# Is the principle of least action a tautology ?

1. Oct 27, 2011

### motion_ar

In classical mechanics, if we consider a force field (uniform or non-uniform) in which the acceleration $\vec{a}_{\scriptscriptstyle \mathrm A}$ of a particle A is constant, then

$$\vec{a}_{\scriptscriptstyle \mathrm A} - \, \vec{a}_{\scriptscriptstyle \mathrm A} = 0$$$${\vphantom{\delta \int_{t_{1}}^{t_{2}}}} \left( \vec{a}_{\scriptscriptstyle \mathrm A} - \, \vec{a}_{\scriptscriptstyle \mathrm A} \right) \cdot \delta \vec{r}_{\scriptscriptstyle \mathrm A} = 0$$$$\int_{t_{1}}^{t_{2}} \left( \vec{a}_{\scriptscriptstyle \mathrm A} - \, \vec{a}_{\scriptscriptstyle \mathrm A} \right) \cdot \delta \vec{r}_{\scriptscriptstyle \mathrm A} \; \, dt = 0$$$$\delta \int_{t_{1}}^{t_{2}} \left( {\textstyle \frac{1}{2}} \; \vec{v}_{\scriptscriptstyle \mathrm A}^{\,2} \, + \, \vec{a}_{\scriptscriptstyle \mathrm A}^{\vphantom{^{\,2}}} \cdot \vec{r}_{\scriptscriptstyle \mathrm A}^{\vphantom{^{\,2}}} \right) \, dt = 0$$$$m_{\scriptscriptstyle \mathrm A}^{\vphantom{^{\,2}}} \; \delta \int_{t_{1}}^{t_{2}} \left( {\textstyle \frac{1}{2}} \; \vec{v}_{\scriptscriptstyle \mathrm A}^{\,2} \, + \, \vec{a}_{\scriptscriptstyle \mathrm A}^{\vphantom{^{\,2}}} \cdot \vec{r}_{\scriptscriptstyle \mathrm A}^{\vphantom{^{\,2}}} \right) \, dt = 0$$$$\delta \int_{t_{1}}^{t_{2}} \left( T_{\scriptscriptstyle \mathrm A} - \, V_{\scriptscriptstyle \mathrm A} \right) \, dt = 0$$$$\delta \int_{t_{1}}^{t_{2}} L_{\scriptscriptstyle \mathrm A} \; \, dt = 0$$where$$T_{\scriptscriptstyle \mathrm A} = {\textstyle \frac{1}{2}} \; m_{\scriptscriptstyle \mathrm A}^{\vphantom{^{\,2}}}\vec{v}_{\scriptscriptstyle \mathrm A}^{\,2}$$$$V_{\scriptscriptstyle \mathrm A} = - \; m_{\scriptscriptstyle \mathrm A}^{\vphantom{^{\,2}}} \; \vec{a}_{\scriptscriptstyle \mathrm A}^{\vphantom{^{\,2}}} \cdot \vec{r}_{\scriptscriptstyle \mathrm A}^{\vphantom{^{\,2}}}$$
If $\vec{a}_{\scriptscriptstyle \mathrm A}$ is not constant but $\vec{a}_{\scriptscriptstyle \mathrm A}$ is function of $\vec{r}_{\scriptscriptstyle \mathrm A}$ then the same result is obtained, even if Newton's second law were not valid.
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