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motion_ar
#5
Nov6-11, 06:28 PM
P: 36
Reformulation:

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]\vec{a}_A - \, \vec{a}_A = 0[/tex]
[tex]\left( \vec{a}_A - \, \vec{a}_A \right) \cdot \delta \vec{r}_A = 0[/tex]
[tex]\int_{t_{1}}^{t_{2}} \left( \vec{a}_A - \, \vec{a}_A \right) \cdot \delta \vec{r}_A \; \, dt = 0[/tex]
[tex]\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[/tex]
[tex]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[/tex]
[tex]\delta \int_{t_{1}}^{t_{2}} \left( T_A - \, V_A \right) \, dt = 0[/tex]
[tex]\delta \int_{t_{1}}^{t_{2}} L_A \; \, dt = 0[/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]