Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Mike Gale on 23/04/2017 00:57:39
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It is well known that the speed of light goes to zero at the event horizon of a Schwarzschild black hole. You can prove this to yourself by solving the metric for a radially in-falling photon, for which proper time is identical to zero. The result is:
c=co(1-rs/r)
The Shapiro effect exemplifies this phenomenon.
What happens to an in-falling mass is less obvious because the speed of proper time is a function of velocity:
dT/dt=sqrt((1-rs/r)(1-(v/c)2))
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As I understand it, what happens depends on the frame of reference of the observer:
- If you are an observer on the falling particle, you cross the event horizon in the blink of an eye.
- if you are an observer distant from the black hole, you never see the particle crossing the event horizon.
It is what happens after the particle crosses the event horizon that is more mysterious.
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The freely falling frame is inertial since it is not supported against gravity. No force is felt. Proper time must continue up to and beyond the horizon. Beyond the horizon our definition of proper time may or may not be correct.
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One expert I consulted on this issue had this to say:
1) The term 'proper time speed' is unconventional and may lead to confusion. The conventional term is 'time dilation'.
2) The expression I quoted for that quantity is obtained by dividing both sides of the metric by 'dt'. That's invalid if dt=0, which may be the case in some contexts.
3) The metric permits trajectories that are not geodesic. Although a geodesic describes a trajectory that is consistent with the metric, the reverse is not necessarily true.
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Even so, this formulation of the metric suggests that dT and dr go to zero at the horizon:
(co*dT)2=((c*dt)2-dr2)/(1-rs/r)-(r*d(angle))2
The question is whether there is a geodesic for which that is the case and, if so, does it approximate Newtonian free fall in the weak field limit?
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Another way to look at it is:
(co*dT)2=(c*dt)2*(1-(v/c)2)/(1-rs/r)-(r*d(angle))2
If d(angle)=0 and (v/c)2=rs/r then:
dT=(1-rs/r)dt
Is that a geodesic? If so, then v=0 at the horizon.
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Wikipedia has a good article on Schwarzschild geodesics: https://en.wikipedia.org/wiki/Schwarzschild_geodesics
In a nutshell, evan_au is correct. Penetration is a mundane occurrence for a local observer. However, the local perspective is inconsequential for the rest of us. From our point of view, the black hole cannot accumulate mass. Furthermore, although Wikipedia gives exact solutions for time dilation in the free fall case, it does not address spatial dilation. Flamm's paraboloid teaches us that radial distances are dilated in the local reference frame. The gravitating mass recedes due to spatial dilation so penetration may well be impossible in any context.
In any case, the distant observer perceives objects to accelerate towards the gravitating mass in a weak field context and stop dead in their tracks at the horizon so there has to be a point in between where they start to decelerate. You can't observe an object impacting the horizon because it is red-shifted into oblivion and eternally hidden by the Shapiro effect. But you can extrapolate its trajectory to predict the speed of impact. Conventional wisdom says impact velocity is finite. I disagree, but I can't find a mathematically vigorous argument to make or break my case. The best I can do is appeal to causality. A reference frame in which mass can out pace light doesn't make any sense.
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The weird thing about SC time keeping is that coordinate time and proper time are both functions of coordinate radius. Coordinate time is a bookkeeping concept. Nobody actually ages at that rate. It's more like an optical illusion.
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I think this is what I'm talking about: https://en.wikipedia.org/wiki/Gullstrand–Painlevé_coordinates
The article confirms the expression I derived for coordinate velocity in the radial free fall case. The in-falling object does indeed appear to decelerate and stall at the horizon. They call this result an "accounting entry" since coordinate velocity is reckoned by the elapsed time between messages, which are delayed by the Shapiro effect. It is an illusion, but a stubbornly persistent one because the horizon is inarguably impenetrable from that perspective.
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The question is, does the raindrop affect the event horizon in coordinate space when it crosses over in the local reference frame? I think the answer is no because changes in the field must not propagate faster than light.
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It is well known that the speed of light goes to zero at the event horizon of a Schwarzschild black hole. You can prove this to yourself by solving the metric for a radially in-falling photon, for which proper time is identical to zero. The result is:
c=co(1-rs/r)
The Shapiro effect exemplifies this phenomenon.
What happens to an in-falling mass is less obvious because the speed of proper time is a function of velocity:
dT/dt=sqrt((1-rs/r)(1-(v/c)2))
While there's a question in the subject line, the answer of which depends on the observer, there's no question in this opening post. What is the question here?
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From our point of view (as distant observers), the black hole cannot accumulate mass.
Due to Newton's shell theorem, any mass closely approaching the black hole's event horizon appears (from the viewpoint of a distant observer) as if the black hole has gained that mass, despite the fact that the observer never sees it actually cross the event horizon.
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Accumulation of mass outside the horizon is not the same as crossing the horizon.
It is well known that the speed of light goes to zero at the event horizon of a Schwarzschild black hole. You can prove this to yourself by solving the metric for a radially in-falling photon, for which proper time is identical to zero. The result is:
c=co(1-rs/r)
The Shapiro effect exemplifies this phenomenon.
What happens to an in-falling mass is less obvious because the speed of proper time is a function of velocity:
dT/dt=sqrt((1-rs/r)(1-(v/c)2))
While there's a question in the subject line, the answer of which depends on the observer, there's no question in this opening post. What is the question here?
I'm trying to figure out how a Schwarzschild black hole grows from the perspective of the bookkeeper, who is infinitely removed from the central singularity. From that perspective, it takes an infinite amount of time for anything (including light) to reach the horizon so it is, in effect, impenetrable and there is no opportunity for growth. Everything piles up just outside. If the pile up is radially symmetric, its centre of mass will be coincident with the central singularity so the bookkeeper will perceive that as growth. A hollow shell of mass looks the same as a central singularity when viewed from the outside so it would seem that the latter is a fiction.
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From our point of view (as distant observers), the black hole cannot accumulate mass.
Due to Newton's shell theorem, any mass closely approaching the black hole's event horizon appears (from the viewpoint of a distant observer) as if the black hole has gained that mass, despite the fact that the observer never sees it actually cross the event horizon.
I think you meant to say Gauss's theorem (of divergence.) Newton's shell theorem addresses the field strength inside the boundary. But I see your point. A sufficiently dense shell of mass will form a black hole and it is entirely possible that the masses are pushed together (e.g. by a collapsing star) rather than being pulled together by a pre-existing singularity. The primordial case would be a perfectly head-on collision between two elemental masses.
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Negative mass puts an interesting twist on the primordial case because:
a) Positive mass attracts negative mass.
b) Negative mass repels positive mass.
There is no net force on either particle as they approach one another so they can get arbitrarily close and annihilate (or rather nullify.) I don't know if that leaves a singularity or just flat space.
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I've just met this assertion by Christoph Galfard (The Universe in your Hand); and am trying to process it in terms of SR.
He says of an asteroid that appears to be stationary on the event horizon of a black hole:
”It is gone………..Only its image remains. That is the distortion of spacetime in action. ,,,,,,,,Our time, yours and mine, is not ticking like the rock’s. The asteroid is beyond the horizon. Its image is still on the horizon. That’s how it is.”
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An image is an apt analogy because the asteroid has no radial extent at the horizon, but the image is undetectable due to infinite redshift. I find the latitude/longitude analogy more instructive. It's like the in-falling object is heading towards a spacetime cliff and it takes an infinite amount of time to reach the brink, where it can no longer make headway in the radial direction. The horizon (at R=2m) is the edge of the perceptible universe in the same sense as R=infinity.
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The object crosses the horizon. The light reflecting off it is slowed by the gravitational field and takes longer to get to the remote observer situated further from the horizon. It appears to slow down.
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An object can cross the horizon in a finite amount of proper time, but the cross-over event occurs at the end of coordinate time. The same is true of light so you can't illuminate objects on the horizon. Kevin Brown breaks it down here: http://mathpages.com/rr/s6-04/6-04.htm, He provides a nice visualization, too.
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Without an actual black hole to play with it is all conjecture.
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a) Positive mass attracts negative mass.
I am not aware of any proven examples of negative gravitational mass for an object in a vacuum.
- Although some weird quantum things can occur for charge carriers in semiconductors
CERN's ALPHA experiment is trying to measure the gravitational attraction of antimatter.
- Because it is antimatter does not necessarily mean that it is anti-mass.
- But their final conclusion has not yet been published.
See: https://en.wikipedia.org/wiki/Gravitational_interaction_of_antimatter
Newton's shell theorem addresses the field strength inside the boundary.
Newton's Shell Theorem also addresses the gravitational field outside the shell: It behaves as if there is a point mass at the center of the shell.
See: https://en.wikipedia.org/wiki/Shell_theorem
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It is well known that the speed of light goes to zero at the event horizon of a Schwarzschild black hole. You can prove this to yourself by solving the metric for a radially in-falling photon, for which proper time is identical to zero. The result is:
c=co(1-rs/r)
The Shapiro effect exemplifies this phenomenon.
What happens to an in-falling mass is less obvious because the speed of proper time is a function of velocity:
dT/dt=sqrt((1-rs/r)(1-(v/c)2))
The mass density of a back hole defines the local space-time profile. This profile is not defined by the observer, any more than the gravity of the sun and the pressures it exerts is dependent on the observer. Mass is an invariant. It defines local space-time, via GR, to be a local invariant. Confusion appears because we measure this with second hand data, stemming from energy output signals. This second hand data can is reference dependent.
The analogy is judging a book by the cover, but never opening the book. The cover or second hand data is where sale pitch rules. A smile does not always mean friendly, but this is how relative reference seeks to interpret it. An acceleration will exert pressure; g-force. This pressure will impact matter but may not show up in the energy profile used by relative reference. Energy does not accelerate.
Gravity can exert pressure, which physically moves mass together and then alters the phases of matter. This has nothing to do with the observer, I might be able to find a reference that appears to alter the space-time profile of the sun to it appears lighter based on its apparent profile. However, this relative reference will not make the sun stop fusion, since the fusion has nothing to do with me or my reference illusion.
For example, if hydrogen atoms were in motion, we might see the emission spectrum red shifted. This does not mean a new type of hydrogen has formed. The hydrogen atom and its spectrum is invariant. We know enough to tell that this is due to motion. But what becomes of exotic phases where data is scarce? Relative reference will use its imagination to make up the plot, based on the book cover and sales pitch on the cover.
If you fell into a black hole, the pressures would kill you long before we could run any test. All your tools would get squished and stop transmitting. Space-time is only half the story. Drop a space probe on Venus and see if spacetime leads. If you fell into a black hole you would become squished and then all your molecules and atoms would undergo phases changes until you become elementary particles and beyond.
One problem this creates, in terms of reference, is although space-time will appear to slow the process, the matter transitions speed up as pressure increases. Matter time is running opposite to space-time. The sun does the same thing. On the surface we have plasma speed transition in matter, but the core is gamma speed transitions which are faster, even though space-time should show slowed time. An acceleration is d/t/t or in this case is expresses as two components; space-time plus time due to matter phases.
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Negative mass experiment: https://en.wikipedia.org/wiki/Negative_mass#Experimentation
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Interaction between positive and negative masses: https://en.wikipedia.org/wiki/Negative_mass#Runaway_motion
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You could define a negative energy, but only for kinetic energy and if the energy were expressed as a vector rather than a scalar. Negative rest energy would be a very bizarre concept indeed.
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The mass density of a back hole defines the local space-time profile. This profile is not defined by the observer, any more than the gravity of the sun and the pressures it exerts is dependent on the observer. Mass is an invariant.
Only rest mass is invariant. The effective mass of a gravitating body is zero for an observer in free-fall.
If you fell into a black hole, the pressures would kill you long before we could run any test.
The (radial) pressure gradient is irrelevant if the observer is arbitrarily small. The point is that the observer reaches the horizon in coordinate space at the end of coordinate time.
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Eventually all this can end up in group theory. Especially considering the Lorentz and Poincare groups. Group theory isn't my favourite pastime.
https://en.m.wikipedia.org/wiki/Fundamental_group
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You could define a negative energy, but only for kinetic energy and if the energy were expressed as a vector rather than a scalar. Negative rest energy would be a very bizarre concept indeed.
Absolutely, but it is allowed by the equations and we've been burned many times by our prejudices in that regard. Best to keep an open mind on the subject.
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Here's a pretty good explanation for black hole growth: http://mathpages.com/rr/s7-02/7-02.htm
In a nutshell, the Schwarzschild solution cannot accommodate black hole growth because it only addresses the one-body problem. Lacking an exact solution for the two-body problem, we can only speculate about the process.
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Eventually all this can end up in group theory. Especially considering the Lorentz and Poincare groups. Group theory isn't my favourite pastime.
https://en.m.wikipedia.org/wiki/Fundamental_group
Yikes.
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You wrote "The best I can do is appeal to causality. A reference frame in which mass can out pace light doesn't make any sense." Mike. Not sure what you're thinking of there?
A red shifted light should pass into oblivion from the far away observers point of view. We don't see the 'mass', we see the (reflected/excited) light of the infalling object, right?
As far as I get it any 'caught' light, inside a 'real event horizon' still has a geodesic it follows, although it's a closed one leading only one way, into some 'center' unless we assume that there also is a possibility of the geodesic to be in some sort of equilibrium relative that center.