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Are everyday objects spacelike?

  1. Feb 2, 2015 #1
    Hey guys,

    Imagine an everyday object, like a glass of water for example. Now imagine two points on the glass, like one near the top and one near the bottom. The interval between these two points, [itex]\Delta X[/itex], is it spacelike because these two "events" are occurring at the same time?
     
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  3. Feb 2, 2015 #2

    PeterDonis

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    If you imagine things that way, then yes. ;) But if, for example, you imagine events corresponding to the top point at some instant of time and the bottom point after an interval of time long enough for light to travel from the top to the bottom, then those two events will not be spacelike separated.
     
  4. Feb 2, 2015 #3

    Vanadium 50

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    Objects are not events, or pairs of events. Intervals are functions of the coordinates of pairs of events.
     
  5. Feb 2, 2015 #4

    Orodruin

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    To be more concrete, a three-dimensional object will sweep out a world-volume in Minkowski space. If you just look at two points on the glass, they will be described by world lines. These are going to have points which are space-like separated, but also other points which are not. Like Peter said, if you wait long enough, light from one end of the glass will reach the other.
     
  6. Feb 2, 2015 #5
    Wow thank you guys for such quick replies!

    So the next thing is...I just wana understand the component structure of four vectors. let's take for example a four vector [itex]n_{\mu}[/itex]. It can either be timelike, spacelike or null. Here's how I understand the component structure in each case:
    1) TIMELIKE - [itex]n_{\mu}=(1,0,0,0)[/itex] - basically all spatial components are 0
    2) SPACELIKE - [itex]n_{\mu}=(0,1,1,1)[/itex] - time component vanishes
    3) NULL - [itex]n_{\mu}=(1,1,0,0)[/itex] - everything else apart from the time component and one spatial component vanishes

    of course the "1" entries can be anything not just 1.

    So: are these correct?

    Thanks guys!
     
  7. Feb 2, 2015 #6

    Orodruin

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    No. A time-like vector may very well have non-zero space-like components. The thing is that it has a positive square (well, this depends on your taste in metrics ...), i.e., the square of the time component is larger than the sum of the squares of the space components. The opposite goes for space-like vectors and for light-like vectors, the square of the time component is equal to the sum of the squares of the space components.

    Edit: That being said, there always exists an inertial frame where a time-like vector has all space components equal to zero and there always exists an inertial frame where a space-like vector has zero time component. A light-like vector always has a non-zero time component and at least one non-zero space component.
     
  8. Feb 2, 2015 #7
    Okay - so in a space where the vector [itex]n_{\mu}[/itex] exists, a Lorentz observer can always find a frame wherein the conditions you wrote are met (depending on whether the vector is timelike / spacelike / lightlike) ?
     
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