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Orthogonality of timelike and null vector

  1. Oct 28, 2013 #1
    Can we show orthogonality of timelike and null vector?
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    Last edited: Oct 28, 2013
  2. jcsd
  3. Oct 28, 2013 #2

    George Jones

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    If X is timelike, and Y is orthogonal to X, then Y is spacelike.
     
  4. Oct 28, 2013 #3

    WannabeNewton

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    I just want to expand upon what George said. Let ##(M,g)## be a space-time and ##p\in M##. Furthermore, let ##X## be a time-like vector in ##T_p M## and ##Y## a non-zero vector in ##T_p M## such that ##g_p(X,Y) = 0##. Finally, let ##\{e_i \}## be an orthonormal basis for ##T_p M## so that ##X = X^i e_i## and ##Y = Y^i e_i##.

    Then ##(X^2)^2 + (X^3)^2 + (X^4)^2 < (X^1)^2## and ##X^1 Y^1 = X^2 Y^2 + X^3Y^3 + X^4Y^4##.
    Note this immediately implies that ##(X^1)^2 > 0## and that ##(Y^2)^2 + (Y^3)^2 + (Y^4)^2 > 0##.

    Therefore [tex](X^1 Y^1)^2 \leq ((X^2)^2 + (X^3)^2 + (X^4)^2)((Y^2)^2 + (Y^3)^2 + (Y^4)^2)< (X^1)^2((Y^2)^2 + (Y^3)^2 + (Y^4)^2)[/tex] thus ##(Y^1)^2 < (Y^2)^2 + (Y^3)^2 + (Y^4)^2## i.e. ##Y## is space-like.
     
  5. Oct 28, 2013 #4

    George Jones

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    WLOG, choose ##X = \{e_0 \}##.
     
  6. Oct 28, 2013 #5

    clem

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    A time-like vector can be Lorentz transformed to be purely time-like:[itex]x^\mu=(t,0)[/itex].
    A null vector always has x=t, so they can't be orthogonal.
     
    Last edited: Oct 28, 2013
  7. Oct 28, 2013 #6
    I think you have to show that:
    [tex](X^1 Y^1)^2 \leq ((X^2)^2 + (X^3)^2 + (X^4)^2)((Y^2)^2 + (Y^3)^2 + (Y^4)^2)[/tex]
     
  8. Oct 28, 2013 #7

    WannabeNewton

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    I had forgotten completely that one has the freedom to Lorentz boost to a frame wherein ##X = e_0## as stated by George and Clem above, which makes the calculation much simpler. Nevertheless, the above inequality follows directly from the Cauchy-Schwarz inequality.
     
  9. Oct 28, 2013 #8
    I am student at geometry so your notation is quite different for me but thanks for everyone for all replies...
     
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