Past exam of the theoretical physics I course of the University of Cambridge / 2018 / 3

If $L^{\prime }=L+dF(q,t)/dt$, the action changes by an endpoint term:In the principle of stationary action with fixed endpoints, $\delta q(t_i)=\delta q(t_f)=0$ and the endpoint times are fixed. Therefore $\delta (S^{\prime }-S)=0$, and the two actions give the same bulk Euler-Lagrange equations.

For a Lagrangian density, adding ${\partial }_{\mu }{K}^{\mu }$ changes the action by ${\int }_{\partial \Omega }{K}^{\mu }{n}_{\mu }d\Sigma$, by the divergence theorem. Its variation vanishes for compactly supported field variations, or when the appropriate boundary data are held fixed. These total divergences in a variational principle can therefore be discarded when deriving bulk equations under those conditions.

The qualification about boundaries matters: if $K^{\mu }$ depends on field derivatives, their variations on the boundary must also be controlled. A total derivative can change the canonical momenta or the chosen representative of a Noether current, even while leaving the bulk dynamics unchanged.

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.

Put $R=h_t-\nu {\nabla }^{2}h$, where the diffusion coefficient $\nu$ is constant. ThenOne temporal and two spatial integrations by parts givewith compactly supported variations. HenceHere ${\nabla }^4=({\nabla }^2{)}^2$ is the biharmonic operator. The cross terms cancel because $\nu$ is constant and spatial and temporal derivatives commute. Equivalently, $P^0=2R$ and $Q^{ij}=-2\nu R{\delta }^{ij}$ in part (b), producing $-2{\partial }_{t}R-2\nu {\nabla }^{2}R=0$.

This squared diffusion-residual action does not have only the first-order diffusion equation as its stationary equation. The zero-residual paths $h_t=\nu {\nabla }^{2}h$ have $S=0$ and are certainly solutions, but the variational equation is the higher-order one above. For a spatial Fourier mode with nonzero wavevector $\mathbf{k}$, it givesBoundary data or an additional physical selection principle are needed to retain only the decaying diffusion branch. The spatially uniform mode instead has $h_0(t)=A_0+B_{0}t$ in the stationary equation.

With constant $\nu$, the bulk action is invariant under time translations, translations in the two spatial coordinates, and rotations or reflections within the substrate plane. The Laplacian has no preferred direction, giving the planar $O(2)$ symmetry. It also has scalar-field shift symmetry $h\mapsto h+a$, expressing the arbitrary origin of height, and the discrete symmetry $h\mapsto -h$ because its residual is squared.

There is an additional exact tilt symmetry $h\mapsto h+\mathbf{b}\cdot \mathbf{x}$: the added slope is time independent and has zero Laplacian. More generally, adding any function $f$ satisfying $f_t=\nu {\nabla }^{2}f$ leaves the residual unchanged. Such bulk transformations need not preserve prescribed boundary conditions or a restricted physical interpretation of the height.

In two spatial dimensions the action also has the anisotropic scale invarianceThe spacetime measure scales as $b^4$ and the squared residual as $b^{-4}$. The time and length scales are therefore related diffusively, rather than by a Lorentz transformation.

Ordinary time reversal needs a boundary qualification. The identityshows that the density is time-reversal invariant up to a total divergence. Thus the bulk stationary equation $h_{tt}-{\nu }^2{\nabla }^{4}h=0$ is invariant under $t\mapsto -t$, although the printed density is not pointwise unchanged. The time reversal of the squared diffusion-residual action exchanges the growing and decaying branches; the selected first-order diffusion equation alone is not time-reversal invariant at fixed positive $\nu$.