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Inverse transformations

  1. Oct 7, 2013 #1

    Mentz114

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    1. The problem statement, all variables and given/known data

    ##{\Lambda_c}^b## is a Lorentz transformation and ##{\Lambda^c}_b## is its inverse, so ##{\Lambda_c}^b {\Lambda^c}_b## gives an identity matrix.

    How can I write this, assuming it's possible, in terms of ##\delta##'s ?


    2. Relevant equations


    3. The attempt at a solution

    ##{\Lambda_a}^b {\Lambda_c}^a =\delta^b_c ##

    Wrong.
     
    Last edited: Oct 7, 2013
  2. jcsd
  3. Oct 7, 2013 #2

    WannabeNewton

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    ##\Lambda^{a}{}{}_{c}\Lambda^{c}{}{}_{b} = \delta^{a}{}{}_{b}##; it's just regular matrix multiplication.
     
  4. Oct 8, 2013 #3

    Mentz114

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    OK, thanks. I wasn't too far off.
    Can I use a ##\delta## to relabel these indexes ? Like

    ##{\Lambda_c}^b {\Lambda^d}_a {\delta_d}^c = {\delta^b}_a\ \Rightarrow\ {\Lambda_c}^b {\Lambda^d}_a {\delta_d}^c{\delta^d}_c = {\delta^b}_a {\delta^d}_c ##
     
    Last edited: Oct 8, 2013
  5. Oct 8, 2013 #4

    WannabeNewton

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    I can't parse what you wrote down because the indices are being summed over more than twice. But if you're asking if something like ##\Lambda^{a}{}{}_{c}\delta_{ab} = \Lambda_{bc}## is true then the answer is yes. Keep in mind that this is technically not index raising and lowering but rather just applying the definition of ##\delta_{ab}##. The two are different mathematical operations in spaces that aren't Euclidean.
     
    Last edited: Oct 8, 2013
  6. Oct 8, 2013 #5

    Mentz114

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    Yes, I edited while you were writing and took out the raising/lowering.

    I can also see now that what I wrote is not sensible. I'm actually trying to establish if for a geodesic ##u^a##

    ##{\Lambda_c}^b {\Lambda^d}_a u^c \nabla_b u_d = 0##

    Messing ##\delta##s looks like a red-herring.
     
  7. Oct 8, 2013 #6

    WannabeNewton

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    Is this an exercise from a textbook?
     
  8. Oct 8, 2013 #7

    Mentz114

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    No. It comes from

    [tex]
    \begin{align}
    \dot{v}_a &=\left({\Lambda_c}^b u^c \right)\nabla_b \left( {\Lambda^d}_a u_d \right)\\
    &= {\Lambda_c}^b {\Lambda^d}_a u^c \nabla_b u_d + {\Lambda_c}^b u^c u_d \nabla_b {\Lambda^d}_a
    \end{align}
    [/tex]
    I reason that if ##\Lambda## is an inertial boost then both terms must be zero. But the first term does not depend in any way on this distinction ( no derivatives of ##\Lambda##) so it must be identically zero. Am I wrong ?
     
  9. Oct 8, 2013 #8

    WannabeNewton

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    I really can't understand what you're doing. What is ##\dot{v}_a##? I assume since you're using ##\nabla## that this is a general curved space-time in which case ##\Lambda## must be a local Lorentz transformation which is different from a global Lorentz transformation of ##(\mathbb{R}^{4},\eta)##.
     
  10. Oct 8, 2013 #9

    Mentz114

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    Sorry, ##v^a = {\Lambda_b}^a u^b## and ##\dot{v}_a## is its proper acceleration.

    So, this is not the correct way to boost in curved spacetime ? I suppose I could take this into the frame field of ##u^\mu## in which case I can boost the basis, which is legitimate. Hmm.
     
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