In the case of diffusion through a thin film it is possible to combine the variables in the following way: $\sigma = \frac{x}{\sqrt{4Dt}}$


  • $x$=length
  • $D$=diffusion coefficient
  • $t$=time
  • $\sigma$=a function of x and t

Thereby, x and t is substituted by the single variable sigma

Is there a similar expression for a sphere?

E. L. Cussler discribes in detail how to solve the equation for a thin film by applying "combination of variables".

  • $\begingroup$ Good, but still unclear. You can combine (say, multiply) any variable with any other variable. What is the significance of this particular combination? $\endgroup$ Feb 19, 2018 at 20:24
  • 1
    $\begingroup$ @Sigils while not off topic here, you might get a better answer on Physics SE. $\endgroup$
    – Tyberius
    Feb 19, 2018 at 20:25
  • $\begingroup$ Are you looking for a solution which gives you for example in how much time a sphere of sugar will dissolve in water by only diffusion? $\endgroup$
    – ParaH2
    Feb 19, 2018 at 22:19

2 Answers 2


Given the diffusion equation


You can put in


And then




Plug these into your diffusion equation and a lot of terms cancel leaving the same equation you would have in rectangular coordinates:


So you then use the same combination of variables as in the rectangular case except you put $z=ry$ as your dependent variable.

  • $\begingroup$ Then I get a result of the type: z(sigma) = C1+erf(sigma) * C2 Which leads to an equation of the following type with respect to c: c(r,t) = (C1 + erf(x/(sqrt(4Dt))) * C2)/(r) Is this correct? $\endgroup$
    – Sigils
    Feb 20, 2018 at 18:18
  • 1
    $\begingroup$ Your boundary conditions on $z$ will not be analogous to those on $y$ in the rectangular case. You can reduce your problem to one independent variable but you might get different functional forms depending on the form of the boundary conditions for $z$. So treat that carefully. Also remember to use $r$ for your coordinate throughout. $\endgroup$ Feb 20, 2018 at 19:04

Yes. Let's improve the wording here: that is not just an expression, it's the definition of a dimensionless variable that eases the understanding of the solution of the transient diffusion is a semi-infinite medium. Because there is no clear characteristic length as in a finite diffusion path problem, Buckingham π theorem will tell us the two dimensionless quantities we need are $$\Pi_1 = \frac{z}{\sqrt{D t}}$$ (Or any function of it, like squaring it or dividing it by 4, such as your $\sigma$)

And $$\Pi_2 = \frac{y-y_{A\infty}}{y_{As} - y_{A\infty}}$$

With $s$ being the surface index and $\infty$ indicating a position "at infinity" (often we use $y_{A\infty} = 0$).

Then we conclude $\Pi_2 =\textrm{function}(\Pi_1)$. This particular function can found by solving Fick's second law with the appropriate boundary and initial conditions. In the most common case, it yields a Gauss error function.

Had we be initially interested in a sphere instead of a planar surface, the only major difference would be the use of spherical coordinates instead of a Cartesian frame, which would result in a different functional solution. But the problem of defining the dimensionless variables would be the same, given we assume perfect angular symmetry. The $z$ position would just be substituted by a radial position $r$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.