Does time slow down when falling into a black hole?

[.:Endeavour:.]

The astronaut, like in my post earlier, to actually know that the astronaut is slowing down is because of the little light that actually escapes from a black hole. That light can be last second ago to a century ago, or more. The astronaut might be already made it out of the black hole, but you are still seeing him falling into the black hole from the time that it takes light to reach the eyes of the observer. Is like seeing light traveling at the speed of sound or at 9 m/s while seeing a diver dive into the pool. The diver might be on the pool, but you are still seeing the diver falling in the pool.

 

[Sheneron]

Ok I know gravity slows the time down I get that. I really do. Everyone forget time for a second hah. My question this whole time is whether his actual velocity slows down along with the time, or whether he still travels at near the speed of light. Does it just seem like he is there forever(optical effect) or is he really there forever?

 

[v2kkim]

The gravity near black hole makes things complicated and interesting. And I am still trying to understand about it, but in general the following seems to be accepted: The far away person will observe the falling astronaut become slow and got frozen forever; but the astronaut will observe himself continuously moving to the center of BH. So, are you ready to accept two realities ?

Now, one interesting question is what the astronaut will observe while crossing event of horizon about the far away outside ?

 

[stevebd1]

Hello Sheneron

Here's a couple of equations from 'Exploring Black Holes' by Taylor & Wheeler (where M=Gm/c^2 and c=1)

velocity of a free-falling object as clocked by the shell observer (v=1 at event horizon)-

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

(minus square root because the expression describes a decreasing radius as the object falls toward the black hole outside the horizon)

where dr_{shell}=dr/\sqrt(1-2M/r) and dt_{shell}=dt\sqrt(1-2M/r)

velocity as viewed from infinity (v=0 at event horizon)-

\frac{dr}{dt}=-\left(1-\frac{2M}{r}\right)\sqrt{\frac{2M}{r}}

which is derived from the energy equation E/m=(1-2M/r)dt/d\tau=1

 

[.:Endeavour:.]

Ok I know gravity slows the time down I get that. I really do. Everyone forget time for a second hah. My question this whole time is whether his actual velocity slows down along with the time, or whether he still travels at near the speed of light. Does it just seem like he is there forever(optical effect) or is he really there forever?

It might seem he is falling forever, but its because the light that you seeing been reflected the astronaut is light that actually escaped from the black hole, which as we know is very little. I think that excerpt that you posted about the observer looking at a continuous falling person, is because light doesn't easily escape a black holes surface. To actually know that he is slowing down, we cannot know. Black hole's gravity is massive on how compact the black hole is. So if a black hole is 2 miles across, it will have a massive gravity in contrast of a black hole that is 50 miles across. What I think, is that the astronaut will be accelerating until he eventually reaches the horizon, just like a free-falling object here on Earth accelerates do to the force of gravity.

 

[monty37]

well,it is said that the laws of physics break in a black hole,and all mostly all laws
of physics are to do with time,mass and velocity.so how can gravitational
dilation occur at all,since it is gravitational force controlling time,this is a law -the
universal law of gravitation.we are so attached to laws of physics to think
of other consequences!

 

[.:Endeavour:.]

Not all laws of physics are broken, for example a black hole has a limit on how much matter it can contain, that's the reason why scientist see jets shooting upward and downward from a black hole. Is like stuffing dirt in a 2 liter container. There is a limit on how much the container can contain until it breaks.

 

[Chronos]

Irrelevant, there is no mass limit on a black hole. Provide references supporting your claim. The infalling observer is at a disadvantage. While you observe him slowing to a crawl, in his reference frame the outside universe becomes increasingly blue shifted, with increasingly undesirable side effects.

 

[.:Endeavour:.]

Irrelevant, there is no mass limit on a black hole. Provide references supporting your claim. The infalling observer is at a disadvantage. While you observe him slowing to a crawl, in his reference frame the outside universe becomes increasingly blue shifted, with increasingly undesirable side effects.

Remember that matter cannot be destroyed. The matter doesn't seemly disappear into thin air, so where is your part of the story that the black hole gobbles up the matter everything and doesn't have a limit? The matter gets crushed with the enormous gravity of a black hole, but does it grow in size or keep a constant mass?

 

[monty37]

matter would rather have a constant mass,why should it grow in size,if it still has
a limit ,is it that we still don't know the limit.what will happen if matter crosses the
limit in a black hole?

 

[v2kkim]

My understanding is that the mass of BH is concentrated at the center of volume almost zero, and of no limit in mass. With mass increase the horizon of event will grow large. Some energy or materials can be released to outside somehow but I think it is very small quantity. Actually I wonder what mechanism can spit out stuff. There should be other mechanism, at least as powerful as gravity collapse, maybe a new physics law on BH ?

 

[Division]

If you venture into a black hole, there won't be a single law of physics that will apply, as we know of.

 

[monty37]

well it is a white hole that spits stuff,and a law as such is out of question
as laws are meant to be broken in a black hole,so actually speaking
do the constant scientific values which we have found for an electron etc.hold true
in black hole?and the only law of physics that applies as one ventures a black hole
is gravitation,it prevails and rules!

 

[Chronos]

Provide a reference paper that acknowledges the exsistence of a 'white hole', monty. I would say you have pinned yourself in a corner

 

[monty37]

when we believe in black holes that have not been observed,why not consider white holes ,and when we believe in antiparticles.

 

[v2kkim]

Black hole friends: Recently I found that the falling astronaut does not freeze on the event horizon to outside observer. I heard the reason is as he approaches event horizon very close the event horizon itself can grow to swallow him. I think our old thinking was too ideally mathematical and simple. So it is time to unfreeze the frozen astronaut, and we see his journey continue.

 

[Klockan3]

Considering that "standing still" in a gravity field is the same thing as moving and in the same way as you can't see anyone moving faster than the speed of light you can't see anyone move beyond the schwarzschild radius.

Technically though if you consider a guy moving at 0.99c but instead of calculating it based on your time you calculate it based on his time he would be moving faster than the speed of light, its the same thing, he can be closer but in your reference frame you don't see it like that. By the way in special relativity time isn't something magical, its just the speed at which every process except direct movement in one direction happens at.

 

[DaleSwanson]

when we believe in black holes that have not been observed,why not consider white holes ,and when we believe in antiparticles.

There is no reason black holes shouldn't exist. They are simply matter dense enough that the escape velocity at the surface exceeds c. We create antiparticles in particle accelerators.

 

[v2kkim]

For falling objects to BH, I agree that time dilation could be a considerable factor to outsider. So to outsider the falling object will look like slow motion movie, but I think the movie will end. The falling object will disappear completely. The duration of the movie will depend on the falling object motion and event horizon situation (I assume event horizon can be changing more or less.).

 

[stevebd1]

Technically though if you consider a guy moving at 0.99c but instead of calculating it based on your time you calculate it based on his time he would be moving faster than the speed of light, its the same thing, he can be closer but in your reference frame you don't see it like that.

Can you elaborate on what you mean here? You might also find the following of interest.

Considering the calcs from post #34 regarding the rain frame (falling from rest at infinity), at about 2.1M (just outside the EH), the velocity of the infalling plunger would be clocked by a shell observer (at rest at 2.1M) to have a velocity of 0.976c. From infinity, the same infalling plunger would be observed as having a velocity of 0.047c at 2.1M


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

 

[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" .