Extended Data Fig. 2: The time course of M(t) (equation [5]) is sensitive to values of D and σ, but insensitive to values of R.

a, The time course of M(t)/M(0) for values of D from 20–120 μm² s⁻¹. b, Corresponding half times of M(t)/M(0) over the same range of D showing that the half times change greatly with D. c, Corresponding half times of M(t)/M(0) over the same range of D showing that the half times change greatly with σ. d, Corresponding half times of M(t)/M(0) over the same range of D showing that the half times change little with R. Derivation of equation [5]. The total number of mols M(t) of fluorescent dye in a hemisphere of radius R, is given by equation [2] multiplied by the area of a hemisphere (2πr²), integrated from 0→R (because we have assumed that the volume being recorded from is a hemisphere of radius R): \(M(t)=C(0,0){[1+\frac{2Dt}{{\sigma }^{2}}]}^{-\frac{3}{2}}\underset{0}{\overset{R}{\int }}2\pi {r}^{2}exp[\frac{-{r}^{2}}{4Dt+2{\sigma }^{2}}]dr\) This can be written as: \(M(t)=a\underset{0}{\overset{R}{\int }}{r}^{2}exp[-b{r}^{2}]dr\), where \(a=2\pi C(0,0){[1+\frac{2Dt}{{\sigma }^{2}}]}^{-\frac{3}{2}}\) and \(b={(4Dt+2{\sigma }^{2})}^{-1}\) Integrating by parts gives: \(M(t)=[-\frac{aR}{2b}exp[-b{R}^{2}]]+\underset{0}{\overset{R}{\int }}\frac{a}{2b}exp[-b{r}^{2}]dr\) Using the standard integral: \(\underset{0}{\overset{R}{\int }}exp[-b{r}^{2}]dr=\sqrt{\frac{\pi }{4b}}erf(\sqrt{b}R)\), we have \(M(t)=\frac{a}{2b}\{\sqrt{\frac{\pi }{4b}}erf(\sqrt{b}R)-Rexp[-b{R}^{2}]\}\) so, finally, substituting in a and b we have Equation [5]: \(M(t)=\frac{2\pi C(0,0){\sigma }^{3}}{\sqrt{(2Dt+{\sigma }^{2})}}\{\sqrt{\frac{\pi (2Dt+{\sigma }^{2})}{2}}erf(\frac{R}{\sqrt{(4Dt+2{\sigma }^{2})}})-Rexp[-\frac{{R}^{2}}{(4Dt+2{\sigma }^{2})}]\}\).