Past exam of the relativity course of the University of Cambridge / 2021 / Section B

For rest mass $m$ and electric charge $q$, the generally covariant Lorentz force law isHere $F_{ab}={\partial }_a{A}_b-{\partial }_b{A}_a$ is the electromagnetic field tensor for the chosen electromagnetic four-potential normalization, and the covariant derivative includes the spacetime Levi-Civita connection. Antisymmetry of $F_{ab}$ ensures $u_{a}D{u}^a/d\tau =0$, preserving the four-velocity norm and the particle's rest mass. An uncharged particle follows a geodesic.

Let a dot denote differentiation with respect to the five-dimensional affine parameter $\lambda =\tilde{u}$, and define $W={\dot{x}}^4+A_a{\dot{x}}^a$. Use the quadratic geodesic LagrangianThe assumed independence of $x^4$ makes it a cyclic coordinate, so the corresponding Euler-Lagrange equation isSince $\partial L/\partial {\dot{x}}^4=W$, this givesGeometrically, this is the conserved momentum associated with the fifth-direction Killing vector field ${\partial }_4$. It is the conserved fibre momentum underlying the Kaluza-Klein geodesic Lorentz force.

An affine geodesic has constant squared tangent norm, but that constant depends on the scale of its parameter. WriteUsing $W=\xi$ in the line element givesThe affine normalization in a Kaluza-Klein geodesic reduction matters: the specific formula requested in the paper uses $\mathcal{N}=1$, namely $\lambda$ equal to the positive five-dimensional arclength. With that normalization,A timelike four-dimensional projection has $d{u}^2=c^{2}d{\tau }^2>0$, requiring $|\xi |<1$ in this normalization. At $|\xi |=1$ the projection is null and has no proper-time four-velocity. For an arbitrary affine parameter the first boxed relation, with $\mathcal{N}$ retained, is the general result.

Differentiate the geodesic Lagrangian before substituting the conserved quantity. For a four-dimensional index $a$,Its Euler-Lagrange equation, together with $W=\xi$ and $\dot{W}=0$, isThus the sign of the projected Kaluza-Klein Lorentz force is fixed by the printed metric:The derivative order in the PDF's proposed equation is reversed. Its right-hand side is the negative of the result above for the stated $\xi$ and metric. This is not repaired by renaming the field tensor, because the two explicit derivative expressions have opposite signs. For example, $A_0=A_0(x^1)$, $A_1=0$ gives ${\ddot{x}}^1=-\xi A_0^{\prime }(x^1){\dot{x}}^0$ directly from the Lagrangian, whereas the printed expression gives the opposite. The corrected equation is used below.

Normalize the five-dimensional tangent as in part (c), and let $s=\sqrt{1-{\xi }^2}>0$. Since $u=c\tau$, one has $d\tau /d\lambda =s/c$, soSubstitution into the corrected equation in part (d) givesThis is the Lorentz force form in four-dimensional Minkowski spacetime. The effective charge-to-mass coefficient is proportional to the conserved fifth-direction momentum: in $c=1$ units and a potential normalization identifying $A$ with the electromagnetic four-potential, $q/m=\xi /\sqrt{1-{\xi }^2}$. The factor converting the geometrical $A$ to an SI potential fixes the physical charge units. Opposite signs of $\xi$ give opposite effective charges, and $\xi =0$ gives a neutral inertial projection. Antisymmetry gives $d({\eta }_{ab}{u}^a{u}^b)/d\tau =0$.

This is a Kaluza-Klein theory mechanism: a higher-dimensional geodesic appears as forced motion after projection. With the explicitly supplied $+---$ base signature, the positive fibre term gives a second positive direction in the five-dimensional metric. The derivation is a valid geometrical reduction of that metric; a conventional single-time spacetime would require a spacelike extra direction and corresponding sign changes. If one instead retains the sign printed in part (d), the inferred effective charge reverses, but that version is not the geodesic equation of the metric supplied here.

Use the metric signature $\eta =\mathrm{diag}(1,-1,-1,-1)$ and $c=1$ throughout this question. The trace-reversed metric perturbation satisfies $\varphi ={\eta }_{\mu \nu }{\varphi }^{\mu \nu }=-h$, so its inverse relation isSet $C_{\nu }={\partial }^{\sigma }{\varphi }_{\nu \sigma }$. Substituting this inverse relation into the linearized field operator cancels the second derivatives of the trace, leavingThe imposed Lorenz gauge in linearized gravity is $C_{\nu }=0$. Therefore the Linearized Einstein equations in vacuum reduce toThe overall sign of the field operator is immaterial for this homogeneous equation.

Let $\Psi =k_{\alpha }{x}^{\alpha }$. For a nonzero symmetric amplitude, substituting the plane wave into the wave equation gives $-(k_{\sigma }{k}^{\sigma }){ϵ}^{\mu \nu }\cos \Psi =0$. The gauge condition gives $-k_{\sigma }{ϵ}^{\nu \sigma }\sin \Psi =0$. Thus a propagating mode satisfiesThe wavevector is null, so the gravitational wave propagates at the vacuum speed of light. The amplitude is transverse as a four-dimensional tensor in Lorenz gauge in linearized gravity. This condition alone does not set all time components or the trace to zero: residual gauge symmetry of linearized gravity remains. Of the ten symmetric components, four Lorenz constraints and four independent residual coordinate choices leave two physical gravitational wave polarizations.

For the supplied plus polarization, the perturbation is traceless, so $h_{\mu \nu }={\varphi }_{\mu \nu }$. At a fixed longitudinal coordinate $z_0$, put $H(t)=h_{\oplus }\cos [\omega (t-z_0)]$. With the $+---$ convention chosen above, the transverse spatial metric measuring positive squared lengths isThe initial coordinate separation is $(\Delta \cos \theta ,\Delta \sin \theta )$ and remains constant in transverse-traceless gauge: $h_{00}=h_{0i}=0$ implies ${\Gamma }^i{}_{00}=0$, so initially stationary particles stay at fixed spatial coordinates. Their measured proper length is insteadThe first expression is understood within the linearized metric; only its first-order expansion is controlled. Separation along the two transverse axes varies oppositely; at $45$ degrees there is no first-order length modulation. This is the angular response of a gravitational-wave separation, despite constant coordinate separation.

If the background is instead chosen as $-+++$ while keeping the same numerical spatial amplitude, the displayed length modulation has the opposite sign. The question does not state a signature here; the convention above agrees with its earlier explicitly specified Minkowski metric. The physical alternating stretch-compress pattern is unchanged by consistently redefining the signed amplitude.

Use the plus-sign transformation rule for $h_{\mu \nu }$ stated in this part. A generator ${\xi }^{\mu }=b^{\mu }\sin \Psi$ with constant $b^{\mu }$ produces the pure-gauge plane wave in linearized gravityBecause $k^2=0$, it obeys both $\square {\xi }^{\mu }=0$ and $k_{\mu }\delta {ϵ}^{\mu \nu }=0$, preserving the imposed Lorenz gauge in linearized gravity.

For an explicit example choose $b^{\mu }=B{k}^{\mu }/(2{\omega }^2)$, with constant small $B$. Since $k\cdot b=0$, this givesIndeed, direct differentiation gives $\delta h_{\mu \nu }=B{k}_{\mu }{k}_{\nu }\cos \Psi /{\omega }^2$, whose trace is zero, so $\delta \varphi =\delta h$. This wave-shaped metric disturbance is only a coordinate description of flat spacetime: its linearized Riemann curvature operator vanishes. All its second-derivative terms are proportional to four factors of $k$, and the antisymmetric curvature combinations cancel.

An independent physical solution is the cross polarizationIt is symmetric, traceless and transverse to $k^{\mu }=(\omega ,0,0,\omega )$. Unlike the pure coordinate mode, it has nonzero linearized Riemann curvature operator: for example, in the convention $R_{0i0j}^{(1)}=-{\partial }_t^2{h}_{ij}/2$,Changing the curvature sign convention reverses this component but cannot make it vanish. Gauge invariance of the linearized Riemann tensor therefore proves that this nonzero mode cannot be removed by a coordinate transformation.

The principal stretch directions are rotated by $45$ degrees relative to the plus polarization. In the same proper-length convention as part (c), its first-order angular response is $\delta \ell /\Delta =-\frac{1}{2}{h}_{\times }\sin (2\theta )\cos [\omega (t-z_0)]$. The two independent transverse patterns are the two propagating gravitational degrees of freedom.

Take $\alpha >0$ and work in a connected patch with $x\ne 0$. Setting the line element to zero givesThese are the two families of Rindler null geodesics. They remain on the same side of $x=0$ and approach it only at infinite coordinate time. $x=0$ itself is not an additional geodesic within this chart, since the metric is degenerate there.

The paths can also be written $t=\pm {\alpha }^{-1}\log |x/C|$. In two dimensions a null direction has only one independent orthogonal null direction, namely itself, so each smooth null curve is a geodesic after reparametrization. More explicitly, the transformation in part (d) makes these paths straight null lines. The coordinate $t$ is generally not an affine parameter.

Put $r_s=2\mu$ and $f(r)=1-r_s/r$. Radial null geodesics of the Schwarzschild metric obeyNear the horizon, $\delta =r-r_s$ satisfies$C>0$ describes the exterior and $C<0$ the interior. The exponential approach in Schwarzschild coordinate time resembles part (a), but $\delta$ is not a proper radial distance.

Outside, set $\rho =2\sqrt{r_s\delta }$ and retain the leading near-horizon terms. The radial metric becomes the Rindler limit of a Schwarzschild horizon:It agrees with part (a) for $x=\rho$, $\alpha =c/(2r_s)$. Accordingly $\rho \propto e^{\pm ct/(2r_s)}$, while $\delta \propto {\rho }^2$ explains the factor two in its exponent.

Inside, write $r=r_s-{\rho }^2/(4r_s)$. The leading radial metric is insteadHere the radial coordinate is timelike and the Schwarzschild $t$ coordinate is spacelike. Thus $x<0$ in the Rindler metric does not represent the black-hole interior: $g_{tt}=(\alpha x{)}^2$ stays positive on both Rindler wedges. In the black-hole interior all future-directed radial light rays move towards decreasing $r$; the two signs of $dr/dt$ do not mean opposite future radial directions because $t$ is no longer time.

The horizon is a coordinate singularity in Schwarzschild coordinates and in the accelerated chart, not a curvature singularity. The full Schwarzschild geometry nevertheless has genuine tidal forces; the flat Rindler metric captures its leading local radial near-horizon structure.

For coordinates $(t,x)$ the inverse metric is $g^{tt}=1/({\alpha }^2{x}^2)$, $g^{xx}=-1$. Only ${\partial }_x{g}_{tt}=2{\alpha }^{2}x$ is nonzero, giving the Christoffel symbolswith all remaining components zero.

On the specified constant-$x_0$ worldline, assume $x_0>0$. The four-velocity is $u^t=c/(\alpha x_0)$, $u^x=0$, and its norm is $c^2$, so $\tau$ is indeed proper time. The coordinate components are constant, but the covariant acceleration is nonzero:The proper acceleration of a static Rindler observer isThe acceleration has constant components in this accelerated chart and constant invariant magnitude. It is not a covariantly constant vector: its inertial-frame components change along the hyperbolic trajectory. For $x_0<0$, the magnitude is $c^2/|x_0|$, with the appropriate choice of future time orientation.

Differentiating the supplied coordinates givesSubtract the squares. The cross terms cancel and ${\cosh }^2-{\sinh }^2=1$, yieldingThe spacetime is flat Minkowski spacetime in Rindler coordinates. $x>0$ covers $X>|cT|$, the right Rindler wedge, while $x<0$ covers the left wedge $X<-|cT|$; neither is the whole spacetime.

The null combinations areFor $x=C{e}^{-\alpha t}$ the first is constant; for $x=C{e}^{\alpha t}$ the second is constant. Thus part (a)'s exponentials are simply the inertial light rays $X\pm cT=\text{constant}$. Their coordinate-time asymptotes reflect the chart's Rindler horizon, not a failure of light to move at $c$ in an inertial frame.

Let $a>0$ be the constant proper acceleration away from the emitter and set $R=c^2/a$. Take the signal emission to be simultaneous, in the initial inertial frame, with the ship being at rest and its clock reading $\tau =0$. Place that initial ship event at $T=0$, $X=R$. Its hyperbolic motion isA photon emitted from $X=R-d$ at $T=0$ towards the departing ship follows $X_{\gamma }=R-d+cT$. At reception,Therefore light reception by a uniformly accelerated observer occurs afterThe same result follows in Rindler coordinates: the ship stays at $x=R$, the emitter starts at $x=R-d$, and the right-moving null ray has $x=(R-d)e^{\alpha t}$. The onboard lapse $d\tau =\alpha Rdt/c$ converts its coordinate catch-up time to the expression above.

If $d=c^2/a$, the ray lies on the limiting null horizon and never intersects it at finite proper time; if $d>c^2/a$, it never reaches the ship. The reception time diverges as the emitter approaches this Rindler horizon. For $ad/c^2\ll 1$, ${\tau }_{\mathrm{rec}}=d/c+a{d}^2/(2c^3)+\cdots$, recovering the inertial light-travel time. A different emission time or initial relative velocity would require specifying that extra datum; the formula uses the initial-rest simultaneous-emission convention stated above.