Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is the principle of least action a tautology ?

  1. Oct 27, 2011 #1
    In classical mechanics, if we consider a force field (uniform or non-uniform) in which the acceleration [itex]\vec{a}_{\scriptscriptstyle \mathrm A}[/itex] of a particle A is constant, then

    [tex]\vec{a}_{\scriptscriptstyle \mathrm A} - \, \vec{a}_{\scriptscriptstyle \mathrm A} = 0[/tex][tex]{\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[/tex][tex]\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[/tex][tex]\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[/tex][tex]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[/tex][tex]\delta \int_{t_{1}}^{t_{2}} \left( T_{\scriptscriptstyle \mathrm A} - \, V_{\scriptscriptstyle \mathrm A} \right) \, dt = 0[/tex][tex]\delta \int_{t_{1}}^{t_{2}} L_{\scriptscriptstyle \mathrm A} \; \, dt = 0[/tex]where[tex]T_{\scriptscriptstyle \mathrm A} = {\textstyle \frac{1}{2}} \; m_{\scriptscriptstyle \mathrm A}^{\vphantom{^{\,2}}}\vec{v}_{\scriptscriptstyle \mathrm A}^{\,2}[/tex][tex]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}}}[/tex]
    If [itex]\vec{a}_{\scriptscriptstyle \mathrm A}[/itex] is not constant but [itex]\vec{a}_{\scriptscriptstyle \mathrm A}[/itex] is function of [itex]\vec{r}_{\scriptscriptstyle \mathrm A}[/itex] then the same result is obtained, even if Newton's second law were not valid.
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted