If looking for more particles or decays that violate CP symmetry can explain why there is so few antimatter in the known universe, I guess finding things that violate CPT symmetry might helps clear up some mystery about the universe. However all physics textbook insist that cpt symmetry must be conserved, why so?

  • $\begingroup$ I think CPT invariance follows from Lorentz invariance, so a violation would mean special relativity breaks down. $\endgroup$ – KF Gauss Dec 12 '18 at 7:27

I would advise you to read carefully this link. As a summary:

1) CPT invariance is an experimental observation, that means all data fitted by the standard model of particle physics have CPT invariance. Thus at the moment it is a law, i.e.axiomatic, as much as for classical gravitation Newtons law is observational and axiomatic.

2) It has been shown that if CPT is violated Lorenz invariance is violated, and there are no experimental observation supporting this, and innumerable measurements that validate Lorenz invariance.

Georg Ludens, Wolfgang Pauli and Julian Schwinger independently showed that invariance under Lorentz transformations implies CPT invariance.

  • $\begingroup$ Exactly what value is Lorentz invariant in this argument? $\endgroup$ – safesphere Dec 12 '18 at 9:15
  • $\begingroup$ All mathematical models of elementary particles are based, among other things, on Lorenz invariance. If there is no Lorenz invariance the models would have been falsified, or at least there would have been an experimental evidence that lorenz invariance is falsified, i.e. data inconsisten with Lorenz transformations, like decays of particles and energy momentum changes between frames. $\endgroup$ – anna v Dec 12 '18 at 9:19
  • $\begingroup$ Sorry, but your comment only repeats your answer without specifying what value is Lorentz invariant. I understand the Standard Model is based on Lorentz transformations, but you are saying, "Lorentz invariance", not "Lorentz transformation". Invariance of what exactly? Of what value, vector, etc.? $\endgroup$ – safesphere Dec 12 '18 at 9:28
  • $\begingroup$ lorenz invariant are th masses of the particles, for example, and all ivariant masses o complex systems too. $\endgroup$ – anna v Dec 12 '18 at 9:37
  • $\begingroup$ for vectors and matrices one speaks of lorenz covariance, i.e. consistency with lorenz transformations see hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html $\endgroup$ – anna v Dec 12 '18 at 9:51

All Lorentz invariant theories are CPT invariant. Hence if you violate CPT invariance, you violate Lorentz invariance, and that's a big no-no.

  • $\begingroup$ @anna v It has been shown that if CPT is violated Lorenz invariance is violated. AFAIK CPT theorem also requires local commutativity. Recently Novikov showed a (non-local) model with Lorentz invariance and CPT violation. And also a reverse situation: Lorentz violation yet CPT invariance. $\endgroup$ – Elio Fabri Dec 12 '18 at 10:33
  • $\begingroup$ All Lorentz invariant theories are CPT invariant. See my comment to anna v. $\endgroup$ – Elio Fabri Dec 12 '18 at 10:34
  • $\begingroup$ @ElioFabri just saw this. You should have commented below the quote you are giving. "non locality " is not an attribute of the standard model. IF cpt violation is found experimentally, this would be a candidate for an extension/modification of the standard model. It is data that demands the need to form a verifiable theory after all. $\endgroup$ – anna v Dec 13 '18 at 6:32
  • $\begingroup$ @safesphere Sorry, I'd missed that warning. $\endgroup$ – Elio Fabri Dec 13 '18 at 16:08
  • $\begingroup$ @annav Motivation of my comment was to clarify relationship between Lorentz invariance and CPT. In several posts the need for local (actually, weakly local is enough) commutativity was not stated. I said nothing about standard model or experimental evidence. My only interest was for logical implications. Now I've read a controversy between Greenberg on one side, Novikov et al. on the other, and can't say my ideas are crystal clear. But connection between Lorentz and CPT should be qualified, IMHO. $\endgroup$ – Elio Fabri Dec 13 '18 at 16:09

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