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Orthogonality of the coefficients of the Lorentz transformation

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data
    For flat spacetime the coefficients of the Lorentz transformation are defined as

    [itex] \alpha^{\nu}_{\mu} = \frac{\partial x^{' \nu}}{\partial x^{\mu}} [/itex]

    Whereas the Lorentz transformation is

    [itex] \begin{pmatrix} x_1' \\ x_2' \\ x_3' \\ x_4' \end{pmatrix} = \begin{pmatrix} \alpha_0^0 & \alpha_1^0 & \alpha_2^0 & \alpha_3^0 \\ \alpha_0^1 & \alpha_1^1 & \alpha_2^1 & \alpha_3^1 \\ \alpha_0^2 & \alpha_1^2 & \alpha_2^2 & \alpha_2^3 \\ \alpha_0^3 & \alpha_1^3 & \alpha_2^3 & \alpha_3^3 \end{pmatrix} \begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} [/itex]

    Here's the problem, show that the coefficients are orthogonal. That is

    [itex] \alpha^{\nu}_{\lambda} \alpha^{\lambda}_{\mu} = \delta^{\nu}_{\mu} [/itex]

    The last term is the Kronicker-Delta symbol


    3. The attempt at a solution

    Let's define some arbitrary four vector [itex] \mathbf{A} [/itex]

    [itex] \delta^{\nu}_{\mu} A^{\mu} = \alpha^{\nu}_{\lambda} \alpha^{\lambda}_{\mu} A^{\mu} = \frac{\partial x^{' \nu}}{\partial x^{\lambda}} \frac{\partial x^{' \lambda}}{\partial x^{\mu}} A^{\mu} [/itex]

    [itex] \delta^{\nu}_{\mu} A^{\mu} = \frac{\partial x^{' \lambda}}{\partial x^{\lambda}} A^{' \nu} [/itex]

    Now according to Einsteins summation rule we should have

    [itex] \mathbf{A} = \sum_{ \nu = 0}^{3} \left( \frac{\partial x^{' 0}}{\partial x^{0}} A^{' \nu} + \frac{\partial x^{' 1}}{\partial x^{1}} A^{' \nu} + \frac{\partial x^{' 2}}{\partial x^{2}} A^{' \nu} + \frac{\partial x^{' 3}}{\partial x^{3}} A^{' \nu} \right) [/itex]

    This is where I get stuck. I have no idea how the sum on the right side should be equal to our 4-vector [itex] \mathbf{A} [/itex].

    Any ideas?
     
  2. jcsd
  3. Oct 21, 2011 #2
    please note that the transpose of [tex] \alpha^{\mu}_{\nu} = \frac{\partial x'^{\mu}}{\partial x^{\nu}} [/tex] is [tex] (\alpha^T)^{\mu}_{\nu} = \frac{\partial x^{\mu}}{\partial x'^{\nu}} [/tex] and you are supposed to show that

    [tex] \alpha^{\mu}_{\nu} (\alpha^T)^{\nu}_{\sigma} = \delta^{\mu}_{\sigma} [/tex]
     
  4. Oct 21, 2011 #3
    This I have solved with your excellent advice my dear sir

    [tex] \mathbf{A} = \delta^{\nu}_{\mu} A^{\mu} = \alpha^{\nu}_{\lambda} (\alpha^T)^{\lambda}_{\mu} A^{\mu} = \frac{\partial x^{\nu}}{\partial x^{' \lambda} } \frac{\partial x^{' \lambda} }{\partial x^{\mu}}A^{\mu} = \frac{\partial x^{\nu}}{\partial x^{\mu}} A^{\mu} = \mathbf{A} [/tex]


    But this is of course different from my original problem where there was no transpose. Could it be that my textbook is wrong?
     
  5. Oct 21, 2011 #4

    dextercioby

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    The standard proof is to start off with the requirement that the norm of a 4-vector be a scalar/invariant wrt a Lorentz transformation. So x'2=x2.
     
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