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A Lorentz invariance of the Heaviside function

  1. Apr 1, 2016 #1
    Consider the Heaviside function ##\Theta(k^{0})##.

    This function is Lorentz invariant if ##\text{sign}\ (k^{0})## is invariant under a Lorentz transformation.

    I have been told that only orthochronous Lorentz transformations preserve ##\text{sign}\ (k^{0})## under the condition that ##k## is a time-like vector.

    I would like to prove this explicitly, to convince myself of the fact. How do I start?
     
  2. jcsd
  3. Apr 1, 2016 #2

    vanhees71

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    Sure, the "direction of time" must be preserved under the Lorentz transformation for the expression ##\Theta(k^0)## to be conserved. Thus you are restricted to Lorentz-transformation matrices ##{\Lambda^{\mu}}_{\nu}## for which ##{\Lambda^{0}}_{0} \geq 1##. Now for a non-like vector ##k^{\mu}##, i.e., ##(k^0)^2-\vec{k}^2 > 0##, i.e., (because of ##k^0>0## ) ##k^0>|\vec{k}|## you have
    $$k^{\prime 0}= {\Lambda^{0}}_{\mu} k^{mu} = {\Lambda^{0}}_0 k^{0} + \vec{\Lambda} \cdot \vec{k} \geq {\Lambda^{0}}_0 k^{0} -|\vec{\Lambda}||\vec{k}|,$$
    where
    $$\vec{\Lambda}=({\Lambda^{0}}_j)|_{j \in \{1,2,3 \}}.$$
    Further for the Lorentz transformation you have
    $$\left ({\Lambda^{0}}_0 \right)^2 - \vec{\Lambda}^2=1 \; \Rightarrow {\Lambda^{0}}_{0}=+\sqrt{1+\vec{\Lambda}^2}>|\vec{\Lambda}|.$$
    Thus we have
    $$k^{\prime 0} \geq {\Lambda^0}_0 (k^0-|\vec{k}|) > 0.$$
    In the same way you can show that if ##k^0<0## also ##k^{\prime 0}<0##, and that was to be shown.
     
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