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Investigations into the infinitesimal Lorentz transformation

  1. Apr 24, 2016 #1
    1. The problem statement, all variables and given/known data

    A Lorentz transformation ##x^{\mu} \rightarrow x'^{\mu} = {\Lambda^{\mu}}_{\nu}x^{\nu}## is such that it preserves the Minkowski metric ##\eta_{\mu\nu}##, meaning that ##\eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\mu\nu}x'^{\mu}x'^{\nu}## for all ##x##. Show that this implies that ##\eta_{\mu\nu} = \eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}##.

    Use this result to show that an infinitesimal transformation of the form ##{\Lambda^{\mu}}_{\nu}={\delta^{\mu}}_{\nu}+{\omega^{\mu}}_{\nu}## is a Lorentz transformation when ##\omega^{\mu\nu}## is antisymmetric: i.e. ##\omega^{\mu\nu}=-\omega^{\nu\mu}##.

    Write down the matrix form for ##{\omega^{\mu}}_{\nu}## that corresponds to a rotation through an infinitesimal angle ##\theta## about the ##x^{3}##-axis.

    Do the same for a boost along the ##x^{1}##-axis by an infinitesimal velocity ##v##.

    2. Relevant equations

    3. The attempt at a solution


    ##\eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\mu\nu}x'^{\mu}x'^{\nu}##

    ##\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\mu\nu}({\Lambda^{\mu}}_{\rho}x^{\rho})({\Lambda^{\mu}}_{\sigma}x^{\sigma})##

    ##\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\rho\sigma}({\Lambda^{\rho}}_{\mu}x^{\mu})({\Lambda^{\sigma}}_{\nu}x^{\nu})##

    ##\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\rho\sigma}{\Lambda^{\rho}}_{\mu}{\Lambda^{\sigma}}_{\nu}x^{\mu}x^{\nu}##

    ##\implies \eta_{\mu\nu}=\eta_{\rho\sigma}{\Lambda^{\rho}}_{\mu}{\Lambda^{\sigma}}_{\nu}##


    Am I correct so far?
     
  2. jcsd
  3. Apr 24, 2016 #2

    TSny

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    Yes, I think so. But someone might want to see explicitly how you get the last line from the next to last line.
     
  4. Apr 24, 2016 #3
    Well, the next to last line consists of a sum of terms ##\eta_{\mu\nu}x^{\mu}x^{\nu}## and ##\eta_{\rho\sigma}{\Lambda^{\rho}}_{\mu}{\Lambda^{\sigma}}_{\nu}x^{\mu}x^{\nu}## with all possible combinations of values ##0,1,2,3## for the indices ##\mu## and ##\nu##.

    However, the last line only contains the terms ##\eta_{\mu\nu}## and ##\eta_{\rho\sigma}{\Lambda^{\rho}}_{\mu}{\Lambda^{\sigma}}_{\nu}## for a specific value of ##\mu## and ##\nu##.

    I thought that this was obvious, so I decided to skip the explanation. Isn't my reasoning sound, though?
     
  5. Apr 24, 2016 #4

    TSny

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    If I have two matrices ##A_{\mu \nu}## and ##B_{\mu \nu}## that satisfy ##A_{\mu \nu}x^{\mu}x^{\nu} = B_{\mu \nu}x^{\mu}x^{\nu}## for all possible ##x^{\mu}##, can I conclude that ##A_{\alpha \beta} = B_{\alpha \beta}## for all ##\alpha## and ##\beta##?
     
  6. Apr 24, 2016 #5
    I think so, yes.
     
  7. Apr 24, 2016 #6

    TSny

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    Try to show explicitly that ##A_{12} = B_{12}##.
     
  8. Apr 24, 2016 #7
    Oh wait. The correct relation is ##A_{12}+A_{21}=B_{12}+B_{21}##.
     
  9. Apr 24, 2016 #8

    TSny

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    Right.
     
  10. Apr 24, 2016 #9
    Well! In that case, I need to rewrite my solution:

    ##\eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\mu\nu}x'^{\mu}x'^{\nu}##

    ##\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\mu\nu}({\Lambda^{\mu}}_{\sigma}x^{\sigma})({\Lambda^{\nu}}_{\tau}x^{\tau})##

    ##\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\sigma\tau}({\Lambda^{\sigma}}_{\mu}x^{\mu})({\Lambda^{\tau}}_{\nu}x^{\nu})##

    ##\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}x^{\mu}x^{\nu}##

    ##\implies \eta_{\mu\nu}+\eta_{\nu\mu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}+\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\nu}{\Lambda^{\tau}}_{\mu}##, since ##x^{\mu}x^{\nu}=x^{\nu}x^{\mu}##

    ##\implies \eta_{\mu\nu}+\eta_{\mu\nu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}+\eta_{\tau\sigma}{\Lambda^{\tau}}_{\mu}{\Lambda^{\sigma}}_{\nu}##, since the metric tensor is symmetric, i.e. ##\eta^{\mu\nu}=\eta^{\nu\mu}##

    ##\implies \eta_{\mu\nu}+\eta_{\mu\nu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}+\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}##

    ##\implies 2 \eta_{\mu\nu}= 2 \eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}##

    ##\implies \eta_{\mu\nu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}##

    Is my solution correct?
     
    Last edited: Apr 24, 2016
  11. Apr 24, 2016 #10

    TSny

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    Yes. The symmetry of the metric tensor is used here.
     
  12. Apr 24, 2016 #11
    Next, I need to use the result ##\eta_{\mu\nu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}## to show that an infinitesimal transformation of the form ##{\Lambda^{\mu}}_{\nu}={\delta^{\mu}}_{\nu}+{\omega^{\mu}}_{\nu}## is a Lorentz transformation when ##\omega^{\mu\nu}## is antisymmetric: i.e. ##\omega^{\mu\nu}=-\omega^{\nu\mu}##:

    ##\eta_{\mu\nu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}##

    ##\implies \eta_{\mu\nu}=\eta_{\sigma\tau}({\delta^{\sigma}}_{\mu}+{\omega^{\sigma}}_{\mu})({\delta^{\tau}}_{\nu}+{\omega^{\tau}}_{\nu})##

    ##\implies \eta_{\mu\nu}=\eta_{\sigma\tau}({\delta^{\sigma}}_{\mu}{\delta^{\tau}}_{\nu}+{\delta^{\sigma}}_{\mu}{\omega^{\tau}}_{\nu}+{\omega^{\sigma}}_{\mu}{\delta^{\tau}}_{\nu}+{\omega^{\sigma}}_{\mu}{\omega^{\tau}}_{\nu})##

    ##\implies \eta_{\mu\nu}=\eta_{\mu\nu}+\eta_{\mu\tau}{\omega^{\tau}}_{\nu}+\eta_{\nu\sigma}{\omega^{\sigma}}_{\mu}+\eta_{\sigma\tau}{\omega^{\sigma}}_{\mu}{\omega^{\tau}}_{\nu}##

    ##\implies \eta_{\mu\tau}{\omega^{\tau}}_{\nu}+\eta_{\nu\sigma}{\omega^{\sigma}}_{\mu} = 0##, where we neglect the term ##\eta_{\sigma\tau}{\omega^{\sigma}}_{\mu}{\omega^{\tau}}_{\nu}## because it is of second order in the infinitesimal ##{\omega^{\mu}}_{\nu}##.

    ##\implies \omega_{\mu\nu}+\omega_{\nu\mu}=0##

    ##\implies \omega_{\mu\nu}=-\omega_{\nu\mu}##

    ##\implies \omega^{\mu\nu}=-\omega^{\nu\mu}##

    so that ##\omega^{\mu\nu}## is antisymmetric.

    Is my solution correct?
     
  13. Apr 24, 2016 #12

    TSny

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    Yes.
     
  14. Apr 24, 2016 #13
    Finally, the remaining parts of the problem:

    The matrix form for ##{\Lambda^{\mu}}_{\nu}## that corresponds to a rotation through a finite angle ##\theta## about the ##x^{3}##-axis is given by

    ## {\Lambda^{\mu}}_{\nu} =
    \left( \begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & \text{cos}\ \theta & -\text{sin}\ \theta & 0 \\
    0 & \text{sin}\ \theta & \text{cos}\ \theta & 0 \\
    0 & 0 & 0 & 1 \end{array} \right) =
    \left( \begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 \end{array} \right) +
    \left( \begin{array}{cccc}
    0 & 0 & 0 & 0 \\
    0 & 0 & -\theta & 0 \\
    0 & \theta & 0 & 0 \\
    0 & 0 & 0 & 0 \end{array} \right) + \cdots ={\delta^{\mu}}_{\nu}+{\omega^{\mu}}_{\nu}+\cdots ,
    ##

    where we used ##\text{sin}\ \theta = \theta - \frac{\theta^{3}}{3!}+\cdots## and ##\text{cos}\ \theta = 1 + \frac{\theta^{2}}{2!}+\cdots ## and we only kept terms up to linear order in ##\theta## in the expansion of ##{\Lambda^{\mu}}_{\nu}##,

    so the matrix form for ##{\omega^{\mu}}_{\nu}## that corresponds to a rotation through an infinitesimal angle ##\theta## about the ##x^{3}##-axis is given by

    ##{\omega^{\mu}}_{\nu}=
    \left( \begin{array}{cccc}
    0 & 0 & 0 & 0 \\
    0 & 0 & -\theta & 0 \\
    0 & \theta & 0 & 0 \\
    0 & 0 & 0 & 0 \end{array} \right).##



    The matrix form for ##{\Lambda^{\mu}}_{\nu}## that corresponds to a boost along the ##x^{1}##-axis by a finite velocity ##v## is given by

    ## {\Lambda^{\mu}}_{\nu} =
    \left( \begin{array}{cccc}
    \gamma & -\gamma v & 0 & 0 \\
    -\gamma v & \gamma & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 \end{array} \right) =
    \left( \begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 \end{array} \right) +
    \left( \begin{array}{cccc}
    0 & -v & 0 & 0 \\
    v & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \end{array} \right) + \cdots ={\delta^{\mu}}_{\nu}+{\omega^{\mu}}_{\nu}+\cdots ,
    ##

    where we used ##\gamma =\frac{1}{\sqrt{1-v^{2}}}=(1-v^{2})^{-1/2}=1+\frac{v^{2}}{2}+\dots ## and ##\gamma v =v\frac{1}{\sqrt{1-v^{2}}}=v(1-v^{2})^{-1/2}=v+\frac{v^{3}}{2}+\dots ## and we only kept terms up to linear order in ##v## in the expansion of ##{\Lambda^{\mu}}_{\nu}##,

    so the matrix form for ##{\omega^{\mu}}_{\nu}## that corresponds to a boost along the ##x^{1}##-axis by an infinitesimal velocity ##v## is given by

    ##{\omega^{\mu}}_{\nu}=
    \left( \begin{array}{cccc}
    0 & -v & 0 & 0 \\
    v & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \end{array} \right).##


    Is this solution correct?
     
  15. Apr 24, 2016 #14

    TSny

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    Check the sign in the first column of the last matrix on the right side.
     
  16. Apr 25, 2016 #15
    It is a typo. We should instead have the following:

    ##{\Lambda^{\mu}}_{\nu} =
    \left( \begin{array}{cccc}
    \gamma & -\gamma v & 0 & 0 \\
    -\gamma v & \gamma & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 \end{array} \right) =
    \left( \begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 \end{array} \right) +
    \left( \begin{array}{cccc}
    0 & -v & 0 & 0 \\
    -v & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \end{array} \right) + \cdots ={\delta^{\mu}}_{\nu}+{\omega^{\mu}}_{\nu}+\cdots ##

    so that

    ##{\omega^{\mu}}_{\nu}=
    \left( \begin{array}{cccc}
    0 & -v & 0 & 0 \\
    -v & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \end{array} \right)##.

    I believe everything else is correct, isnt it?
     
  17. Apr 25, 2016 #16

    TSny

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    Looks good.
     
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