Does time slow down when falling into a black hole?

[skeptic2]

As you can see, regardless of what your velocity is at infinity, you still cross the event horizon at the same speed, c.

Agreed that infalling objects must reach c at the event horizon but doesn't that mean that objects can't fall through the event horizon for the same reason they can't escape, because doing so means their speed must exceed c? Furthermore one wonders how an infalling object ever reaches the event horizon. Due to the contraction of space there is an infinite distance between an infalling object and the event horizon.

 

[stevebd1]

The velocity local to the infalling object doesn't exceed c but once past the EH, space becomes space-like so you might say the space the object is in begins to accelerate towards the singularity also, hence the proper velocity (i.e. relative to infinity) exceeds c. The equation for velocity (for a static black hole) changes from shell velocity which is limited to the EH (i.e. r>2M)-

v_{shell}=\frac{dr_{shell}}{dt_{shell}}=-\sqrt{\frac{2M}{r}}

to http://en.wikipedia.org/wiki/Proper_velocity" -

\frac{dr}{d\tau_{rain}}=-\sqrt{\frac{2M}{r}}

where

\tau_{2\ rain} - \tau_{1\ rain}\ =\ \frac{1}{3}\sqrt{\frac{2}{M}}\left(r_1^{3/2} - r_2^{3/2}\right)

which can apply to r<2M.

You'll find that proper velocity simply equals -\sqrt(2M/r) all the way to the singularity but the above is the full interpretation.


Regarding your second point, there is the http://en.wikipedia.org/wiki/Lorentz_factor" .

 

[v2kkim]

Considering the hail frame (hurled inward at speed v_far from a great distance), the equation for velocity as clocked by a shell observer would be-

v_{shell}=\left[\frac{2M}{r}+v_{far}^2\left(1-\frac{2M}{r}\right)\right]^{1/2}

which is derived from E/m=(1-v^2_{far})^{-1/2}

if the hail plunger was already at a velocity of 0.95c at a great distance from the BH, their velocity clocked by the shell observer at 2.1M would be 0.998c. From infinity, the velocity of the hail plunger would be-

\frac{dr}{dt}=\left(1-\frac{2M}{r}\right)\left[\frac{2M}{r}+v_{far}^2\left(1-\frac{2M}{r}\right)\right]^{1/2}

which is 0.0475c for the hail plunger at 2.1M as observed from infinity

As you can see, regardless of what your velocity is at infinity, you still cross the event horizon at the same speed, c.


source- 'Exploring Black Holes' by Wheeler & Taylor, 3-25

I have some doubt about this equation. Is it valid really up to very close to 2M ?

Near distance 2M to BH, with some error, the distance and time relation (from far away observer),
is obtainable from the 2nd eq. above,
\frac{dr}{dt}=\left(1-\frac{2M}{r}\right)
dt =dr/\left(1-\frac{2M}{r}\right)
Now we can try integration to get the time passed from far way observer, with r: R0 --> 2M
then we get infinity as r approaches 2M, so physically we say infinite time. But I am not sure it is right to apply r-2M can be arbitrary small. It is my imagination but we may avoid this r-2M~0 in the integration, by introducing some other principle like space quantization or uncertainty principle etc. In this way the velocity does not have to be 'c' right after crossing the event horizon (in falling object frame) but may be just very close to 'c'.

 

[skeptic2]

The velocity local to the infalling object doesn't exceed c but once past the EH, space becomes space-like so you might say the space the object is in begins to accelerate towards the singularity also, hence the proper velocity (i.e. relative to infinity) exceeds c. The equation for velocity (for a static black hole) changes from shell velocity which is limited to the EH (i.e. r>2M)-

v_{shell}=\frac{dr_{shell}}{dt_{shell}}=-\sqrt{\frac{2M}{r}}

to http://en.wikipedia.org/wiki/Proper_velocity" -

\frac{dr}{d\tau_{rain}}=-\sqrt{\frac{2M}{r}}

where

\tau_{2\ rain} - \tau_{1\ rain}\ =\ \frac{1}{3}\sqrt{\frac{2}{M}}\left(r_1^{3/2} - r_2^{3/2}\right)

which can apply to r<2M.

You'll find that proper velocity simply equals -\sqrt(2M/r) all the way to the singularity but the above is the full interpretation.

Regarding your second point, there is the http://en.wikipedia.org/wiki/Lorentz_factor" .

My understanding may be wrong but I believe that in a gravitational field the escape velocity replaces v in the Lorentz Transformation resulting in a contraction of space in the radial direction instead of an increase in length. My guess is that the sum of two effects is found by adding the velocity of the infalling object and the escape velocity at the position of the object relativistically.

s = v+u /(1+(v/c)(u/c))

where v is the infalling velocity
u is the escape velocity
and s is the sum of the two velocities

Thus at the event horizon space is infinitely contracted resulting in an infinite distance between any object and the horizon. Given the infinite distance and the velocity reaching c at the event horizon, how can an object ever pass through it?

 

[stevebd1]

http://en.wikipedia.org/wiki/Length_contraction" in SR and length expansion in GR are synonymous-

\gamma=\frac{1}{\sqrt{1-(v/c)^2}}=\frac{1}{\sqrt{1-(2M/r)}}

where M=Gm/c^2, free-fall velocity is v_{rel}=\sqrt(2M/r)c and 2M=EH

basically length contraction induced by free-fall velocity (which in turn is induced by curvature) balances out the length expansion induced by gravity.

For the infalling object at rain velocity, space is locally flat up to and beyond the event horizon (dr=1). Observed from infinity they come to a standstill at the event horizon, observed at very close to the EH, they are observed as traveling very close to c and appear contracted just before they appear to freeze at the EH (depending on how close the shell observer is to the EH). Regarding curved spacetime and distance, nothing appears to change for the infalling object locally traveling at rain velocity, but they would witness the outside universe speed up and have massive blue-shifted radiation at the EH and imminent tidal forces to contend with (depending on how big the BH was).

Here's a web page regarding the rain-frame and inside the horizon of a BH-
http://www.bun.kyoto-u.ac.jp/~suchii/inH.html

and some descriptive lecture notes regarding BH's-
http://physics.ucsd.edu/students/courses/winter2007/physics161/

The important thing to note is that spacetime remains locally flat for the free-falling object.

 

[stevebd1]

It would probably more accurate to say the free-falling object would follow the Kretschman curvature scalar (invariant under coordinate transformations), which for a static black hole is-

R_{abcd}R^{abcd}=\frac{48M^2}{r^6}

where R_{abcd} is the Riemann curvature tensor which is proportional to tidal forces, \Delta g=2Gm/r^3 (the tensor sometimes written as R_{abcd} \propto M/r^3). The invarient curvature scalar is smooth at r=2M but becomes singular at r=0.

 

[v2kkim]

Will the light propagating near BH event horizon slow down too as falling objects do, to the far away observer ?

 

[stevebd1]

Will the light propagating near BH event horizon slow down too as falling objects do, to the far away observer ?

It will appear to slow down and redshift proportionally to the time dilation (though nothing will appear to change locally for the infalling object) to the outside observer. As the object gets closer to the event horizon, the object (i.e. light from the object) will appear to slow down and become red to the external observer, it would then shift into infrared and infrared equipment would be required to see the object, likewise as it shifted through microwaves and into radiowaves, the electromagnetic wave eventually flatlining at the event horizon. It might be said that the highly redshifted image perceived of the object frozen close to the EH is trapped light slowly making its way out of the extreme gravitational field (still traveling at c locally) and what we are seeing is the memory of the object. The opposite would apply to the person falling into the BH who would see the outside universe blueshift and speed up.

Here's a good description of what happens-
http://cfpa.berkeley.edu/Education/BHfaq.html#q4

 

[monty37]

what is the life of a black hole,till when will it continue to swallow matter,at one point
when the limits exceed,does it rupture,due to gravitational overdose,its gravitatonal capacity reaches such an extent that it can't hold itself??

 

[DaleSwanson]

As far as I know there is no maximum size limit for a black hole. The only thing that will end a black hole is the slow process of http://en.wikipedia.org/wiki/Hawking_radiation" .

 

[monty37]

what exactly is this hawking radiation??in what way,can that supposedly end
a black hole?

 

[DaleSwanson]

http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html

 

[monty37]

the black hole's gravitational pull is tremendous ,so how does it allow hawking radiation
to radiate out of a black hole,even if it is a slow process

 

[Chronos]

Hawking radiation is a quantum effect. It occurs at the 'surface' of the event horizon. Some photons escape effectively causing the black hole to radiate and eventually [very slowly] decrease in mass. For black solar mass and larger black holes, it will be eons before they actually start to 'shink'. They will absorb more mass than they emit for many billions of years.

 

[George Jones]

what exactly is this hawking radiation??in what way,can that supposedly end a black hole?

http://www.physics.ucdavis.edu/Text/Carlip.html#Hawkrad

is the best non-technical description of Hawking radiation of which I know.

 

[monty37]

so will the mass of a black hole finally become 0,as it shrinks and disappears??can
black holes be considered as perfect black bodies

 

[Chronos]

Techinically, black holes should explode when they fall below Planck mass. Albeit, a pretty wimpy end to a former bully. A tiny poof of gamma radiation is their legacy. It will, however, take dang near eternity for any such body to evaporate.