# Contradiction between Lagrangian and Hamiltonian formalism

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?

• It is not always true that $H=T+V$. are you sure that if $V=0$ and $K$ is a function of $q$ that $H=T+V$? Commented Aug 7, 2021 at 16:24

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 $$\frac{ \partial T_L}{\partial r} = mr \dot{\theta}^2$$ $$- \frac{\partial T_H}{\partial r} = \frac{l^2}{mr^3} = \frac{m^2 r^4 \dot{\theta}^2}{mr^3} = mr \dot{\theta}^2$$
• It helps to make explicit what's held constant, i.e. to equate two formulae for $\dot{p}_j$ viz. the notation $\left(\frac{\partial L}{\partial q_j}\right)_{\dot{q}_j}=-\left(\frac{\partial H}{\partial q_j}\right)_{p_j}$.