= Charge lattice inference from oil drops
{title2=$q_i\simeq n_i e$}
An <integer> pattern must survive the stated <measurement uncertainty>. For an original simulated example, choose a reference <electric charge> $Q_*>0$, define $y_i=|q_i|/Q_*$, and stipulate state means
$$
\mathbf y=(2.03,\ 2.96,\ 5.02,\ 3.99),\qquad
\sigma_i=0.08.
$$
Here each $\sigma_i$ is an independent <standard uncertainty> of a state mean, before shared calibration effects. The generator constructs repeated noisy observations with these means; the numbers are deliberately chosen, not historical data. There is no conversion of $Q_*$ to <coulombs> and no measured value of the <elementary charge> in this example.
For a candidate spacing $a>0$ in units of $Q_*$ and proposed <integers> $n_i$, use <weighted least squares>:
$$
\chi^2(a,\mathbf n)=\sum_i\frac{(y_i-n_i a)^2}{\sigma_i^2},\qquad
\widehat a=\frac{\sum_i n_i y_i/\sigma_i^2}{\sum_i n_i^2/\sigma_i^2},\qquad
u(\widehat a)=\left(\sum_i\frac{n_i^2}{\sigma_i^2}\right)^{-1/2}.
$$
The last two expressions follow by differentiating the quadratic in $a$ and propagating the independent errors. They are conditional on fixed, correct <integer> assignments and the stated error model. They do not include uncertainty from choosing those assignments. For $\mathbf n=(2,3,5,4)$, they give $\widehat a=1$, $u(\widehat a)\simeq0.01089$, and $\chi^2=0.46875$. Such a small residual is compatible with this chosen error scale; it is not evidence that the errors vanish. A <chi-squared distribution> interpretation additionally needs the appropriate <normal distribution> error model and allowance for the fitted parameter; searching many assignments changes that interpretation.
\Image[/oil-drop-charge-lattice-identifiability.png]
{title=Finite noisy <electric charges> fit a candidate unit and its subdivisions}
{id=oil-drop-charge-lattice-identifiability}
{description=Original calculation for the simulated state means above. Horizontal bars show one <standard uncertainty>. Both spacings $a=1$ and $a=1/2$ have exactly the same fitted locations and residuals when the <integer> labels double. The lower curve minimizes over <integers> separately at every candidate spacing; a low residual alone cannot identify the <elementary charge>.}
The obstruction is <identifiability>: replacing $(a,n_i)$ by $(a/m,mn_i)$ for any positive <integer> $m$ leaves every predicted reading unchanged. In particular, an $e$-spaced pattern also fits an $e/2$ lattice using even indices. Even the coarsest well-fitting spacing need not be the physical unit: a dataset containing only even charge states can have twice that unit. Worse, if arbitrarily fine spacings and unrestricted <integers> are allowed, rounding each $y_i/a$ gives
$$
\min_{\mathbf n}\chi^2(a,\mathbf n)
\leq\frac{a^2}{4}\sum_i\frac1{\sigma_i^2}\longrightarrow0
\quad\text{as }a\longrightarrow0.
$$
Consequently a finite fit cannot prove a minimal transferable <electric charge>. Reproducible smallest observed jumps supply additional evidence; interpreting them as single-carrier transfers requires independent physical support. One visibly abrupt transition could contain several unresolved transfers. In the simulation the successive differences are $0.93$, $2.06$, and $-1.03$, each with <standard uncertainty> $\sqrt2(0.08)\simeq0.1131$. These support changes near $a,2a,-a$ conditional on that unit; they equally support $2,4,-2$ units of $a/2$. Neighboring differences share a state reading and are therefore correlated, even though the state means were stipulated independent.
A shared scale error can preserve every apparent <integer> relation. To first order, an independent fractional scale <standard uncertainty> $s$ adds $s^2a^2n_i n_j$ to the <covariance matrix> of the $y_i$. For the simulation, stipulating $s=0.02$ changes the conditional uncertainty of $\widehat a$ to approximately $\sqrt{0.01089^2+0.02^2}=0.02277$, despite the same residual pattern. Repeating timings cannot average away this common scale contribution. In the uncorrected continuum model in <elementary charge>, holding the other measured inputs fixed gives $|q|\propto\eta^{3/2}$, so a small fractional error in air <dynamic viscosity> causes $3/2$ times that fractional error in inferred <electric charge>. The <gas slip correction to Stokes drag> requires its own propagation. Pattern agreement, drop-stability controls and absolute <measurement calibration> together make the inference persuasive within resolution.
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