Past exam of the theoretical physics I course of the University of Cambridge / 2018 / 4

The derivatives of the Lagrangian density are ${\mathcal{L}}_{\varphi }=0$, ${\mathcal{L}}_{{\varphi }_t}={\varphi }_t$, and ${\mathcal{L}}_{{\varphi }_x}=-{\varphi }_x$. The Euler-Lagrange field equation is thereforeThis is the one-dimensional wave equation with wave speed $c_s=\sqrt{\kappa /\rho }=1$. Both signs of propagation are allowed: $\varphi (x,t)=F(x-t)+H(x+t)$ for sufficiently regular profiles.

Use $x^{\mu }=(t,x)$ and Minkowski metric ${\eta }^{\mu \nu }=\mathrm{diag}(1,-1)$, with the wave speed equal to one. Regarding $\varphi$ as a scalar field, the symmetric canonical stress-energy tensor isIt obeys ${\partial }_{\mu }{T}^{\mu \nu }=0$ on the wave equation. For a scalar field the intrinsic spin current vanishes. The total orbital Lorentz current and its integrated angular-momentum tensor areBecause there is only one spatial dimension, its only independent possibly nonzero component iswhere $P=\int pdx$. The corresponding flux is ${\mathcal{J}}^{101}=t\mathcal{E}-xp$. Symmetry and conservation of $T$ giveThus $J^{01}$ is conserved provided the boundary flux vanishes and the integrals exist. For example, localized wave packets on an infinite line have this property. One can also check directly that $d\int x\mathcal{E}dx/dt=P$, so the two terms in $J^{01}$ have cancelling derivatives.

The significance is the boost charge of a one-dimensional scalar field. There is no independent spatial rotation or ordinary spatial angular-momentum component in one dimension. Instead, the action has the mathematical Lorentz invariance of a massless wave field with invariant speed $c_s=1$, and $J^{01}$ generates its boost. Its conservation says that the center of energy moves uniformly. This is a symmetry of the ideal wave-field model, rather than an assertion of fundamental vacuum Lorentz symmetry for a material rod. Fixed boundaries can break it, and a general solution with nonvanishing boundary flux need not conserve the integrated charge.

With the PDF's inverse Fourier transform convention, the derivatives act asso its forced differential operator becomes ${\omega }^2-k^2+i\gamma \omega k^2$. ThusFor the physical impulse response choose the retarded fundamental solution, vanishing before $t=0$. For $k\ne 0$ and $\gamma >0$, its poles lie in the lower half-plane. The retarded Fourier pole prescription $\omega \mapsto \omega +i0$ specifies the zero mode and the zero-damping limit as well; a Green function without a temporal or boundary prescription would otherwise permit homogeneous additions.

This is the Kelvin-Voigt rod Green function. Multiplying the homogeneous real-space equation by $-1$ givesA spatial Fourier mode therefore satisfies ${\ddot{\varphi }}_k+\gamma k^2{\dot{\varphi }}_k+k^2{\varphi }_k=0$: positive $\gamma$ really damps the motion. For sufficiently decaying fields its energy satisfiesThis checks the sign of the imaginary term in $\hat{G}$ independently of the transform calculation.

For $0<k^2<4/{\gamma }^2$, defineThe simple poles are ${\omega }_{\pm }=\pm {\Omega }_k-i{a}_k$. In the inverse transform $\tilde{G}(k,t)=\int e^{-i\omega t}\hat{G}d\omega /(2\pi )$, close the contour below for $t>0$. Its clockwise orientation gives $-i$ times the sum of residues. That sum isFor $t<0$, close above; there are no poles. The residue theorem therefore givesHere $\Theta$ is the Heaviside step function. At $k=0$ the continuous retarded limit is $-t\Theta (t)$. At critical damping, ${\Omega }_k\to 0$ likewise replaces $\sin ({\Omega }_{k}t)/{\Omega }_k$ by $t$; beyond the stated oscillatory range an imaginary ${\Omega }_k$ converts it to a hyperbolic sine. These continuations retain decaying temporal exponents.

For each fixed $k$, taking $\gamma \to 0^{+}$ givesFor $t>0$, invert in space and set $s=kt$:The correctly normalized sinc function transform isFor instance, $\sin s/s=\frac{1}{2}{\int }_{-1}^1{e}^{isy}dy$, and Fourier inversion immediately gives the factor $\pi$. The PDF's prefactor $\sqrt{2/\pi }$ is inconsistent with its top hat of unit height; it would give height $\sqrt{2\pi }$. Using the normalized identity yieldsas a distribution. The values assigned precisely at a wave front do not change that distribution. This is the negative of the standard one-dimensional retarded wave Green function, because the PDF's differential operator is $-{\partial }_t^2+{\partial }_x^2$.

The sign and initial data can be checked without any contour calculation. Integrating the spatial-Fourier equation through the impulse givesIn real space these are $G_R(x,0^{+})=0$ and ${\partial }_t{G}_R(x,0^{+})=-\delta (x)$. In particular,for a smooth test function $f$, confirming the negative delta-function initial velocity. Before the impulse the displacement and velocity are zero. The disturbance fills the forward causal cone for the constant-speed wave equation, whose fronts move at unit speed, consistent with $\rho =\kappa =1$. The final question's printed $\rho =k$ is interpreted as $\rho =\kappa$, referring to the elastic constant, not to the Fourier wavevector. No initial conditions were otherwise specified, so the causal choice has been made explicitly.