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Orbits of the Lorentz group

  1. Sep 30, 2013 #1
    I want to determine the orbits of the proper orthochronous Lorentz group [itex] SO^{+}(1,3) [/itex].

    If I start with a time-like four-momentum [itex] p = (m, 0, 0, 0) [/itex]

    with positive time-component [itex]p^{0} = m > 0 [/itex],

    the orbit of [itex] SO^{+}(1,3) [/itex] in [itex] p [/itex] is given by:

    [tex] \mathcal{O}(p) \equiv \lbrace \Lambda p \mid \Lambda \in SO^{+}(1,3) \rbrace [/tex]

    Now the point is: how do you show that

    [tex] \mathcal{O}(p) = \lbrace q \mid q^{2} = m^{2}, q^{0} > 0 \rbrace [/tex] ?

    Essently, the question is: why does a Lorentz transformation [itex] \Lambda \in SO^{+}(1,3) [/itex] exist
    such that two four-vectors [itex] p [/itex] and [itex] q [/itex] with [itex] p^{2} = q^{2} = m^{2} [/itex] and [itex] p^{0}, q^{0} > 0 [/itex] are related via [itex] q = \Lambda p [/itex] ?

    In fact, it is possible to answer my question(s) by brute-force calculations. But I am searching for an "elegant way", e.g. with the help of group theory.

    I already searched in the literature, but in most cases it seems to be "trivial" for the authors
    and I see now explicit proof.

    Does anyone know of anything like that?
  2. jcsd
  3. Sep 30, 2013 #2
    Isn't that the very definition of the group SO+(1,3)?
  4. Sep 30, 2013 #3


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    I agree with the above. It's trivial according to those authors because it's the definition.
  5. Oct 1, 2013 #4
    No, it is not the definition of [itex] \mathrm{SO}^{+}(1,3) [/itex]. It is:

    [tex] \mathrm{SO}^{+}(1,3) = \lbrace \Lambda \in \mathrm{GL}(4, \mathbb{R}) \mid \Lambda^{t} \eta \Lambda = \eta, \mathrm{det} \, \Lambda = 1, \Lambda^{0} \, _{0} \geq 1 \rbrace [/tex]
    The elements of this group will not change [itex] p^{2} [/itex] and the sign of [itex] p^{0} [/itex] by definition.

    So we can say that the set
    [tex] \lbrace q \in \mathbb{M} \mid q^{2} = m^{2}, q^{0} > 0 \rbrace [/tex] remains invariant under Lorentz transformations.

    But this does not necessarily mean that for any two fourvectors [itex]p, q [/itex] with [itex]p^{2} = q^{2} = m^{2}[/itex] and [itex] p^{0}, q^{0} [/itex] there exists a transformation [itex] \Lambda \in \mathrm{SO}^{+}(1,3) [/itex] with [itex] q = \Lambda p [/itex]. In other words this would mean that [itex] \mathrm{SO}^{+}(1,3) [/itex] acts transitivly on [itex] \lbrace q \in \mathbb{M} \mid q^{2} = m^{2}, q^{0} > 0 \rbrace [/itex]. And you need to show this in order to proof that
    [tex] \mathcal{O}_{p} = \lbrace q \in \mathbb{M} \mid q^{2} = m^{2}, q^{0} > 0 \rbrace [/tex]

    Actually I also don't see how to do this without using brute force. I think it is not that trivial.
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