1
$\begingroup$

According to this Physics.SE comment, it is gravitationally allowed, though very unlikely, for a proton and an electron to annihilate yielding two photons.

  1. Is that correct?
  2. If so, why? (In particular, why does semiclassical gravity allow nonconservation of baryon number?)

Do gravitational waves violate conservation of baryon number? is somewhat related, but the currently accepted answer discusses only black holes and the cosmological baryon asymmetry problem, neither of which is relevant to this question.

Improve this question
$\endgroup$

4 Answers 4

Reset to default
1
$\begingroup$

At first I thought that Ron Maimon is talking of the equivalent of the $β+$ decay which happens in proton rich nuclei, the energy taken from the binding energy balance:

$p -> n e^+ ν_e$

In the case of the answer to the question about Hawking and Unruh like radiation from large gravitational bodies that are not black holes, this should also be taken into account, with the energy provided by the gravitational field.This also is very improbable because of the tiny size of the gravitational coupling entering the necessary Feynman diagrams. In this case there is no baryon annihilation.

But the answer's "This eventually may happen when the proton decays," is based on GUTS theories. Protons do not decay in the standard model.

Ron is commenting on this sentence, and this implies GUTS.

So it is not effective quantized gravity which allows such a process , in GUTS protons decay, so there is no baryon number conservation. The process that Ron refers to (as also proton decay) can only occur within a GUTS theory.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Thanks. I'm accepting this answer because it clarifies for me what Ron might have been thinking about, even though GUTs have not been confirmed (or conclusively disproven) by available experimental evidence. $\endgroup$
    – Thorondor
    Feb 22, 2019 at 6:33
  • 2
    $\begingroup$ I have left a comment in the answer , that a clarification is necessary. Despite his being suspended I have found Ron;s answers educational and usually clear $\endgroup$
    – anna v
    Feb 22, 2019 at 6:40
  • $\begingroup$ I think Ron would be talking about formation and decay of a micro black hole, not a GUT process. $\endgroup$
    – Mitchell Porter
    Feb 22, 2019 at 7:12
  • 1
    $\begingroup$ @MitchellPorter he is responding that the answer mentions proton decay, and gives an equivalent baryon number non-conserving decay, which exists olnly in GUTS, did you read the answer the question refers to? $\endgroup$
    – anna v
    Feb 22, 2019 at 7:19
  • 1
    $\begingroup$ @MitchellPorter This is my interpretation, I do not know if Ron can comment and clarify his meaning.I found this arxiv.org/pdf/1311.5285.pdf , where it explicitly states that quantized gravity does not respect baryon number and lepton number ". It is suspected strongly that quantum gravity willnot respect any global symmetry such as baryon number." But "suspect" is not at the level of GUTS diagrams which for sure violate B. $\endgroup$
    – anna v
    Feb 22, 2019 at 13:43
1
$\begingroup$

Such a decay would entail violation of both baryon number and lepton number. Experimentally, nobody has ever observed any violation of either of these conservation laws.

It's possible that these quantities are in fact nonconserved. This happens, for example, in some unified theories. Since there is no experimental evidence to support such possibilities, this is entirely speculative.

This answer neglects insanely improbable p+e to two photons which is gravitationally allowed.

"Gravitationally allowed" is not a term that people use, so we can only speculate about what was meant by this comment.

It's true that there are generic reasons to believe that conservation of baryon and lepton number may not hold in theories of quantum gravity. This is because there are no-hair theorems suggesting you can't tell how many baryons and leptons went into a black hole. But this is entirely speculative, since we don't have a theory of quantum gravity or any empirical evidence whatsoever about how quantum gravity works.

So in summary, the comment you refer to was posted by someone who wanted to sound wise and oracular, but was in fact making a misleadingly authoritative pronouncement.

Improve this answer
$\endgroup$
0
1
$\begingroup$

Suppose that the semi-classical theory predicting Hawking radiation is correct and black holes do indeed evaporate. A black hole that was formed by a collapsing star consisting of a vast number of baryons will then slowly evaporate emitting mostly very low energy photons. Only at the very last stages of the evaporation process will the Hawking temperature be large enough for baryons to be emitted. While on average the same number of baryons and anti-baryons will be emitted, there is the possibility of thermal fluctuations causing more baryons to be emitted than anti-baryons, and what looks like randomness in the semi-classical treatment could perhaps be less than random according to a more rigorous theory.

However, what is clear is that there is no way the evaporating black hole could emit the same number of baryons as went into the black hole when it formed. So, whatever one may assume about the validity of the no-hair theorem regarding baryon number conservation in general, it's physically impossible for baryon number to be conserved if black holes can evaporate via the Hawking process.

A fully exact quantum mechanical treatment of electron proton scattering can in principle be given in the path integral formulation. The amplitude for an initial state of a proton and an electron to evolve into a final state of two photons is given as a path integral over all possible evolutions of fields connecting the two states. If Hawking radiation exists then this implies that nonzero contributions to such a path integral do indeed exist. It may be that the amplitude could be astronomically smaller than what one may naively estimate based on virtual micro-black holes, as they may couple to matter in such a way as to almost not violate baryon number. But, as I pointed out above, baryon number cannot be respected by larger evaporating black holes.

The conclusion must therefore be that there must be path integral contributions to the evolution of an initial state of a proton and an electron to a final state of two photons.

Improve this answer
$\endgroup$
0
$\begingroup$

We're all idiots, he was just talking about the well-known electroweak instanton, no micro black holes required.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.