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 P: 36 Reformulation: In classical mechanics, if we consider a force field (uniform or non-uniform) in which the acceleration $\vec{a}_A$ of a particle A is constant, then $$\vec{a}_A - \, \vec{a}_A = 0$$ $$\left( \vec{a}_A - \, \vec{a}_A \right) \cdot \delta \vec{r}_A = 0$$ $$\int_{t_{1}}^{t_{2}} \left( \vec{a}_A - \, \vec{a}_A \right) \cdot \delta \vec{r}_A \; \, dt = 0$$ $$\delta \int_{t_{1}}^{t_{2}} \left( {\textstyle \frac{1}{2}} \; \vec{v}_A^{\;2} \, + \, \vec{a}_A^{\vphantom{^{\;2}}} \cdot \vec{r}_A^{\vphantom{^{\;2}}} \right) \, dt = 0$$ $$m_A^{\vphantom{^{\;2}}} \; \delta \int_{t_{1}}^{t_{2}} \left( {\textstyle \frac{1}{2}} \; \vec{v}_A^{\;2} \, + \, \vec{a}_A^{\vphantom{^{\;2}}} \cdot \vec{r}_A^{\vphantom{^{\;2}}} \right) \, dt = 0$$ $$\delta \int_{t_{1}}^{t_{2}} \left( T_A - \, V_A \right) \, dt = 0$$ $$\delta \int_{t_{1}}^{t_{2}} L_A \; \, dt = 0$$ where: $$T_A = {\textstyle \frac{1}{2}} \; m_A^{\vphantom{^{\;2}}}\vec{v}_A^{\;2}$$$$V_A = - \; m_A^{\vphantom{^{\;2}}} \; \vec{a}_A^{\vphantom{^{\;2}}} \cdot \vec{r}_A^{\vphantom{^{\;2}}}$$ If $\vec{a}_A$ is not constant but $\vec{a}_A$ is function of $\vec{r}_A$ then the same result is obtained, even if Newton's second law were not valid.${\vphantom{aat}}$