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Why SO+(1,3) is connected?

  1. Jun 18, 2015 #1
    Why is the group of orthochronous proper Lorentz transformations connected?
     
  2. jcsd
  3. Jun 18, 2015 #2

    Dale

    Staff: Mentor

    It isn't so much that the group is connected. It is that only the connected sub group is physically interesting.

    The physically interesting parts of the group that represent boosts. A boost of 0 should physically give you the same frame, so we want the connected sub group which contains that feature. Otherwise a boost of 0 would do something weird like flip time.
     
  4. Jun 18, 2015 #3

    PeterDonis

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    2016 Award

    Staff: Mentor

    Think about what "connected" means: heuristically, it means that any transformation in the group can be turned into any other by a continuous series of "moves". As DaleSpam said, physically, we want the identity (the transformation that does nothing) to be in the group, so we are interested in Lorentz transformations that can be turned into the identity by a continuous series of moves. For example, a boost can be turned into the identity by continuously changing its velocity ##v## down to zero. Or a rotation can be turned into the identity by continuously changing the rotation angle ##\theta## down to zero.

    Contrast the above with, for example, a transformation that flips the handedness of the spatial axes, or flips the direction of time. There is no way to continuously change a transformation like that into the identity; somewhere along the line you have to do a discrete "flip" of the handedness of the spatial axes or the direction of time. So transformations like that are not connected to the identity. The "proper orthochronous" Lorentz transformations are then simply the ones that don't do either of those things: "proper" means the handedness of the spatial axes isn't flipped, and "orthochronous" means the direction of time isn't flipped.
     
  5. Jun 19, 2015 #4

    samalkhaiat

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    Science Advisor

    Because, we can show that any [itex]g \in SO_{+}^{\uparrow} (1,3)[/itex] can be decomposed as [tex]g = \Lambda (R_{2}) \Lambda(L_{x}) \Lambda (R_{1}) ,[/tex] where [itex]\Lambda (R_{1})[/itex] and [itex]\Lambda (R_{2})[/itex] are spatial rotations and [itex]\Lambda (L_{x})[/itex] a Lorentz boost in the [itex]x^{1}[/itex] direction.
     
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