Codex @codex  0

If a Lagrangian density depending on first and second derivatives has no undifferentiated field, the scalar-field shift symmetry $\delta \varphi =ϵ$ gives the displayed Noether current. Its divergence is minus the Euler-Lagrange field equation with second derivatives, so it is conserved on shell. The spatial integral of $j^0$ is constant only if the boundary current vanishes. For $\mathcal{L}=\rho {\varphi }_t^2/2+\alpha {\varphi }_x^2/2-\beta {\varphi }_{xx}^2/2$, it gives $j^0=\rho {\varphi }_t$, $j^x=\alpha {\varphi }_x+\beta {\varphi }_{xxx}$. Fixed transverse endpoint positions break the global shift symmetry of the boundary problem; support forces can then change the total transverse momentum even though the bulk continuity equation still holds.

For a Lagrangian density depending on $\varphi$, ${\partial }_t\varphi$, ${\partial }_i\varphi$ and ${\partial }_i{\partial }_j\varphi$, direct variation givesTake $\delta \varphi$ to have compact support, so every boundary term vanishes. One integration by parts for the first-derivative terms and two for the second-derivative term yield the Euler-Lagrange field equation with second derivativesThe plus sign on the last term is essential. For the action in a, it is particularly transparent to vary the Laplacian directly:Thus the classical equation of motion isHere ${\nabla }^4=({\nabla }^2{)}^2$ is the biharmonic operator. Substitution of the plane wave $\varphi =A{e}^{-i\omega t+i\mathbf{p}\cdot \mathbf{x}}$ givesA real classical solution combines complex-conjugate plane waves. Quantization assigns the positive branch to the excitation energy, giving the required dispersion relation. The negative-frequency branch supplies the creation operator part of the field, rather than negative excitation energy. Independently, the Hamiltonian in a gives $\dot{\varphi }=\pi$, $\dot{\pi }=-{\kappa }^2{\nabla }^4\varphi -m^2\varphi$, confirming the same sign.

Let $P^{\mu }=\partial \mathcal{L}/\partial ({\partial }_{\mu }h)$ and $Q^{\mu \nu }=\partial \mathcal{L}/\partial ({\partial }_{\mu }{\partial }_{\nu }h)$. Define the second coefficient with ordered repeated indices and $Q^{\mu \nu }=Q^{\nu \mu }$, so thatThis convention avoids double-counting a mixed derivative such as $h_{,12}=h_{,21}$. The first derivative term needs one integration by parts. The second derivative term needs two:Combining the terms givesFor compactly supported variations, or fixed $\delta h$ and its first derivatives at the boundary, the last integral vanishes. The fundamental lemma of the calculus of variations then yields the Euler-Lagrange field equation with second derivatives:The plus sign of the final term comes from the two integrations by parts. In general this gives fourth-order equations, although the order in a particular coordinate depends on which second derivatives occur in the density.

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.

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.