Learn Time Dilation and Redshift for a Static Black Hole
The following is an overview of the time dilation and gravitational redshift effects of a static (Schwarzschild) black hole. In accordance with general relativity, a strong gravitational field can slow downtime. The closer you get to the event horizon of a black hole (if you can survive the gravity gradients, g-forces and have some means of propulsion that will keep you stable in the overwhelming gravity), there will appear no different to yourself but time will pass slower for you and quicker for people outside the field (with no apparent difference to them either). On returning from the edge of the black hole, depending on how close you were to the event horizon and how long you were there for, you would have aged less than the people outside the black hole’s strong gravitational field. This can be calculated using the following equation-
[tex]\tau=t\sqrt{1-\frac{2Gm}{rc^2}}[/tex]
which can be rewritten as
[tex]\tau=t\sqrt{1-\frac{r_s}{r}}[/tex]
where [itex]\tau[/itex]- actual (proper) time experienced within black holes gravitational field, [itex]t[/itex]- time observed from infinity, [itex]r_s[/itex]- Schwarzschild radius of black hole (where [itex]r_s=2Gm/c^2[/itex]), [itex]r[/itex]- the coordinate radius time was spent at ([itex]G[/itex]- gravitational constant, [itex]m[/itex]- mass, [itex]c[/itex]- speed of light).
For example, if an observer was to spend an ‘observed’ 3 days at 1 km from the event horizon of a 12,000* solar mass black hole, they would only experience-
[tex]\tau=3\sqrt{1-\frac{35,444,396}{35,445,396}}[/tex]
[tex]\tau=0.01594\ \text{days = 23 minutes}^{**}[/tex]
where t=3 days, [itex]r_s=2Gm/c^2[/itex] which for a 12,000 solar mass black hole equals 35,444,396 m and to be 1 km from the event horizon would mean r=rs+1000 m=35,445,396 m
*A 12,000 solar mass BH was chosen due to the tolerable tidal forces (dg) near the event horizon, [itex]dg=(2Gm/r^3)dr[/itex] where dr=2 (approx. height of a person), which means dg=71.5 g which is 7.3 earth g from head to toe which is just about manageable.
**That is, if they instantaneously travel to the mark and travel instantaneously back again, otherwise time dilation would need to be taken into account for the various stages of the trip.
Likewise, due to the immense gravity, the electromagnetic information being sent out by person would be extremely redshifted. This can be calculated using the following equation-
[tex]z=\frac{1}{\sqrt{1-\frac{2Gm}{rc^2}}}-1[/tex]
(black hole event horizon = surface of infinite redshift)
and to calculate the actual change in wavelength-
[tex]z=\frac{\lambda_o – \lambda_e}{\lambda_e}[/tex]
which becomes
[tex]\lambda_o=(z \cdot \lambda_e) + \lambda_e[/tex]
where [itex]\lambda_o[/itex]- wavelength observed from infinity and [itex]\lambda_e[/itex]- wavelength emitted from the object.
Therefore, the redshift at 1 km from the event horizon would result in z=187.27. Information at a wavelength of 550 nm (the color green in the visible spectrum) [itex](\lambda_e)[/itex] would be stretched to a wavelength of 103,549 nm (0.10355 mm or 103.55 um) [itex](\lambda_o)[/itex] which would put it in the FIR wavelength (far-infrared light waves) (in comparison, the sun has a surface gravitational redshift of z=0.0000021 which has near to zero effect on the electromagnetic information. A white dwarf with a 1.14 sol mass and a 4.5×10^6 m radius has a gravitational redshift of z=0.00037. For a neutron star with a 2 sol mass and a 12 km radius, z=0.4034.). The strong gravitational field would have the opposite effect on light going the other way as it was pulled in towards the event horizon. Viewed from outside the strong gravitational field, the individual would appear to slow down and disappear over time as the light coming from them was redshifted towards radio waves, but from the inside, the observer might see the outside 3 days pass in the 23 minutes they spent at 1 km from the event horizon. It’s very likely that this light would be blue shifted into the x-ray and gamma-ray spectrum which would no doubt have fairly drastic consequences for the observer.
Imagine a spaceship is at 1,000,000 km from a 12,000 solar mass black hole (even here, the craft would have to have powerful engines in order for it not to be pulled in by the gravity of the black hole which would still be 1,511,711 m/s^2*** at this point). Gravitational redshift would be minimal (550 nm [itex](\lambda_e)[/itex] would be 560 nm [itex](\lambda_o)[/itex], green would still be green). Here, time dilation would already be beginning to take effect (~59 minutes would pass compared to every hour in regular space outside the black holes gravitational field). A scout ship with the capacity to travel over 1,000,000 km almost instantaneously (an average velocity of 30,000 km/s, 0.1c, covering the 1 million km in ~35 seconds. Time dilation due to this velocity is virtually undetectable) is sent out towards the black hole. Due to the velocity of the craft, it would appear orange to the main ship for the majority of the journey but as it reached the 100,000 km mark from the event horizon, it would begin to turn deep red (light wavelengths from the scout ship would be stretched from 600 nm to 650 nm due to gravitational redshifting). At around the 50,000 km mark, it would disappear to the naked eye and the observer ship would have to use infrared equipment in order to monitor the scout ship’s progress. As the scout ship passed the 10,000 km mark it would appear to slow down as time dilation meant that 30 seconds for the scout ship would appear as 1 minute for the main ship (the scout ship would appear to half its speed). By the time the ship reached the 1 km mark from the event horizon, it would be in far-infrared. It would be seen to hover for 3 days, visible only to infrared equipment on the main ship, the crew had plenty of time to monitor the scout ship’s position. After a full 3 days at the 1 km mark, the scout ship would at first appear to slowly make its way back to the ship, picking up speed as it went through far, mid, and near-infrared, back into visible light, 100,000 km from the event horizon, appearing nearly full speed now as the time dilation here is only 0.85 (1 minute for the main ship would be ~52 seconds for the scout ship). Over the remaining 900,000 km, the ship would appear blue due to its velocity. Once the scout ship had returned, it would inform the crew of the main ship that it had been traveling at maximum speed all the way, taking only 35 seconds to arrive at the 1 km mark where it hovered for just 23 minutes before accelerating away at full power to arrive back at the main ship.
***Gravity for a static black hole is [itex]g=Gm/(r^2\sqrt(1-r_s/r))[/itex].
Time dilation and gravitational redshift become more extreme the closer you get to the black hole, at 1 m from the event horizon (r=rs+1 m), you would only experience-
[tex]\tau=3\sqrt{1-\frac{35,444,396}{35,444,397}}[/tex]
[tex]\tau=0.0005\ \text{days = 43 seconds}[/tex]
The redshift would be z=5,957.83 (550 nm (green) [itex](\lambda_e)[/itex] would be at 3,277,358 nm or 3.277 mm [itex](\lambda_o)[/itex], which would put it in EHF, Extremely high frequency radio waves, detectable only by radio receivers)
and at 5 mm from the event horizon (r=rs+0.005 m)-
[tex]\tau=3\sqrt{1-\frac{35,444,396}{35,444,396.005}}[/tex]
[tex]\tau=0.000036\ \text{days = 3 seconds}[/tex]
The redshift would be z=104,962.52 (550 nm (green) would be at 57,729,938 nm or 57.730 mm, which would put it in SHF, Super high frequency radio waves).
Within the last 5 mm as you reached the event horizon, you would shift through medium to a very low frequency to infinite wavelength, and time dilation would go from 3 days passing in 3 seconds to centuries passing in mere fractions of a second. A lifetime would pass for anyone observing you from the outside with sensitive radio equipment.
At exactly on the event horizon, where gravity is considered to be infinite if a person was to look outwards from the black hole, they might see some portion of the future of the universe flash before them before they free fall at close to the speed of light towards the singularity, the time of this fall (for a static Schwarzschild black hole) can be calculated by the following (which seems derivative of the antipode distance of a hypersphere based on the gravitational radius [itex](M=Gm/c^2)[/itex])-
[itex]M \pi/c = Gm \pi/c^3 = 1.548×10^{-5}\text{ x solar mass = time in seconds to singularity}[/itex]
For a 12,000 solar mass black hole, this would be a full 0.1857 seconds to admire the view before reaching the singularity.
Note: As stated, the spaceship would experience a slight time dilation of its own. If the spaceship was hovering at r=10^6 km, then 3 days at infinity would be 2 days, 22 hours, 45 minutes on the spaceship. If it was in orbit at r=10^6 km, then the time dilation would be 2 days, 22 hours, 7 minutes where-
[tex]\tau_o=t\sqrt{1-\frac{r_s}{r}}\cdot\sqrt{1-v_s^2}[/tex]
where [itex]\tau_o[/itex] represents the time dilation for an object in stable orbit and [itex]v_s[/itex] is velocity required for stable orbit where [itex]v_s=\sqrt(M/(r(1-r_s/r)))[/itex].
The above equation is equivalent to
[tex]\tau_o=t\sqrt{1-\frac{3}{2}\cdot\frac{r_s}{r}}[/tex]
where [itex]\frac{3}{2}r_s[/itex] is considered the absolute last stable orbit, it also defines the photon sphere.
Another interesting equation is the proper distance to the event horizon for a hovering observer-
[tex]\Delta r’=\frac{\Delta r}{\sqrt{1-\frac{r_s}{r_s+\Delta r}}}[/tex]
where [itex]\Delta r'[/itex] is the proper distance as observed by the hovering observer and [itex]\Delta r[/itex] is the coordinate distance as observed from infinity.
At 1 km from the EH (as observed from infinity), the hovering observer will see himself to be 188 km from the EH, at 1 m, the hovering observer will see himself to be 5.954 km from the EH and at 5 mm, 421 m from the EH.
Also note, the equations shown for time dilation are for an object hovering at a specific r, for an object that has fallen from rest at infinity, the time dilation would be-
[tex]\tau=t\left(1-\frac{r_s}{r}\right)[/tex]
where [itex]\tau=t\sqrt(1-r_s/r)\cdot\sqrt(1-v^2)[/itex] where for an object that has fallen from rest at infinity, [itex]v=\sqrt(r_s/r)[/itex]
Early life spent working and studying in York UK, 3 year architecture degree at Oxford polytechnic, 2 year architecture diploma at Oxford polytechnic, part-time in US. Worked in both York and London within architectural profession.
Thread closed for moderation.
Edit: Thread will remain closed.
[QUOTE=”John Duffield, post: 5259140, member: 573828″]Can you give a reference to back that up?[/QUOTE]Lemaitre appears to be the first to recognize that the different spherically symmetric metrics which had been published previously were simply coordinate transforms of each other and not physically different spacetimes. That would be Lemaitre’s 1933 paper on the Expanding Universe.
[QUOTE=”John Duffield, post: 5259140, member: 573828″]But when you adopt Gullstrand–Painlevé coordinates you start running into problems.[/QUOTE]Nonsense. Any invariant (including the outcome of any physical measurement) can be calculated in any coordinate system, including GP coordinates.
[QUOTE=”John Duffield, post: 5258990, member: 573828″]what we are dealing with ia still a black hole because it grows from the centre out, like a hailstone. Imagine you’re a water molecule. You alight on the surface but you can’t pass through it. However you get surrounded and buried by other water molecules, so the surface passes through you.[/QUOTE]
This is a mishmash of somewhat correct and incorrect statements. It is true that, in the case of a black hole that forms by gravitational collapse of a massive object like a star, the [I]horizon[/I] of the hole first forms at the center ##r = 0##, and “grows” in radius until it reaches ##r = 2M##, at which point it intersects the surface of the collapsing matter falling inward; after that, it stays at ##r = 2M## forever. But even at that point, the singularity has not yet formed; it doesn’t form until the surface of the collapsing matter falls in to ##r = 0##.
However, the horizon of a black hole is not a “surface” that you can “alight on”–it is an outgoing lightlike surface, so only a ray of light, one that is moving in a perfectly radially outgoing direction, can stay there. And once the collapsing matter has passed through the horizon, it is vacuum; there isn’t anything there. And the growth of the horizon from ##r = 0## to ##r = 2M## is not a matter of “molecules” of the collapsing matter falling in and stopping; the collapsing matter never stops collapsing, and a particular piece of it that happens to pass through the horizon when the horizon is between ##r = 0## and ##r = 2M## (it is inside the collapsing matter all during this period) just passes right through–which it must, because, as above, the horizon is an outgoing lightlike surface, so no piece of the collapsing matter can stop there.
[QUOTE=”John Duffield, post: 5258990, member: 573828″]I think the “frozen star” interpretation is correct because the coordinate speed [I]does[/I] have physical meaning.[/QUOTE]
No, it doesn’t. Sorry, John, but you are simply mistaken.
[QUOTE=”John Duffield, post: 5259140, member: 573828″]when you adopt Gullstrand–Painlevé coordinates you start running into problems.[/QUOTE]
No, you don’t. The fact that someone, even Einstein, was “unhappy” about them decades ago is irrelevant; that’s a matter of history, not physics. G-P coordinates–and Eddington-Finkelstein coordinates, and Kruskal coordinates–are in common use by relativists, and there are no problems with them at all. In fact, they all [I]avoid[/I] the key problem with Schwarzschild coordinates, namely the coordinate singularity at the horizon. That’s why relativists like them.
[QUOTE=”Mike Holland, post: 5258976, member: 429699″]In other words we very quickly get an Eternally Collapsing Object, or Almost-Black Hole, much as you seem to dislike those names.[/QUOTE]
Yes, because they are highly misleading, for reasons which have already been given ad nauseam in this thread.
[QUOTE=”Mike Holland, post: 5258976, member: 429699″]This what Oppenheimer, Snyder, Landau et al described when they first calculated the collapse for a remote observer.[/QUOTE]
They (O-S at least, I’m not familiar with Landau’s paper) said that the collapse would [I]appear[/I] to a distant observer to slow down more and more; but they were also very clear that, to an observer falling in with the collapsing matter, the collapse would not appear to slow down; it would take about an hour (IIRC) for a horizon to form, according to an observer falling in, for a mass of a few solar masses.
Orodruin: that’s what Einstein said, and Professor Ned Wright: “[I]In a very real sense, the delay experienced by light passing a massive object is responsible for the deflection of the light”.[/I]
[IMG]http://www.astro.ucla.edu/~wright/Einstein-wavelets-75.gif[/IMG]
[QUOTE=”DaleSpam, post: 5259047, member: 43978″]Yes, the fact that the coordinate speed of light is not physical was not recognized in Einstein’s day.[/quote]Can you give a reference to back that up? And see for example [URL=’http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html’]this[/URL] by Don Koks where you can read this:
[SIZE=2]”Given this situation, in the presence of more complicated frames and/or gravity, relativity generally relinquishes the whole concept of a distant object having a well-defined speed. As a result, it’s often said in relativity that light always has speed [I]c[/I], because only when light is right next to an observer can he measure its speed— which will then be [I]c[/I]. When light is far away, its speed becomes ill-defined. But it’s not a great idea to say that in this situation “light everywhere has speed [I]c[/I]”, because that phrase can give the impression that we [I]can[/I] always make measurements of distant speeds, with those measurements yielding a value of [I]c[/I]. But no, we generally can’t make those measurements. And the stronger gravity is, the more ill-defined a continuum of observers becomes, and so the more ill-defined it becomes to have any good definition of speed. Still, we can say that light in the presence of gravity does have a position-dependent “pseudo speed”. In that sense, we could say that the “ceiling” speed of light in the presence of gravity is higher than the “floor” speed of light.[/SIZE]
[SIZE=2]Einstein talked about the speed of light changing in his new theory. In the English translation of his 1920 book “Relativity: the special and general theory” he wrote: [I]”according to the general theory of relativity, the law of the constancy of the velocity[/I] [Einstein clearly means speed here, since velocity (a vector) is not in keeping with the rest of his sentence] [I]of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity[/I] […] [I]cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity[/I] [speed] [I]of propagation of light varies with position.”[/I] This difference in speeds is precisely that referred to above by ceiling and floor observers.”[/SIZE]
[QUOTE=”DaleSpam, post: 5259047, member: 43978″]Both of which can be explained in other coordinates where the coordinate speed of light either is c or differs from the Schwarzschild coordinate speed of light.[/QUOTE]In GR we tend to say all coordinate systems are equally valid. But when you adopt Gullstrand–Painlevé coordinates you start running into problems. I think there’s a reference to Einstein being unhappy with them somewhere, I’ll try to dig it out.
[QUOTE=”John Duffield, post: 5259043, member: 573828″]I’m afraid it does, because a variation in it is what makes light curve and your pencil fall down. See those links, and the [URL=’https://en.wikipedia.org/wiki/Shapiro_delay’]Shapiro delay[/URL] and [URL=’http://www.astro.ucla.edu/~wright/deflection-delay.html’]this[/URL].[/QUOTE]
No it does not. You will be able to explain this in any coordinate system of your choice. It does not make coordinate speed any more physical. That some things are easier to see in one coordinate system than another is an issue of being able to select a coordinate system which is well suited to your problem.
There are a whole LOT of comments here from a bunch of smart people who know much more that I do about this stuff. (not false modesty) Could one of you please cite the observations / measurements (the primary astronomy articles) you are relying on to validate you claims about any of the physics inside the event horizons of black holes? This is intended to be a gentle reminder that any claim about what happens inside the EH is speculation (based on Physics which is expected to be incorrect there, if not near the EH, then near the singularity). It actually would be interesting (for me, at least) to find out to what extent our Physics has been validated exterior to EHs. Have we any reason, that is OBSERVATIONALLY VALIDATED reason, to believe that our Physics at r = 2 x the Schwartzchild radius is correct? How about at 10X Rs,? At 100X or 1000X? As I understand it, we’ve observed the accretion disc, and are only weakly? moderately?, certainly NOT “strongly”, confident that we understand – roughly – what is happening there and there are some claims (are they believed by most mainstream experts?) of (possible) detection of Hawking radiation, so how close to the EH have we “seen”? Anyway, this thread is great, although I’m not impressed with the OP, (which is a bit over my head). Constructive suggestions: 1. “Black hole’s” is singular possesive, “black holes” is plural, and “black holes’ ” is plural possesive. get it right, please. 2. Don’t use both meters and kilometers when discussing the distance (radius) of astronomical objects, I suggest anything over 1 km be in units of km. 3. Oh, please, please don’t use “quicker” and “slower” in describing the passage of time! IMHO, starting out on the wrong foot, ie with false and misleading pedagogy, should be avoided. (as a sidebar, I’d suggest mentioning that we can measure the difference in two clocks which differ in height at Earth’s surface of about 1 foot. This might be one way of introducing the fact that each and every observer travels to the beat of their own clock and their own time…call it, say, their proper time. 4. It may be my browser’s font, but tau and r in the text are very hard to distinguish. Despite its common use to symbolize proper time, if a better font can’t be found, I’d suggest a different letter be used 5. r is used in an equation before it is defined. In fact the reader has to infer what it means, iirc. 6. I also have a problem with instantaneous changes in velocity and location. They are neither necessary nor physically attainable, nor are they a “reasonable” approximation for any possible (foreseeable) spaceship, manned or unmanned. I’d suggest instead of using fantasy propulsion, that clocks be synchronized by the near and far observer once position/orbit/velocity is attained. (the two clocks ‘talk’, count down, and both reset to time 0 at the times necessary for both to receive the reset confirmation signal after the expected transmission time lag…) 7. I have some doubt about the light (starlight?) being blue-shifted to the point it is a problem for the near observer. Is this based on calculations of (UV, X-ray, gamma) energy INTENSITY hitting that observer’s ship? 8. The OP doesn’t (probably intentionally) clearly distinguish between gravitational effects (force of mass at center point of object) and tidal effects. Whether 7g for 20 minutes is survivable is questionable to me, and shouldn’t the 2 meter tall astronaut’s orientation be mentioned (speaking of tidal forces…) Anyway, cudos to all who have taken the time to post and comment on this thread.
[QUOTE=”John Duffield, post: 5259015, member: 573828″]What Einstein referred to as the speed of light is now referred to as the coordinate speed of light, whilst the locally-measured speed of light is now referred to as the speed of light.[/QUOTE]Yes, the fact that the coordinate speed of light is not physical was not recognized in Einstein’s day.
[QUOTE=”John Duffield, post: 5259043, member: 573828″]I’m afraid it does, because a variation in it is what makes light curve and your pencil fall down. See those links, and the [URL=’https://en.wikipedia.org/wiki/Shapiro_delay’]Shapiro delay[/URL] and [URL=’http://www.astro.ucla.edu/~wright/deflection-delay.html’]this[/URL].[/QUOTE]Both of which can be explained in other coordinates where the coordinate speed of light either is c or differs from the Schwarzschild coordinate speed of light.
I’m afraid it does, because a variation in it is what makes light curve and your pencil fall down. See those links, and the [URL=’https://en.wikipedia.org/wiki/Shapiro_delay’]Shapiro delay[/URL] and [URL=’http://www.astro.ucla.edu/~wright/deflection-delay.html’]this[/URL].
This does not make the coordinate speed of light any more physical.
Dale/Orodruin: see my post 50 above and follow the links. What Einstein referred to as the speed of light is now referred to as the coordinate speed of light, whilst the locally-measured speed of light is now referred to as the speed of light.
[QUOTE=”DaleSpam, post: 5258991, member: 43978″]The coordinate speed of light isn’t even c.[/QUOTE]
To emphasise this point, you can make the coordinate speed of light whatever value you want by simply redefining your coordinates.
[QUOTE=”John Duffield, post: 5258990, member: 573828″]the coordinate speed [I]does[/I] have physical meaning.[/QUOTE]No, it doesn’t. The coordinate speed of light isn’t even c.
See Kevin Brown’s [URL=’http://mathpages.com/rr/s7-02/7-02.htm’]Formation and Growth of Black Holes[/URL] on mathpages. This used to refer to Weinberg and Wheeler until I queried it with Weinberg and he obviously got on to Kevin Brown, who changed it to Einstein and quantum field theories. I’m not too happy about that but anyway, see this:
[I]”Incidentally, we should perhaps qualify our dismissal of the “frozen star” interpretation, because it does (arguably) give a servicable account of phenomena outside the event horizon, at least for an eternal static configuration. Historically the two most common conceptual models for general relativity have been the “geometric interpretation” (as originially conceived by Einstein) and the “field interpretation” (patterned after the quantum field theories of the other fundamental interactions). These two views are operationally equivalent outside event horizons, but they tend to lead to different conceptions of the limit of gravitational collapse. According to the field interpretation, a clock runs increasingly slowly as it approaches the event horizon (due to the strength of the field), and the natural “limit” of this process is that the clock asymptotically approaches “full stop” (i.e., running at a rate of zero). It continues to exist for the rest of time, but it’s “frozen” due to the strength of the gravitational field. Within this conceptual framework there’s nothing more to be said about the clock’s existence. In contrast, according to the geometric interpretation, all clocks run at the same rate, measuring out real distances along worldlines in curved spacetime…”[/I]
I think the “frozen star” interpretation is correct because the coordinate speed [I]does[/I] have physical meaning. Sorry Peter. Mike, IMHO what we are dealing with ia still a black hole because it grows from the centre out, like a hailstone. Imagine you’re a water molecule. You alight on the surface but you can’t pass through it. However you get surrounded and buried by other water molecules, so the surface passes through you.
[QUOTE=”PeterDonis, post: 5258681, member: 197831″]This is a fair point. A more precise way of saying it would be that a remote observer will see the apparent “speed” (I put this in scare-quotes because it’s a coordinate speed and doesn’t really have a physical meaning) of an infalling object get closer and closer to zero, without ever quite reaching it, and he will see the apparent position of the object get closer and closer to the horizon, without ever quite reaching it.
In practical terms this will happen fairly quickly, yes. Discussions of the theory involved usually ignore this point, but it will certainly come into play if we ever try to run any actual experiments along these lines.[/QUOTE]
In other words we very quickly get an Eternally Collapsing Object, or Almost-Black Hole, much as you seem to dislike those names. This what Oppenheimer, Snyder, Landau et al described when they first calculated the collapse for a remote observer.
I was going to ask what you thought Sagittarius A was, but that answers my question
[QUOTE=”Boing3000, post: 5256818, member: 559056″]A remote observer will [I]never[/I] see anything “effectively coming to a stop”. Only the event horizon is at time infinity. Everything outside is at finite time and is not stopped.[/QUOTE]
This is a fair point. A more precise way of saying it would be that a remote observer will see the apparent “speed” (I put this in scare-quotes because it’s a coordinate speed and doesn’t really have a physical meaning) of an infalling object get closer and closer to zero, without ever quite reaching it, and he will see the apparent position of the object get closer and closer to the horizon, without ever quite reaching it.
[QUOTE=”Boing3000, post: 5256818, member: 559056″]Beside the object will effectively be out of “sighting” range way before due to the extreme red shifting of the signals[/QUOTE]
In practical terms this will happen fairly quickly, yes. Discussions of the theory involved usually ignore this point, but it will certainly come into play if we ever try to run any actual experiments along these lines.
[QUOTE=”DaleSpam, post: 5258631, member: 43978″]do you know if the technical definition of “event horizon” excludes the Rindler horizon for any reason?[/QUOTE]
Yes, it does. The technical definition of an event horizon is that it is the boundary of the causal past of future null infinity. The Rindler horizon does not meet that definition; all of Minkowski spacetime, regardless of which side of any Rindler horizon it is on, is in the causal past of future null infinity.
[USER=197831]@PeterDonis[/USER] do you know if the technical definition of “event horizon” excludes the Rindler horizon for any reason? If it does not, then I would say that we are always passing through event horizons.
[QUOTE=”Mike Holland, post: 5258547, member: 429699″]There are no event horizons in the past of anything we can see.[/QUOTE]
Yes, this is true. So what?
[QUOTE=”Mike Holland, post: 5258547, member: 429699″]there are no event horizons in the past light cone of any lump of rock or mote of dust in the universe that is outside a collapsing almost-black hole.[/QUOTE]
Again, this is true. There can’t be an event horizon in the past light cone of any event outside the horizon. That’s obvious from the definition of an event horizon. But again, so what? You appear to think this means something profound; what?
I can tell you what it doesn’t mean: it doesn’t mean you don’t have to worry about never being able to escape again if you free-fall into one of these objects that you cal a “collapsing almost-black hole”. If you do that, you will fall below the horizon and be unable to escape in a finite, fairly short time by your clock. You certainly won’t get an infinite length of time to enjoy your existence, simply because events on the horizon happen to have a time coordinate of ##+ infty## in a certain coordinate chart. You can, as you said, avoid this fate by not falling into such an object; but, as I said, that very qualification underscores the fact that “it takes an infinite amount of coordinate time to reach the horizon” doesn’t actually mean anything physical. If it did, it would imply that you wouldn’t have to worry even if you didn’t avoid falling into such an object, because it would take an infinite amount of time by your own clock to free-fall to the horizon. But it won’t.
So again, since you keep on reiterating that there are no event horizons in your past light cone, why? What do you think it means?
[QUOTE=”PeterDonis, post: 5257136, member: 197831″]
As Nugatory said, events on and inside the horizon can never be in your past light cone if you are outside the horizon, even if you live forever, and that statement is invariant, independent of coordinates.
.[/QUOTE]
Going back to this statement, my past light cone is not just a pair of lines dropping away at 45 degrees in a space-time diagram. It is the observable universe, right out to the most distant galaxies the Hubble telescope can see. There are no event horizons in the past of anything we can see. We can observe the Andromeda galaxy, which is 2.9 million LY away. So prior to 2.9 million years ago, there were no event horizons in the Andromeda galaxy. Similarly, there was no event horizon around Saggitarius A prior to 26,000 years ago. So what is it? An eternally collapsing object, or a naked singularity?
[QUOTE=”PeterDonis, post: 5257739, member: 197831″]
Yes. But you also can’t see beyond the observable universe. Do you believe the universe ends at the boundary of the observable universe?
[/QUOTE]
I am not the only observer in the universe. We have no clue how common life is in the universe, but I am sure there would be other observers in every other galaxy. Anyway, a past light cone exists for any point in space, not just for living observers. So there are no event horizons in the past light cone of any lump of rock or mote of dust in the universe that is outside a collapsing almost-black hole.
This looks interesting. I would urge all posters to look at the [URL=’http://einsteinpapers.press.princeton.edu/vol7-trans/156?highlightText=%22speed%20of%20light%22′]Einstein digital papers[/URL] which came online fairly recently. I think there’s one or two things in there that help to shape one’s thought’s about black holes. Also see [URL=’http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html’]this article[/URL] written by PhysicsFAQ editor and relativist Don Koks. It replaces [URL=’http://www.desy.de/user/projects/Physics/Relativity/SpeedOfLight/speed_of_light.html’]this article[/URL] by Carlip and Gibbs. With your permission I’ll start a thread on what I think is a moot point.
[QUOTE=”Mike Holland, post: 5256773, member: 429699″]for all the observers I know of, event horizons don’t exist yet.[/QUOTE]
[QUOTE=”Mike Holland, post: 5257711, member: 429699″]That means there are no event horizons in the [I]visible [/I]universe, [I]as far as I can see. [/I][/QUOTE]Notice that these two statements are very different. The first one is talking about simultaneity, and the second one is talking about past light cones.
[QUOTE=”Mike Holland, post: 5257711, member: 429699″]There are no event horizons in my past light cone[/QUOTE]
There are not event in your past light cone past that horizon. That’s why it is called an event horizon.
[QUOTE=”Mike Holland, post: 5257711, member: 429699″]That means there are no event horizons in the [I]visible [/I]universe, [I]as far as I can see.[/I][/QUOTE]
There is an “horizon” each time you cannot see past “it”. There are many types of horizons…
[QUOTE=”Mike Holland, post: 5257711, member: 429699″]So any supermassive objects we [I]can [/I]see are presumably eternally collapsing objects.[/QUOTE]
No because there is not physics of “eternally collapsing objects”. The real/invariant physics describing those objects also predict the real/invariant distortion that happens to signal coming from these object to a distant observer. These distortions are as real as the optical distortions of your fish size by your fish bowl. They are not illusion. Yet the fish real size do [B]not[/B] change.
[QUOTE=”Mike Holland, post: 5257711, member: 429699″]And as I plan to live forever, that will be the case forever (provided I can avoid falling into one of these objects[/QUOTE]
Correct you will never see past this horizon, and yet you will still [B]experience[/B] this horizon, this black spot in the sky. You will see it evolve like a EH, not like a “frozen star”. You will see it evaporate (well, if that part is confirmed), you will see it merge with other BH and whatnot.
Eternity would be a cool vantage point to experience a BH.
[QUOTE=”Mike Holland, post: 5257711, member: 429699″]that sums up what I have been trying to say all along. There are no event horizons in my past light cone.[/QUOTE]
That’s part of what you have been saying, but not all of it.
[QUOTE=”Mike Holland, post: 5257711, member: 429699″]That means there are no event horizons in the [I]visible [/I]universe, [I]as far as I can see.[/I][/QUOTE]
Yes. But you also can’t see beyond the observable universe. Do you believe the universe ends at the boundary of the observable universe?
[QUOTE=”Mike Holland, post: 5257711, member: 429699″]any supermassive objects we [I]can [/I]see are presumably eternally collapsing objects.[/QUOTE]
No, they are not. We’ve been over this.
[QUOTE=”Mike Holland, post: 5257711, member: 429699″]as I plan to live forever, that will be the casew forever (provided I can avoid falling into one of these objects).[/QUOTE]
Saying that you can avoid falling into a black hole indefinitely is not the same as saying the hole will take an infinite time to form. The very fact that you put in that qualifier–that you’ll be ok as long as you don’t fall in–shows the difference. If the hole was going to take “forever” to form in any absolute sense, you wouldn’t have to worry about falling in. Yet you do.