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Show that this vector is timelike?

  1. Feb 16, 2015 #1
    Hi guys,

    So I have a vector which I need to show is timelike. The vector is

    [itex] v^{\mu}=t^{\mu}-\frac{t^{\mu}\cdot X^{\mu\prime}}{X^{\mu\prime}\cdot X^{\mu\prime}}X^{\mu\prime} [/itex],

    where [itex]t^{\mu}[/itex] is a timelike vector and [itex]X^{\mu\prime}[/itex] is spacelike, however these two vectors are not perpendicular so their dot product does not vanish.

    I understand that in order to show that [itex]v^{\mu}[/itex] is timelike, I need to find [itex]v_{\mu}v^{\mu}[/itex] and show that this is greater than 0. So:

    [itex]v_{\mu}v^{\mu}=t^{2}-2\frac{(t\cdot X')^{2}}{X^{\prime}\cdot X^{\prime}}+(t\cdot X^{\prime})^{2}[/itex]

    So first of all I dont know if this is correct (yes I dont know how to take a dot product obviously lol) and secondly, even if it is, how do I show that this is greater than 0?

    Thanks guys!
     
  2. jcsd
  3. Feb 16, 2015 #2

    Orodruin

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    Your last term is wrong. Try to redo the inner product.

    You should also look over your use of indices in the original problem statement.
     
  4. Feb 16, 2015 #3
    Yes doctor that is wrong - fixed it and now I get this:

    [itex]v_{\mu}v^{\mu}=t^{2}-\frac{(t\cdot X^{\prime})^{2}}{X^{\prime}\cdot X^{\prime}}[/itex] and... is this equal to 0? :O
     
  5. Feb 16, 2015 #4

    Orodruin

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    No, if it was zero then v would be light-like. I suggest you try to figure out the signs of both of these contributions.
     
  6. Feb 16, 2015 #5
    I dont know how the signs could change. But is the maths right? I mean is that the true expression?
     
  7. Feb 16, 2015 #6
    Nevermind I think i got it -- because X is spacelike, its square is negative right :O so i get something positive overall?
     
  8. Feb 16, 2015 #7

    Orodruin

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    Correct. Also ##(t\cdot X)^2## is just the square of a number and thus also positive.
     
  9. Feb 16, 2015 #8
    Thank you doctor once again :) while we're at it could you please also confirm / deny the following statement:

    [itex](t\cdot X^{\prime})^{2} = t^{2}X^{\prime 2}[/itex]?
     
  10. Feb 16, 2015 #9

    Orodruin

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    Not true. The left hand side is positive and the right hand side is negative.
     
  11. Feb 16, 2015 #10
    I thought so. Thank you doctor :D you're a life saver lol ! (no pun intended)
     
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