Thirty exercises that turn recognition into understanding

These are not speed tests. Several are designed so that the tempting answer is wrong. Try the problem before reading the solution. If you obtain a different sign, first compare conventions; if you obtain a different physical prediction after doing so, investigate.

A.1 Transforming a vector and a covector#

Problem. In two dimensions, let $x'=2x+y$ and $y'=y$. A vector has components $(V^x,V^y)=(3,4)$ and a covector has components $(\omega_x,\omega_y)=(5,-2)$. Transform both and check their contraction.

Solution. The Jacobian is

A vector transforms with $J$, so $V'=(10,4)$. A covector, represented as a row, transforms with $J^{-1}$, giving $\omega'=(5/2,-9/2)$. Thus

The components changed while the pairing did not. Using the same transformation rule for both objects would lose this invariance. The upper and lower indices encode how the two kinds of components compensate each other.

A.2 An inverse metric is a matrix inverse#

Problem. For the positive-definite metric

and $V^i=(1,2)$, find $g^{ij}$, $V_i$, and $g(V,V)$.

Solution. Since the determinant is $3$,

Lowering gives $V_i=(4,5)$, and the norm squared is $V^iV_i=1\cdot4+2\cdot5=14$. Raising again returns $(1,2)$. Replacing each matrix entry by its reciprocal would fail even the identity check $g^{ik}g_{kj}=\delta^i{}_j$.

A.3 Energy depends on the observer, even when it is a scalar#

Problem. A massive particle moves at $v=0.6c$ relative to an inertial observer. Find its energy and momentum. What energy does an observer moving with the particle assign to it?

Solution. $\gamma=1/\sqrt{1-0.36}=1.25$. The first observer measures $E=1.25mc^2$ and $p=\gamma mv=0.75mc$. The comoving observer measures $mc^2$ and zero spatial momentum.

There is no contradiction with $E_{(U)}=-p_\mu U^\mu$ being a scalar. A coordinate change transforms both $p$ and the same observer $U$, leaving that number unchanged. Choosing a different observer replaces $U$ by a different vector and defines a different measurement.

A.4 The Hessian knows about polar coordinates#

Problem. On the Euclidean plane with $ds^2=dr^2+r^2d\theta^2$, let $f=r$. Although $\partial_\theta\partial_\theta f=0$, show that $\nabla_\theta\nabla_\theta f=r$ and $\nabla^2f=1/r$ for $r>0$.

Solution. The first derivative of a scalar is the covector $df$. Differentiating it requires a connection:

The radial Hessian component vanishes. Contracting with $g^{\theta\theta}=1/r^2$ gives $\nabla^2r=r/r^2=1/r$. The second covariant derivative of a scalar is not generally just a matrix of ordinary second derivatives.

A.5 A divergence-free field can hide a puncture#

Problem. On the same plane, take $V^r=1/r$ and $V^\theta=0$. Compute its divergence for $r>0$. Why is the flux through a circle nonzero?

Solution. Since $\sqrt{g}=r$,

away from the origin. But a circle of radius $r$ has outward flux $(1/r)(2\pi r)=2\pi$.

The field is undefined at the origin. Applying the divergence theorem to a disk without accounting for that singularity violates its smoothness assumptions. On an annulus, the outer and inner boundary fluxes cancel with their appropriate orientations. In a distributional extension to the whole plane, the divergence is $2\pi\delta^{(2)}(\mathbf{x})$.

The lesson applies far beyond this example: local equations plus domain assumptions determine whether an integral inference is legitimate.

A.6 Constant Cartesian arrows have changing polar components#

Problem. A constant Cartesian vector pointing in the $x$ direction has polar components $V^r=\cos\theta$ and $V^\theta=-\sin\theta/r$. Show that $\nabla_\theta V^r=0$.

Solution. The partial derivative is $\partial_\theta V^r=-\sin\theta$, which by itself falsely suggests the vector changes. The connection contribution is

They cancel. This is a concrete example of a covariant derivative correcting the change in a coordinate basis. No physical bending of space was needed.

A.7 The connection can vanish while curvature survives#

Problem. At an event $P$, use normal coordinates so that $\Gamma^\rho{}_{\mu\nu}(P)=0$. Why does this not imply $R^\rho{}_{\sigma\mu\nu}(P)=0$?

Solution. At $P$, the terms quadratic in $\Gamma$ vanish, but

can remain nonzero. A function vanishing at a point need not have vanishing derivatives there. In normal coordinates the metric’s first derivatives vanish at $P$, but its second derivatives contain curvature information.

A.8 Measuring curvature with a circumference#

Problem. On a sphere of radius $A$, a geodesic circle at geodesic distance $s$ from the north pole has circumference $C(s)=2\pi A\sin(s/A)$. Expand it for small $s$ and compare it with a flat circle.

Solution. Taylor expansion gives

It is smaller than $2\pi s$. Intrinsic measurements of radius and circumference reveal positive curvature without any view from an embedding space. The Gaussian curvature is $1/A^2$; in two dimensions the scalar curvature is twice that, $R=2/A^2$.

A.9 Newtonian tides know the sign convention#

Problem. For $\Phi=-G_NM/r$, derive the radial and transverse relative accelerations of nearby freely falling particles in the Newtonian limit.

Solution. In local Cartesian directions aligned with the radial axis, the Hessian of the potential has eigenvalues

Relative acceleration is $\delta\ddot x^i=-\delta^{ik}\partial_k\partial_j\Phi\,\delta x^j$, so the corresponding tidal eigenvalues are

Radial separations stretch; transverse separations compress. The trace vanishes outside the source, consistent with the vacuum Poisson equation. In the relativistic convention used here, $R^i{}_{0j0}\simeq\delta^{ik}\partial_k\partial_j\Phi/c^2$, and geodesic deviation supplies the minus sign.

A.10 Trace reversal produces the missing factor of two#

Problem. With $\Lambda=0$, contract the four-dimensional Einstein equation and rewrite it as an equation for $R_{\mu\nu}$. Evaluate its $00$ source for slowly moving dust.

Solution. The trace is $R-2R=\kappa T$, so $R=-\kappa T$, where $\kappa=8\pi G_N/c^4$. Therefore

For slow dust, $T_{00}\simeq\rho c^2$, $T\simeq-\rho c^2$, and $g_{00}\simeq-1$. The parenthesis in the $00$ component becomes $\rho c^2/2$. Thus $R_{00}\simeq4\pi G_N\rho/c^2$, consistent with $R_{00}\simeq\nabla^2\Phi/c^2$ and Newton’s Poisson equation.

A.11 What zero scalar curvature implies#

Problem. Does $R=0$ imply vacuum? Does vacuum with $\Lambda=0$ imply no gravitational waves?

Solution. Both answers are no. A classical electromagnetic field in four dimensions has a trace-free stress tensor. Einstein’s trace equation gives $R=0$ when $\Lambda=0$, even though $T_{\mu\nu}$ and $R_{\mu\nu}$ may be nonzero. Vacuum implies $R_{\mu\nu}=0$, but the Weyl tensor can still describe gravitational waves or an exterior tidal field.

The three conditions are called scalar-flat, Ricci-flat, and Riemann-flat, respectively. Each sets a different tensor or contraction to zero.

A.12 Pressure enters the local curvature source#

Problem. In a perfect fluid’s local orthonormal rest frame, find $T$ and the trace-reversed $00$ source.

Solution. With $T_{\hat\mu\hat\nu}=\operatorname{diag}(\epsilon,p,p,p)$,

and

That is one precise sense in which pressure gravitates. It does not license adding $3p/c^2$ to the mass of every isolated box while ignoring stresses in its walls. A complete system includes its stresses, binding, and appropriate global mass definition.

A.13 The determinant supplies the trace term#

Problem. Starting from $\delta\ln|\det M|=\operatorname{tr}(M^{-1}\delta M)$, show why varying $g^{\mu\nu}$ produces a minus sign in $\delta\sqrt{-g}$.

Solution. Varying $g^{\mu\alpha}g_{\alpha\nu}=\delta^\mu{}_\nu$ gives

Then

The sign is a consequence of inverse-matrix variation. It is not an arbitrary special rule for gravity.

A.14 A constant Lagrangian can affect gravity#

Problem. In natural units $c=1$, shift the matter Lagrangian by a constant: $\mathcal L_m\mapsto\mathcal L_m-V_0$. Find the stress tensor change.

Solution. The action changes by $-V_0\int\sqrt{-g}\,d^4x$. Its variation is

Comparing with $\delta S_m=-\tfrac12\int\sqrt{-g}\,T_{\mu\nu}\delta g^{\mu\nu}\,d^4x$ yields

Every timelike observer measures energy density $V_0$ and isotropic pressure $-V_0$; this stress tensor selects no preferred rest frame. A constant can be dynamically irrelevant to nongravitational equations on a fixed background while affecting the metric equation when geometry is dynamical.

A.15 Why a boundary condition on the metric is not automatically enough#

Problem. Suppose $\delta g_{\mu\nu}=0$ on a boundary. Must the normal derivative of $\delta g_{\mu\nu}$ vanish there too?

Solution. No. The elementary function $f(x)=x$ satisfies $f(0)=0$ but $f'(0)=1$. The same distinction applies componentwise to a field variation. The Einstein–Hilbert bulk action produces boundary terms containing derivatives of the metric variation. For a suitable fixed-induced-metric problem on a non-null smooth boundary, the Gibbons–Hawking–York term supplies the required cancellation. Setting more boundary data to zero than the problem calls for can conceal rather than solve this issue.

A.16 Constructing a conserved current#

Problem. Let $T^{\mu\nu}$ be symmetric and covariantly conserved. If $\xi^\mu$ obeys $\nabla_{(\mu}\xi_{\nu)}=0$, prove that $J^\mu=T^{\mu\nu}\xi_\nu$ is conserved.

Solution. The product rule gives

The first term vanishes. A symmetric tensor contracts only the symmetric part of the second factor, so the second becomes $T^{\mu\nu}\nabla_{(\mu}\xi_{\nu)}=0$. Integrating this local conservation law into a conserved charge also requires an appropriate region, hypersurfaces, and control of boundary fluxes. The Killing field supplies an actual symmetry, not merely a preferred-looking coordinate label.

A.17 A height difference changes a clock’s rate#

Problem. Estimate the fractional clock-rate difference between stationary clocks separated vertically by $100\,\mathrm m$ in a weak approximately uniform field with $g_{\mathrm{local}}=9.8\,\mathrm{m/s^2}$. Estimate the accumulated difference over one day.

Solution. The potential difference is $\Delta\Phi\simeq g_{\mathrm{local}}h=980\,\mathrm{m^2/s^2}$. Thus

Over $86{,}400$ seconds, the higher clock gains approximately $9.4\times10^{-10}$ seconds, or $0.94$ nanoseconds. This estimate isolates the gravitational height effect; Earth’s rotation, the actual potential, motion, and the comparison protocol matter in a precision experiment.

A.18 A horizon is not a place where every clock stops#

Problem. In Schwarzschild spacetime, a static clock obeys $d\tau=\sqrt{1-2m/r}\,dt$. Why can’t we infer that a freely falling clock physically stops at $r=2m$?

Solution. The formula applies to worldlines with fixed Schwarzschild $r,\theta,\phi$. Such worldlines require increasing proper acceleration as $r$ approaches $2m$ from outside. The limiting static worldline is not a timelike observer sitting on the horizon. A falling observer follows a different worldline, has a finite proper-time crossing for ordinary infall, and can use a regular horizon-crossing chart. A relation between one coordinate time and one family of clocks is not a universal claim about all clocks.

A.19 Doubling a black hole changes horizon tides#

Problem. The Schwarzschild Kretschmann scalar is $K=48m^2/r^6$, with $m=G_NM/c^2$. Evaluate it at $r=2m$ and determine its mass scaling.

Solution. At the horizon,

Representative orthonormal curvature components scale as $m/r^3\sim M^{-2}$ at the horizon. Larger nonrotating black holes can therefore have weaker horizon-scale tidal curvature. “Bigger black hole” does not mean “more violent local horizon crossing.” This says nothing by itself about the singularity deeper inside.

A.20 Gravitational-wave strain is a relative measurement#

Problem. A plus-polarized wave has $h_+=10^{-21}$. For an ideal freely falling separation $L=4\,\mathrm{km}$ along one principal axis, estimate the leading displacement amplitude. What is the ideal differential change between orthogonal axes?

Solution. To leading order in the long-wavelength approximation,

The differential change is $\delta L_x-\delta L_y\simeq h_+L=4\times10^{-18}\,\mathrm m$. A real interferometer measures optical phase with a frequency-dependent response; this elementary estimate captures the geometric strain scale, not the full instrument transfer function.

A.21 Why an isolated source has no leading mass-dipole gravitational radiation#

Problem. In the slow-motion weak-field approximation, define $D^i=\int\rho x^i\,d^3x$. Explain why its second time derivative vanishes for an isolated system and what that implies.

Solution. The first derivative is total momentum, $\dot D^i=P^i$, assuming suitable mass conservation and vanishing boundary flux. For an isolated system $\dot P^i=0$, hence $\ddot D^i=0$. A putative leading radiative term based on that second derivative cannot carry changing dipolar structure. Together with conservation of the monopole at the relevant leading order, this helps explain why the mass quadrupole is the first generic radiative contribution in GR’s slow-source expansion. The claim has an approximation and isolation regime; it is not a statement about arbitrary additional fields in modified gravity.

A.22 An expanding box gives the fluid equation#

Problem. For a homogeneous perfect fluid in a comoving cell with physical volume $V\propto a^3$, use $d(\epsilon V)=-p\,dV$ to derive the continuity equation and the scaling for $p=w\epsilon$, with constant $w$.

Solution. Expand the differential:

Divide by $Vdt$ and use $\dot V/V=3H$:

For constant $w$, $d\ln\epsilon=-3(1+w)d\ln a$, giving $\epsilon\propto a^{-3(1+w)}$. Dust has $a^{-3}$, radiation $a^{-4}$, and a cosmological-constant fluid $a^0$. This equation is a local covariant balance law specialized to the symmetry, not evidence for a universally defined conserved total cosmic energy.

A.23 Derive a universe’s power-law growth#

Problem. In a spatially flat, single-fluid universe with $\Lambda=0$ and constant $w>-1$, combine $H^2\propto\epsilon$ with the previous result to derive $a(t)$.

Solution. We have

Integrating the expanding branch gives $a^{3(1+w)/2}\propto t-t_0$, hence

Dust gives $t^{2/3}$, radiation $t^{1/2}$. The $w=-1$ case must be solved separately: constant positive energy density gives constant $H$ and exponential expansion. Substituting $w=-1$ into the power-law exponent is not a valid limiting derivation.

A.24 When does pressure accelerate expansion?#

Problem. With no separately written cosmological constant and $p=w\epsilon$, determine when a positive-density FLRW fluid drives $\ddot a>0$.

Solution. The acceleration equation is

For $\epsilon>0$, acceleration requires $w<-1/3$. Ordinary positive pressure contributes toward deceleration in this equation. Sufficiently negative pressure reverses the sign. This cosmological statement must not be replaced by the indiscriminate slogan “pressure is repulsive” or “pressure is always attractive.”

A.25 Four constraints do not mean four lost metric components#

Problem. Explain why GR has two local propagating metric degrees of freedom in four spacetime dimensions using ADM phase space.

Solution. The spatial metric has six independent components, and its conjugate momentum adds six, for twelve phase-space variables per spatial point. Four first-class constraints each remove one constrained phase-space direction and one associated gauge direction. Thus $12-2\times4=4$ physical phase-space dimensions remain, corresponding to two configuration degrees of freedom and their conjugate momenta. Lapse and shift act as gauge multipliers rather than additional propagating fields in this count. This is a local count for ordinary GR; boundaries and global sectors require further care.

A.26 The ADM constraint recognizes an expanding universe#

Problem. Use $c=1$, a flat FLRW slice, and the convention $K_{ij}=-\tfrac12\mathcal L_n\gamma_{ij}$. Show that the Hamiltonian constraint reproduces the first Friedmann equation.

Solution. For comoving proper-time slicing, $K_{ij}=-H\gamma_{ij}$. Hence $K=-3H$ and $K_{ij}K^{ij}=3H^2$. Since the slice has ${}^{(3)}R=0$,

Set this equal to $16\pi G_N\epsilon+2\Lambda$ to obtain

The expansion lives in the extrinsic curvature of the slices even when their intrinsic spatial curvature vanishes.

A.27 Raychaudhuri gives a deadline for focusing#

Problem. For a geodesic, hypersurface-orthogonal timelike congruence in four dimensions with $R_{\mu\nu}u^\mu u^\nu\ge0$, use $c=1$ and show that negative initial expansion $\theta_0<0$ must diverge to negative infinity within proper time at most $3/|\theta_0|$, provided the congruence remains defined up to that point.

Solution. Raychaudhuri gives

because shear contributes nonpositively and vorticity vanishes. For $\theta<0$,

Starting from $1/\theta_0<0$, the reciprocal must reach zero no later than $3/|\theta_0|$, corresponding to focusing. A caustic is not by itself a spacetime singularity: geodesics can cross in perfectly regular spacetime. Singularity theorems require additional global assumptions to infer incompleteness.

A.28 A pure frame rotation is not curvature#

Problem. On the flat plane, use the orthonormal coframe $e^1=dr$, $e^2=r\,d\theta$. Why can the spin connection be nonzero while the curvature two-form vanishes?

Solution. Since $de^2=dr\wedge d\theta$, the torsion-free Cartan equation requires $\omega^2{}_1=d\theta$ and $\omega^1{}_2=-d\theta$. These one-forms record the rotation of the polar frame. But $d(d\theta)=0$ on a regular angular chart, and the relevant connection wedge products vanish in this two-dimensional example. Thus $\Omega^a{}_b=d\omega^a{}_b+\omega^a{}_c\wedge\omega^c{}_b=0$. The polar chart and frame fail at the origin; their behavior there must not be confused with a curved plane.

A.29 Why black-hole entropy has the right dimensions#

Problem. Verify that $S_{\mathrm{BH}}=k_BA/(4\ell_P^2)$ has entropy units, where $\ell_P^2=\hbar G_N/c^3$. For a Schwarzschild black hole, derive the $M^2$ dependence of entropy and check that $T_HdS=d(Mc^2)$.

Solution. $A/\ell_P^2$ is dimensionless, so $S$ has units of $k_B$. With $A=16\pi G_N^2M^2/c^4$,

Multiply by $T_H=\hbar c^3/(8\pi G_NMk_B)$ to obtain $T_HdS=c^2dM$. This is a consistency check within the semiclassical result for an uncharged, nonrotating hole, not a derivation of its microscopic degrees of freedom.

A.30 Estimating an effective-theory correction#

Problem. In units $c=\hbar=1$, consider a schematic gravitational Lagrangian

with dimensionless coefficient $a$ of order unity. On a slowly varying geometry with typical curvature scale $R\sim L^{-2}$, estimate the relative size of the correction. State why $R=0$ alone is not a sufficient validity check.

Solution. Relative to the $R$ term, the displayed correction scales as $a/(M_*L)^2$. It is small when $M_*L\gg1$. This is generic power counting, not a prediction that this particular $R^2$ term changes a Ricci-flat vacuum solution: its variation vanishes on that solution, as discussed in Section 23.2. An effective action can also contain higher curvature contractions and derivative operators. The scalar $R$ may vanish while Riemann or Weyl curvature is nonzero, as in a Schwarzschild exterior. Validity requires control of the physically relevant curvature components, invariant scales, frequencies, and state-dependent effects, not a single convenient scalar. The effective theory can be predictive below its cutoff without claiming validity at arbitrarily short distances.

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