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Poincare Group and Generators

  1. Dec 17, 2011 #1
    Alright, so I was reading Ryder and he defines the generator corresponding to [itex]a^{\alpha}[/itex] as the following

    [itex]X_{\alpha}[/itex]=[itex]\frac{\partial x'^{\mu}}{\partial a^{\alpha}}[/itex][itex]\frac{\partial}{\partial x^{\mu}}[/itex] ([itex]\alpha =1,...r[/itex]) for r-parameter group of transformations

    Now this makes sense for
    [itex]a^{\alpha}[/itex]=θ and we get Rotation...but he then says he applies it to "pure" Lorentz transformations:

    and I'm not even sure what parameter he's going after here ._. I feel really dumb asking, but what does he do to get from there to here:
    Kx=i[itex]\left(t\frac{\partial}{\partial x}+x\frac{\partial}{\partial t}\right)[/itex]

    It feels like he loses a factor of γ somewhere or something, too.

    Thanks for any and all help, this has been bothering me for a while.
  2. jcsd
  3. Dec 17, 2011 #2


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    The generator applies to infinitesimal transformations. For example a rotation is

    x' = x cos θ + y sin θ
    y' = y cos θ - x sin θ

    but you don't use the full form, you let θ be infinitesimal and write

    x' = x + θ y
    y' = y - θ x

    Likewise for a Lorentz transformation you use the infinitesimal form, in which γ ≈ 1.
  4. Dec 17, 2011 #3
    How is that infinitesimal? I understand what's done with rotation, but γ ≈ 1...wait the velocity is infinitesimal...and hence our parameter? So, for K_x we'd have say...
    and then when we take partial of say, the 0 (t') component we'd get x[itex]\frac{\partial}{\partial t}[/itex] :D

    I think I get what's going on now. . .

    This is fantastic, thank you.
  5. May 9, 2012 #4
    I don't understand how you get this. What would be the infinitesimal transformation on the rotation in this case?

    When I do the calculation, when I expand the rotation matrix
    cos θ -sinθ
    sin θ cosθ

    I get
    x' = x cos θ + y sin θ = x-ydθ
    y' = y cos θ - x sin θ = xθ +y
    Can you show me the calculation that you used?
    Also I'm having a hard time in finding the transformation of
    x' = x cos θ + y sin θ
    y' = y cos θ - x sin θ
    when θ is infinitesimal
    Last edited: May 9, 2012
  6. May 9, 2012 #5


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    Just use the Taylor expansion sin x =x +O(x^3) and cos x=1+O(x^2).
    In case of the Lorenz trafos expand in v: gamma=1+O(v^2)
  7. May 9, 2012 #6


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    You simply use the Taylor expansion wrt. the parameter of the Lie group. How to choose this parameter is in principle not so important since a Lie group is a manifold, where you can choose any parameter you like, but it's very convenient to choose parameters, which are additive for the Abelian one-parameter subgroups.

    For rotations, the natural one-parameter subgroups are given by the rotations around a fixed axis, and you can parametrize any rotation by the direction of its axis (two parameters for the corresponding normalized vector) and the rotation angle around this axis.

    In your example you consider the fundamental representation of the rotation group SO(3) and look at the rotations around the three axis,

    [tex]\vec{x}'=D_3(\varphi) \vec{x}=\begin{pmatrix}
    \cos \varphi & \sin \varphi & 0 \\
    -\sin \varphi & \cos \varphi &0 \\
    0 & 0 & 1

    You go over from the Lie group to the corresponding Lie algebra by expanding this law to first order around the group identity, which is here chosen in the usual way to correspond to [itex]\varphi=0[/itex]. Then you use Taylor's series for the trigonometrix functions up to first order,

    [tex]\cos \varphi=1+\mathcal{O}(\varphi^2), \quad \sin \varphi=\varphi + \mathcal{O}(\varphi^3).[/tex]

    This you plug into the equation above to find

    [tex]\vec{x}'=\vec{x}+\mathrm{i} J_z \varphi \vec{x}+\mathcal{O}(\varphi^2)=\begin{pmatrix}
    1 & \varphi & 0 \\
    -\varphi & 1 & 0 \\
    0 & 0 & 0

    This gives you

    [tex]J_z=-\mathrm{i} \begin{pmatrix}
    0 & 1 & 0 \\
    -1 & 0 & 0\\
    0 & 0 & 0
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