Linearising a relation

The human eye can judge a straight line accurately and a curve only poorly. And a straight line has just two parameters, slope and intercept, which you can read and give errors for. So whenever a model is $y = f(x; A, B)$, look for new variables $X(x)$ and $Y(y)$ such that

with $m$ and $c$ simple functions of the unknowns $A$ and $B$.

• Power law $y = Ax^n$ (Kepler's third law, mass–luminosity, the Cepheid $L\propto P^\beta$): $\log y = n\log x + \log A$. Plot $\log y$ against $\log x$; the slope is the exponent.
• Exponential $y = Ae^{kx}$ (decays, $N = N_0e^{-t/\tau}$): $\ln y = kx + \ln A$.
• Magnitudes are already logarithms. $M = \beta'\log P + C'$ is a straight line in $M$ against $\log P$, with $\beta' = -2.5\beta$. The same applies to atmospheric extinction, $m = m_0 + kX$ against airmass $X$.
• Inverse relations such as Wien's law $\lambda_{\max} = b/T$: plot $\lambda_{\max}$ against $1/T$.
• Kepler's third law $a^3 = \dfrac{GM}{4\pi^2}P^2$: plot $a^3$ against $P^2$, or $\log a$ against $\log P$ (slope 2/3).
• Hubble's law $v = H_0d$: already linear, and should pass through the origin.

Recent problems often hand you an awkward model and expect you to find the variable that makes it linear.

2017 D2 (cored dark matter halo). $M_{\rm DM}(r) = 4\pi\rho_0r_C^2\left[r - r_C\arctan(r/r_C)\right]$. Once $r_C$ is known, define $X = r - r_C\arctan(r/r_C)$. Then $M_{\rm DM} = (4\pi\rho_0r_C^2)X$, a straight line through the origin whose slope gives $\rho_0$.

2023 DA2 (astrometric microlensing). $\Delta\theta = \theta_E\,u/(u^2+2)$. With $X = u/(u^2+2)$ this is $\Delta\theta = \theta_EX$: the slope is the Einstein radius.

2018 D1 (UV slope). $f_\lambda\propto\lambda^\beta$ becomes $m_{\rm AB} = -2.5(\beta+2)\log\lambda + {\rm const}$, so a line in $m_{\rm AB}$ against $\log\lambda$ has slope $-2.5(\beta+2)$.

The recipe: isolate the unknown you want as a multiplicative constant, then call everything it multiplies $X$.

Transform the error bars too. If $Y = \log_{10}y$, then $\Delta Y = \Delta y/(y\ln10)$. Error bars that were equal in $y$ become unequal in $\log y$, which is a reason to use a weighted fit afterwards.

Put the quantity with the largest relative error on the vertical axis. Ordinary least squares assumes $x$ is exact.

Finally, check the result is straight. If the transformed points still curve, the model or a parameter is wrong. In 2017 D2, using the fallback $r_C = 3.2$ kpc makes the $M_{\rm DM}$–$X$ plot visibly bent, which tells you that value is wrong.