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Orthochronous lorentz transf.

  1. Aug 2, 2006 #1
    hi,
    I've find ( Weinberg: one particle state ) this :
    .. functions of p^mu that are invariant by orthocronous lorentz transformation are square p^2 and ( for p^2 <=0 ) the sign of p^0 ...
    for the square it's ok, but I don't understand why the sign of p^0 is "invariant" !! and is it realy a "function" of p^mu ??
     
  2. jcsd
  3. Aug 2, 2006 #2

    robphy

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    What does "orthochronous" mean?
    Can you express the "component of a vector" in terms of the vector itself?
     
  4. Aug 2, 2006 #3

    George Jones

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    p^2 < 0 means that p is a timelike 4-vector. If p is a timelike 4-vector then: p^0 > 0 means that p is future-directed; p^0 < 0 means that p is past-directed. Orthochronous Lorentz transformations preserve the time sense of timelike 4-vectors, i.e., orthochronous Lorentz transformations take future-directed timelike 4-vectors to future-directed timelike 4-vectors and past-directed timelike 4-vectors to past-directed timelike 4-vectors.

    In other words, orhtochronous Lorentz transformations don't switch the past and the future.
     
  5. Aug 2, 2006 #4

    robphy

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    My first question was meant to be a hint.
     
  6. Aug 3, 2006 #5

    George Jones

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    Yes, I probably shoud have used your pedagogically better Socratic approach.

    Didn't see your post before I posted.
     
  7. Aug 3, 2006 #6
    thanks George Jones but if you can, write all this in equations. I mean :
    - Why orthochronous Lorentz transformations preserve the time sense of timelike 4-vectors ( I know that Lambda(0,0) > 0 et det = 1)
     
  8. Aug 3, 2006 #7
    Because that's their defintion. You can devide the Lorentz group into two disjoint sets - those transformations which do preserve the sign of the timelike component, and those which reverse it (there is no other possibility). We call the first group "orthochronous", and the second group "nonorthochronous"; the etymology is "ortho", 'correct', and "chronous', time, e.g., "correct time".

    You can verify (using an additional Lorentz transformation) that if a lorentz transformation [tex]\Lambda[/tex] preserves the sign of the time component of any one timelike vector, it does so for all timelike vectors, and vice-versa; which is why it's a useful definition.
     
  9. Aug 3, 2006 #8
    thanks Rach3
     
  10. Aug 8, 2008 #9
    Is the following argument valid to identify which of the [tex]\mbox{O(1,3)}[/tex] subsets is indeed a Subgroup?

    Again, using Weinberg, we split [tex]\mbox{O(1,3)}[/tex] into the following subsets

    [tex]L^{\uparrow}_{+} = \{ \mbox{O(1,3)} | \Lambda^{0}_{0} \geq+1 , \mbox{det{\Lambda}}=+1 \}[/tex]

    [tex]L^{\uparrow}_{-} = \{ \mbox{O(1,3)} | \Lambda^{0}_{0} \geq+1 , \mbox{det{\Lambda}}=-1 \}[/tex]

    [tex]L^{\downarrow}_{+} = \{ \mbox{O(1,3)} | \Lambda^{0}_{0} \leq-1 , \mbox{det{\Lambda}}=+1 \}[/tex]

    [tex]L^{\downarrow}_{-} = \{ \mbox{O(1,3)} | \Lambda^{0}_{0} \leq-1 , \mbox{det{\Lambda}}=-1 \}[/tex]

    This takes into account all possibilities.

    The only subsets that could possibly be groups would be the identity connected components so that [tex]L^{\uparrow}_{-}[/tex] and [tex]L^{\downarrow}_{-}[/tex] can be immediately ruled out as the identity couldn't possible be in these sets under matrix multiplication

    The only two subsets that might be subgroups are [tex]L^{\uparrow}_{+}[/tex] and [tex]L^{\downarrow}_{+}[/tex].

    The Proper Orthochronous, and the Proper non-orthochronous subsets. However, [tex]\Lambda^{0}_{0}[/tex] needs to be allowed to take on the value 1.

    This would rule out [tex]\Lambda^{0}_{0}\leq -1[/tex].

    So the only possibility of subgroup is the Proper Orthochronous subset.

    Now proving that this indeed is a subgroup is another few lines. But my post is purely about my process of elimination.
     
  11. Aug 8, 2008 #10

    George Jones

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    The identity has [itex]\Lambda^{0}_{0} = 1[/itex], so [itex]L^{\downarrow}_{+}[/itex] also doesn't contain the identity.
     
  12. Aug 8, 2008 #11
    Thank you George. I was unsure of my reasoning. However, I did mention implicitly that [itex]L^{\downarrow}_{+}[/itex] doesn't contain the identity when I said:

    Thank you again. Much appreciated.
     
  13. Dec 6, 2008 #12
    Hallo! I don't understand why lambda(0,0) needs to be allowed to take on the value 1. The identity has all lambda (i,i) =1. Why only lambda(0,0) must be =1?
     
  14. Dec 6, 2008 #13

    George Jones

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    I don't think I understand your question. The identity is a Lorentz transformation that has [itex]\Lambda^0_0=1[/itex], not a Lorentz transformation that has only [itex]\Lambda^0_0=1[/itex].
     
  15. Dec 6, 2008 #14
    My question is:
    The only subsets that could possibly be groups would have det=1 becouse the identity has det=1. But why I must impose also that lambda(0,0)=1 ,for all the trasformation of the subsets, to form the identity connected component of the lorentz group?
    why the identity component of the Lorentz group is the set of all Lorentz transformations that have det=1 and also lambda(0,0)=1?
     
  16. Dec 6, 2008 #15

    George Jones

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    I think we're talking past each other.

    No one has said that you should impose the condition [itex]\Lambda^0_0 = 1[/itex] on all transformations [itex]|\Lambda[/itex] in the component of the Lorentz group connected to the identity. In particular, this is not what is said in posts #10 and #11.
    It isn't. Clearly, there are transformations [itex]\Lambda[/itex] in the component of the Lorentz group connected to the identity that have [itex]\Lambda^0_0 \neq 1[/itex] (e.g., boosts).
     
  17. Dec 6, 2008 #16
    Sorry , lamb(0,0)>=1, not lamb(0,0)=1. the trasformations in the subset 1 (post 9) have all lamb(0,0)>=1.
     
  18. Dec 8, 2008 #17

    Fredrik

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    It's easy to see both that the 00-component must be either ≥1 or ≤-1 and that the Lorentz transformations with 00-component ≥1 is a subgroup if we first identify which components correspond to the velocity of the Lorentz transformation. I'm writing all indices downstairs and using a -+++ metric.

    [itex]\Lambda[/itex] must map the 0 axis to the line

    [tex]\tau\mapsto\tau\begin{pmatrix}1\\ -v_1\\ -v_2\\ -v_3\end{pmatrix}[/tex]

    and

    [tex]\Lambda\begin{pmatrix}1\\ 0\\ 0\\ 0\end{pmatrix}=\begin{pmatrix}\Lambda_{00}\\ \Lambda_{10}\\ \Lambda_{20}\\ \Lambda_{30}\end{pmatrix}=\Lambda_{00}\begin{pmatrix}1\\ \Lambda_{10}/\Lambda_{00}\\ \Lambda_{20}/\Lambda_{00}\\ \Lambda_{30}/\Lambda_{00}\end{pmatrix}[/tex]

    so we must have

    [tex]v_i=-\Lambda_{i0}/\Lambda_{00}[/tex]

    Now consider the 00-component of the condition [itex]\Lambda^T\eta\Lambda=\eta[/itex]:

    [tex]-1=\eta_{00}=\Lambda^T_{0\mu}\eta_{\mu\nu}\Lambda_{\nu 0}=-\Lambda_{00}^2+\Lambda_{i0}\Lambda_{i0}=-\Lambda_{00}^2(1-\vec v^2)[/tex]

    [tex]\Lambda_{00}^2=\frac{1}{1-\vec v^2}\geq 1[/tex]

    This proves that the 00-component is either ≥1 or ≤-1.

    Proving that Lorentz transformations with the 00-component ≥1 is a subgroup isn't hard either if we use that

    [tex]\Lambda^{-1}_{i0}=-\Lambda_{0i}[/tex]

    and

    [tex]\Lambda^{-1}_{00}=\Lambda_{00}[/tex]

    Both results follow from the formula [itex]\Lambda^{-1}=\eta\Lambda^T\eta[/itex], which tells us that the inverse transformation can be obtained by transposing [itex]\Lambda[/itex] and then changing the sign of the zeroth row and the zeroth column.

    [tex](\bar\Lambda\Lambda)_{00}=\bar\Lambda_{00}\Lambda_{00}+\bar\Lambda_{0i}\Lambda_{i0}=\bar\Lambda_{00}\Lambda_{00}\bigg(1-\frac{\bar\Lambda^{-1}_{i0}}{\bar\Lambda^{-1}_{00}}\frac{\Lambda_{i0}}{\Lambda_{00}}\bigg)=\bar\Lambda_{00}\Lambda_{00}(1-\vec u\cdot\vec v)\geq 0[/tex]

    where [itex]\vec u[/itex] is the velocity associated with [itex]\bar\Lambda^{-1}[/itex]. We wanted to show that [itex](\bar\Lambda\Lambda)_{00}[/itex] is ≥1, but it's sufficient (and easier) to show that it's ≥0 since we already know that it must be ≥1 or ≤-1. The ≥0 must hold because the norms of the velocity vectors are <1.
     
    Last edited: Dec 8, 2008
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