If I have a solid sample material inside the atmosphere of oxygen-16, after some time diffusion will make the concentration of oxygen-16 uniform inside the sample. If then oxygen-18 (the tracer) is introduced for shorter time compared to oxygen-16, diffusion profile of oxygen-18 will develop and can be measured.

My colleagues claim that this a self diffusion process since there aren't any concentration gradients when oxygen-18 is introduced and we have the same component present, oxygen (only different isotopes).

While this argument seems reasonable, to get the tracer diffusion coefficient we fit the measured profile to the appropriate solution of the diffusion equation (2nd Fick's law).

This is the part that's confusing to me. If there aren't any concentration gradients and thus no diffusion, how can Fick's 2nd law describe the process of self diffusion?

This law assumes that concentration change in time requires at least for there to be a non-zero value of net molar flux at the point. More precisely, it requires that its divergence is non-zero and this can't be the case if the system is in equilibrium (zero concentration gradient).


2 Answers 2


It's important here to distinguish between chemical differences and chemical potential differences. The tracer experiment exploits the latter in the (near) absence of the former, allowing an irreversible process—namely, the dispersion of labeled particles—to serve as a surrogate for a reversible process—namely, the self-diffusion of particles—because the diffusion mechanism is (near) identical at the particle level.

The tracer isotope $^{18}\mathrm{O}$ is assumed to be chemically identical to the default isotope $^{16}\mathrm{O}$; this removes the qualifier "near" from the above paragraph and allows investigation of the self-diffusion of oxygen in materials because some atoms can be labeled—very convenient—but the labeling doesn't affect the process of interest.

However, $^{18}\mathrm{O}$ and $^{16}\mathrm{O}$ are still distinct species, so each has its own chemical potential (i.e., its own partial molar Gibbs free energy).

Movement of matter is driven by chemical potential differences; this is a more general statement than Fick's Law. In the absence of interactive effects and external fields, though, the concentration $C$ can be used as a surrogate for the chemical potential $\mu$. Put another way, the activity coefficient is assumed to be 1; the solution is assumed to be Raoultian. In this way, the broad framework that fluxes of matter $J$ are driven by chemical potential gradients,

$$J=f(-\nabla \mu),\tag{1}$$

is linearized and more conveniently expressed as the flux scaling with the concentration gradient,

$$J=-D^\star\nabla C,\tag{2}$$

where $D^\star$ is the self-diffusivity or self-diffusion coefficient. This is Fick's first law, and a mass balance on an infinitesimal region gives Fick's second law, $\dot C=D^\star\nabla^2 C$. See Balluffi, Allen, and Carter's Kinetics of Materials (Chapters 2 and 3) for more detail on the context of and links between Eqs. (1) and (2).

(A subtle point: $\mu$ and $C$ are related as $\mu=\mu_0+RT\ln C$ in a Raoultian solution, with reference zero $\mu_0$, gas constant $R$, and temperature $T$. So the transition from $-\nabla \mu$ to $-\nabla C$ seems hand-wavy, as $-\nabla \mu = -\nabla (RT\ln C)=-(RT/C)\nabla C$ in actuality at uniform temperature; what happened to the $1/C$ factor? Here, we can look at the precise diffusion mechanism. For a tracer atom to change locations in a lattice, it must trade spaces with a vacancy (unlikely, since vacancy concentrations are typically low) or with a default atom. Therefore, the fluxes $J_\mathrm{tracer}$ and $J_\mathrm{default}$ are nearly equal and opposite, as are the concentration gradients $\nabla C_\mathrm{tracer}$ and $\nabla C_\mathrm{default}$. We can move from Eq. (1) to Eq. (2) above more specifically by writing $J_\mathrm{tracer}\propto -(1/C_\mathrm{tracer})\nabla C_\mathrm{tracer}$, but this is $J_\mathrm{tracer}\propto -(1/C_\mathrm{default})\nabla C_\mathrm{tracer}$ by the balances above, and $C_\mathrm{default}$ is nearly constant for low tracer concentrations. This constant ends up going into what we call $D^\star$, as does $RT$.)

Given all this, one might still ask: Why are we calling this self-diffusion when it involves two different isotopes interdiffusing? How can we assume chemical equilibrium in one sense (a lattice uniformly filled with oxygen, plus some vacancies) but not in another sense (a nonuniform and nonequilibrium distribution of tracer)?

Imagine having no tool on hand to distinguish isotopes. If the isotopes are indeed suitably similar chemically, then their mixture is indistinguishable from a sample of all one isotope or another. We would look at all three samples and conclude that self-diffusion (no more, no less) is occurring in all three equivalently. That is why the tracer experiment is an acceptable surrogate for pure self-diffusion: All conclusions about the diffusivity remain the same. (Another subtle point: Conclusions regarding entropy, reversibility, and spontaneity are not the same, but this doesn’t affect the diffusivity measurement. Observers can agree on entropy measurements only if they agree on what types of work can be done on a system, and an observer who doesn’t know about isotopes also doesn’t know about how to do thermodynamic work to separate isotopes. An observer who knows about isotopes but can’t distinguish them can give only a range of estimated entropies and can’t conclude whether equilibrium has been reached. Again, this doesn’t affect the diffusivity model.)

This raises another interesting point: It may be that $^{16}\mathrm{O}$ actually consists of two types of species—call them $^{16}\mathrm{O}_\alpha$ and $^{16}\mathrm{O}_\beta$—and we don’t know it (yet). We unknowingly say that what appears to be pure $^{16}\mathrm{O}$ is undergoing self-diffusion and assign it a single chemical potential, but in reality, $^{16}\mathrm{O}_\alpha$ and $^{16}\mathrm{O}_\beta$ are interdiffusing, each following Fick’s law. As with the case of the researcher without isotopic analysis tools, the two scenarios are indistinguishable to us. (We also unknowingly calculate the entropy incompletely, and we don’t know if the system is truly at equilibrium—only whether it’s at equilibrium regarding the parameters we’re aware of. For a related discussion, see Section 5 of Jaynes' "The Gibbs paradox.") Importantly, whether we know or don’t know to assign one chemical potential to $^{16}\mathrm{O}$ or to assign a chemical potential each to $^{16}\mathrm{O}_\alpha$ and $^{16}\mathrm{O}_\beta$ has no bearing on the behavior of the physical world.

Finally, it may be useful to summarize the differences between self-diffusion and tracer diffusion.

  • Self-diffusion: Homogeneous system at equilibrium. Thermodynamically reversible. No entropy is generated (to our knowledge; see the above discussion of possible $^{16}\mathrm{O}_\alpha$ and $^{16}\mathrm{O}_\beta$). Characteristic diffusion coefficient (the self-diffusivity) that depends on the kinetics of a constituent particle moving from one lattice point to another.

  • Tracer diffusion: Heterogeneous system out of equilibrium and evolving toward equilibrium. Thermodynamically irreversible, with entropy generation. If an isotope is used as the tracer, any isotope effect is assumed to be zero or is otherwise accounted for. Characteristic diffusion coefficient (the self-diffusivity) that still depends on the kinetics of a constituent particle moving from one lattice point to another.

Thus, the two processes, although not identical, are taken as functionally equivalent specifically when considering the diffusion coefficient.

  • $\begingroup$ I agree that these are distinct species and thus that chemical potential gradient for oxygen-18 is in fact non-zero. But, that means that the system isn't in equilibrium as my colleagues claim. Net mole flux of matter can't happen without electrochemical potential gradient. $\endgroup$ Commented Mar 6 at 19:28
  • $\begingroup$ My problem is that self-diffusion by definition is the statistical, Brownian motion of atoms, there are no chemical potential gradients. How can we study this phenomenon if by adding an isotopic label, we create chemical potential gradient of that label? It's not the same phenomenon anymore. $\endgroup$ Commented Mar 6 at 19:46
  • $\begingroup$ As the label moves along the non-zero chemical potential gradient while in self-diffusion this gradient is equal to zero. $\endgroup$ Commented Mar 6 at 19:54
  • $\begingroup$ Thank you for your comments; I’ve edited my answer to try to address them. $\endgroup$ Commented Mar 6 at 21:29
  • $\begingroup$ Thank you for the update. I'll reply on my post on Reddit as it's more convenient. $\endgroup$ Commented Mar 7 at 8:23

After some discussions, the answer boils down to understanding Gibbs mixing paradox. Whether the system is at equilibrium depends on our point of view. If we can't distinguish between the particles, system is at equilibrium and there is one, equilibrium value of chemical potential.

With adding isotopic label, we are now able to distinguish between the particles which are chemically speaking the same (or at least almost the same). Because we can distinguish them from other species (oxygen-16), the value of chemical potential changes for these particles as well as it does for other oxygen-16 particles. From this perspective, system isn't at equilibrium and diffusion profile develops.

It's a self-diffusion process in sense that system is at equilibrium when particles can't be distinguished from each other. In that sense, chemical potential gradient is zero as this is taken as a definition of self-diffusion.

This is an interesting point which shows that chemical potential (along with other thermodynamic quantities) have a subjective character. Gibbs mixing paradox is solved with this point.

Whether the entropy of mixing of two ideal gases of the same type (A and B) is positive or equal to zero depends on whether we're able to distinguish particles of gas A from particles of gas B. If we can, entropy of mixing is positive and zero otherwise.

The same goes for chemical potential, does chemical potential of gas A go down when it's mixed with gas B? Yes, if we can distinguish the two after mixing is taken place. If not, chemical potential doesn't change at all.


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