Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 9

The generalized momentum of this Lagrangian isThe Euler-Lagrange equation therefore givesBecause $\mathcal{L}$ has no explicit time dependence, the energy form of the Beltrami identity gives a conserved quantity $E=\dot{x}\partial \mathcal{L}/\partial \dot{x}-\mathcal{L}$. HereFor a direct check of conservation of energy, differentiate:This is the rest-plus-kinetic energy of a relativistic particle, in units with light speed one, plus its potential energy.

Write $\Delta V=V_2-V_1>0$ and ${\gamma }_i=(1-v_i^2{)}^{-1/2}$, the initial Lorentz factor. The conserved energy from part (a) is $E=m{\gamma }_i+V_1$. To emerge on the higher plateau with positive speed, the particle needs $E>m+V_2$. HenceAbove this threshold, conservation of energy gives ${\gamma }_f={\gamma }_i-\Delta V/m>1$. The transmitted velocity is positive, soBelow threshold there is a turning point within the steep transition, where $\dot{x}=0$ and $V=E-m<V_2$. The particle returns to the lower plateau. Its final potential energy equals its initial value, so its final Lorentz factor equals ${\gamma }_i$ but its direction is reversed:At exact threshold the limiting speed on the upper plateau is zero. This is the marginal case, not passage with positive outgoing speed. The ideal discontinuous potential does not specify the detailed trajectory at equality; a smooth transition can make the particle approach the top asymptotically. The final-speed formulas above cover the strictly transmitted and reflected regimes requested.