Past exam of the physics A course of the University of Cambridge / 2021 / Paper 1 / Section C

Take the natural logarithm of the exponential decay model. To keep the logarithm dimensionless, choose the reference rate $R_{\ast }=1\mathrm{count}\mathrm{da}{\mathrm{y}}^{-1}$ and defineThen $y=mx+c$. With time measured in picoseconds, the slope has units ${\mathrm{ps}}^{-1}$ and the intercept is dimensionless. A straight-line linear regression in these variables therefore estimates the inverse mean lifetime and logarithmic normalization.

Use $x=t$ in picoseconds and $y=\ln (R/R_{\ast })$. Since the time uncertainty is negligible, only the vertical errors enter. First-order propagation of uncertainty givesThe transformed measurements areThus the logarithmic transformation converts constant fractional rate uncertainties into approximately equal absolute log-rate uncertainties. The equality is the linear error approximation; finite error bars would become slightly asymmetric under an exact logarithm.

Assume independent approximately Gaussian log-rate errors with known common standard deviation $s=0.15$. The weighted least squares objective isSince all weights agree, the normal equations for linear least squares reduce toFor these data, $\bar{x}=1.5\mathrm{ps}$, $\bar{y}=4.1596533$, $S_{xx}=4.375{\mathrm{ps}}^2$ and $S_{xy}=-10.3782409\mathrm{ps}$. Hence $\hat{m}=-2.3721693{\mathrm{ps}}^{-1}$ and $\hat{c}=7.7179073$.

The covariance matrix of the fitted slope and intercept follows from the inverse weighted design matrix product, or directly by writing both estimates as linear combinations of the independent $y_i$:This givesThe parameter conversion and first-order propagation of uncertainty areConsequently,These are one-standard deviation errors. The estimates are correlated, since $\mathrm{Cov}(\hat{m},\hat{c})=-0.0077143{\mathrm{ps}}^{-1}$. The minimum is ${\chi }_{\min }^2=4.6537$ for four residual degrees of freedom. Because $s$ is supplied as a known measurement uncertainty, it should not be rescaled by the observed reduced ${\chi }^2$. This is a log-linear exponential-decay fit.

For an independent count with a Poisson distribution, collected for exposure $T_i$ at rate $R_i$, the expected count is $N_i=R_i{T}_i$ and its standard deviation is $\sqrt{N_i}$. The fractional rate uncertainty is consequentlyA target of $0.15$ requires $N_i\simeq (0.15{)}^{-2}=44.44$ counts at each proper-time setting. The exposure time for a Poisson rate at fixed fractional precision is therefore $T_i=1/(0.15^2{R}_i)$, givingSince the settings are acquired independently rather than simultaneously,Nearly three quarters of the exposure is spent at the latest, lowest-rate setting. Proper time in picoseconds labels the decay distribution; it is not the laboratory acquisition time in days. This estimate assumes counting statistics dominate, with no additional background-subtraction uncertainty or acquisition overhead.

A chi-squared test compares the sum of squared standardized regression residuals with the appropriate chi-squared distribution. With both line parameters fitted, the data give ${\chi }^2=4.6537$ on $6-2=4$ degrees of freedom, with upper-tail P-value about $0.325$: the fitted exponential decay model describes the data adequately.

First consider fixing the alternative lifetime at its central value ${\tau }_a=0.478\mathrm{ps}$. Its slope is $m_a=-1/{\tau }_a=-2.09205{\mathrm{ps}}^{-1}$. One must refit the normalization rather than retain the earlier intercept. Minimization over the intercept gives $c_a=\bar{y}-m_a\bar{x}=7.29773$, andon $6-1=5$ degrees of freedom, with upper-tail P-value about $0.00130$. Thus that fixed central lifetime gives a substantially worse fit than $\hat{\tau }=0.4216\mathrm{ps}$. Equivalently, the increase after profiling over the intercept isThis fixed-value comparison does not yet include the uncertainty of the alternative measurement.

For the compatibility of independent lifetime measurements, include both errors. Under the approximately Gaussian lifetime-error model, equality of the two true lifetimes is tested byThis has one degree of freedom and P-value approximately $0.032$, equivalent to a $2.15$-standard deviation discrepancy. The two measurements are therefore in mild tension: equality is rejected at a $5\%$ significance threshold, but not at $1\%$. There is no unique yes-or-no compatibility verdict without choosing such a threshold. Treating the alternative central value as exact would overstate the disagreement between the experiments.

The lifetime that provides the best fit to these six rates remains $\boxed{\tau =0.4216\mathrm{ps}}$, not $0.478\mathrm{ps}$. The compatibility calculation above assumes the two experiments have independent uncertainties; a shared systematic error would require its covariance.