View Full Version : Acceleration vs Velocity - Black Hole Event Horizon

xylophobe

2012-Feb-28, 08:18 PM

If an object has an acceleration = c/1000 m/s2 then according to Newtonian calculations it would reach the speed of light in 1000 seconds.

But then there is the time dilation effect sqrt(1- v2/c2) which when multiplied by the above Newtonian calculations results in the velocity = 0.0 m/s at time = 1000 s

According to my Microsoft Excel chart (below) the maximum speed the object can attain is 0.5c at time = 707 seconds.

Is this the braking effect of GR because I read somewhere on this forum that GR actually begins to brake an in-falling object so that the object stops at the event horizon?

Or maybe I'm just not calculating it correct ....

16358

Cougar

2012-Feb-29, 04:05 AM

I read somewhere on this forum that GR actually begins to brake an in-falling object so that the object stops at the event horizon?

No, unless I'm mistaken, you misunderstood. By my logic, it continues to accelerate and reaches c at the event horizon (or at least almost), after which it quickly crosses. It's the distant observer who sees things differently.

quotation

2012-Feb-29, 01:47 PM

but apparently not everything crosses the event horizon? I just saw this yesterday regarding ultra-fast outflows (real "ufo's") that in some cases may serve as a "feedback mechanism between a galaxy's black hole and its star-formation processes." I don't have access to the original paper (Feb. 27 issue of Monthly Notices of the Royal Astronomical Society) but the gee-whiz version is here: http://www.sciencedaily.com/releases/2012/02/120227162801.htm and unless i'm mistaken does sort of suggest some kind of braking mechanism?

antoniseb

2012-Feb-29, 02:22 PM

... and unless i'm mistaken does sort of suggest some kind of braking mechanism?

Magnetic fields are the usually assumed culprit. There are a lot of very cool looking simulation studies that show how a disk of charged particles around a rotating black hole will turn into relativistic jets.

xylophobe

2012-Feb-29, 04:54 PM

No, unless I'm mistaken, you misunderstood. By my logic, it continues to accelerate and reaches c at the event horizon (or at least almost), after which it quickly crosses. It's the distant observer who sees things differently.

When I multiplied the velocity by sqrt(1-v2/c2) I forgot that both the meters and the seconds receive the same relativistic factoring which means the sqrt(1-v2/c2) for the meters cancels out the sqrt(1-v2/c2) for the seconds and thus we end up with just m/s.

For acceleration m/s2 because the seconds are squared then the acceleration becomes sqrt(1-v2/c2)*m/s2 which causes a graph like this:

16362

xylophobe

2012-Feb-29, 05:41 PM

An interesting fact to note when looking at that graph (http://www.bautforum.com/attachment.php?attachmentid=16362&d=1330534411) is that gravitational acceleration ceases when an object reaches the speed of light.

So if an in-falling object that starts at zero velocity from an infinite distance plunges towards a blackhole then at the event horizon it reaches c-velocity - the blackhole no longer pulls the object towards itself ... but the inward velocity remains so that the object slams into the singularity at the speed of light.

This is also an anti-gravity mechanism because if we can get a spacecraft to c-velocity, instantly, then the earth's gravity no longer affects the spacecraft (other than by redshifting the energy of the spacecraft).

whimsyfree

2012-Mar-07, 12:14 AM

An interesting fact to note when looking at that graph (http://www.bautforum.com/attachment.php?attachmentid=16362&d=1330534411) is that gravitational acceleration ceases when an object reaches the speed of light.

I don't think that graph tells us anything about gravity, or what happens to objects travelling at the speed of light.

So if an in-falling object that starts at zero velocity from an infinite distance plunges towards a blackhole then at the event horizon it reaches c-velocity - the blackhole no longer pulls the object towards itself ... but the inward velocity remains so that the object slams into the singularity at the speed of light.

I don't think an object can fall an infinite distance in a finite amount of time.

caveman1917

2012-Mar-11, 11:48 PM

Or maybe I'm just not calculating it correct ....

You're indeed not calculating it correctly. For an object under constant acceleration you have to use the rindler metric, you can't just multiply your newtonian result by \gamma, for one thing accelerations don't transform like velocities. And if you want this acceleration to apply to gravity (such as a black hole), you have to use GR, specifically the schwarzschild metric in your case.

Jeff Root

2012-Mar-12, 07:55 AM

Is this the braking effect of GR because I read somewhere on

this forum that GR actually begins to brake an in-falling object

so that the object stops at the event horizon?

There is no braking effect. Distant observers would see a

falling object appear to slow to a stop at the event horizon,

but it would not "actually" slow at all. An object free-falling

from infinite distance reaches c at the event horizon, relative

to the event horizon. It does not stop accelerating. The

acceleration continues to increase as it approaches the

mass at the center of the black hole. But the speed of the

object is never observed by anyone to be greater than c,

because it is never greater than c relative to any "local"

observer. All observers at or inside the event horizon are

falling toward the center, though they can be falling at

different speeds. All objects in free-fall in any gravity

field are pulled farther and farther apart because the

objects closer to the gravitating mass are accelerated

more than those farther away.

-- Jeff, in Minneapolis

Andrew D

2012-Mar-15, 06:00 AM

The reason the particle slows to a stop at the horizon is not due to the acceleration of the particle but the "change in the shape of space" above the horizon, which depends on the Ricci tensor in the coordinates of the distant observer.

Also, the statement that the particle accelerates to c at the horizon is not valid, a particle can cross the horizon (in its own coodrinates) with an arbitrarily small velocity. It seems to me that a massive particle starting from infinite distance will reach a velocity which depends on the mass of the black hole. If a particle from infinitely far reaches c at the horizon, it has an infinite kinetic energy there, so the amount of work done by the black hole is infinite. However, the integral of the force acting on the particle over the path from the point at infinity to the horizon (the work) is finite since the force achieves a maximum at the horizon and goes to zero at infinity.

w =\int_{P}\mathbf{F}d\mathbf{x} = \int_{P}\frac{M}{\mathbf{r}^2}d\mathbf{r} = M\int_{R_s}^{\infty}\frac{1}{\mathbf{r}(s)^{2}}\ma thbf{r}'(s)ds = M\int_{R_s}^{\infty}\frac{1}{s^{2}}ds =\frac{M}{R_s}

Where P is the path the particle travels, r(s) = s (since the path is a line) is a parameterization of P, M = GmBHmp, and Rs is the Swarzchild radius.

Notice that if we substitute 0 for Rs, then the value of the integral is infinity. With this substitution the calculation is for a path all the way to the singularity.

astromark

2012-Mar-15, 07:53 AM

[QUOTE=Roobydo;1998484]. . .

w =\int_{P}\mathbf{F}d\mathbf{x} = \int_{P}\frac{M}{\mathbf{r}^2}d\mathbf{r} = M\int_{R_s}^{\infty}\frac{1}{\mathbf{r}(s)^{2}}\ma thbf{r}'(s)ds = M\int_{R_s}^{\infty}\frac{1}{s^{2}}ds =\frac{M}{R_s}

Its very interesting. Eloquent even., but for me it is filled with sadness.

My Father was a mathematics teacher and spoke often of the beauty of mathematics..

Trouble with this, is this.. (As I see it.) It's wrong.

The reality I talk of has that area inside the event horizon as a rather crowded and dense ..

Right on down to the singularity. If there is one.. Mathematics can be so clean. reality is filled with stuff. Lots of stuff...

Try and see it as I do. It's a very clear image from my point of view.

Watching that object in falling.. yes it would appear to slow and stop.. but being that object you accelerate to oblivion.

Being ripped to bits and consumed completely ... without the very nice maths imagery.

astromark

2012-Mar-15, 08:04 AM

Write the following.. ( Maths can be both right and good looking.) accelerating mass to reach velocity c.,

and the destruction of the said mater to a particle soup and plasma. Gone to become part of the mass object of the BH.

Am I witnessing the arrival of mathematics to be greater than science.. No. I did not think so either.

The maths must always be a tool. We just do not know the equation. Clearly.

So I return to add.. that I do agree 100% with the conclusions reached by the Maths .. I do concede it as correct.

It's only the simplicity of it I challenge. BH are messy. The rules of the unknown are unknown.

caveman1917

2012-Mar-15, 11:37 AM

The reason the particle slows to a stop at the horizon is not due to the acceleration of the particle but the "change in the shape of space" above the horizon, which depends on the Ricci tensor in the coordinates of the distant observer.

Also, the statement that the particle accelerates to c at the horizon is not valid, a particle can cross the horizon (in its own coodrinates) with an arbitrarily small velocity. It seems to me that a massive particle starting from infinite distance will reach a velocity which depends on the mass of the black hole. If a particle from infinitely far reaches c at the horizon, it has an infinite kinetic energy there, so the amount of work done by the black hole is infinite. However, the integral of the force acting on the particle over the path from the point at infinity to the horizon (the work) is finite since the force achieves a maximum at the horizon and goes to zero at infinity.

w =\int_{P}\mathbf{F}d\mathbf{x} = \int_{P}\frac{M}{\mathbf{r}^2}d\mathbf{r} = M\int_{R_s}^{\infty}\frac{1}{\mathbf{r}(s)^{2}}\ma thbf{r}'(s)ds = M\int_{R_s}^{\infty}\frac{1}{s^{2}}ds =\frac{M}{R_s}

Where P is the path the particle travels, r(s) = s (since the path is a line) is a parameterization of P, M = GmBHmp, and Rs is the Swarzchild radius.

Notice that if we substitute 0 for Rs, then the value of the integral is infinity. With this substitution the calculation is for a path all the way to the singularity.

You can't calculate this with Newton, that approximation will only work far away from the black hole, it won't give you the correct answer for a particle crossing the event horizon. You really do need to work on the schwarzschild metric directly, for a suitable definition of "velocity" (which becomes ill-defined globally, the easiest is to define velocity as local wrt a stationary observer at that point). If you do it that way you'll see that a massive particle dropped from rest at infinity accelerates to c at the horizon.

Andrew D

2012-Mar-15, 02:28 PM

[QUOTE=Roobydo;1998484]. . .

w =\int_{P}\mathbf{F}d\mathbf{x} = \int_{P}\frac{M}{\mathbf{r}^2}d\mathbf{r} = M\int_{R_s}^{\infty}\frac{1}{\mathbf{r}(s)^{2}}\ma thbf{r}'(s)ds = M\int_{R_s}^{\infty}\frac{1}{s^{2}}ds =\frac{M}{R_s}

Its very interesting. Eloquent even., but for me it is filled with sadness.

My Father was a mathematics teacher and spoke often of the beauty of mathematics..

Trouble with this, is this.. (As I see it.) It's wrong.

The reality I talk of has that area inside the event horizon as a rather crowded and dense ..

Right on down to the singularity. If there is one.. Mathematics can be so clean. reality is filled with stuff. Lots of stuff...

Try and see it as I do. It's a very clear image from my point of view.

Watching that object in falling.. yes it would appear to slow and stop.. but being that object you accelerate to oblivion.

Being ripped to bits and consumed completely ... without the very nice maths imagery.

Without the maths, we wouldn't know the first thing about any of this stuff. You don't have to know it, but you have to respect it. Read my signature, I don't claim that there is any perfect mathematical description of nature, but there's extremely close models.

You can't calculate this with Newton, that approximation will only work far away from the black hole, it won't give you the correct answer for a particle crossing the event horizon. You really do need to work on the schwarzschild metric directly, for a suitable definition of "velocity" (which becomes ill-defined globally, the easiest is to define velocity as local wrt a stationary observer at that point). If you do it that way you'll see that a massive particle dropped from rest at infinity accelerates to c at the horizon.

Well damn. That's why I keep you around.

Andrew D

2012-Mar-15, 10:47 PM

[QUOTE=Roobydo;1998484]. . .

w =\int_{P}\mathbf{F}d\mathbf{x} = \int_{P}\frac{M}{\mathbf{r}^2}d\mathbf{r} = M\int_{R_s}^{\infty}\frac{1}{\mathbf{r}(s)^{2}}\ma thbf{r}'(s)ds = M\int_{R_s}^{\infty}\frac{1}{s^{2}}ds =\frac{M}{R_s}

Its very interesting. Eloquent even., but for me it is filled with sadness.

My Father was a mathematics teacher and spoke often of the beauty of mathematics..

Trouble with this, is this.. (As I see it.) It's wrong.

The reality I talk of has that area inside the event horizon as a rather crowded and dense ..

Right on down to the singularity. If there is one.. Mathematics can be so clean. reality is filled with stuff. Lots of stuff...

Try and see it as I do. It's a very clear image from my point of view.

Watching that object in falling.. yes it would appear to slow and stop.. but being that object you accelerate to oblivion.

Being ripped to bits and consumed completely ... without the very nice math

I don't want to start a splinter discussion here, but do you find mathematical descriptions of nature "full of sadness" because they are ultimately incomplete, or because you think it is preferable to forsake mathematics and rely solely on thought experiment?

The former I think is silly, the latter I find quite offensive.

I do not wish to see it as you do, but as Einstein saw it. There was a reason he travelled to Italy before developing his great theory, and it was certainly not to learn new methods for thought experimentation, it was to learn differential geometry. He had the incredible insight to see that the mathematical framework of differential geometry was necessary to develop his theory beyond thought and conjecture. I hope to understand his theory beyond thought and conjecture also, so I study mathematics. You may choose to fumble in the dark if you wish, but mathematics is surely the lamp which shines brightly to uncover the mysteries of nature.

Jeff Root

2012-Mar-16, 12:15 AM

Roobydoo,

I think that Mark finds mathematical descriptions of nature

"full of sadness" because they are so cryptic and so far

removed from the direct experience of nature.

Did I get that right, Mark? Is there more to it?

-- Jeff, in Minneapolis

Selfsim

2012-Mar-16, 12:22 AM

I don't want to start a splinter discussion here, but do you find mathematical descriptions of nature "full of sadness" because they are ultimately incomplete, or because you think it is preferable to forsake mathematics and rely solely on thought experiment?

The former I think is silly, the latter I find quite offensive.

I do not wish to see it as you do, but as Einstein saw it. There was a reason he travelled to Italy before developing his great theory, and it was certainly not to learn new methods for thought experimentation, it was to learn differential geometry. He had the incredible insight to see that the mathematical framework of differential geometry was necessary to develop his theory beyond thought and conjecture. I hope to understand his theory beyond thought and conjecture also, so I study mathematics. You may choose to fumble in the dark if you wish, but mathematics is surely the lamp which shines brightly to uncover the mysteries of nature.I whole heartedly agree, Roobydo !

Dismissal of basic concepts like mathematics, using appeals to emotions like 'sadness' in a science forum, is unfathomable for me also.

(This is my last off-topic post on this front ..).

Regards

astromark

2012-Mar-16, 01:21 AM

No 'Roobydo' I do not see it as off topic. Acceleration vs Velocity is exactly what we are speaking of.

Your beautiful mathematics equation might be absolutely correct.. Describing the curve of energies witnessed or seen.

and Yes 'Jeff' thats how I see it.

It becomes a complex understanding when a realization of mass distortion of space time is written as a acceleration..

I think we agree more is happening than that.

Utter catastrophic destruction of the very structures of mater and chemical bonding lost...

When a Star's mass exceeds stability and a collapse begins.. All the rules we know are broken.

I can not be so conclusive as fact. That a singularity is to be found.

What might be mathematically correct might not be reality.

Selfsim

2012-Mar-16, 03:46 AM

No 'Roobydo' I do not see it as off topic. Acceleration vs Velocity is exactly what we are speaking of.Ok .. so if its not OT, then I'll comment further ..

It becomes a complex understanding when a realization of mass distortion of space time is written as a acceleration..

Objects don't fall into black holes at c, (as per the usual Special Relativity constraints).

For an observer outside the Event Horizon (EH), if an object is released from rest, and falls towards a Schwarzschild BH, it will reach a maximum co-ordinate velocity of 0.39c at a distance of 3 times the Schwarzschild radius, after which it appears to slow down, (from the outside observer's perspective), due to the effects of gravitational redshift. At the EH, the object is stationary, or the co-ordinate velocity equals zero. This is totally different to Newtonian gravity, where the velocity will continue to increase monotonically.

In the object's frame of reference, the object reaches an infinite geodesic acceleration at the EH .. it won't escape the BH.

The geodesic acceleration is based on the object's proper time, but the co-ordinate velocity is based on the observer's time. The object has an infinite acceleration, but zero velocity at the EH. This is because the acceleration is based on the object's frame of reference, and the velocity is based on the outside observer's frame of reference.

The way the above description is derived, is by substituting the Schwarzschild metric into the Einstein (vacuum) Field equations, and solving the resultant equations. (Ie: by 'performing the maths' !!!).

The solutions to the above, provide a testable theory for either stationary or slow rotating BHs.

I think we agree more is happening than that.How do 'we' come to that 'agreement' ?

Utter catastrophic destruction of the very structures of mater and chemical bonding lost…How is this conclusion reached ?

When a Star's mass exceeds stability and a collapse begins.. All the rules we know are broken.How does one know this ?

I can not be so conclusive as fact. That a singularity is to be found.Why not ?

What might be mathematically correct might not be reality.Then upon what bases did your above statements originate ?

Regards

Andrew D

2012-Mar-16, 05:43 AM

No 'Roobydo' I do not see it as off topic. Acceleration vs Velocity is exactly what we are speaking of.

Your beautiful mathematics equation might be absolutely correct.. Describing the curve of energies witnessed or seen.

and Yes 'Jeff' thats how I see it.

I think caveman's correction pointed out what you're getting at. Regardless of the validity of the equation, it fails to adequately capture the dynamics of the system we wish to describe, ultimately because it relies on notions we now know to be incorrect.

I think I understand what you mean by "full of sadness" (thanks Jeff), but I have to say that I disagree. I think that a rigorous mathematical description of nature does not remove us from the experience of nature but enriches our experience. Mathematics is indeed cryptic and difficult, but is an incredibly powerful tool once the hard work of learning is behind you; it is surely the only tool capable of a complete description of nature, if it is. As a note: any complete formal system must also be inconsistent (per Godel), which I find interesting, and not discouraging.

There are far too many instances of abstract mathematical ideas finding perfectly suited applications as descriptions of natural phenomena for me to deny that 1) the fundamental principles of natural phenomenon are completely mathematical, or 2) the fundamental principles of mathematics are completely natural.

My favorite example is the Reimann zeta function. Its purpose was to accept integer arguments and describe the distribution of prime numbers, but it was soon discovered that the function could be extended to be defined over rational and complex arguments, and it turns out zeta(3/2) is used in the calculation of the critical temperature for a Bose–Einstein condensate in a container with periodic boundary conditions, and for spin wave physics in magnetic systems. zeta(3) arises naturally in the second- and third-order terms of the electron's gyromagnetic ratio obtained in quantum electrodynamics. zeta(4) is the denominator of the constant of proportionality between the total energy radiated per unit surface area of a black body per unit time and the fourth power of the black body's thermodynamic temperature.

There have been (very) recent developments in field called "representation theory" (specifically the Langland's correspondence) which show that the qualities of numbers are not just useful for describing nature, they are the same.

It becomes a complex understanding when a realization of mass distortion of space time is written as a acceleration..

I think we agree more is happening than that.

Utter catastrophic destruction of the very structures of mater and chemical bonding lost...

When a Star's mass exceeds stability and a collapse begins.. All the rules we know are broken.

I can not be so conclusive as fact. That a singularity is to be found.

What might be mathematically correct might not be reality

What seems to be your main objection is that mathematics is "black and white" and reality has "grey area". However, this is only your qualification of the way you believe math to be, since you're not a mathematician, and the way you believe nature to be, since you're not... well... you know who.

I do agree with you about the singularity.

astromark

2012-Mar-16, 06:16 AM

Then I have nothing to add.. You asked me a question and then answered it for me.. Thats great.

Mathematics does provide a clarity of proof I find compulsive.. and like. It tidies it all up.

As you rightly assert. I am not a mathematics head.. I do understand most of what I see.. I do not know it all..

I was when talking of the reality of the inside components of a BH which I 'imagine' as chaotic..

Perspective and point of observation critical.

Selfsim

2012-Mar-16, 06:52 AM

The description of infalling matter into a Scwarzschild BH Event Horizon IS mathematical.

Proofs have nothing to do with it.

Verbal or textual descriptions, only have veracity AFTER the mathematical descriptions have been constructed and analysed.

There seems to be some kind of perception that mathematics is kind of an optional part of understanding Astrophysics.

This is a misconception, as if mathematics did not exist, our abilities to repeatedly describe what is going on around us, would also not exist.

As far as the field of 'Mathematical Physics' is concerned, we should pay particular attention to what it has to say, also .. for exactly the same reasons just described.

I don't see mathematics in physics as being optional, in even the slightest way .. nor is its interpretation.

Regards

astromark

2012-Mar-16, 07:03 AM

No. I can not agree. Theorem * and theories can and are formulated. Proposed. Tabled, presented.

and some times later mathematics is applied to support or de construct. Testing as science does.

That is the work of mathematics as a tool for the work of confirmation of principal proposed.

It is most certainly not the mathematics that formulates a proposal. It might add weight to a principal.

Who said it was optional ? Mathematics is a vital tool in our understandings of physics.

Andrew D

2012-Mar-16, 07:08 AM

Then I have nothing to add.. You asked me a question and then answered it for me.. Thats great.

Mathematics does provide a clarity of proof I find compulsive.. and like. It tidies it all up.

As you rightly assert. I am not a mathematics head.. I do understand most of what I see.. I do not know it all..

I was when talking of the reality of the inside components of a BH which I 'imagine' as chaotic..

Perspective and point of observation critical.

I couldn't agree more with your last statement. To the external observer: there is no interior of the black hole. The singularity does not exist, the mass of the black hole is completely contained in the horizon.

If at the horizon the in-falling observer reaches c in his coordinate system as many here have said, then his coordinate system collapses to a two dimentional compression of space to a plane normal to the velocity, which coincidentally, because of the relativistic curvature of space at the horizon, means that the entirety of his space-like coordinates are the points on the horizon itself which, interestingly, is closed.

Now, do the proponents of luminal velocity at the horizon suppose that the infalling observer continues to accelerate to superluminal velocities as he continues falling?

Also, I'd like to add that we are limited (unless someone here has 'moved past' the Schwarzchild metric) to discussing uncharged, non-rotating black holes in a vacuum universe, and our limitation is not due to limitations of mathematics as a discipline.

Selfsim

2012-Mar-16, 07:45 AM

No. I can not agree. Thorium and theories can and are formulated. Proposed. Tabled, presented.

and some times later mathematics is applied to support or de construct. Testing as science does.

That is the work of mathematics as a tool for the work of confirmation of principal proposed.

It is most certainly not the mathematics that formulates a proposal. It might add weight to a principal.

Who said it was optional ? Mathematics is a vital tool in our understandings of physics.

'Thorium' is a radioactive element .. it appears in the periodic table.

If you mean 'theorems', then you are speaking about 'Pure Mathematics'. Mathematics used in Physics, becomes known as 'Applied Mathematics'.

There is a big difference between 'Pure Mathematics', and 'Applied Mathematics'.

Both progress independently, but Applied Mathematics also depends on the evolution of Pure Mathematics.

Physics is dependent on Applied Mathematics for its descriptive purposes.

Integration of mathematics into Physics, also acts as an inbuilt test of self-consistency. This in itself, is called for by the scientific process.

To turn a 'blind eye' on mathematics, or to discredit it, is to willingly accept ignorance of the real universe.

astromark

2012-Mar-16, 08:09 AM

and yes.. you will see I have altered the spelling as meant. Know body is or wants to turn a blind eye to your maths..

I thought I had covered that..point. Its called the scientific method.

Maths IS a important tool in the understanding and testing of all and any.. no problems with that.

and by way of a question for 'Roobydo' have we observed or have knowledge of any non rotating BH ?

I just wish to add a point. That to the reference frame of the in falling observer.. No event hor., is observed..

That you passed it a moment ago you will not have noticed.

Jeff Root

2012-Mar-16, 12:13 PM

If at the horizon the in-falling observer reaches c in his

coordinate system as many here have said, then his coordinate

system collapses to a two dimentional compression of space to

a plane normal to the velocity, which coincidentally, because of

the relativistic curvature of space at the horizon, means that the

entirety of his space-like coordinates are the points on the

horizon itself which, interestingly, is closed.

Now, do the proponents of luminal velocity at the horizon

suppose that the infalling observer continues to accelerate to

superluminal velocities as he continues falling?

As I said in post #9, and as caveman1917 said in post #13,

an object free-falling from rest at infinite distance would cross

the event horizon at c relative to the event horizon. No observer

could be at rest at the event horizon, but must also be falling, so

the relative speed between any local observer and the free-falling

object is always less than c. The object continues to accelerate

at an ever-increasing rate as it approaches the mass at the black

hole's center. At no time can its speed be observed by anyone

anywhere to be greater than c, even though it may be greater.

In this regard it is very much like distant galaxies receeding from

us at speeds greater than c due to cosmic expansion.

-- Jeff, in Minneapolis

xylophobe

2012-Mar-16, 05:23 PM

... The object continues to accelerate

at an ever-increasing rate as it approaches the mass at the black

hole's center. ...

-- Jeff, in Minneapolis

In another post it is mentioned that the relative mass increases to infinity as the infalling object approaches "c" so the acceleration must decrease else the infalling object would attain an infinite relative mass. The acceleration is at an ever-decreasing rate. Neither blackholes nor space-thrusters can accelerate an object with rest mass to "c".

astromark

2012-Mar-16, 07:29 PM

A point possible relevant and maybe missed... The in falling object does never attain velocity c.

Just as the expansion of the universe does.. It is the space between that has the motion. In both cases.

The gravity of the BH is consuming space at c. At the event horizon.

Jeff Root

2012-Mar-17, 01:57 AM

In another post it is mentioned that the relative mass

increases to infinity as the infalling object approaches "c"

so the acceleration must decrease else the infalling object

would attain an infinite relative mass.

Some years after Einstein introduced his theory of relativity,

but still several decades ago, it was realized that the notion

of increasing mass due to relative motion ("relativistic mass")

was unnecessary and not a good idea. The total energy of

a massive body accounts for the relativistic effect without

using the confusing concept of relativistic mass. Proper

mass ("rest mass") doesn't change with changes in motion,

and is always a "local" measurement. It is almost always

what is meant nowdays when a physicist says "mass".

The kinetic energy of an object falling at c relative to the

event horizon of a black hole would be infinite relative to

distant observers if it could be observed, but it cannot be

observed. Instead, distant observers see the falling object

redshift and fade out at the event horizon. Its kinetic energy

measured by any observer is always finite. Nothing can

stop its acceleration as it falls closer to the attracting mass

at the black hole's center.

-- Jeff, in Minneapolis

caveman1917

2012-Mar-17, 03:33 AM

Now, do the proponents of luminal velocity at the horizon suppose that the infalling observer continues to accelerate to superluminal velocities as he continues falling?

Yes, see below.

If at the horizon the in-falling observer reaches c in his coordinate system as many here have said, then his coordinate system collapses to a two dimentional compression of space to a plane normal to the velocity, which coincidentally, because of the relativistic curvature of space at the horizon, means that the entirety of his space-like coordinates are the points on the horizon itself which, interestingly, is closed.

Remember that an observer never reaches any velocity in his own coordinate system (that's what it means to have its own coordinate system after all, he carries the origin of his coordinate system with him so to speak), it's always a velocity relative to someone else's coordinate system.

In another post it is mentioned that the relative mass increases to infinity as the infalling object approaches "c" so the acceleration must decrease else the infalling object would attain an infinite relative mass. The acceleration is at an ever-decreasing rate. Neither blackholes nor space-thrusters can accelerate an object with rest mass to "c".

Where your argument works in special relativity, it fails in general relativity, for the reason i gave earlier:

a suitable definition of "velocity" (which becomes ill-defined globally, the easiest is to define velocity as local wrt a stationary observer at that point).

There is no global definition of "velocity" over the entire spacetime, just like there is no global definition of energy. This is in fact the general situation, notions such as velocity and energy are local, it is only in the special case of minkowski spacetime that these can be extended globally (because there is no curvature, or stricter, because the entire spacetime has the same tangent space).

What this means for our black hole is that the schwarzschild coordinates don't function on questions such as these, the schwarzschild metric works fine, its standard representation in terms of schwarzschild coordinates doesn't.

What we need to do is consider a series of stationary observers all the way from infinity down to the singularity, and the velocity of the "infaller" at each point is his velocity as measured by the stationary observer at that point. That's also why a newtonian "layer" over the entire schwarzschild coordinate chart as roobydo did won't work, though it will work if you integrate it over the series of stationary observers, then it will correctly give you the velocity of the infalling particle at some point as equal to the newtonian escape velocity at that point. Which is to be expected, since escape velocity means the velocity to "come to rest at infinity", which is the exact opposite of the velocity you have when you fall in from rest at infinity.

A full derivation can be found in the wiki article on Gullstrand-Painlevé coordinates (http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates).

ETA: note that these stationary observers are only physical outside the event horizon (their proper acceleration to stay stationary diverges below the EH), they represent a means to convert from the infaller's frame to a notion of velocity of that infaller, not physical observers per se.

Andrew D

2012-Mar-23, 10:25 PM

A full derivation can be found in the wiki article on Gullstrand-Painlevé coordinates (http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates).

Thank you!

xylophobe

2012-Mar-25, 03:20 PM

...Remember that an observer never reaches any velocity in his own coordinate system (that's what it means to have its own coordinate system after all, he carries the origin of his coordinate system with him so to speak), it's always a velocity relative to someone else's coordinate system.

...

What we need to do is consider a series of stationary observers all the way from infinity down to the singularity, and the velocity of the "infaller" at each point is his velocity as measured by the stationary observer at that point. That's also why a newtonian "layer" over the entire schwarzschild coordinate chart as roobydo did won't work, though it will work if you integrate it over the series of stationary observers, then it will correctly give you the velocity of the infalling particle at some point as equal to the newtonian escape velocity at that point. Which is to be expected, since escape velocity means the velocity to "come to rest at infinity", which is the exact opposite of the velocity you have when you fall in from rest at infinity.

A full derivation can be found in the wiki article on Gullstrand-Painlevé coordinates (http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates).

ETA: note that these stationary observers are only physical outside the event horizon (their proper acceleration to stay stationary diverges below the EH), they represent a means to convert from the infaller's frame to a notion of velocity of that infaller, not physical observers per se.

You state that one can not reach a velocity in his/her own frame but then doesn't that mean that one can not have an acceleration in his/her own frame - but one can feel a force (acceleration)?

There is also the specification of "stationary" observers but then with what are they stationary because the infalling participant is stationary in his/her frame?

These descriptions sound like stationary is defined as "with respect to the center of the blackhole" which seems like a logical choice because it is with respect to this location that the infaller changes location whilst the so called stationary observers maintain their respective locations to the same location.

And wouldn't clocks be excellent indicators of the state of the observers?

caveman1917

2012-Mar-25, 05:01 PM

You state that one can not reach a velocity in his/her own frame but then doesn't that mean that one can not have an acceleration in his/her own frame - but one can feel a force (acceleration)?

There are two notions of velocity, coordinate velocity and proper velocity. If we have a schwarzschild observer at infinity, he has a radial coordinate (call it r) and a clock that gives him a time coordinate (call it t). The coordinate velocity of an object in radial motion is then the usual dr/dt the speed with which its radial coordinate changes per unit time. Note that we can change this value at will by changing our coordinates. The proper velocity is the change in radial coordinate per unit time on the clock of the object itself. Call that time \tau and the proper velocity then is dr/d\tau. Note that this also depends on the radial coordinate the observer at infinity, we can change this value by changing his radial coordinate.

The thing that is invariant is proper acceleration. It is the change of proper velocity of an object to the coordinate time of a local inertial observer. In a sense it measures the deviation of the object from being inertial. All observers will give the same value to the object's proper acceleration. It is then also equal to the actual force felt by the object. There is unsurprisingly also a thing called coordinate acceleration, which is the change in coordinate velocity to coordinate time. This value can be changed at will by choosing different coordinates. So the accelerating object, in his own coordinates, doesn't have any coordinate acceleration, but it does have a proper acceleration and thus feels a force like expected.

There is also the specification of "stationary" observers but then with what are they stationary because the infalling participant is stationary in his/her frame?

These descriptions sound like stationary is defined as "with respect to the center of the blackhole" which seems like a logical choice because it is with respect to this location that the infaller changes location whilst the so called stationary observers maintain their respective locations to the same location.

Stationary with respect to the black hole indeed. Specifically, they are stationary in the coordinates of the schwarzschild observer at infinity.

And wouldn't clocks be excellent indicators of the state of the observers?

Sure. Basically an observer is nothing more than three orthonormal measuring rods and a clock.

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