(13) 變分對稱與守恆 ─ 拓展時空地位相等 Extended Configuration Space
上一章節提到將Lagrangian分別對時間和空間微分時，可以找到對應的守恆量。 不過我們早就從狹義相對論知道，時間也是一個維度，可以和空間等同視之。 現在我們也來做一樣的事情，將時間地位調到和空間相等，這個就稱做
Extended Configuration Space。
設另一個共同的參數$\lambda$
定
\[\left\{ \begin{array}{ll} t = t(\lambda)\\ q = q(\lambda) \end{array} \right. \quad\lambda:0\rightarrow 1\]邊界條件
\[\left\{ \begin{array}{ll} q(0) = q_0,\quad q(1) = q_f\\ t(0) = 0,\quad t(1) = T \end{array} \right.\]所以
\[\left\{ \begin{array}{ll} \dot{t}\equiv\frac{dt}{d\lambda}\\ \dot{q}\equiv\frac{dq}{d\lambda} \end{array} \right.,\qquad \left\{ \begin{array}{ll} Q\equiv(t,q)\\ \dot{Q}\equiv\left(\frac{dt}{d\lambda},\frac{dq}{d\lambda}\right) \end{array} \right.\]推導
\[\int_{0}^{T}L\left(q,\frac{dq}{dt},t\right)\,dt=\int L\frac{dt}{d\lambda}d\lambda\] \[=\int_0^1 \left(q(\lambda),\left(\frac{dq}{d\lambda}\cdot\frac{1}{dt/d\lambda}\right),t\right)\left(\frac{dt}{d\lambda}\right)\,d\lambda\] \[=\int_{0}^{1}L\left(q,\dot{q}\left(\frac{1}{\dot{t}}\right),t\right)\dot{t}\,d\lambda\]定
\[L'\equiv L(\cdots)\cdot\frac{dt}{d\lambda}, \quad q'\equiv\frac{\dot q}{\dot t}\]所以
\[=\int_{0}^{1}(L(q,q',t)\cdot\dot{t})\,d\lambda\] \[=\int_{0}^{1}L'(q,q',t)\,d\lambda\]得到了新的一組$L’$，注意這裡是以$\lambda$做積分。
比較：新的Euler-Lagrange Equation
\[A: \frac{d}{d\lambda}\left(\frac{\partial L'}{\partial \dot{q}}\right)=\frac{\partial L'}{\partial q}\] \[B:\frac{d}{d\lambda}\left(\frac{\partial L'}{\partial \dot{t}}\right)=\frac{\partial L'}{\partial t}\]空間項驗證A：一樣得Euler-Lagrange
\[\begin{align*} \frac{d}{d \lambda}\left[\frac{\partial}{\partial\dot{q}}(L(q,q',t)\dot{t})\right]&=\frac{\partial L(q,q',t)\dot{t}}{\partial q}\\ \frac{d}{d \lambda}\left(\frac{\partial L}{\partial q'}\frac{\partial q'}{\partial \dot{q}}\dot{t}\right)&=\frac{\partial L}{\partial q}\dot{t}\\ \frac{d}{d \lambda}\left(\frac{\partial L}{\partial q'}\left(\frac{1}{\dot{t}}\dot{t}\right)\right)&=\frac{\partial L}{\partial q}\dot{t}\\ \frac{d}{d \lambda}\left(\frac{\partial L}{\partial q'}\right)&=\frac{\partial L}{\partial q}\frac{dt}{d\lambda}\\ \frac{d}{dt}\left(\frac{\partial L}{\partial q'}\right)&=\frac{\partial L}{\partial q}\\ \frac{d}{dt}\left(\frac{\partial L\left(q,\frac{dq}{dt},t\right)}{\partial \left(\frac{dq}{dt}\right)}\right)&=\frac{\partial L}{\partial q} \end{align*}\]所以新的
\[\frac{d}{d\lambda}\left(\frac{\partial L'}{\partial \dot{q}}\right)=\frac{\partial L'}{\partial q}\]的確和第八節推的Euler-Lagrange
\[\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)=\frac{\partial L}{\partial q^*}\]是類似的。
時間項驗證B：一樣可得守恆量
\[\begin{align*} \frac{d}{d\lambda}\left[\frac{\partial}{\partial\dot{t}}\left(L\left(q,\frac{\dot{q}}{\dot{t}},t\right)\dot{t}\right)\right]&=\frac{\partial\left(L\left(q,\frac{\dot{ q}}{ \dot{t}},\dot{t}\right)\dot{t}\right)}{\partial t}\\ &=\frac{\partial L}{\partial t}\dot{t}\\ \frac{d}{d\lambda}\left(\frac{\partial L}{\partial q'}\frac{\partial q'}{\partial \dot{t}}\dot{t}+L\cdot{1}\right)&=\frac{\partial L}{\partial t}\dot{t}\\ \frac{d}{d\lambda}\left[\frac{\partial L}{\partial q'}\cdot\left(-\frac{ \dot{q}}{ \dot{t}^2}\right)\dot{t}+L\right]&=\frac{\partial L}{\partial t}\dot{t}\\ \frac{d}{d\lambda}\left(-\frac{\partial L}{\partial q'}\frac{\dot{q}}{\dot{t}}+L\right)&=\frac{\partial L}{\partial t}\dot{t}\\ \frac{d}{d\lambda}\left(-\frac{\partial L\left(q,\frac{dq}{dt},t\right)}{\partial \frac{dq}{dt}}\frac{dq}{dt}+L\right)&=\frac{\partial L}{\partial t}\dot{t}\\ \frac{d}{d\lambda}\left(p\cdot\frac{dq}{dt}-L\right)&=-\frac{\partial L}{\partial t}\dot{t}=-\frac{\partial L}{\partial t}\frac{dt}{d\lambda}\\ \frac{d}{dt}\left(p\cdot\frac{dq}{dt}-L\right)&=-\frac{\partial L}{\partial t} \end{align*}\]所以新的
\[\frac{d}{d\lambda}\left(\frac{\partial L'}{\partial \dot{t}}\right)=\frac{\partial L'}{\partial t}\]如果L不是時間t的函數，一樣可以得出
\[p\cdot\frac{dq}{dt}-L\]是不變量！
小結
之所以想把時間拉到和空間等同視之，是因為前面有提過，如果$L$不是$q_j$的函數，會得出一個守恆量，那時間呢？若一樣等同視之，經過以上推導，若$L$不是$t$的函數，也能得出一個守恆量！因此當$q$和$t$等同地位後，我們可以事情看得更清楚了！