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Representation of Lorentz algebra

  1. Aug 3, 2015 #1

    naima

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    i find here a representation of the Lorentz algebra.
    Starting from the matrix representation (with the ##\lambda## parameter) i see
    how one gets the matrix form of ##iJ_z##
    I am less comfortable with the ## -i y\partial_x + x \partial_y## notation
    Where does it come from? They say that it is a killing vector on ##R^4##
    I suppose that this is basic but as i read it in a physics paper i have not the complete background.
     
  2. jcsd
  3. Aug 3, 2015 #2

    fzero

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    Let's look at an infinitesimal rotation around the ##z##-axis, by a positive infinitesimal angle ##\alpha##. We can just draw a diagram of the x-y plane and rotate the axes a bit counter clockwise. Then we'll see that
    $$\begin{split}
    & x \rightarrow x' = x + \alpha y, \\
    & y \rightarrow y' = y - \alpha x, \\
    & z \rightarrow z' = z. \end{split}$$
    Now check that
    $$\begin{split}
    & - \alpha ( x \partial_y - y \partial_x) x = \alpha y, \\
    & - \alpha ( x \partial_y - y \partial_x) y = - \alpha x. \end{split}$$
    Therefore
    $$ \delta x^k = i \alpha J_z x^k$$
    if we define
    $$ J_z = i ( x \partial_y - y \partial_x).$$

    Further, we can show that ##J_z## is a Killing vector for the Minkowski metric. The Killing condition is
    $$ (J_z)_{\mu ; \nu} + (J_z)_{\nu ; \mu} =0.$$
    Since the Minkowski metric is flat, the covariant derivatives reduce to ordinary derivatives and the Killing condition becomes:
    $$0 = (J_z)_{x ,y} + (J_z)_{y,x} = \partial_y ( -i y) + \partial_x ( i x),$$
    which is obviously satisfied.

    One can derive the rest of the generators by examining the other infinitesimal rotations and boosts and then verifying that the Killing equation is satisfied as I did above. Or else one can just compute the rest of the Killing vectors and then see that they have an interpretations as generators.
     
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