This month’s installment completes the program begun last month by showing how setting the physical quantity $Q$ to be $m$, $m {\vec v}$, and $1/2 m v^2$ reproduces the conventional equations of fluid mechanics. We will be comparing the results with the earlier blog (General Equations of Fluid Mechanics) but some care must be take when looking at the work and heat as that blog followed looked at work done on and heat flowing into the the fluid. Kinetic theory generates objects that represent the work done by or the heat flowing out of the fluid (i.e. the collection of the interacting particles) and so some minus signs will result when making the comparisons.

Setting $Q = m$ and recognizing that $\rho({\vec r},t) = m n({\vec r},t)$, $\langle m \rangle = m$ and $\langle {\vec v} \rangle = {\vec U}({\vec r},t)$ (often called the bulk velocity) yields

\[ \partial_t \rho + \nabla \cdot (\rho \, {\vec U} ) = 0 \; , \]

which is the equation of continuity. It is important to recognize two things. First, the spatial dependence in this equation is carried entirely by the phase space density and that it is the averaging process that imparts that dependence to the first moment of the velocity, ${\vec U}$. Second, the average of the collisional term on the right-hand side is zero due to the conservation/symmtery considerations since the intra-particle collisions leave the number of particles unchanged.

Next we set $Q = m {\vec v}$ and employ the same set of rules as before to get

\[ \partial_t (\rho \, {\vec U}) + \nabla \cdot (\rho \langle {\vec v}\,{\vec v} \rangle ) – \rho {\vec g} = 0 \; .\]

Next we interpret the term $\langle {\vec v} {\vec v} \rangle$ by first defining

\[ {\vec v} = {\vec w} – {\vec U} \; \]

to get that

\[ \langle {\vec v} {\vec v} \rangle = \langle ( {\vec w} – {\vec U})({\vec w} – {\vec U}) \rangle = \langle {\vec w} {\vec w} \rangle – 2 {\vec U} \langle {\vec w} \rangle + {\vec U }{\vec U} \; . \]

If we assume that the deviations, ${\vec w}$, from the bulk velocity average to zero (which is a far more subtle point than Weinberg discusses in his write-up) then

\[ \langle {\vec v} {\vec v} \rangle = \langle {\vec w} {\vec w} \rangle + {\vec U }{\vec U} \; . \]

The first term, which is the variance in the velocity distribution, captures the spread in the marginal distribution over the time and space degrees-of-freedom, from the bulk average.

This same assumption suggests that the tensor $\rho \langle {\vec w} {\vec w} \rangle$ can be decomposed into a diagonal piece and a symmetric traceless piece as

\[\rho \langle {\vec w} {\vec w} \rangle = P \stackrel{\leftrightarrow}{1} – \stackrel{\leftrightarrow}{\sigma} \; ,\]

where $P$ is the scalar pressure ($Tr$ is the trace)

\[ P = \frac{1}{3} Tr \left( \langle {\vec w} {\vec w} \rangle \right) \; \]

and $\sigma$ is the viscous stress tensor

\[ \sigma_{ij} = P \delta_{ij} – \rho \langle w_i w_j \rangle \; .\]

The form of the reduce Boltzmann equation is now

\[ \partial_t ( \rho {\vec U} ) + \nabla \cdot (\rho {\vec U} {\vec U} ) = – \nabla P \cdot \stackrel{\leftrightarrow}{1} + \rho {\vec g} + \nabla \cdot \stackrel{\leftrightarrow}{\tau} \; , \]

where we assumed, due to conservation/symmetry considerations that the collisional term integrates to zero since the intra-particle collisions conserved momentum.

This form of the reduced Boltzmann equation is the usual (up to variations of notation) form of the momentum equation in fluid mechanics (see e.g. https://underthehood.blogwyrm.com/?p=1250 noting the notational matchup of $\stackrel{\leftrightarrow}{T} = – \rho \langle {\vec w}{\vec w} \rangle$ (note our first minus sign), ($\stackrel{\leftrightarrow}{T}$ is sometimes also called $\stackrel{\leftrightarrow}{\tau}$), which is the full Cauchy stress tensor).

Setting $Q = (1/2) m v^2$ yields

\[ \partial_t \left( \frac{1}{2} \rho \langle v^2 \rangle \right) + \nabla \cdot \left( \frac{1}{2} \rho \langle {\vec v} v^2 \rangle \right) – 1/2 \rho {\vec g} \cdot \langle \nabla_v v^2 \rangle = 0 \; . \]

Simplifying and expanding requires more work than the two examples above and will be best handled term-by-term and, often, in index notation.

Starting with the time-derivative term, we need to deal with

\[ \langle v^2 \rangle = \langle ({\vec w} + {\vec U}) \cdot ({\vec w} + {\vec U}) = \langle w^2 \rangle + 2 \langle {\vec w} \rangle \cdot {\vec U} + U^2 \; . \]

The middle term vanishes since $\langle {\vec w} \rangle = 0$. We relate the specific internal energy $e$ (sometimes also denoted as $u$) to $\langle w^2 \rangle$ via

\[ \frac{1}{2} \rho \langle w^2 \rangle \equiv e \; . \]

Substituting back in gives the time derivative as

\[ \partial_t \left( \frac{1}{2} \rho \langle v^2 \rangle \right) = \partial_t \left( e + \frac{U^2}{2} \right) \; . \]

We next move onto the spatial derivative where we need to deal with

\[ \langle {\vec v} v^2 \rangle = \langle ({\vec w} + {\vec U}) v^2 \rangle = \langle {\vec w} v^2 \rangle + \langle {\vec U} v^2 \rangle \; . \]

The first term on the right-hand side becomes

\[ \langle {\vec w} v^2 \rangle = \langle {\vec w} (w^2 + 2 {\vec w} \cdot {\vec U} + U^2) \rangle \; . \]

Now we employ index notation

\[ \langle {\vec w} (w^2 + 2 {\vec w} \cdot {\vec U} + U^2 \rangle = \langle w_i w^2 \rangle + 2 \langle w_i w_j U_j \rangle + \langle w_i U^2 \rangle \; . \]

For the first term, we identify the heat flux $q$ as (our second instance of a minus sign)

\[ \frac{1}{2} \rho \langle w_i w^2 \rangle \equiv -q_i \; . \]

As establish above, the second term is proportional to the total stress tensor

\[ \langle w_i w_j \rangle U_j = -{\tau}_{ij} U_j \; .\]

As before, the third term, which is linear in ${\vec w}$, averages to zero.

The second term of the spatial derivative becomes

\[ {\vec U} \langle v^2 \rangle = U_i \left ( \langle w^2 \rangle + U^2 \right) = U_i \left( 2 \rho u + U^2 \right) \; . \]

Putting these pieces together gives the spatial derivative as

\[ \nabla \cdot \left( \frac{1}{2} \rho \langle {\vec v} v^2 \rangle \right) = \partial_i \left[ -q_i – \tau_{ij} U_j + \left( e + \frac{1}{2} \rho U^2 \right) U_i \right] \; . \]

Assembling the time and spatial derivatives into the reduced Boltzmann equation gives

\[ \partial_t \left[ e + \frac{1}{2} \rho U^2 \right] + \partial_i \left[ \left(e + \frac{1}{2} U^2 \right) U_i \right] = \partial_i q_i + \tau_ij U_j + \rho g_i \; , \]

where we again used the conservation/symmetry consideration that intra-particle collisions conserved energy.

This final form matches the energy equations derived in General Equations of with the same notational change as noted above and the additional assumptions that the volumetric heating source (denoted, unfortunately, as $\dot Q$) is zero.

So, it should be apparent that the moment approach serves as bridge between the Boltzmann equation in abstract phase space and the classical equations of fluid mechanics.