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How does gravitational time dilation increase the spacetime interval?

  1. Dec 3, 2011 #1
    I (think I) understand that:

    1. With the Schwarzschild metric, the ratio of proper time to coordinate time decreases ("clock runs slower") with decreasing radial distance. (And ratio of proper distance to coordinate distance increases.)

    2. The geodesic path followed by a freely falling object maximizes the magnitude of the spacetime interval between two points. (Books all say interval is "extremal" because sometimes it's a minimum rather than maximum, but no one ever explains the minimal case to me.)

    I am confused because these two ideas seem contradictory. It seems that when your clock is ticking more slowly, your spacetime interval will be smaller than if it ran up a larger number of ticks.

    Of course if I am comparing a worldline that bends toward the gravitational mass to a worldline that doesn't, I don't have the same endpoints, so it's apples and oranges. The principle of maximizing the interval ("principle of extremal aging") only tells me that if I picked an alternative route with the same endpoints, the interval would be smaller because it would involve more distance, and the interval is time-squared minus distance-squared.

    Nevertheless I try to compare an object released from rest (apple just beginning to fall from tree) to one that stays where it is (thus not following a geodesic): I say to myself, the apple that refuses to fall is racking up more clock ticks while traveling zero distance; surely its interval (if I could find a way to compare the different worldlines) must be larger than that of the apple that falls.

    I've seen all the embedding diagrams and visual aids, showing that time is "stretched out" by gravity and that the geodesics bend into the more stretched-out region. I focus in particular on the representation where a sheet with one time and one space dimension is rolled into a cylinder, with the time dimension running around it. Spacetime curvature is then shown as a fattening of the cylinder, and the geodesics bend in the direction of this fattening. Here an object at rest in a region without curvature would just follow a circle around the cylinder perpendicular to its axis; but as soon as it encounters a bulging of the cylinder, it bends into the bulge.
    Here my confusion is the same: I expect the trajectory to bend away from the bulge, not toward it.

    The bulging cylinder picture at first seemed to help me, because I thought, yes, it's traveling a longer distance in spacetime. But then I said wait, this "longer distance" through stretched-out time involves fewer clock ticks, not more. The time is dilated so that there is "more of it" between ticks, but it's the ticks that matter. Isn't it?
     
  2. jcsd
  3. Dec 3, 2011 #2

    Dale

    Staff: Mentor

    The curvature of spacetime is a rank-4 tensor. It is simply not possible to replace it by a scalar density field.
     
  4. Dec 3, 2011 #3

    Dale

    Staff: Mentor

    Actually, you can have a stationary clock and a geodesic clock that are at the same endpoints simply by tossing the geodesic clock upwards and letting it free-fall up and down. When it goes up past the stationary clock that is one endpoint, and when it goes down past the stationary clock, that is the other endpoint.

    As you mention, the geodesic between those two endpoints is the maximal proper time. There are two competing effects here.

    First, a path which goes high will maximize the proper time by running in a region with less gravitational time dilation. The stationary clock is not a maximal path because it stays down in the region with high gravitational time dilation.

    Second, a path which goes slow will maximize the proper time by having less velocity time dilation. So a third clock which would shoot up to the moon and back in the time it takes the geodesic clock to free-fall is not a maximal path because it has too much velocity time dilation (despite having very little gravitational time dilation).

    The geodesic path is the optimal balance between getting as high as possible and going as slow as possible.
     
    Last edited: Dec 3, 2011
  5. Dec 3, 2011 #4

    Dale

    Staff: Mentor

    That is still a scalar, and not a rank 4 tensor. It is simply mathematically impossible to replace a rank-4 curvature tensor with a scalar density.
     
  6. Dec 3, 2011 #5

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    Ok, look at this one:
    http://www.adamtoons.de/physics/relativity.swf
    Set:
    coortinate time : 1s
    intial speed: 0c
    gravity : max

    The object in free fall needs 1s coortinate time to fall from x=0 to x=0.62 and accumulates 0.55s proper time. Now try do find another path from x=0 to x=0.62 with the same length (1s coortinate time) that ends up further along the proper time dimension than those 0.55s.

    I think it is easy to see that the straight path (geodesic) maximizes the proper time for a path of certain length (1s coortinate time) and certain space displacement.


    Seems like you need a intuitive analogy for geodesics and why they "tend" towards the "fatter" region. Advancing on geodesics means advancing locally straight ahead like a car without steering wheels. Imagine (or try it) you have such a toy car going on a curved surface:

    hatstarter.jpg

    From: http://www.relativitet.se/spacetime1.html

    Another way to model a geodesic is to stick adhesive tape around curved surfaces (vases, bottles, toilets). You have to stick it without tearing, or folding the edges, then it is locally straight.

    I assume you refer to this:
    http://www.adamtoons.de/physics/gravitation.swf

    Note that this it a space-propertime diagram. The length of the path is the coordinate time, so all objects (including light) are advancing at the same rate in this diagram. There is no "traveling a longer distance in spacetime" here. All world lines have the same length in this diagram. The proper time is not the distance but the angular displacemnt around the cylinder. And that can be different for different objects.

    To understand this go back to the simpler version:
    http://www.adamtoons.de/physics/relativity.swf
    Set:
    coortinate time : 1s
    check the box: stationary object
    gravity : max

    Now play around with the position of the stationary object and see how it affects the object's proper time after 1s of coordinate time. That is gravitational time dilation.

    For more information check this:
    http://www.relativitet.se/Webtheses/lic.pdf (space-propertime diagrams are in chapter 6)
    http://www.relativitet.se/Webtheses/tes.pdf


    Lets say you are hovering near a black hole using your rockets engines. Then you send a light pulse to go around the black hole and return to you. You experience some proper time between the the two events. The light going on a geodesic world line experiences zero proper time so it minimizes it.
     
    Last edited: Dec 3, 2011
  7. Dec 4, 2011 #6
    Dale: Thanks for the example of clock tossed up in the air. Can you show me the equation expressing the tradeoff between faster clock further from earth (GR effect) and faster clock due to lower velocity (SR effect)?

    A.T.: This is helpful. It helps me subordinate the role of proper time to the straightness of the worldline, which is the primary thing.

    If the object is stationary, its worldline is curved by the force holding it up. But the freely falling object "offsets" the spacetime curvature by moving in the direction that balances the time dilation with coordinate distance (although this is also the direction of increasing curvature).

    How can I see why this is the right direction to "offset" the time dilation? The time-coordinate lines bend away from gravity, so straightening the path requires moving toward it. The straight line against (locally) constant curvature signifies constant acceleration (and increasing curvature implies increasing acceleration).
    I think I've got it.

    (I suspected that the "minimal case" of extremal aging had to do with light. Thanks for spelling it out.)
     
  8. Dec 4, 2011 #7

    Dale

    Staff: Mentor

    You simply minimize (or maximize) the following action using the Euler Lagrange approach:
    [tex]S=\int g_{\mu\nu}\frac{dx^{\mu}}{ds}\frac{dx^{\nu}}{ds} ds[/tex]
     
  9. Dec 4, 2011 #8

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    There is an alternative way to see it. Instead of imagining the space-time as stretched with geodesics bending towards the stretched areas, you can see space time having varibale density and the free-fall world-lines bending towards the denser areas, just like light beams in a medium with a variable optical density.

    This method has the advantage that you don't need to embed a curved space-time diagram in higher dimensional space to visualize the curvature. So you can show one space-time dimension more: 2+1 space-time.
     
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