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Is the half life of a material only accurate as long as you are still in a macroscopic regime? If I had 8 particles in a box would I observe a fluctuation in half lives, and what would occur within the 4th half life?

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    $\begingroup$ It helps to realize that a single particle cannot half-decay. At t=0 you observe the original state and at some t>T you observe the decayed state. This time T is just the time it took the particle to decay from the moment you started observing it, and it's not the half life. This shows that you cannot extrapolate the half life concept to a single observation. $\endgroup$ – MSalters Jun 9 '15 at 10:51
  • $\begingroup$ But what if the particles where created just in the moment I start to observe them @MSalters ? (e.g. by decay of another particle) $\endgroup$ – Josef Jun 9 '15 at 15:25
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    $\begingroup$ @Josef: Particles have no memory. "Just created" doesn't give them a longer life expectancy. $\endgroup$ – MSalters Jun 9 '15 at 20:08
  • $\begingroup$ @MSalters at least in their rest frame $\endgroup$ – Skyler Jun 10 '15 at 4:53
  • $\begingroup$ @Skyler: The "no memory" property does apply in all rest frames. $\endgroup$ – MSalters Jun 10 '15 at 6:54
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Half life is, by definition, the amount of time until half of an infinitely large sample would decay. That's precisely equivalent (according to the frequentist interpretation of probability, if that matters to you) to the time until an individual particle's probability of decay reaches one half. The half life is a theoretical quantity that doesn't depend on the actual number of particles you're dealing with.

If you actually put 8 particles in a box and watch how long it takes for half of them to decay, you could consider that a measurement of the half life of the particles. As with any measurement, the value you measure will not, in general, be the same as the true (theoretical) value. So yes, there will be fluctuations, and once the number of particles remaining drops to two or one or zero, those fluctuations will be very very large. But what is fluctuating is your measurement of the half life, not the true theoretical half life itself.

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  • $\begingroup$ Thanks. If you have anymore input on the decay time of small sets of particles I would be happy if you could elaborate on that. I guess I should reorient my question to focus more on what I'm interested in, the decay dynamics of small systems. $\endgroup$ – Skyler Jun 9 '15 at 8:30
  • $\begingroup$ @Skyler what exactly would you want to know? If you're curious about the distribution of measurements of the half life, I could edit that in, but if it goes beyond that, it might be the case that you should be asking a separate question. $\endgroup$ – David Z Jun 9 '15 at 8:47
  • $\begingroup$ Who defines half life in this matter. What does it mean that half of an infinitely large sample has decayed? $\endgroup$ – Taemyr Jun 9 '15 at 13:38
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    $\begingroup$ @Taemyr - It's one of many consequences of the law of large numbers (google that phrase). Suppose you put two particles in a box and measure the time until one has decayed. Repeat that experiment and you'll get a very different answer. You'll get a much smaller variance on repeated experiments if you start with eight particles in the box and measure the time until four have decayed. You'll get an even smaller variance with one hundred particles, and even smaller variance with a million particles. The variance becomes vanishingly small in the limit of an infinitely large sample. $\endgroup$ – David Hammen Jun 9 '15 at 15:36
  • $\begingroup$ Given that a cubic cm of stuff is close to 6.02*10^23 isn't that enough to consider a fair sample? By infinite, do you just mean that single digits are too small to consider, and we need lots of zeros to have an observable qty? $\endgroup$ – JoeTaxpayer Jun 9 '15 at 18:17
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Yes, it is a statistical average in the sense that the measured half life will approach a single value of a true half life if you do lots of measurements.

In other words, if you did the experiment many, many times you would find that on average you had 4 particles left after a half-life had passed.

For any individual experiment, the results would vary.

Each atom has a probability of surviving intact after a time $t$ according to $$p = \exp(-\lambda t)$$ where $\lambda$ is the decay constant and the half life $t_{1/2} = \ln 2/\lambda$.

If you wait 4 half lives then $t = 4\ln 2/\lambda$ and the probability of an individual particle surviving is $\exp(-4\ln 2) = 0.0625$.

In practice, you have to have an integer number of particles, so the most likely outcomes are either 1 or zero intact atoms remain.

If you have 8 atoms and the probability that any of them will have decayed is $p=0.0625$, then one can use the binomial probability distributionto work out the probability that any number $n$ will survive from a population of $N$ is $$ P(n) = \frac{N!}{n! (N-n)!} p^{n}(1-p)^{N-n}$$

So $P(0)= 0.597$, $P(1) = 0.318$, $P(2)= 0.037$ and so on.

Now, if your aim is to estimate the half life based on a single experiment with these 8 atoms, then I see (at least) two possibilities.

(i) If you measure the time it takes for the 4th decay to occur, then you can calculate $P(4)$ as above, but calculate it for a range of possible values of $\lambda$. This will give you a probability distribution for $\lambda$ from which you can find the maximum likelihood value or a confidence interval.

(ii) If you have the individual decay times of each decay, then for each atom you can calculate a probability that it would have decayed in less than its observed decay time, given an assumed $\lambda$, which is $P_i(\lambda) = (1- \exp[-\lambda t_i])$. You can also include any atoms that haven't decayed, $P_i(\lambda) = \exp[-\lambda t_i]$. You then form the product of these probabilities $P(\lambda)= \prod P_i(\lambda)$ to give you an overall likelihood distribution for $\lambda$, from which you can estimated a maximum likelihood value for $\lambda$ and a confidence interval.

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    $\begingroup$ It looks like the answer is confusing likelihoods and posterior distributions in (ii). $P_i(\lambda)$ is the probability that $i$th data point is what was observed assuming that the decay constant is $\lambda$. Thus $P(\lambda)$ is the likelihood of $\lambda$, i.e., probability of data assuming that the decay constant is $\lambda$. It is however true that with (improper) uniform prior for $\lambda$, the posterior pdf for $\lambda$ would be $\mathrm{const.}\times P(\lambda)$, but either state it explicitly or don't call it "probability distribution for $\lambda$". $\endgroup$ – JiK Jun 9 '15 at 14:08
  • $\begingroup$ @Jik Yes, I'm sure you are right, the distribution would have to be normalised. $\endgroup$ – Rob Jeffries Jun 9 '15 at 16:24

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