Objects falling into a black hole

aademarco
Messages
11
Reaction score
1
I have a question pertaining to objects falling through the event horizon of a black hole.

It is my understanding that due to the immence gravity of a black hole and the way gravity affects the flow of time, that from the point of view of an observer at a safe distance from a black hole watching any object fall through the event horizon you can never see an object fall through due to how its flow of time changes relative to the observer. Its time will appear to become slower and slower until it appears to freeze infinitely close to the event horizon. From the point of view of the object falling through the event horizon it continues to fall all the way to the singularity. If that object could turn and look back out through the event horizon it would see time speed up infinitely.

Im having a hard time understanding, if the above is true, why black holes are not glowing bright with images of everything they have ever 'eaten' just above or at the event horizon. From our perspective, all the matter which has ever fallen into it had a clock which slowed to a stop the moment it hit the horizon despite the fact that from their own perspectives they went right through.
 
Physics news on Phys.org
aademarco said:
Im having a hard time understanding, if the above is true, why black holes are not glowing bright with images of everything they have ever 'eaten' just above or at the event horizon. From our perspective, all the matter which has ever fallen into it had a clock which slowed to a stop the moment it hit the horizon despite the fact that from their own perspectives they went right through.

We do see them glow brightly at the event horizon, in the form of an accretion disk, as well as at the poles, where large quantities of x-rays can be emitted. Observing the actual black hole is not possible with visible light, but measuring the orbital speeds of visible matter and calculating the central mass about which they are falling towards is.
 
aademarco said:
Im having a hard time understanding, if the above is true, why black holes are not glowing bright with images of everything they have ever 'eaten' just above or at the event horizon. From our perspective, all the matter which has ever fallen into it had a clock which slowed to a stop the moment it hit the horizon despite the fact that from their own perspectives they went right through.
The light from a collapsing object persists, but it becomes increasingly faint and increasingly redshifted. See this paper by Ames and Thorne for a complete analysis.

Ames and Thorne said:
The star is brightest and bluest at its rim, where the spectrum and intensity are independent of time; but the width of this time-independent region decays exponentially. Nearer the center of its disk, the star appears darker and redder; the redshift increases exponentially in time, and the intensity decays exponentially. The total, integrated radiation is dominated by the contribution from the rim: The spectrum is nearly time-independent and is peaked at its high-frequency end; but the total luminosity decays exponentially with time.
 
Last edited:
aademarco said:
If that object could turn and look back out through the event horizon it would see time speed up infinitely.
This is not quite correct, although it's a bit tricky to explain. Take a look at this earlier PF post https://www.physicsforums.com/showpost.php?p=2336347&postcount=4 which shows a Kruskal diagram copied from Misner, Thorne and Wheeler, the one labeled Fig 31.4(b).

The event horizon is the diagonal line labeled r = 2M, t = +∞. The falling particle is the curve with labels A, A', A''. As the particle approaches the event horizon, the coordinate time t runs through all its values and becomes infinite.

However the incoming light that the particle would see looking back is represented by left-sloping diagonal lines such as B, B', B''. Even when the particle hits the horizon, the incoming photons do not pile up. Although the retarded time as measured by the outgoing light rays becomes infinite, the "advanced time" as measured by the incoming rays remains finite.
 
OK, so this has bugged me for a while about the equivalence principle and the black hole information paradox. If black holes "evaporate" via Hawking radiation, then they cannot exist forever. So, from my external perspective, watching the person fall in, they slow down, freeze, and redshift to "nothing," but never cross the event horizon. Does the equivalence principle say my perspective is valid? If it does, is it possible that that person really never crossed the event horizon? The...
ASSUMPTIONS 1. Two identical clocks A and B in the same inertial frame are stationary relative to each other a fixed distance L apart. Time passes at the same rate for both. 2. Both clocks are able to send/receive light signals and to write/read the send/receive times into signals. 3. The speed of light is anisotropic. METHOD 1. At time t[A1] and time t[B1], clock A sends a light signal to clock B. The clock B time is unknown to A. 2. Clock B receives the signal from A at time t[B2] and...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...

Similar threads

Back
Top