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## Main Question or Discussion Point

In classical mechanics, if we consider a force field (uniform or non-uniform) in which the acceleration [itex]\vec{a}_A[/itex] of a particle A is constant, then

[tex]{\vphantom{\int}} \vec{a}_A = \vec{a}_A[/tex][tex]\int \vec{a}_A \cdot d\vec{r}_A = \int \vec{a}_A \cdot d\vec{r}_A[/tex][tex]{\vphantom{\int}} \Delta \; {\textstyle \frac{1}{2}} \; \vec{v}_A^{\;2} = \Delta \; \vec{a}_A \cdot \vec{r}_A[/tex][tex]{\vphantom{\int}} \Delta \; {\textstyle \frac{1}{2}} \; \vec{v}_A^{\;2} - \, \Delta \; \vec{a}_A^{\vphantom{^{\;2}}} \cdot \vec{r}_A^{\vphantom{^{\;2}}} = 0[/tex][tex]{\vphantom{\int}} m_A^{\vphantom{^{\;2}}} \left( \Delta \; {\textstyle \frac{1}{2}} \; \vec{v}_A^{\;2} - \, \Delta \; \vec{a}_A^{\vphantom{^{\;2}}} \cdot \vec{r}_A^{\vphantom{^{\;2}}} \right) = 0[/tex][tex]{\vphantom{\int}} \Delta \; T_A + \, \Delta \; V_A = 0[/tex][tex]{\vphantom{\int}} T_A + \, V_A = constant[/tex]

where

[tex]T_A = {\textstyle \frac{1}{2}} \; m_A^{\vphantom{^{\;2}}}\vec{v}_A^{\;2}[/tex][tex]V_A = - \; m_A^{\vphantom{^{\;2}}} \; \vec{a}_A^{\vphantom{^{\;2}}} \cdot \vec{r}_A^{\vphantom{^{\;2}}}[/tex]

If [itex]\vec{a}_A[/itex] is not constant but [itex]\vec{a}_A[/itex] is function of [itex]\vec{r}_A[/itex] then the same result is obtained, even if Newton's second law were not valid.

[itex]{\vphantom{aat}}[/itex]

[tex]{\vphantom{\int}} \vec{a}_A = \vec{a}_A[/tex][tex]\int \vec{a}_A \cdot d\vec{r}_A = \int \vec{a}_A \cdot d\vec{r}_A[/tex][tex]{\vphantom{\int}} \Delta \; {\textstyle \frac{1}{2}} \; \vec{v}_A^{\;2} = \Delta \; \vec{a}_A \cdot \vec{r}_A[/tex][tex]{\vphantom{\int}} \Delta \; {\textstyle \frac{1}{2}} \; \vec{v}_A^{\;2} - \, \Delta \; \vec{a}_A^{\vphantom{^{\;2}}} \cdot \vec{r}_A^{\vphantom{^{\;2}}} = 0[/tex][tex]{\vphantom{\int}} m_A^{\vphantom{^{\;2}}} \left( \Delta \; {\textstyle \frac{1}{2}} \; \vec{v}_A^{\;2} - \, \Delta \; \vec{a}_A^{\vphantom{^{\;2}}} \cdot \vec{r}_A^{\vphantom{^{\;2}}} \right) = 0[/tex][tex]{\vphantom{\int}} \Delta \; T_A + \, \Delta \; V_A = 0[/tex][tex]{\vphantom{\int}} T_A + \, V_A = constant[/tex]

where

[tex]T_A = {\textstyle \frac{1}{2}} \; m_A^{\vphantom{^{\;2}}}\vec{v}_A^{\;2}[/tex][tex]V_A = - \; m_A^{\vphantom{^{\;2}}} \; \vec{a}_A^{\vphantom{^{\;2}}} \cdot \vec{r}_A^{\vphantom{^{\;2}}}[/tex]

If [itex]\vec{a}_A[/itex] is not constant but [itex]\vec{a}_A[/itex] is function of [itex]\vec{r}_A[/itex] then the same result is obtained, even if Newton's second law were not valid.

[itex]{\vphantom{aat}}[/itex]