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Understanding Time and Space as One entity.

  1. Mar 3, 2008 #1
    I ( Swaminathan) started studying theory of relativity. I need some help in understanding the following:

    Special relativity says that what looks like time to one observer(in a moving frame) looks like space to another observer(in a stationary frame).

    I assume that Two events are happening. An observer views these two events happening at same space but at different times. Another observer views these two events happening at same time but at different space.

    Is my assumption is correct. If so, can someone give real world example for the above situation. I searched for examples in text books and it is not available.

    - Swaminathan.
     
  2. jcsd
  3. Mar 4, 2008 #2

    pam

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    That can't happen. Your first example describes a "space-like" interval (dx>dt).
    Your second example is a "time-like" interval (dt>dx).
    A Lorentz transformation (a rotation in space-time) cannot transform from a space-like to a time-like interval.
     
  4. Mar 4, 2008 #3
    I don't think that's correct.

    Why can't a rotation in space-time change the projection of a 4-space vector onto the time axis? Are all rotations about the time axis???

    Regards,

    Bill
     
  5. Mar 4, 2008 #4
    I think what pam is saying is true strictly within within the “Minkowski spacetime” coordinate system, if you refer to the 2nd paragraph of Wikipedia Lorentz transformations. Minkowski coordinates are “flat spacetime”, an approximation not taking into account the GR effects of gravity. The Lorentz transformation is sort of embedded in the coordinate system which is the reason for the special property pam is talking about.

    I think what Bill is saying is sort of true generally within any 4-space, since it isn't usually as straightforward to classify intervals as spacelike or timelike in that case.

    dswam you may be interested in [THREAD=215019]this thread[/THREAD], in which I tried to discuss the effects of relativity in non-technical, mostly non-mathematical terms with a film student.
     
  6. Mar 4, 2008 #5

    dx

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    dswam, the idea of spacetime as one entity is embodied in what is called the spacetime interval. say (t,x) are the coordinates of an event E according to observer A. Then the coordinates of the same event according to another observer B (t',x') must satisfy

    t^2 - x^2 = t'^2 - x'^2

    This spacetime interval is invariant. It is more fundamental than any particular set of coordinates assigned to it by an observer.
     
    Last edited: Mar 4, 2008
  7. Mar 4, 2008 #6

    tiny-tim

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    Observers have time-like trajectories

    I'm with Pam on this. :smile:

    dswam asked about the view of the same things by two different observers.

    Observers (who move slower than light) have time-like trajectories.

    A "rotation in space-time" which rotates one time-like vector onto another (one observer's trajectory onto another's) cannot rotate any time-like vector onto any space-like vector (including the t-axis). :smile:

    Different observers (who move slower than light) will always agree on what is time-like and what is space-like
     
  8. Mar 4, 2008 #7

    Fredrik

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    A space-time rotation (i.e. a homogeneous Lorentz transformation) around the time axis in Minkowski space is just an ordinary rotation around an arbitrary axis in space. So no, they (homogeneous Lorentz transformations) are not all rotations around the time axis.

    I don't know what you mean by "change the projection of a 4-space vector onto the time axis". If you mean "why can't a space-time rotation take the t axis to the x axis?", the short answer is that a function that does that doesn't preserve the "length" of all vectors as a "rotation" must do by definition. It would take 4-vectors with positive "length" to 4-vectors with negative "length".

    Note that the "length" in this case is defined in a different way than what (I guess) you're used to. For example, the "distance" between the origin and (t,x,y,z) isn't defined as t2+x2+y2+z2. It's defined as -t2+x2+y2+z2.

    The fact that "length" is something different than we're used to implies that rotations are different too. (Because a rotation is defined as a linear map that preserves the length of vectors). In particular, a homogeneous Lorentz transformation in 1+1 dimensions (1 time and 1 space dimension) that rotates (in the usual sense of that word) the t axis down a bit towards the x axis also rotates the x axis in the opposite direction by the same amount.
     
  9. Mar 4, 2008 #8

    Dale

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    pam is correct

    It doesn't matter what axis the rotation is about, all rotations preserve the length of a vector. Spacelike and timelike intervals have different lengths, so they cannot be exchanged through a rotation.
     
  10. Mar 4, 2008 #9
    Pile on Bill day

    Sorry, but I still disagree.

    I never said that length wasn't preserved - what I tried to say was: To assume that a 4-space vector never changes its' projection along the time axis implies that all rotations of that vector are about the time axis.

    If I were to suggest that a 3-vector rotated about a vector in 3-space never changes its' projection along the z-axis, would I not imply that all rotations are about z?

    I'm afraid we're talking apples and oranges. A space-time interval in Minkowski space is not a 4-space vector - it's a scalar.

    Regards,

    Bill
     
  11. Mar 4, 2008 #10

    George Jones

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    While it is true that a spacetime "rotation" can change the projection onto the time axis (of a particular observer), it is also true that a spacetime rotation, i.e., a (restricted) Lorentz transformation, cannot, by definition, transform a spacelike interval into a timelike interval.

    [edit]I somehow missed seeing StatusX's post before I posted.[/edit]
     
    Last edited: Mar 4, 2008
  12. Mar 4, 2008 #11
    AntennaGuy (and anyone else who's interested): this page: http://casa.colorado.edu/~ajsh/sr/sr.shtml is the start of one of my favorite expositions of Relativity and Minkowski Space. There are some very nice illustrations and animations of Lorentz transformations, i.e. the "rotations" of Minkowski space-time, that really make it clear how to think about these things.

    If I understand you correctly, your question about projections of vectors onto axes still retains the usual Euclidean ideas about vector transformations. If you read through the pages I linked above, I think you'll find your answer.
     
  13. Mar 5, 2008 #12
    The speed of light is constant only in relation to events happening in the same frame.
     
  14. Mar 5, 2008 #13

    Fredrik

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    That statement doesn't make sense.
     
  15. Mar 5, 2008 #14

    robphy

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    Note: "Rotations" are about "an axis" only in a 3-dimensional space.
    (In a similar vein, the cross-product of two vectors is more about the parallelogram determined by the two vectors rather than a "vector perpendicular to the parallelogram".)
     
  16. Mar 5, 2008 #15
    It means that for there to be an observed, there needs to be an observer, and both need to be in the same frame. The speed of light is the same for all observers but not to all frames. In other words, its constancy is local.
     
  17. Mar 5, 2008 #16
    That is not consistent with my understanding. I'm pretty sure that the speed of light is constant in all frames and it's time and space that vary between frames, via time dilation and length contraction.
     
  18. Mar 5, 2008 #17

    tiny-tim

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    observer = frame

    Still doesn't make sense. :frown:

    An observer has one frame - like an astronomer has one telescope.

    But an observed can be seen in every frame, just as a star can be seen in every telescope.

    A frame is simply the choice of x y z and t which the observer makes for himself.

    Basically, observer = frame.

    "The speed of light is the same for all observers" is the same as "The speed of light is the same for all frames."

    Only observers have frames, not the observed. :smile:
     
  19. Mar 5, 2008 #18
    Right, c is constant in all frames... but not to all frames at once.
     
  20. Mar 5, 2008 #19
    For there to be an observer there needs to be an observed... in any frame.

    But whatever is happening in one frame does not affect the speed of light on another. The speed of light does not need to be constant among frames, only within them.
     
  21. Mar 5, 2008 #20

    Dale

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    I think you miss pam's point. The OP wanted to find a case where two co-located events would transform into two simultaneous events. In other words, for coordinates s = (ct,x) he wanted to find some Lorentz transform, L, where Δs = (c dt,0) transforms to Δs' = L.Δs = (0,dx').

    This is not possible for any L since any L preserves the interval c²t² - x² and c²dt² ≠ -dx'² for all real dt and dx' (except zero).

    The "projection along the time axis" does change. That is time dilation. But it does not change in the way requested by the OP, as pam correctly stated. In fact, by looking at the interval you can see that the projection must always lengthen along the time axis (relative to the "proper" frame). This is because the rotation is hyperbolic along that axis rather than circular.
     
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