# Transforming a lagrangian to hamiltonian and vice versa

I am not refering to Legendre transform, but to something more simple.

In analytical mechanics, the Lagrangian can be described as $L=T-V$, and the Hamiltonian is if the Lagrangian doesn't explicitly depend on time, then $H=T+V$.

There a simple change of functions which I am contemplating here, basically if I write:

$U=i \sqrt{V}$, the the Lagrangian becomes: $L=T+U^2$, and the Hamiltonian becomes $H=T-U^2$.

I know it looks like meaningless, but also going from Minkowskian metric from Euclidean metric and vice versa doesn't seem like such a big deal to me, but physicist use it.

So is this change of variables between Lagrangian and Hamiltonian being used in theoretical physics?

Does it have any meaningful applications?

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Note that there are Lagrangians not of the form kinetic minus potential energy, cf. e.g. this Phys.SE post. – Qmechanic Dec 28 '13 at 19:45

$U=i \sqrt{V}$, the the lagrangian becomes: $L=T+U^2$, and the hamiltonian becomes $H=T-U^2$.
Many Lagrangians/Hamiltonians are not of the $T\pm V$ form. If there are velocity dependent potential terms or similar, this breaks down and you have to use a Legendre transform to switch between the two.
For more complicated systems (such as the ones considered these days), it may not even be immediately evident (without carrying out a Legendre transform) if the Lagrangian/Hamiltonian can be written as $T\pm V$. Since this method involves first verifying that the system is of the $T\pm V$ form, which requires a Legendre transform, I don't really see this method being of any use without the Legendre transform.
Not really, I have no deep understanding for why some things work and other don't. (PS: don't be surprised I deleted my comment, suggesting to work out the difference between the proposed transformation $V\mapsto i\sqrt{V}$ and $t\mapsto it$, because I made a crucial typo in my example.) – NikolajK Dec 28 '13 at 20:00