Is always a Lagrangian L=T-V ?

[Karlisbad]

That's my question..although in more general cases $$L=T-V$$

$$H=T+V$$ however there're several important exceptions..for example:

a) Classically (Non relativisitc) the Gravitational "Energy" (=Hamiltonian for a time-independent Potential) is:

$$H=(1/2)\int_{V}\rho (\gra \phi)^{2}$$

b) Einstein-HIlbert Lagrangian $$L=\sqrt (-g) R$$ -g is the

determinant of the metric and R is Ricci scalar.

Is there always a kind of "transform" so you can always split te Lagrangian into a Kinetic and a potential terms...

 

[StatMechGuy]

First of all, it is not sufficient that the hamiltonian have a time-independent potential for it to equal the energy. That's only for the hamiltonian to be conserved in time (actually, it's that $$dH/dt = \partial_t L$$ that you need). Also, you might want to look up the situation of a charged particle in the magnetic field. the lagrangian is decidedly NOT T - V, but the hamiltonian does equal thte energy of the charged particle (does not include the energy of the field).