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Space and Time question

  1. Jun 23, 2004 #1
    I was reading "Black Holes and Time Warps", by Kip Thorne. I was puzzled a bit on a certain subject. Einstein implies that your space and time is a mixture of my space and time. I'm confused there. Can somebody help me out.
  2. jcsd
  3. Jun 23, 2004 #2


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    The distance between two points (ds) in 3 dimensional space is (dx is difference along the x axis, etc.):

    [tex]ds^2 = dx^2 + dy^2 + dz^2[/tex]

    (this is just the Pythagorean theorum)
    If you rotate the x-axis/y-axis/z-axis or move the orgin of the particular co-ordinate system ds always has the same value.

    I'm sure your at least partially famliar with the concepts of time dialtion and length contraction in special relativity, well because of length contraction if my co-ordinate system is travelling relative to another co-ordinate system ds is no longer the same in both co-ordinate systems, i.e. the two obsrevers measure diffrence distances between objects.

    But if we amend our equation for ds so that:

    [tex]ds^2 = dx^2 + dy^2 + dz^2 - c^2dt^2[/tex]

    Once again all diffreent co-ordinate systems measure the same values for ds.

    In this way a change in velcoity is analogus to a rotation of our axes, so two obsrevers might no longer have the same x-axis or 'time-axis'.
  4. Jun 23, 2004 #3
    Ok, thanks for trying to explain, but I had no idea what you said.:lol: Whats a x-axis, and all the other stuff you said. I could always use a physics teacher. I am only 14, and am pretty new to physics. Yes, I do know that motion dilates time. However, we as people do not notice it, because it is only obviously noticable or effective when approaching the speed of light.
  5. Jun 23, 2004 #4


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    Think of x, y, and z as length, width, & height. The "d" thing just means "a change in" (like "a change in x"). Now imagine a 3D grid. That grid has length (x), width (y), and height (z). Now imagine you're at one point in the grid and jcsd is at another. If you draw a line between you and jcsd, you can describe the distance of that line as a difference in your x, y, and z coordinates.

    If jcsd is 10 ft away from you at the same ground level, then dx is 10 ft. And your distance is 10 ft.

    [tex]ds^2 = 10^2 + 0 + 0[/tex]

    If jcsd is 10 ft away from you sitting on top of a 10 ft tall basketball net, then he's 14 ft away from you (imagine the slanted part of the triangle if you draw the horizontal, vertical, and connecting lines between you and him).

    [tex]ds^2 = 10^2 + 0 + 10^2[/tex]

    Then once you start inserting the effects of Relativity, things get screwy and you need that additional factor in the equation that jcsd mentioned.
  6. Jun 23, 2004 #5
    Ok, thanks. I got most of that. :smile:
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