science
Thermal conductivity
Ref: McQuarrie “Statistical Physics”
[Diffusion equation]
\[\tilde{E} \equiv U - \left<U\right>, \tag{1} \\\\ U: \rm{Internal\:energy}\:[J]\] \[\rho\tilde{E} = \tilde{E}/V: \rm{Energy\:density} \tag{2}\] \[\rho \frac{\partial \tilde{E}}{\partial t}+ \nabla \cdot \mathbf{h} = 0: \rm{Continuity\:equation} \tag{3} \\\\ \mathbf{h}: \rm{heat\:current\:vector}\] \[\mathbf{h} = - \lambda \nabla T : \rm{Fourier's\: law} \tag{4}\] \[\lambda : \rm{thermal\:conductivity} \left[\frac{W}{m \cdot K}\right]\] \[\tilde{E} = c_vT + \rm{const.} \tag{5} \\\\ c_v: \rm{specific\: heat\: at\: constant\: volume}\](4) $\rightarrow$(3)
\[\rho \frac{\partial \tilde{E}}{\partial t}-\lambda \nabla^2T=0 \tag{6}\](5)$\rightarrow$ (6)
\[\rho c_v \frac{\partial T}{\partial t}- \lambda\nabla^2T = 0 \\\\ \rm{or} \\\\ \rho \frac{\partial \tilde{E}}{\partial t}- \frac{\lambda}{c_v}\nabla^2E = 0 \\\\ \rm{Heat\:diffusion\:equation} \tag{7}\]Fourier transformation
\[L(\mathbf{k}, t) \equiv \frac{1}{\sqrt{2\pi}^3}\iiint_\mathbf{r} \tilde{E}(\mathbf{r},t)\exp(-i\mathbf{k}\cdot\mathbf{r})d\mathbf{r} \tag{8}\]Inverse Fourier transformation
\[\tilde{E}(\mathbf{r},t) = \frac{1}{\sqrt{2\pi}^3}\iiint_{\mathbf{k}}L(\mathbf{k},t)\exp(i\mathbf{k}\cdot\mathbf{r})d\mathbf{k} \tag{9}\](9)$\rightarrow$(7)
\[\rho c_v\frac{\partial L}{\partial t} = - \lambda k^{2}L \tag{10}\]For a system consisting of $N$ particles,
\[\tilde{E}(\mathbf r,t)=\sum_{j=1}^{N}\tilde{E}\delta(\mathbf{r}-\mathbf{r}_j) \tag{11}\]Without loss of generality, we can take $\mathbf{k}$ in the $z$-direction (in other words, we assume that heat flows along the $z$-direction), then
\[\begin{eqnarray} L(\mathbf{k}, t)&=&\sum_{j=1}^{N}\tilde{E_j}e^{i\mathbf{k}\cdot\mathbf{r}_j(t)} \\\\ &=& \sum_{j=1}^{N} \tilde{E_j}e^{ikz_j(t)} \tag{12} \end{eqnarray}\](11)$\rightarrow$(10) and evaluate $\left<\rm{L.H.S.}\times{L^(\mathbf{k},0)}\right>=\left<\rm{R.H.S.}\times{L^(\mathbf{k},0)}\right>$, then we get:
\[\left<\sum_{j=1}^{N}\sum_{l=1}^{N}\tilde{E}_j(0)\tilde{E}_l(t)e^{ik[z_l(t)-z_j(0)]}\right> = \left<\tilde{E}_j(0)\tilde{E}_l(0)e^{ik[z_l(0)-z_j(0)]}\right>\exp\left(\frac{-\lambda k^2 t}{\rho c_v}\right) \tag{13}\]In equilibrium ensemble, {$\tilde{E}$}s and {$z$}s are independent in the pre-expnential factor in R.H.S. of Eq. (13), therefore,
\[\begin{align*} \left<\sum_{j=1}^N\sum_{l=1}^{N}\tilde{E}_j(0)\tilde{E}_l(0)e^{ik[z_l(0)-z_j(0)]}\right> &= \left<\sum_{j=1}^N\sum_{l=1}^{N}\tilde{E}_j(0)\tilde{E}_l(0)\cdot\delta_{jl}\cdot e^{ik[z_l(0)-z_j(0)]}\right> \\\\ &= \left<\sum_{j=1}^N\sum_{l=1}^{N}\tilde{E}_j(0)\tilde{E}_l(0)\right> \\\\ &= \left< (U-\left<U\right>)^2 \right> k_B T^2c_v \tag{14} \end{align*}\]Here we used the folloing relation:
\[\begin{eqnarray} c_v &=& \frac{\partial U}{\partial T} = \frac{\partial}{\partial T} \left( \frac{ \sum_j \epsilon_j e^{-\beta \epsilon_j} } { \sum_{j} e^{-\beta \epsilon_j} } \right) \\\\ &=& \frac{ \left( \frac{\partial}{\partial T} \left(\sum_j \epsilon_j e^{-\beta \epsilon_j} \right) \right) \left( \sum_{j} e^{-\beta \epsilon_j} \right) - \left(\sum_j \epsilon_j e^{-\beta \epsilon_j} \right) \frac{\partial}{\partial T}\left( \sum_{j} e^{-\beta \epsilon_j} \right) }{ (\sum_{j} e^{-\beta \epsilon_j})^2 } \\\\ &=& \left( \frac{-1}{k_B T^2} \right) \frac{ -\sum_j \epsilon_j^2 e^{-\beta \epsilon_j}\left(\sum_{j} e^{-\beta \epsilon_j} \right) +\left( \sum_j \epsilon_j^2 e^{-\beta \epsilon_j} \right)^2 }{ (\sum_{j} e^{-\beta \epsilon_j})^2 } \\\\ &=& \frac{-1}{k_B T^2}\left<U^2-\left<U\right>^2\right> \\\\ &=& \frac{-1}{k_B T^2}\left< \left(U-\left<U\right>\right)^2\right> \tag{15} \end{eqnarray}\]where,
\[\frac{\partial}{\partial T} = \frac{\partial}{\partial \beta}\frac{\partial \beta}{\partial T} = -\frac{-1}{k_B T^2}\frac{\partial}{\partial T}.\]L.H.S. of Eq. (13) can be expanded as :
\[\begin{equation} \left< \sum_{j=1}^{N}\sum_{l=1}^{N}\tilde{E}_j(0)\tilde{E}_l(t)\right> + ik \left< \sum_{j=1}^{N}\sum_{l=1}^{N}\tilde{E}_j(0)\tilde{E}_l(t)[z_l(t)-z_j(0)] \right> + \frac{k^2}{2}\left< \sum_{j=1}^{N}\sum_{l=1}^{N}\tilde{E}_j(0)\tilde{E}_l(t)[z_l(t)-z_j(0)]^2 \right> + \cdots \tag{16} \end{equation}\]By using energy conservation $(\sum_{l=1}^{N}\tilde{E}l(t))=\sum{l=1}^{N}\tilde{E}_l(0))$, the first term of (16) becomes:
\[\left< \sum_{j=1}^{N}\sum_{l=1}^{N}\tilde{E}_j(0)\tilde{E}_l(t)\right> = \left< \sum_{j=1}^{N}\sum_{l=1}^{N}\tilde{E}_j(0)\tilde{E}_l(0)\right> = k_B T^2 c_v \tag{17}\]The second term of (16) is expanded as:
\[ik \left< \sum_{j=1}^{N}\sum_{l=1}^{N}\tilde{E}_j(0)\tilde{E}_l(t)z_l(t) \right> - ik \left< \sum_{j=1}^{N}\sum_{l=1}^{N}\tilde{E}_j(0)\tilde{E}_l(t)z_j(0) \right> \tag{18}\]Due to energy conservation, we can substitute $\tilde E_{j(0)}$ of the first term in (18) with $\tilde E_{j(t)}$. Furthermore, due to stationaity of the ensemble, the first term in (18) becomes $ik \left< \sum_{j=1}^{N}\sum_{l=1}^{N}\tilde{E}j(0)\tilde{E}_l(0)z_l(0) \right>$. Likewise, the second term in (18) can be written as $ik \left< \sum{j=1}^{N}\sum_{l=1}^{N}\tilde{E}_j(0)\tilde{E}_l(0)z_l(0) \right>$ because we can substitute $\tilde{E}_l(t)$ with $\tilde{E}_l(0)$ due to energy conservation. Consequently, we see that the second term is (18) vanishes.
The third term in (16) is expanded as:
\[\left< \sum_{j=1}^{N}\sum_{l=1}^{N}\tilde{E}_j(0)\tilde{E}_l(t)(z_l(t))^2 \right> \- 2\left< \sum_{j=1}^{N}\sum_{l=1}^{N}\tilde{E}_j(0)\tilde{E}_l(t)z_l(t)z_j(0) \right> \+ \left< \sum_{j=1}^{N}\sum_{l=1}^{N}\tilde{E}_j(0)\tilde{E}_l(t)(z_j(0))^2 \right> \tag{19}\]The first term of (19) can be written as:
\[\begin{eqnarray} \left< \sum_{j=1}^{N}\sum_{l=1}^{N}\tilde{E}_j(t)\tilde{E}_l(t)(z_l(t))^2 \right> &=& \left< \sum_{j=1}^{N}\sum_{l=1}^{N}\delta_{jl}\tilde{E}_j(0)\tilde{E}_l(t)(z_l(t))^2 \right> \\\\ &=& \left< \sum_{j=1}^{N}\sum_{l=1}^{N}(\tilde{E}_j(t))^2(z_j(t))^2 \right> \end{eqnarray}\] \[\because \sum_{j=1}^N\tilde{E}_j(0) = \sum_{j=1}^N\tilde{E}_j(t) \rm{\:\: energy\: conservation}\]The second term of (19) can be written as :
\[-2\left< \sum_{j=1}^{N}\sum_{l=1}^{N}\delta_{jl}\tilde{E}_j(0)\tilde{E}_l(t)z_j(0)z_l(t) \right> = -2 \left< \sum_{j=1}^{N}\tilde{E}_j(0)\tilde{E}_j(t)z_j(0)z_j(t) \right>\]And the third term of (19) becomes:
\[\left< \sum_{j=1}^{N}(\tilde{E}_j(0))^2(z_j(0))^2 \right>\]Consequently, (19) is written as:
\[\left< \left[\tilde{E}_j(t)z_j(t) - \tilde{E}_j(0)z_j(0) \right]^2\right> \tag{20}\]where we denote $\sum_{j=1}^{N}\tilde{E}(t)z_j(t)$ as $A(t)$ which corresponds to “energy distribution”. Then,
\[A(t) - A(0) = \int_0^t dt'J(t')\]where $J(t)$, heat current, is defined as the time-derivative of energy distribution.
\[\left(A(t)-A(0)\right)^2 = \int_0^t dt'\int_0^t dt'' J(t')J(t'')\] \[\left< \left[ A(t)-A(0) \right]^2 \right> = \int_0^t dt'\int_0^t dt''\left<J(t')J(t'')\right>\]where we denote $\left<J(t’)J(t’’)\right>$ as $\phi(t’-t’’)$ which only depends on the time difference $t’-t’’$. Then, (20) can be written as:
\[\begin{eqnarray} \int_0^t dt'\int_0^t dt'' \phi(t'-t'') &=& \int_0^t dt' \int_0^{t'} dt'' \phi(t'-t'') + \int_0^t dt' \int_{t'}^t dt'' \phi(t'- t'') \\\\ &=& \int_0^t dt' \int_0^{t'} dt'' \phi(t'-t'') + \int_0^t dt'' \int_{0}^{t''} dt' \phi(t'- t'') \\\\ &=& \int_0^t dt' \int_0^{t'} dt'' \phi(t'-t'') + \int_0^t dt' \int_{0}^{t'} dt'' \phi(t''- t) \\\\ &=& \int_0^t dt' \int_0^{t'} dt'' [\phi(\tau) + \phi(-\tau)] \\\\ &=& 2\int_0^t dt' \int_{t'}^{0} (-d\tau) \phi(\tau) \\\\ &=& 2\int_0^t dt' \int_0^{t'} d\tau \phi(\tau) \\\\ &=& 2\int_0^t d\tau \int_{\tau}^{t} dt' \phi(\tau) \\\\ &=& 2\int_0^t d\tau (t-\tau)\phi(\tau) \tag{21} \end{eqnarray}\], where $\tau \equiv t’-t’’$ and the respective domain of integration are
(I) L.H.S. of the first line:
(II) the first term of R.H.S. of the first line:
(III) the second term of R.H.S. of the first line:
(IV) the second term of R.H.S. of the second line:
(V) R.H.S. of the sixth line:
(VI) R.H.S. of the seventh line:
We swapped the integration variable $t’ \leftrightarrow t’’$ in the second term of R.H.S. of the third line, and used the time reversibility of $\phi(\tau)$ in R.H.S. of the fourth line. We changed the variable of integration from $t’’$ to $\tau$ from the fourth to the fifth line. Note that $t’$ and $\tau$ are independent variables and therefore $\phi(\tau)$ does not depend on $t’$.
Consequently, Eq. (13) can be written as:
\[k_B T^2 c_v - \frac{k^2}{2}\cdot 2\int_0^{\tau}d\tau (t-\tau)<J(0)J(\tau)> = k_B T^2 c_v \left(1 - \frac{\lambda k^2 t}{\rho c_v} + \cdots \right) \tag{22}\], where R.H.S. of Eq. (22) is the Taylor expansion of the exponential term of R.H.S. of Eq. (13). By rewriting Eq. (22) with taking the limit of $t \rightarrow \infty$, we obtain:
\[\begin{eqnarray} \lambda &=& \frac{\rho}{k_B T^2} \lim_{t \rightarrow \infty} \int_0^t \left(1 - \frac{\tau}{t} \right) \left<J(\tau)J(0) \right> d\tau &=& \frac{1}{Vk_B T^2} \lim_{t \rightarrow \infty} \int_0^t \left<J(\tau)J(0) \right> d\tau \tag{23} \end{eqnarray}\], where we assumed fast decay of $\left<J(t)J(0) \right>$. Here we considered three dimensional isotropic fluid, and the heat current$J = \frac{d}{dt} \sum_{j=1}^{N}\tilde{E}jz_j$ corresponds to the $z$-component, $h_z$, of three dimensional heat current vector $(h_x, h_y, h_z) = \mathbf{h} \equiv \frac{d}{dt} \sum{j=1}^{N}\tilde{E}_j \mathbf{r}_j$, where $h_x = h_y = h_z$ due to isotropy. Accordingly, we can rewrite Eq. (23) as:
\[\lambda = \frac{1}{3Vk_BT^2}\lim_{t\rightarrow\infty}\int_0^t\left<\mathbf{h}(0)\cdot \mathbf{h}(\tau) \right>d\tau \tag{24}\]