feeding a black hole (APOD 27 Jun 2008)

[apodman]

Usually, the more I read the more I understand.

With some of the posts in this topic, the more I read the less I understand. Some are speaking in their own terms, speaking of something more conceptually advanced than I have been familiar with heretofore (newer than "modern" physics), or speaking phrases that simply can not be understood in terms of the Physics I was taught and still see published as the conventional wisdom of the day - I can't tell which.

On the whole, I don't know whether I know more or less than I did before. If a list of points of confusion counts as knowledge, I know more.

[harry]

G'day apodman

In general the more we read the more info we gain but it allows us to find how little we know. When you think you know is when you stop the research to gain more info.

[henk21cm]

Usually, the more I read the more I understand.

With some of the posts in this topic, the more I read the less I understand.

The subject was a phrase i learned in ancient Greek and it is rather appropriate when you encounter Quantum Mechanics. The world of QM is so alien, that our brains are not trained to grasp it. QM takes place in the world we usually do not observe. Our experience is based on the observations in the dayly macroscopic world, in which cats, cars, clippers and contracts are usual. The inventors of QM struggled with exactly the same concepts as you do. About entanglement Einstein spoke of "Ich glaube nicht in spukhafte Fernwirkung", I do not believe in ghostly influence over (large) distances. I'm more inclined to the modern mainstream and follow Heisenbergs pragmatic approach: QM as a tool to do successfull calculations on the microscopic world. He just used the matrix calculus to solve his problems.

Note: quoting my vision between the vision of great man like Einstein and Heisenberg does not mean that my value for the world is nearly the same as of these impressive men. Their intellect outshines mine by orders of magnitude.

The previous discussion over entanglement at the event horizon is not the kind of the pragmatic approach. Even Hawking struggles with these concepts. This places us in the right context and pose the question: "What are we in this discussion?"

Just Amateurs

[astrolabe]

Hello henk21cm,

I agree. We are just amateurs and if it wasn't for this Forum I think it doubtful that a lot of us would take the time to have even the smallest questions answered. Being amateurs, there are limits to our knowledge and for me it can be a little frustrating as there doesn't yet appear to be any limits to my questions.

There are lots of thinking forums out there and this is most definitely one of them. I think when one "hits the wall" with what one knows and then asks more than what one knows then they're, most assuredly, on the right road.

[Martin]

Correct me if I am wrong but before anything even approaches the event horizon it has already fell victim to spegitification. All of it's atoms have been stretched and torn apart before even reaching the event horizon. Due to the extreme gravitational forces. This is what you would see =String cheese!

Light can escape short of the EH but you and I would not for we cannot achieve 186,000 miles per second. The EH only marks the known boundary for light to escape not mass.

Regarding DM & a BH - Although we have observed evidence of DM having a gravitational influence on observable mass we have yet to observe any likewise reaction on DM. As far as we know there may be nothing in our realm of existence that can gravitationally act upon DM. Or everything does and we cannot observe it.

However, due to the extreme gravitational forces that are caused by a BH I can easily imagine a freakish event if the two meet. The existing relationship between observable mass and DM may appear harmonic but mix in a gravity well with all of its mass and power and well.. I think it is a bit presumptuous to entertain the thought that no unusual reaction would occur.

[Henning Makholm]

Correct me if I am wrong but before anything even approaches the event horizon it has already fell victim to spegitification. All of it's atoms have been stretched and torn apart before even reaching the event horizon. Due to the extreme gravitational forces.

Depends on which black hole we're talking about. The heavier the black hole gets, the flatter will space be at the horizon.

It is not the strength of gravity per se that tears stuff apart (and by the way, gravitational strength is not easily definable in non-Newtonian territory such as black holes), but tidal forces, i.e. spatial variation of gravity. That doesn't have to be very large at the horizon, e.g. for a galaxy-mass black hole.

[henk21cm]

G'day Henning,

It is not the strength of gravity per se that tears stuff apart (and by the way, gravitational strength is not easily definable in non-Newtonian territory such as black holes), but tidal forces, i.e. spatial variation of gravity. That doesn't have to be very large at the horizon, e.g. for a galaxy-mass black hole.

I'll give it a Newtonian try, since relativity theory is not my daily "push of the button".

The radius of the event horizon is called the Schwarzschild radius:

     R_s = 2GM/c²

where G is the gravitational constant, M the mass of the object inside the event horizon and c the velocity of light. The Newtonian accelleration is defined by:

     a = GM/r²

with a the accelleration, r the distance between the point where the accelleration is measured and the center of the mass.

As you mentioned, the spatial variation in accelleration is important. So we need a kind of "sniffer" sensor to find the sensitivity of the accelleration with its distance to the object inside the event horizon. So we differentiate the accelleration with respect to distance r:

     ∂a/∂r = -2GM/r³ = -2a/r

Next question is: how sensitive are we at the event horizon? For r we have to fill in r_s:

     ∂a/∂r (@r_s) = -2GM/{(2GM)/c²}³ = - {c³/(2GM)}²

The result is that the spatial sensitivity is inversely proportional to the square of the mass of the object inside the event horizon.

Now it's time for some numbers. Lets assume a massive black hole of 100000 solar masses, as is suggested in some galaxies. The mass of the object inside the event horizon M = 1E5 * 2E30 kg = 2E35 kg. The spatial sensitivity of the accelleration at the EH ∂a/∂r = -1 [1/s²]. When expressed in terms of earth' accelleration of gravity: ∂N_g/∂r = -0.1 [1/m].

Conclusion: when you are near the event horizon of this massive (1E5 solar masses) black hole, the difference in relative accelleration between your head and your feet is about 20%. That is not a solid cause for 'spaghettification', if you are in a free fall towards the object inside the event horizon. Decrease its mass by 100, and the sensitivity increases by 10000. That is sufficient for a result comparable to quartering a traitor in the middle ages.

Your point is prooved!

[Henning Makholm]

I'll give it a Newtonian try, since relativity theory is not my daily "push of the button".

Neither is it mine, but a Newtonian calculation should give errors on the order of unity, purely from dimensional considerations.
The relative error in the result, being dimensionless, has no right to vary with the size of the hole when the only inputs to the calculation are G, c, and the mass of the hole.

Now it's time for some numbers. Lets assume a massive black hole of 100000 solar masses, as is suggested in some galaxies.

Sounds reasonable. Current estimates of the mass of Sagittarius A* seem to be about 30-40 times as heavy again.

The spatial sensitivity of the accelleration at the EH ∂a/∂r = -1 [1/s²]. When expressed in terms of earth' accelleration of gravity: ∂N_g/∂r = -0.1 [1/m].

Conclusion: when you are near the event horizon of this massive (1E5 solar masses) black hole, the difference in relative accelleration between your head and your feet is about 20%. That is not a solid cause for 'spaghettification', if you are in a free fall towards the object inside the event horizon.

I don't think it makes much sense to express the difference as a percentage. Being in free fall eliminates the common acceleration from your experience and leaves only the difference, measured in m/s². By your numbers, a traveler will be stretched less by those tides than he will by hanging from his hands in Earth gravity, which certainly does not spaghettify most people.

In this particular case, expressing the differential acceleration as a percentage of the Newtonian acceleration of gravity is certainly a fallacy. Viewed from GR, what in the Newtonian picture is "the acceleration of gravity" is just minus the proper acceleration of a particle with constant space coordinates. However, as we approach the event horizon, the proper acceleration of the Schwarzschild lattice points become infinte, so that is the point where the Newtonian acceleration of gravity definitely loses its value as approximation to GR, because the thing in GR it used to approximate does not even exist here.

[henk21cm]

G'day Henning,

First of all: Welcome on this forum. I forgot to say that in my first reply.

... a Newtonian calculation should give errors on the order of unity, purely from dimensional considerations.

I was aware of the fact that the Newtonian approximation is not correct. When pressed for an aswer "how much?" maybe a factor two. Your reply confirms my worst suspicion.

I don't think it makes much sense to express the difference as a percentage. Being in free fall eliminates the common acceleration from your experience and leaves only the difference, measured in m/s².

The common accelleration is huge. It would flatten any of earth' lifeforms, maybe with the exception of a virus. In order to get rid of the huge (constant) accelleration i introduced 'free falling' and focus on the differences between head and feet. The answer is about 2m/s². In order to put a value into perspective, the comon technique is to normalize it by some reference value, hoc loco the standard accelleration of gravity on the earth' surface. The normalized value gives the impression that the differential accelleration was small, less than when a person is hanging on his/her feet, head down during Yoga excercises. Similarly you referred to "hanging from his hands".

Then you continue:

In this particular case, expressing the differential acceleration as a percentage of the Newtonian acceleration of gravity is certainly a fallacy. Viewed from GR, what in the Newtonian picture is "the acceleration of gravity" is just minus the proper acceleration of a particle with constant space coordinates. However, as we approach the event horizon, the proper acceleration of the Schwarzschild lattice points become infinte, so that is the point where the Newtonian acceleration of gravity definitely loses its value as approximation to GR, because the thing in GR it used to approximate does not even exist here.

I hope that you understand the reason for normalizing the differential accelleration, and i hope that you are convinced that i did not want to mislead anybody. Then, all after "fallacy" is rather a puzzle to me. I see words, i recognise them, however their meaning is as alien to me as a computer hardware manual to Vergilius or Plato. In other words, could you explain what you have written in more detail, preferably with some background information or references? Especially "the Schwarzschild lattice points becoming infinite" give me the idea that part of my brain is missing.

[Henning Makholm]

First of all: Welcome on this forum. I forgot to say that in my first reply.

Thanks. Let me take this opportunity to join the "I'm just an amateur" chorus.

... a Newtonian calculation should give errors on the order of unity, purely from dimensional considerations.

I was aware of the fact that the Newtonian approximation is not correct. When pressed for an aswer "how much?" maybe a factor two. Your reply confirms my worst suspicion.

Worst suspicion? I agree that a factor two is likely, or perhaps 2pi or 4ln(2), but at least something in that ballpark. What I tried to say was "I agree that a Newtonian calculation should be acceptable for our back-of-the-envelope purposes here. This dimensional reasoning is the reason I agree".

The comonn accelleration is huge. It would flatten any of earth' lifeforms, maybe with the exception of a virus. In order to get rid of the huge (constant) accelleration i introduced 'free falling' and focus on the differences between head and feet. The answer is about 2m/s².

Um, it seems I misunderstood you. You wrote:

the difference in relative accelleration between your head and your feet is about 20%

I can see now that you meant "about 20% of g" -- which I don't disagree with -- but somehow I got into my head that you meant "about 20% of the common acceleration". If I had thought a bit more I'd have noticed how absolutely preposterous that percentage would be and realized you couldn't possibly have meant it that way. But I never got past the "argh, dimensionless number! nonsense! nonsense!" stage. Sorry.

Then, all after "fallacy" is rather a puzzle to me. I see words, i recognise them, however their meaning is as alien to me as a computer hardware manual to Vergilius or Plato. In other words, could you explain what you have written in more detail, preferably with some background information or references? Especially "the Schwarzschild lattice points becoming infinite" give me the idea that part of my brain is missing.

As it turns out, the "fallacy" was mine alone; I apologize for pinning it on you. My point is moot now (I see that nothing you said contradicts it), but let me try to rephrase it with less jargon nonetheless:

The huge (flatten-any-earthly-lifeform) common acceleration we compute by the Newtonian is not a good order-of-magnitude approximation to the GR reality at the horizon, because GR says that the true value actually becomes infinite at the horizon.

How can the acceleration of gravity be infinite at the horizon, yet the tides finite? The answer is that "tides is the spatial derivative of the acceleration of gravity" is itself a Newtonian approximation which fails when general-relativistic effects begin to dominate. The tidal effect is a quantifiable physical reality at every point in GR spacetime (it's described by the "Weyl tensor", which can be derived from the curvature of spacetime). But "acceleration of gravity" has no intrinsic meaning in the GR world. It ought to mean "the acceleration of a freely-falling particle relative to a non-accelerating observer" -- but GR teaches that there is no such thing as an objectively non-accelerating observer; or rather, the best non-accelerating observer we can have IS a freely-falling particle.

Well, how can I say that the acceleration of gravity becomes infinite when I've just argued that it doesn't exist at all? I extrapolate from the definition that makes acceleration of gravity a useful concept in the Newtonian approximation. This says that the acceleration of gravity is the acceleration of a freely-falling particle A with respect to a point B that stays in the same position with respect to the sources of gravity. In the Newtonian picture, B is a good standard of non-acceleration, whereas GR says that A is the best standard of non-acceleration we have. However, we get the same value for "acceleration of gravity" if we define it in GR as minus the acceleration of B with respect to A.

That only leaves the trouble of defining what it means for the point B to be stationary with respect to the sources. In GR that is done by requiring that B has constant spatial coordinates in a "stationary" coordinate system. Being "stationary" is a rather special condition on a coordinate system; essentially it requires that the curvature of spacetime is the same at all events that share the same spatial coordinates. Most physical situations do not allow for stationary coordinates; they fail to exist even for two massive bodies orbiting each other in a non-circular obit. However, there IS a stationary coordinate system for a universe containing exactly one gravitating point-mass, namely the Schwarzchild solution to the GR equations. Those are the standard coordinates for doing one-body problems in GR.

Now, I may define the "acceleration of gravity" outside the black hole to be minus the acceleration of a Schwarzchild coordinate point with respect to a free particle. In more everyday terms, this corresponds to the gravitational field one would feel standing on a non-rotating solid shell that surrounded the black hole concentrically. Far from the central body, this is well approximated by the Newtonian value, but the GR value goes towards infinity when we approach the event horizon. On and inside the event horizon even the Schwarzchild solution does not provide stationary coordinates, and it can be shown that no stationary coordinates CAN exist inside the event horizon.

As for references, I can warmly recommend the textbook that taught me all this, "Relativity" by Wolfgang Rindler (Oxford University Press, 2001). I have found it an excellent self-study resource.

[henk21cm]

G'day Henning,

I have read the first part of your reply. Our slight misunderstanding has been resolved. No hard feelings! I'll focus on the Schwarzschild explanation, which is more intriguing and opens a world to previously unknown facts.

... because GR says that the true value actually becomes infinite at the horizon. How can the acceleration of gravity be infinite at the horizon, yet the tides finite?

In the Newtonian approximation where the tides are the derivative of the "acceleration" it is hard to find a function which is infinite at some value z0, whereas the derivative of that function with respect to z at z0 is finite. tan(z) at z=π/2 does not qualify, just like any function in the family ln(z) and 1/z^n.

The tidal effect is a quantifiable physical reality at every point in GR spacetime, it's described by the "Weyl tensor", which can be derived from the curvature of spacetime.

Correct me if i'm wrong, first picture a perfectly flat space, no curvature. Now introduce a mass, which curves space. In a 1D analogon: a flat straight line and some curve, e.g. a Gaussian bell, upside down. The vertical distance between a point on the flat line and the corresponding point on the curve is a measure for the gravitational force between the introduced mass and a very small test mass. The slope of the curve would be a measure for the tide. Is this what you mean?

Next you explained the problem with the stationary reference frame, "B". Based on equivalence, the acceleration of A with respect to B is the same as minus the acceleration of B with respect to A.

In GR that is done by requiring that B has constant spatial coordinates in a "stationary" coordinate system. Being "stationary" is a rather special condition on a coordinate system; essentially it requires that the curvature of spacetime is the same at all events that share the same spatial coordinates.

First idea was: that should not be too difficult. Any point on a sphere that surrounds a mass has a constant space-time curvature. But then: that means trouble. Since the original mass is not alone in this universe, there more spheres, intersecting the original sphere. That leaves us with a deformed sphere, which in a mathematical sense has entirely lost its spherical symmetry (although within a few percent may be a sphere). Your elucidation on the two bodies, moving in non-circular orbits, triggered this thought. If its wrong, do correct me.

...but the GR value goes towards infinity when we approach the event horizon. On and inside the event horizon even the Schwarzchild solution does not provide stationary coordinates, and it can be shown that no stationary coordinates CAN exist inside the event horizon.

There must be a reason why the acceleration of B with respect to A (=free falling) reaches infinity at the event horizon and i fail to see it. The only reason i can think of is "by definition". At the event horizon there exists no escape to the outside world. No force or energy is sufficient to increment the distance between A and the center of mass. That means that the force must be infinte. Is that the background for "infinity"?

Wolfgang Rindlers book on Relativity: located on the web and on my list.

[Henning Makholm]

In the Newtonian approximation where the tides are the derivative of the "acceleration" it is hard to find a function which is infinite at some value z0, whereas the derivative of that function with respect to z at z0 is finite. tan(z) at z=π/2 does not qualify, just like any function in the family ln(z) and 1/z^n.

Correct. So what I was trying to say is that according to GR the tides are real, but the Newtonian gravitational field whose gradient the tides are supposed to be is not.

The tidal effect is a quantifiable physical reality at every point in GR spacetime, it's described by the "Weyl tensor", which can be derived from the curvature of spacetime.

Correct me if i'm wrong, first picture a perfectly flat space, no curvature. Now introduce a mass, which curves space. In a 1D analogon: a flat straight line and some curve, e.g. a Gaussian bell, upside down. The vertical distance between a point on the flat line and the corresponding point on the curve is a measure for the gravitational force between the introduced mass and a very small test mass. The slope of the curve would be a measure for the tide. Is this what you mean?

I don't think I follow you here. Tides in GR are just a fact of how space is curved; they are not related to why space is curved. The standard gedankenexperiment (IIUC) is to deposit a small ball of test particles (with negligible gravity of their own) somewhere in space, initially at rest with respect to each other, and watch how they start to drift under gravity. In general it may start to shrink or enlarge, and/or to deform into an ellipsoid. The rate at which the ball deforms into an ellipsoid is a measure of the strength of the tides.

According to this nice explanation by John Baez, Einstein says that the rate at which our test ball changes its volume is proportional to the energy density at its location, measured from the rest frame of the ball. If the ball floats in a vacuum, its volume will stay constant. Thus, a tendency to become ellipsoidal will look as if something is tugging it apart in one direction and squeezing in another -- presto, tides!

When I speak of "rate" here, I mean the second-degree coefficient in a Taylor expansion of the behavior of the ball as a function of time. The first-degree coefficient is zero because we stipulate that the test particles initially have no mutual velocity. Accordingly, the natural dimension of tides becomes 1/s², consistent with the Newtonian idea of acceleration per meter.

In GR that is done by requiring that B has constant spatial coordinates in a "stationary" coordinate system. Being "stationary" is a rather special condition on a coordinate system; essentially it requires that the curvature of spacetime is the same at all events that share the same spatial coordinates.

First idea was: that should not be too difficult. Any point on a sphere that surrounds a mass has a constant space-time curvature. But then: that means trouble. Since the original mass is not alone in this universe, there more spheres, intersecting the original sphere. That leaves us with a deformed sphere, which in a mathematical sense has entirely lost its spherical symmetry (although within a few percent may be a sphere).

It sounds like you're misunderstanding me here. You don't need spherical symmetry to have stationarity. Different points of space can have different curvatures, but each point has to keep its curvature as time passes, forever. So stationarity in this sense is possible only in a changeless universe.

In a general two-body system the distance between the bodies varies periodically, so it is not changeless, and perfect stationarity is not possible. (We can have approximate stationarity. An ordinary coordinate system for the solar system, including the interior of the sun and the planets, is close enough to stationarity for the Newtonian approximation to be useful. Inside a black hole, however, stationarity cannot even be approximated, because any timelike path in it ends up at the singularity, and thus passes through regions with arbitrarily large curvature).

There must be a reason why the acceleration of B with respect to A (=free falling) reaches infinity at the event horizon and i fail to see it. The only reason i can think of is "by definition". At the event horizon there exists no escape to the outside world. No force or energy is sufficient to increment the distance between A and the center of mass. That means that the force must be infinte. Is that the background for "infinity"?

Yes, I think this is good, but it is perhaps more convincing to think in terms of "no particle world-line with finite proper acceleration can touch the horizon and escape". Eliminating appeals force and energy gives a cleaner and more geometric picture.

If point B is just outside the horizon, it could be the worldline of a rocket hovering just above the horizon. Another rocket with more thrust (i.e. a greater acceleration) could be descending towards the black hole, decelerating to come to a standstill just beside B, and going off into space. In fact, since the other rocket is stronger, it can afford to come to its standstill a bit further down than B. If the acceleration a rocket a B needs to exert to stand still were to approach a finite limit as B comes closer to the horizon, the lowest point from which a rocket with twice as good an acceleration as this limit can escape ought to be below the horizon, contradicting the definition of the horizon.

[henk21cm]

G'day Henning,

It sounds like you're misunderstanding me here. You don't need spherical symmetry to have stationarity. Different points of space can have different curvatures, but each point has to keep its curvature as time passes, forever. So stationarity in this sense is possible only in a changeless universe.

Well, i partly misunderstood you. I came to the conclusion that, in order to keep a test mass (with negligible gravitation, like the ones in the ball of beads) at a constant curvature, it has to orbit another significant mass, in a circular orbit. That orbit could be any great circle on a sphere, concentric with the center of the (significant) mass. If the test mass would not be moving, it would follow the curvature of space in the direction of the steepest gradient: towards the large mass, which originated the curvature of space.

I found and downloaded the article by Baez and Bunn. I'll have to study that in detail before it makes sense to continue this discussion.

[Henning Makholm]

I came to the conclusion that, in order to keep a test mass (with negligible gravitation, like the ones in the ball of beads) at a constant curvature, it has to orbit another significant mass, in a circular orbit.

Ah, but these points (the lattice points in stationary coordinate systems) are not test masses; they are just abstract points. Such a point in an earth-centered coordiante system might be "400 m above the tip of the Eiffel tower and 3.6 m to the north-east". Nothing special has to inhabit that point for us to speak about whether spacetime curvature there is constant in time (which it is, to the extent that we can ignore the gravity of the moon, the sun, etc). We don't need to imagine any test mass staying there, or a machinery for keeping it there.

[astrolabe]

Hello Henning Makholm,

My apologies, I must have been a coma. Tell me again about rocket A and rocket B. As I understand you, world-line means real-world? If so, then 1) we wouldn't be able to see Rocket B and 2) every particle, massless or not, would be subjected extreme forces at their QM levels all the way to Plank length and Plank time.

I don't qualify as JAA so I can be guilty of serious speculation and conjecture but really try to keep all the "what if's" within the concepts that I know are theory. And I do appreciate very much that you are able to couch your ideas in common language that I can understand and, as a result of this quality, I was able to follow your line of thinking. However, I ran into some difficulty when your analogies went from real-world to perfect-world examples. For instance, when you said that a stationary point doesn't have to have anything important in it where I think there will always be something important there. And ,for the sake of ideas, for all we know there ARE real stationary points.

All in all, I enjoyed your post as well as your logic along these lines as I do the other members as well. At least when you utilize "could", "if", and the like, It tells me you're not trying to state things as fact whiich IMHO a good sign. Welcome, grab a thick-skinned coat on your way, we all wear 'em!

[Henning Makholm]

On reflection, I was wrong to introduce stationarity into the discussion. Intuitively, I'm still fairly certain that the right thing to measure "acceleration of gravity" against is the points of constant Schwarzschild space coordinates, but I'm no longer sure why. I certainly isn't just because they are lattice points of a stationary coordinate system; rotating coordinates could form a stationary system too, but would not be the right thing to measure a pseudo-Newtonian gravity field against.

It didn't help that I erred in explaining what stationarity means. It is actually not enough that the curvature at every lattice point is time-invariant; the entire metric tensor must be. Unfortunately, I don't know where to start explaining "metric tensor" without writing a book.

Tell me again about rocket A and rocket B.

There's no rocket A -- A was a free-falling particle. There was a rocket B and a second imaginary rocket, which I artfully left unnamed :-)

As I understand you, world-line means real-world?

This does not make much sense to me, so no. I think.

"World-line" is standard jargon in releativity. It means the set of all the points ("events") in four-dimensional space-time where the particle in question exists. This set makes up a one-dimensional trace through space time. Now that you mention it, I'm not quite sure what the word "world" is doing in the term, but it's traditional.

If so, then 1) we wouldn't be able to see Rocket B and 2) every particle, massless or not, would be subjected extreme forces at their QM levels all the way to Plank length and Plank time.

Oh, no. That would be true if we were playing at the middle of the black hole, where a singularity is supposed to be found and everything becomes infinite. But here we're only speaking about the event horizon, which locally looks like just an ordinary place. It is a horizon only by virtue of how it fits into the scenario as a whole.

In particular, rocket B hovers just outside the horizon, and should be perfectly visible from outside, if more or less redshifted.

For instance, when you said that a stationary point doesn't have to have anything important in it where I think there will always be something important there. And ,for the sake of ideas, for all we know there ARE real stationary points.

The operative idea (which I now think was not actually pertinent, see above) is not one of a stationary point, but of a stationary coordinate system. Stationarity means that every point of the coordinate system keeps the same distance to its neighbour points as time passes (i.e. when we increase the time coordinate and hold the other three coordinates constant). This has to be true for every point, so it does not make one of the points special. There cannot be something important everywhere; that would belie the importance.

[astrolabe]

,Hello Henning Makholm,

I sincerely thank you for your reply and clarifications. It's sometimes hard , of course, to visualize this stuff but if successful another small piece of the puzzle falls (mostly) into place. It's always a good experience to see another's views and ideas. Hope you stick around!