Relativistic Hamiltonian mechanics

Source: ChatGPT.
To transition from relativistic Lagrangian mechanics to Hamiltonian mechanics, we follow the same fundamental procedure as in non-relativistic mechanics: compute the canonical momentum, express the Hamiltonian, and reformulate the equations of motion. However, special care is taken to handle the relativistic forms of momentum and energy.

1. Relativistic Lagrangian:
   The Lagrangian for a free relativistic particle of mass $m$ is:

   $$L=-m{c}^2\sqrt{1-\frac{{\mathbf{v}}^2}{c^2}},$$

   where $\mathbf{v}=\frac{d\mathbf{r}}{dt}$ is the velocity.

2. Canonical Momentum:
   The canonical momentum $\mathbf{p}$ is obtained as:

   $$\mathbf{p}=\frac{\partial L}{\partial \mathbf{v}}.$$

   Substituting $L$:

   $$\mathbf{p}=\frac{-m{c}^2}{\sqrt{1-\frac{{\mathbf{v}}^2}{c^2}}}\cdot (-\frac{\mathbf{v}}{c^2})=\frac{m\mathbf{v}}{\sqrt{1-\frac{{\mathbf{v}}^2}{c^2}}}.$$

   This is the relativistic momentum:

   $$\mathbf{p}=\gamma m\mathbf{v},$$

   where $\gamma =\frac{1}{\sqrt{1-\frac{{\mathbf{v}}^2}{c^2}}}$.

3. Relativistic Hamiltonian:
   The Hamiltonian $H$ is defined as:

   $$H=\mathbf{p}\cdot \mathbf{v}-L.$$

   Substituting $L$ and expressing $\mathbf{v}$ in terms of $\mathbf{p}$, we get:

   $$\mathbf{v}=\frac{\mathbf{p}}{\gamma m},$$

   and $\gamma$ can be expressed as:

   $$\gamma =\sqrt{1+\frac{{\mathbf{p}}^2}{m^2{c}^2}}.$$

   Therefore:

   $$H=\mathbf{p}\cdot \frac{\mathbf{p}}{\gamma m}+m{c}^2\gamma =\frac{{\mathbf{p}}^2}{2\gamma m}+m{c}^2\gamma .$$

   After simplifying:

   $$H=\sqrt{{\mathbf{p}}^2{c}^2+m^2{c}^4}.$$

   This is the total relativistic energy of the particle. This expression agrees with the modulus of the four-momentum.

4. Hamilton's Equations:
   The relativistic Hamiltonian dynamics is governed by Hamilton’s equations:

   $$\frac{d\mathbf{r}}{dt}=\frac{\partial H}{\partial \mathbf{p}},\frac{d\mathbf{p}}{dt}=-\frac{\partial H}{\partial \mathbf{r}}.$$

   Since the Hamiltonian does not depend explicitly on $\mathbf{r}$ for a free particle, $\mathbf{p}$ is conserved, and:

   $$\frac{d\mathbf{r}}{dt}=\frac{\partial H}{\partial \mathbf{p}}=\frac{\mathbf{p}{c}^2}{\sqrt{{\mathbf{p}}^2{c}^2+m^2{c}^4}}.$$

   This expresses the velocity in terms of the relativistic momentum.