5 ms·

"On their own, neutrons are very unstable and will convert into protons — positively charged subatomic particles — after ten minutes. " ... "For the tetraneutr

"On their own, neutrons are very unstable and will convert into protons — positively charged subatomic particles — after ten minutes. " ... "For the tetraneutron, this lifetime is only 5×10^(-22) seconds (a tiny fraction of a billionth of a nanosecond). " 10 minutes is an eternity and 5×10^(-22) seconds is closer to what I'd consider 'very unstable'

Dylan16807 10y ago
Compared to the almost-forever you can get inside a nucleus, or from protons, it's quite unstable.
- bonzini 10y ago
  Inside a nucleus---yes; from protons, not really. Helium-2 is as unstable or possibly more unstable than the tetraneutron; lithium-3 and beryllium-4 don't exist at all.
  - Dylan16807 10y ago
    They will break apart, but the stability of the protons themselves is enormous.
    - bonzini 10y ago
      Yes, but the article is about multi-particle systems.
      - Dylan16807 10y ago
        The discovery is. The comment about neutrons decaying in ten minutes isn't.
twic 10y ago
> The research in Japan used a beam of Helium-8, Helium with 4 extra neutrons, colliding with a regular Helium-4 atom. The collision breaks up the Helium-8 into another Helium-4 and a tetraneutron in its brief resonance state, before it, too, breaks apart, forming four lone neutrons. I don't know how fast these particles were going in the experiment, but if they were going at the speed of light, then the tetraneutron would make it 150 femtometres from the site of the collision before decaying. By way of comparison, neutrons and protons are 1 - 2 fm across, and a uranium nucleus is 15 fm across [i]. So if the tetraneutron was actually going at a tenth of the speed of light, it would barely make it a nucleus's diameter away before exploding. [i] https://en.wikipedia.org/wiki/Atomic_nucleus https://en.wikipedia.org/wiki/Atomic_nucleus
- windsurfer 10y ago
  That's not how relativistic speeds work. If this was going near enough to the speed of light, it could travel almost any distance before decaying. Time dilation would be observed of the particle, where it would appear that time is ticking slower for the particle compared to our own frame of reference.
  - chunky1994 10y ago
    Actually what the parent comment said would be correct. Time dilation and length contraction do not arbitrarily change the proper distances that particles travel at near light speeds. As observers, the detectors would see precisely what we expect if we calculate the distance using the detector's proper time when compared with the distance we experimentally measure.
    - amelius 10y ago
      Please explain why you think this is so.
    - russdill 10y ago
      This is super basic stuff https://en.wikipedia.org/wiki/Time_dilation_of_moving_particles https://en.wikipedia.org/wiki/Time_dilation_of_moving_partic...
      - nitrogen 10y ago
        The linked wiki article says that rapidly moving particles take longer to decay.
    - gus_massa 10y ago
      Mmm... I have the strange feeling that you both agree, let's write this with more details to avoid confusion. The half life of a static tetraneutron is (theoreticaly) 5E-22s. If you are at a laboratory and the tetraneutron is moving at relativistic speed, the apparent half life in the laboratory frame will increase substantially. To calculate the distance that the tetraneutron will travel you must multiply the velocity x the apparent half life. I.E d ~= v * 1/(1-v^2/c^2) * t_hl where t_hl = 5E-22s = the half life of the tetraneuton not the classic version that is d_clasic_wrong = v * t_hl For a more detailed explanation, with numbers and graphic you all can see the analogue experiment with muons: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html -- Anyway, IIRC a half life of 5E-22s is still very small. IIRC you can use a bubble chamber or something similar to see the trajectory of the particles. You can only see a collision and examine the debris and you will note that when the collision has some specific energy you will get an unexpected excess of collisions. You only can examine the direction and energy of the debris to make a good guess of the intermediate steps of the collision and discover that for a very short time you had a tetraneutron.
gpvos 10y ago
You forgot the sentence just before the last one. The weird thing is that they call the 5×10^(-22) seconds "stable for a period of time", but the 10 minutes "very unstable". >four neutrons together can form a resonance, a structure stable for a period of time before decaying.
- ianai 10y ago
  You're adding context to "a period of time" that's not there. 1 ns, 1ms, 1 day, and 1 millennia are all "periods of time." As for stable vs unstable you have to contrast lifetimes across other particles.
  - gpvos 10y ago
    Oh, I understand that; I just think that in the interest of clarity they should have added a bit more explicit context. Now it looks odd.

Dylan16807 10y ago

Compared to the almost-forever you can get inside a nucleus, or from protons, it's quite unstable.

bonzini 10y ago
Inside a nucleus---yes; from protons, not really. Helium-2 is as unstable or possibly more unstable than the tetraneutron; lithium-3 and beryllium-4 don't exist at all.
- Dylan16807 10y ago
  They will break apart, but the stability of the protons themselves is enormous.
  - bonzini 10y ago
    Yes, but the article is about multi-particle systems.
    - Dylan16807 10y ago
      The discovery is. The comment about neutrons decaying in ten minutes isn't.

twic 10y ago

> The research in Japan used a beam of Helium-8, Helium with 4 extra neutrons, colliding with a regular Helium-4 atom. The collision breaks up the Helium-8 into another Helium-4 and a tetraneutron in its brief resonance state, before it, too, breaks apart, forming four lone neutrons. I don't know how fast these particles were going in the experiment, but if they were going at the speed of light, then the tetraneutron would make it 150 femtometres from the site of the collision before decaying. By way of comparison, neutrons and protons are 1 - 2 fm across, and a uranium nucleus is 15 fm across [i]. So if the tetraneutron was actually going at a tenth of the speed of light, it would barely make it a nucleus's diameter away before exploding. [i] https://en.wikipedia.org/wiki/Atomic_nucleus https://en.wikipedia.org/wiki/Atomic_nucleus

windsurfer 10y ago
That's not how relativistic speeds work. If this was going near enough to the speed of light, it could travel almost any distance before decaying. Time dilation would be observed of the particle, where it would appear that time is ticking slower for the particle compared to our own frame of reference.
- chunky1994 10y ago
  Actually what the parent comment said would be correct. Time dilation and length contraction do not arbitrarily change the proper distances that particles travel at near light speeds. As observers, the detectors would see precisely what we expect if we calculate the distance using the detector's proper time when compared with the distance we experimentally measure.
  - amelius 10y ago
    Please explain why you think this is so.
  - russdill 10y ago
    This is super basic stuff https://en.wikipedia.org/wiki/Time_dilation_of_moving_particles https://en.wikipedia.org/wiki/Time_dilation_of_moving_partic...
    - nitrogen 10y ago
      The linked wiki article says that rapidly moving particles take longer to decay.
  - gus_massa 10y ago
    Mmm... I have the strange feeling that you both agree, let's write this with more details to avoid confusion. The half life of a static tetraneutron is (theoreticaly) 5E-22s. If you are at a laboratory and the tetraneutron is moving at relativistic speed, the apparent half life in the laboratory frame will increase substantially. To calculate the distance that the tetraneutron will travel you must multiply the velocity x the apparent half life. I.E d ~= v * 1/(1-v^2/c^2) * t_hl where t_hl = 5E-22s = the half life of the tetraneuton not the classic version that is d_clasic_wrong = v * t_hl For a more detailed explanation, with numbers and graphic you all can see the analogue experiment with muons: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html -- Anyway, IIRC a half life of 5E-22s is still very small. IIRC you can use a bubble chamber or something similar to see the trajectory of the particles. You can only see a collision and examine the debris and you will note that when the collision has some specific energy you will get an unexpected excess of collisions. You only can examine the direction and energy of the debris to make a good guess of the intermediate steps of the collision and discover that for a very short time you had a tetraneutron.

gpvos 10y ago

You forgot the sentence just before the last one. The weird thing is that they call the 5×10^(-22) seconds "stable for a period of time", but the 10 minutes "very unstable". >four neutrons together can form a resonance, a structure stable for a period of time before decaying.

ianai 10y ago
You're adding context to "a period of time" that's not there. 1 ns, 1ms, 1 day, and 1 millennia are all "periods of time." As for stable vs unstable you have to contrast lifetimes across other particles.
- gpvos 10y ago
  Oh, I understand that; I just think that in the interest of clarity they should have added a bit more explicit context. Now it looks odd.