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Four vector properties.

  1. Jun 1, 2012 #1
    Hey I'm reading about relativity and I stumble across statements such as if a four-vector is timelike then it's always possible to find a reference frame for which the space-components vanish.

    Further if a four-vector is 'four-orthogonal' to a timelike four vector it must be spacelike. I wondered if someone know how to prove these statements or guide me to avaiable text that do. Are there any 'intuitive' ways to see these statements?
     
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  3. Jun 1, 2012 #2

    ghwellsjr

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    If a four-vector is timelike, then it is possible for an inertial clock to be present at the two events defining its endpoints. The accumulated time on that clock is a measure of the length of the four-vector (the same as the a timelike spacetime interval). In the rest frame of this inertial clock, it's spatial compontents are the same so it has no components in any spatial direction.
    If we take the four-vector length and consider a vector that is orthogonal to it, it will have no time component but only a spatial component making it a spacelike spacetime interval. In this case, it is not possible for a clock (or any other object) to be present at both endpoint events because it would have to travel at faster then the speed of light. Instead, we can think of an inertial ruler placed between the two endpoints in a frame in which the two events have the same time component.

    I haven't really answered your question because I haven't offered a formal proof but I think my explanation may help you to visualize the situation in a way that you may not require further proof.
     
  4. Jun 1, 2012 #3

    PeterDonis

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    I don't know about "intuitive", but you may be able to get some insight from considering what the components of a 4-vector look like in an arbitrary inertial frame, and how the components relate to whether the vector is timelike or spacelike (or null), and how the components of two 4-vectors relate to whether they are orthogonal.

    A generic 4-vector has 4 components in an inertial frame; the components are usually called (t, x, y, z). The invariant "length" of the 4-vector is then given by the formula:

    [tex]t^{2} - x^{2} - y^{2} - z^{2}[/tex]

    If this formula gives a positive number, the 4-vector is timelike; if it gives a negative number, the 4-vector is spacelike; if it gives zero, the 4-vector is null (or "lightlike"). In the timelike case, the above may help you to see that you can always find a Lorentz transformation that makes x = y = z = 0.

    Similarly, if I have two 4-vectors (t1, x1, y1, z1) and (t2, x2, y2, z2), then they are orthogonal if and only if:

    [tex]t_{1} t_{2} - x_{1} x_{2} - y_{1} y_{2} - z_{1} z_{2} = 0[/tex]

    Putting this together with the above may help in seeing that any 4-vector that is orthogonal to a timelike 4-vector must be spacelike.
     
  5. Jun 1, 2012 #4
    That is ofcourse the ideal place to be. Thank you for the reply, it was illuminating.
     
  6. Jun 1, 2012 #5

    George Jones

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    Do you see how the second statement follows from the first?
     
  7. Jun 1, 2012 #6
    Yes, because of the argument above I am convinced that there is a frame for which a time like vector has no space components, namely the frame of an inertial clock with the a worldline going along the four vector. Considering the timelike vector A in this frame and taking the scalar product with an orthogonal four vector B in the same frame, this product will only contain the product of their time components (since the space components of A are zero), and this product must be zero, which can only happen if B's time component is zero. Implying it is spacelike.
     
  8. Jun 1, 2012 #7

    George Jones

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    Yes.

    I like to use this second statement as one of the defining properties of Minkowski space.
     
  9. Jun 1, 2012 #8
    Alright, instead of using the metric?

    However it does not seem to follow from this argument that for the same frame for which the timelike vector only has a time component, the orthogonal spacelike vector only has space components. This does seem to be the case, for example for four acceleration and four velocity.

    Can one argue that this is always true?
     
  10. Jun 1, 2012 #9

    George Jones

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    No, it allows for a for the definition of the "metric" and its signature without using any bases.
    Because Lorentz transformations preserve the metric (by definition!), the causal character (spacelike, lightlike, or timelike) of a 4-vector is independent of reference frame. For example, if a 4-vector is spacelike in one reference frame, then it is spacelike in all reference frames.
     
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