Can Light Move Inside a Black Hole Event Horizon?

Italian_Mike
Messages
3
Reaction score
0
The question is as follows: suppose I throw a metal bar 1m long inside the event horizon of a supermassive black hole of 1 million solar masses. At both ends of the metal bar there is a light source.

(I chose a supermassive black hole to rule out any spaghettification process: with some quick calculations tidal forces on the metal bar would be just 1/1000th of g - so they are negligible - when the bar crosses the event horizon).

So, let's consider the following scheme:

H--------B++++A------------S

Where B++++A is the metal bar, H is the (just crossed) event horizon an S is the singularity in the middle of the black hole.

My question is:

Is it possible for light emitted in A to reach the other end point B of the bar?

From what I know once inside the event horizon all possible space-time curves always lead closer to the singularity, never farther. My opinion is that is not possible for light in A to reach point B.

Now, substitute the metal bar with your head, and A and B with two neurons in your head and the light signal with an electric impulse sent between them. If it is not possible to send signals from A to B then you'll immediately lose conscoiusness and die as soon as you cross the event horizon (you'll die because what happens to the electric signals in your brain will also happen to all your bodily fluids: they'll be never be able to reach the back of your body).
So you'll be quite aware of when the event horizon is located as you'll immediately die.

A mechanical probe powered by electric signals will stop working as well.

Is this picture correct, in your opinion?
Thanks in advance for your reply!
 
Physics news on Phys.org
Italian_Mike said:
The question is as follows: suppose I throw a metal bar 1m long inside the event horizon of a supermassive black hole of 1 million solar masses. At both ends of the metal bar there is a light source.

(I chose a supermassive black hole to rule out any spaghettification process: with some quick calculations tidal forces on the metal bar would be just 1/1000th of g - so they are negligible - when the bar crosses the event horizon).

So, let's consider the following scheme:

H--------B++++A------------S

Where B++++A is the metal bar, H is the (just crossed) event horizon an S is the singularity in the middle of the black hole.

My question is:

Is it possible for light emitted in A to reach the other end point B of the bar?
Yes. A local inertial frame still behaves just like SR. Tidal forces simply rapidly decrease the scale (space and time) over which 'locally SR' is meaningful approximation. In a typical global coordinate system, this operation would be described as A and B heading for S faster than the outward directed light signal is heading for S, so B catches up to it. But in a local frame, the description would be as per SR.
Italian_Mike said:
From what I know once inside the event horizon all possible space-time curves always lead closer to the singularity, never farther. My opinion is that is not possible for light in A to reach point B.
Incorrect. See above for the resolution.
Italian_Mike said:
Now, substitute the metal bar with your head, and A and B with two neurons in your head and the light signal with an electric impulse sent between them. If it is not possible to send signals from A to B then you'll immediately lose conscoiusness and die as soon as you cross the event horizon (you'll die because what happens to the electric signals in your brain will also happen to all your bodily fluids: they'll be never be able to reach the back of your body).
So you'll be quite aware of when the event horizon is located as you'll immediately die.

A mechanical probe powered by electric signals will stop working as well.

Is this picture correct, in your opinion?
Thanks in advance for your reply!
All the last parts are also incorrect.
 
Thanks for the reply.

If I understood correctly, as no significant tidal forces are acting on the bar, then it is possible to associate a local inertial frame to this body.
So, for an observer in A the light will reach B in approx. 3.3 x 10-9 sec, just as if the inertial frame was outside the event horizon. So nothing special happens during the crossing of the horizon itself.
The event "B is reached by the light signal" should also happen for any other observer inside the event horizon so, as it is not possible for the light signal to actually move outwards with respect to the central singularity, then I guess that something similar to the following should be seen by such observers:

Time 0: H--------B+++PA------------S
Time 1: H-----------B++P+A---------S
Time 2: H--------------B+P++A------S
Time 3: H-----------------BP+++A---S

Where P is a photon belonging to the ray of light.
A, B, and P are always moving closer and closer to the singularity, but P definitely manages to reach B.

Is this picture correct?
 
That is the correct description in what I called 'typical global coordinates'. If A or B set up Fermi-Normal coordinates (the GR analog of Minkowski coordinates, usable only for a limited space-time region), the description would be just like SR up to second order effects (for 'soon after horizon crossing' for a quiescent supermassive BH). My quiescent, I mean no significant matter around producing intense radiation as it infalls. I also mean considering only classical effects, as many physicist argue the quantum requirements produce a firewall at the horizon.
 
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...
Back
Top