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I have two questions regarding protons

1) Wikipedia says

Mean lifetime of a proton $>2.1×10^{29}$ years (stable)

Obviously this means practically nothing happens to a proton, but what does this mean lifetime signify ? What does the proton decompose into ? Or what particles is it emitting which will ultimately make it non-existant ?

2) Since proton is made up of quarks which though do not individually exist do exist together and have charge, is it possible that under extremely high power electric fields, even proton or electron or neutrons may also get induced charge, i.e. down quarks move to one side and up quarks to one ? Do in their native states they exist as kind of an electrical tripole ?

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closed as too broad by David Z Jan 24 '18 at 8:57

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what does this mean lifetime signify ?

This is the experimental lower limit on the proton lifetime. What it means is just because we havn't seen a proton decay, it doesn't mean that it can't.

What does the proton decompose into?

This probably depends what theoretical model you look at, but it seems that crucially it must conserve the quantity $B-L$, rather than $B$ and $L$ (baryon and lepton number) which are individually conserved in the Standard Model. What scientists tend to search for is the decay of a proton as:

$p^{+} \rightarrow e^{+} \pi^{0}$

and subsequently

$\pi^{0} \rightarrow \gamma \gamma$

Where $\gamma$ is a photon, $e^{+}$ is a positron and $\pi^{0}$ is a neutral pion.

2) Since proton is made up of quarks which though do not individually exist do exist together and have charge, is it possible that under extremely high power electric fields, even proton or electron or neutrons may also get induced charge, i.e. down quarks move to one side and up quarks to one ? Do in their native states they exist as kind of an electrical tripole ?

Perhaps this is more easy to see in terms of neutrons, which are also composed of quarks and are electrically neutral but have a magnetic moment due to the quark content as you mention. This definitely wouldn't happen for electrons as they are not composite particles. It should also be noted that the electron does have a magnetic moment, but this is because it is has intrinsic spin.

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  • $\begingroup$ Just one little query, so is it possible to induce charges in protons/neutrons ? $\endgroup$ – Rijul Gupta Jan 16 '14 at 16:06
  • $\begingroup$ Protons already are charged. But I don't think this is what you mean. I think you mean "can you polarise the charge distribution of protons/neutrons?". You should remember that the strong interaction (which binds the quarks in the proton/neutron) is many times stronger than EM interaction- so good luck trying! $\endgroup$ – kd88 Jan 16 '14 at 16:21
  • $\begingroup$ I know it is, and I know it will take unimaginably high potential to create such a field, but the question is can you polarise it ? Can you polarise even a fundamental unit of charge (too be dramatic ;) ) ? $\endgroup$ – Rijul Gupta Jan 16 '14 at 16:26
  • $\begingroup$ I honestly don't know, I suspect it would be very hard to prove/disprove. $\endgroup$ – kd88 Jan 17 '14 at 9:09
  • $\begingroup$ @RijulGupta It definitely has a magnetic moment we can measure (only somewhat weaker than proton's, and with opposite sign - the neutron seems to have charge distribution that leaves it negative on the "skin" with a positive "core"). It also has a predicted electric dipole moment, though the standard model's prediction is well below the sensitivity of our instruments - there are alternate models that give the value much larger, and there are experiments trying to verify that. I've heard of some neutron polarization experiments, but can't find many details about that. $\endgroup$ – Luaan Oct 18 '18 at 11:42
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Just my opinion, but I think the quoted statement is inconsistent.

An estimate of a half-life as greater than some number is another way of saying that "We looked for decays; if the half-life had been short enough we would have seen some decays with our equipment; we didn't, so it isn't."

OTOH, stability is a theoretical concept: "is there an energetically possible reaction that will lead to the decay in question.

The best example is the isotope Bismuth-209. There was no measurable decay of this isotope: it was listed as stable in most references. However, the alpha-decay of bismuth-209 can easily be shown, from measured isotope masses, to be energetically possible. In 2003, the radiation was detected, and a half-life of 600 yottaseconds was measured for bismuth-209.

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    $\begingroup$ ""OTOH, stability is a theoretical concept: "is there an energetically possible reaction that will lead to the decay in question"" --> but with the proton this is something that no one knows. According to the PDG this result is at 90% confidence level. How else would you make a statement about the lifetime of the proton? This measurement of a confidence limit is more useful for theorists than no measurement at all. $\endgroup$ – kd88 Jan 17 '14 at 9:03
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Regarding the polarizability of the nucleon--it's best to look at the neutron (since it's neutral). It has a magnetic dipole moment. Any electric dipole moment would violate time reversal and parity symmetry.

CP-violation in the Standard Model is expected to contribute:

$$d_n^{SM} \approx 10^{-31}e\cdot cm,$$

which is well below the experimental limit:

$$ d_n \le 3.0\times 10^{-26}e\cdot cm.$$

Since the available charge is roughly $e$, that means the positive and negative charge are not offset by roughly more than $10^{-26} cm$, which is 1 part in $10^{-13}$ relative to the 1 fm neutron.

Regarding an induced polarization, the electric field would most certainly by above the pair-production threshold. This is sometimes called the "sparking of the vacuum"--if nucleus (or ion collision) with $Z > 1/\alpha \approx 137$ is created, the electric field is so strong that it becomes energetically favorable to produce an $e^+e^-$ pair--which then screens the field.

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For the first part of the question, this means that the experimental evidence is sufficient to rule out decay with a lifetime of up to the figure given of ~$10^{29}$ years (~$10^{36}$ s, or $10^{12}$ Ys), and is consistent with a lifetime even longer - including up to infinity. In particular, we have observed NO spontaneous decay whatsoever with our current instruments (such as Super Kamiokande, which is designed to look for this kind of thing)... but because we've only been observing so many protons for so long in them, we can't rule out that it may undergo a decay with a lifetime longer than this, that is, that none have been seen to decay so far, but that doesn't mean one won't be seen to decay in the future. And empirically, this is actually true of all particles - we can never prove a particle (or anything else, for that matter) to be absolutely stable since this would require observing it for an infinitely long period of time, which is impossible (if for no other reason than the Universe will cease to be able to support life due to increasing entropization via the second law of thermodynamics at some point in the distant future - that is, the dissipation and dispersal of all concentrated, available-to-do-work energy into dispersed, unavailable-to-do-work forms, and "work" includes "biological life processes" and "doing experiments", not just force over a distance. But even without that, there's just the simple fact we "won't reach infinity" - something that itself gives rise to interesting philosophical paradoxes if you imagine a state "when infinity has been reached" and how the being would have accounted for this when it couldn't have reached it before...). That is, it's even possible the electron could decay after some ungodly amount of time, and we will never be able to prove it otherwise, though the theory we have so far says it shouldn't.

Which of course makes one wonder then why we'd be looking at proton decay anyways, and the reason for this is some theories suggest that it could happen. None of them have any experimental evidence to back them up yet; these experiments are a way to try and find some (namely, if we observed a proton to decay it would be exactly that evidence we were looking for). In particular, so-called "grand unified theories" suggest that on a scale of $10^{32}$ years or so (so higher than the experimental bound so far) the proton will decay to a positron and a neutral pion ($\pi^0$), the latter of which would instantly decay to gamma ray photons, thus leaving only positrons and photons, and the positron may annihilate with an electron, the end result being the whole universe will eventually be degraded to a bath of photons alone (of incomprehensibly long wavelength due to cosmic expansion). This answers your other question. Now this is not the only possible avenue by which decay could happen - another possibility is decay through higher-order processes related to the gravitational interaction, in particular so-called "virtual black holes", which would (I believe) lead to the same decay result (a positron and pion) but with a much, much longer half-life, here $10^{200}$ years, making it intractable to test since you'd need to be observing around $10^{200}$ protons to see one decay, and there's only on the order of $10^{80}$ protons in the entire observable Universe, a full $10^{120}$ too few (to imagine this, imagine that all the observable Universe's worth of protons were inside a proton, and this proton was one in another observable Universe each of whose protons was a whole observable Universe of protons, and then imagine this Universe was a proton inside a planetoid-sized object of protons each of Universes of protons of protons of this type.).

Regarding your other question, the quarks on a proton have fractional charges where two have $+\frac{2}{3} e$ and one have $-\frac{1}{3} e$. So in theory you should be able to separate them with a field, BUT you cannot isolate quarks, and the reason for this is the strong force or color interaction between them fails to let go with distance and as a result you could pull them as far as you like and they'd still be attracting each other as madly as ever ... but as you do this eventually you build up so much potential energy you rip a new quark/antiquark pair out (you have put in enough energy doing work against the strong force that it equals the mass-energy of those particles and then quantum uncertainty puffs one out of the vacuum), and it "snaps" into diquark particles, so no lone quarks. (And according to @JEB, it might also be that the electric field breaks down the vacuum before you begin to significantly polarize the proton anyways so you can't even get that far.)

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  • $\begingroup$ "Regarding your other question, the quarks on a proton have fractional charge $+\frac{1}{3}e$. So all of them add to $+1e$, the charge on the whole proton." Uhm. The 2-up valence content has a charge of $+\frac{2}{3}e$ each, while the 1-down valence content has a charge of $-\frac{1}{3}e$ making a total of $+1e$. $\endgroup$ – dmckee Jan 24 '18 at 2:25
  • $\begingroup$ @dmckee : Ah, I kind of was wondering if there was something like that, I couldn't quite remember if there was also a $\frac{2}{3} e$ charge, that sounded familiar but I wasn't sure (and didn't bother to look it up). I guess in that case then JEB's answer covers the polarization bit - yes you could in theory polarize it, but you'll rip a particle from the vacuum before you get enough field to make that happen given the strength of the forces involved, so you can't really. $\endgroup$ – The_Sympathizer Jan 24 '18 at 3:30
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Since the mean life time of proton is >2.1×1029 years and since all stars, suns and matter are made of protons, then the big bang did not happen 13.7 billions years ago. when a star is dead, it means that no enough hydrogen (proton) is there for burning

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    $\begingroup$ The quoted value is for the free decay of a proton $p \to \text{stuff}$ and has nothing to do with the fusion rate of protons inside stellar cores. $\endgroup$ – dmckee Jul 16 '17 at 0:52

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