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About spacetime curvature

  1. Aug 9, 2008 #1
    hey folks,

    as far as I understand, according to Einstein's general theory of relativity, any mass that exists in spacetime causes a curvature in it, right?! now, my question is: does this curve take place in the time dimension (ct) or in spacetime (ct,x,y,z) itself?
  2. jcsd
  3. Aug 9, 2008 #2


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    What is "time dimension" to one observer is (partially) "space dimension" to another one. The reason we call it spacetime, is because space and time cannot be clearly separated.
  4. Aug 9, 2008 #3
    If we lived in a 2D space, we wouldn't recognize a curvature that takes place in a 3rd dimension.
    likewise, in our 3D space, the curvature that is caused by mass/energy takes place in a 4th dimension, which is spacetime (ct). Is this correct??
  5. Aug 9, 2008 #4


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    No, because - if I understand what you are saying - then we would only see the effects in time observations. But it turns out that, for example, light is actually bent ("in space") by a strong gravitational field (search term: gravitational lensing).
  6. Aug 22, 2008 #5
    CompuChip, thanks for the clear explanations. however, you said earlier that time may look (partially) as space to a certain observer. can I ask what do you mean by (partially) and could you give me an example of such case? (has this got to do with the so-called Lorentz Boosts?)
    Last edited: Aug 22, 2008
  7. Aug 22, 2008 #6


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    If you label events with temporal and spatial coordinates (t,x), then "space, at time T" to you is the line t=T. However, if I'm moving with velocity v in the positive x direction then "space" to me is a line parallel to the line t=vx. Each such line is "space" to me, at different times of course. The change from your coordinates to mine is a Lorentz boost.
  8. Aug 23, 2008 #7


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    So let me try to explain the principles of relativity in basic terms.

    Consider this simple (and famous) example. We have a train of length L with an observer (e.g., you) inside it. The train is moving through the landscape at a constant velocity v. Because the windows are only one-way see-through, you cannot reference any point outside the train and because you are not accelerating, your gut feeling will tell you the train is at rest. In fact, there is no experiment that you could to that would tell you if you were actually moving or not (this is one of the postulates of special relativity).
    Now we flash a light source in the middle of the train in all directions, at time 0. You measure the time it takes the light to reach the ends of the train and you will find that the light flash reaches both the front and the back end at some time t simultaneously.
    Now I look through the window from outside the train, which is moving relative to me at velocity v. Since the speed of light is the same to me as it is to you (the other postulate of relativity) I will see the following: the front of the train moves away from the light source as the light travels toward it; the back of the train moves toward it. Therefore, the light flash will reach the front of the train at some time larger than t (the one you measured), and the back at some time smaller than t.
    Whichever way you adopt your coordinate systems, the two events (arrival of the light flash at the front of the train, and at the back of the train) will never have the same time and space "coordinates". For me, the flash to the front actually does take longer to reach it.

    Looking at a single flash (for example, the one to the front of the train), the distances x for me and x' for you are different, as are the times t (for me) and t' (for you). However, if I calculate a particular combination of the time and space coordinates, [tex]c t^2 - x^2[/tex] and you calculate [tex]c (t')^2 - (x')^2[/tex] we will get the same answer. Note that this is just the Pythagorean theorem, but with a strange sign in it: it basically means something like "the distance in space and the "distance" (duration) in time are different, but the "distance" in spacetime is the same for both of us".
    Now it turns out that there are formulas involving t, t', x, x', the relative velocity v and the speed of light c, which allow to go from one frame of reference to another. For example, if I measured time t and the velocity v of the train, I can calculate which time t' you will have found. Conversely, if I messaged you somehow that I was moving at velocity -v relative to you (that's a cryptical one, huh? :smile:) and you measured t', you could calculate which time t I had measured. Such conversions are called Lorentz boosts.
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