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Casimir operators and rest mass

  1. Oct 31, 2011 #1
    Penrose says in “Cycles of Time” that rest mass isn't exactly a Casimir operator of the de Sitter group, so a very slow decay of rest mass isn't out of the question in our universe.
    If rest mass is strictly conserved, should it be a Casimir operator of the de Sitter group?
    Decay of rest mass is crucial for Penrose's “conformal cyclic cosmology” theory, so how strong is this argument that rest mass isn't exactly a Casimir operator?
    thanks
    Laura
     
  2. jcsd
  3. Nov 8, 2011 #2

    Bill_K

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    Laura, I have the highest respect for Penrose, and his Cyclic Conformal Cosmology is a very interesting and innovative idea. Nevertheless it is difficult to support. There is no evidence for his belief that all rest mass will eventually decay. It's well established that the Poincare group is the symmetry group of the laws of physics. There is no reason why the symmetry group of the cosmology we live in (de Sitter group) should be the same, or have any influence on local physics.
     
  4. Nov 8, 2011 #3

    dextercioby

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    The de Sitter group has quadratic Casimir operators none of which is exactly the rest mass. So his statement is theoretically correct. However, there's still no experimental evidence that the laws be invariant to de Sitter and not Poincare.
     
  5. Nov 8, 2011 #4

    tom.stoer

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    the only indication for deSitter I am aware of is the accelerated expansion which could indicate a hidden deSitter invariance
     
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