Why does a neutron star not decay, as it is composed of neutrons, and free neutrons should decay in 15 minutes?
Is it because the neutrons are in an energetic "well" and to decay out would actually require energy? In collapsing under gravity to neutron degeneracy, did the neutrons say, radiate away their ability to decay any more?
But it all comes down to lots of things being possible in such a crowded place and neutrons not being “free” in a neutron star. Nuclear chemistry is full of crazy stuff happening constantly in stable situations that all kind of balances out.
A free neutron decays spontaneously into a proton, electron and neutrino, because decaying provides energy, because the mass of a neutron is higher than the sum of the masses of the decay products.
A free proton does not decay because none of the possible decay modes can produce particles with a lesser mass.
This is the same reason why your body does not fragment spontaneously in separate parts, but some external energy is required for that, e.g. someone wielding a meat cleaver.
The neutrons forming a neutron star are bound together by the gravitational force. When the neutron star has formed, the energy equal to the binding energy has been lost, so the average mass of a neutron in a neutron star, i.e. the mass of the star divided by the number of neutrons, is less than the mass of a free neutron and it is also less than the mass of a free proton and even less than the average mass of a nucleon inside the nucleus with the highest binding energy (iron 56).
Otherwise the star would have remained composed of ordinary nuclei instead of becoming a neutron star.
To extract a free neutron from a neutron star you must provide an energy at least as large as corresponding to the difference in mass between a free neutron and the mass of a neutron bound in the neutron star.
To make it "decay" (of course, that is not decay, because it is not spontaneous) while remaining in the neutron star, you need to provide some lower energy, which could convert a neutron into a proton, electron and neutrino, creating an excited state of the star, like an excited state of a nucleus or atom. Soon after that, the difference in energy will be radiated, either when the proton and electron would recombine again, or the proton will spontaneously decay into a neutron and a positron (which will later annihilate with the electron).
So a neutron star should behave like any other bound system. The state with the lowest energy is the state when all the nucleons are neutrons, unlike the state with the lowest energy of an ordinary nucleus, where a part of the nucleons must be protons.
This being the state with the lowest energy, no decay processes can exist. External energy can produce excited states, where a few protons, electrons and positrons may exist, but these other particles will decay, combine or annihilate, so the base state will be reached again.
The same happens with atomic nuclei, which are bound by strong nuclear forces instead of gravitational forces. The neutrons in stable nuclei or in nuclei with excess protons do not decay. On the contrary, the protons in nuclei with excess protons over the corresponding stable nucleus decay into neutrons and positrons (or they capture electrons).
So a neutron star behaves in the same way as a nucleus where the state with the lowest energy happens to be the one with no protons.
When the W- boson decays into an electron and antineutrino it happens millions of times faster than the life of the neutron itself. What makes that trigger point happen when it does?
Fermi's Golden Rule in principle allows you to calculate the decay rate. In practice we dont know how to calculate all the relevant quantities since QCD is hard.
So you can control neutrons with magnetic fields??
> ...detected sparks of light each time a neutron decayed.
Detecting a spark of light would also require photon(s) to be emitted, right? Is this not called out because it is a byproduct of the decay and not part of the decay reaction itself?
They let the neutrons decay for a while and then measure all the remaining neutrons in the trap by lowering there a detector.
This detector has a scintillator so when a neutron is captured it emits some photons and those are converted to an electric signal.
So no. There is no requirement for the neutron to emit an additional photon during the decay that is measured
The decay products of neutrons is a proton, and electron and electron antineutrino. Light emissions would most likely be in the from acceleration or impingement of charge particles in the “bottle” magnetic field.
Clearly the article claims they are detecting “sparks”.
The reason the photons are not part of the primary emission is related to the momentum / energy balance in the decay. There are other conserved quantities such as lepton number.
As you say the light is a byproduct of the acceleration/ detections scheme of the charged particles emitted.
Isn’t this due to relativistic effects? What percentage of the speed of light are these beams?
There is probably a Nobel for whoever solves this.
Capturing some from a beam into a magnetic trap seems like a good start.
If the discrepancy persists, then there really will be a problem.
There is plenty of theoretical speculation, especially in the last couple of years, about what such a discrepancy, if true, might mean. Again, it would be a real surprise, but if neutrons have another decay pathway other than the known one, then the bottle method measures the total decay rate, while the beam method measures the electron/beta-decay rate only.
The emerging state of the art, should this discrepancy persist, will become experiments that measure both channels simultaneously.
Nature seems to love tricks like this: Ha, you thought you understood how something is? Wrong, it just pretends to be like that! It is really much more eldritch, and you are damned. (I.e., the maths are too hard for actual people to work.) Good luck figuring out how it is only pretending to be that, too.
That is not a universally reasonable assumption. Cosmologists apparently never really tried adjusting their expectations for galaxy rotation for general relativity. When a plasma fluid dynamicist facile with the maths looked, the rotation anomaly evaporated. (Cosmologists were assiduously ignoring this, last I checked.) GR maths are considered hard, except by plasma fluid dynamicists, who have to do actually hard maths just to graduate.
https://doi.org/10.3390/atoms6040070
(Thermal meaning the neutrons scatter lots of times against atoms in a solid material, until they reach thermal equilibrium. ~km/s is a typical Boltzmann velocity for atom-size things at room temperature).
(Not a domain expert).
If the speed of the neutrons is less than 1e-5 c, the correction is less than 1e-10.
The lifetime of neutrons is approximately 15 minutes lifetime, so the correction is less than 1e-7 seconds. But they are measuring with a precision of only only a few tenths of a second, so the corrections is negligible.
I expected an almost comical consideration of relativity and the locations he lived.
I'm a little disappointed that wasn't the case.
https://en.m.wikipedia.org/wiki/Urca_process this does seem to be going on though, which tbh I just don't understand so it may contradict what I just said, or not.
Then I thought his notion likely balmy or eccentric, now I've no opinion as my brain tends to overheat whenever I think about it. ;-)
Reckon that stands to reason, same goes for the physical constants, c, µ, ε, α, etc.
But why remains the question.
However, in relativity a photon cannot decay: because it travels at the speed of light and has infinite time dilation, it does not subjectively experience the passage of time in which the possibly of decay could exist.
And what would happen in some theoretical meta material where say, values for say µ and ε were lower than their vacuum counterparts?
OK, it's a red herring, but interesting to contemplate.
If you have a bunch of neutrons and put them in a box, and look again 10 minutes later, you will see that you have only half of them.
If pick all the neutrons that survived for 3 minutes, and put them in a box, and look again 10 minutes later, you will see that you have only half of the neutrons that survived for 3 minutes.
If pick all the neutrons that survived for 7 minutes, and put them in a box, and look again 10 minutes later, you will see that you have only half of the neutrons that survived for 7 minutes.
If pick all the neutrons that survived for 42 minutes, and put them in a box, and look again 10 minutes later, you will see that you have only half of the neutrons that survived for 42 minutes.
When the time is too long, you need to create a really big number of neutrons initially, so enough survive until you start the experiment.
So ... the waiting time until the experiment start doesn't matter.
It's easier to understand with a discrete model with coins. You have perfectly balanced coins with 50% chance of head and 50% chance of tail. Each minute you flip all the coins at the same time and remove all the "heads". So you can repeat this, and each time you have less coins. You start the experiment, flip the coins 10 times (and remove the heads), and you get 1000 coins that survived. How many additional times should you flip them to remove half of them and have only 500?
These are thermalized neutrons, that means they have bounced a few times in random directions. You can model this and have some kind of convolution of the result, and then deconvolute the experimental data or fit it. But they are trying to measure 1/100 of seconds, and this is possible but very noisy, so it's strange. Wait a minute, it's a exponential decay ... so it doesn't matter ...
When you look for it, you either find a neutron that has not decayed yet, or you find the decay products.
The quoted decay time is just the average time. A free neutron might decay after 1 millisecond or after 5 hours.
Like for any other decay process, it is unpredictable when an individual neutron will decay. Nevertheless the decay probability in a time interval is predictable so if you start with a large number of free neutrons, you can predict very accurately how many will remain after 1 minute or after 10 minutes or after an hour, because the number will decrease exponentially with a known time constant, which was measured more precisely in this new experiment.
In my opinion the word 'virtual' doesn't help as it's a catch-all word for when we've no clearer description. That's certainly not a criticism of you for using it, it's just a bit vague or general when we also apply the name in connection with Zero Point Energy/Quantum Vacuum, Casimir and static electric/magnetic fields etc. My point is that a 'virtual W' is significantly different to the others I've mentioned.
That said, you'll note in my reply to adrian_b that I'm no angel in such matters either in that I've postulated somewhat by repurposing a Feynman diagram as a graph. But then, Wiki led the way by providing the axis!
Is the total time taken for a neutron to fully complete its decay longer but still comparable to the time taken for the W- boson to decay into an electron and antineutrino or is the latter's decay time much, much shorter than the overall process? That is, is the following statement true or otherwise?
[time (total) for n0 → (p+) + (e-) + (-ve)] >> [time W- → (e-) + (-ve)]
...and if so, then do we know by how much; if not then what is it? Alternatively, if the Feynman diagram for neutron beta decay shown in the following link were to actual scale then what would the scale on the vertical (time) axis be? https://en.wikipedia.org/wiki/Free_neutron_decay
I'm not really trying to be deliberately pedantic or dispute orthodoxy here but my question was in response to these and similar recent stories:
https://scitechdaily.com/zeptoseconds-new-world-record-in-sh...
https://www.quantamagazine.org/quantum-tunnel-shows-particle...
If the info therein is all or in part factual, or if similar measurement methodologies were applicable to other particles, then the ballgame may change, hence the initial reason for my question (similarly so for my first/initial post).
The second (Quanta magazine) link was the subject of a HN story going on about a year ago and it generated many comments (they resolved nothing but many were interesting nonetheless); unfortunately the time for comments was up before discussion had finished (my last, rather prolix comment was still in draft and missed the deadline). In my opinion, controversial topics like this should sometimes be left open to give one time to dwell upon them.
In hindsight, it seems the techniques mentioned in those links to measure a particle's time are unlikely to be applicable or adaptable here. That then begs the question about how did we initially determine that the W- boson's decay is much faster than the overall process.