# Neutrino Antineutrino annihilation possible?

1. Jul 20, 2012

### Claustral

As far as I know, the anihilation of an electron by a positron is an electromagnetic process described by QED. Neutrinos, however, do not participate in the elektromagnetic interaction. Does that mean a Antineutrino will not anihilate a Neutrino of the same kind? Is there an interaction between Neutrino and Antineutrinofields and how does it look like? Maybe the weak interaction can also cause anihilation processes?

2. Jul 20, 2012

### Bill_K

Yes, any particle can annihilate with its antiparticle. As you say, it would be caused by the weak interaction.

3. Jul 20, 2012

### vanhees71

Sure, an electron neutrino and an electron antineutrino can be annihilated into an electron-positron pair. In the electroweak standard model the leading-order diagram is a W-boson exchange diagram.

4. Jul 20, 2012

### Dickfore

No, it's a $Z^{0}$-boson exchange, since the total incoming (and outgoing) charge is zero.

5. Jul 20, 2012

There is a key difference between low-energy electron/positron and neutrino/antineutrino pairs, though. The former can annihilate into pairs of photons because the latters' masses (zero) are less than the electrons'. In the case of anti/neutrinos, however, no such lighter particles exist. They can bump into one another and form a Z0, but the latter is off mass shell so has to decay almost immediately, and the only pair it has enough energy to create when decaying is another anti/neutrino pair. In effect, therefore, everyday anti/neutrinos such as ones from the cosmic background cannot annihilate.

6. Jul 20, 2012

### Staff: Mentor

It is true that low-energetic neutrinos cannot annihilate similar to electrons/positrons, but with a W-boson exchange there are allowed Feynman diagrams. I would assume that the cross-section is negligible, but not 0.

7. Jul 20, 2012

### Parlyne

Neutrinos have non-zero mass; and, as such can, at least in principle, annihilate to photons. The problem with this is that the annihilation involves a loop diagram with a highly off-shell W. This suppresses the cross-section significantly, making it essentially cosmologically irrelevant. But, it isn't correct to say that annihilation to photons is not allowed at all.

8. Jul 20, 2012

Good point - thanks.

9. Jul 20, 2012

Staff Emeritus
There is also t-channel W exchange. I don't know for sure, but suspect that it's the dominant process.

10. Jul 20, 2012

### Dickfore

Ah, yes. The difference in cross-sections may be explained by the properties of the corresponding branching ratios for the decay of the corresponding weak vector boson.

11. Jul 23, 2012

### Claustral

Thank you for your answers. You helped me a lot .