Contradiction between Lagrangian and Hamiltonian formalism

Question

I am currently studying an introductory course to theoretical physics. I stumbled upon something I can't seem to understand, namely:

Suppose that we have a system without any potential energy, and thus only kinetic energy $T$. The Lagrangian $\mathcal{L} = T - V = T $ is then equal to the Hamiltonian $ \mathcal{H} = T + V = T$.

The generalised momenta $p_{q_j}$ can be defined in a Lagrangian or Hamiltonian formalism:

1. Lagrangian $$ p_{q_j} = \frac{\partial \mathcal{L}}{\partial \dot{q}_j} \enspace \to \enspace \dot{p_{q_j}} = \frac{\partial \mathcal{L}}{\partial q_j} $$

2. Hamiltonian $$ \dot{p_{q_j}} = -\frac{\partial \mathcal{H}}{\partial q_j} $$

Now because of the fact that $\mathcal{L} = \mathcal{H} = T $, we get

$$ \frac{\partial T}{\partial q_j} = - \frac{\partial T}{\partial q_j} $$

This is clearly not correct, but where am I wrong in my reasoning?

Answer 1

The functional form of T is different in Hamiltonian (H) and Lagrangian (L) formalisms. In L formalism, T is a function of $q_j$ and $\dot{q}_j$, while in H formalism it's a function of $p_j$ and $q_j$. As an example take T in 2d polar coordinates $$T_L = \frac{1}{2} m\dot{r}^2 + \frac{1}{2} m r^2 \dot{\theta}^2$$ $$T_H = \frac{p_r^2}{2m} + \frac{l^2}{2mr^2}$$ Here $$\frac{ \partial T_L}{\partial r} = mr \dot{\theta}^2$$ $$- \frac{\partial T_H}{\partial r} = \frac{l^2}{mr^3} = \frac{m^2 r^4 \dot{\theta}^2}{mr^3} = mr \dot{\theta}^2 $$

Answer 2

H is not always the total energy; see the earlier comments by @CedricL and @knzhou. For example, in a moving coordinate system H is not the total energy. See the texts Symon Mechanics and Goldstein Classical Mechanics for details.