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Non-static spacetime effects

  1. Jan 21, 2015 #1
    First of all, I'm having a difficulty in defining what a static spacetime is. Does it presume that the objects with mass in the system are just sitting around and doing nothing, with no motion, relative to our frame, so there can be no motion and change in curvature of spacetime through time?

    Secondly, if objects with motion curve space time in a way that they change the curvature along their path, does this also imply that free-falling objecs also curve space time in the same way?

    Thirdly, in Newtonian physics there is an earth and apple example where Earth also accelerates towards the apple because of the same force with opposite directions. Does the same hold in GR?
     
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  3. Jan 21, 2015 #2

    PeterDonis

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    The technical definition is that a static spacetime has a timelike Killing vector field ("static" means the field is also hypersurface orthogonal, but I don't think we need to go into that part here). Now I'll translate that into English. ;)

    If a spacetime has a timelike Killing vector field, physically this means that there is some family of worldlines in this spacetime along each of which the metric doesn't change. We can therefore use these worldlines to define "points in space", such that each point in space has an unchanging spacetime geometry in its vicinity. In principle we can then have a family of observers in this spacetime each of whom always sees the same spacetime geometry (each observer follows one of the worldlines in the family), so we can think of each such observer as "standing still" at a particular point in space.

    Note that the spacetime geometry can still change if we move from one point in space to another. See below.

    You appear to be making an incorrect distinction between "objects with motion" and "free-falling objects". You also are confusing "test objects", the objects that follow particular worldlines in a given spacetime, from objects that act as "sources" of spacetime curvature; the two are different.

    Let me give a concrete example to illustrate what I mean with the first point (this will also illustrate what "moving from one point in space to another" means in what I said above): Schwarzschild spacetime, the solution of the Einstein Field Equation that describes the vacuum spacetime around a spherically symmetric gravitating object, is static. That means there is a family of worldlines along each of which the metric doesn't change. An observer following one of these worldlines will be "hovering" at a constant altitude above the gravitating object, and will not be revolving about the object at all. This observer is static, i.e., at the same point in space for all time. Note that such an observer is not freely falling; he must either use rockets to maintain altitude, or be standing on something (like a platform, or indeed the surface of the gravitating body), and in either case he will have nonzero proper acceleration and will feel weight.

    Any other observer in this spacetime is "moving" from one point in space to another. However, although any freely falling observer must be moving (because, as above, observers who are static cannot be freely falling), not all moving observers will be freely falling. An observer who uses rockets or stairs or a ladder to climb from one altitude to another is not freely falling, but is moving (not staying that the same point in space). An observer who jumps off a high platform, or out of his rocket, and falls toward the gravitating body is both moving and freely falling, of course. All of these moving observers see a changing spacetime geometry as they move; but note that they themselves do not cause the changes, they just observe them (see below).

    (Note also that an observer can be freely falling without changing altitude, if he is in a circular orbit about the gravitating body. In this particular case, a moving observer will not see any change in spacetime geometry. That is because this particular spacetime has an additional symmetry, spherical symmetry, over and above being static. A static spacetime does not have to have any additional symmetry, although I don't have a handy simple example of one that doesn't.)

    Now for the second point: In the example above, the spherically symmetric gravitating object is the "source". The other objects (i.e., all the different observers I described) are all test objects and are assumed to have a mass that is so small, compared to the mass of the source, that they don't affect the spacetime geometry. So test objects, whether they are freely falling or not, don't curve spacetime (more precisely, they don't curve it enough to be significant in the scenario under discussion).
     
  4. Jan 22, 2015 #3
    Thanks for another great post Peter. I've understood many things you said, but I still don't understand the main question: Is motion allowed in a static spacetime?

    we have a hovering Schwarzschild observer and his definition of space, is any kind of motion allowed relative to him? Or do all objects in the static spacetime have to just sit where they are without any motion? There are many types of motion in general, and all of the objects involved seem to curve nearby spacetime while they are travelling, so I am wondering what is the answer to my first question?
     
  5. Jan 22, 2015 #4

    PeterDonis

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    Of course it is. "Static" refers to the spacetime geometry; it doesn't put any restrictions about how test objects, which are too small to affect the spacetime geometry, can move.

    What makes you think this? As above, test objects are too small to curve spacetime. In the scenario I described with one central gravitating object, that object is the only thing that curves spacetime; all the observers I described, some of which are moving (relative to the notion of "space" defined by the static spacetime), do not curve spacetime, because they are too small.
     
  6. Jan 22, 2015 #5
    What about bigger objects with greater mass? Why can't the space change curvature relative to a Schwarzschild static (hovering) observer
     
  7. Jan 22, 2015 #6

    PeterDonis

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    If you have more than one gravitating object, the spacetime won't be static. The two gravitating objects will attract each other, so they will either fall together, or orbit around their common center of mass. (Adding more gravitating objects makes it even more complicated.) The Schwarzschild solution does not describe this; it describes one gravitating object only, with nothing else present that can produce spacetime curvature.
     
    Last edited: Jan 22, 2015
  8. Jan 22, 2015 #7

    Orodruin

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    Great post Peter, I just want to make one addition/query: If I do not misremember, the Schwarzschild solution is only static outside of the event horizon. Once inside the event horizon, any observer will have a changing metric as any future directed timelike world line invariably has a decreasing r-coordinate.
     
  9. Jan 22, 2015 #8

    PeterDonis

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    Correct. In this discussion, we are only considering ordinary gravitating bodies like a planet or star, which are not black holes and don't have an event horizon. The vacuum region of such a spacetime is entirely static (and the non-vacuum region is too, although we aren't really discussing that here). But a black hole is only static outside the horizon, yes; the Killing vector field that is timelike outside the horizon is null on the horizon and spacelike inside it, so the condition for a static spacetime is not met, and any observer following a timelike worldline will see a changing metric.
     
  10. Jan 22, 2015 #9
    Thanks for the response Peter, so can we calculate gravitational time dilation for some region of spacetime that changes its spacetime curvature from let's say, flatter to curvier (sorry if my English is bad). Since the Schwarzschild formula is for static situation, will it work for a dynamic case?
     
  11. Jan 22, 2015 #10

    PeterDonis

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    No. The concept of "gravitational time dilation" only makes sense in a static spacetime. (More precisely, it only makes sense in a stationary spacetime, which is a slight generalization of a static spacetime to cover things like uniformly rotating objects, where the source of gravity is not entirely motionless but there is still a sense in which things "are not changing".)
     
  12. Jan 22, 2015 #11
    So what can be said of a portion of spacetime which ic changing its curvature? Isn't it logical to say, for instance, that when spacetime goes from flatness to curvature the rate of clocks also slows down?
     
  13. Jan 22, 2015 #12

    PeterDonis

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    Um, that it's changing its curvature?

    No. Spacetime curvature is not the same as "strength of gravity" in the Newtonian sense, or "depth of gravitational potential well".

    For one thing, spacetime curvature includes time as well as space. For example, according to our best current model, our universe is spatially flat; all of the spacetime curvature due to the matter and energy in the universe as a whole (i.e., on the scale of the universe as a whole) is along the time dimension (when we say that the universe is "expanding", we are really referring to it being "curved along the time dimension"). But this same model of the universe tells us that the rates of "comoving" clocks (clocks that see the universe as homogeneous and isotropic on a large scale) are constant--they do not change as the universe expands.
     
  14. Jan 22, 2015 #13
    But shouldn't space also be curved? I mean, I watched some Brian Greene videos where it's clearly visualized how matter and planets curve space like a ball curves a trampoline. Almost every video that I watched about GR mentiones that planets curve the spatial dimension by their mass and that the ticking of clocks depends on how strong gravity/ spacetime curvature is.
     
  15. Jan 22, 2015 #14

    PeterDonis

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    Not on the scale of the universe as a whole.

    This model can be misleading for several reasons. One is that it's coordinate-dependent, because the split of spacetime into "space" and "time" is coordinate-dependent. The visuals you are talking about assume that you are using Schwarzschild coordinates. If you use Painleve coordinates, the space around a gravitating body is flat.

    Another reason that model can be misleading is that it is based on Schwarzschild spacetime, which, as I said before, assumes a single gravitating body that is alone in the universe, surrounded by empty space. That model does not apply to our universe, not just because there are multiple gravitating bodies, but because the masses in the universe are everywhere--the universe is not a bunch of masses surrounded by empty space.

    This is a good illustration of why you should not try to actually learn science from pop science presentations, even if they're fronted by actual scientists like Brian Greene. This model, as above, is based on Schwarzschild spacetime, which is not a good model for the universe as a whole. The scientists in the videos, like Greene, know that, and would never try to get away with such a statement in an actual peer-reviewed scientific paper. But in videos like these, as far as I can see, they are not actually trying to teach you the science, in the sense of giving you a logically consistent model that you can use to make predictions. They are just trying to get you to say "oh, wow, neat!" and then go on to the next sound bite without actually trying to reason about what they are telling you. That's a shame, but that's how it seems to me.

    Also you are equating "gravity" with "spacetime curvature" (and I suspect that the videos explicitly did that, so it's not just your interpretation), which, as I said in an earlier post, is not correct. The correct statement is that spacetime curvature is the same as tidal gravity.
     
  16. Jan 23, 2015 #15
    Then on what does gravitational time dilation depend on, if not on the spacetime curvature?
     
  17. Jan 23, 2015 #16

    pervect

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    space-time curvature is generally considered to be the Riemann curvature tensor, ##R_{abcd}##, the most important components of which have a physical interpretation as tidal forces. Gravitational time dilation would be one particular component ##g^{tt}## of the metric tensor. Because it's only one component of the tensor, gravitational time dilation is coordinate dependent, without the rest of the components it can't be treated as a tensor.

    You can construct the Riemann curvature tensor as a function of the metric tensor and it's first and second derivatives. So the highest order terms would involve the second derivatives of the metric tensor. Sorry if this is a bit technical, I don't know how to put it more simply, perhaps someone else has a way.
     
  18. Jan 23, 2015 #17
    I still somehow believe that clocks on some spacetime portion that is changing its curvature must elapse less proper time than clocks in flat spacetime. After all, changing curvature is still curved spacetime which implies clocks running slow when influenced by mass, which also curves spacetime. That seems logical, if it isn't like that then how is it? Nothing can be said about gravitational time dilation while spacetime doesn't have a constant curvature?
     
  19. Jan 23, 2015 #18

    Nugatory

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    What exactly does it mean to say that one clock shows less time passed than another?

    A clock near me and at rest relative to me ("my clock") reads 12:00:00. At the same time that my clock reads 12:00:00, some other clock at a distance from me also reads 12:00:00. Later I look at my clock and it reads 12:00:05. At the same time, the other clock reads only 12:00:03, so I conclude that the other clock is running slow.

    But notice that this conclusion depends on having a definition of "at the same time". In curved spacetime there is no natural way of defining "at the same time", and by choosing different definitions I can come to different conclusions about which clock is running slow.
     
  20. Jan 23, 2015 #19

    PeterDonis

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    If the spacetime is not static (technically, stationary--see post #10), there is no such thing as gravitational time dilation. That is a very important point that you appear to be missing. See below.

    If the spacetime is static (technically, stationary), then gravitational time dilation depends on where you are in space--where "space" is defined using the family of worldlines along each of which the metric doesn't change, as I described in post #2.

    No, it does not. This implication is simply wrong, no matter how intuitive it seems to you. Once again, if spacetime is not static (technically, stationary), there is no such thing as gravitational time dilation, because there is no invariant way to tell which spatially separated clocks are "running slow" relative to which others. There just isn't.
     
  21. Jan 23, 2015 #20

    PeterDonis

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    This is true in general, but note that in a static/stationary spacetime, the timelike Killing vector field gives a natural way of defining "at the same time", and this in turn gives a way of defining gravitational time dilation in an invariant sense. Of course this way of defining "at the same time" is still not the only possible one, even in a static/stationary spacetime. It's just that the timelike KVF gives a way of picking out a definition of "at the same time" that does at least have an invariant property associated with it, whereas in a non-stationary spacetime there is no definition of "at the same time" for which that is true.
     
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