The time-averaging operator definition gives us


From my understanding, this should mean that the Reynolds stress tensor, $\overline{u_iu_j}$, is a function of space only, and not of time.

But the Reynolds Stress Equations contain partial derivatives of the stress tensor with respect to time, i.e. $\frac{\partial \overline{u_iu_j}}{\partial t}$.

Since the stress tensor is time-averaged (and hence independent of time), why doesn't it follow that these time-derivative terms are zero?


1 Answer 1


This answer just depends on how you define your averaging. The Reynolds stress transport equation defines it abstractly, so it can have both time and space variation.

The equation you give is for time averages, but you can also consider space averages. For example, for a homogeneous turbulent flow, it makes sense to define the averages as space averages that are averaged throughout the entire domain. In that case, the Reynolds stresses and other statistics vary with time. Space averaging can also be done in a periodic manner called "pattern averaging" too.

There are also "ensemble averages" that allow for both time and space variation. How? Well, you average over different "realizations" of the flow. That is, you collect data from one "run" (either experimentally or numerically), and you then repeat the test as many times as you see fit. Given slight differences in the initial conditions, the data are different in each run. You then average over the "ensemble" of runs and form an ensemble average and then varies in both space and time.

This is why the Reynolds stress transport equation can have both time and space variation, since it considers averaging in a very abstract form. For some applications, you can consider the time derivatives to be zero, but for others, they may not be. It all just depends on your application and how you define your averages.


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.