Some personal notes from particle to fluid

Here are notes the momentum equation of the fluid from particle scale

Basic Kinetic Theory

If we know the form of the distribution, \(f(\vec{x},\vec{v})\), the total integral gives the total particle number,

\[N=\int f(\vec{x},\vec{v})\rm d^3x d^3v.\]

Thus, the local number density is defined as,

\[n(\vec{x})=\int f(\vec{x},\vec{v})\rm d^3x.\]

From the Boltzmann equation,

\[\partial_tf+\nabla_{\bf x}\cdot({\bf v}f) + \nabla_{\bf v}\cdot({\bf a} f) = \mathcal{C}(f).\]

If we only consider the \textit{Lorentz} and gravity forces, \({\bf a}=\frac{q}{m}({\bf E}+{\bf v} \times {\bf B})+{\bf g}\).

\[\partial_tf+{\bf v}\cdot\nabla_{\bf x}f + {\bf a}\cdot\nabla_{\bf v}f = \mathcal{C}(f).\]

The macro-scale averaged, mass density, velocity, and energy can be defined as follows,

\[\rho=m\int f(\cdot) \rm d^3v,\\{\bf v}=n^{-1}\int f(\cdot) {\bf v}\rm d^3v,\\{\bf e}=n^{-1}\int f(\cdot) \frac{1}{2}mv^2\rm d^3v\]

The conservation of probability can be convert to the mass density conservation,

\[\begin{aligned} \partial_t\rho &=m\int \partial_tf \rm d^3v\\ &=-\int m{v}\cdot\nabla f\rm d^3v,\\ &=-\nabla\cdot\int mf{\bf v}\rm d^3v,\\ &=-\nabla\cdot(\rho{\bf v}). \end{aligned}\]

The momentum equation can be written as,

\[\begin{aligned} \partial_t(\rho u_i)&=\int mv_i\partial_t f\rm d^3 v\\ &=-\int m v_i(v_j\partial_jf+a_j\partial_{v_j}f)\rm d^3 v\\ &=-\partial_j\int m v_iv_jf{\rm d^3 v} + \int \frac{\partial v_i}{\partial v_j}ma_jf{\rm d^3v}\\ &=-\int m(u_i+\delta v_i)(u_j+\delta v_j)\partial_jf{\rm d^3 v} + \int m a_i f{\rm d^3 v}\\ &=-\partial_j\int m u_iu_jf{\rm d^3 v} - \partial_j\int m\delta v_i\delta v_jf{\rm d^3v}+\rho\bar{a}_i\\ &=-\partial_j(\rho u_iu_j)- \partial_j P_{ij} + \rho\bar{a}_i, \end{aligned}\]

where, \(P_{ij}=\int m\delta v_i\delta v_jf{\rm d^3v}\), known as the pressure tensor. Different distribution can give a different form of the pressure tensor. It is obvious that the pressure tensor is a symmetric tensor, \(P_{ij}=P_{ji}\). Thus the tensor can be written as \(P\delta_{ij}+\pi_{ij}\), \(\pi_{ij}\) is a trace-less symmetric tensor. Then the momentum equation can be re-written as,

\[\partial_t(\rho u_i)+\partial_j(\rho u_iu_j)=- \partial_j P - \partial_j\pi_{ij} + \rho\bar{a}_i.\]

One notation should be clarified is that \(u_i=\int v_i f{\rm d^3 v}\), thus the velocity can be written as \(v_i=u_i+\delta v_i\).

The energy is highly related with the pressure tensor. The energy density can be written as,

\[\begin{aligned} \rho e &= \int \frac{1}{2}m|v|^2 f {\rm d^3 v}\\ &=\int \frac{1}{2}m({\bf u}+\delta {\bf v})^2 f {\rm d^3 v}\\ &=\frac{1}{2}m|{\bf u}|^2+\frac{1}{2}m\int \delta{\bf v}\delta{\bf v}f{\rm d^3 v}\\ &=\frac{1}{2}m|{\bf u}|^2 + \frac{\sum P_{ii}}{2}\\ &=\frac{1}{2}m|{\bf u}|^2 + \frac{3}{2}P. \end{aligned}\]

Thus the energy equation can be written as,

\[\begin{aligned} \partial_t(\rho e)&=\frac{1}{2}m\int |{\bf v}|^2 \partial_t f{\rm d^3 v}\\ &=-\frac{1}{2}m\int |{\bf v}|^2 (v_i\partial_if + \partial_{v_i}(a_if)) {\rm d^3 v}\\ &=-\frac{1}{2}m\partial_i \int |{\bf v}|^2v_i f {\rm d^3 v}+\text{integral by part for the second term}\\ &=-\frac{1}{2}m\partial_i \int |{\bf v}|^2v_i f {\rm d^3 v} + m\int v_ia_if{\rm d^3 v}\\ &=-\frac{1}{2}m\partial_i \int |{\bf v}|^2v_i f {\rm d^3 v} + \rho u_i\bar{a}_i. \end{aligned}\]

Separately, consider the first term in the energy equation.

\[\begin{aligned} m\int |{\bf v}|^2v_i f {\rm d^3 v}=&m\int (u_j+\delta v_j)^2(u_i + \delta v_i)f{\rm d^3 v}\\ =&m\int \{ |u|^2u_i f + |\delta v|^2u_i f + 2u_j\delta v_ju_if\\ &+|u|^2\delta v_i f + |\delta v|^2\delta v_i f + 2u_j\delta v_j\delta v_if\}{\rm d^3 v}\\ =&\rho u^2u_i+3u_iP+2P_{ij}u_j + \int|\delta v|^2\delta v_i f{\rm d^3 v}\\ =&\rho u^2u_i+3u_iP+2(P\delta_{ij}+\pi_{ij})u_j + \int|\delta v|^2\delta v_i f{\rm d^3 v}\\ =&\rho u^2u_i+5u_iP+2\pi_{ij}u_j + \int|\delta v|^2\delta v_i f{\rm d^3 v}\\ =&\rho u^2u_i+5u_iP+2\pi_{ij}u_j + 2Q_i, \end{aligned}\]

where, \(Q_i=\frac{1}{2}\int|\delta v|^2\delta v_i f{\rm d^3 v}\) is the heat flux. Therefore, the energy equation can be written as,

\[\partial_t(\rho e)+\partial_i(\frac{1}{2}\rho u^2u_i+\frac{5}{2}Pu_i)=-\partial_i(\pi_{ij}u_j)-\partial_iQ_i+\rho u_i\bar{a}_i\]

Till now, we need the further information to enclose the equation. Now, the equations are now depend on the specific distribution, expect the isotropic hypothesis. The higher order information is hidden in the heat flux term, which is needed to enclose the equation.

Collision

So far we did not consider the collision term \(\mathcal{C}(f)\). \subsection{Heat Flux} It is obvious that under the even symmetric distribution without collision, e.g.~Maxwell and \(\kappa\) distributions, the heat flux is zero, \(Q_i=\frac{1}{2}\int \delta v_j\delta v_j\delta v_i{\rm d^3 v}\), each term is in odd order inside the integral.

If we have the BGK collision operator.

Maxwell Distribution

Using the Maxwell-Boltzmann distribution,

\[\begin{aligned} f({\bf v})&=N\Pi_i f(v_i),\\ &=N\Pi_i\frac{1}{\sqrt{2\pi}v_{th}}\exp(-\frac{v_i^2}{2v_{th}^2}),\\ &=\frac{N}{(2\pi)^{3/2}v_{th}^3}\exp(-\frac{v^2}{2v_{th}^2}), \end{aligned}\]

where, \(v_{th}^2=k_BT/m\).

The pressure tensor is,

\[\begin{aligned} P_{ij}&=m\int\delta v_i\delta v_j f{\rm d^3 v}\\ &=m\int \frac{n}{(2\pi)^{3/2}v_{th}^3}\exp(-\frac{\delta v^2}{2v_{th}^2}) \delta v_i\delta v_j {\rm d^3 v}\\ &= \frac{m}{3}\delta_{ij}\int_0^{\infty} \frac{n}{(2\pi)^{3/2}v_{th}^3}\exp(-v^2/2v_{th}^2)v^2 4\pi v^2{\rm dv}\\ &= \frac{mn}{3}\delta_{ij}\int_0^{\infty} \frac{8v_{th}^2}{(\pi)^{1/2}}\exp(-(\frac{v}{\sqrt{2}v_{th}})^2)(\frac{v}{\sqrt{2}v_{th}})^4 {\rm d(\frac{v}{\sqrt{2}v_{th}}})\\ &= \frac{mn}{3}\frac{8v_{th}^2}{\sqrt{\pi}}\delta_{ij}\int_0^{\infty}\exp(-x^2)x^4dx\\ &= \frac{mn}{3}\frac{8v_{th}^2}{\sqrt{\pi}}\frac{3\sqrt{\pi}}{8}\delta_{ij}\\ &=mnv_{th}^2\delta_{ij}\\ &= nk_BT\delta_{ij}. \end{aligned}\]

Thus the state equation is \(P=nk_BT\).

\(\kappa\) Distribution

TBC.