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Why a timelike vector and a null vector cannot be orthogonal?

  1. Jan 23, 2005 #1
    Why a timelike vector and a null vector cannot be orthogonal?
    Isn't a null vector orthogonal to any vector, by definition? Anyway, each component of a vector is multiplied by zero, so in the end the sum is zero.
     
  2. jcsd
  3. Jan 23, 2005 #2

    dextercioby

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    Null 4-vector:
    [tex] n^{\mu};n^{2}=:n^{\mu}n_{\mu}=0 [/tex] (1)

    Timelike 4-vector:
    [tex] l^{\mu};l^{2}=:l^{\mu}l_{\mu}<0 [/tex] (2)

    Prove that
    [tex] l^{\mu}n_{\mu} \neq 0 [/tex](3)

    HINT:Use components and the property of the 'cosine' function.


    Daniel.

    P.S.Esti varza... :yuck:
     
  4. Jan 24, 2005 #3

    dextercioby

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    HINT:'cosine' appears in the expression of the scalar product between those vectors (space components).Pay attention with the metric...

    Daniel.
     
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