well-defined LaGrangian

You WiLL SET this VERY well-defined LaGrangian STANDARD:
GUESS LaGrangian L = E - m*c^2 - nA*hbar*fA
= E - m*c^2 - pA*fA
= E - eM - eV
= E - eG
= pL*c
= h*fL.

GUESS Hamiltonian ENTHALPY = E.
Hope this helps.
```Brian


Juan R. wrote:
It is not the same. The 'correct' lagrangian is

L = square-root[g_ab (dx^a/dTau)(dx^b/dTau)]

Juan R. Center for CANONICAL |SCIENCE)
You FORGOT to dot ONE i after the OTHER.!!