Massless electrons stable if the Universe approaches a de Sitter Universe?

  • #1
Hi,

All cosmological models with a non-zero cosmological constant will approach a de Sitter universe in the far future. In theory this can means that the most basic group of particle physics should be the de Sitter rather than Poincaré. Mass is a Casmir operator of the Poincare but not of the Sitter.

If we suppose above, will a massless electron be stable if the universe approach a de Sitter universe in the far future ? According my information a massless electron cannot be stable according our current physic model.
What will happen with the electron if the universe approach a de Sitter universe, and we assume that this means the basic group of particle physics will be the de Sitter ?

I did some research but I am not sure if below URL is relevant for my question or not.
URL: https://arxiv.org/abs/hep-th/0612184

Thank you very much,
 

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  • #2
strangerep
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All cosmological models with a non-zero cosmological constant will approach a de Sitter universe in the far future. In theory this can means that the most basic group of particle physics should be the de Sitter rather than Poincaré.
Lots of people have researched this idea. Try googling for "de Sitter relativity". Prominent authors are Aldrovani, Almeida, Pererira. Also Cotaescu.

Cacciatori, Gorini and Kamenshchik authored the review paper Special Relativity in the 21st Century. I studied this subject some years ago (sorry, my knowledge of it is now out of date, and my RAM is rather more volatile). But if you follow the citations on Google Scholar, you'll get lots of references.

Mass is a Casmir operator of the Poincare but not of the Sitter.
Yes, but this just means that we should be constructing our basic quantum fields as unitary irreducible representations of de Sitter, not Poincare. The corresponding mass-like Casimir in de Sitter involves an extra constant term, presumably related to the cosmological constant.

So although one might interpret this as "Poincare mass is not stable", that's only because we're using the wrong formula for invariant mass.

I did some research but I am not sure if below URL is relevant for my question or not.
URL: https://arxiv.org/abs/hep-th/0612184
I've only skimmed that paper, but one common mistake (imho) in such attempts is that people forget that the de Sitter translation operators are noncommutative (unlike Poincare). Therefore, the naive momentum eigenstates cannot be used in the same way as in ordinary QFT. Recall: to construct the quantum space we normally need a maximal set of mutually commuting operators from which to construct a state basis. For Poincare, the ordinary translation operators (identifiable with ordinary momentum) commute, but not for de Sitter. I'm not aware of any attempts to confront this issue rigorously, but I haven't looked at the more recent literature. Maybe you'll find something if you search hard enough. :oldwink:

It's even more complicated when you consider that the noncommutativity of the de Sitter translation generators involves the rotation (and boost) generators, so the standard Wignerian "little group" techniques might not apply without modification.

HTH.
 
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  • #3
Hi strangerep / Hi all,

Thank you very much for the detailed answer. I would like to take an example to understand above.

I am not a physicist, so if below is very "strange" please correct me:

According our model based on Poincaré a massless electric charge particle cannot exist because a massless electric charge particle moving through space at the speed of light would radiate into Cherenkov radiation.

Now, if we suppose de Sitter will be the the most basic group of particles, what will be the consequence ? I think the consequence should be that a massless electric charge particle can exist because mass is not a Casimir in de Sitter.

What do you think ?
 
  • #4
strangerep
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What do you think ?
I think you're getting very close to a speculative personal theory, which is not appropriate for discussion on PF.
 
  • #5
PeterDonis
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According our model based on Poincaré a massless electric charge particle cannot exist because a massless electric charge particle moving through space at the speed of light would radiate into Cherenkov radiation.
Where are you getting this from? Do you have a reference?
 
  • #6
I think you're getting very close to a speculative personal theory, which is not appropriate for discussion on PF.

Sorry for not being clear.
Penrose conformal cyclic cosmology seems to suggest massless electric charged particles:

https://link.springer.com/article/10.1007/s10701-013-9763-z

Assuming charge conservation to be an absolute law (and if it were not, we would have a difficulty with the photon itself acquiring mass), then we would be ultimately stuck with the least massive charged particles, namely electrons and positrons. Present-day observations (particularly in relation to pair annihilation) assures us that no massless charged particles exist now, so the remaining possibility for eliminating all rest-mass in the very remote future of each aeon, appears to be the ultimate evaporation of rest-mass.

He solved the rest-mass with de Sitter universe, that seems (in theory) be possible because mass is not a Casimir of de Sitter and because of the cosmological constant the universe maybe reach this state in the far future (still speculative, but not impossible).

But what regarding the electric charge ? Is Cherenkov radiation not a problem in conformal cyclic cosmology ? Can a massless charged particle exist because mass is not a Casimir in de Sitter ?
 
  • #7
Vanadium 50
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According our model based on Poincaré a massless electric charge particle cannot exist because a massless electric charge particle moving through space at the speed of light would radiate into Cherenkov radiation.
Who is "our"? This is why people think you are promoting a personal theory.

In conventional physics, there is no Cerenkov radiation from massless charged particles in a vacuum.
 
  • #8
PeterDonis
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Cherenkov radiation
I have already asked you for a reference for your claim that a massless charged particle in ordinary flat Minkowski spacetime would emit Cherenkov radiation (at least, that's what I take you to be claiming in post #3). Do you have one?
 
  • #9
Where are you getting this from? Do you have a reference?

Hi All,

Thank you very much. John G. Cramer claims that a massless charged particle in space would emit Cherenkov, but based on above discussion I think he made a mistake.

Reference:
There are no known stable massless charged particles. They cannot exist because an electric charge moving through space at the speed of light would radiate away all its energy into Cherenkov radiation.
https://www.npl.washington.edu/av/altvw199.html

It still feels weird. John G. Cramer is a physicist (https://faculty.washington.edu/jcramer/) and I do not understand why he made so weird mistake. How is this possible ?

@Vanadium 50, thank you for making clear "our".
 
  • #10
PeterDonis
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Reference
Note that, while Cramer is a physicist, this is not a peer-reviewed paper; it's an article in a science fiction magazine. So his statements were not checked by other physicists, as they would have been in a peer-reviewed paper. People do make mistakes sometimes; one of the reasons for peer review is to try to catch them before a paper is published in a scientific journal.
 
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  • #11
Good Day,

I sent Mr. Cramer an email and he replied to my email !

I sent the following question to Mr. Cramer:
According to Albert Einstein, nothing can travel faster than light in vacuum. Because of this, it is assumed that the Cherenkov radiation cannot occur in vacuum. So why would a massless charged particle produce Cerenkov radiation ?

Mr. Cramer replied:
Cerenkov radiation is the electromagnetic equivalent of the sonic-boom shock wave made by an airplane traveling at or above the speed of sound. The shock wave front forms a cone with an angle that is 90 degrees to the path (a flat plane) when the speed is exactly at the critical speed and folds backward to a smaller angle as the speed increases beyond that.

Charged particles passing through transparent media where the speed of light is less than c can make such Cerenkov shock waves, and one can measure their speed by measuring the cone angle. The charged particle does not have to travel faster than c; even at c there will be a shock wave. But since there exist no massless charged particles, this does not happen in vacuum.
 
  • #12
Vanadium 50
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Uh, it's Prof. Cramer or Dr. Cramer.

Anyway, note that he said "transparent media". Not space.
 
  • #13
PeterDonis
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Mr. Cramer replied
The critical statement in his reply seems to me to be the statement that "even at c there will be a shock wave".

The normal result of Cherenkov radiation for a massive charged particle is that the particle quickly slows down to a speed that is slower than the speed of light in the medium, and the Cherenkov radiation ceases.

However, if we imagine a massless charged particle in vacuum, and Cherenkov radiation can occur even at c, then the particle would emit Cherenkov radiation, but since it is massless, it can't slow down (since massless particles can only travel at c), so the radiation won't stop; it will continue until all of the energy of the particle is radiated away. But if the particle is massless and also charged, it can't radiate all of its energy away, because then there would be nothing left for the charge to be attached to (heuristically speaking).
 
  • #14
Vanadium 50
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One can calculate the power radiated by Cerenkov radiation of a massless charged particle in vacuum. It's zero. (It's also colinear with the particle's motion, if this is even a sensible thing to write.
 

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