# Why heavier particles decay faster than lighter ones?

1. Aug 2, 2006

### BuckeyePhysicist

Why heavier particles decay faster than lighter ones?

Does Uncertainty Principle explain? How?
Does Fermi's Golden Rule explain? How?
I am sure QFT does explain much, but how?

Could somebody WHO REALLY KNOW this at least give me some hint to get me started? Thank you in advance!

2. Aug 2, 2006

### mathman

No one can give you an explanation, because it is not true!
Examples: U238 is extremely long lasting (half life over 4 billion years), while H3 (Tritium) has a half life of 12.33 years.

3. Aug 2, 2006

Firstly welcome to the forums, and secondly... Could you explain where you got the information that heavy particles decay faster than lighter ones? Because I'd certainly like to see it...

4. Aug 2, 2006

### BuckeyePhysicist

Not always, but among similar'' type of particles, usually heavier ones decay faster. For example, top quark as combared to bottom quark.

It is common sense in high energy physics that heavy particles have a short life.

5. Aug 3, 2006

### nrqed

It's because there is more "phase space" available (and channels of decays).

Basically, there are two effect.s First, the heavier an elementary particle is, the more particles it can decay into (people say that there are more "decay channels"). A muon can decay into an electron plus a neutrino but an electron can't decay into a muon.

But even if you look at a specific decay mode (let's say the decay of the tau or of the muon into an electron), the more massive particle will decay faster because there is more phase space available. There is more energy to distribute among the final particles and if there is more energy to distribute, there is more ways for it to be distributed, which increases the decay rate and hence shortens the lifetime. (A way to think about it is to picture the calculation as a sum over final states where a given final state corresponds to a certain split of the energy among the final particles. More energy means more possibilities, i.e. more final states to sum over)

6. Aug 3, 2006

### Orion1

Is there an equation in QCD that predicts the decay lifetimes of quark-gluon based particles?

Last edited: Aug 3, 2006
7. Aug 3, 2006

### BuckeyePhysicist

8. Aug 11, 2006

### arivero

It is very interesting to draw a log log plot of mass versus total decay rate. Weak decays have a well known quintic dependence. Electromagnetic decays have an unexplained cubic dependence.

9. Aug 12, 2006

### cterence_chow

No,
the heavier the particle,the more unstable it is compared to its products.
Hence, it wil undergo decay faster as it is "dieing" to become stable.
*Refer to the binding energy per nucleon vs nucleon number graph.

10. Aug 12, 2006

### ZapperZ

Staff Emeritus
That isn't right either. The binding energy per nucleon does not increase nor decrease monotonically with increasing number of nucleon. See, for example

http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.html

There is a peak in the binding energy. So if you argue that everything tends to the highest binding energy, then the lighter atoms do not follow that trend you just described.

Zz.

11. Aug 12, 2006

### Astronuc

Staff Emeritus
I don't think that is the right question. Others have provided examples of long-lived 'large' or 'heavier' particles.

Another example - a free neutron of mass (rest energy) 939.573 MeV/c2 has a mean half-life of approximately 886 seconds, as compared to a neutral pion $\pi^o$ (rest mass = 135.0 MeV/c2), which has a mean lifetime of 0.84×10-16 s, or the charged pions ($$\pi^\pm$$ with rest mass = 139.6 MeV/c2 and mean lifetime of 2.60×10-8 s. Free neutrons last a long time compared to pions.

Certainly baryons heavier than the proton or neutron, e.g. $$\Lambda ,\, \Sigma ,\, \Xi$$ particles have lifetimes on the order of 10-10s or less.

And the muon (rest mass = 105.6 MeV/c2) has a mean life-time on the order of 2 x 10-6 s

I think the appropriate question is - why do some particles decay faster than others?

Last edited: Aug 12, 2006
12. Aug 12, 2006

### cterence_chow

No, regarding ur reply, there is fusion and fission. At higher nucleon number, it tends to undergo fisson(decay) rather than fission as products are more stable

13. Aug 12, 2006

### ZapperZ

Staff Emeritus
Yes, I know what fission and fusions are. However, note your EXPLANATION to the original question:

My rebuttal is that your explanation does not work all the time! This is why I point out to the lighter atoms. Thus, this is not a universal explanation and might not be the fundamental reason why things are not stable, i.e. you can't simply point to the binding energy and leave it at that.

Zz.

14. Aug 12, 2006

### cterence_chow

whats a perfectly elastic collision?Does it exist in real life?

15. Aug 12, 2006

### cterence_chow

ok i get ur point, but the guys asking bout heavier particles here.So lets chill it.k

16. Aug 12, 2006

### ZapperZ

Staff Emeritus
Please do not hijack a thread. This is against our Guidelines that you have agreed to. And don't tell me to "chill it" when all I did was to correct a wrong explanation.

Zz.

17. Aug 14, 2006

### arivero

Please check figure 1 in page 3 of http://arxiv.org/abs/hep-ph/0603145 and then come back to tell me if it is not appropiate to ask about the dependence of decay width versus mass.

The plot includes every particle having a lifetime or width listed in the particle data group, (and/but only the fundamental state of each particle). It is evident that in the SU(2) and U(1) decays there is an increasing with mass as the author of the post claimed. For SU(3) disintegrations, the situation, and the error bars, are somehow fuzzy.

Furthermore, the U(1) decay line, with the cube of mass, is unreported in the textbooks, as far as I can tell. So it is not so rare you guys have jumped with a "No"... but you are wrong.

Last edited: Aug 14, 2006
18. Aug 14, 2006

### ZapperZ

Staff Emeritus
In all fairness to Astronuc, do you think the OP had THIS particular scenario in mind? I doubt it. And I think it is a valid comment to say "maybe it's more appropriate to ask why something is unstable and decay first, and then establish the relationship, if any, on the stability of a particle with mass".

I'll also ask you this: Do you really believe that the scattering of the data in the top clump of your fig. 1 actually has some monotonic trend that you can describe with your line fit? I see no such trend, and it is even more washed out here since it is a logarithmic plot. I showed this to a one high energy physicist here who performs data analysis of the CDF data, and his first question to me was "where are the error bars?" I am sure you already know that in this field, by its nature of statistical cuts and background subtraction, no experimental data worth its salt would be presented without one, even when they came from the particle data book.

Zz.

19. Aug 14, 2006

### arivero

I think this scenario was nearer to the starting question than the initial answers about nuclei. The OP asked about particles, someone told nuclei, then he insisted he was asking for elementary particles with some likeness. The likeness he asked for is decay force.

Of course not. And it is not only that I do not believe it, it is that I DO NOT claim it.

The top clump are the particles decaying via SU(3), as I told above (I told it in the edited parragraph, so perhaps you were too fast to read Physicsforums after my posting). There is cubic for U(1) and quintic for SU(2). The quintic one are actually two paralell quintic dependences very well known, namely muon decay on one side and decay via the so called spectator model on hte another. The cubic one is intriguing but it is there; it is dimensionally corrrect because it disposed of the Fermi constant respect to the weak decays, so a dimension two diference is expected.

You can tell your friend (please tell him, as now you have left him in a disbelief state) that the error bars in weak and electromagnetic decays are mostly too small for the plot, so I disposed of them for clarity, but it is just (except perhaps for eta prime) a gnuplot of the data from particle data group and it can be replotted with the errors provided there. The real problem with error bars, as you can guess, is in the aforementionated SU(3) decays, with very short lifetimes and very broad masses, mostly resonance business.

Last edited: Aug 14, 2006
20. Aug 14, 2006

### arivero

Let me add, I think it is just no a good idea to quote the neutron as a counterexample. For the same token, I could quote the proton, whose lifetime is known to be greater than 10E30 years and it is more massive than the pion.

Ah -you will tell me- but the proton is stable, or in any case it decays via a different interaction from the GUT group!.

Well, this is the point: the question has sense where examined for each decay path.

Honestly, a huge part of the question is answered by nrqed: phase space. But a hidden part is: coupling constant. Or better: dimensionality of the coupling constant.

21. Aug 14, 2006

### ZapperZ

Staff Emeritus
But that was the point that I was making. Based on simplified assumption (i.e. such as nuclear binding energy), one CANNOT make such easy statement as "higher mass, more unstable". No one is saying that a larger mass can decay, and a smaller mass can also decay, with rates that are governed by several different channels and rules. It is just that one simply cannot make that naive of a "rule" that larger mass has shorter lifetime than smaller mass and leave it at that. I certainly cannot make that claim as easily even after looking at your Fig 1 the way you insist Astronuc look at it. It isn't clear.

And oh, btw, of course the error bars will look too small, especially for the upper clump of data. Try plotting it in linear scale and they'll show up!

Zz.

22. Aug 15, 2006

### arivero

By the way, I also think that nuclear binding energy does not sound good as an example, because the question was about particles, not nuclei. But it could be an interesting exersice to plot the decay rate as a function not of the mass but of the available energy for each decay path.

Yes, and in post #4 the original poster already corrected its statement to mean "similar type of particles". People overlooked this, and it made me a bit fourious yesterday; now I am more calm .

I am astonished it is not clear. You are invited into the data file, http://pdg.lbl.gov/2006/mcdata/mass_width_2006.csv if you want to try other visualisations (caveat: the 2004 version had some bugs and I have not checked yet the 2006 version)

Yep, and try to discriminate between a cubic and a quintic dependence in a linear plot covering some orders of magnitude.

I really do not know which is the standard methodology to give a correlation coefficient for a minimum square fit when error bars are involved. I could invent one such methodology on my own, but it should not be very informative.

23. Aug 15, 2006

### ZapperZ

Staff Emeritus
But do you really think that the top clump of your data actually show any clear dependence of any kind? I mean, if I'm refereeing this paper and see this kind of scatter, and the author fit it with a curve that somehow represents some kind of a monotonic trend, I'd send it back to the publisher in a kaon decay minute with a big rejection. I've seen arguments here among high energy physicists regarding data that are 10 times clearer than that!

When a certain trend fits only a limited scope of observation but is not seen in others, then one simply cannot make a generalized rule. This is a given, and this is what I've been trying to emphasized all along.

Zz.

24. Aug 16, 2006

### arivero

But no, I do not think so. I have already told you this, a couple of messages above.

I only claim to observe a dependence for the NON STRONG decays. Strong decays have huge experimental errors and extra dependency on channel selection, for instance via the OZI rule.

25. Aug 16, 2006

### ZapperZ

Staff Emeritus
Then you should never have challenged Astronuc to ".... come back to tell me if it is not appropiate to ask about the dependence of decay width versus mass...", because such a rule is misleading at best! That was what I objected to in the first place when you used that figure.

BTW, you never did reference where this paper was published in.

Zz.