Diffusion, Only If...

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...