Viscosity arise because of fluid deformation, which causes shear within it. Momentum is then diffused down the velocity gradient. What I can't seem to understand, is how the heating arises.

I'm not 100% certain of what I'll be saying here. Let us consider a shear flow in the x direction with a velocity gradient in the z direction, with velocity increasing in the increasing z direction. Within an infinitesimally thin slab of fluid perpendicular to the z direction, the fluid is instantenously in thermal equilibrium with a certain Maxwellian distribution, but from what I understand, it has a certain independant velocity superposed to it in the x direction. For simplicity's sake, let us consider a monoatomic ideal fluid. At any time, an atom from above can penetrate into this slab. This atom comes from a region where the Maxwellian distribution might be the same, but the mean velocity is higher. When this atom arrives in the slab, the difference in the mean velocities makes it have an excess velocity in the x direction. The atom then suffers elastic collisions with the atoms of the slab, increasing the x component of their velocities. And analogously for atoms coming from under. There is energy interchange. What I don't understand, is where the losses come from. Intuitively, I'd say some of it gets absorbed by the thermal (Maxwellian) motions. But how? Isn't the mean velocity vector of a Maxwellian distribution supposed to be 0? If momentum is conserved, then isn't the x-velocity excess supposed to be 100% converted into fluid mass motion?

EDIT (To make sure I understand Thomas' answer.

Let's consider 2 contiguous slabs of fluid in the above shear flow. So the upper one is faster, and the lower one slower. They have the same number of atoms and so the same total mass $M$ (where $m$ is the mass of an individual atom). Let's separate the total momentum (in the x direction) of the individual slabs in 2 contributions: $$p_1=\sum p_{v1}+\sum p_{u1}$$ and similarly for 2. 1 being the upper faster fluid, and 2 the lower slower fluid. The sum is over all particles in the slab. The first term is the momentum due to the random velocities, and the second one is the momentum due to the fluid flow. The total momentum of the system is then: $$p=p_1+p_2.$$ Suppose that the total momentum due to the random velocities of both slabs is equal to 0. That is: $$\sum p_{v1}=\sum p_{v2}=0.$$ We also have: $$\sum p_{u1}=Mu_1$$ and similarly for slab 2. The total momentum of the system is then: $$p=M(u_1+u_2).$$ Now let's introduce one atom from slab 1 into slab 2, and one atom from slab 2 into slab 1. Without taking into account collisions, the equations of the contributions of the total momentum of slab 2 becomes: $$p_{v2}=m(v_1-v_2)$$ $$p_{u2}=Mu_2+m(u_1-u_2)$$ and similarly for slab 1. And when the contributions to the total momentum of both slabs are added, momentum is obviously conserved as we get our previous expressions. Here, it is pure exchange of momentum from both slabs and slab 2 is accelerated. Now, collisions randomize part of the flow velocity associated to the atom from slab 1 that penetrated into slab 2, so we have a transfer of momentum from the $p_{u2}$ to the thermal contribution $p_{v2}$, let's say by a factor $a$: $$p_{v2}=m(v_1-v_2)+am(u_1-u_2)$$ $$p_{u2}=Mu_2+(1-a)m(u_1-u_2).$$ $a$ would depend on the viscosity of the medium. The greater it is, the greater the heat loses. But it would at the same time minimize momentum transfer? And a low viscosity would maximize momentum transfer? That doesn't seem right. Where did I go wrong?

  • $\begingroup$ I'm not sure whether you can really understand on the level of a single atom, because the essential observation is that there is a collective component $u$ shared by all atoms (which determines the net momentum of the fluid), and a random component $v$. Correlations between $v$ and $u$, absent in a homogeneous fluid, lead to momentum transport, energy diffusion, and viscous heating. I added a postscript that hints at how that works. $\endgroup$
    – Thomas
    Sep 2, 2019 at 1:30
  • $\begingroup$ Thanks a lot for the precisions! $\endgroup$ Sep 2, 2019 at 18:44

1 Answer 1


The Boltzmann distribution in a moving fluid is $$ f(v,x,t)= \exp\left(\frac{\mu(x)}{T(x)}-\frac{m(v-u(x))^2}{T(x)}\right) $$ where $\mu$ is the chemical potential, $T$ is the temperature, and $u$ is the fluid velocity. All are slowly varying functions of $x$ (this is needed for fluid dynamics to be applicable). For the present purpose we can assume that $\mu$ and $T$ are constant, but $u_x$ is a function of $y$. This is called a ``shear flow''.

If you compute the total energy density of the fluid $$ {\cal E}(x) = \int d^3v \, E(v) f(v,x,t) $$ (in an almost ideal gas $E(v)=\frac{1}{2}mv^2$) then we find two contributions $$ {\cal E}(x) = {\cal E}_{int}(x)+\frac{1}{2}\rho u^2. $$ The first is the density of internal energy (in an ideal gas $\frac{3}{2}nT$), and the second is the kinetic energy of the flow with local velocity $u(x)$. Note that the momentum density of the fluid is $$ \pi(x) = \int d^3v \, mv f(v,x,t)=\rho u $$ (completely determined by the collective velocity $u$).

What happens in a sheared flow is that atoms from regions of the flow where $u$ is larger diffuse (by Brownian motion, that is random collisions) to regions where the mean velocity $u$ is lower. Because diffusion randomizes the velocity, the energy stored in the collective velocity $u$ is converted to internal energy stored in the random velocity $v$. (Note that the total momentum of the fluid is conserved, so slow fluid layers speed up, and fast layers slow down. Alternatively, the shear flow can be sustained by an external force.).

In this process kinetic energy $\frac{1}{2}\rho u^2$ is converted to internal energy, so the fluid heats up. This is called viscous heating, and the heating rate is controlled by viscosity $$ \dot Q = \frac{1}{2}\int d^3x \, \eta(\nabla u)^2 $$ where $\eta$ is the viscosity.

Postscript: If you want to understand this more microscopically, you have to look at solutions of the Boltzmann equation. This is discussed in text books on kinetic theory, so I will only give a very brief outline. It is easiest to consider a stationary shear flow, sustained by an external force (this is called Poiseuille flow). Because there is momentum diffusion and viscous heating the solution must show small departures from the equilibrium distribution given above $$ f_{neq}(x,v,t) = f(x,v,t)\left( 1 + \psi (x,v,t) + \ldots \right) $$ For shear flow $\partial_y u_x\neq 0$ the solution for $\psi$ is of the form $$ \psi(x,v) = a\eta v_x v_y (\nabla_y u_x) $$ where $a$ is a constant. This induces a very specific correlation between the velocity components, where (depending on the sign of $a$ and $\nabla_y u_x$), up/down going atoms have preferentially positive/negative $v_x$.

We can now check that this corresponds to a non-zero momentum flux $T_{xy}$, and a non-zero energy current into the fluid, $\jmath^\epsilon_y \neq 0$. Here, momentum flux is defined by $$ T_{xy} = \int d^3v\, mv_x v_y f(x,v,t) $$ and energy current is $$ \jmath_i^\epsilon =\int d^3v\, \frac{1}{2}v_i\, mv^2 f(x,v,t) $$

  • $\begingroup$ excellent exposition. my answer is deleted! $\endgroup$ Sep 1, 2019 at 3:43
  • $\begingroup$ Thanks for the exposition! I think I got it. I want to explicitate something to make sure I understand. I'll make an edit to my post. $\endgroup$ Sep 1, 2019 at 9:42

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