A relationship between Lagrangian formalism and Hamiltonian formalism

Question

In the Lagrangian formalism, The Lagrangian L = T(kinetic energy) - V(potential energy). The equations of motion for a given system is given by minimizing the action functional which is a integration of L. [1]

In the Hamiltonian formalism, The Hamiltonian H = T(kinetic energy) + V(potential energy). The equations of motion for a given system is described by Hamilton’s equations which are differential equations of H. [1]

Say it in an informal way: one formalism is 'subtract (T-V)' then 'integrate', the other formalism is 'add (T+V)' then 'differential'.

There is an interesting relationship between them: 'add' is the inverse of 'subtract', and 'differential' is the inverse of 'integrate'.

My question is : is there any deep understanding behind this relationship?

References:

1. No-Nonsense Classical Mechanics, JAKOB SCHWICHTENBERG, 2019. page 114-115.

Answer 1

There are, of course, relationships between the two but not exactly the way you described. Also In general, it's not true, $$\mathcal{L}=T-V \ \ \ \ \text{Not true in general}\ !!!$$ $$\mathcal{H}=T+V \ \ \ \ \text{Not true in general}\ !!!$$

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The first relationship can be seen as Legendre transform

$$\mathcal{H}=p\dot{q}-\mathcal{L}$$ So both Hamiltonian and Lagrangian follow the differential equations.

Another can be derived from the same $$\delta\int\mathcal{L}dt=0$$ $$\delta \int (p\dot{q}-\mathcal{H})dt=0$$ In this respect, both follow the least action principle.