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I Lorentz group

  1. Mar 24, 2016 #1
    I don't understand why we can write the elements of the lorentransformation in the form

    ## {\Lambda}^{\mu}\:_{\nu} = [exp(-\frac{i}{2}{\omega}^{\rho\sigma}M_{\rho\sigma})]^{\mu}\:_{\nu} ##

    I know that we can write it in the form

    ## {\Lambda} = exp(t\Theta) ##

    where
    ## \Theta ##
    are elements of the Lie algebra
     
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  3. Mar 24, 2016 #2

    haushofer

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    Well, the contraction of omega with M indicates a linear combination of M's. These M's span the algebra. So the group element you write there is general. Your second expression seems like you pick one particular element from the algebra, Theta, and multiply it with one parameter t.

    So your first expression is just the most general group element you can write when the M's span your algebra. In your second expression you apparently pick one particular M, call it Theta, and you pick one particular omega, and call it t.

    Maybe this comparison helps: I can always choose coordinates in R3 such that a vector is written as

    V = (0,0,c)

    i.e. has only a component along the z-axis, which I call c. But in general such a vector has components

    V= (a,b,c).
     
  4. Mar 24, 2016 #3
    ok, so the ## M_{\rho\sigma} ## represents a matrix and not an element of the matrix? i.e., ## M_{10} ## is a matrix? But where is the time parameter in the first expression?
     
  5. Mar 24, 2016 #4

    haushofer

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    Yes. Your M is a Lorentzboost in the x-direction (M_{10} ). For rotations, omega is an angle; for boosts, omega is the rapidity,

    https://en.m.wikipedia.org/wiki/Rapidity

    I'm not sure where you expect time to appear.
     
  6. Mar 24, 2016 #5

    samalkhaiat

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    If the Lie algebra [itex]\mathcal{L}(G)[/itex] can be exponentiated to give the Lie group [itex]G[/itex], then
    [tex]G = e^{\mathcal{L}(G)} .[/tex]
    So,
    [tex]\mathcal{L}(SO(1,3)) = - \frac{i}{2} \omega^{\mu\nu}M_{\mu\nu} \equiv X .[/tex]

    [itex]e^{t \mathcal{L}(G)} \equiv e^{tX}[/itex] is a one-parameter subgroup of [itex]G[/itex].
     
  7. Mar 25, 2016 #6

    haushofer

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    Maybe this helps: the group elements only depend on parameters which define the angle of rotation or rapidity. They don't depend on spacetime coordinates (!). In the fundamental representation the group elements act on the spacetime coordinates, giving the familiar Lorentz transformations of the spacetime coordinates.

    Maybe you're confusing the fundamental representation on the coordinates with more general representations. E.g., a Lorentz transfo on a vector field or spinor does not contain spacetime coordinates.
     
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