布朗運動

布朗運動就是愛因斯坦的博士論文，講的主要就是花粉會被水中粒子碰撞而不規則運動的行為，但是他寫的論文實在是沒什麼人看得懂，所以後來朗之萬(Langevin)寫了一個說人話的版本，就是利用Langevin equation推出布朗運動的運動行為。

Langevin equation

\[\begin{aligned} m \dot{v}=-\alpha v +F(t) \end{aligned}\] 其中\(F(t)\)表示的是表面的隨機力，而\(\alpha\)的部分是代表阻尼常數，這裡先看看如果忽略\(F(t)\)會發生甚麼事 \[\begin{aligned} m \frac{dv}{dt}&=-\alpha v\\ \frac{1}{v}dv&=-\frac{\alpha}{m}dt\\ lnv&=-\frac{\alpha}{m}t+C\\ v(t)&=v_0e^{-\frac{\alpha}{m}t} \end{aligned}\] 可以看出來，當忽略\(F(t)\)讓\(F(t)=0\)時\(v(t)\)最終會趨近於\(0\)，這就表示花粉在水中是不會隨機運動的，但現實顯然不是這樣，因此考慮\(F(t)\)的影響就是必要的。這裡先將Langevin equation左右同乘\(x\)開始 \[\begin{aligned} mx \ddot{x}&=-\alpha x\dot{x} +xF(t)\\ m\frac{d}{dt}\left ( x\dot{x} \right ) &= m \dot{x}^2-\alpha x\dot{x} +xF(t) \end{aligned}\] 接著因為\(\left \langle xF(t) \right \rangle = \left \langle x \right \rangle \left \langle F(t) \right \rangle\)的緣故 \[\begin{aligned} m\frac{d}{dt}\left \langle x\dot{x} \right \rangle &= m \left \langle \dot{x}^2\right \rangle-\alpha \left \langle x \dot{x}\right \rangle \end{aligned}\] 接著令\(\frac{1}{2}m\left \langle \dot{x}^2 \right \rangle = \frac{1}{2}k_BT\) \[\begin{aligned} m\frac{d}{dt}\left \langle x\dot{x} \right \rangle &= k_BT-\alpha \left \langle x \dot{x}\right \rangle\\ \left \langle x\dot{x} \right \rangle &= \frac{k_BT}{\alpha}+Ce^{-\frac{\alpha}{m}t} \end{aligned}\] 可以藉由邊界條件，在\(t=0\)時\(\left \langle x\dot{x} \right \rangle = 0\)來解出積分常數\(C=-\frac{k_BT}{\alpha}\)，得到 \[\begin{aligned} \left \langle x\dot{x} \right \rangle &= \frac{k_BT}{\alpha}-\frac{k_BT}{\alpha}e^{-\frac{\alpha}{m}t} \end{aligned}\] 接著代入\(\frac{1}{2}\frac{d}{dt}\left ( x^2 \right ) = x \dot{x}\)後將上面的東西積分 \[\begin{aligned} \left \langle x^2 \right \rangle &= \frac{2k_BT}{\alpha}\left [ t+\frac{m}{\alpha}e^{-\frac{\alpha}{m}t} \right ]+C \end{aligned}\] 可以發現 \[\left \langle x^2 \right \rangle = \begin{cases} \frac{2k_BT}{\alpha}t &,t>>\frac{\alpha}{m}\\ \frac{k_BT}{m}t^2 &,t<<\frac{\alpha}{m} \end{cases}\] 對這就布朗運動在時間尺度微觀和巨觀時的運動方式了。

粒子的機率分佈(Probability distribution of the particle)

我們想知道在相同具來範圍內\(\{\overset{\rightharpoonup}r,\overset{\rightharpoonup}r+d\overset{\rightharpoonup}r\}\)在時間\(t\)時找到粒子的機率\(P(\overset{\rightharpoonup}r,t)d^3r\) \[\begin{aligned} P\left(\overset{\rightharpoonup}r,t \right)&=P\left(\overset{\rightharpoonup}r,t-dt \right)+\int P\left(\overset{\rightharpoonup}r-\overset{\rightharpoonup}\Delta,t-dt\right)\xi\left(\overset{\rightharpoonup}\Delta,\overset{\rightharpoonup}r-\overset{\rightharpoonup}\Delta,dt\right) \, d^3 \Delta \\ &-\int P\left(\overset{\rightharpoonup}r,t-dt\right)\xi\left(\overset{\rightharpoonup}\Delta,\overset{\rightharpoonup}r,dt\right) \, d^3 \Delta \end{aligned}\] 其中 \[\begin{aligned} \xi\left(\overset{\rightharpoonup}\Delta,\overset{\rightharpoonup}r,dt\right)=\mbox{在}dt\mbox{下的粒子位移}\left \{ \overset{\rightharpoonup}\Delta,\overset{\rightharpoonup}\Delta+\overset{\rightharpoonup}d\Delta \right \}\mbox{的機率} \end{aligned}\] 當然這會符合 \[\begin{cases} \int \xi\left(\overset{\rightharpoonup}\Delta,\overset{\rightharpoonup}r,dt\right) \, d^3\Delta=1 \\ \xi \mbox{比較大而} \left \lvert \overset{\rightharpoonup}\Delta \right \lvert \mbox{比較小} \end{cases}\]

例子

\[\begin{aligned} P\left(\overset{\rightharpoonup}r,t \right) = \int P\left(\overset{\rightharpoonup}r,t-dt\right)\xi\left(\overset{\rightharpoonup}\Delta,\overset{\rightharpoonup}r,dt\right) \, d^3 \Delta - \int \overset{\rightharpoonup}\Delta \cdot \overset{\rightharpoonup}\nabla \left ( P\xi \right ) \, d^3 \Delta + \frac{1}{2} \int \Delta_i\Delta_j\partial_i\partial_j\left ( P\xi \right ) \, d^3 \Delta \end{aligned}\] \[\begin{cases} \mbox{第}1\mbox{部分} = P\left(\overset{\rightharpoonup}r,t-dt \right)\\ \mbox{第}2\mbox{部分} = -\partial_i\left ( p\int \overset{\rightharpoonup}\Delta \cdot \overset{\rightharpoonup}\nabla \xi \, d^3 \Delta \right ) = -\overset{\rightharpoonup}\nabla \cdot \left ( \langle \overset{\rightharpoonup}\Delta \rangle P \right )\\ \mbox{第}3\mbox{部分} = \frac{1}{2}\partial_i\partial_j\left ( P \int \Delta_i\Delta_j\xi \, d^3 \Delta \right ) = \frac{1}{2}\partial_i\partial_j\left ( \left \langle \Delta_i\Delta_j \right \rangle P \right ) \end{cases}\] 所以 \[\begin{aligned} \partial_tP=-\overset{\rightharpoonup}\nabla\cdot\left ( \frac{ \langle \overset{\rightharpoonup}\Delta \rangle}{dt}P \right )+\frac{1}{2}\partial_i\partial_j\left ( \frac{\left \langle \Delta_i\Delta_j \right \rangle}{dt} P \right ) \qquad dt\to0 \end{aligned}\] \[\begin{aligned} \frac{\langle \overset{\rightharpoonup}\Delta \rangle}{dt} = \frac{\overset{\rightharpoonup}F_{ext}}{\gamma} \end{aligned}\] \[\begin{aligned} \overset{\rightharpoonup}\Delta = \frac{\overset{\rightharpoonup}F_{ext}}{\gamma}dt + \int_t^{t+dt} \frac{\overset{\rightharpoonup}F_{ran}(t^\prime)}{\gamma} \, dt^\prime \end{aligned}\]

\[\begin{aligned} \frac{\left \langle \Delta_i\Delta_j \right \rangle}{dt} &= \left ( \sim F_{ext}^2 \right )+ \left ( \sim F_{ext\,i} \left \langle F_{ran\,j} \right \rangle\right )+\delta_{ij}\frac{A_{ran}}{\gamma^2}\\ &=\delta_{ij}\frac{A_{ran}}{\gamma^2}\\ &=2D\delta_{ij} \end{aligned}\]

\[\begin{cases} \partial_tP = -\overset{\rightharpoonup}\nabla \cdot \overset{\rightharpoonup}j\\ \overset{\rightharpoonup}j = \frac{\overset{\rightharpoonup}F_{ext}}{\gamma}P-\overset{\rightharpoonup}\nabla \left ( DP \right ) \end{cases}\]

確定\(P\)的初始條件和邊界條件就可以確定\(P\)的值

原子的平衡和穩定(steady state and equilibrium; atomic reality)

是不是穩定的看\(\partial_tP\)就知道了\(\partial_tP = 0\)就表示機率不會隨時間改變，那就平衡的，同時也表示物質的密度不會隨便亂動，因此也可以知道 \[\begin{aligned} \overset{\rightharpoonup}\nabla \cdot \overset{\rightharpoonup}j = 0 \end{aligned}\] 但是有意思的是這有表示它的旋度也等要於零嗎?顯然沒有，所以 \[\begin{aligned} \overset{\rightharpoonup}\nabla \times \overset{\rightharpoonup}j \not= 0 \end{aligned}\] 如果\(\overset{\rightharpoonup}j = 0\)並且系統達到平衡，此時的\(P=P_{eq}\) \[\begin{aligned} \frac{\overset{\rightharpoonup}F_{ext}}{\gamma}P_{eq}-\overset{\rightharpoonup}\nabla \left ( DP \right )=0 \end{aligned}\] 可以知道\(D\)對於位置\(\overset{\rightharpoonup}r\)來說是獨立的變數，並且由\(\overset{\rightharpoonup}F_{ext} = -\overset{\rightharpoonup}\nabla U\)可以得到 \[\begin{aligned} -\frac{\overset{\rightharpoonup}\nabla U}{D\gamma} &= \overset{\rightharpoonup}\nabla ln P_{eq}\\ P_{eq}&=Ce^{-\frac{U(\overset{\rightharpoonup}r)}{D\gamma}} \end{aligned}\] 而實驗上我們知道平衡時 \[\begin{aligned} P_{eq}&=Ce^{-\frac{U(\overset{\rightharpoonup}r)}{k_BT}} \end{aligned}\] 因此合併兩式就可以得到結果 \[\begin{aligned} D\gamma = k_BT \end{aligned}\]