The energy of a free, relativistic particle of rest mass \( m \) moving along the \( x \)-axis in one dimension, is denoted by \( T \). When moving in a given potential \( V(x) \), its Hamiltonian is \( H = T + V(x) \). In the presence of this potential, its speed is \( v \), conjugate momentum \( p \), and the Lagrangian \( L \). Then, which of the following option(s) is/are correct?
Show Hint
- \( H = c^2 \sqrt{m^2 + \frac{p^2}{c^2}} + V(x) \)
- \( v = \frac{pc}{\sqrt{p^2 + m^2 c^2}} \)
- \( L = m c^2 \sqrt{1 - \frac{v^2}{c^2}} - V(x) \)
- \( L = -m c^2 \sqrt{1 - \frac{v^2}{c^2}} - V(x) \)
The Correct Option is A, B, D
Solution and Explanation
Step 1: The relativistic Hamiltonian for a particle is given by: \[ H = \sqrt{m^2 c^4 + p^2 c^2} + V(x). \] This represents the total energy, where \( m \) is the rest mass and \( p \) is the momentum of the particle.
Step 2: The velocity \( v \) of the particle is related to the momentum \( p \) by the relativistic relation: \[ v = \frac{pc}{\sqrt{p^2 + m^2 c^2}}. \] This expression gives the speed in terms of the relativistic momentum and mass.
Step 3: The Lagrangian \( L \) is derived from the Hamiltonian. The relativistic Lagrangian is given by: \[ L = -m c^2 \sqrt{1 - \frac{v^2}{c^2}} - V(x), \] which accounts for both the kinetic energy and the potential energy.
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