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In a couple months, I'll do the Millikan experiment. Then, I'll end up with a number of charge measurements and their errors $$((q_i, \Delta q_i))_{i \in \mathbb N}.$$ The idea is that all those $q_i$ can be represented as a multiple of a fixed elementary charge $e$ like $$q_i = n_i e.$$

Now I do not know $e$ in advance. Is there some method to get $(e, \Delta e)$ out of the sequence of measurements?

I found this question on StackOverflow, but it does not have any answers that solve the problem in a stable way.

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    $\begingroup$ No you can't. Assume you have measurments (6,0.1) (12,0.1), (24,0.1). The only thing you can conclude is that this values are multiples of $3$, $2$ or $6$ $\endgroup$
    – Norbert
    Aug 8, 2012 at 22:04
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    $\begingroup$ @Norbert That's not true. $e$ doesn't have to be an integer. For example, if $e=1/2$, then $n_1=12$, $n_2=24$, and $n_3=48$. $\endgroup$ Aug 8, 2012 at 22:29
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    $\begingroup$ I don't know, but this is actually a question I wondered about when I first learned about the Millikan experiment. It's possible that you can just get enough data not to worry about it. If you're really curious you might go and find the original paper and see whether Millikan worried about it. $\endgroup$ Aug 8, 2012 at 23:21
  • $\begingroup$ It turns out that $e$ is not the elementary charge but that $e/3$ is it. But that is just the physical interpretation. I am just worried with the math here. $\endgroup$ Aug 9, 2012 at 14:01
  • $\begingroup$ You may be interested in this paper: tandfonline.com/doi/abs/10.1080/0020739910220602 $\endgroup$
    – student
    Jan 20, 2013 at 20:07

2 Answers 2

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This is how you would do it if there were no experimental error: Let $q_0$ be your smallest measured value for the charge and enumerate the other charges $q_1,\dotsc,q_n$. By assumption, $q_k=en_k$ for some positive real $e$ and $n_k$ an integer. Thus, $\frac{q_k}{q_0}$ is rational for all $k$. Write all of these rationals is reduced form and let $\ell$ be the least common multiple of all the denominators. Take $e=q_0/\ell$. Then, by construction, $q_k/e$ is an integer for all $k$ and $e$ is the smallest positive real number such that this is true.

The problem with this, however, is that, when you calculate $\frac{q_k}{q_0}$, you are only getting an estimate, and so $\ell$ is going to be completely off. In practice, I think you just have to hope that $\frac{q_k}{q_0}$ is (approximately) an integer for all $k$, and if that's not the case, go back to the lab and take more data until this is the case.

Of course, as pointed out, if $e$ works, so will $e/2$, and $e/3$, etc. The best you will ever be able to do here is calculate the probability that you measured your value of $e$ (call it $e_0$) was indeed the correct value of $e$ under the assumption the true value of $e$ is $e_0/2$, $e_0/3$, etc.

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  • $\begingroup$ Okay, this makes sense for exact data. The problem is that the $q_0$ has to be really exact, otherwise all the $q_k/q_0$ ratios will be way off. I'll try this approach, maybe it works with the data that I will be gathering. $\endgroup$ Aug 9, 2012 at 13:59
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Generally, computing approximate integer gcds is hard. In fact, a couple of years ago an encryption scheme was introduced whose security depends on such hardness, see Fully Homomorphic Encryption over the Integers by van Dijk, Gentry et al. Chasing references to this should turn up much of interest.

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