Given,
\(L= kinetic\, energy-potential\, energy = T-V\)
Consider,
\(\cfrac { \partial \, L }{ \partial \, \dot { x } } =\cfrac { \partial \, L }{ \partial \, \, x } \cfrac { \partial \, x }{ \partial \, \dot { x } } +\cfrac { \partial \, L }{ \partial \, t } \cfrac { \partial \, t }{ \partial \, \dot { x } } \)
as \(L=f(x,\dot{x},t)\)
Take derivative wrt \(t\),
\(\cfrac { \partial }{ \partial \, t } \{ \cfrac { \partial \, L }{ \partial \, \dot { x } } \} =\cfrac { \partial }{ \partial \, t } \{ \cfrac { \partial \, L }{ \partial \, x } \cfrac { \partial \, x }{ \partial \, \dot { x } } +\cfrac { \partial \, L }{ \partial \, t } \cfrac { \partial \, t }{ \partial \, \dot { x } } \} \)
Consider the second term,
\(\cfrac { \partial }{ \partial \, t } \{ \cfrac { \partial \, L }{ \partial \, t } \cfrac { \partial \, t }{ \partial \, \dot { x } } \} =\cfrac { \partial ^{ 2 }\, L }{ \partial \, t^{ 2 } } \cfrac { \partial \, t }{ \partial \, \dot { x } } -\cfrac { \partial \, L }{ \partial \, t } \cfrac { 1 }{ \{ \cfrac { \partial \dot { x } \, }{ \partial \, t } \} ^{ 2 } } \cfrac { \partial ^{ 2 }\dot { x } \, }{ \partial \, t^{ 2 } } \)
Consider the first term,
\(\cfrac { \partial \, }{ \partial \, t } \left\{ \cfrac { \partial \, L }{ \partial \, \, x } \cfrac { \partial \, x }{ \partial \, \dot { x } } \right\} =\cfrac { \partial \, }{ \partial \, t } \left\{ \cfrac { \partial \, L }{ \partial \, \, x } \right\} \cfrac { \partial \, x }{ \partial \, \dot { x } } +\cfrac { \partial \, L }{ \partial \, \, x } \cfrac { \partial \, }{ \partial \, t } \left\{ \cfrac { \partial \, x }{ \partial \, \dot { x } } \right\} \)
\( =\cfrac { \partial ^{ 2 }\, L }{ \partial \, t\partial \, \, x } \cfrac { \partial \, x }{ \partial \, \dot { x } } +\cfrac { \partial \, L }{ \partial \, \, x } \cfrac { \partial }{ \partial \, \dot { x } } \left\{ \cfrac { \partial \, x }{ \partial \, t } \right\} \)
\( =\cfrac { \partial ^{ 2 }\, L }{ \partial \, t\partial \, \, x } \cfrac { \partial \, x }{ \partial \, \dot { x } } +\cfrac { \partial \, L }{ \partial \, \, x } \)
For the Euler-Lagrange equation to be true,
\( \cfrac { \partial }{ \partial \, t } \{ \cfrac { \partial \, L }{ \partial \, \dot { x } } \} =\cfrac { \partial \, L }{ \partial \, x } \)
to be true, we have,
\(\cfrac { \partial ^{ 2 }\, L }{ \partial \, t\partial \, \, x } \cfrac { \partial \, x }{ \partial \, \dot { x } } +\cfrac { \partial ^{ 2 }\, L }{ \partial \, t^{ 2 } } \cfrac { \partial \, t }{ \partial \, \dot { x } } -\cfrac { \partial \, L }{ \partial \, t } \cfrac { 1 }{ \{ \cfrac { \partial \dot { x } \, }{ \partial \, t } \} ^{ 2 } } \cfrac { \partial ^{ 2 }\dot { x } \, }{ \partial \, t^{ 2 } } =0\)
But,
\(\cfrac { \partial ^{ 2 }\, L }{ \partial \, t\partial \, \, x } =\cfrac { \partial ^{ 2 }\, L }{ \partial \, t\partial \dot { x } } \cfrac { \partial \, \dot { x } }{ \partial \, x } +\cfrac { \partial ^{ 2 }\, L }{ \partial \, t\partial t } \cfrac { \partial \, { t } }{ \partial \, \dot { x } } \cfrac { \partial \, \dot { x } }{ \partial \, x } \)
We have,
\(\cfrac { \partial ^{ 2 }\, L }{ \partial \, t\partial \dot { x } } +2\cfrac { \partial ^{ 2 }\, L }{ \partial \, t^{ 2 } } \cfrac { \partial \, t }{ \partial \, \dot { x } } -\cfrac { \partial \, L }{ \partial \, t } \cfrac { 1 }{ \{ \cfrac { \partial \dot { x } \, }{ \partial \, t } \} ^{ 2 } } \cfrac { \partial ^{ 2 }\dot { x } \, }{ \partial \, t^{ 2 } } =0\)
\(\cfrac { \partial \, L }{ \partial \, x } +2\cfrac { \partial ^{ 2 }\, L }{ \partial \, t^{ 2 } } \cfrac { \partial \, t }{ \partial \, \dot { x } } -\cfrac { \partial \, L }{ \partial \, t } \cfrac { 1 }{ \{ \cfrac { \partial \dot { x } \, }{ \partial \, t } \} ^{ 2 } } \cfrac { \partial ^{ 2 }\dot { x } \, }{ \partial \, t^{ 2 } } =0\)
\(\cfrac { \partial \, L }{ \partial \, x } =\cfrac { 1 }{ \ddot { x } } \left\{ \cfrac { \dddot { x } }{ \ddot { x } } \cfrac { \partial \, L }{ \partial \, t } -2\cfrac { \partial ^{ 2 }\, L }{ \partial \, t^{ 2 } } \right\} \)
If \( \dddot { x }=0 \), ie the forces in the fields do not change with time,
\(\cfrac { \partial ^{ 2 }\, L }{ \partial \, t^{ 2 } } =-\cfrac{\ddot { x } }{2}\cfrac { \partial \, L }{ \partial \, x } \)
That going through a \(L\) gradient of \( \cfrac { \partial \, L }{ \partial \, x } \) with acceleration \(\ddot{x}\), the second rate of change in \(L\) experienced by the body is given by the expression above.
Obviously when \(\sum F=0\), \(\ddot{x}=0\) and \(\cfrac { \partial ^{ 2 }\, L }{ \partial \, t^{ 2 } } =0\) this implies,
\(\cfrac { \partial \, L }{ \partial \, t } = A(x,\dot{x})\)
\(A\) is a function in \(x\) and \(\dot{x}\) not necessarily zero. Consider the total energy of the system,
\(E=L+2V\)
\( \cfrac { \partial\,E }{ dt } =\cfrac { \partial\,L }{ \partial\,t } +2\cfrac { \partial\,P }{\partial\,t } \)
when \(\ddot{x}=0\), \(\dot{x}\) is constant, \(T\) is a constant,
\( \cfrac { \partial\,E }{ \partial\,t } =0\)
\(\cfrac{\partial\,\,T}{\partial\,\,t}=0\)
the change in \(L\) is solely due to the change in \(P\)
\(\cfrac { \partial\,P }{ \partial\,t }=-\cfrac{1}{2}\cfrac { \partial\,L }{ \partial\,t }=-\cfrac{1}{2}A(x,\dot{x})=\cfrac{1}{4}\int{\ddot { x } \cfrac { \partial \, L }{ \partial \, x } }dt\) ---(**)
An expression in \(\ddot{x}\) is different from the value of \(\ddot{x}\).
\(\cfrac { d^2P }{ dt^2 }=\cfrac{1}{4}\ddot { x } \cfrac { \partial \, L }{ \partial \, x }=0\)
as we started with \(\ddot{x}=0\)
This suggests that when a body is in circular motion (for example electrons, protons in orbit), \(\ddot{x}\) in the direction of \(\dot{x}\) is zero, the centripetal force does no work and \(\dot{x}\) is a constant, there can still be a change in potential energy with time given by expression (**). When
\(\cfrac { \partial\,P }{ \partial\,t }\gt0\) energy is absorbed,
\(\cfrac { \partial\,P }{ \partial\,t }\lt0\) energy is radiated
Is this the explanation why energy tend to diffuse throughout a material? Because the orbiting particles are constantly absorbing and radiating energy? Only if,
\(A(x,\dot{x})\ne0\).
In general this is possible when the system is periodic as in circular motion where over one period, the net gain in energy is zero. Which brings us to \(\phi\) from the post "Catastrophe! No Small Change"...until next time...
