epsilon expansion of derivatives

Dear all

I’m a Lattice Boltzmann newbie. And when I try to derive the Navier-Stokes equations I get a bit lost in the algebra.

As I understand, we generally assume that the macroscopic velocity, u, is small (order epsilon). We assume that we can expand the particle distribution function in the same small parameter, epsilon, around the equilibrium particle distribution. We also expand the derivatives in epsilon. And we expand the time derivative up to epsilon^2 and the spacial derivative to epsilon. Right?

What I do not understand is why we don’t also expand the spacial derivative to second order? This seems arbitrary to me. Is there a physical reason for not doing so?




the reason not to decompose the gradient is that all relevant physics (advection, diffusion) happens on the same length scale. This is not the case for the time scales, though.


Thanks for answer Timm. But I still do not understand.

I agree that diffusion and advection happens at two different time scales. But I thought that the scheme of lattice boltzmann was to make a ‘lattice game’ which reproduces the dynamics of a real system. How do we know that the second order of the spacial derivative won’t affect the other terms to the same order? Or am I missing something fundamental?


Indeed, I am also looking for a more convincing explanation. Waiting for someone to post it…


actually the Chapman-Enskog does not need to be made on two different time scales. In order to recover the Navier-Stokes equations the separation of the time scales in two parts in not needed (see in the book by Huang for example or in my thesis). Since from the mathematical point of view this expansion is not really well posed, it is not really important that many paths can lead you to the same result.

But, the fact that you do not need to expand in different lengthscales is related with the local nature of the interactions considered. In the Boltzmann equation you consider that the time evolution is only governed by local collisions and usually there are no long range interactions involved. Therefore expanding over many spatial scales would have no meaning.


Thanks for answer Malaspin. But what do you mean by “from the mathematical point of view this [Chapman-Enskog] expansion is not really well posed”?

I’ll definitely have a look at your thesis.


Hi Jon,

The short answer is we expand the time derivatives to preserve the ordering. Note that the Chapman-Enskog expansion does not actually expand the spatial derivatives - the epsilon you normally often see in front of such terms is a rescaling parameter.

Let’s think about the (continuous) Boltzmann equation. The macroscopic properties are found by taking moments of the distribution function, but if we take successive moments of the Boltzmann equation we find ourselves with an infinite hierarchy of equations. Therefore, we need to find a way of closing the system, and this is where the Chapman-Enskog expansion helps us.

If we assume the collision interval is small compared with the time-scale of variation of the macroscopic properties then we can expand f as f=f^0+epsilon*f^1+… Now pop this expansion into the Boltzmann equation, take moments again, and have a look at one of them to see what you get.

As an example, let’s look at the momentum equation. Before expanding, it will look like this:
(temporal derivative of momentum)+(flux)=0
and after expanding f it looks like this:
(temporal derivative of a momentum)+(flux of momentum)+epsilon*(something)+epsilon^2*(somethingelse)…=0

Now, we need to look at things order-by-order to “solve” the system (or rather, we look for successive approximation) so the temporal bit in the equation above needs to be expanded too because, due to the expansion of f, the flux is not known completely. This is a very crude way of looking at it, but it may help you see why dx isn’t expanded and why dt is like it is - the time derivative of the conserved quantities determine that of the equilibrium function, which in turn affects the equation from which f^1 (which is a function of gradients and gives the viscous stress) is determined. The expansion also satisfies the solvability conditions dictated by the theory of integral equations.

As the LBE is a discretisation of the Boltzmann equation (I prefer not to think of it like any lattice games - we are solving a PDE) we’d expect the same things to apply.

Books on kinetic theory, such as Chapman and Cowling (1970), can be helpful here.
Good luck!