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Sinking instead of expanding universe?

  1. Oct 16, 2014 #1
    We all know the concept of the universe expanding. Would it be possible that the universe is not expanding at all, but the spacetime between objects is increasing? My question comes from the idea that spacetime is changed due to the presence of gravitational objects and the idea that gravity is pretty weak, at least weaker than one might expect it to be. What if the gravitational forces are actually a lot bigger, but most of the gravitational force is being converted into the actual stretching of spacetime. From the point of view of an observer this would result in the universe seemingly expanding.
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  3. Oct 16, 2014 #2


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    What you are describing IS the standard understanding of the universe expanding.
    Yes, the spacetime between objects is indeed what is expanding. Thus, objects will naturally trend to drift apart, and the farther away they are from one another, the greater the drift speed, and no matter where you will see them drifting away from you in all directions just as if you were at the center of an expansion. Objects like the cat on the table in front of me, the earth, the solar system, galaxies and even galactic clusters don't drift apart because the forces holding their component parts together are strong enough to pull them together more rapidly than the drift (which is negligible except at cosmological distances) can separate them.

    There's a pretty good sticky thread at the top of the cosmology forum... you might want to take a look at it.
  4. Oct 17, 2014 #3
    As far as I understand it is supposedly the dark matter, which still cannot be proven, what makes the universe expand. My vision would be that gravity itself would be creating sinkholes in the fabric of spacetime. The distance between objects would become greater by stretching the fabric of spacetime itself, not so much by objects pushing each other away.

    Is this indeed what the current accepted standard is for the universe expanding?

    I would not consider this expansion per se, but I would call this stretching of spacetime. In essence, all objects in the galaxy would remain in the same place, but the spacetime would just change shape around these objects.

    What would the role of dark matter still be? As far as I understand, the only reason for dark matter to exist on paper, is to explain the mismatch in gravity present in the galaxy and to explain the expansion itself. If part of the gravitational forces are basically used up by stretching spacetime, would that not disqualify the existence of dark matter?

    Forgive me for nagging, but this issue is nagging me. I hope I am asking good questions and I am not boring or annoying anyone too much with this...
  5. Oct 17, 2014 #4
    By "using up" the gravitational force, I mean that we would not be able to measure part of the effects of gravity simply because we cannot measure the effect on spacetime in all dimensions.
  6. Oct 17, 2014 #5


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    I remind everyone posting in this thread to stay on topic and to refrain from personal speculation or this thread will have to be locked permanently.

    I do not know what you are referring to when you say sinkholes in the fabric of spacetime. Perhaps you are referring to the analogy with a rubber sheet being stretched out? What you need to understand about such analogies is that they are just that, analogies, which offer a popularized image of how things are working. It does not correspond to the actual theory.

    Dark matter is not particularly related to the expansion of spacetime, but rather to how large objects affect each other gravitationally. The current explanation for the accelerated expansion of the universe is dark energy. However, the universe could still be expanding without it, just not with an accelerated expansion.

    The expansion is the stretching of spacetime without it necessarily changing shape. Even if you do not consider this expansion, this is the nomenclature that is used within the scientific community. This expansion is not due to objects pushing each other apart.

    See above, dark energy is used to explain the acceleration of the expansion. Expansion in itself is possible without both dark matter and dark energy.
  7. Oct 17, 2014 #6


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    I've had some trouble reconciling the idea of "expanding space" with "objects moving apart in space". How do you distinguish these views? I read a published article about this (unfortunately I have to hunt for the link) which suggested that "expanding space" can be a misleading model.

    In the "expanding space" interpretation, it appears that the solar system and galaxies are held together against some "force of expansion of space". Ignoring Dark Energy here, in the current ongoing expansion, galaxies are simply drifting apart inertially, except that the gravitational potential increases as the galaxies separate slowing the expansion. If you take have two particles that are not moving relative to one another, the expansion will not cause them to move apart.

    I'm not sure, but I think in a curved universe (one with Dark Energy), the expansion may be claimed to cause particles with no initial relative motion to drift apart, but in a flat universe, this does not happen. Galaxies just coasts apart and there is no need to interpret this as caused by some "expansion of space".

    Dark Energy (acceleration of expansion) may cause acceleration of particles. In the absence of Dark Energy things move apart because they have always moved apart since the Big Bang. Everything is just coasting apart.
  8. Oct 17, 2014 #7


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    Yeah, I just can't see space as "something" that bends / stretches / expands / etc. I see posts here that espouse that point of view but I think it's just a popularization / simplification that doesn't really describe metric expansion, in which things are moving apart withing a framework (space) that is just a measure of distance. Not everyone here agrees w/ me on that.
  9. Oct 17, 2014 #8


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    Ignoring dark energy, this is true--but note that the expansion will not leave them at relative rest either. If the universe contained only ordinary matter, the two particles that started out at relative rest would begin moving closer together, because the expansion would be decelerating (see below). The only way they would stay at relative rest is if the universe were completely empty.

    This is not quite right. First, it's important to distinguish between curvature of spacetime, which is invariant, and curvature of space, which depends on how you split up spacetime into space and time. The two are not the same: for example, in our current best-fit model of the universe, space is flat (more precisely, if you split up spacetime into space and time the way "comoving" observers--observers who see the universe as homogeneous and isotropic--would, space is flat); but spacetime is curved.

    What does "spacetime is curved" actually mean? It means that freely falling test objects that start out at rest with respect to each other, don't stay at rest with respect to each other. Both ordinary matter (like galaxies) and dark energy cause spacetime curvature; the difference is that ordinary matter causes freely falling test objects that start out at rest relative to each other to fall towards each other, while dark energy causes freely falling test objects that start out at rest relative to each other to "fall" away from each other.

    In the context of cosmology, this means that the effect of ordinary matter is to slow down (i.e., decelerate) the universe's expansion (where here "expansion" means the average motion of galaxies relative to each other), while the effect of dark energy is to accelerate the expansion. In other words, neither one will leave the expansion "the same", in the sense of the relative motion of galaxies not changing with time.

    Whether the average motion of galaxies relative to each other can be called "expansion of space" is, IMO, a question about words, not physics. We don't have any way of detecting "space" apart from the motion of objects, so saying the expansion is "expanding space" and saying it's the motion of galaxies relative to each other are, physically speaking, the same thing--they both make the same predictions for what we will observe.
  10. Oct 21, 2014 #9


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    I think I understand (at a layman level) what you are saying and I agree except for your last paragraph.

    The problem I see with "expanding space" is that it leads people to believe that space itself grows causing particles to move apart. So they go on to conclude that galaxies stay together because gravity counteracts the expansion of space. This concept further leads people to imagine that the "expansion of space" is pushing things apart.

    For example, using the concept of "expanding space" people may imagine that if two test particles are placed at rest wrt to one another, then the effect of gravity will be partially offset by "expanding space". They might also believe that you could measure the expansion of space in an experiment. E.g. they might conclude that two nearly massless particles placed at rest (at a sufficient distance from one another to minimize gravity) would spontaneously move apart due to the universal "expansion of space".

    In the case of a cosmological constant this view of "expanding space" may be correct, but otherwise it appears incorrect. I brought it up because it so common to hear this wrong idea that gravity counteracts expansion of space when in reality there is nothing about space itself that is counteracted by gravity. Gravity transforms the inertial motion of expansion into potential energy, orbits, collisions. The observed inertial motion of galaxies apart from one another in BBT was imparted by primordial expansion (again ignoring the existence of Dark Energy). At early times, the pressure of the dense energy that existed resulted in the moving apart of the particles that formed shortly after the BB.

    This idea of "expanding space" mislead my own understanding of BBT for some time. There is no momentum imparted to particles by the "expansion of space" (without Dark Energy). Therefore I feel that invoking "expanding space" (in the absence of a cosmological constant) is misleading, if not outright incorrect.

    If there is some equivalence of "expanding space" and "motion of galaxies apart" in the Universe as a whole in BBT, this equivalence cannot sensibly be extended to other cases of motion, such as the motions of molecules of gas in a container.

    As you know I'm no expert on GR so if this view is wrong, I trust you to correct it.
  11. Oct 21, 2014 #10


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    I agree, although I might rephrase it something like this: in the absence of a cosmological constant, there is no term describing "expansion of space" in the equations that govern the dynamics of objects.

    I think this is a good illustration of the limitations of using ordinary language to talk about physics, instead of math. The math is unambiguous, as my rephrasing just above suggests: if you look at the equations, there is no "expansion of space" in there (unless there is a nonzero cosmological constant).

    This is not really correct; it's another illustration of the limitations of ordinary language in discussions about physics. The concept of potential energy only makes sense in a stationary spacetime, and the universe is not a stationary spacetime; the math makes this evident, but it's easy to lose sight of in a discussion using ordinary language.
  12. Oct 21, 2014 #11


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    Can you explain in ordinary language what you mean by "a stationary spacetime". Stationary in what sense? And also why is the universe a non-stationary spacetime (I suppose because it's expanding)?

    When you refer to "a spacetime" are you referring to a particular coordinate system with 3 spatial and one temporal dimension? Is there something special about such a coordinate system to qualify as a spacetime?

    People talk about "curved spacetime". I'll conjecture that "a curved spacetime" can be illustrated as such a 4D coordinate system superimposed on a Euclidean spacetime diagram. We say it is curved because the spatial and/or temporal grid of the system is curved relative to the Euclidean coordinate system. I have a feeling that you are going to tell me this is a wrong conception, at which point I'll feel like George Clooney in "Gravity" - nothing left to hold on to.

    Having only a rudimentary education in classical physics, I'm not sure what has become of concepts like potential energy in GR. Classical concepts must sometimes fail, but on the other hand does GR deny that the KE of expansion is converted into gravitational potential energy (whatever that is) over time?

    Also I've been told that conservation of energy/matter no longer holds in GR. That's a rather disturbing idea because of the implication that energy/matter can just appear or disappear. Maybe GR implies that perpetual energy generators are possible? (Infinite Energy Magazine would like that;)).

    On the other hand, the concept of total energy, e.g. in a system of particles, has no particular meaning even in SR nor in classical mechanics, since the total observed energy depends on the motion of the observer. The only way I can think of to have a concept of total energy is assume an absolute space in which to uniquely measure it.
  13. Oct 21, 2014 #12

    George Jones

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    You should take George Harrson's and Ringo Starr's to heart,

    i.e., in order not to appear foolish, you have to spend time in David Clayton-Thomas's band.
  14. Oct 21, 2014 #13
    Question is silly. Spacetime itself is a part of the universe.
  15. Oct 21, 2014 #14


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    First let me give a somewhat technical definition; then I'll try to unpack it.

    The technical definition is that, in a stationary spacetime, you can find at least some worldlines along which the geometry of spacetime is not changing. This doesn't mean the geometry has to be unchanging along every worldline; but there must be some family of worldlines that meets the requirement. In a non-stationary spacetime, it is not possible to find any worldlines along which the geometry is not changing.

    Now for some unpacking. The most important thing is to be able to think of spacetime as a geometric object, independently of any coordinates we might use to describe it; so, for example, when you ask...

    ...you are mixing up two things that it's important to keep separate. Spacetime is a geometric object with 3 spatial dimensions and one temporal dimension; but the same spacetime, the same geometric object, can be described by many different coordinate systems, just as the surface of the Earth, the same geometric object, can be described by many different coordinate systems. The geometry is described by invariants--things that are the same regardless of what coordinates you choose. For example, the radius of curvature of a 2-sphere (like the Earth's surface--or at least an idealized "Earth's surface" that is a perfect sphere) is such an invariant; it's the same regardless of what coordinates you use to describe the surface.

    So when we say that, in a stationary spacetime, there are at least some worldlines along which the geometry is unchanged, what we mean is that, along such a worldline, all of the invariants that describe the geometry are unchanged. Since a worldline is just a curve in spacetime that describes a possible trajectory for a piece of ordinary matter, like you or me, that means that if you or I were to travel along that worldline, we could repeatedly measure, over time, all the invariants describing the geometry of spacetime--the spacetime equivalents of things like the radius of curvature of a 2-sphere--and find them unchanged.

    In our universe, which is expanding, there is no worldline that has this property--there is no trajectory that we, or anyone, can travel through spacetime that will allow us to repeatedly measure any invariants describing the geometry and find them unchanged. They will all change as time passes, along every worldline.

    Yes. Meaning, the fact that the universe is "expanding" is one way of expressing the fact that it's impossible to find any worldlines in the universe along which the geometry is not changing.

    It's not wrong, but it's limited, because you're still implicitly relying on some underlying Euclidean space in which the curved grid is embedded. To really understand GR, you have to be able to let go of that and think of curvature as something intrinsic to a geometric object. The best way I know of to do that is to think about how you would detect whether a geometric object was curved, if you were restricted to only making measurements within the object.

    For example, suppose we live on an idealized, perfectly spherical Earth, and we want to prove that it is, in fact, a sphere, rather than a flat plane. But we are not allowed to "cheat" by using any phenomena external to the surface of the Earth (so, for example, we can't use Eratosthenes' method and measure the difference in angle of the noon Sun's rays at different locations). How could we do it? Here's one way: pick out two meridians of longitude that are nearby, and start at the point where they intersect the equator. Notice that they both intersect the equator at right angles: that is, they are parallel at the equator. Then follow them north, and notice that they don't stay parallel: they get closer and closer together, until at the North Pole, they intersect.

    What does this tell us? The key is that meridians, and the equator, are great circles, and great circles are the analogues, on a 2-sphere, of straight lines on a flat plane. The general term for curves that are the analogues of straight lines is "geodesics"; and if you actually look at the axioms of Euclidean geometry, you will see that the term "straight line" can be generalized to "geodesic", and the same axioms--with one exception--will be applicable to the intrinsic geometry of a curved surface, like a 2-sphere. The one exception is the parallel postulate, which (at least in one version) says that, on a flat plane, a pair of straight lines that are parallel anywhere (as shown by their both intersecting a third straight line at right angles) are parallel everywhere. On a curved surface, this no longer holds, as the meridians above illustrate.

    We can detect curvature of 4-D spacetime the same way; all we need is to find out what are the analogues of straight lines--the geodesics. It turns out that, at least in the timelike direction, these are the worldlines of freely falling objects--objects which are weightless, feeling no force. (Sometimes we say that are moving "solely due to the force of gravity", but saying it that way can be misleading; it's better, IMO, to focus on the fact that they feel no force and are weightless, since that's the direct observable.) So to see whether spacetime is flat or curved, we simply pick two nearby geodesics--two worldlines of nearby freely falling objects--that are parallel at some point--i.e., the objects are at rest relative to each other at some instant of time--and see whether they stay parallel--i.e., whether the objects stay at rest relative to each other. In flat, empty space, they do; but in curved spacetime, i.e., where gravitating bodies are present, they don't. For example, if at some instant two rocks are at rest high above the Earth, at slightly different altitudes, they won't stay at rest relative to each other; the one that's lower will fall slightly faster, so the rocks will move apart with time. That is spacetime curvature; and since what I've described is just tidal gravity, we can say that spacetime curvature is tidal gravity.

    As I said before, potential energy only makes sense in a stationary spacetime. The reason is that potential energy is energy that depends on an object's position; and if spacetime is not stationary, then there is no way to define "position" in a way that's invariant (i.e., that is independent of coordinates). In a stationary spacetime, you can use the worldlines along which the geometry is not changing to mark out "positions"; but in a non-stationary spacetime, there are no such worldlines.

    Yes, because the universe is not stationary. There are ways to interpret the dynamics of a closed universe as expansion converting kinetic energy to "potential energy", but that requires defining "potential energy" a different way, and IMO doesn't really add anything to understanding because it doesn't generalize.

    That's not quite right. Local conservation of energy holds just fine; the mathematical expression of this is that the covariant divergence of the stress-energy tensor (the mathematical object that describes "matter and energy" in GR) is zero. In ordinary language, this is saying that matter/energy can't be created or destroyed at any location in spacetime.

    The problem is trying to come up with a definition of "energy" that is global, not local. Again, this only really works in certain kinds of spacetimes, and the definitions don't generalize well.

    It most certainly does not. That would violate the local conservation of stress-energy that I described above.
  16. Oct 21, 2014 #15


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    Excellent answer Peter. I'll be thinking about it in detail. I once asked Dale if all manifolds (probably with reasonable restrictions) can be embedded in a higher dimensional euclidean space. If I understand what I asked correctly, he said yes. In GR can we embed any physical spacetime in 4D Euclidean space so that we can visualize what's going on? I realize such a representation is not unique, but I could pick the local inertial frame of the CMB for example. Perhaps I could drop one spatial dimension to get a picture of this and play with it in my mind. When you talk about non-Euclidean geometry without embedding it in a higher dimension, it's hard to have a overall visualization. It's like following directions to someplace without a map but trying to imagine what your path looks like on a map.

    A friend and I were discussing a notion of "flowing space" that apparently comes up sometimes in threads about GR to "explain" the gravity field of an object. We ruled out a simple version of flowing space in which space moves like vertical falling objects. This discussion left me with a challenge to come up with a spacetime diagram for a point gravitational mass in which all free falling objects move at constant velocity and in straight lines (along geodesics) since such a spacetime would serve as a "inertial frame" for the gravitation field of an object. Is this even possible?
  17. Oct 21, 2014 #16

    George Jones

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    Doing this isometrically can require more than 100 dimensions.
  18. Oct 21, 2014 #17


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    There is such a model (more precisely, it's an interpretation), called the "river model", which was invented as a way of describing the spacetime around a black hole in terms of space "flowing" inward like a river. You can read about it here:


    This interpretation can also be used for the vacuum region around any spherically symmetric gravitating body. It has some limitations, but it does give an easy way to visualize many phenomena in this particular kind of spacetime. Unfortunately, it's one of those things that doesn't generalize well beyond that limited set of spacetimes.

    Not globally, no. You can construct freely falling coordinates locally, centered on a particular event, in which all freely falling objects move at constant velocity and in straight lines, but those only cover a small patch of spacetime around that event, and there is no way to fit all the different local coordinate charts together into a single global one.

    It should be evident why this is if you stop to think about it. Consider two freely falling objects near Earth: one is falling radially inward above the North Pole, the other is falling radially inward above the South Pole. Globally, these objects are moving in opposite directions, and their speeds relative to Earth are not constant; so it's obviously impossible to construct a single inertial frame in which they both move in a straight line at a constant speed.
  19. Oct 22, 2014 #18


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    I don't mean to rain on anybody's parade, but, is there any compelling reason to work this out in more than 4 dimensions?
  20. Oct 22, 2014 #19


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    The river model works fine in this case (objects falling straight down move at constant velocity in the river) until you get to the center of the earth where you hit a singularity in the river where it abruptly changes direction. In contradiction, the local inertial space is smooth (locally flat) at the center. The river model barely works at all. If you consider a ballistic object launched upward from earth, the model immediately fails to give the object inertial motion, instead it decelerates. An object on the ground feels acceleration but moves at constant speed in the river. I thought, maybe what's missing here is the proper treatment of time?

    I was just guessing that there might actually be a spacetime in which motion in a gravitation field is globally inertial and that it is exactly this spacetime that describes gravity. In this proposed spacetime, all free-falling objects move without acceleration. That is, spacetime is distorted by gravity (relative to the Euclidean spacetime that exists in the absence of the gravitational field).

    So, I thought GR explained gravity in terms of the geometry of spacetime around a gravitating body. You seem to be saying that there is no global representation of this geometry in Euclidean spacetime. It's some sort of mysterious geometry that has no representation in a 4D space?

    I'm totally lost but that's not surprising because I don't have the mathematical background for Riemann geometry, tensors and whatever else is required to get GR.

    Ultimately I assume GR is described by equations that can relate spatial and time (x, y, z, t) coordinates in some way to a gravitating mass. If so, what coordinates are these? What possible meaning do coordinates have in the absence of a global coordinate system? Are these coordinate systems only defined on infinitesimal volumes? E.g. the equations only say something about (dx, dy, dz, dt).

    Are you suggesting that the only way to describe this GR geometry is to feel your way around it? E.g. along some surface, I walk two steps in orthogonal directions and then measure the the number of steps between two points reached and conclude that the surface is not flat and yet there is no way that I can embed this entire surface in a Euclidean space so that I can have a global picture of it? I cannot say that there is such a mapping?

    It's a concept that is hard to grasp, but perhaps that's why GR is so impenetrable to the uninitiated.
  21. Oct 22, 2014 #20


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    This is not correct. Remember that in this model, "inertial motion" does not mean moving at a constant speed relative to the "river bed". Relative to the river bed, the object decelerates, yes; but the river bed is not physical, it's just "scaffolding" in the model that is supposed to help with visualization.

    Think of it as the object having to "swim upstream" against the river in order to stay in the same place relative to the river bed. If the object stopped "swimming" (i.e., stopped accelerating and fell freely), the river would carry it downstream, meaning it would fall inward.

    There isn't; at least, not if "globally inertial" means there is a single global inertial frame. There isn't.

    That's true in real spacetime, but you have to define "acceleration" properly, as acceleration that is actually felt, rather than acceleration relative to some set of coordinates. Acceleration that is actually felt (which is in fact called "proper acceleration" if you want to be precise) is a direct observable and is an invariant; all observers will agree on it. Acceleration relative to coordinates ("coordinate acceleration" in GR terminology) is not--change the coordinates and you change the acceleration. That means, in GR, that coordinate acceleration is not a good thing to focus on; the things to focus on are the invariants.

    It does.

    Not in any 4-D Euclidean space, no. So what?

    No, it's a perfectly well-defined 4-D geometry (actually, there are many different possible geometries, which one you use depends on what particular situation you are trying to model) that is not Euclidean. Again, so what? There's no reason why all geometries have to be Euclidean, or representable in a 4-D Euclidean space. If you feel like they all should be, that's just a mistaken intuition that you are going to have to learn to let go of. There's no way around it.

    Basically, yes. The central equation in GR is the Einstein Field Equation, which looks like this:

    G_{\mu \nu} = 8 \pi T_{\mu \nu}

    ##G_{\mu \nu}## is the Einstein tensor, which describes spacetime curvature (actually just a portion of it, the portion that is directly linked to gravitating mass), and ##T_{\mu \nu}## is the stress-energy tensor, which describes gravitating mass (actually not just mass, but energy, momentum, pressure, and other stresses, all of which contribute to generating spacetime curvature). The Einstein tensor is composed of derivatives of the metric tensor ##g_{\mu \nu}##, which describes how coordinate intervals are related to actual physical measurements of space and time.

    Any coordinates you like. Coordinates by themselves have no physical meaning; all of the physical meaning is in the geometric invariants. The metric tensor, and the tensors derived from it, tell you how to compute invariants from coordinate intervals. So if all you know is coordinates, you don't know enough to know the physics. You have to also know the metric tensor and the tensors derived from it, as well as the stress-energy tensor and the invariants derived from it.

    I'll turn that question around: what possible meaning do coordinates have in the presence of a global coordinate system? Once again, if all you know is the coordinates, you don't know any of the actual physics. If you think you do, it's because you are implicitly assuming a metric tensor--because you are implicitly assuming a particular way of converting coordinate intervals to actual physical measurements of space and time.

    For example, in the simplest case of flat spacetime, in special relativity, we say that an interval of space or time ##ds^2## is given by the equation (called a "line element")

    ds^2 = - dt^2 + dx^2 + dy^2 + dz^2

    That means we have adopted a metric tensor ##\eta_{\mu \nu}## that is diagonal (considered as a 4 x 4 matrix), with elements (-1, 1, 1, 1). In other words, we have implicitly assumed that we have chosen coordinates such that ##t## directly measures time intervals, and ##x##, ##y##, and ##z## directly measure space intervals in three mutually perpendicular directions. But there's nothing that requires us to do this; we could adopt different coordinates and still describe the same flat spacetime. We would just use a different metric tensor ##g_{\mu \nu}## with different elements in the matrix, to convert coordinate intervals to space and time intervals.

    No, but as you can see from the line element I wrote down above, the strictly correct way of using the metric tensor uses infinitesimal coordinate intervals. To compute finite intervals of space and time, you have to integrate the line element along some curve, or over some surface. In many cases, that complication can be glossed over; for example, in flat spacetime with the standard coordinates I used above, we can express the interval along any straight line as ##s^2 = - t^2 + x^2 + y^2 + z^2##, without worrying about the infinitesimals. But notice that, even in those coordinates, if you want to compute the arc length along a curved line (such as the worldline of an accelerating object), you will have to actually do the integral.

    Ultimately, yes, that's the only option we have in our actual universe, because we have no way of stepping outside it. We have to make measurements entirely within spacetime, so if we are going to detect whether it's flat or curved, we have no other option but to do it by "feeling our way around it", as you say.

    Not in a 4-dimensional Euclidean space, no. There are ways to do it in a higher-dimensional Euclidean space, but how much higher depends on the particular spacetime geometry (as George Jones pointed out in an earlier post). I'm not sure why that's a problem, since even if you could embed it in a 4-dimensional Euclidean space, you can't visualize 4 dimensions anyway. Any visualization at all is going to have to leave out information.

    Not if you want to be correct. ;)
  22. Oct 23, 2014 #21


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    By "river bed", I think you mean the moving river itself?

    OK. Here's where I got confused. If you imagine a particle moving with the river passing by a fish who is swimming to be stationary, that particle is accelerating. Turning that around from the point of view of the river, the fish is accelerating. But the odd thing is the velocity of the fish in the river never changes which seems like a contradiction.

    OK. I was having a hard time trying to invent one. I thought maybe I was missing a possibility involving curvature in time.

    I understand proper acceleration, but I'm not sure how all observers can agree on it, except by reading an accelerometer attached to the accelerating body. For example if I am on the earth, I see a falling object accelerate, but it has no proper acceleration. So how can I observe that this acceleration is 0? It is only trivially observer independent when you observe the accelerometer that correctly measures it. In that sense you can say that everything about a body is observer independent. That is all "proper" measurements are done by observer attached to the body. For example the proper color of an object is fixed but others will see different colors in their own frame and not agree.

    So I cannot have a "picture" of this space, that's all. I'm not trying to claim anything.

    Yes. Well I would have to understand this idea of geometry better to see why this is so. For the simplest examples it's no problem. Suppose I measure a surface and find that it has constant curvature. In that case I can say, oh, I know something that has that property, it's a sphere. I know that a sphere has a continuous surface. I have a global picture of it's nature. In the "feel you way around" approach, it's hard to get a global picture. I'm not even sure how would you would prove, for example, that a surface with constant curvature is closed and continuous. You are not using global coordinates with which you can say, if I keep going in a straight line I will end up exactly where I started. It would be interesting to know how such things are proved in this approach to geometry.

    This is a bit over my head.

    I guess what you are saying is that these "metrics" are not considered coordinate systems? Rather they are "proper" measurements?

    Are you saying that

    ds^2 = - dt^2 + dx^2 + dy^2 + dz^2

    only applies in SR?

    I here what you are saying but I have trouble with the idea that no global 4D description exists. It's like saying there is a path to get from A to B and I can tell you how to move along it, but no map of this path can exist. It's just a little mysterious, but most likely some training in non-Euclidean geometry would dispel the mystery.

    Not if you want to be correct. ;)[/QUOTE]

    I can always drop one spatial dimension and try to understand what happens in a 2D slice of space.

    All I was looking for was a visualization of a simple case of a uniform spherical mass. But there is no possible mapping to 4D.

    I found that Minkowski space gives me a tool for visualizing SR. Of course in that case you start with a system of linear orthogonal coordinates for some IRF. You can visualize wordlines of some object. You can superimpose a coordinate system for another inertial reference frame and see how the same worldline looks in that frame. I believe you can even plot an accelerating frame, but apparently not a frame for a gravitating object.

    So, I was looking for a coordinate system (obviously not linear) superimposed in that 4D space that might give some insight to GR (for a very simple gravitating mass), but it is not to be, an entirely different perspective on geometry is needed.

    Riemann geometry is a prerequisite and I'm clueless about the basics except a little insight that Dale provided about measure distance and angles.

    Thanks for your efforts Peter, at least I understand what it is that I don't understand. :)
  23. Oct 23, 2014 #22


    Staff: Mentor

    No. I mean the underlying flat "space" that is used, in the model, as a reference to describe the motion of the river. Don't overthink this; the river/river bed distinction is just like you would expect based on the words being used.

    This is why the distinction between coordinate and proper acceleration is so important. You have just described how coordinate acceleration changes depending on how you choose coordinates. But proper acceleration is invariant; see below.

    Exactly. All observers must agree on the reading of that particular accelerometer attached to that particular accelerating body, regardless of their state of motion. In the fish vs. particle example you give, the fish's accelerometer will read nonzero, and the particle's accelerometer will read zero, and all observers will agree on this, regardless of the coordinates anyone chooses.

    Not quite. It only applies globally in SR; that is, only in globally flat spacetime (SR) can you find a single coordinate chart in which the line element looks like this everywhere. But you can do it locally in any spacetime, even a curved one, just as you can draw a flat map of a small local patch of the Earth's surface that accurately reflects distances using the ordinary Euclidean metric ##ds^2 = dx^2 + dy^2##, even though the Earth's surface is globally curved. (More precisely, the flat map will reflect distances using the ordinary Euclidean metric with some error; the smaller you need the error to be, the smaller the size of the local patch that the flat map can describe.)

    Correction: no 4-D flat map of this path can exist. But there are other ways to draw maps.

    Yes, as long as you don't mind leaving out some information. But you also have to include time. For example, a common visualization of a 2D slice of space around a spherically symmetric gravitating body is the Flamm paraboloid:


    This is an embedding of a 2D slice of space (at an instant of Schwarzschild coordinate time) in a 3D Euclidean space. But it raises at least as many questions as it answers; we have frequent threads here on PF where people are asking questions based on misconceptions caused by this visualization, and those misconceptions often arise from the fact that this visualization does not include time. So, while visualizations like this can be helpful, you have to be careful to recognize their limitations. (There are other visualizations that include one, or sometimes two, spatial dimensions and one time dimension, which can also be helpful. Unfortunately I don't have any handy links at the moment.)

    There are a number of different coordinate charts that are used for this case; I think the problem is not so much that there aren't any as that there are multiple ones that seem to say different things. The main ones I'm familiar with are Schwarzschild coordinates, isotropic coordinates, Painleve coordinates, Eddington-Finkelstein coordinates, and Kruskal-Szkeres coordinates. (The links below describe these coordinates in relation to a spherically symmetric black hole; but if you limit attention to values of the radial coordinate ##r## larger than some particular value that represents the surface of an ordinary gravitating body like a planet or star, these coordinates describe the empty space around that body just as well.)



    As a warmup exercise, you might want to try looking at descriptions of the Earth's surface in different charts: latitude/longitude, Mercator, stereographic projection, etc. Seeing how the same geometry can "look" very different in these different charts might help when you try the more complex case of spacetime.
  24. Oct 27, 2014 #23


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    Just a misunderstanding. When you used "inertial" I thought you were referring to something moving with the river.

    OK. Coordinates that accelerate have strange properties. In this coordinate system you can accelerate without gaining velocity.

    This measurement of proper acceleration has little to do with coordinates, except perhaps the local coordinates that might be needed to read an instrument like an accelerometer.

    In Newtonian physics, acceleration is a measurement of rate of change in velocity relative to some IRF. Does GR give us a new definition of acceleration? Maybe the definition is the same if you choose the right coordinates. Suppose I'm in an accelerating spaceship. I can release a mass (so that it moves inertially) and then measure it's acceleration over a very short distance (essentially in an instantaneous comoving IRF) to determine my proper acceleration.

    When you say "regardless of the coordinates anyone [any observer] chooses" it seems to imply that observers can choose some coordinates but will come up with the same value, but that's not so. In the case of the accelerometer reading, they are in a sense agreeing that the proper coordinates to use are local to object.

    This brings to mind a problem. Acceleration is a vector. While it's easy to measure a proper magnitude that can be agreed upon by all observers, how can we express a "proper direction"? With only magnitude the notion of proper acceleration doesn't seem useful.

    Velocity I suppose is not proper (in SR and GR) because there is no particular coordinate system in which we can consider it as a local property of a particle.

    OK. Even though the error vanishes locally, you cannot use it globally. (dx, dy) is measured in a local coordinate system, a tangential plane which changes as you move along the surface.

    Hmm... does that leave the door open for the representation of a gravitational field?

    In this visualization, the 3rd (vertical) dimension is artificial. If I understand correctly this is the projection of a flat polar system (a 2D IRF centered on the mass) onto a surface devised such that geodesic distances along the surface represent distances that we would measure in the gravitational field? The flat polar coordinates are projected onto the surface to show the difference between flat measurements and field measurements?

    Referring to another recent question in the SR & GR forum, the projection of the flat polar space onto this surface, shows that circumferential measurements are unaffected by the field, but radial distances are. Strong "curvature" is seen near the event horizon. Radial distances in the field are longer near the (the Schwarzschild radius or event horizon). The ratio of gravitational ##dr## to the flat ##dr## approaches infinity toward event horizon. Is this the ##r_s## singularity that bothered people? Nevertheless, both distances to the event horizon are finite.

    It's nice to know that even some really smart people had much trouble deciding what these coordinates say, e.g. about the apparent singularity at the event horizon.

    Can we take this funnel-shaped surface and map it, for example, onto a plane? In that mapping, it looks like a polar chart, except that the radial units get closer to one another as you approach ##r_s##. Although that mapping certainly works for radial and circumferential measurements, I'm unsure whether it works in general. On the plane, geodesics (a shortest path) are normally straight lines, but in this planar coordinate system they are not.

    I tried this planar mapping of the Flamm surface in order to work toward a coordinate system to which I could add time coordinates to illustrate the gravitational field in 3D (x, y, t). However, I couldn't figure out how to add time. Are these spatial coordinates in the gravitational field are constant? The measurements of space shouldn't change as time progresses. But time runs at different rates.

    You say that the Flamm embedding is "at an instance of Schwarzschild coordinate time". This time is read by a distance clock which one could think of as a clock in the absence of the field. Do the coordinates on the surface change at the next instant? Does it flow, like the river?

    It's interesting that Einstein formulated GR without even solving the differential equations for the simple case of point or spherical mass. Wow! How could he know his equations would work (e.g. that they approached the Newtonian theory).

    Each of the systems (shown in the second link) seem to use different types of coordinates. In the Schwarzschild case, we have polar coordinates. In the Isometric case we have (x, y, z, t). In some of the others I can't tell what the variables are. It says certain coordinate systems have transformations. Some don't?

    Here it's pretty obvious that mappings of 3D surfaces onto 2D are distorted and in this case have discontinuities. For example, in some maps of the world, Greenland is huge. We usually use longitude and latitude coordinates which I suppose are accurately mapped (excluding the poles which are singular) in this common "projection". In the projection (that I'm thinking of) lines of latitude are horizontal and lines of longitude are vertical and equally spaced by degrees. But measuring the great circle distance between two points seems difficult. I cannot find a coordinate system that allows this. I'm not sure that one is possible.

    GR is largely about geometry. To understand GR would take a lot of studying. It's not something that you can explain to someone in an hour, even if you really do understand it. Look how much more complicated it is than Newton's theory! It's ironic that this very simple theory of Newton's works so well. Perhaps nature likes to trick us?
  25. Oct 27, 2014 #24


    Staff: Mentor

    Actually, that wasn't quite a misunderstanding. The term "inertial" can be used to refer to something moving with the river. ;) Objects moving with the river are free-falling into the black hole from rest at infinity, so they are moving inertially. But of course there are many other possible states of inertial motion that do not just move with the river.

    Yes, and yes, that is weird if you look at it like that.

    I would say it is independent of coordinates. See further comments below.

    Not really a "new" definition, but a clearer treatment of the distinction between coordinate acceleration and proper acceleration. Proper acceleration can be modeled in Newtonian mechanics, but it is not emphasized because Newtonian mechanics doesn't make a clear distinction between direct, coordinate-free observables and coordinate-dependent quantities, the way GR does.

    No, they aren't. The accelerometer reading can be obtained, and talked about, without anyone ever adopting any coordinates at all. And different people using different coordinates can still compute a correct prediction of what a given accelerometer will read. This is true of any physical observable in GR, not just proper acceleration; a correct prediction of the observable's value can be computed in any coordinate system you like.

    Good question! The answer is that, when we say a given observer's proper acceleration vector points in a particular direction, we are really saying that the vector points in the same direction as something else that's observable. For example, say an observer is "hovering" above a black hole, and there is a star directly overhead with respect to him. Then the "direction" of his proper acceleration can be specified as being parallel to incoming light rays from the star. Or, the observer could carry a gyroscope with him, which he orients to point radially outward at some instant; then the direction of his proper acceleration can be specified as being parallel to the gyroscope (and the fact that both continue to point in the same direction is itself a direct physical observable, which can be predicted using GR in any coordinates you like).

    There is something called "proper velocity" that some sources use, but it's not really what we're talking about here. The key velocity for this discussion is 4-velocity, which is simply a unit vector that is tangent to an object's worldline. Proper acceleration is the derivative of 4-velocity with respect to proper time. Both 4-velocity and proper acceleration are indeed local properties of the object.

    I'll respond to the rest of your post separately because this is getting long and I'm pressed for time right now.
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