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Null vector help!

  1. Jan 4, 2007 #1
    I read this in the notes:

    Show that any vector that is orthogonal to a null vector must be either be:-
    i) parallel to a null vector
    ii) space-like

    How??
     
  2. jcsd
  3. Jan 4, 2007 #2

    robphy

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    Should this be in the howework section?
    What's your starting point? What does it mean for two vectors to be orthogonal?
     
  4. Jan 4, 2007 #3
    Orthogonal means normal, I guess.
    This is no homework. I am doing it as a hobby.
    This comes out from GR notes describing the light cone, and a general description of the four linearly independent vectors that may or may not lie parallel to the light cone, which is a null surface, ie may or may not be null vectors, and the properties of the vectors that are not null vectors.
     
  5. Jan 4, 2007 #4

    George Jones

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    Let {e_0, e_1, e_2, e_3} be an arbitary orthonormal basis. Then, up to a constant multiple,

    n = e_0 + e_1

    is an arbitrary null vector. Let

    v = v^0 e_0 + v^1 e_1 + v^2 e_2 + v^3 e_3

    be an arbitrary 4-vector. If n and v are orthogonal, what does this give you?
     
  6. Jan 4, 2007 #5
    Thanks. I see.
    <n,n>^2=-1^2+1^2=0
    <n,v>=0 => =-v^0+v^1=0 => V^1=v^0
    Therefore, <v,v>^2 = -v^0^2 + v^1^2 + V ^2^2 + v ^3^2 = -v^0^2 + v^0^2 + V ^2^2 + v ^3^2 = V ^2^2 + v ^3^2 >=0
    If =0 , then null-like.
    If >0 then space-like.
    Correct?
    But why can n = e_0 + e_1 for any arbitrary n such that <n,n>=0 where {e^v} is orthonormal basis?
     
  7. Jan 5, 2007 #6

    George Jones

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    Yes.

    For this case, there is one other, trivial, possibility. :smile:

    Consider an arbitrary vector in a 2-dimensional spatial plane.

    We are free to choose a basis that helps us. For example, we might choose: a basis such that the vector is in the direction of e_1; a basis such that the vector is in the direction of e_2; a basis such that the vector is halfway between e_1 and e_2, i.e., in the direction e_1 + e_2.

    None of these choices restricts us.
     
  8. Jan 5, 2007 #7
    Thank you so much.
    I guess you mean a vector subspace.
    If we have 4 orthonormal vectors {e} that span a vector space, we can always find a 2 -D subspace which contains n, spanned by orthonormal basis f1, f2.
    Hence if n is null, then <n,n>=0 => -n0^2 + n1^2 =0 => n1=n2.
    And should be able to find f3 and f4 to complete the {f} for the total vector space T(M).
     
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