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Proof of a Lorentz Transformation

  1. Oct 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that the following is a Lorentz Transform:
    [itex] \Lambda _{j}^{i}=\delta _{j}^{i}+v^iv_j\frac{\gamma -1}{v^2}[/itex]
    [itex] \Lambda _{j}^{0}=\gamma v_j , \Lambda _{0}^{0}=\gamma , \Lambda _{0}^{i}=\gamma v^i[/itex]

    where [itex]v^2 =\vec{v}\cdot \vec{v}[/itex], and [itex]\delta _{j}^{i}[/itex] is the Kronecker Delta.
    2. Relevant equations
    [itex]\eta_{\mu\nu}=\eta_{\mu'\nu'}\Lambda_{\mu}^{\mu'} \Lambda_{\nu}^{\nu'}[/itex]
    [itex]\eta = \Lambda^T \eta \Lambda[/itex]

    3. The attempt at a solution
    I know how to go about proving a transform is a Lorentz transform, based on my "relevant equations", but I'm having a hard time setting the [itex]\Lambda[/itex] matrix up correctly. When I set up the matrix, I have terms in every cell, such as
    [itex]\Lambda_{1}^{1}=1+v^1 v_1 \frac{\gamma -1}{v^2}[/itex]
    and
    [itex]\Lambda_{1}^{2}=v^2 v_1 \frac{\gamma -1}{v^2}[/itex]

    and so on and so forth, but this feels wrong. I end up having to multiply two exceedingly complicated matrices along the way, which I know to be wrong (the professor hinted that excessive matrix multiplication was a sign you were doing the problem wrong.) How do I set things us? What I really want to know is, what is [itex]\Lambda_{j}^{i}[/itex]? How do I handle the vector indices (vi, vj)?
     
  2. jcsd
  3. Oct 17, 2012 #2
    I think you should do this, let [itex]g_{\mu\nu}=\eta_{\mu'\nu'}\Lambda_{\mu}^{\mu'} \Lambda_{\nu}^{\nu'}[/itex] then you sholud show that [itex]g_{\mu\nu}=\eta_{\mu\nu}[/itex]. This you can do calculating for each case [itex]g_{00},g_{0k}\, and\, g_{kl}[/itex] using
    [tex]g_{\mu\nu}=\eta_{\mu'\nu'}\Lambda_{\mu}^{\mu'} \Lambda_{\nu}^{\nu'}=\eta_{00}\Lambda_{\nu}^{0}\Lambda_{\mu}^{0}+\eta_{ij}\Lambda_{\nu}^{i}\Lambda_{\mu}^{j}[/tex]
    [tex]=-\Lambda_{\nu}^{0}\Lambda_{\mu}^{0}+\delta_{ij}\Lambda_{\nu}^{i} \Lambda_{\mu}^{j}=-\Lambda_{\nu}^{0}\Lambda_{\mu}^{0}+\Lambda_{\nu}^i\Lambda_{\mu}^i[/tex]
     
    Last edited: Oct 17, 2012
  4. Oct 17, 2012 #3
    Ok, that makes some sense to me. I'll give it a try, thank you!

    The only part I couldn't follow is where you came up with the [itex]\delta_{ij} \Lambda_{\nu}^{i} \Lambda_{\mu}^{j}[/itex]. Where did the delta come from? Sorry, I'm quite new at this sort of math.
     
  5. Oct 18, 2012 #4
    assuming that latin indices take on values 1,2,3 while greek indices 0,1,2,3 then [itex]\eta_{ik}=\delta_{ik}[/itex] while [itex]\eta_{00}=-1[/itex]
     
  6. Oct 19, 2012 #5
    Ah, right, thank you!
     
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