Why Aren't Event Horizons Bright?

1. Nov 28, 2013

EskWIRED

My understanding is that an outside observer sees objects undergo time dilation as they approach sources of gravity, such that at the event horizon of a black hole, a distant observer would see time stop for the infalling object, with it seemingly suspended at the event horizon forever.

If that is correct, then it seems to me that some event horizons should appear to be clusters of stars which have fallen in over the eons, or more likely, bright areas, with the brightness of millions of stars which have fallen in over billions of years.

Are event horizons bright? If not, why not?

2. Nov 28, 2013

George Jones

Staff Emeritus
Think of a star as emitting so many photons per second, with the energy of the photons inversely proportional to wavelength(s) of the emitted photons.

As the the star nears the event horizon, because of gravitational time dilation:

1) the number of photons we receive per second drops;

2) the wavelengths of the photons increase (redshift), thus dropping the energy of each received photon.

Consequently, we see a light-source rapidly dim as it approaches an event horizon.

3. Nov 29, 2013

Naty1

except that nothing can pass out of an event horizon once crossed. So far as is known, low energy Hawking radiation, too little to be detected, is all that is 'emitted' via quantum fluctuations.

If a BH actually explodes after shrinking via HAwking radiation, towards the end of the universe, when the horizon is smaller than an atomic nucleus, and the BH becomes extremely hot, you'll get your 'bright' result.

4. Nov 29, 2013

WannabeNewton

If a Schwarzschild black hole was formed through spherically symmetric gravitational collapse of a star then radially outgoing radiation emitted at $r = 2M$ from the surface of the star will get trapped at $r = 2M$ indefinitely (ignoring the dynamics of the event horizon due to infalling matter) so will not an observer freely falling into the black hole see a bath of radiation outgoing at $c$ in his local Lorentz frame the moment he reaches the event horizon?

5. Nov 29, 2013

Staff Emeritus
You cannot make that approximation. You and the star fell into the BH at different places and times, and the geometry of the BH will not change that interval to 0.

6. Nov 29, 2013

Bill_K

Let worldline S represent the surface of the collapsing star, and let T be the worldline of the observer. Both S and T are ingoing. T will continually receive radiation from S, both before and after reaching r = 2M, but there will not be an explosive burst of radiation received as the horizon is crossed.

Let s, t be the proper time along S and T, and at each point of S draw the outgoing null geodesic. Radiation emitted from S at time s will be received by T at some time t. If the radiation is emitted at a constant rate as measured by s, then it will be received by T at a rate dt/ds. For specific worldlines the rate could be calculated, but I think it's clear from looking at a Kruskal diagram that dt/ds is not singular at the horizon.

7. Nov 29, 2013

DrGreg

A rhetorical question: In Minkowski spacetime, any outgoing radiation emitted at $x - ct = x_0$ will get trapped within the plane $x - ct = x_0$ indefinitely so will not an inertial observer see a bath of radiation outgoing at $c$ in his local Lorentz frame the moment he reaches the plane $x - ct = x_0$ ?

8. Nov 29, 2013

Naty1

Wannabe: I'm not quite sure what assumptions you have subtley wrapped up in your question, but in general there are numerous caveats about 'what's real' :

PAllen: general comment:
Dalespam:
Pervect:

{PeterDonis}

HOVERING OBSERVER:

Well, one could ask the question 'How hot does the hovering observer measure the temperature near the event horizon"?
Classically, he measures zero temperature.

And someone in these forums previously posted these rather extensive computer simulations and visual representations collaboration between UV ands UC....also with lotsa caveats:

9. Nov 29, 2013

Naty1

yes, that is an idealized static Schwarszchild coordinate view....no rotation, no charge.....what's actually observed may be another matter. I suspect, but do not know, that we have yet to actually observe such an effect experimentally, let alone find a non rotating BH. Seems like either the horizons are obscured from observation by accretion disks, or are 'black' and we don't know where to find them.....either way, the BH will likely be affected by spacetime ripples ....gravitational waves....and spacetime ripples.

The simulations of the sort I linked to may be our best insights so far. Anybody know?

10. Nov 29, 2013

pervect

Staff Emeritus
I think some of the aspects referred to by Dr. Greg and WannabeeNewton can be illustrated by a specific example.

Lets consider a spaceship falling into an event horizon. The bow has already passed. The stern is "just at" the event horizon now. The hole is very massive, so the tidal forces aren't extreme.

An observer in the stern of the spaceship won't see any change in the appearance of the bow. If he measures the brightness of the bow as a function of time, he will see that this brightness will be constant as the ship falls through the event horizon. Any realistic measurement of brightness will take a finite amount of time - but there's no reason to say that crossing the event horizon changes the brightness, it never flickers or dims, not even instantaneously.

As we make the spaceship longer and longer, the brightness of the bow as seen from the stern will naturally decrease - just because it's further away.

There will be no significant gravitational redshift from the bow to the stern of the spaceship - only tidal forces will cause such redshift, and we've assumed they are small. There will be no significant doppler shift due to relative motion of the stern and the bow, because there are no tidal forces and because we've assumed the spaceship wasn't stretching as it fell into the black hole.

We'll comment, in passing that the notion of "time stopping" at the event horizon of a black hole may be true in some specific coordinate-dependent sense, but is generally misleading. The reason the notion is misleading is because "time stopping" is interpreted as if absolute time were stopping. And the time in GR, like SR, is not "absolute time". So when one thinks of "time stopping" at the surface of a black hole, and one uses one's Newtonian intuition, one generally gets bad results. This is why I recommend against "the time stopping" point of view.

There are some alternatives, albeit with problems of their own, such as the "waterfall" or "river" model due to Hamiltion, http://jila.colorado.edu/~ajsh/insidebh/waterfall.html or http://arxiv.org/abs/gr-qc/0411060. I haven't seen people try to use this enough to see how badly they get confused relative to how badly they get confused due to the "time stopping" model. The main thing I don't like about the waterfall model is that there is no physical way to measure a "velocity of space".

Sorry for the digression on time stopping. Now lets try and generalize the "spaceship" result.

In the spaceship case, both the bow an the stern had the same "energy-at-infinity" when they feel into the hole. If we consider two general observers, they won't have the same "energy-at-infinity". We can incorporate this into the spaceship model by having the the rear observer in a "virtual spaceship", but having the object at the front of the spaceship moving relative to the bow. Then we have a space-ship model with added doppler shift, due to the relative motion of the particle at the bow of the spaceship to the ship.

When we include tidal forces, the analysis becomes much more complex. If we've got a long enough space-ship, or if two objets have fallen into the BH at greatly different times, we'll unfortunately have to take the tidal forces into account to get a good answer. I don't have a fast-and-dirty way of doing this though except to say that the sign of the effect is pretty clear, the tidal forces cause the bow to redshift.

I'll mention that the tidal forces at the event horizion are finite to an infalling observer and approach zero for a sufficiently massive black hole. This is a textbook result, but it may be surprising to people who aren't familiar with the detailed mathematics. I won't go into more details on this point unless it is asked as its very tangential - but it may be a source of some confusion to some readers.

11. Nov 30, 2013

WannabeNewton

Thanks for the replies everyone! Yeah I see where I got myself confused. I was originally thinking of a point source emitting radiation radially outwards at regular intervals so that on a Kruskal diagram we have pulses going out along the $+ 45^{\circ}$ lines passing through the worldline of the point source including the $r = 2M$ line bounding regions $I$ and $II$. Then as $t\rightarrow \infty$ an observer falls freely into the black hole. For some reason I originally pictured all the radiation except for that going out along $r = 2M$ as long gone as far as this observer is concerned although I can't recall why I insulated $r = 2M$