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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.

In the end, the Langrangian and Hamiltonian are different quantities with different origins, behaviors, and differential equations. It is unreasonable to expect all but the simplest to follow from addition.

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@NickKidman I can try, but it really seems like you're in a better position to write an answer like that :P – Manishearth Dec 28 '13 at 19:46
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

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