Relativity course of the University of Cambridge (Core lecture)

Use angular size under a collinear relativistic boost. For a small distant source, the two angular diameters have ratio $e^{|\eta |}$, where $\eta$ is the relative rapidity. Thus $e^{|\eta |}=2$, and the relative velocity satisfiesThe speed is therefore $0.6c$. The observers are momentarily coincident, so this is relativistic aberration, rather than a difference in their distance from the planet. The observer moving faster towards the planet measures the smaller angular diameter under the source-direction aberration convention.

The metric tensor and its inverse are diagonal: $g_{xx}=1$, $g_{yy}=f^2$, $g^{xx}=1$, $g^{yy}=f^{-2}$ on a patch where $f\ne 0$. Only ${\partial }_x{g}_{yy}=2f{f}^{\prime }$ is nonzero. The Christoffel symbol formula therefore giveswith all other components zero. Using the stated Riemann curvature tensor convention,so $\boxed{R^x{}_{yxy}=-f{f}^{\prime \prime }}$. This is the curvature of a two-dimensional warped metric; its Gaussian curvature is $-f^{\prime \prime }/f$.

For a grazing null geodesic, take the impact parameter as $b\simeq R=6.0\times 10^6\mathrm{m}$. The leading Schwarzschild light deflection isWith the gravitational constant $G\simeq 6.6743\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$ and speed of light $c\simeq 2.998\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$,Here $GM/(R{c}^2)\simeq 2.5\times 10^{-4}\ll 1$, so the weak-field approximation is appropriate. A ray passing at a larger impact parameter has a correspondingly smaller deflection.

A fundamental observer follows the coarse-grained cosmological matter, with fixed comoving coordinates. The cosmological principle asserts spatial homogeneity and cosmological isotropy on large scales for this family of observers. A moving observer generally sees a preferred direction, for example a Doppler dipole in the cosmic microwave background, even when the underlying cosmology is isotropic.

In an FLRW metric, these worldlines are orthogonal to homogeneous spatial slices. Synchronize their clocks to define cosmic time $t$, which is their proper time. Along a fixed spatial label, $d{s}^2=c^{2}d{t}^2$, giving $g_{tt}=c^2$. Orthogonality of the spatial coordinate directions to the observers' four-velocity gives $g_{ti}=0$. Thus, in the $+---$ metric signature,where the spatial matrix $g_{ij}$ is negative definite. Spatial homogeneity and cosmological isotropy further restrict it to $-a(t{)}^2{\gamma }_{ij}(\mathbf{x})$, with $\gamma$ a constant-curvature spatial metric.

For radial motion, use the geodesic LagrangianThe Euler-Lagrange equation for $t$ gives $c^2\ddot{t}+a(da/dt){\dot{\chi }}^2=0$. Since $\chi$ is a cyclic coordinate, its conjugate momentum is constant. HenceFor a massive particle, take $\tau$ as proper time; the four-velocity normalization is $c^2{\dot{t}}^2-a^2{\dot{\chi }}^2=c^2$. For a photon, use an affine parameter rather than proper time, and the corresponding tangent is a null vector. ThereforeThe normalization of the null affine parameter remains arbitrary.

Choose the null affine parameter so that the photon tangent is its four-momentum $P^{\mu }$. A fundamental observer has $U^{\mu }=(1,0,0,0)$ in cosmic-time coordinates, so the measured energy is $E=P_{\mu }{U}^{\mu }=c^2{P}^t$. The null normalization and conserved radial quantity in part (b) giveThus redshifting of peculiar momentum gives $E\propto a^{-1}$. Since $E=h\nu$, with $h$ the Planck constant, the received-to-emitted frequency ratio is ${\nu }_R/{\nu }_E=a(t_E)/a(t_R)$. By the definition of cosmological redshift,

A nearby fundamental observer measures radial proper distance $d\ell =ad\chi$ and proper time $dt$. Thus the signed radial peculiar velocity isSubstitute $a^2\dot{\chi }=C$ and the massive normalization from part (b):Equivalently, for velocity $v_0$ at scale factor $a_0$,This is the exact law of redshifting of peculiar momentum. In the nonrelativistic limit, $v\propto a^{-1}$; an initially relativistic speed stays below $c$ and decreases as the universe expands.

Write $f(r)=1-2\mu /r$. In the equatorial plane, the geodesic Lagrangian is $\mathcal{L}=\frac{1}{2}(f{c}^2{\dot{t}}^2-f^{-1}{\dot{r}}^2-r^2{\dot{\varphi }}^2)$. The cyclic coordinates $t$ and $\varphi$ give the conserved quantitiesThe squared tangent norm is constant along an affinely parametrized geodesic, since metric compatibility gives $d[g(\dot{X},\dot{X})]/d\tau =2g(D\dot{X}/d\tau ,\dot{X})=0$. HenceFor a massive particle parametrized by proper time, $E=c^2$; a photon has a null tangent, so $E=0$. This $E$ denotes the tangent norm, not the photon's physical energy. A future-directed exterior photon has $k>0$, proportional to its positive Killing energy. Its affine parameter can be rescaled, so there is no mass-shell normalization fixing $k$ to a particular positive value.

Set $E=0$ in part (a), substitute the two conserved quantities, and multiply the normalization by $f$:Consequently the null effective potential in Schwarzschild spacetime isOnly radii where the total radial energy is at least this effective potential are accessible.

For $h\ne 0$, the effective potential vanishes at $r=2\mu$, rises to a maximum, and then falls to zero as $r\to \infty$. Its derivative is $V_{\mathrm{eff}}^{\prime }=h^2(3\mu -r)/r^4$, so the only circular null geodesic hasThe second derivative there is $V_{\mathrm{eff}}^{\prime \prime }=-h^2/(81{\mu }^4)<0$, proving instability by the effective potential stability criterion. These circular orbits make up the photon sphere. If the formula is continued inside the horizon, the potential is negative and diverges to $-\infty$ as $r\downarrow 0$. For purely radial light, $h=0$, the potential is identically zero and there is no circular orbit.

A photon arriving from infinity crosses the barrier precisely when $c^2{k}^2/2$ exceeds the maximum in part (c):At infinity its spatial tangent has speed in affine-parameter units $ck$, while $h$ is its angular momentum per unit affine normalization. A straight asymptotic ray with impact parameter $b$ therefore has $|h|=bck$. ThusAt equality, the incoming ray approaches the unstable photon sphere asymptotically and is not captured. A larger impact parameter gives an outer turning point and scattering back to infinity.

Measure $\alpha$ in the orthonormal tetrad of a static observer at $r>2\mu$. This specifies the local frame needed to define an angle. If $E_{\infty }$ is the conserved photon energy, gravitational redshift gives $E_{\mathrm{loc}}=E_{\infty }/\sqrt{1-2\mu /r}$. The tangential local momentum is $E_{\mathrm{loc}}\sin \alpha /c$, so the impact parameter isInside the photon sphere, an outward ray must have $b<3\sqrt{3}\mu$ to cross the potential barrier. The Schwarzschild photon escape cone therefore has critical half-angleThe upper bound is strict: a critical ray only approaches the circular orbit. At $r=3\mu$, the critical half-angle is $\pi /2$, and exactly tangential light is circular. As $r\downarrow 2\mu$, the cone shrinks to zero opening. No static observer exists exactly on the Schwarzschild event horizon, and no future-directed photon emitted there escapes to exterior infinity; the zero-angle result is an exterior limit.

At any event on a massive observer's worldline, choose an orthonormal tetrad with timelike basis vector $e_{(0)}=U/c$. Its components obey $g(e_{(a)},e_{(b)})={\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$, so locally $d{s}^2={\eta }_{ab}d{X}^{(a)}d{X}^{(b)}$: the Minkowski metric. A local inertial frame can also remove the Christoffel symbols at that event. Curvature cannot be removed on a finite region, and a frame following an accelerated observer need not remain inertial in its neighbourhood.

In general coordinates, $X^{\mu }(\tau )$ labels the trajectory, while the four-velocity and four-acceleration areCoordinate values $X^{\mu }$ are not themselves a vector under arbitrary nonlinear coordinate changes; their tangent $U$ is. For fixed rest mass $m$, the four-momentum is $P^{\mu }=m{U}^{\mu }$ and the four-force is $f^{\mu }=D{P}^{\mu }/d\tau =m{A}^{\mu }$. With the $+---$ metric signature,where $a$ is the proper acceleration. The middle identity follows by differentiating the constant velocity norm; in the instantaneous rest frame, $U=(c,0)$ and $A=(0,\mathbf{a})$, proving the last identity.

The inverse transformation is $t=(p+q)/2$, $x=(q-p)/2$. ThusThe metric tensor in these light-cone coordinates is thereforeBoth off-diagonal entries are needed, since the quadratic line element sums over both index orders.

The Lorentz transformation is $t^{\prime }=\gamma (t-vx)$, $x^{\prime }=\gamma (x-vt)$ in units $c=1$. Adding and subtracting givesSince $\gamma (1+v)=\sqrt{(1+v)/(1-v)}$ and the other factor is its reciprocal,Here $\eta$ is the boost rapidity. Their product is unchanged, as required by invariance of the Minkowski interval.

The proper time normalization of the astronaut's four-velocity gives ${\dot{t}}^2-{\dot{x}}^2=1$. In the light-cone coordinates this isThe four-acceleration has squared norm ${\ddot{t}}^2-{\ddot{x}}^2=-a^2$, because $a$ is the constant proper acceleration. The same coordinate change therefore givesEquivalently write $\dot{q}=e^{\eta }$, $\dot{p}=e^{-\eta }$. Their second derivatives give $\ddot{p}\ddot{q}=-{\dot{\eta }}^2$, so the rapidity form of one-dimensional proper acceleration makes $\dot{\eta }=\pm a$, with sign fixing the acceleration direction.

For the stated outward-accelerating hyperbola, take positive signed proper acceleration and an initially stationary spaceship. Then $\eta =a\tau$, so $\dot{p}=e^{-a\tau }$ and $\dot{q}=e^{a\tau }$. The initial values are $p(0)=-1/a$, $q(0)=1/a$. Integration givesTaking their half-sum and half-difference yields the hyperbolic motionThe printed position data alone do not specify the initial velocity. If the initial rapidity is an arbitrary ${\eta }_0$, the equally valid positive-acceleration solution isThus the requested special expression implicitly selects ${\eta }_0=0$.

Use the specified hyperbolic trajectory. The emitted right-moving photon follows $x=L+t-t_0$, so its retarded null coordinate is $p=t_0-L$. Interception at astronaut proper time $\tau$ requiresThe astronaut has velocity $v=\tanh (a\tau )$ relative to the star and $\gamma =\cosh (a\tau )$. The collinear relativistic Doppler effect givesThis is the Doppler shift for a uniformly accelerated receiver. A finite reception time exists precisely when $t_0<L$, so the limiting emission time is $L$, approached from below. At $t_0=L$ the photon lies on the Rindler horizon and never reaches the astronaut. If reception is restricted to $\tau \ge 0$, the full emission range is $L-1/a\le t_0<L$; for emission after $t=0$, it is $0\le t_0<L$.

Take $\alpha >0$, $t\ne 0$, and the negative-norm convention for a timelike curve used by the printed metric. Writing $v=dx/dt$ and $d{s}^2=-c^{2}d{\tau }^2$ givesSpatial translation is a Killing vector field. Equivalently, the geodesic Lagrangian has the cyclic coordinate $x$, so its conserved momentum isTherefore the correct first integral for this metric isThe first-integral assertion in the PDF is false. Its dimensionless square root requires $v/c$, and even after setting $c=1$ it is missing the factor $|t|$. For example, the nonvertical geodesics below have $v/\sqrt{1-v^2/c^2}$ proportional to $1/|t|$.

Set $b=\sqrt{\alpha }P/c$. Solving for the velocity, with increasing $t$ as the time orientation, gives $v=cb/\sqrt{t^2+b^2}$. The complete unparametrized family of timelike geodesics of a quadratic-time conformal metric isEach curve is restricted to a connected positive- or negative-$t$ patch; $t=0$ is excluded because the metric becomes degenerate. For $t\ne 0$, $|v|<c$, and the two independent constants are $x_c$ and $b$. If an explicit affine parametrization is desired,while the vertical family has $d\tau =|t|dt/\sqrt{\alpha }$. A reversal of the time orientation reverses the parameter without changing these paths.

Put $\kappa =8\pi G/c^4$ and $T=T_{\nu }{}^{\nu }$. The contracted Bianchi identity and metric compatibility giveTaking the trace of the field equation in four dimensions gives $(1-4\alpha )R=-\kappa T$. For $\alpha \ne 1/4$, eliminate $R$:Ordinary covariantly coupled matter obeys stress-energy conservation, ${\nabla }_{\nu }{T}_{\mu }{}^{\nu }=0$. Its trace need not be constant, so the trace coefficient fixed by stress-energy conservation isThis restores the Einstein tensor and is the only coefficient compatible with arbitrary conventional matter without adding a new constraint or changing the matter conservation law. Other coefficients can work for specially restricted sources with constant trace, but do not give the same universal matter coupling.

The excluded case $\alpha =1/4$ must be treated before dividing: its trace equation requires $T=0$. If stress-energy is also conserved, its divergence equation requires ${\partial }_{\mu }R=0$. It can describe a restricted traceless-source sector with constant scalar curvature, not general matter such as ordinary dust. In particular it is not the general trace-free Einstein equation with source $T_{\mu \nu }-g_{\mu \nu }T/4$, which is a different equation.

Interpret the orbit in the Schwarzschild metric as an unforced circular timelike geodesic and choose its plane as $\theta =\pi /2$. Put $f=1-2\mu /r$. The radial geodesic equation at constant radius uses the Christoffel symbolsand gives the Angular velocity of a circular Schwarzschild geodesicThe proper-time normalization then yieldsOver one revolution, the Proper period of a circular Schwarzschild geodesic is thereforeBoth the gravitational and orbital kinematic contributions to time dilation are included. Circular timelike geodesics exist for $r>3\mu$; they are stable for $r>6\mu$ and unstable for $3\mu <r<6\mu$, with marginal stability at $6\mu$.

If “circular orbit” instead means a forced trajectory with an externally chosen angular velocity, the radius alone does not fix the answer. Its elapsed time is $\Delta \tau =\Delta t\sqrt{1-2\mu /r-r^2{\Omega }^2/c^2}$, or $2\pi \sqrt{1-2\mu /r-r^2{\Omega }^2/c^2}/|\Omega |$ per revolution. The geodesic interpretation supplies the missing angular velocity.

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.