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A couple basic spacetime questions.

  1. Jan 19, 2014 #1
    Hey everyone, for anyone who saw my thread in the chemistry section you know I'm changing my view point to that of the "there are no dumb questions", and with that I have a couple of things I've been curious about for a while regarding spacetime.

    First off, from reading and research I am fairly familiar with the fact that mass curves spacetime, which if I understand correctly is responsible for gravity... If these basic assumptions are correct, they lead to my first question; When spacetime curves, is it a fairly flimsy medium, or is it ridged? In other words, How difficult is the act of curving spacetime?

    The second is more of just an assumption I've had, that I want to verify... If you watch anything about the cosmos on Discovery Channel, Science Channel, NOVA, etc; they give the visual of curved spacetime similar to a bowling ball on a bed sheet, I've always assumed that this was an inaccurate picture, it seems like spacetime should have a 3rd spacial dimension, it should have depth as well as length and width. Is this correct?

    As I mentioned these very well may be dumb questions, but I'm just trying to educate myself.
    Thanks for any insight!
  2. jcsd
  3. Jan 19, 2014 #2


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  4. Jan 19, 2014 #3


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    The Einstein field equations describe how much energy / mass you need to cause which effect on spacetime geometry.

    Space-time has 4 dimensions. You can omit 2 of them to visualize the curvature. The rolling balls on the sheet is bad, because it omits the time dimension, which is crucial at low speeds (much less than ligthspeed). This analogy includes the time dimension:

  5. Jan 19, 2014 #4


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    I'm not sure how to answer this, or if it has an answer.

    What I can say:

    The curvature of space-time can be broken down into "sectional curvatures", one for every plane. The sectional curvature can be interpreted as the curvature of the surface that we'd need to draw the space-time diagram of the specified plane on in order to have it undistorted. I.e, if we choose the r-t plane, we compute the sectional curvature of the r-t plane, then our space axis on our space-time diagram is r, our time axis is t, and our diagrams will be undistorted when the surface has the correct sectional curvature.

    We are here describing the abstract concept of curvature via an embedding diagram, which is I think the easiest way to describe it without a lot of math.

    I'm not sure how more precise I can be about the distortion issue, except to note that the surface of the Earth is accurately represented on a globe, which is to a high degree of approximation a sphere of constant radius, and that if you try to draw maps of the Earth on a flat sheet of paper (or the wrong radius of globe), the shapes won't come out right, the map cannot be everywhere to scale.

    For positive curvatures, we need to draw our space-time graphs on a surface of positive curvature (like a sphere), for negative curvatures , we'd need a saddle surface.

    Sectional curvatures have units of 1/distance^2, so for positive sectional curvatures, the radius of the sphere we'd use to model the surface would be sqrt(1 / curvature).

    For more details try the wiki http://en.wikipedia.org/wiki/Sectional_curvature

    If we take the r-t plane of the Schwarzschild geometry representing a single mass, the sectional curvature (in units where c=1) will be -2GM/r^3, which has units of 1/seconds^2 - and is equal to the tidal force. It's negative, which means we'd need to draw our space-time diagrams on a saddle surface.

    If we take the theta-t or the phi-t plane, the sectional curvature will be GM/r^3, with the same units. Since it's positive, it's a bit easier to interpret. The radius of the sphere that we'd need to draw our space-time diagrams on to have them appear "to scale" would be sqrt(r^3 / GM) if the curvature were everywhere constant. Since this has units of seconds, it might be easier to visualize if you multiply it by "c", even though we originally set c=1.

    Note that the radius of the space-time diagram sphere isn't constant, so we'd need a more complex shape than a sphere. I'm not aware of any one who has worked out the exact shape you'd have. The radius of the sphere required approaches infinity as r approaches infinity, which implies zero curvature.

    Yes, it's not a very good picture, though it could be a good picture if they described it differently, as the purely spatial part of the space-time curvature (i.e the r-theta plane). You're right about needing more dimensions as well, but there's no practical way to describe a curved surface of more than two dimensions with a simple picture.
  6. Jan 21, 2014 #5
    Thanks everyone for the answers!

    I'm so confused, but still determined to understand lol
  7. Jan 26, 2014 #6
    Imagine a net, now add another dimension, a cube of net or a 3d matrix. Now imagine stuffing a bead into one of the spaces.. Now picture a bigger bead displacing some rows of your net, putting the spaces around it closer together. This is your gravitational field.
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