical prediction of relativity ( Black Holes ) to disprove one of the basic tenets of relativity.

Edited June 15, 2017 by MigL

[imatfaal]

The only stable place for photons near black hole is the photon sphere at 3/2 of the Schwarzchild radius - but as MigL says they are not stopped; they orbit at that distance. Others not in orbit either escape or loop back in. But describing what happens to an infalling test particle is fraught with trouble - what is position of observer, which coordinate system etc; gets conceptually harder when there is no inertial frame for the infaller

[beecee]

You will never see a photon 'hover'.

It will be red-shifted to a near infinite wavelength, but will always move at c.

 

You can't use a mathematical prediction of relativity ( Black Holes ) to disprove one of the basic tenets of relativity.

From any local frame associated with any photon that is directed radially away from just this side of the EH, within that frame, that photon will appear to "hover"never quite escaping, and never falling in.

From a remote frame of reference, that photon will be gradually redshifted beyond view.

[swansont]

From any local frame associated with any photon that is directed radially away from just this side of the EH, within that frame, that photon will appear to "hover"never quite escaping, and never falling in.

From a remote frame of reference, that photon will be gradually redshifted beyond view.

"frame associated with a photon"?

 

How can such a photon be redshifted to a remote observer, if the photon never gets to the observer?

[beecee]

"frame associated with a photon"?

 

How can such a photon be redshifted to a remote observer, if the photon never gets to the observer?

 

 

Good point! Off goes my head and on goes a pumpkin!

The photon though emitted directly radially away, just on the EH, will certainly appear to "hover," never escaping and never falling in.

http://casa.colorado.edu/~ajsh/singularity.html#disappearingview

Photons do not orbit in circles at the horizon, just skimming the surface. The place where photons orbit in circles is the photon sphere, at 1.5 Schwarzschild radii. Photons emitted at the horizon fall in; except that if a photon is emitted exactly vertically outward exactly at the horizon, then it will hover at the horizon, not moving at all.

Edited June 15, 2017 by beecee

[Strange]

 

 

Good point! Off goes my head and on goes a pumpkin!

The photon though emitted directly radially away, just this side of any BH EH, will certainly appear to "hover," never escaping and never falling in.

http://casa.colorado.edu/~ajsh/singularity.html#disappearingview

Photons do not orbit in circles at the horizon, just skimming the surface. The place where photons orbit in circles is the photon sphere, at 1.5 Schwarzschild radii. Photons emitted at the horizon fall in; except that if a photon is emitted exactly vertically outward exactly at the horizon, then it will hover at the horizon, not moving at all.

 

 

They say "exactly at" while you say "just this side". Very different things.

[beecee]

 

 

They say "exactly at" while you say "just this side". Very different things.

Again, point taken, and obviously correct and edited to reflect that..

Edited June 15, 2017 by beecee

[beecee]

When light is outside of a black hole's event horizon and is moving away from the event horizon - I understand that the light is red shifted, but can someone just confirm that the light still moves at the speed of light at all times?

 

I was reading another post on black holes, and it talked about light 'hovering' near the event horizon and wondered what that meant.

That was my mistake....

I should have said light directed radially away on the EH.

Here is a link from that other thread.

http://casa.colorado...isappearingview

Photons do not orbit in circles at the horizon, just skimming the surface. The place where photons orbit in circles is the photon sphere, at 1.5 Schwarzschild radii. Photons emitted at the horizon fall in; except that if a photon is emitted exactly vertically outward exactly at the horizon, then it will hover at the horizon, not moving at all.

::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

Just to add...It mentions also the photon sphere: If you could exist there and shone a torch directly in front of you, the light would go in a circle and be reflected off the back of your head.

[MigL]

I'll try this again...

In any valid frame the speed of light is measured to be c .

Once you start making measurements 'across' frames, things tend to get quickly screwed up.

 

Since energy is also frame dependant, I prefer to think of the photon being infinitely red-shifted to zero energy.

To a semi-local observer ( you cannot be local with a photon ), it would still have some energy, but to a non-local observer it would simply disappear.

[beecee]

I'll try this again...

In any valid frame the speed of light is measured to be c .

Once you start making measurements 'across' frames, things tend to get quickly screwed up.

 

 

That's OK and certainly correct, but also every frame of reference is as valid as any other frame.

[MigL]

There is no valid frame for light, nor for events at, or inside, the event horizon.

[beecee]

There is no valid frame for light, nor for events at, or inside, the event horizon.

My point though is as I have said and referenced.....a photon emitted just on the EH, directly radially away, will never be swallowed and never get away.

http://casa.colorado.edu/~ajsh/singularity.html#disappearingview

My point though is as I have said and referenced.....a photon emitted just on the EH, directly radially away, will never be swallowed and never get away.

http://casa.colorado.edu/~ajsh/singularity.html#disappearingview

Obviously I speak of the simple garden variety of Schwarzchild BH.

[Mordred]

Thats probably true or would be except a Kerr rotating BH has two photon spheres.

 

https://arxiv.org/abs/gr-qc/0005050

 

Though the garden variety is rotating lol

Edited June 16, 2017 by Mordred

[beecee]

Thats probably true or would be except a Kerr rotating BH has two photon spheres.

 

https://arxiv.org/abs/gr-qc/0005050

 

Though the garden variety is rotating lol

Agreed. Which brings up the interesting scenario of the photon spheres travelling in opposite directions!

https://universe-review.ca/R15-17-relativity04.htm

 

 

And of course the interesting concept concerning some advanced civilisation extracting useful energy from within the Ergosphere!

[Mordred]

Ah yes Hawking Berkenstein radiation from the ergosphere. One of my favourite BH articles is this blackhole Accretion Disk'' -Handy article on accretion disk measurements provides a technical compilation of measurements involving the disk itself. There is some coverage on the radiation within.

 

http://arxiv.org/abs/1104.5499

Edited June 16, 2017 by Mordred

[imatfaal]

But Hawking radiation is predicted to be emitted - and that is precisely radiation which you talk about ie photons created at EH and outwards. If Hawking radiation stayed within the confines of the BH then blackhole evaporation etc would not function, BH would only get bigger, heat death of universe would not happen...

 

Perhaps the HR calc's take this into account in the measure of luminosity - but I have never come across this idea of a stationary photon

[imatfaal]

My point though is as I have said and referenced.....a photon emitted just on the EH, directly radially away, will never be swallowed and never get away.

http://casa.colorado.edu/~ajsh/singularity.html#disappearingview

...

 

Hadn't realised we were on two separate threads - I might have to merge

[beecee]

But Hawking radiation is predicted to be emitted - and that is precisely radiation which you talk about ie photons created at EH and outwards. If Hawking radiation stayed within the confines of the BH then blackhole evaporation etc would not function, BH would only get bigger, heat death of universe would not happen...

 

Perhaps the HR calc's take this into account in the measure of luminosity - but I have never come across this idea of a stationary photon

The special case of a photon emitted directly radially away, occurs just on the EH.....

http://casa.colorado...isappearingview

As far as I know HR as generally accepted, occurs just this side of the EH, in particle pair creation event, where the negative falls in and the positive escapes to become real.

https://en.wikipedia.org/wiki/Hawking_radiation

extract:

Hawking-Zel'dovich radiation^[1] is blackbody radiation that is predicted to be released by black holes, due to quantum effects near the event horizon. It is named after the physicist Stephen Hawking,

The special case of a photon emitted directly radially away, occurs just on the EH.....

http://casa.colorado...isappearingview

As far as I know HR as generally accepted, occurs just this side of the EH, in particle pair creation event, where the negative falls in and the positive escapes to become real.

https://en.wikipedia.org/wiki/Hawking_radiation

extract:

Hawking-Zel'dovich radiation^[1] is blackbody radiation that is predicted to be released by black holes, due to quantum effects near the event horizon. It is named after the physicist Stephen Hawking,

I would guess that the orientation of the particle pair creation would be crucial also.....

Edited June 16, 2017 by beecee

[imatfaal]

The special case of a photon emitted directly radially away, occurs just on the EH.....

http://casa.colorado...isappearingview

As far as I know HR as generally accepted, occurs just this side of the EH, in particle pair creation event, where the negative falls in and the positive escapes to become real.

https://en.wikipedia.org/wiki/Hawking_radiation

extract:

Hawking-Zel'dovich radiation^[1] is blackbody radiation that is predicted to be released by black holes, due to quantum effects near the event horizon. It is named after the physicist Stephen Hawking,

 

"As far as I know HR as generally accepted, occurs just this side of the EH, in particle pair creation event, where the negative falls in and the positive escapes to become real." This is very much a heuristic (ie not really true) explanation - the reality is vastly more recondite but I am not sure the details matter.

 

But if the leeway for photon hovering is that small (ie that a particle/anti-particle pair are created with a separation beyond its realm) then how is a photon ever created AT the event horizon - they don't just pop into existence without something else happening. And the "something else happening" will affect the size of the EH and mean that the photon is now not AT the event horizon, just close.

!

Moderator Note

Sections of original thread ("BH Question") - and this one started by Robin to double-check statements within original have been merged

[beecee]

 

 

But if the leeway for photon hovering is that small (ie that a particle/anti-particle pair are created with a separation beyond its realm) then how is a photon ever created AT the event horizon - they don't just pop into existence without something else happening. And the "something else happening" will affect the size of the EH and mean that the photon is now not AT the event horizon, just close.

!

Moderator Note

Sections of original thread ("BH Question") - and this one started by Robin to double-check statements within original have been merged

Anything actually crossing the EH, could emit a photon directly radially away: A valid point though made by another, is that we cannot see this from a remote frame of reference...In fact we see nothing cross the BH EH...We don't even see a BH form! as far as I know for the same reasons.

[imatfaal]

Anything actually crossing the EH, could emit a photon directly radially away: A valid point though made by another, is that we cannot see this from a remote frame of reference...In fact we see nothing cross the BH EH...We don't even see a BH form! as far as I know for the same reasons.

 

Which said action would change the mass-energy of the BH - changing the position of the EH.

 

On another tack:

From a local inertial frame the photon must travel at c, there is no frame of the photon (regardless to your comment above that "every frame of reference is as valid as any other frame" - if you posit an inertial frame in which the speed of light is not locally c then you have contradicted one of the postulates of einstein's relativity and henceforward you cannot rely on the calculations of relativity), and for a distant accelerated frame (ie someone being held in position a distance away) the photon will either be redshifted to obscurity (but still moving) or by your claim will never reach them. So in what frame does it hover

 

Final questions:

What is a stationary photon - ie one not travelling through a medium at the local speed of light, what is its frequency if both c and lambda equal zero(undefined?), what is its energy from planck's equation if frequency is not clear