Say you have a vial of tritium and monitor their atomic decay with a geiger counter. How does an atom "know" when it's time to decay? It seems odd that all the tritium atoms are identical except with respect to their time of decay.
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1$\begingroup$ I found some answers here PSE-is-there-a-way-to-decrease-the-rate-of-nuclear-beta-decay $\endgroup$– Helder VelezCommented May 12, 2011 at 9:16
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$\begingroup$ Does it have anything to do with langevin noise forces ? $\endgroup$– New HorizonCommented Jun 3, 2011 at 18:23
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$\begingroup$ I think you would be interested in the first part of this Feyman lecture where he explains the idea that nature is not deterministic (with typical Feynman bluntness: "if you don't like it, go find another universe.") - vega.org.uk/video/programme/45 $\endgroup$– KutuluMikeCommented Aug 7, 2012 at 20:41
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6 Answers
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Actually, all the atoms are identical. The time at which it is observed to decay is not an intrinsic property of a given atom, but rather an effect of quantum mechanics. For any given time bin, there is some finite amplitude for a transition to a decayed state, which in turn corresponds to a finite probability, since the particle(s) emitted will escape from the system once such a state is reached. This also means that the process is in irreversible, due to the open nature of the system. This works in the same way as atomic transitions when atoms emit photons (see the relevant Wikipedia page).
For each undecayed atom, in each time bin $T$ there is a probability of transitioning to the decayed state given by a fixed probability $p$ (which is independent of $T$, and depends only on the binning size). Thus between the time $t$ and $t+\Delta t$ there is a fixed probability $\Delta p = \lambda \Delta t$ of transitioning to the decayed state for any given atom. So if we have $N(t)$ undecayed nuclei at time $t$, then at time $t+\Delta t$ we should have $N(t+\Delta t) = (1-\lambda\Delta t)N(t)$. Rearranging thisa and taking the limit $\Delta t \to 0$ we obtain $dN/dt = -\lambda t$. Solving this equation yields the total number of nuclei left undecayed at time $t$ as $N(t) = N(0) e^{-\lambda t}$.
Anyway, the point to take from all this is simply that the atoms are all identical and decay by a purely random process.
UPDATE: I forgot to mention that decay probability can be increased, for example via collision with another particle for the right energy, and this is exactly how fission based nuclear bombs work. Here though, again, there is nothing special about the particular atom decaying, and it is simply the particles involved in the collision that have the increased decay probability. (I must admit that I have pared this picture right down to the basics as otherwise it would need to be a far more technical discussion).
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4$\begingroup$ Thanks Joe. I realize that decay rate can be modeled by assuming a random decay process. But making that assumption doesn't really explain why the process is "random" in the first place. I guess Nature simply likes to play dice...and I need to get used to the odds! $\endgroup$ Commented Jan 18, 2011 at 11:04
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2$\begingroup$ Great answer, Joe. And yes, BuckyBadger, the fact that processes occur randomly and only the probabilities may be predicted from the inherent properties of the physical systems is one of the main lessons of quantum mechanics. There are no "hidden variables" that would secretly decide when a particle should decay. A wave function of a neutron evolves into a wave function that also contains a "proton+electron+antineutrino" state multiplied by complex number $c$. Whenever you look whether the neutron is still there, the probability is $|c|^2$ that it has already decayed. $\endgroup$ Commented Jan 18, 2011 at 11:08
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$\begingroup$ I'd suggest being careful with your language here in two ways. 1) "Intrinsic property" risk sounding like there is a number store in the nucleus (i.e. a hidden variable) but experiments on Bell's inequality show clear that there are no local hidden variables; 2) you don't really "increase the probability that the system will change with strong fields or neutron interactions, etc... rather you change the system to a different state that has a shorter halflife. Just semantic nits. $\endgroup$ Commented Jan 18, 2011 at 16:46
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2$\begingroup$ Not sure I agree with your fission argument. Absorbing a neutron changes the isotope into another one, which is unstable to fission, but that is different from increasing the probability of say alpha decay. B.T.W we could build a classical box with a small hole in it, put a gas molecule into the box, and the box in a vacuum, and the molecule escaping would be a probablistic event, so QM isn't even needed to get such an effect. $\endgroup$ Commented Jan 18, 2011 at 19:21
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2$\begingroup$ @dmckee: I said it was -not- an intrinsic property, and I had hidden variables specifically in mind. As regards the increased probability, your point is exactly the reason for the disclaimer immediately following it. To explain it properly would require quite a technical discussion about nuclear structure and interaction cross-sections. $\endgroup$ Commented Jan 18, 2011 at 19:38
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You can trigger decay of certain nuclei with gamma rays, just like you can stimulate emission of photons from excited atoms with incoming radiation. You can even make a bomb if that is your kind of thing. Induced emission
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I think if even there were a "trigger variable" for each atom, one would need to randomize it anyway to describe an ensemble of decaying atoms.
On the other hand, in case of atoms there is a stimulated emission - with help of photons coherent with the "future" photon. This shows that the "environment" is somewhat important. As soon as the environment is complicated and is hard to control, one can loosely think that the random character of decays is due to random character of the "triggering QM environment".
answered Jan 18, 2011 at 12:54
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$\begingroup$ Yes, thanks Vladimir. And Joe mentioned above that "stimulated radioactive decay" is also possible! $\endgroup$ Commented Jan 18, 2011 at 13:08
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There are more reasons then simple random time as we can see in this image.
(M. Yamamoto et. al. Journal of Environmental Radioactivity, 2006, 86, 110-131) 
(from WP )
EDIT add 1
This data is based in sediments. As sediments trace the environment conditions, they act as a proxy.
The next image is a variation of The decay rate of the radioactive isotope 32Si, and is found in
"“Evidence for Correlations Between Nuclear Decay Rates and Earth-Sun Distance” by Jenkins et al. 2008.
Later the distance from Sun connection was dismissed, but data is still valid.
Similar data about neutrino and Wimp seasonal variarance can be found.
EDIT add 1 end
EDIT add 2
a related paper, in the opposite direction can be found here
"Evidence against correlations between nuclear decay rates and Earth-Sun distance"
"We have reexamined our previously published data .. We find no evidence for such correlations"
They used ratios and this null result is expectable if both the samples and the 'presumed' reference are affected by the same nuclear processes. IMO, in this situation a counting procedure is better than to use ratios.
If the Jenkins proposal were correct, it is very unlikely that the alpha, beta-minus, beta-plus, and electron-capture decays of all radioactive isotopes would be affected in quantitatively the same way. Thus the ratios of counts observed from two different isotopes would also be expected to show annual variations.
In order to minimize the influence of any changes in detector and/or electronics performance, we analyzed ratios of gamma-ray peak areas from the isotope of interest and those from a reference isotope whose half life was well known.
EDIT add 2 end
There are seasonal variations (diurnal and annual) in radioactive processes:
- Radon, Lead, etc ...
- Neutron production on reactors (at lab and in space missions RTGs)
- Neutrino
- DM - WIMPs
I dont know what the experts say about the actual explanations. I think that they dont know the whys.
I'm in the chase of data labeled with the timedate and geographical local of the 'crime'.
Can someone help, pls?
Are the atoms of the same isotope equal?
I can not agree with the most voted answer, by Joe:
"Actually, all the atoms are identical. The time at which it is observed to decay is not an intrinsic property of a given atom, but rather an effect of quantum mechanics."
Since when quantum mechanics has effects? QM does not produce any effect, QM describes what we see at a statistical level.
We see that at a particular time moment, a particular atom decayed, and not the other. It has to be an intrinsic property of that particular atom that made it decay at that precise moment.
I do not know of any experiment that tried to measure how much equal or distinct can be a similar group of atoms. The community has the hope that they are identical. I'm skeptical about this issue and I take nothing from granted, that I can say 'They are different one from the others'.
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$\begingroup$ Think of the fermi exclusion principle. If there were individuality in the electrons, for example, it would be at some level assigned by a quantum number different for each one, and therefore there could be no fermi exclusion principle, something against experimental evidence. $\endgroup$– anna vCommented May 12, 2011 at 11:56
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$\begingroup$ @anna : yes I know. But an extreme tiny difference in mass can happen without trouble. (Really I'm not sure that such a difference exist, but as a test was not attempted, AFAIK, a skeptical person as I am will have to do the Devil's Avocate work) $\endgroup$ Commented May 12, 2011 at 13:08
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3$\begingroup$ The diagram from WP above is the variation of Pb210 concentration in a sediment core!. No conection to some seasonal influence on half life! (Which would be 1 : 10 by the way in this example!) Do You misuse anything for Your crackpottery? $\endgroup$– GeorgCommented May 12, 2011 at 13:11
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$\begingroup$ I droped this sentence from the Answer: (I've theoretical reasons to think that they are more different than expected). I've added another graph about Si-32 and Ra-226 variability, direct measures I suspect. @Georg I'm presenting DATA, and asking for more data. Before this data was gathered by patient experimentalists it was a strong beleif that such variability could not exist. Instead of calling me 'a crackpot', which I dont like, you can assume a much more interesting position: find data to invalidate those studies, or the data that I'm trying to find or candidate explanations. $\endgroup$ Commented May 12, 2011 at 14:41
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1$\begingroup$ This answer is lacking some accuracy, but the idea in itself is not crackpot at all. Firstly, disregard the first graph, the second graph is the real evidence and sufficiently makes the argument. The thinking is that the distance from the sun, and thus the neutrino flux (which is known to reach the sample) is possibly correlated with a change in measured decay rate. This isn't some correction to "radioactive decay", but just another level of physical complexity. A reaction we THOUGHT was just decay is now decay+(infrequent interaction). Nothing controversial here. $\endgroup$ Commented May 31, 2011 at 3:30
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It seems to be widely assumed that the observed click in the geiger counter corresponds to the instantaneous decay of a particular tritium atom. I don't know if I'm just pointing out the obvious, but I'm quite sure this correspondence has never been explicitly demonstrated. Quantum mechanics tells us there is a certain flux of electrons emanating from the vial of tritium; that there are is a certain frequency of clicks in a geiger counter; and that if analyzed, the sample of tritium may be separated into two streams, one of which turns out to be helium. These are three different phenomena, none of which can be easily correlated with any of the others.
To put it plainly, all we can say about your sample of tritium is that the atoms are in a superposition of states. When they are observed individually, they are found to be in one or the other atomic state - tritium or He3. There is no experiment I know of where we can identify the moment when a particular tritium atom changed state.
answered Jun 4, 2011 at 1:37
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$\begingroup$ Marty explains more in this recent post: The streaks in the cloud chamber $\endgroup$ Commented Jan 8, 2013 at 17:33
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$\begingroup$ Thanks, Helder. It's too bad that this website doesn't promote more discussion. I feel like after the first day or two, the topics just disappear from the radar. $\endgroup$ Commented Jan 8, 2013 at 20:12
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Sir Isaac Newton struggled with exactly this question, in the context of optics, and the best he could come up with was a theory of superluminal waves associated with light particles.
(The question Newton asked was essentially the same question as yours: If a light beam is 20% absorbed into a glass surface and 80% reflected, and if light is particular, then how can one particle acting alone make a correct decision?)
The whole business of selecting one from a possible set for no apparent reason can be used for quantum computing. It can effectively compute a "diagonalisation of a matrix".
Lets look more closely at the meaning of random.
Random is the limit of compressibility of a pattern, the removal of all predictability.
The pattern as a whole has this character, so it is a feature of the set.
A radioactive particle does not have the knowledge of which you speak, in fact it is missing information. It is forbidden to carry predictive knowledge. No information can be projected from any part of the sequence, present or future, to determine any part of the sequence. In this sense, every event in the sequence has no extractable knowledge about its position in the sequence. It is a safe spy, unable to betray its fellows.
This does not even answer the question, but gives a better way to think about it. What we see about randomness as contrived should more be seen as a natural default. Einsten protests that God does not play dice, but he is possibly complaining if God does plays dice then that means that God will always be with-holding information.
