Symmetry in Space Implies Conservation of Momentum

Theorem
The total derivative of the action $S_{12}$ from states 1 to 2 with regard to position is equal to the difference in momentum from states 1 to 2.


 * $\dfrac{d S_{12}}{dx} = p_2 - p_1.$

Proof
From Definition:Generalized Momentum and Euler-Lagrange Equations,


 * $\displaystyle 0 = \frac{d}{dt} \frac{\partial\mathcal L}{\partial \dot x} - \frac{\partial\mathcal L}{\partial x}=\dot{p}_i - \frac{\partial\mathcal L}{\partial x}$


 * $\dot{p}_i = \dfrac{\partial\mathcal L}{\partial x}$

Therefore, via Definition:Action, Differentiation Under Integral Sign and Fundamental Theorem of Calculus


 * $\displaystyle \frac{d S_{12}}{dx} = \frac{d}{dx} \int_{t_1}^{t_2}\mathcal{L}\mathrm{d}{t} = \int_{t_1}^{t_2}\frac{\partial\mathcal{L}}{\partial x}\mathrm{d}{t} =\int_{t_1}^{t_2}\dot{p}_i\mathrm{d}{t} = p_2 - p_1$