Casimir operators and rest mass

Click For Summary
Penrose's assertion in "Cycles of Time" suggests that rest mass may not be a strict Casimir operator of the de Sitter group, allowing for the possibility of its slow decay in our universe. This idea challenges the notion of strict conservation of rest mass, raising questions about its implications for Penrose's theory of conformal cyclic cosmology. Critics argue that while Penrose's theory is innovative, it lacks empirical support for the decay of rest mass. The Poincare group, which governs the laws of physics, is well-established, and there is no definitive evidence that the de Sitter group influences local physics. Overall, the discussion highlights the theoretical complexities surrounding rest mass and its relationship to cosmological symmetries.
lark
Messages
157
Reaction score
0
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
 
Physics news on Phys.org
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.
 
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.
 
the only indication for deSitter I am aware of is the accelerated expansion which could indicate a hidden deSitter invariance
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 34 ·
2
Replies
34
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 0 ·
Replies
0
Views
4K
  • · Replies 6 ·
Replies
6
Views
7K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
29
Views
5K
  • · Replies 62 ·
3
Replies
62
Views
11K