Massive observers with light speed when crossing the event horizon

by Lapidus
Tags: event, horizon, light, massive, observers, speed
 P: 290 Is it true that massive observers travel with c when passing an event horizion? I know that light cones get tilted at the event horizon. But every observers travels at light speed there? thanks in advance
P: 15,319
 Quote by Lapidus Is it true that massive observers travel with c when passing an event horizion? I know that light cones get tilted at the event horizon. But every observers travels at light speed there? thanks in advance
No. No massive object can reach c, even at the event horizon of a BH.

The EH simply means that the escape velocity at that point is c, meaning that that any object (such as a photon) will never achieve escape velocity. It is incidental that any other object moving slower than c cannot escape either.
P: 1,555
 Quote by Lapidus Is it true that massive observers travel with c when passing an event horizion?
Only an instantly short period.

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Massive observers with light speed when crossing the event horizon

 Quote by Passionflower Only an instantly short period.
No. Not for any length of time.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 No, that is not true. A massive object can never have speed c, even for an "instantly short period" (whatever that means). The speed of an object crossing the event horizon has nothing to do with the escape velocity. Saying that the escape velocity of a black hole is c simply means that no massive object can escape from the black hole, going awasy from the black hole. That has nothing to do with crossing the event horizon going toward the black hole.
PF Gold
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 Quote by Lapidus Is it true that massive observers travel with c when passing an event horizion?
Actually, it's the other way round. When a massive observer crosses the event horizon, the speed of the horizon relative to the observer is c.

You may think that's the same thing but it isn't. The reason for this is the same reason that, although all observers measure the local speed of light to be c, the concept of the speed of an observer relative to light isn't even defined. (See FAQ: Rest frame of a photon.)
 P: 290 Thanks for all the answers! Another related question: What does the world (the outside of the black hole, the universe) look like to an observer crossing the EH?
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 Quote by Lapidus Thanks for all the answers! Another related question: What does the world (the outside of the black hole, the universe) look like to an observer crossing the EH?
50% redder.
 P: 3,967 Another related question: What does the world (the outside of the black hole, the universe) look like to an observer crossing the EH?[/QUOTE] The universe from just inside the EH looks pretty much like the universe looked when the observer was just outside the EH which follows from the frequent claim that an observer will see nothing remarkable at the EH. There is no sudden change. It is also often claimed that a falling observer will see the whole future of the universe evolve as he falls through the EH but that is not true. Light rays are blue shifted as they fall towards the EH from the point of view of an observer that is static with respect to the black hole, but in the case of a free falling observer there will be an additional Doppler shift due to his falling motion away from the source, but I am not sure if that is enough to change the blue shift into a red shift. So redder or bluer?
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"But every observers travels at light speed there?"

Your speed in free fall depends on your initial conditions ....like free fall height from above earth where g is 32ft/sec/sec....but is speed is never 'c'...A free falling observer passes the event horiozon without incident....it is invisible to such an observer....In the example I give below, radiation from the horizon would be detected...virtual particles become 'real'...

DrGreg:

 Actually, it's the other way round. When a massive observer crosses the event horizon, the speed of the horizon relative to the observer is c.
Can you explain this a bit further? If you mean that photons are passing in a circular fashion in the photon sphere, I would understand that...but I never heard of that referred to as a 'horizon'...it's close, but outside the horizon.

Yuiop:

 Light rays are blue shifted as they fall towards the EH from the point of view of an observer that is static with respect to the black hole, but in the case of a free falling observer there will be an additional Doppler shift due to his falling motion away from the source, but I am not sure if that is enough to change the blue shift into a red shift. So redder or bluer?
Depends on your free fall speed relative to the gravitational potential outside the horizon [which varies with the size of the black hole].....here is a blueshift example:

Kip Thorne has the most complete description of observations to be expected from a space ship nearing a black hole horizon that I have seen ..outside the horizon of course...in the prologue of his book BLACK HOLES AND TIME WARPS....here are a few from several pages of text: [ near exact quotes]

Thorne uses a 10g acceleration in the following....

At first sight it appears the black hole blots out light from all the stars and galaxies behind the hole..but you discover the black hole acts as a lense...deflecting some light from the 'hidden' stars and galaxies into a thin bright ring [initially] at the edge of the black disc revealing images of some obscured stars....another image of light rays pulled into one orbit around the hole and then released in your direction, another by rays that orbited twice, and so on....You expect the hole to stop growing as would a black floor spreading out beneath your ship, but no it keeps growing up and around you leaving a circular opening directly above you through which you can see the external universe....A galaxy that was previously near the horizon now appears almost vertically overhead...but the colors of all the stars and galaxies are 'wrong' ....gravitational attraction has made the radiation [light] more energetic by decreasing its wavelength....
a local experiment reveals the speed of light is 'c'... everything is normal aboard, absolutely the same as if the ship had been resting on a massive plant with 10g gravity...The most interesting external observation is the bizzare spot of light overhead and the engulfing blackness all around.....
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 Quote by Naty1 Depends on your free fall speed relative to the gravitational potential outside the horizon [which varies with the size of the black hole]
If you free fall radially at escape velocity then at the EH all light coming from far away is 50% of the original frequency, this is independent of the size of the black hole.
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 Quote by HallsofIvy No, that is not true. A massive object can never have speed c, even for an "instantly short period" (whatever that means).
If I was to ask you what the vertical velocity of a free falling object relative to the ground was at its apogee, you would probably say 0 m/s. I would then ask you how you propose to measure this on the principle that we only believe what we can directly measure. You might propose that we measure its position 1 femto second before apogee and one femto second after apogee and calculate its average velocity and obtain zero. However, I would say that is not good enough and point out that we both accept its velocity is continually changing and I want to know its exact velocity at the instant the object is exactly at its apogee and I think you would have to concede that is impossible to measure because it requires a finite time interval to measure a velocity. Demanding to know the the instantaneous velocity at an exact location or point in time requires the concept, that I think passionflower was getting at, of a time interval with a duration of exactly zero time units and that is not practical or useful. This might seem like extreme nit picking, but it is important in relativity because as we all know, being able to get arbitrarily close to the speed of light is still a world away from actually travelling at the speed of light. Similarly, if we accept the idea that it is impossible to have a stationary observer exactly at the event horizon, (along with the notion that there is no such thing as an instant - See paper by Peter Lynds) then we have to accept that there is no practical way to measure the the velocity of an object exactly at the EH. Even though we can establish that the velocity of a free falling object can be as arbitrarily close to the speed of light as measured by an observer hovering arbitrarily close to the event horizon, this does not establish that the object reaches light speed exactly at the the EH.

 Quote by HallsofIvy The speed of an object crossing the event horizon has nothing to do with the escape velocity. Saying that the escape velocity of a black hole is c simply means that no massive object can escape from the black hole, going away from the black hole. That has nothing to do with crossing the event horizon going toward the black hole.
I think it is only fair to point out that generally the escape velocity at a given point in a gravitational field has everything to do with the free fall velocity of an object at that point falling towards the massive body. The EH (and all points below it) may be an exception, because it is not possible to have a stationary observer at or below the EH as mentioned above. Of course to prove this claim we need to prove that it is not possible to have a stationary observer at the EH.

To investigate this further we need to check the GR equations for free fall velocity and acceleration in the region of the EH. THe local free fall velocity is given by:

$$\frac{v}{c} = \sqrt{\frac{2GM}{rc^2}}$$

Interestingly this equation for the escape velocity (or free falling velocity) of an object is the same in Newtonian and GR physics. It gives us the superficial result that the free fall velocity at the event horizon when $r = 2GM/c^2$ is the speed of light. However this is the velocity as measured by a stationary local observer at the EH and we have not established that it is possible to have such an observer. An alternative equation:

$$\frac{v}{c} = \sqrt{\frac{2GM}{rc^2}} \left( 1-\frac{2GM}{rc^2} \right)$$

is the coordinate velocity and this suggests that the free fall velocity at the event horizon is zero and does not support the idea that it is impossible to be stationary at the EH, but in turn it does not support the idea that it is possible to free fall at the speed of light at the EH. So the coordinate point of view gives stationary at the EH and the local point of view gives essentially undefined at the EH. Next, we have to look at the equations for gravitational acceleration to shed some light on this issue. The local acceleration is given by:

$$g = \frac{GM}{r^2\sqrt{1-2GM/rc^2} }$$

supporting the view that it is impossible to be stationary at the EH because the acceleration becomes locally infinite there. However, the coordinate equation for the acceleration is:

$$g = \frac{GM}{r^2} \left(1-\frac{2GM}{rc^2}\right)$$

which gives the result that the coordinate gravitational acceleration at the EH is zero supporting the equation that the coordinate free fall velocity is also zero and in coordinate terms it is possible to have a stationary observer at the EH. It all depends upon whether you believe the local equations or the coordinate equations are a better reflection of reality, but it is only fair to state that the overwhelming conventional wisdom is that the local view is the better reflection of reality (infinite acceleration at the EH, no stationary observers at the EH and undetermined free fall velocity at the EH).
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 Quote by HallsofIvy No, that is not true. A massive object can never have speed c, even for an "instantly short period" (whatever that means). The speed of an object crossing the event horizon has nothing to do with the escape velocity.
An object radially free falling from infinity travels at escape velocity, the escape velocity exactly at the EH is c.

This can be made clear with math so I do not know what you are arguing against.
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 Quote by Passionflower If you free fall radially at escape velocity then at the EH all light coming from far away is 50% of the original frequency, this is independent of the size of the black hole.
You may well be right, but can you support that with an equation?
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 Quote by yuiop You may well be right, but can you support that with an equation?
The Doppler factor wrt to far away light for a radially free falling from infinity test observer in a Schwarzschild solution depends on two factors:

1. The gravitational based Doppler effect:

$${\Large \frac {1}{\sqrt {1-{\frac {{\it rs}}{r}}}}}$$

2. The velocity based Doppler effect, in case for free falling from infinity we have:

$$\Large \sqrt {1-{\frac {{\it rs}}{r}}} \left( 1+\sqrt {{\frac {{\it rs}}{r}}} \right) ^{-1}$$

If we multiply the two factors and play around with the math to simplify the formula and normalize to rs=1 we get:
$$\Large \left( 1+\sqrt {{r}^{-1}} \right) ^{-1}$$

Which is 0.5 if r goes to 1.

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 Quote by Passionflower If we multiply the two factors and play around with the math to simplify the formula and normalize to rs=1 we get: $$\Large \left( 1+\sqrt {{r}^{-1}} \right) ^{-1}$$ Which is 0.5 if r goes to 1. (rs is the Schwarzschild radius)

The velocity based relativistic Doppler shift can be written as:

$$\frac{\sqrt{1-v^2/c^2}}{(1+v/c)}$$

where v is the velocity of the observer away from the source.

When combined with the gravitational Doppler effect this gives:

$$\frac{\sqrt{1-v^2/c^2}}{(1+v/c)}* \frac{1}{\sqrt{1-r_s/r}}$$

Earlier I gave the free falling velocity as:

$$\frac{v}{c} = \sqrt{\frac{2GM}{rc^2}} = \sqrt{\frac{r_s}{r}}$$

Substituting this value for v/c into the above equations gives the total Doppler shift as:

$$\frac{1}{(1+ \sqrt{R_s/r)}}$$

so it appears your equation is right and that the redshift due the falling velocity exceeds the blueshift due to gravity. Good work :)
Emeritus