Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 309 / 1

The proper time elapsed along the smooth timelike curve isThe expression is invariant under every orientation-preserving reparametrization of the curve.

Write ${\dot{x}}^{\mu }=d{x}^{\mu }/ds$ and $L=\sqrt{-g_{\mu \nu }{\dot{x}}^{\mu }{\dot{x}}^{\nu }}$. Varying $T=\int Lds$ with fixed endpoints and then choosing proper time as parameter, for which $L=1$, gives the Euler-Lagrange equationExpanding the derivative, raising the free index with the inverse metric, and using the symmetry of ${\dot{x}}^{\alpha }{\dot{x}}^{\beta }$ gives the geodesic equationThese are the Christoffel symbols of the Levi-Civita connection.

For the quadratic geodesic Lagrangianthe Euler-Lagrange equation isRaising $\mu$ and symmetrizing the coefficient of the two velocities produces exactlyThe quadratic action fixes an affine parameter; proper time is an affine parameter for a timelike geodesic.

Put $u=t-z$. Wherever $|u|>U$, the profile $A(u)$ vanishes and the displayed metric is exactly the Minkowski metric. Thus the two flat regions in the $(t,z)$-plane are the half-planes $t-z>U$ and $t-z<-U$, separated by the stripAt fixed $t$ this strip is $t-U\le z\le t+U$, so it has longitudinal width $2U$. A surface $u=\text{constant}$ obeys $z=t-u$ and travels in the positive $z$ direction at the speed of light. The curvature is confined to that moving strip, identifying the solution as a gravitational-wave pulse represented by a plane-fronted gravitational wave.

With $u=t-z$, the quadratic geodesic Lagrangian isIts transverse Euler-Lagrange equations areThe longitudinal equations areafter using the difference of those same equations. In particular,This conserved quantity also follows from the Killing vector field ${\partial }_t+{\partial }_z$ and the geodesic conserved quantity from a Killing vector.

The initial data and $\ddot{u}=0$ give $u=\tau$. To first order in $ϵ$, replace $x$ and $y$ on the right-hand sides of the transverse equations by $x_0$ and $y_0$. Twice integrating with the initial rest conditions givesFor $\tau >U$, define $A_0={\int }_{-\infty }^{\infty }a(s)ds$ and $A_1={\int }_{-\infty }^{\infty }s,a(s)ds$. ThenHenceThese permanent changes are forms of displacement memory and velocity memory.

There is a discrepancy in the question's final instruction. The longitudinal equation actually givesand thereforeThus $z$ does not vanish to first order for a general allowed profile. It vanishes after the pulse under the additional hypothesis $A_0=0$ used in part (iv), but it need not vanish while that pulse is passing. The proper-time normalization independently gives the same relation $\dot{z}=xyA+O({ϵ}^2)$.

Now $A_0=0$, so every mass is again at rest after the pulse and $z=0+O({ϵ}^2)$. Put $k=ϵA_1>0$. The final transverse coordinates areThe unit vectors along $x=y$ and $x=-y$ are eigenvectors with eigenvalues $1-k$ and $1+k$, respectively. Consequently an initial circle becomes, to first order, an ellipse compressed along the line $x=y$ and stretched along $x=-y$. This persistent deformation is displacement memory.