GATE 2025 PH – Question 54
The energy of a free relativistic particle of rest mass $m$ moving along x is $T$. In a potential $V(x)$, $H=T+V(x)$. With speed $v$, momentum $p$ and Lagrangian $L$, which option(s) is/are correct?
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Correct answer: (A) $H=c^2\sqrt{m^2+p^2/c^2}+V(x)$; (B) $v=pc/\sqrt{p^2+m^2c^2}$; (D) $L=-mc^2\sqrt{1-v^2/c^2}-V(x)$
Explanation
**Energy and Hamiltonian.** The relativistic energy of a free particle is $T=\sqrt{m^2c^4+p^2c^2}=c^2\sqrt{m^2+p^2/c^2}$, so
$$H=c^2\sqrt{m^2+\frac{p^2}{c^2}}+V(x).\quad(\text{A true})$$
**Velocity** from Hamilton's equation:
$$v=\frac{\partial H}{\partial p}=\frac{pc^2}{\sqrt{m^2c^4+p^2c^2}}=\frac{pc}{\sqrt{p^2+m^2c^2}}.\quad(\text{B true})$$
**Lagrangian.** The relativistic Lagrangian of a free particle is $L=-mc^2\sqrt{1-v^2/c^2}$, with the potential subtracted:
$$L=-mc^2\sqrt{1-\frac{v^2}{c^2}}-V(x).\quad(\text{D true, C has the wrong sign})$$
(Check: for small $v$, $L\approx-mc^2+\tfrac12mv^2-V$, which is the usual $T-V$ plus a constant.)
Answer **A, B and D**.