B Shape of Hubble sphere at relativistic velocity

[Chris Miller]

TL;DR Summary

Would distance to Hubble horizon (distance over which expansion = c) increase, decrease or remain the same when traveling near c.

Consider the Hubble horizon as the proper distance over which Hubble expansion equals c, so that you are in the center of a Hubble sphere with a radius of about 13.5 billion light-years. As you approach light speed in any direction, does the Hubble horizon draw closer in that direction due to length contraction? Or does time dilation increase your measure of the Hubble constant? I'm imagining zipping through galaxies in milliseconds while their inhabitants watch me come and go for eons. Where is my Hubble horizon? What is the shape of my Hubble sphere?

 

[anorlunda]

I think you know the answer yourself if you think about it.

All motion is relative. In the reference frame of your non-accelerating spaceship, it is sitting still. So what is the Hubble Horizon for a stationary object?

 

[Chris Miller]

Thank you @anorlunda. Bit isn't this hypothetical non-accelerating craft moving relative to the CMB? In other words, it has direction relative to the cosmic microwave background, and so velocity as well.

 

[anorlunda]

Thank you @anorlunda. Bit isn't this hypothetical non-accelerating craft moving relative to the CMB?

No. The CMB is not a place.

https://en.wikipedia.org/wiki/Cosmic_microwave_background

The cosmic microwave background radiation is an emission of uniform, black body thermal energy coming from all parts of the sky. The radiation is isotropic to roughly one part in 100,000: the root mean square variations are only 18 µK,[8] after subtracting out a dipole anisotropy from the Doppler shift of the background radiation. The latter is caused by the peculiar velocity of the Sun relative to the comoving cosmic rest frame as it moves at some 369.82 ± 0.11 km/s towards the constellation Leo (galactic longitude 264.021 ± 0.011, galactic latitude 48.253 ± 0.005).[9] The CMB dipole as well as aberration at higher multipoles have been measured, consistent with galactic motion.[10]

Remember that you are right now moving at near light speed relative to the reference frame of some object somewhere. What do you observe when you look at the CMB with your radio telescope?

 

[jbriggs444]

No. The CMB is not a place.

It does, however, establish a local standard of rest -- a local frame in which it is observed to be isotropic. A "moving" radio telescope will observe blue-shifted CMBR in one part of the sky and red-shifted CMBR in another.

 

[Chris Miller]

Thank you @jbriggs444 and @anorlunda for your replies. I was afraid my reference to the CMB would derail my question. Let me try to put it a different way.

You are approaching a planet at exactly the same near-c velocity that Hubble expansion is causing it to recede, so the distance between you and this planet is constant. Is this distance foreshortened by your measurement? (I'm thinking not at all.)

Suppose the planet's proper distance is only a few light-years (so that Hubble expansion is negligible), and you are approaching it at the same velocity. Do you now measure it to be closer? (I'm thinking yes, a lot.) Since velocity is relative, do the inhabitants of this planet also measure this distance as contracted? (I'm thinking no, but can't say why.)

 

[jbriggs444]

Thank you @jbriggs444 and @anorlunda for your replies. I was afraid my reference to the CMB would derail my question. Let me try to put it a different way.

You are approaching a planet at exactly the same near-c velocity that Hubble expansion is causing it to recede, so the distance between you and this planet is constant. Is this distance foreshortened by your measurement? (I'm thinking not at all.)

Suppose the planet's proper distance is only a few light-years (so that Hubble expansion is negligible), and you are approaching it at the same velocity. Do you now measure it to be closer? (I'm thinking yes, a lot.) Since velocity is relative, do the inhabitants of this planet also measure this distance as contracted? (I'm thinking no, but can't say why.)

I think we need somebody smart to answer this one. But my gut says that the notion of "distance" here is problematic. "Distance" is a measure that depends on a coordinate system. You want an observer-dependent ("foreshortened") distance measure, but you are working within a curved space-time.

In flat space-time, the motion of a single observer pretty well determines what coordinate system you will use. In curved space-time, it does not.

 

[Chris Miller]

@jbriggs! I really appreciate your expressed (mild) confusion, and your gut's answer, which seems to apply more to the first scenario. But aren't time and distance always observer (frame of reference) dependent? How do these questions differ from textbook relativity problems? (I doubt I'd understand someone smarter than you's answer.)

 

[jbriggs444]

@jbriggs! I really appreciate your expressed (mild) confusion, and your gut's answer, which seems to apply more to the first scenario. But aren't time and distance always observer (frame of reference) dependent? How do these questions differ from textbook relativity problems? (I doubt I'd understand someone smarter than you's answer.)

I appreciate the back-handed compliment.

Textbook relativity problems (in the texts that I'd find understandable anyway) would be posed within the background of special relativity. In those environments, life is simple. Observer, frame of reference, coordinate system -- you can treat them all as pretty much interchangeable concepts.

In those problems, time is usually observer dependent. Though we also have proper time which is the invariant length of a time-like trajectory between two events. Obviously you can measure proper time directly with a wristwatch.

Distance is (to the best of my knowledge) almost always coordinate system dependent. Technically you have available an invariant notion of distance as the total length of a particular space-like trajectory between two events.

In flat space-time, the invariant distance between two space-like separated events or the invariant time between two time-like events is unambiguous. In curved space-time, the invariant distance or time between two suitably nearby events is still unambiguous. But over greater separations, there may be more than one unaccelerated (geodesic) trajectory between the two events. [It seems that "geodesic" can apply to the space-like case].

In the case at hand, we are trying to assess the "distance" between two world-lines rather than between two events. So that makes things messy. We are, by design, after a frame-relative measure rather than an invariant measure. So that makes things worse.

But...

Given a global standard of simultaneity, we could pick out a particular event on the far-away planet to match up with an event where we are located. We have a reasonably simple global topology so there should be a unique space-like geodesic between those two events. And we could, I expect, do a parallel transport of our local peculiar velocity along that geodesic to compute our velocity relative to any particular path segment. So we could integrate the contracted "length" of each path segment relative to our peculiar velocity and obtain a cumulative distance measure for the path. One would expect that distance measure to then be massively length contracted relative to a more simplistic computation of invariant distance.

Alternately, we could use a light-like trajectory between the world lines. I am not 100% sure how one would associate a "distance" metric with the incremental segments of a light-like geodesic. Possibly it is as simple as dotting the parallel-transported four-velocity with the incremental segment displacement.

 

[Chris Miller]

Didn't intend it to be back-handed, laughed when I realized how it was. You'll be proud to know, then, that I don't completely understand your very carefully worded answer. Particularly, (after looking up "geodesic") your

But over greater separations, there may be more than one unaccelerated (geodesic) trajectory between the two events.

How can there more than one "shortest possible line between two points on a sphere or other curved surface"...?

Ah. Duh. Of course. Could be infinitely many. But, still, equal.

Been revisiting Minkowski spacetime, which so far clears up a little relativity confusion.

Appreciate the "But..." and your highlighted conclusion. Even if wrong, it satisfies my need not to have my SF present as too ignorant. So... at near c, the Hubble sphere might not be spherical, with the forward radius greatly foreshortened? Still some confusion around time dilation causing the H "constant" to be greatly increased in my FoR (cancelling distance contraction). I read yesterday somewhere that gravitational time dilation caused H to be 20,000 times greater in the early universe.

Thanks very much for taking the time here @jbriggs444 . Most grateful. I think I have a better grasp now.

 

[jbriggs444]

How can there more than one "shortest possible line between two points on a sphere or other curved surface"...?

You've understood the case for two spatial dimensions on a spherical shell. Good.

Now consider that I stand on the surface of an airless moon. I can fire a bullet in a flat trajectory so that it completes an orbit and arrives back at my firing position. That is one geodesic trajectory.

Or I could aim my gun straight up and fire the bullet in a vertical trajectory so that it eventually falls back down to my position. If I adjust my muzzle velocity for the second shot, I can arrange for the elapsed time (by my pocket watch) for the two trajectories to be equal -- the two bullets can be fired simultaneously and arrive back at my position simultaneously.

[Possibly I should take one step to the left before awaiting the simultaneous arrival]

Consider the elapsed proper time experienced by the two projectiles. The horizontal trajectory is low in a gravity well and is moving rapidly. Lots of time dilation. Low elapsed proper time. The vertical trajectory is high in a gravity well and can spend a lot of time moving slowly. Little time dilation. High elapsed proper time.

Voila -- two geodesics between the same two events with different time-like path lengths.

 

[Chris Miller]

@jbriggs444 Your atmosphere-less moon geodesic example isn't non-inertial, if that matters.

I've never quite understood the importance of measurement (ability to observe) on theoretical reality. The universe is what the universe is, whether you are a blind gnat or a technologically advanced AI physicist in the year 4040). At near c, the Hubble radius (I prefer horizon) is or is not foreshortened. Hubble expansion could be measured over much shorter distances when traveling at near c, could it not? Or "chased" at far slower velocities where length contraction, though negligible if in effect, might be more measurable.) I'm also still not sure Hubble expansion effects time dilation and length contraction the way velocity does (actually, I'm pretty sure it does not). But that you are also a little confused is more than good enough for me. I really appreciate your thoughts.

 

[jbriggs444]

@jbriggs444 Your atmosphere-less moon geodesic example isn't non-inertial, if that matters.

I am not sure what you mean. The two trajectories are both inertial by design. An object following a time-like geodesic has zero proper acceleration throughout.

If your point is that the shooter is experiencing non-zero proper acceleration, that is true. But his trajectory is not under consideration.

 

[Nugatory]

The universe is what the universe is, whether you are a blind gnat or a technologically advanced AI physicist in the year 4040).

Yes, but...

At near c, the Hubble radius (I prefer horizon) is or is not foreshortened.

The physical fact, what you mean by "the universe is what it is" and more often called "invariant" or "frame-independent" or "coordinate-independent", is whether there is a light-like geodesic between an event on the distant planet's worldline and yours. If such a geodesic exists, then that event is at the Hubble radius from you at the moment that the geodesic intersects your worldline. Take the collection of all events for which such a geodesic exists and passes though a given event on your worldline; collectively these form what you're calling the Hubble horizon at that event. And yes, for every single event in the entire universe, we can say that either it is inside that Hubble radius or it is not.

But does this collection of events form a sphere in space, or is it foreshortened? Light from one star but not another reaches you; is this because you are moving towards one and away from the other, or because they were at different distances from you, or because the light was emitted at different times? These answers aren't "the universe is the way it is" things, they are the consequences of the coordinates that you're using to define distances, positions, "at the same time", and relative velocities. Different choices, different answers.

 

[Chris Miller]

@jbriggs444 But aren't they impacted by different gravitational influences?

 

[Chris Miller]

@Nugatory I agree with, and think I actually understand, what you say. I thought, perhaps wrongly, that I'd specified my choices sufficiently in my original question.

Actually, I intended my admittedly too trite "it is what it is" to assume frame dependence. Like at some relative velocity the distance(s) to something (in my frame of reference) will be what it is (they are), whether or not I have the tools to measure it (them).

 

[PeterDonis]

The physical fact, what you mean by "the universe is what it is" and more often called "invariant" or "frame-independent" or "coordinate-independent", is whether there is a light-like geodesic between an event on the distant planet's worldline and yours. If such a geodesic exists, then that event is at the Hubble radius from you at the moment that the geodesic intersects your worldline. Take the collection of all events for which such a geodesic exists and passes though a given event on your worldline; collectively these form what you're calling the Hubble horizon at that event.

No, that's not the Hubble horizon. That's the particle horizon (or boundary of the observable universe) "now".

The Hubble horizon is the set of events "now" (in the same spacelike hypersurface of constant FRW coordinate time as us here on Earth "now") that have an FRW coordinate recession velocity "now" of c relative to us here. I'm not aware of a simple definition of this in terms of light cones or other obvious invariants.

This paper by Davis and Lineweaver is an excellent reference for the different kinds of horizons:

https://arxiv.org/abs/astro-ph/0310808

Note that, as the paper says, the Hubble "horizon" is not actually a horizon of any kind; it's different from both the particle horizon and the event horizon (the latter is the boundary of the region of spacetime that can never send light signals to us here at all). The paper calls it the "Hubble sphere" rather than the "Hubble horizon" for that reason.

 

[PeterDonis]

You are approaching a planet at exactly the same near-c velocity that Hubble expansion is causing it to recede, so the distance between you and this planet is constant. Is this distance foreshortened by your measurement? (I'm thinking not at all.)

Suppose the planet's proper distance is only a few light-years (so that Hubble expansion is negligible), and you are approaching it at the same velocity. Do you now measure it to be closer? (I'm thinking yes, a lot.) Since velocity is relative, do the inhabitants of this planet also measure this distance as contracted? (I'm thinking no, but can't say why.)

As @jbriggs444 says, none of these questions are answerable until you specify what notion of "distance" you are using in each case. In flat spacetime, distance according to the inertial frame in which the observer is at rest is the natural assumption; but there is no such thing in curved spacetime. The Hubble expansion is not the same as the intuitive concept of "things moving away from you" that you are implicitly using, because of spacetime curvature.

At near c, the Hubble radius (I prefer horizon) is or is not foreshortened.

No. "Distance" is not an invariant. There are invariant notions of "distance" you could use: the most obvious would be proper distance in a spacelike surface of constant time for comoving observers. But this distance would not be foreshortened since changing your state of motion does not change this spacelike surface; it's defined with respect to a fixed family of observers, whether you are one of them or not.

Also, as noted in my post in response to @Nugatory a little bit ago, the Hubble sphere is not a horizon and "horizon" is not a good term for it. I strongly suggest reading the Davis and Lineweaver paper I linked to in that post.

 

[PeterDonis]

Given a global standard of simultaneity, we could pick out a particular event on the far-away planet to match up with an event where we are located. We have a reasonably simple global topology so there should be a unique space-like geodesic between those two events. And we could, I expect, do a parallel transport of our local peculiar velocity along that geodesic to compute our velocity relative to any particular path segment. So we could integrate the contracted "length" of each path segment relative to our peculiar velocity and obtain a cumulative distance measure for the path. One would expect that distance measure to then be massively length contracted relative to a more simplistic computation of invariant distance.

No. One would expect exactly the opposite. The distance measure you are describing is just the proper distance in a spacelike surface of constant time for comoving observers that I referred to in my post in response to @Chris Miller just now. That proper distance is an invariant; it is independent of your state of motion.

Perhaps it's worth commenting here on what "length contraction" in flat spacetime actually means. It does not mean two observers in relative motion use different "distance measures" along the same spacelike curve. It means they pick two different spacelike curves as their "distance measures" between two worldlines (for example, the worldlines of the two ends of a ruler which is at rest relative to one of them, but moving relative to the other).

One of the reasons I like spacetime diagrams as a tool for problems like this is that it puts things like the above right in front of you in the diagram.

 

[jartsa]

As you approach light speed in any direction, does the Hubble horizon draw closer in that direction due to length contraction? Or does time dilation increase your measure of the Hubble constant? I'm imagining zipping through galaxies in milliseconds while their inhabitants watch me come and go for eons. Where is my Hubble horizon? What is the shape of my Hubble sphere?

Well, all the galaxies in our visible universe form a 10 cm thick layer, if you move at suitable velocity.

By suitable velocity I mean a very high velocity relative to the galaxies in our visible universe. Or the center of mass of those galaxies.

 

[PeterDonis]

all the galaxies in our visible universe form a 10 cm thick layer, if you move at suitable velocity.

Please specify the coordinates on which you are basing this claim. Remember that the spacetime of our universe is not flat, so "the inertial frame in which you are at rest" is not a correct answer, since there are no global inertial frames in curved spacetime.

 

[jbriggs444]

@jbriggs444 But aren't they impacted by different gravitational influences?

Gravitational influences? Gravity is not a force. We have two inertial trajectories. They start at the same event. They end at the same event. They show different elapsed times. That is the point.

 

[jartsa]

Please specify the coordinates on which you are basing this claim. Remember that the spacetime of our universe is not flat, so "the inertial frame in which you are at rest" is not a correct answer, since there are no global inertial frames in curved spacetime.

I don't know almost anything about coordinates.

But, I was thinking about the observer seeing galaxies being one millimeter thick, the voids between galaxies being 999 millimeters wide, and the path through the visible universe consisting of of 1000 of those voids, so the thickness of the visible universe would then be somehow equal to 1000 m.

I said the thickness was 10 cm. But I did not say what 10 cm means. But the observer would probably say that the visible universe is not very thick.

 

[Ibix]

But, I was thinking about the observer seeing galaxies being one millimeter thick, the voids between galaxies being 999 millimeters wide, and the path through the visible universe consisting of of 1000 of those voids, so the thickness of the visible universe would then be somehow equal to 1000 m.

But that's not what the observer sees. That's how he interprets what he sees on the assumption he's in flat spacetime and there's an obvious way to interpret things. But on cosmological scales that assumption is not valid, so he has to decide how he's going to interpret what he sees - that means picking a foliation (i.e. a notion of what 'space, now' means) even if not a complete set of coordinates.

Ultimately that's the problem with the question in this thread. The answer depends on the choice of what you mean by "now". (Edit: or equivalently, what you mean by "space".)

 

[pervect]

I don't know almost anything about coordinates.

But, I was thinking about the observer seeing galaxies being one millimeter thick, the voids between galaxies being 999 millimeters wide, and the path through the visible universe consisting of of 1000 of those voids, so the thickness of the visible universe would then be somehow equal to 1000 m.

I said the thickness was 10 cm. But I did not say what 10 cm means. But the observer would probably say that the visible universe is not very thick.

You are probably analyzing this using SR and not GR, then. While what you say would be true in SR with the Lorentz transform ,howverGR is not SR and does not have a Lorentz transform. So you'd need to provide a reference for your statement, and/or valid calculation to defend your statement.

 

[Chris Miller]

@jartsa: That was kind of my take on the observable universe (particle horizon), but not the whole universe, only the forward part, and its being convex (bowl shaped) with the thinnest part most directly ahead, and thickening out to the sides. But from others' comments I can see how curved spacetime (and I still maintain, Hubble expansion) over these distances leads to ambiguity. I'm just now trying to think about whether time dilation and length foreshortening brings forward objects/events closer also in time...

 

[Ibix]

I'm just now trying to think about whether time dilation and length foreshortening brings forward objects/events closer also in time...

Time dilation is a coordinate effect in flat spacetime, so also doesn't apply to curved spacetime. Again, you need to specify a foliation before you can interpret your measurements, and what effects similar to time dilation that you might get (or not) depend on that choice. Not sure what you mean by "length foreshortening", but if that's just length contraction then the same comment applies.

 

[Chris Miller]

@PeterDonis Thank you for sharing your understandings. I now understand better the problem of asking about the shape/appearance/presentation of the universe at near c.

You replied to @jbriggs444

No. One would expect exactly the opposite.

in regard to distance contraction to the Hubble horizon. You're not saying it could be expanded? Only unchanged?

I just downloaded the paper you recommend, will see how much is accessible to me. The discussions here are most interesting. My personal objective is maximize credibility in my SF with minimal sacrifice of understandability. Though balancing act. You have been a great help (which I understand might not come across as the compliment it's intended as).

 

[Chris Miller]

Time dilation is a coordinate effect in flat spacetime, so also doesn't apply to curved spacetime. Again, you need to specify a foliation before you can interpret your measurements, and what effects similar to time dilation that you might get (or not) depend on that choice. Not sure what you mean by "length foreshortening", but if that's just length contraction then the same comment applies.

Thanks for the concise answer. Had to look up "foliation." Seems almost synonymous with "unit of measurement" but not quite. I'm getting the feeling from your and other remarks here that time dilation and length contraction only apply in shorter distances where spacetime curvature is negligible (i.e. flat). This gives rise to all sorts of other questions for me. Like where is the boundary? Is it a cut-off or does the effect wane gradually? I know it's asking a lot, but I'd really like to see someone "describe" their frame of reference's universe at a velocity of near c relative to the CMB. I know this is not possible without the language of math. So something poetic might have to suffice.

 

[PeterDonis]

Had to look up "foliation." Seems almost synonymous with "unit of measurement"

No, it isn't. A foliation of spacetime is a "slicing" of spacetime into an infinite set of spacelike slices (3-dimensional spacelike hypersurfaces), which are ordered by time.

I'm getting the feeling from your and other remarks here that time dilation and length contraction only apply in shorter distances where spacetime curvature is negligible (i.e. flat).

The concepts of "time dilation" and "length contraction" as you are intuitively using them only make sense if spacetime is flat. They are limited concepts, and if you want to understand what happens in a curved spacetime you need to discard them since they won't work. That doesn't mean that "time" and "length" are fixed in a curved spacetime; it means that the intuitive definitions you are using for "time" and "length", namely, time and length in some inertial frame, can't even be used in a curved spacetime since there are no global inertial frames in curved spacetime. That means that "time dilation" and "length contraction" lose the only connection they had with something with physical meaning; all you have left in curved spacetime is coordinate labels that have no physical meaning. So the best thing to do is to forget about them and focus on things that do have physical meaning.

I'd really like to see someone "describe" their frame of reference's universe at a velocity of near c relative to the CMB

There is no such "frame of reference" in the sense you mean. The concept of "frame of reference" as you are familiar with it in special relativity simply does not work in a curved spacetime.

I know this is not possible without the language of math.

It's not even possible with the language of math. The only thing the language of math will give you is a global coordinate chart, and a global coordinate chart in a curved spacetime simply fails to satisfy the properties of a "frame of reference" that you are looking for. So the thing you are asking for does not exist.

 

[Ibix]

Had to look up "foliation."

It's a technical term for slicing. Take a cheese slicer and slice a block of cheese - that's a foliation (well, if the slices were infinitely thin) of the block of cheese.

I'm getting the feeling from your and other remarks here that time dilation and length contraction only apply in shorter distances where spacetime curvature is negligible (i.e. flat).

It only applies between inertial frames on flat spacetime. As long as you can approximate the spacetime as flat, you can use global inertial frames on it.

Like where is the boundary? Is it a cut-off or does the effect wane gradually?

My kitchen floor is flat and I can tile the floor easily, but the Earth is round and I can't build a square grid over all of it. Where's the cutoff? There isn't an answer to that. In principle, you could detect the curvature of your kitchen floor - it's just that it doesn't matter when you are tiling. If you are mapping a city, though, you might need to care. Or you might not if you're just producing a street atlas. It depends how precise you can be and how precise you need to be.

I know it's asking a lot, but I'd really like to see someone "describe" their frame of reference's universe at a velocity of near c relative to the CMB.

The problem is that there isn't a single answer to this. You can certainly describe your direct observations (which will differ from those of a co-moving observer), but "how far away is the CMB/particle horizon/whatever" isn't a direct observation. It's an interpretation of your observations, and the answer you get depends on your choices. The simplest choice, in many ways, is to adopt the co-moving coordinate system that we always use and interpret everything that way.

 

[Chris Miller]

@Ibix I like your kitchen floor analogy. It sounds like you're saying relativistic effects wane over distance but never completely disappear, and so your "simplest choice" is always to some degree valid. @PeterDonis seems to say not, that there are distances over which you cannot at all approximate spacetime as flat? On a sphere I can measure the distance between any opposite poles as its diameter or half its circumference. Is this the choice you refer to? What would the analogous spacetime measurement choice be?

 

[PeterDonis]

It sounds like you're saying relativistic effects wane over distance but never completely disappear

No, that's not at all what he's saying. What he's saying is that "relativistic effects" in a spacetime that is not flat cannot be described by the terms like "time dilation" and "length contraction" that you are used to from special relativity.

@PeterDonis seems to say not, that there are distances over which you cannot at all approximate spacetime as flat?

I have said no such thing. But the distances over which we can approximate spacetime as flat are miniscule in cosmological terms.

On a sphere I can measure the distance between any opposite poles as its diameter or half its circumference. Is this the choice you refer to?

The analogue of the diameter of the sphere does not even exist for spacetime; spacetime is not embedded in any larger space.

 

[Ibix]

@Ibix I like your kitchen floor analogy. It sounds like you're saying relativistic effects wane over distance but never completely disappear, and so your "simplest choice" is always to some degree valid.

Not really. The point is that curvature is always present, and you are always (strictly speaking) wrong to use special relativity at any scale, just as you are (strictly speaking) wrong to use Euclidean geometry to plan the tiling of your kitchen. But the errors from doing so are far, far, smaller than anything you could hope to measure. Your inaccuracy in laying the tiles is much, much larger.

But if you keep on tiling a larger and larger area you will, eventually, find that you can't continue to lay the tiles in a square grid because a square grid won't fit on the surface of a sphere. And that's analogous to what happens if you continue trying to use SR concepts on larger and larger regions of spacetime. The square (well, hypercubical) grid of an inertial reference frame won't fit on the curved spacetime, and the errors from just squashing it a bit eventually become noticeable.

On a sphere I can measure the distance between any opposite poles as its diameter or half its circumference.

In this analogy, there is no interior or exterior - there's only the surface of the Earth. Everything else is a higher dimensional space in which the Earth's surface is embedded. There's no evidence of a higher dimensional space in which spacetime is embedded.

Is this the choice you refer to?

No. I was just saying that it would make perfect sense to adopt the usual co-moving coordinates and describe the universe in the same way as it is in every cosmology text. It's kind of like driving a car into a tree. It's perfectly physically valid to say the tree hit you at 30mph and accelerated your car rapidly from rest to 30mph. That's a description of what happens in the inertial frame where your car was initially stationary. But most people, including you, would normally adopt the Earth's surface rest frame and say you hit the tree doing 30mph and came to a rapid stop. Similarly, on cosmological scales, it makes sense to adopt the usual cosmological coordinates even if you then end up not describing yourself as "at rest".

 

[Chris Miller]

@PeterDonis

I have said no such thing. But the distances over which we can approximate spacetime as flat are miniscule in cosmological terms.

Isn't that the same as saying, "there are distances over which you cannot at all approximate spacetime as flat"?

Isn't the universe considered to be infinite, and so any finite distance would be infinitesimal in cosmological terms? How far, in your understanding, in light-years would the distance over which you meaningfully can be?

 

[PeterDonis]

Isn't that the same as saying, "there are distances over which you cannot at all approximate spacetime as flat"?

Sorry, I misread your earlier post. Yes, your statement here is correct.

Isn't the universe considered to be infinite

Spatially infinite, according to our best current model, yes.

so any finite distance would be infinitesimal in cosmological terms?

No. To approximate spacetime as flat, you must limit your analysis to a small enough patch of spacetime that the effects of spacetime curvature (not space curvature) are not observable. For the universe, "spacetime curvature" means the expansion of the universe, so to approximate the spacetime of the universe as flat, you must limit your analysis to a small enough patch of that spacetime that any effects associated with the expansion of the universe are not observable. But that assumes you are only looking at large scale effects to begin with. See below.

How far, in your understanding, in light-years would the distance over which you meaningfully can be?

It depends on what you are modeling. If you are modeling the orbits of satellites around the Earth, a patch of spacetime large enough to include the Earth cannot be approximated as flat. If you are modeling the orbits of planets in the solar system, a patch of spacetime large enough to include the solar system cannot be approximated as flat; but a patch that just covers the Earth (for a short enough period of time--remember we are talking about a patch of spacetime, not just space) might be if you ignore all effects of the Earth's gravity and just approximate the Earth as a point particle. And so on up the scale.

 

[Chris Miller]

@PeterDonis Again, very much appreciate your explaining in ways I believe I can (sort of) understand. Particularly,

For the universe, "spacetime curvature" means the expansion of the universe...

since my thread question here concerned the effect of Hubble expansion on near-c time dilation and length contraction. I see now that, as I sort of suspected, they're incompatible. I thought they'd eventually cancel each other out, but it seems more complicated than that.

@Ibix I've been thinking about your tile analogy:

But if you keep on tiling a larger and larger area you will, eventually, find that you can't continue to lay the tiles in a square grid because a square grid won't fit on the surface of a sphere. And that's analogous to what happens if you continue trying to use SR concepts on larger and larger regions of spacetime.

I could tile the Earth (some equivalent perfect sphere) no problem. Every tile 2x2' tile would fit flat and snugly with those around it, seemingly level. The minuscule errors would not be apparent or measurable in any local region. Analogizing to to spacetime's "hypercubal" grid, I imagine traveling at very near c from star to star, perhaps only seconds (by my clock) separating, and think, where am I in a year? But then, if I understand you and PeterDonis, their proper distances are affected by Hubble expansion (spacetime curvature) which messes with the placement of my tiles (Minkowski coordinates).

Given the seconds in a year (31,536), light-years in a megaparsec (3,261,564), kilometers in a megaparsec (30,856,775,714,409,000,000) and Hubble expansion (73 km/sec) , I calculate that in the one second it takes me (by my clock) to cross a megaparsec, it will have expanded by 1/4109 (0.024 percent) which seems still pretty flat and wouldn't seem to have had that much bearing on my trip? At c, am I staying equidistant from my original, proper, (at rest relative to CMB) Hubble horizon?

 

[jbriggs444]

I could tile the Earth (some equivalent perfect sphere) no problem. Every tile 2x2' tile would fit flat and snugly with those around it, seemingly level. The minuscule errors would not be apparent or measurable in any local region.

You can tile any small region to a decent fit, yes. You cannot tile the whole thing with one square tesselation.

Try it.

Start the layout of your square grid with the 2x2' resolution at the north pole. You lay a strip of some 30 million tiles south along the prime meridian until you hit the equator. You lay another strip of some 30 million tiles south at 90 degrees to the first. You lay another strip of some 30 million tiles along the equator from the first strip to the second. Those can all mesh cleanly together if you are willing to adjust your 2 foot tile size by a micron or two.

But now you have a triangular region which you have to tile in a square pattern. That will be an interesting challenge. [Actually, you have no freedom here. Once you place the first tile, the rest of the pattern is fixed. And contradictory].

 

[PeterDonis]

in the one second it takes me (by my clock) to cross a megaparsec, it will have expanded by 1/4109 (0.024 percent)

More precisely, if we assume that the two objects you are traveling between are both moving with the Hubble flow (i.e., they are not gravitationally bound to each other), it will take you, not one second, but slightly longer--1 plus 0.024 percent seconds. Actually, it will be an average value of the amount of expansion over the length of your trip; if we assume that the expansion is linear to a good enough approximation during your trip, the elapsed time by your clock would be 1 plus 0.012 percent seconds.

However, you were not talking about this in your OP, but about how your Hubble sphere would "appear" to you if you were in this state of motion (one second by your clock to travel 1 megaparsec according to comoving observers--observers moving with the Hubble flow). Those are different questions. The elapsed time question answered above can be answered by just looking at the patch of spacetime that contains your worldline for the one second that elapses by your clock. The question about how the Hubble sphere "appears" to you cannot; to answer that question, you need to look at a much larger patch of spacetime, that includes your entire Hubble sphere for the time it took light to reach you from objects that are on your Hubble sphere "now". The latter patch is nowhere near flat even if the former patch can be approximated reasonably well as flat over the time period you used.

 

[PeterDonis]

The minuscule errors would not be apparent or measurable in any local region.

But the errors build up; they don't cancel out. So your tiling would stop working outside the local region in which you started. And if you started tilings in two different local regions, they would not match up when they met up with each other.

 

[Ibix]

I could tile the Earth (some equivalent perfect sphere) no problem

Not on a square grid, though. And that's the point. You can start a grid anywhere and cover part of the sphere, but you can't extend anyone grid to cover the whole of the sphere.

Consider your two inch tiles. Make a row of them circling the equator. Now make another row next to it, touching on the north side of the first one. That can't quite work because this row is slightly shorter - at latitude $\theta$ the distance around the world (a sphere of radius $R_E$) is $2\pi R_E\cos\theta$. For the first row off the equator $\theta\approx 10^{-8}$ radians and you can easily absorb the error - it's about a nanometer. But by the time you get to 60° north, there is only half the distance around the world that there is at the equator. So you've only got space for half the tiles and you can't possibly line them up with the equatorial ones.

 

[pervect]

I've been thinking about your tile analogy: I could tile the Earth (some equivalent perfect sphere) no problem. Every tile 2x2' tile would fit flat and snugly with those around it, seemingly level. The minuscule errors would not be apparent or measurable in any local region.

I don't think you can. Here's an equivalent problem. Take a single flat piece of graph paper, that isn't stretchy. Real paper will be a bit stretchy - if you tug on the edges of the paper, something drawn on it will distort a tiny bit as the fibers stretch - but it's not very pronounced. That's what I mean by stretchy. A certain amount of stretchiness is necessary, though, for the paper to bend at all if it has finite thickness, by the way. If the fibers making up the paper weren't a little stretchy, they'd break

Now, try and wrap this flat sheet of graph paper around a ball. You'll find that you always have excess material.

This is easy enough to try if you're skeptical. You don't need graph paper - some reasonably rigid (in the above sense) wrapping paper will do. In this version, it becomes "gift wrapping a baseball".

 

[PeterDonis]

Now, try and wrap this flat sheet of graph paper around a ball. You'll find that you always have excess material.

A good contrasting exercise is to wrap the flat sheet of graph paper around a cylinder--which can be done without any excess material. This is a good illustration of what "flat" means in geometry (and of the difference between extrinsic and intrinsic curvature).

 

[pervect]

Spherical triangles gives a slightly more rigorous illustration of the difficulties of tiling the sphere, with the concept of "excess angle". It's a bit different from the issue of tiling the sphere with squared, but related.

The key result is that the sum of the angles of a spherical triangle is always greater than 180 degrees. See for instance https://en.wikipedia.org/wiki/Spherical_geometry#Relation_to_Euclid's_postulates

A statement that is equivalent to the parallel postulate is that there exists a triangle whose angles add up to 180°. Since spherical geometry violates the parallel postulate, there exists no such triangle on the surface of a sphere. The sum of the angles of a triangle on a sphere is 180°(1 + 4f), where f is the fraction of the sphere's surface that is enclosed by the triangle. For any positive value of f, this exceeds 180°.

The angular excess is proportional to the area of the triangle, so halving the length of a side causes the excess angle to drop by a factor of 4.

In the limit, one has 360 degrees in a circle at any point, and greater than 360 degrees for the sum of six finite sized equilateral spherical triangles. So they simply can't fit.

Angular excess can be directly related to the concept of the Riemann curvature in two dimensions. For higher dimensions, one needs something more sophisticated. Riemann curvature reduces to a single number in two dimensions, but not for higher dimensions.

 

[Chris Miller]

Thank you all. You've made it abundantly clear, and I understand, one cannot perfectly tile a perfect sphere with perfectly square tiles of any size. And not just squares, but any polygon? Or collection of polygons? But if locally immeasurable curvatures and distortions were permitted (say in the grouting), then?

@PeterDonis

However, you were not talking about this in your OP, but about how your Hubble sphere would "appear" to you if you were in this state of motion (one second by your clock to travel 1 megaparsec according to comoving observers--observers moving with the Hubble flow).

Yes, that's the question. Say, for "simplicity," even a nanosecond (or Planck unit) instead of a second, i.e., c in the sense that .9999... = 1.

The elapsed time question answered above can be answered by just looking at the patch of spacetime that contains your worldline for the one second that elapses by your clock. The question about how the Hubble sphere "appears" to you cannot; to answer that question, you need to look at a much larger patch of spacetime, that includes your entire Hubble sphere for the time it took light to reach you from objects that are on your Hubble sphere "now". The latter patch is nowhere near flat even if the former patch can be approximated reasonably well as flat over the time period you used.

So ahead, does one "now" remain equidistant from the "former" Hubble sphere? Had the Hubble sphere "now" formerly been receding at 2c? How about behind? Is there "now" only empty space? Isn't time dilation still a thing? Do the countless megaparsec-separated (by their measure) observers you pass (almost simultaneously by your clock) watch you come and go for cumulative quadrillions of years by their clocks?

 

[jbriggs444]

one cannot perfectly tile a perfect sphere with perfectly square tiles of any size. And not just squares, but any polygon? Or collection of polygons?

Polygons are probably not the right idea to be chasing here. One can tile a sphere with triangles -- in a way we call a tetrahedron [or an octahedron or an icosahedron]. One can tile it with squares -- in a way we call a cube. One can tile it with pentagons -- in a way we call a a dodecahedron.

But these approaches do not give you a tiling that translates smoothly between a flat piece of paper and the surface of a sphere. A square tiling on a piece of paper would typically have four squares meeting at each vertex. On a cube there are just three.

There is an intrinsic difference between a flat piece of paper and a sphere. That difference is intrinsic curvature.

You might find it both entertaining and educational to read Sphereland.

 

[Ibix]

icosahedron

And you can further subdivide the icosahedron's triangles, which is an excellent way of building "spheres" in 3d modelling. But the triangles are typically different sizes and some of those triangles meet 6 at a corner and some 5 at a corner.

 

[PeterDonis]

does one "now" remain equidistant from the "former" Hubble sphere?

There is no well-defined answer to this question, or to all the other questions you are asking, because there is no well-defined "now" for you. It depends on how you choose your coordinates. If you are a "comoving" observer, there is a "natural" choice of coordinates, which is just the standard FRW coordinates that cosmologists use. But if you are not a comoving observer, and you can't be if you are traveling between galaxies, then there is no "natural" choice of coordinates for you.

 

[PeterDonis]

Do the countless megaparsec-separated (by their measure) observers you pass (almost simultaneously by your clock) watch you come and go for cumulative quadrillions of years by their clocks?

This question, at least, can be answered since it is about direct observables. The answer is yes (assuming you do "come and go", i.e., change direction to pass by the same megaparsec-separated observers multiple times).

 

[Chris Miller]

All your replies to my questions and (misguided) conjectures are informative, and make me wish I knew more, understood more.

@jbriggs444

One can tile a sphere with triangles ...

Doh! Of course! Why didn't I see that? Six equilateral triangles with apexes at either pole meeting at the equator. Not a sphere, but the more you subdivide them into smaller equilateral triangles the closer you can approximate (adhere to) a sphere. So maybe there is a way to tile the hypercube with small regions of locally flat spacetime?

@PeterDonis

This question, at least, can be answered since it is about direct observables. The answer is yes (assuming you do "come and go", i.e., change direction to pass by the same megaparsec-separated observers multiple times).

I appreciate the clarification, but don't fully get why I must bounce back and forth as if between two mirrors to observe for myself a quadrillions of years old universe. I've also given some thought to what exactly "no well defined answer" means, especially since I intend to plagiarize/paraphrase it in my SF. I know this is wrong, but it gives off an undecidedness vibe that I, for some reason, associate with QM.