12
$\begingroup$

Plan an experiment as follows: A neutrino source provides only neutrinos, and a detector is sensitive only to antineutrinos. If you get a signal, that proves that neutrinos are their own antiparticles. If something as simple would work, I reckon it would have been done already. So why does this not work (or does it?), and we have to rely on neutrinoless double beta decay searches?

$\endgroup$
  • $\begingroup$ For one thing, neutrinos are extremely hard to detect. So hard, in fact, that false signals are a real problem. And our sun, stars and supernovae are the only sources big enough for us to have a chance of detecting any. Add conditions to the source and the detectors and you may not have signal left. $\endgroup$ – RBarryYoung Apr 30 '18 at 19:39
  • 1
    $\begingroup$ @RBarryYoung, Neutrinos from artificial sources can be and have been detected. en.wikipedia.org/wiki/OPERA_experiment $\endgroup$ – Solomon Slow Apr 30 '18 at 22:05
19
$\begingroup$

We know that neutrinos and antineutrinos exist, and it's possible to tell the difference between them. For example in a charged current detector electron neutrinos produce this reaction:

$$ \nu + n \to e^- + p $$

while electron antineutrinos produce this reaction:

$$ \bar{\nu} + p \to e^+ + n $$

And detectors can easily distinguish between electrons and positrons so they can easily tell neutrinos and antineutrinos apart.

In your experiment you emit a beam of neutrinos, and the detector would detect only neutrinos not antineutrinos, but that isn't proof the two are different particles. The neutrinos your detector emits all have left handed chirality while antineutrinos all have right handed chirality. You can change the chirality by having your detector move in the same direction as the emitted neutrinos at a speed faster than the neutrinos are travelling. That would mean the left handed neutrino in your frame would be right handed in the detector frame. If you did this and your detector started detecting antineutrinos you'd have proved the two are the same particle.

But for obvious reasons this is not a practical experiment. Because neutrinos are so light they travel at nearly the speed of light even when they have small kinetic energies. Designing an experiment where the detector moved faster than the neutrinos would be challenging at best!

$\endgroup$
  • 1
    $\begingroup$ Love the last sentence. :) But I bet that experiment would get results faster than one where you fire a beam of neutrinos at a beam of antineutrinos and watch for annihilation reactions. ;) $\endgroup$ – PM 2Ring Apr 30 '18 at 13:03
  • 3
    $\begingroup$ You can change the helicty of a neutrino by moving faster than it, but it's chirality remains the same. $\endgroup$ – safesphere Apr 30 '18 at 13:37
  • $\begingroup$ Why is not a practical experiment? In principle if the neutrons are moving fast enough along the same trajectory as the neutrinos, in their frame they should look as left handed, and the cross section drop at zero. If they are still neutrinos, then the cross section should remain at some non-zero value $\endgroup$ – lurscher Apr 30 '18 at 15:08
  • $\begingroup$ the same should apply to proton moving close enough to the speed of light so that anti-neutrinos looks right handed, and if cross section becomes non-zero then you know they are still anti-neutrinos $\endgroup$ – lurscher Apr 30 '18 at 15:10
  • $\begingroup$ @lurscher Good "in principle" observation. Now compute the approximate $\gamma$ of available neutrino beams and point at the proton beam we can use to run the experiment... $\endgroup$ – dmckee Apr 30 '18 at 19:50
4
$\begingroup$

Good point, such experiments have been done and they see nothing. But there's something more.

The neutrino source provides only left-handed neutrinos, their spin pointing against their direction of travel, because the weak interaction does that.

The detector can be sensitive only to right-handed antineutrinos, their spins pointing in their direction of travel, because the weak interaction does that.

We see nothing - but that could be just because the neutrinos have the wrong handedness, and not because of a neutrino/antineutrino difference.

So it's not proven, and we need the double-beta-decay experiments.

$\endgroup$
  • 1
    $\begingroup$ The direction of spin does not define if a neutrino is right or left handed. If you move faster than a left handed neutrino, it does not become right handed. The direction of spin defines helicity, but whether a neutrino is right or left handed is defined by chitality. Helicity and chirality are related in the massless limit, but are different properties for massive particles. $\endgroup$ – safesphere Apr 30 '18 at 13:43
  • 1
    $\begingroup$ While @safesphere is right in principle, real world experiments are conducted with MeV or high energy neutrinos whose mass-states are all less than $0.25\,\mathrm{eV}$, which pushes the measurable consequences of the difference between helicity and chirality below experimental thresholds. For now, the distinction is experimentally uninteresting. $\endgroup$ – dmckee Apr 30 '18 at 16:26
  • 2
    $\begingroup$ @safesphere is technically correct, but the difference between chirality and helicity is graduate-level detail for an undergrad-level question. Let's not bring $(1-\gamma_5)$ into this. $\endgroup$ – RogerJBarlow Apr 30 '18 at 17:01
  • $\begingroup$ @dmckee You make it sound as if the difference between helicity and chirality is the difference in quantity, like an apple and a very slightly larger apple. However, chirality and helicity are two different things, like apples and oranges. While current experiments are limited, the scientific creativity is not. The distinction between chirality and helicity is experimentally fascinating. $\endgroup$ – safesphere Apr 30 '18 at 23:46
  • $\begingroup$ @RogerJBarlow Yeah, let's mislead undergraduates. They'll never graduate anyway. Except, your answer also breaks down if the neutrino is a Majorana fermion and its own antiparticle. $\endgroup$ – safesphere Apr 30 '18 at 23:56
1
$\begingroup$

Because the only way to be sure is to nuke it from orbit is to observe them annihilate each other.

The traditional way for this is to take some particles, preferably at rest, or at least in a beam, and some anti-particles, and put/throw them together. You then get to observe them annihilating each other, emitting two photons with the restmass of the particle/anti-particle.

However, this is very hard to do for neutrinos. You can not effectively stop them (at least not with less than a light-year or so of lead). You can barely focus them into a beam (only by focusing particles into a beam which then decay into neutrinos). In general you barely get them to interact with ordinary material, let alone with each other (there are trillions of neutrinos passing through your body every second, the vast majority of which never interact with any part of your body).

In other words, what you need is a process where you know that two neutrinos should be coming out (two neutrinos, not one neutrino and one anti-neutrino) and then observe less of those processes than you would naively assume if the neutrinos were not their own anti-particle. Because if they are, then they can annihilate before you can observe them. (Actually, the quantum mechanical amplitudes interfere destructively, so they don't even generate photons with the restmass, they become entirely virtual particles.)

One of these processes (the easiest to understand theoretically and the one that happens most often in nature, because of all those radioactive decays going on) is the double-beta decay, where two electrons, and two anti-electron-neutrinos, are emitted.

$\endgroup$

protected by Qmechanic May 1 '18 at 5:22

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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