Past exam of the theoretical physics I course of the University of Cambridge / 2022 / 2

Vary the action $S=\int dtdx\mathcal{L}$ with compactly supported $\delta \varphi$, or impose vanishing $\delta \varphi$ and its first derivatives on the spacetime boundary. The first variation isOne integration by parts on the first-derivative term and two on the second-derivative term remove all derivatives from $\delta \varphi$. The boundary terms vanish, leavingThe fundamental lemma of the calculus of variations gives the Euler-Lagrange field equation with second derivatives:Repeated indices run over time and space. The coefficient of the symmetric second derivative is defined by the displayed variation; if ${\varphi }_{tx}$ is instead treated as a single independent mixed derivative variable, it is counted once rather than twice.

Interpret $\varphi$ as the small transverse displacement of the elastic beam. The first term is the positive kinetic energy density $\rho {\varphi }_t^2/2$, with $\rho$ the mass per unit length. The Lagrangian density is kinetic minus potential energy, so the potential density encoded by the other two terms isFor a compressed rod, transverse deflection releases compression energy: the small-slope geometric change in length is proportional to ${\varphi }_x^2/2$, so the applied compressive load lowers the potential by $\alpha {\varphi }_x^2/2$. This explains the positive sign of the $\alpha$ term in the Lagrangian. Tension would give the opposite sign and a restoring gradient term.

Bending costs elastic energy. At small slopes the curvature is ${\varphi }_{xx}$, so positive bending stiffness $\beta$ gives potential $\beta {\varphi }_{xx}^2/2$, and hence a negative term in the Lagrangian. For a uniform slender beam $\beta$ is its flexural rigidity. The competition between destabilizing compression and stabilizing bending is responsible for Euler buckling of an elastic filament.

A constant shift $\varphi \mapsto \varphi +ϵ$ leaves every derivative unchanged, and the density has no undifferentiated $\varphi$. Thus it is exactly invariant under this scalar-field shift symmetry. Localize the shift parameter temporarily to $ϵ(x,t)$:Integrating the last term by parts once identifies the Noether current for a constant shift with second derivatives:The Euler-Lagrange field equation is $\rho {\varphi }_{tt}+\alpha {\varphi }_{xx}+\beta {\varphi }_{xxxx}=0$, precisely the continuity equation ${\partial }_t{j}^0+{\partial }_x{j}^x=0$. Therefore the conserved quantity, when the endpoint current vanishes or the field has suitable periodic/decaying conditions, isthe total transverse momentum. More explicitly $dQ/dt=-[\alpha {\varphi }_x+\beta {\varphi }_{xxx}{]}_{\mathrm{ends}}$. The fixed endpoints introduced later can exert transverse forces and break the global shift symmetry of the boundary problem, so vanishing endpoint current must not be inferred merely from fixed displacement.

For this density the Euler-Lagrange field equation with second derivatives givesInsert a plane wave $\varphi =A{e}^{i(kx-\omega t)}$. The second and fourth spatial derivatives give $-k^2$ and $k^4$, respectively, soFor propagating modes, $k^2>\alpha /\beta$ and the phase velocity has magnitude $|\omega /k|=\sqrt{(\beta k^2-\alpha )/\rho }$, which depends on the wave number. Thus the waves are dispersive. For $0<k^2<\alpha /\beta$, the squared frequency is negative: the corresponding solutions grow or decay exponentially rather than propagate with a real frequency. The uniform $k=0$ displacement is a zero-frequency shift mode.

The dispersion relation shows that an infinite rod has unstable long-wave modes. A finite rod excludes sufficiently long wavelengths, so stability depends on its length and the precise end conditions. The fourth spatial derivative requires two conditions at each end; fixing transverse positions alone does not specify whether slopes are constrained.

First take displacement-fixed ends with free rotation. Setting $\varphi (0)=\varphi (l)=0$ leaves slope variations free; the bending boundary term in the first variation then gives the corresponding natural boundary conditions on the slope, ${\varphi }_{xx}(0)={\varphi }_{xx}(l)=0$. The spatial normal modes are $\sin (n\pi x/l)$, andThe first mode destabilizes first. ConsequentlyAt equality, the first mode has zero restoring force and is marginal, with a static component and a possible linear-in-time component in the linear theory. Every integer $n<l\sqrt{\alpha /\beta }/\pi$ gives an exponentially growing mode.

If "fixed" instead means clamped, the clamped boundary condition is $\varphi ={\varphi }_x=0$ at both ends. The critical compression is thenFor example, the static critical shape $1-\cos (2\pi x/l)$ obeys the four clamp conditions and the zero-frequency equation at this load. To establish that it is the first threshold, write $v={\varphi }_x$: it has zero endpoints and zero mean, and periodic extension gives ${\int }_0^l{v}_x^{2}dx\ge (2\pi /l{)}^2{\int }_0^l{v}^{2}dx$ by the zero-mean periodic Poincaré inequality, with equality at $v\propto \sin (2\pi x/l)$. This proves the critical Rayleigh quotient. Both cases are instances of the compressed Euler-Bernoulli beam and its endpoint-dependent buckling threshold. The source does not specify the endpoint rotation, so both interpretations are stated explicitly.