(9) 變分入門 ─ 引入電磁場的Lagrangian

前面第二節和第五節

都是在探討將位能場表示成梯度即為力的方式，但實際的電磁場沒有這麼簡單，因為電場和磁場在時間和空間上都有變化，勞倫茲力和速度也有關係，所以接下來要更廣義地引入電磁場討論。

引入電磁場的EOM

\[m\frac{d^2 \vec{r}}{dt^2}=-\nabla{V}+q\vec{E}+q{\vec{v}}\times\vec{B}\]

Note:
- Second derivative identities $$ \nabla\cdot(\nabla\times\bf{F})=0 $$ $$ \nabla\times(\nabla\phi)=0 $$ - Cross product rule $$ \bf{A}\times(\nabla\times{\bf{B}})=\nabla_{\bf{B}}(\bf{A}\cdot{\bf{B}})-(\bf{A}\cdot\nabla)\bf{B} $$ - Magnetic vector potential $$ \nabla\cdot{\vec{B}}=0\Rightarrow{\vec{B}}=\nabla\times{\vec{A}} $$ - Electromagnetic field $$ \nabla\times{E}=-\frac{\partial \vec{B}}{\partial t}=-\nabla\times\frac{\partial \vec{A}}{\partial t}\Rightarrow\nabla\times\left(\vec{E}+\frac{\partial \vec{A}}{\partial t}\right)=0 $$ - Scalar potential $$ \vec{E}+\frac{\partial A}{\partial t}=-\nabla\phi $$

把以上的note結合到EOM中

\[-m\frac{d^2\vec{r}}{dt^2}-\nabla{V}-q\nabla\phi-q\frac{\partial \vec{A}}{\partial t}+q\vec{v}\times(\nabla\times\vec{A})=0\]

仿照第八節的方式，設有加*號是真實的軌跡，$\delta\vec{r}$是任意微小變化量，\(\nabla\)是對空間的梯度

\[\int_{0}^{T}\delta{\vec{r}}\cdot\left(-m\frac{d^2\vec{r}^*}{dt^2}-\nabla{V^*}-q\nabla\phi^*-q\frac{\partial \vec{A}^*}{\partial t}+q\vec{v}^*\times(\nabla\times\vec{A}^*)\right)\,dt = 0\] \[\nabla{V^*}=\nabla{V}(\vec{r})\big |_{\vec{r}\rightarrow\vec{r}^*}{}\]

前面三項

按照之前做法，得

\[\Rightarrow\delta\left(\int_{0}^{T}\frac{m}{2}\left(\frac{d\vec{r}}{dt}\right)^2-{V(r)}-q\phi(t)\,dt\right)\]

後面三項

\[\int_{0}^{T}\delta{\vec{r}}\cdot\left(-\frac{\partial\vec{A}^*}{\partial t}+\vec{v}^*\times(\nabla\times\vec{A}^*)\right)\,dt\\ =-\int_{0}^{T}\delta{\vec{r}}\cdot\left(\frac{\partial\vec{A}^*}{\partial t}-\nabla(\vec{v}^*\cdot\vec{A}^*)+(\vec{v}^*\cdot\nabla)\vec{A}^*) \right)\,dt\\ =-\int_{0}^{T}\delta\vec{r}\left(\frac{\partial\vec{A}^*}{\partial t}+(\vec{v}^*\cdot\nabla)\vec{A}^*\right)\,dt+\int_{0}^{T}\delta\vec{r}\cdot\nabla(\vec{v}^*\cdot\vec{A}^*)\,dt\\ =-\int_{0}^{T}\delta\vec{r}\frac{d\vec{A}^*}{dt}\,dt+\int_{0}^{T}\delta\vec{r}\cdot\nabla(\vec{v}^*\cdot\vec{A}^*)\,dt\\ =\int_{0}^{T}\left(-\frac{d}{dt}(\delta\vec{r}\cdot\vec{A}^*)+\frac{d\delta{\vec{r}}}{dt}\cdot\vec{A}^*\right)\,dt+\int_{0}^{T}\delta\vec{r}\cdot\nabla(\vec{v}^*\cdot\vec{A}^*)\,dt\]

第一項：邊界任意變化量$\delta\vec{r}=0$，所以積分完消掉。

第二項：

\[\frac{d\delta{\vec{r}}}{dt}=\delta\frac{d\vec{r}}{dt}=\delta\vec{v}\]

統整得

\[=\int_{0}^{T}\left(\delta\vec{v}\cdot\vec{A}^*+\delta\vec{r}\cdot\nabla(\vec{v}^*\cdot\vec{A}^*)\right)\,dt\]

又跟空間有關的是vector potential A，以及

\[\delta \vec{r}\cdot\nabla f=\delta f\]

所以

\[=\int_{0}^{T}\left(\delta\vec{v}\cdot\vec{A}^*+\vec{v}^*\cdot\delta\vec{A}\right)\,dt\] \[=\int_{0}^{T}\delta(\vec{v}\cdot{A})\,dt=\delta\int_{0}^{T}\vec{v}\cdot\vec{A}\,dt\]

統整

\[\int_{0}^{T}\delta{\vec{r}}\cdot\left(-m\frac{d^2\vec{r}^*}{dt^2}-\nabla{V^*}-q\nabla\phi^*-q\frac{\partial \vec{A}^*}{\partial t}+q\vec{v}^*\times(\nabla\times\vec{A}^*)\right)\,dt = 0\] \[\Rightarrow\delta\int_{0}^{T}\left(\frac{m}{2}\left(\frac{d\vec{r}}{dt}\right)^2-{V(r)}-q\phi(\vec{r},t)+q\frac{d\vec{r}}{dt}\cdot\vec{A}(\vec{r},t)\right)\,dt\] \[\Rightarrow\delta\int_{0}^{T}L\left(\vec{r},\frac{d\vec{r}}{dt},t\right)\,dt=0\]