RC, with all respect, that's not the point at all. Farsight's mistake is much worse, and much more fundamental, than mishandling an asymptote.
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RC, with all respect, that's not the point at all. Farsight's mistake is much worse, and much more fundamental, than mishandling an asymptote.
W.D.Clinger
Philosopher
some homework problems for Farsight, part 3
This is the third set of homework problems.
In this part 3, we will use Lemaître coordinates to define charts on a larger manifold M' that includes all of M plus the event horizon and a region of spacetime inside the event horizon that extends all the way down to the essential singularity at the center of the black hole. That will prove there is no real singularity at the event horizon. Another exercise will establish that particles falling radially into the black hole pass through the event horizon with no special fanfare and reach the essential singularity in finite proper time.
[size=+1]Schwarzschild coordinates and charts[/size]
With the (-,+,+,+) signature convention and natural units in which c=G=1, the Schwarzschild coordinates are (t, r, θ, φ), and the Schwarzschild metric is
where
is the usual metric on a 2-sphere in spherical coordinates and
[size=+1]Lemaître coordinates and charts[/size]
The Lemaître coordinates are (τ, ρ, θ, φ), and the Lemaître metric is
where dΩ2 is defined as above,
will turn out to coincide with the r coordinate of Schwarzschild coordinates, and
Exercises 17 through 20 will prove that R(τ, ρ) can be identified with the r coordinate of a Schwarzschild chart. For now, mathematically scrupulous readers will eliminate r from the Lemaître metric above by writing R(τ, ρ) wherever r appears.
Define a Lemaître chart to be of the above form, and let M' be the manifold that's covered by the set of all Lemaître charts. It takes two Lemaître charts to cover all of M' (see exercises 12 and 13).
Exercise 17. Prove: If 0 < x < 1, then
Exercise 18. Prove: If
Exercise 19. Prove: If dτ and dρ are defined as in exercise 18, then dR/dr = 1.
Exercise 20. Prove: Given any Schwarzschild chart on M, there exists a Lemaître chart on M' whose restriction to (4/3)M < (ρ - τ) is isometric to the Schwarzschild chart.
Exercise 21. Prove: M is (isometric to) a proper submanifold of M'.
Now that we know M' contains a submanifold that's isometric to M, we might as well identify M with that submanifold. (That gets rid of the pedantic "isometric to" stuff.)
Exercise 22. Prove: If ρ, θ, and φ are held constant, then
If (dτ, dρ, dθ, dφ) are infinitesimal displacements from some point p = (τ, ρ, θ, φ), then ds2 is the square of the pseudo-distance between p and q = (τ+dτ, ρ+dρ, θ+dθ, φ+dφ), and is calculated by applying the metric tensor at p to two copies of (dτ, dρ, dθ, dφ).
ds2 is a pseudo-metric because (unlike Euclidean metrics) it can be negative. The physical significance of a negative value is that a massive particle's world line can go through both of the points p and q. According to exercise 22, the timelike separation between points p and q is exactly the same as the difference between the values of their timelike coordinate τ. That means τ measures the proper time of an observer that considers himself to be at rest with respect to the Lemaître chart (because his spatial coordinates aren't changing).
In particular, consider a point-like observer who considers himself to be at Lemaître coordinates (τ, ρ, θ, φ), is not moving in a way that would change his spatial coordinates ρ, θ, and φ, and is not experiencing any sensation of acceleration. Since his ρ coordinate is not changing, his proper time coordinate τ will eventually equal ρ, which means he will encounter the central τ=ρ singularity in finite proper time (proportional to ρ-τ). So long as (ρ-τ) > (4/3)M, he will be outside the event horizon but falling radially toward the center of the black hole. Nothing special happens to him when his proper time coordinate increases to τ = (ρ - (4/3)M) and he passes through the event horizon.
When his proper time coordinate increases to τ = ρ, however, he leaves the spacetime manifold M' and encounters the central singularity. I don't know what happens to him at that moment, or whether that even counts as a moment.
Exercise 23. Prove: If the Schwarzschild coordinates r, θ, and φ are held constant, then
The physical significance of exercise 23 is that the Schwarzschild time coordinate does not correspond to the proper time of an observer who considers himself to be at rest with respect to the spatial coordinates of the Schwarzschild chart. What's more, an observer cannot remain at rest with respect to the spatial Schwarzschild coordinates without undergoing a sensation of acceleration away from the black hole. The magnitude of that acceleration is a monotonic function of 2M/r, so the acceleration is small when r is large. That's why the Schwarzschild time coordinate t approximates the proper time of observers who are far from the black hole. That approximation becomes worse as observers approach the event horizon, and goes off the charts at the event horizon, where Schwarzschild coordinates fail entirely due to a coordinate singularity.
To understand what happens to an observer falling into a black hole, we must switch to a chart that doesn't have the Schwarzschild coordinate singularity at the event horizon. Lemaître coordinates are suitable, as was shown by exercises 20 and 21 above. Many other coordinate systems could be used instead, such as Eddington-Finkelstein, Novikov, or Kruskal-Szekeres coordinates.
Kruska-Szekeres coordinates are especially interesting because they cover an even larger manifold M*. It's my impression that M* can be regarded as the manifold M' that's covered by Lemaître coordinates, glued to a time-reversed copy of M'.
But I'm not a physicist. I'd be interested to hear what the physicists and mathematicians have to say about M*.
This is the third set of homework problems.
- Part 1 went over the basic definitions of manifolds and charts, using the 2-sphere as an example. (ctamblyn made an important correction to one of my inequalities.)
- Part 2 defined Schwarzschild charts on the spacetime manifold M outside the event horizon of an isolated black hole.
In this part 3, we will use Lemaître coordinates to define charts on a larger manifold M' that includes all of M plus the event horizon and a region of spacetime inside the event horizon that extends all the way down to the essential singularity at the center of the black hole. That will prove there is no real singularity at the event horizon. Another exercise will establish that particles falling radially into the black hole pass through the event horizon with no special fanfare and reach the essential singularity in finite proper time.
[size=+1]Schwarzschild coordinates and charts[/size]
With the (-,+,+,+) signature convention and natural units in which c=G=1, the Schwarzschild coordinates are (t, r, θ, φ), and the Schwarzschild metric is
[latex]
\[
ds^2 = - (1 - 2M/r)dt^2 + (1 - 2M/r)^{-1} dr^2 + r^2 d\Omega^2
\]
[/latex]
\[
ds^2 = - (1 - 2M/r)dt^2 + (1 - 2M/r)^{-1} dr^2 + r^2 d\Omega^2
\]
[/latex]
where
[latex]
\[
d \Omega^2 = d \theta^2 + \sin^2 \theta d \phi^2
\]
[/latex]
\[
d \Omega^2 = d \theta^2 + \sin^2 \theta d \phi^2
\]
[/latex]
is the usual metric on a 2-sphere in spherical coordinates and
[latex]
\[
\begin{align*}
r &> 2M \\
0 &< \theta < \pi \\
- \pi &< \phi < \pi
\end{align*}
\]
[/latex]
\[
\begin{align*}
r &> 2M \\
0 &< \theta < \pi \\
- \pi &< \phi < \pi
\end{align*}
\]
[/latex]
[size=+1]Lemaître coordinates and charts[/size]
The Lemaître coordinates are (τ, ρ, θ, φ), and the Lemaître metric is
[latex]
\[
ds^2 = - d \tau^2 + \frac{2M}{r} d \rho^2 + r^2 d\Omega^2
\]
[/latex]
\[
ds^2 = - d \tau^2 + \frac{2M}{r} d \rho^2 + r^2 d\Omega^2
\]
[/latex]
where dΩ2 is defined as above,
[latex]
\[
r = R(\tau, \rho) = \left[ \frac{3}{2} (\rho - \tau) \right]^{2/3} (2M)^{1/3}
\]
[/latex]
\[
r = R(\tau, \rho) = \left[ \frac{3}{2} (\rho - \tau) \right]^{2/3} (2M)^{1/3}
\]
[/latex]
will turn out to coincide with the r coordinate of Schwarzschild coordinates, and
[latex]
\[
\begin{align*}
\tau &< \rho \\
0 &< \theta < \pi \\
- \pi &< \phi < \pi
\end{align*}
\]
[/latex]
\[
\begin{align*}
\tau &< \rho \\
0 &< \theta < \pi \\
- \pi &< \phi < \pi
\end{align*}
\]
[/latex]
Exercises 17 through 20 will prove that R(τ, ρ) can be identified with the r coordinate of a Schwarzschild chart. For now, mathematically scrupulous readers will eliminate r from the Lemaître metric above by writing R(τ, ρ) wherever r appears.
Define a Lemaître chart to be of the above form, and let M' be the manifold that's covered by the set of all Lemaître charts. It takes two Lemaître charts to cover all of M' (see exercises 12 and 13).
Exercise 17. Prove: If 0 < x < 1, then
[latex]
\[
(1/x)^{1/2} (1 - x)^{-1} - x^{1/2} (1 - x)^{-1} = (1/x)^{1/2}
\]
[/latex]
\[
(1/x)^{1/2} (1 - x)^{-1} - x^{1/2} (1 - x)^{-1} = (1/x)^{1/2}
\]
[/latex]
Hint: trivial algebra.
Exercise 18. Prove: If
[latex]
\begin{align*}
d \tau &= dt + \left[ \frac{2M}{R(\tau, \rho)} \right]^{1/2}
\left[ 1 - \frac{2M}{R(\tau, \rho)} \right]^{-1} dr \\
d \rho &= dt + \left[ \frac{R(\tau, \rho)}{2M} \right]^{1/2}
\left[ 1 - \frac{2M}{R(\tau, \rho)} \right]^{-1} dr
\end{align*}
[/latex]
then\begin{align*}
d \tau &= dt + \left[ \frac{2M}{R(\tau, \rho)} \right]^{1/2}
\left[ 1 - \frac{2M}{R(\tau, \rho)} \right]^{-1} dr \\
d \rho &= dt + \left[ \frac{R(\tau, \rho)}{2M} \right]^{1/2}
\left[ 1 - \frac{2M}{R(\tau, \rho)} \right]^{-1} dr
\end{align*}
[/latex]
[latex]
\[
\frac{d \rho}{dr} - \frac{d \tau}{dr} = \sqrt{ \frac{R(\tau, \rho)}{2M} }
\]
[/latex]
\[
\frac{d \rho}{dr} - \frac{d \tau}{dr} = \sqrt{ \frac{R(\tau, \rho)}{2M} }
\]
[/latex]
Hint: trivial algebra; use exercise 17.
Exercise 19. Prove: If dτ and dρ are defined as in exercise 18, then dR/dr = 1.
Hint: basic calculus; substitute and simplify, using exercise 18 once.
Exercise 20. Prove: Given any Schwarzschild chart on M, there exists a Lemaître chart on M' whose restriction to (4/3)M < (ρ - τ) is isometric to the Schwarzschild chart.
Hint: Guess how R(τ, ρ), dτ, and dρ might be translated into Schwarzschild coordinates. To confirm your guesses, substitute them for R(τ, ρ), dτ, and dρ within the Lemaître metric, simplify, and compare your simplified result against the Schwarzschild metric.
Exercise 21. Prove: M is (isometric to) a proper submanifold of M'.
Hint: Use exercise 20 to prove the Lemaître charts cover M, and show that Lemaître charts also cover the event horizon and its interior (excluding the essential singularity at τ=ρ).
Now that we know M' contains a submanifold that's isometric to M, we might as well identify M with that submanifold. (That gets rid of the pedantic "isometric to" stuff.)
Exercise 22. Prove: If ρ, θ, and φ are held constant, then
[latex]
\[
ds^2 = - d \tau^2
\]
[/latex]
\[
ds^2 = - d \tau^2
\]
[/latex]
Hint: trivial algebra/calculus.
If (dτ, dρ, dθ, dφ) are infinitesimal displacements from some point p = (τ, ρ, θ, φ), then ds2 is the square of the pseudo-distance between p and q = (τ+dτ, ρ+dρ, θ+dθ, φ+dφ), and is calculated by applying the metric tensor at p to two copies of (dτ, dρ, dθ, dφ).
ds2 is a pseudo-metric because (unlike Euclidean metrics) it can be negative. The physical significance of a negative value is that a massive particle's world line can go through both of the points p and q. According to exercise 22, the timelike separation between points p and q is exactly the same as the difference between the values of their timelike coordinate τ. That means τ measures the proper time of an observer that considers himself to be at rest with respect to the Lemaître chart (because his spatial coordinates aren't changing).
In particular, consider a point-like observer who considers himself to be at Lemaître coordinates (τ, ρ, θ, φ), is not moving in a way that would change his spatial coordinates ρ, θ, and φ, and is not experiencing any sensation of acceleration. Since his ρ coordinate is not changing, his proper time coordinate τ will eventually equal ρ, which means he will encounter the central τ=ρ singularity in finite proper time (proportional to ρ-τ). So long as (ρ-τ) > (4/3)M, he will be outside the event horizon but falling radially toward the center of the black hole. Nothing special happens to him when his proper time coordinate increases to τ = (ρ - (4/3)M) and he passes through the event horizon.
When his proper time coordinate increases to τ = ρ, however, he leaves the spacetime manifold M' and encounters the central singularity. I don't know what happens to him at that moment, or whether that even counts as a moment.
Exercise 23. Prove: If the Schwarzschild coordinates r, θ, and φ are held constant, then
[latex]
\[
ds^2 = (-1 + 2M/r) d t^2
\]
[/latex]
\[
ds^2 = (-1 + 2M/r) d t^2
\]
[/latex]
Hint: trivial.
The physical significance of exercise 23 is that the Schwarzschild time coordinate does not correspond to the proper time of an observer who considers himself to be at rest with respect to the spatial coordinates of the Schwarzschild chart. What's more, an observer cannot remain at rest with respect to the spatial Schwarzschild coordinates without undergoing a sensation of acceleration away from the black hole. The magnitude of that acceleration is a monotonic function of 2M/r, so the acceleration is small when r is large. That's why the Schwarzschild time coordinate t approximates the proper time of observers who are far from the black hole. That approximation becomes worse as observers approach the event horizon, and goes off the charts at the event horizon, where Schwarzschild coordinates fail entirely due to a coordinate singularity.
To understand what happens to an observer falling into a black hole, we must switch to a chart that doesn't have the Schwarzschild coordinate singularity at the event horizon. Lemaître coordinates are suitable, as was shown by exercises 20 and 21 above. Many other coordinate systems could be used instead, such as Eddington-Finkelstein, Novikov, or Kruskal-Szekeres coordinates.
Kruska-Szekeres coordinates are especially interesting because they cover an even larger manifold M*. It's my impression that M* can be regarded as the manifold M' that's covered by Lemaître coordinates, glued to a time-reversed copy of M'.
But I'm not a physicist. I'd be interested to hear what the physicists and mathematicians have to say about M*.
I don't think Kruskal-Szekeres says anything about the time orientation of the glued copy by itself, but not time-reversing it creates a conical singularity at (u,v) = (0,0), where spacetime failing to be smooth violates the equivalence principle in the sense that there is no well-defined local inertial frame there. So your interpretation is in a sense more natural, but both should be consistent with the Kruskal-Szekeres chart.
Just to be clear, here's a Penrose diagram:
The stretched horizon does destroy stuff relative to an external observer, but your statement is not quite correct in the sense that even completely classically (with no hot surface whatsoever), you will not see anything fall into the horizon.
I'm not sure if it's helpful, but take a look at the Penrose diagram above on the left, illustrating the causal structure of Schwarzschild spacetime. Outgoing light rays go diagonally upper-right, ingoing light rays diagonally upper-left. The black hole is the region ABC, outside of it is the region ABED. Any massive particle must stay within its local light cone, i.e., between those diagonal light rays. So once you're anywhere on AB, your future is the singularity, and any signal you send stays within the black hole. Only someone who has already crossed the horizon can see things crossing the horizon, because every light ray from AB either stays on AB (outgoing) or goes inside the hole (ingoing).
All our measurements are local. How we label events is arbitrary, and irrelevant to those measurements. For example, for events not connected by a causal (lightlike or timelike) signal, the choice of whether to consider them simultaneous or not is completely arbitrary. So "our frame" is not objectively defined. It's a matter of choice.
In special relativity, there is a convention of Einstein synchronizationWP to extend an observer's local inertial inertial frame to all of spacetime. But there's nothing unique about it; it's just convenient.
It isn't like that, it's much simpler. My logic is this:
We all know that the coordinate speed of light varies in a non-inertial reference frame, and we know that super-accurate optical clocks lose synchronisation when separated by only a foot of vertical elevation. Simplify them to parallel-mirror light clocks and we can say that the actual speed of light varies, like so:
|----------------|
|----------------|
When we lower the bottom clock to the black-hole event horizon, then if we accept the assertion that gravitational time dilation is said to be infinite at that location according to "distant observers", then we have to assert that the actual speed of light at that location is zero, like so:
|----------------|
Hence rather than the observer not noticing anything unusual, he doesn't notice anything at all. He can't see, he can't think. His situation is akin to the hypothetical SR observer travelling at c. And a light beam directed straight upwards can't get out, not because it curves back down to the event horizon, but because the speed of light is zero.
I'm not advocating some wild speculation here, merely the original "frozen star" black hole interpretation.
Some people who appear to know mathematics and physics will tell you that the infalling observer crosses the event horizon in finite "proper time", but they won't tell you that in this entire universe, all the observers who have ever fallen into a black hole haven't crossed the event horizon yet. You can wait for a year, a thousand years, a billion years, and they still haven't crossed the event horizon yet. This is the point.
I already knew this.
They're imagining a line beyond eternity, and employing a coordinate system that does a hop skip and a jump over the end of time. See how the top of the left-hand chart below is omitted? The curve on the right of this chart goes vertical, and has no upper bound. That's the end of time. The light cone hasn't gotten past it yet, and never ever will, so everything to the left of the vertical dashed line is in never-never land. Sadly some people who appear to know mathematics and physics will duck and squirm and say anything other than admitting this to you because they fear that fallibility will erode their status, and the status of people like Hawking and Susskind.
I should interject here I suppose:
Not really. It's more like stepping back from any particular reference frame to see the bigger picture. Don't forget that a reference frame is an artefact of measurement. It's not as if you can point up to the night sky and say "look, there's a reference frame". See above, we can point to two beams of light at different elevations and say one's going slower than the other, hence the optical clocks lose synchronisation. Then we take the lower beam to a location where it doesn't move and clocks don't tick, whence there can be no measurment, and thus no reference frame at all.
I don't do this. I talk about light moving through space, not time moving in frames. It's not as if you can open up a clock and see time flowing through it. Be it a mechanical clock, a quartz wristwatch, or an atomic clock, all clocks "clock up" some kind of regular cyclic motion, and if the clock stops, that means the motion has stopped. So when we're dealing with light clocks, light has stopped. And if light has stopped, so has the observer, because all electromagnetic propagation stops, including the nerve impulses in his brain. So the observer can't think, light doesn't move towards his eye, hence he doesn't see his stopped clock carry on ticking. Instead he's like the hypothetical SR observer travelling at c. He doesn't see anything at all. Forever.
It's no mistake, I'm the one getting down to the fundamentals here. And let's not forget: you can't offer any counter-argument.
Vorpal: I'll get back to you.
Clinger: Give it a rest. Deal with the argument.
sol invictus
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Yes, although I prefer the above wording.
In special relativity, one considers inertial frames. The property of being inertial is very restrictive, and it's useful because the laws of physics, written in any inertial frame, look exactly the same. You can usefully talk about "your" frame in special relativity, because (up to translations of the origin and rotations of the coordinate axis) there is a unique inertial frame in which you are at rest.
Unfortunately, all of that goes out the window in general relativity. In a curved spacetime there are no inertial frames at all, and that means that there is a huge continuous infinity of frames in which you are at rest, but which differ sharply from each other far away from you. A nice example are the Schwarzschild coordinates, in which "time stops" at the horizon, and Kruskal-Szekeres coordinates, which extend smoothly through and beyond the horizon. Both coordinates can perfectly well be regarded as "the" frame of an observer far from the hole.
Physically, the right way to think about this is that we should only expect physics to predict the results of definite experiments. All coordinate systems should give identical predictions for the same physical experiment, so we're free to use any we like. In special relativity we get used to using intertial coordinates, but then again we often use spherical or polar coordinates (which are not inertial) when it's convenient, since it's obvious that doing so doesn't change anything physical. It's the same in GR - you can use any coordinates you like, they all give identical predictions, and that's the fundamental insight Einstein built the theory on.
W.D.Clinger
Philosopher
some homework problems for Farsight, epilogue 1
Your favorite chart is not the manifold. There are infinitely many charts, all of them equally correct. Depending on your purpose, some of those charts will be more convenient or more useful than others. Farsight insists upon using the Schwarzschild family of charts, which are completely useless if you want to understand what happens when things cross the event horizon. Schwarzschild charts literally don't go there.
Had Farsight done his homework, he'd know that Lemaître charts are perfectly smooth, with no discontinuities. Lemaître charts cover every part of spacetime that's covered by Farsight's beloved Schwarzschild charts, and then some.
Farsight is saying the spacetime manifold ends where his Schwarzschild chart ends. He's wrong. His chart is not the manifold.
I have no status here that I could lose. I'm not a physicist, and have never pretended to be a physicist. So far as I know, I'm the only person who cares whether I understand relativity. Furthermore, I know I don't understand relativity very well, and I know that quite a few of the contributors to this thread understand relativity better than I do. I came up with the homework exercises and worked through them myself because I wanted to improve my own understanding of the manifolds M, M', and M*.
Farsight isn't a physicist either, but he's told us his real name and told us about a book he's written that has "relativity" in its title. If any participant in this thread is motivated by "fear that fallibility will erode their status", it would be Farsight.
Thank you.
What Farsight isn't telling you, and appears not even to know, is that relativity has no well-defined concept of "yet". Your concept of "yet", like your concept of time itself, is observer-dependent; it depends on the chart (aka coordinate system/frame/patch) you choose to use. That's why the theory of relativity is called the theory of relativity.
Your favorite chart is not the manifold. There are infinitely many charts, all of them equally correct. Depending on your purpose, some of those charts will be more convenient or more useful than others. Farsight insists upon using the Schwarzschild family of charts, which are completely useless if you want to understand what happens when things cross the event horizon. Schwarzschild charts literally don't go there.
That's the sort of dismissive but vague language we often see from people who haven't done their homework.
Had Farsight done his homework, he'd know that Lemaître charts are perfectly smooth, with no discontinuities. Lemaître charts cover every part of spacetime that's covered by Farsight's beloved Schwarzschild charts, and then some.
Farsight is saying the spacetime manifold ends where his Schwarzschild chart ends. He's wrong. His chart is not the manifold.
Sounds as though Farsight is projecting.
I have no status here that I could lose. I'm not a physicist, and have never pretended to be a physicist. So far as I know, I'm the only person who cares whether I understand relativity. Furthermore, I know I don't understand relativity very well, and I know that quite a few of the contributors to this thread understand relativity better than I do. I came up with the homework exercises and worked through them myself because I wanted to improve my own understanding of the manifolds M, M', and M*.
Farsight isn't a physicist either, but he's told us his real name and told us about a book he's written that has "relativity" in its title. If any participant in this thread is motivated by "fear that fallibility will erode their status", it would be Farsight.
I've dealt with it. You haven't.
OK, thank you.
Although I have never seen a Penrose diagram before, it seems to say what I thought, although more clearly.
Thank you. I think that yours and Vorpals descriptions make sense to me.
But my question about GPS satellites makes me wonder about something else: As seen from my GPS unit the time signals from the satellites is going slower (I thought it was faster, but in another thread I saw somebody claim they actually go slower). Suppose there is a clock on board that was synchronised with our clocks at launch. This clock would then drift behind our time. Suppose that the satellite could be retrieved and taken back to Earth. Would the satellite clock still be behind our clocks, or would it now be synchronous with our clocks like before?
sol invictus
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There are two effects - the speed of the satellite relative to a point on the surface makes its clocks run slow, but the fact that it's higher up in the gravitational potential makes them run fast. I think for GPS the second effect wins and they run fast.
In any case, to answer your question - the satellite clock would still be ahead of our clocks. That experiment has been done (many times) and the results are in accord with the predictions of relativity. Time dilation is a real, physical effect in exactly that sense - it affects the results of physical experiments. What's surprising at first is that you can calculate those same results in any frame.
Last edited:
So, just like we see the clock of an astronaut go slow, an astronaut would also see our clocks go fast, not slow? I would have thought that the astronaut would feel himself to be stationary while the Earth spinned around under him very fast, and that would make Earth clocks go slow as seen from space.
Yes, it is surprising that this can be calculated to give the same result in any frame !
sol invictus
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To be clear, let's suppose we're talking about two spaceships, A and B, in outer space (no gravity) and in relative motion. In the rest frame of A, B's clocks run slow, and in the rest frame of B, A's clocks run slow. Even so, all calculations of the results of physical experiments will agree.
With gravity and/or rotation involved things get more complex. Any rest frame of the satellite is going to be pretty complicated - the earth and the rest of the galaxy is whirling around it very fast, but there are also very large "fictitious" forces (like centrifugal force) that affect time dilation (in the same way "ordinary" gravitational fields do).
Fortunately you can prove mathematically that it's true, which is much easier than trying to check calculations using complicated rotating frames.
Steenk: if you took two identical clocks and gave one to an astronaut, then fired him up into space, his clock is running faster than ours. We see this, and he sees that our clocks are running slower than his. When he comes back, his clock is ahead of our clocks, not behind.
Sol: have a look at the picture below. It's by Phil Fraundorf, associate professor at the University of Missouri in St Louis, see http://www.umsl.edu/~fraundorfp/.
See the red line? Clocks go slower because of the speed of the satellite. But also see the green line, clocks go faster because of the elevation. The latter is the bigger effect. The blue line shows the net.
Sol: have a look at the picture below. It's by Phil Fraundorf, associate professor at the University of Missouri in St Louis, see http://www.umsl.edu/~fraundorfp/.
See the red line? Clocks go slower because of the speed of the satellite. But also see the green line, clocks go faster because of the elevation. The latter is the bigger effect. The blue line shows the net.
sol invictus
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Not necessarily. It depends on how fast we fire him.
Yes, that looks right. I didn't remember where the crossover was (as you can see from that plot, for a very low orbit the net effect would be a slowing), but I checked another source and it is indeed the case that GPS clocks run fast.
No we don't. If "speed of light is zero" were a real property of that location, it would be seen as such by anyone at that location. It's not.
sol invictus
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Exactly.
Suppose someone falls into a black hole, holding a "light clock" (an arrangement of two mirrors held a fixed distance apart by a rigid piece of metal, with a light pulse bouncing back and forth between the mirrors).
Q1: Assuming our victim is inside a sealed container and cannot see out, will she notice anything when she crosses the horizon? Specifically, measured with respect to the time scale of her own biological processes (or any other clock or process inside the container), will the time it takes for the light in the light clock to bounce back and forth change?
A1: Nope - not if the black hole is large enough that tidal forces are small at the horizon, at least.
Q2: After having passed through the horizon, our intrepid investigator opens a porthole. Can she still see stars etc. that are outside the hole? Can she receive signals sent by her friend in a spaceship hovering above the horizon?
A2: Yes to both. They might be distorted, but they are still visible. That would be quite hard to understand if time really stopped at the horizon...
The truly fascinating this is how this is compatible with the description of the horizon as a hot membrane that I just gave above. Exercise for the reader: figure out how and why that's possible.
Good catch. Farsight's saying "nothing has fallen into the black hole yet", and pretending it's an absolute statement, is basically the same mistake as saying "the back of the ladder hasn't entered the barn yet" in the famous thought-experiment.
Reality Check
Penultimate Amazing
How dumb- that is what I have been saying. Not only do they never cross the event horizon, they never even reach it
! This is the point about you making that mistake in your previous posts. What happens at the event horizon does not matter because they never reach it according to the external observer. Their clock is never observed to stop.
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