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The intensity of light varies as the inverse square of distance, since the light you see from an object is spread across a sphere which has area proportional to the square of distance (= radius of the sphere).

I've wondered if there is a similar power law for scent. I'm sure that other factors (wind, etc.) dominate in practice, but suppose one is in an volume with still (up to thermal vibrations), initially homogeneous air, when at time $t_0$ an unmoving spherical source produces a fixed (small) amount of a volatile chemical per unit time uniformly across its surface which evaporates as quickly as it is produced. (At $t_0$ the concentration of the chemical is 0 except on the source.) I'm interested in the asymptotic behavior of the concentration at a certain distance $d$ which is small compared to the enclosing volume of still air. For whatever reason I can't convince myself whether it should follow a similar behavior or if the persistence of the chemical makes a difference.

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The distribution for a diffusing chemical is not an inverse square, but it's still a power law. The situation is different from that of light for the reason you point out: it's a dynamic equilibrium formed by persistent molecules constantly moving back and forth.

In equilibrium, the density $\rho(\mathbf{r})$ satisfies the diffusion equation $$\nabla^2 \rho = 0.$$ By spherical symmetry, $\rho$ is only a function of the radial distance $r$. Working in spherical coordinates, the diffusion equation reduces to $$\frac{1}{r^2} \partial_r (r^2 \partial_r \rho) = 0.$$ Multiplying by $r^2$ and integrating, we have $$r^2 \partial_r \rho = \text{const} \quad \to \quad \partial_r \rho \propto \frac{1}{r^2} \quad \to \quad \rho \propto \frac{1}{r}.$$ The density falls off only as $1/r$, not as $1/r^2$.

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  • $\begingroup$ Thanks! By the way, do you think I chose reasonable tags for this question? I really wasn't sure what to go with. $\endgroup$ – Charles Nov 1 '16 at 0:39
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Care! The $1/r$ solution is OK for a STATIONARY situation (continuously emitting source). Otherwise the diffusion equation leads to an exponential distribution.

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  • $\begingroup$ IMHO, this is too brief to be useful. Can you please elaborate? $\endgroup$ – AccidentalFourierTransform Apr 10 '18 at 15:14
  • $\begingroup$ I don't think it's too brief to be useful, +1 and welcome to physics.SE. But yes, more detail would be nice. $\endgroup$ – Ben Crowell Apr 10 '18 at 17:08

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