# Do photons decay?

1. Apr 11, 2007

### brianthewhitie7

[Mentor's note: this post was moved from a thread about gravitons where it was off topic.]

I know this kind of off topic but do photons decay?

Last edited by a moderator: Apr 11, 2007
2. Apr 12, 2007

### Meir Achuz

No................

3. Apr 12, 2007

### mormonator_rm

Even if they were unstable, they are traveling at the speed of light and therefore, by the very nature of time dilation in Special Relativity, they cannot decay in any reference frame.

4. Apr 14, 2007

### abbottsys

Well, in so much as there is a valid Feynman diagram for virtual photons:
p--> e+, e- -->p
(a virtual photon spends some of its time as an electron/positron pair)
I suppose you could think of that as a "decay".

5. Apr 14, 2007

### humanino

$$\gamma^*\rightarrow e^+e^-$$ should not be considered as "photon decay". At most you can name it "virtual photon transition". When you say something "decays", one would immeditely associate the context of "half-lives" (decay rate) in radioactive decays of nuclear physics, or in particle physics "mean half life" (or width). Those refer to real particles, whose mass are well defined. I do not think anybody would talk about the decay of a virtual particle, which would be rather odd.

6. Apr 14, 2007

### abbottsys

Totally agreed.
BTW: While we're on the topic of virtual particle transitions, would you care to speculate on possible G --> ??? -->G transitions, where G is a Graviton.

Last edited: Apr 14, 2007
7. Apr 15, 2007

### Meir Achuz

In graviton theories, G interacts with matter, so any pair could appear in a Feynman diagram.

8. Apr 15, 2007

### josh1

The translation-invariance of the vacuum means that momentum conservation disallows this transition - at least at the tree-level - since the three-momentum of the virtual pair must be zero while that of a photon is always nonvanishing.

9. Apr 15, 2007

### humanino

what ?
this transition is actually observed, and I think tree-level or loop(s) diagrams have nothing to do with globally conserved current such as momentum (for real/fully calculated processes)

10. Apr 15, 2007

### josh1

At tree-level we can have two-photon annihilation of an electron and positron. But one photon cannot convert directly into a virtual pair.

11. Apr 15, 2007

### humanino

The star on my photon indicated he is the virtual guy. The pair is real.

Do you agree that, tree-level or loops, three-momentum conservation is respected anyway ?

Do you agree that a virtual photon has a finite mass, and thus a rest frame ?

I think you do.

12. Apr 15, 2007

### josh1

That explains it. Sorry for wasting your time.

13. Apr 17, 2007

### Meir Achuz

A virtual photon is virtual (as in not real) and has E^2-p^2<0.
Try to find a rest frame for that.

14. Apr 17, 2007

### humanino

There are virtual photons in the time-like region as well. That's what I had in mind, and I admit that I have been sloppy. Typically, vitual photons produced in $$e^+e^-$$ annihilation are in the timelike region, whereas virtual photons produced in lepton scattering are in the spacelike region.

15. Apr 19, 2007

### neu

In the Manchester series Particle physics textbook it mentions that it has been asserted controvercialy by some that the lifetime of a photon is ~10^31 years

I havent got it with me, i'll check later

16. Apr 19, 2007

### humanino

The authority reference in the community is the Review of particle physics, which is freely available on the web...
They do not seem to me to report on any "photon decay" experiment however.
Maybe because nobody is even checking it

No seriously, photon do not decay...

17. Apr 19, 2007

### mormonator_rm

Yes... and why bother checking it? No one will ever witness the decay of a photon no matter how long they wait since they travel at the speed of light in all reference frames, and hence time dilation in essence dooms the photon to immortality. The very concept of searching for photon decay is inherently flawed.

18. Apr 20, 2007

### neu

Haha, i checked and they were actualy talking about protons. Still thats mental, protons are unstable?

19. Apr 20, 2007

### humanino

Well, the proton stability is much more surprising than that of the photon ! In Grand Unified Theories, particulary SU(5) scheme, it is predicted that the proton must decay. Someone said, I don't remember who, but "I would feel it in my bones"

Anyway, so far the proton does not seem to decay, but contrary to the photon, there are good reasons to check it.

20. Apr 21, 2007

### Haelfix

Protons actually are predicted to decay in the standard model as well via dimension 6 operators.

21. Apr 21, 2007

### humanino

I never heard of that ! :surprised
Hadronic number non-conservation within the standard model !?
Do you have any reference I could read, I would be quite interested

edit

I found it :
Symmetry Breaking through Bell-Jackiw Anomalies
G. 't Hooft
Phys. Rev. Lett. 37, 8 - 11 (1976)
Seems serious but kind of old. Thank you very much for pointing me to that
Please let me know if you have a more recent reference.

Last edited: Apr 21, 2007
22. Apr 21, 2007

### AlphaNumeric

Quite a number of attempts at explaining baryogenesis and leptogenesis in the early universe require a lack of conservation in that regard. Can't remember any references off the top of my head, but searching for 'leptogenesis' on ArXiv will bring up the relevent material I think.

Though if the operators are 6 dimensional, would it still be 'the Standard Model'? 6 dimensional operators are non-renormalisable so you'd have to put in a cut off to make it viable.

23. Apr 22, 2007

### Meir Achuz

"Standard Model" means different things to different people.
In my terminology, which I think is fairly "standard", GUT theories that require proton
decay are considered beyond the standard model, which is part of their allure.

24. Apr 23, 2007

### mormonator_rm

I think most people's definitions of the "Standard Model" would not include proton decay. Afterall, in order to conserve baryon number in the "Standard Model", the proton is naturally constrained to not decay because it is the lightest baryonic resonance...

Which brings up a further question... If physics beyond the Standard Model is capable of baryon number non-conservation, then wouldn't it be interesting if physics beyond the Supersymmetric Model would allow for R-parity violation? In that way, the lightest R-negative SUSY resonance would also be capable of decay, despite the fact that it would be constrained from decay in a SUSY model...

I'm not saying that I think that would be true, but its just a hypothetical thought...

25. Apr 24, 2007

### Haelfix

Hi Meir, I am not talking about GUT physics, but rather the standard model lagrangian.

Lepton and Baryon number conservation is an accidental symmetry of the perturbative 'renormalizable' standard model lagrangian. Weinberg, Wilcek and others showed years ago that this is broken by dimension 6 operators (suppressed by M^2, where M is some heavy mass scale). If you take the point of view of the standard model as an effective theory, then there is nothing that prevents these terms from being realized.

Also, T'Hooft and others pointed out that nonperturbative effects can also break baryon number conservation.

The problem with all these proposals is they typically output proton stability on the order of 10^94 years or something ridiculous like that. Eg the effects are so ridiculously small, we already expect something else to swamp them ipso facto.

For instance, from cosmology we have the Sakharov conditions to output the observed matter-antimatter asymmetry, which requires baryon number non conservation. There is absolutly no way these somewhat esoteric considerations could possibly lead to sufficient baryon nonconservation, ergo we know that some other physics must control the protons fate. Enter GUTs, and other proposals