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For instance, the average 'mean' lifetime of a muon is just over 2 microseconds....

If I plotted many, many muon lifetimes after careful experimentation, would the chart show a Gaussian distribution?

Or would it show some other type of statistical distribution?

Kurt Hikes
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    If you are also interested in half-life model of unstable non-primordial isotopes,- you can check my draft. It seems that half-life of short-lived nucleus follows an exponential law in the form : $$ t_{1/2} = \tau e ^{-\beta n} $$ Where $n$ is nucleon amount in unstable nucleus. However I'm not sure how it relates to half-life law of elementary particles themselves. Though it can hint that some dependence on particle mass can exist too- if it's charged one. – Agnius Vasiliauskas Jan 03 '22 at 08:31
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    Not directly related, but perhaps helpful: https://physics.stackexchange.com/a/632945/247642 – Roger V. Jan 03 '22 at 08:53

2 Answers2

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Yes, indeed, the average lifetime will tend to the normal distribution, as required by the Central limit theorem. A caveat: one has to distinguish between the distribution of lifetime and the distribution of the average lifetime.

Distribution of lifetimes can be taken as exponential: $$ w(t)=\frac{1}{\tau}e^{-\frac{t}{\tau}},$$ where $\tau$ is the mean lifetime. This distribution is obviously not Gaussian/normal.

If we perform $n$ measurements, with results $\{t_1, t_2,...,t_n\}$, we can define the average lifetime $$ \bar{t}=\frac{1}{n}\sum_{i=1}^nt_i,$$ where each of the $t_i$ is distributed according to the above exponential law. $\bar{t}$ is itself a random variable which varies from an experiment to an experiment, and we can calculate its distribution as: $$ w_{ave}(\bar{t})=\left\langle \delta\left(\frac{1}{n}\sum_{i=1}^nt_i-\bar{t}\right)\right\rangle=\\ \int_0^{+\infty}dt_1...\int_0^{+\infty}dt_nw(t_1)...w(t_n)\delta\left(\frac{1}{n}\sum_{i=1}^nt_i-\bar{t}\right)=\\ \left(\frac{n}{\tau}\right)^n\frac{\bar{t}^{n-1}}{(n-1)!}e^{-\frac{n\bar{t}}{\tau}},$$ (See the full derivation in the Appendix.) which is a Gamma distribution with mean $\tau$ and standard deviation $\frac{\tau}{\sqrt{n}}$. In the limit of large $n$ it approaches Gaussian/normal distribution with these parameters.

Remarks
To answer the comments

  • The distribution approaches the normal distribution in the sense that it can be approximated by the Gaussian to any order of accuracy (provided that we choose $n$ sufficiently high). It however does not converge to the normal distribution in strictly mathematical sense.
  • According to the central limit theorem, distribution of a sum of many i.i.d. quantities (independent and identically distributed) approaches normal distribution. A product of many such quantities approaches a log-normal distribution - simply because the log of this product is a sum of many i.i.d. quantities approaching the normal.

Appendix: Gaussian/normal limit
$$ w_{ave}(\bar{t})= \left(\frac{n}{\tau}\right)^n\frac{\bar{t}^{n-1}}{(n-1)!}e^{-\frac{n\bar{t}}{\tau}}= \left(\frac{n}{\tau}\right)^n\frac{1}{(n-1)!}e^{-\frac{n\bar{t}}{\tau}+(n-1)\log\bar{t}} $$ The "phase" $\phi(\bar{t})=-\frac{n\bar{t}}{\tau}+(n-1)\log\bar{t}$ has a minimum at $$\phi'(\bar{t})=-\frac{n}{\tau}+\frac{n-1}{\bar{t}}=0\Rightarrow \bar{t}^*=\frac{n-1}{n}\tau.\\ $$ Its second derivative at the minimum is $$\phi''(\bar{t})=-\frac{n-1}{\bar{t}^2}=-\frac{n^2}{(n-1)\tau^2}=-\frac{1}{\sigma^2}, $$ and the Taylor expansion around the minimum is $$\phi(\bar{t})\approx -(n-1)+(n-1)\log\left(\frac{n-1}{n}\tau\right)-\frac{(\bar{t}-\bar{t}^*)^2}{2\sigma^2}\\ $$ We thus can write $$ w_{ave}(\bar{t})\approx \left(\frac{n}{\tau}\right)^n\frac{1}{(n-1)!}e^{-(n-1)+(n-1)\log\left(\frac{n-1}{n}\tau\right)-\frac{(\bar{t}-\bar{t}^*)^2}{2\sigma^2}}=\\ \left(\frac{n}{\tau}\right)^n\frac{1}{(n-1)!}\left(\frac{(n-1)\tau}{ne}\right)^{n-1}e^{-\frac{(\bar{t}-\bar{t}^*)^2}{2\sigma^2}}= \\ \frac{n}{\tau}\frac{1}{(n-1)!}\left(\frac{n-1}{e}\right)^{n-1}e^{-\frac{(\bar{t}-\bar{t}^*)^2}{2\sigma^2}}. $$ We can now make use of the Stirling approximation $$ (n-1)!\approx \sqrt{2\pi(n-1)}\left(\frac{n-1}{e}\right)^{n-1},$$ which gives us $$ w_{ave}(\bar{t})\approx \frac{1}{\sqrt{2\pi(n-1)\tau^2/n^2}}e^{-\frac{(\bar{t}-\bar{t}^*)^2}{2\sigma^2}}= \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(\bar{t}-\bar{t}^*)^2}{2\sigma^2}}, $$ which is the normal distribution.

Roger V.
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  • How can it approach Gaussian as, accordingly, there will be finite probability that particle will have negative lifetime? It seems to me that asymptotic distribution would be log-normal ... – i_prob_should_know_this Jan 03 '22 at 19:14
  • @i_prob_should_know_this it is a legitimate concern - when approximating this distribution by normal one has to watch that the probability of negative values is negligeably small. It is however not approaching the log-normal, since it is a sum of many i.i.d. quantities rather than a product. – Roger V. Jan 03 '22 at 20:10
  • @i_prob_should_know_this Why log-normal? There are lots of other non-negative continuous distributions. – J.G. Jan 04 '22 at 07:31
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    @i_prob_should_know_this It approaches Gaussian. It's never exactly equal. As $n$ gets larger, the amount of probability mass of the Gaussian that's negative decreases quickly. The number of standard deviations needed to get to negative goes as $\sqrt n$, and the probability decreases very quickly with higher standard deviations. – Acccumulation Jan 04 '22 at 07:35
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    One of your integrals should be over $dt_n$ rather than $dt_1$ – Ruslan Jan 04 '22 at 09:36
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    @Ruslan Thanks - I corrected that and also expanded the answer, to address the above comments. – Roger V. Jan 04 '22 at 09:45
  • @J.G. It was just a guess. If I had anything beyond that, I'd answer ;) – i_prob_should_know_this Jan 05 '22 at 08:27
  • @Acccumulation You are using "approaches" in an unusual way. I understand it an "limit is exactly equal". And if it approaches any Gaussian, then the probability of negative lifetime tends to some finite value. If the limit has finite difference from any Gaussian form, then it does not approach that distribution. Roger's appendix is very clear what he was demonstrating, but I am not sure what are you saying. – i_prob_should_know_this Jan 05 '22 at 08:37
  • @i_prob_should_know_this No, I'm using it in the normal mathematical sense. The CLT does not exactly say that the limit is equal to a fixed Gaussian. It says that the limit of the normalized z-score is a Gaussian distribution. Perhaps you should ask a question on a SE such Cross Validated to get a more detailed answer. This isn't really the sort of thing the comments are meant for. – Acccumulation Jan 05 '22 at 23:11
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A particle's lifespan is has a memoryless and hence exponential distribution. If $\tau$ is the mean lifetime, the lifetime has CDF $1-e^{-t/\tau}$ (so the half-life is $\tau\ln2$) and PDF $\frac{1}{\tau}e^{-t/\tau}$ for $t\ge0$. Therefore, the proportion of surviving particles after a time $t$ is on average $e^{-t/\tau}$, and the mean lifetime of one particle is indeed$$\int_0^\infty\frac{1}{\tau}e^{-t/\tau}tdt\stackrel{x:=t/\tau}{=}\int_0^\infty\tau xe^{-x}dx=\tau.$$Muons have one value of $\tau$; other species will have another.

J.G.
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