
#1
Feb2312, 01:25 PM

P: 270

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 



#2
Feb2312, 01:29 PM

P: 15,325

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. 



#3
Feb2312, 01:30 PM

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#4
Feb2312, 01:34 PM

P: 15,325

massive observers with light speed when crossing the event horizon 



#5
Feb2312, 01:35 PM

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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.




#6
Feb2312, 02:43 PM

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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.) 



#7
Feb2412, 01:33 AM

P: 270

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? 



#8
Feb2412, 02:26 AM

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#9
Feb2412, 04:05 AM

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? 



#10
Feb2412, 09:00 AM

P: 5,634

"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: Yuiop: 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..... 



#11
Feb2412, 10:24 AM

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#12
Feb2412, 01:04 PM

P: 3,967

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: [tex] \frac{v}{c} = \sqrt{\frac{2GM}{rc^2}}[/tex] 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 [itex]r = 2GM/c^2[/itex] 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: [tex] \frac{v}{c} = \sqrt{\frac{2GM}{rc^2}} \left( 1\frac{2GM}{rc^2} \right)[/tex] 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: [tex] g = \frac{GM}{r^2\sqrt{12GM/rc^2} }[/tex] 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: [tex] g = \frac{GM}{r^2} \left(1\frac{2GM}{rc^2}\right)[/tex] 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). 



#13
Feb2412, 01:21 PM

P: 1,555

This can be made clear with math so I do not know what you are arguing against. 



#14
Feb2412, 01:27 PM

P: 3,967





#15
Feb2412, 01:49 PM

P: 1,555

1. The gravitational based Doppler effect: [tex]{\Large \frac {1}{\sqrt {1{\frac {{\it rs}}{r}}}}}[/tex] 2. The velocity based Doppler effect, in case for free falling from infinity we have: [tex]\Large \sqrt {1{\frac {{\it rs}}{r}}} \left( 1+\sqrt {{\frac {{\it rs}}{r}}} \right) ^{1}[/tex] If we multiply the two factors and play around with the math to simplify the formula and normalize to rs=1 we get: [tex] \Large \left( 1+\sqrt {{r}^{1}} \right) ^{1}[/tex] Which is 0.5 if r goes to 1. (rs is the Schwarzschild radius) 



#16
Feb2412, 03:22 PM

P: 3,967

The velocity based relativistic Doppler shift can be written as: [tex]\frac{\sqrt{1v^2/c^2}}{(1+v/c)}[/tex] where v is the velocity of the observer away from the source. When combined with the gravitational Doppler effect this gives: [tex]\frac{\sqrt{1v^2/c^2}}{(1+v/c)}* \frac{1}{\sqrt{1r_s/r}}[/tex] Earlier I gave the free falling velocity as: [tex]\frac{v}{c} = \sqrt{\frac{2GM}{rc^2}} = \sqrt{\frac{r_s}{r}}[/tex] Substituting this value for v/c into the above equations gives the total Doppler shift as: [tex]\frac{1}{(1+ \sqrt{R_s/r)}} [/tex] so it appears your equation is right and that the redshift due the falling velocity exceeds the blueshift due to gravity. Good work :) 



#17
Feb2412, 09:25 PM

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The event horizon is a lightlike, or null surface. In fact, you can think of the event horizon as being the worldines of "trapped light'. As a consequence of the fact that the worldline of a stationary observer on the event horizon is lightlike, there isn't any such thing as a frame for a stationary observer at the horzion. See any of the FAQ questions about why a photon doesn't have a "point of view" for why. Note that not having a frame doesn't mean one can't use more general concepts of coordinates, such as null coordinates. It's just that the coordinates can't be broken down into one time and three space coordinates as a frame demands  because a coordinate that parameterizes the light beam (or the event horizon) is a null coordinate, it is neither a time coordinate or a space coordinate. The infalling observer does have a valid frame. In that frame, the speed of the lightlike event horzion will always be 'c'. 


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