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A Lot of Basic Question About Lorentz-Minkowski Geometry

  1. Oct 29, 2013 #1
    Hi there, i have a lot of question about Lorentz-Minkowski geometry:

    Is Lorentz metric degenere or non-degenere? Why?

    2) In spacelike subspaces only spacelike vectors live in it there is not problem here but how can

    we say that timelike subspaces include null and spacelike vectors and lightlike subspaces include

    spacelike vector?

    3) Let E1 spacelike and E2 timelike vector then E1+E2 is null vector, proof...?

    4) If Sp{E1,E2} spacelike, Sp{E1,E3} and Sp{E2,E3}

    are timelike planes Sp{E1,E2+E3} is lightlike plane?

    5) E1+E2+E3 is spacelike vector but

    Sp{E1,E1+E2+E3} is lightlike plane?

    6) E2+E3 is lightlike vector but Sp{E2+E3,E3} is timelike plane?
  2. jcsd
  3. Oct 29, 2013 #2


    Staff: Mentor

    That is a lot of questions.
    This is not true in general. In units where c=1 and using the (-+++) signature, then the sum of
    E1=(0,1,0,0) and E2=(1,0,0,0) is null
    E1=(0,1,0,0) and E2=(5,0,0,0) is timelike
    E1=(0,5,0,0) and E2=(1,0,0,0) is spacelike

    Hopefully other people will address the other questions.
  4. Oct 29, 2013 #3
    Thanks, i agree with you, these questions are from the Rafael Lopez Minkowski Lectures, about question 3 the author able to think the unit vectors, the original is E2+E3 is null so if we take E2=(0,1,0) and E3=(0,0,1) then E2+E3=(0,1,1) which is a null vector, here the signature is (+,+,-) in E13...
  5. Oct 29, 2013 #4


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    2016 Award

    It's non-degenerate, because the representing matrix [itex](\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)[/itex] (sorry for changing the convention within the thread, I'm used to the west-coast convention; if I'd change it, most probably I'd get confused myself).

    I don't know, what you mean by "subspace". In the usual sense of subspaces of a vector space, indeed there exist three-dimensional subspaces of time-like vectors. They can be defined as to be 3D subspaces Minkowski-perpendicular to an arbitrary time-like vector.

    Proof: Let [itex]n[/itex] a time-like unit vector, i.e., [itex]n \cdot n=\eta_{\mu \nu} n^{\mu} n^{\mu}=1[/itex]. Then you can build the following Lorentz transformation
    n^0 & \vec{n}^t \\
    \vec{n} & 1+\frac{n^0-1}{\vec{n}^2} \vec{n} \otimes \vec{n}
    in obvious notation. Then
    [tex]E_0'=n=\Lambda E_0, \quad E_j'=\Lambda E_j[/tex]
    build a new Minkowski-orthonormal system, and [itex]\text{span}(E_1',E_2',E_3')[/itex] is a 3D subspace containing only space-like vectors.

    On the other hand subspaces with only time-like vectors are necessarily one-dimensional, because suppose you have two linearly independent space-like vectors [itex]n_1[/itex] and [itex]n_2[/itex] you can construct the vector
    [tex] a=n_1-\frac{(n_1 \cdot n_2)}{n_2 \cdot n_2} n_2 \neq 0[/tex]
    within the subspace. It's Minkowski-perpendicular to the space-like vector [itex]n_2[/itex], but then it must be either spacelike or lightlike.

    To prove this you just take [itex]a \cdot n_2=0[/itex] and write it out in components
    [tex]a^0 \cdot n_2^0-\vec{a} \cdot \vec{n}=0 \; \Rightarrow \; |a^0|=\frac{|\vec{a} \cdot \vec{n}|}{n^0} \geq \frac{|\vec{a} \cdot \vec{n}|}{|\vec{n}|} \geq |\vec{a}|,[/tex]
    and thus
    [tex]a \cdot a=(a^0)^2-\vec{a}^2 \leq 0.[/tex]

    In contradiction to the assumption the subspace necessarily contains at least one spacelike or lightlike vector. In conclusion, subspaces that contain only time-like vectors are one-dimensional.

    is wrong as shown by the counter example by DaleSpam as are the other claims, or are their additional restrictions on the vectors [itex]E_j[/itex]?
  6. Oct 29, 2013 #5
    Sorry but i can't understand anything with physics notation, i think Lorentz metric be a non-degenerate because of the definition of degenere metric...I mean subspaces by subset of Lorentz space, timelike subspaces are space which has a non-degenerate metric on it and lightlike subspace are space which has a degenerate metric and different from {0}, but how can we guarantiate the lightlike space has a spacelike vector, our teach told this with tangent plane to the timeconi but i am not sure of the truth, the question is if any subspace contain null vector is this set must be a lightlike subspace?
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