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Successive boosts

  1. Nov 2, 2006 #1
    Hi Friends,

    From time to time I have seen transformations replaced by a succession of infinitesimal transformations. The end result ends up being an exponent.

    My knowledge of this is vague and I would like to look into it more seriously. Particularly I am interested in describing the lorentz transformation as a sequence of infinitesimal boosts.

    Can someone point me to the first thing I need read to understand this?

    Thanks
     
  2. jcsd
  3. Nov 2, 2006 #2
    Hi, ActionIntegral,

    I'll be more than happy to take a stab at it! Just expand the lorentz transform in a taylor series in v about v=0 keeping only the first order.

    You can raise this operator to the nth power. Somehow this becomes an exponent but I haven't figured that part out yet.
     
  4. Nov 2, 2006 #3
    That's ok - I wouldn't want you to do all the work for me! :smile:
     
  5. Nov 2, 2006 #4

    pervect

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    What I think you're trying to do is find the "generator" of the Lorentz group.

    http://en.wikipedia.org/wiki/Generating_set_of_a_group

    But you want to do this for a continuous group. The Lorentz transformation is a "group" in the abstract algebra sense with the group operation being the successive application of transforms, because the result is associative f x (g x h) = (f x g) x h, and has an inverse. However, the Lorentz group is an infinite group. This is called a "Lie group".

    So what you need to do is to read up on Lie groups (specifically the generators of Lie groups). Or see if you can find a mathemetician.

    Hope this helps.
     
  6. Nov 2, 2006 #5

    robphy

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    Let [tex]\Theta=\left[ \begin {array}{cc} 0&1\\\noalign{\medskip}1&0\end {array} \right]\theta[/tex].
    Formally, write [tex]\exp(\Theta)=I+\Theta+\Theta^2/2!+\Theta^3/3!+\ldots[/tex]. Do you recognize [tex]\exp(\Theta)[/tex]?
     
  7. Nov 2, 2006 #6
    Both good answers - i'll read up on it. thanks to all three of you!
     
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