Light goes out and back at the same speed, so the reflection sits at the temporal midpoint of emission and reception, a distance equal to half the round-trip light time:

so radar_coordinates(2, 8)$=(5,3)$ and $x/t=0.6$. For a probe at constant $\beta$ each leg is Doppler-stretched by one Bondi factor $k$, hence $t_\text{echo}/t_\text{send}=k^2$. With $\beta=0.6$, $k=\sqrt{1.6/0.4}=2$, so $k^2=4$ (bondi_k(0.6)**2) — the echo at $8=4\cdot2$ — and $\beta=(k^2-1)/(k^2+1)=3/5=0.6$ (beta_from_k(2.0)).

"Synchronized in their own frame" means the two zero-settings are simultaneous there, $\Delta t'=0$ at rest-separation $\Delta x'=L$. A clock fixed in that frame reads, in the lab, $t'=\gamma(t-\beta x)$, so at one lab instant two clocks differ in reading by $\Delta t'=-\gamma\beta\,\Delta x_\text{lab}$. The lab sees their separation contracted, $\Delta x_\text{lab}=L/\gamma$, giving

with the front (larger $x$) clock reading behind — "leading clocks lag." So leading_clocks_lag(0.8, 3.0)$=2.4$; reversing the motion swaps which clock leads, leading_clocks_lag(-0.8, 3.0)$=-2.4$.

The time row of the boost gives $\Delta t'=\gamma(\Delta t-\beta\,\Delta x)$. Simultaneous in S means $\Delta t=0$, so with $\gamma=1/\sqrt{1-0.6^2}=1/0.8=1.25$,

They are not simultaneous in S′. Taking $\Delta x=x_2-x_1=2>0$, the negative $\Delta t'=t'_2-t'_1$ means the event at the larger $x$ happens earlier in S′. This is simultaneity_breakdown(0.6, 0.0, 2.0)$=-1.5$, which collapses to $0$ only in the trivial cases $\beta=0$ or $\Delta x=0$.

In one half-tick of lab-duration $\tau$ the clock slides horizontally by $\beta\tau$, while the light still covers a path of length $\tau$ (speed $c=1$, postulate 2) along the hypotenuse over the transverse gap $D$:

At $\beta=0.6$ this is $1/\sqrt{1-0.36}=1/0.8=1.25$, so the dilation factor is $\gamma=1.25$ — read off the triangle, never assumed. The ratio $\tau/D$ contains no $D$, so it is gap-independent: light_clock_gamma(0.6)$=1.25=1/(1-0.6^2)^{0.5}$, and light_clock_gamma(0.6, 3.7) gives the same $1.25$.

For small $\beta$, $\gamma=(1-\beta^2)^{-1/2}=1+\tfrac12\beta^2+\cdots$ and $k=\sqrt{\tfrac{1+\beta}{1-\beta}}=1+\beta+\cdots$, while the simultaneity offset $-\gamma\beta\,\Delta x=O(\beta)$. So dilation and contraction are second order in $\beta$, but the relativity of simultaneity is first order. At $\beta=10^{-3}$,

i.e. light_clock_gamma(1e-3)$=1.0000005$ and bondi_k(1e-3)$=1.0010005$. The first-order term survives: simultaneity_breakdown(1e-3, 2.0, 3.0)$=1.99700$, short of $\Delta t=2$ by exactly $\beta\,\Delta x=0.003$. Kill the separation and absolute time returns, simultaneity_breakdown(1e-3, 2.0, 0.0)$=2.000001\approx\Delta t$ (only the $O(\beta^2)$ dilation remaining).

What you see is the photon's arrival time; what you observe (assign as the event's coordinate time) is that arrival minus the light-travel delay. For a source approaching at angle $\theta$, photons emitted later leave from nearer points, so their arrivals bunch up and the apparent transverse speed $\dfrac{v\sin\theta}{1-\beta\cos\theta}$ can exceed $c$ — but only the raw reception rate is superluminal, not the source. Einstein synchronization removes exactly this delay through the radar assignment

Placing an event at $(t,x)=(5,3)$ via $t_\text{send}=t-x=2$, $t_\text{echo}=t+x=8$: you see the echo at $t_\text{echo}=8$ but observe the event at radar_coordinates(2, 8)$=(5,3)$ — the seen/observed split of §12.1.2, with no real faster-than-light motion.

The boost is $ct'=\gamma(ct-\beta x)$, $x'=\gamma(x-\beta ct)$. Solve for the unprimed pair (eliminate, then use $\gamma^2(1-\beta^2)=1$):

This is the same transformation with $\beta\to-\beta$ — physically obvious, since S moves at $-\beta$ relative to S′. In matrix form $\Lambda(\beta)=\begin{pmatrix}\gamma&-\gamma\beta\\-\gamma\beta&\gamma\end{pmatrix}$ has $\det=\gamma^2(1-\beta^2)=1$, and its inverse just flips the off-diagonal sign: $\Lambda(\beta)^{-1}=\Lambda(-\beta)$. Hence inverse(boost(0.6)) equals boost(-0.6), and compose(inverse(L), L) is the identity.

Boost into one ship's rest frame; the other's velocity composes by

not $1.8c$ — the gap closes at $1.8c$ as a coordinate rate in your frame, but neither ship measures the other above $c$. The cap is automatic because $\tanh$ (hence the rule) saturates at $1$: feeding in a light ray, velocity_add(0.9, 1.0)$=(0.9+1)/(1+0.9)=1$ — light stays at light speed in every frame. So velocity_add(0.9, 0.9)$=0.99447$.

In the $(ct,x)$ plane a boost is a hyperbolic rotation $\Lambda(\phi)=\begin{pmatrix}\cosh\phi&-\sinh\phi\\-\sinh\phi&\cosh\phi\end{pmatrix}$ with $\phi=\operatorname{artanh}\beta$. Rotations about the same axis add their angles, $\Lambda(\phi_1)\Lambda(\phi_2)\Lambda(\phi_3)=\Lambda(\phi_1+\phi_2+\phi_3)$, exactly like ordinary rotations. So three collinear boosts make a single boost of rapidity $\phi_1+\phi_2+\phi_3$, i.e.

For $\beta_i=(0.4,0.5,0.6)$ the rapidities sum to $1.6661$ and $\beta=\tanh(1.6661)=0.9310$. Thus compose(boost(b1), boost(b2), boost(b3)) equals boost(beta_from_rapidity(rapidity(b1)+rapidity(b2)+rapidity(b3))) — velocities don't add, rapidities do.

The boost $\begin{pmatrix}\gamma&-\gamma\beta\\-\gamma\beta&\gamma\end{pmatrix}$ sits against the Galilean $\begin{pmatrix}1&0\\-\beta&1\end{pmatrix}$ in $(ct,x)$. For $\beta=10^{-4}$, $\gamma=(1-10^{-8})^{-1/2}=1+\tfrac12\beta^2+\cdots$, so the diagonal entries exceed the Galilean $1$ by

the corrections the Check locates on the diagonal. The space-row entry $-\gamma\beta=-\beta(1+\tfrac12\beta^2+\cdots)$ reproduces the Galilean $-\beta$ to relative order $\beta^2$, so $x'=\gamma(x-\beta ct)$ matches $x'=x-\beta ct$ to that order. The one genuinely new entry is the $-\gamma\beta$ in the time row — the $-\beta x$ (relativity-of-simultaneity) term that $t'=t$ drops — and it too vanishes as $\beta\to0$. So boost(1e-4) has diagonal $1.000000005$ and collapses to the Galilean map in the $c\to\infty$ limit.

With $\eta=\mathrm{diag}(-1,1,1,1)$ the squared interval between $A=(0,0,0,0)$ and $B=(5,3,0,0)$ is

so the separation is timelike. Every Lorentz transformation obeys $\Lambda^{\mathsf T}\eta\Lambda=\eta$, hence leaves $s^2=x^{\mathsf T}\eta x$ unchanged for all events: boosting $B$ by any $\boldsymbol\beta$ returns the same $-16$. Concretely interval2([5,3,0,0])$=-16$ equals interval2(apply(general_boost((0.4,-0.2,0.5)), [5,3,0,0])), and classify([5,3,0,0]) is "timelike".

The photon worldline $(t,t,0,0)$ has 3-speed $x/t=1$ and is null, $s^2=-t^2+t^2=0$. Boost at $\beta=0.8$ ($\gamma=1/\sqrt{1-0.64}=5/3$):

so the new 3-speed is $x'/t'=1$ again — both components shrink by the same Doppler factor, leaving the ratio fixed. Thus apply(boost(0.8), [1,1,0,0])$=[\tfrac13,\tfrac13,0,0]$ gives out[1]/out[0]$=1.0$ and classify(out) is "null": light moves at $c$ in every frame, the postulate the module is built to honour.

The muon covers $d=33$ light-μs in lab time $d/\beta$ (with $c=1$), surviving a fraction $\exp(-t_\text{life}/\tau)$ where $t_\text{life}$ is read on its own clock. (a) Ignoring dilation its mean life stays $\tau_0$, so

(b) Time dilation stretches the lab mean life to $\gamma\tau_0$ with $\gamma=1/\sqrt{1-0.98^2}=5.025$, so

In the muon's frame nothing is dilated — the atmosphere is instead contracted to $d/\gamma$, giving the same exponent. This reproduces muon_fraction(2.2, 0.98, 33.0) → (0.0476, 2.25e-07), a ratio $\sim10^5$.

With $\beta=0.6$, $\gamma=1/\sqrt{1-\beta^2}=1/\sqrt{1-0.36}=1/0.8=1.25$. (a) The measured length is $L=L_0/\gamma=1/1.25=0.8$ m; to measure a moving rod you must mark both ends at the same instant of your frame ($\Delta t=0$), else it shifts between the markings. (b) Inverting, $L_0=\gamma L=1.25\cdot0.8=1.0$ m. (c) The same $\gamma$ dilates a clock riding the rod: a 1 s proper tick stretches to $\Delta t=\gamma\,\Delta\tau=1.25$ s — contraction and dilation are one boost read two ways ($\Delta x'=0$ vs $\Delta t=0$). This gives length_contraction(1.0, 0.6) → 0.8, rest_length(0.8, 0.6) → 1.0, and time_dilation(1.0, 0.6) → 1.25.

Synchronised in their own frame, both clocks read $t'=0$. Reading $t'=\gamma(t-\beta x)$ at one instant of your frame (fixed $t$) gives $\Delta t'=-\gamma\beta\,\Delta x$; with the contracted separation $\Delta x=L_0/\gamma$,

The leading clock (in front, along the motion) reads $\beta L_0=5$ behind, so the trailing clock reads ahead by $5$. At $\beta=0$ the offset vanishes — clocks synced in one frame stay synced in every frame. This matches leading_clocks_lag(10.0, 0.5) → 5.0 and leading_clocks_lag(10.0, 0.0) → 0.

With $\beta=0.8$, $\gamma=1/\sqrt{1-0.8^2}=1/\sqrt{0.36}=1/0.6=1.667$. (a) The home twin ages by the coordinate time $T=30$ yr; the traveller's clock runs slow the whole trip, logging proper time $T/\gamma=30/1.667=18$ yr — younger by $12$. (b) It is not symmetric because only the traveller changes inertial frame at the turnaround; there her line of simultaneity sweeps across the home worldline, ageing the home twin suddenly by exactly the missing $12$ yr (a $\beta L_0$-type jump). (c) As $\beta\to0$, $\gamma\to1$ and $T/\gamma\to T$, so the gap closes. Hence twin_ages(0.8, 30.0) → (30.0, 18.0) and twin_ages(1e-4, 30.0) ≈ (30, 30).

With $\beta=0.6$, $\gamma=1.25$. (a) In the barn frame the pole contracts to $L_0/\gamma=10/1.25=8\le10=L_\text{barn}$, so it momentarily fits and both doors can shut simultaneously ($\Delta t_\text{barn}=0$) with the pole inside. (b) In the pole frame the barn is contracted to $L_\text{barn}/\gamma=8<10$, so the pole cannot fit. Those same two door-shut events — simultaneous for the barn — are separated in the pole frame by

the far door shutting (and reopening) before the near one, so the pole is never enclosed. Same events, different "now": pole_in_barn(10, 10, 0.6) → contracted_pole = 8.0, fits_in_barn_frame = True, door_gap_barn = 0.0, door_gap_pole = 7.5.

The $ct'$-axis is the worldline of the moving origin $x'=0$, i.e. $x=\beta ct$, with direction $(ct,x)=(1,\beta)$ and slope $ct/x=1/\beta=1/0.6=1.667$. The $x'$-axis is the simultaneity locus $ct'=0$, i.e. $ct=\beta x$, direction $(\beta,1)$, slope $ct/x=\beta=0.6$. (b) The slopes $1/\beta$ and $\beta$ are reciprocals (product $1$), so the axes are mirror images about the light line $ct=x$ — they scissor toward it through equal angles. (c) That common tilt is $\arctan\beta=\arctan0.6=30.96^\circ$. So boosted_axes(0.6) → ([1.0, 0.6], [0.6, 1.0]), whose slopes $ct/x$ are $1/0.6$ and $0.6$ with product $1.0$.

With $A$ at the origin, $B-A=(ct,x,0,0)$ and the invariant interval is

(a) $s^2=-5^2+3^2=-16<0$ — time wins, timelike: a sub-light worldline links the events. (b) $s^2=-3^2+5^2=+16>0$ — space wins, spacelike: no causal contact. (c) $s^2=-4^2+4^2=0$ — null, joined only by light. Because $s^2$ is Lorentz-invariant these labels hold in every frame, matching classify([5,3,0,0])→timelike, classify([3,5,0,0])→spacelike, classify([4,4,0,0])→null.

A boost along $x$ transforms the time gap as $\Delta t'=\gamma(\Delta t-\beta\,\Delta x)$ with $|\beta|<1$. If $B$ is in $A$'s future the separation is timelike with $\Delta t>0$ and $|\Delta x|<\Delta t$, so

Thus $\Delta t'>0$ in every frame — cause precedes effect, always (the null boundary $|\Delta x|=\Delta t$ gives $\Delta t'=\gamma\Delta t(1\mp\beta)>0$ too). For a spacelike pair $|\Delta x|>|\Delta t|$, so any boost with $\Delta t/\Delta x<\beta<1$ makes $\Delta t-\beta\,\Delta x<0$ and reverses the order — exactly why no signal may travel spacelike. test_causal_order_is_frame_invariant_for_timelike shows both: the timelike $B$ keeps ${B'}^0>0$ under random boosts, while a spacelike event's $\Delta t$ flips.

With $U^\mu=\gamma(1,\mathbf v)$ and $\eta=\mathrm{diag}(-1,+1,+1,+1)$,

But $\gamma=1/\sqrt{1-|\mathbf v|^2}$ gives $\gamma^2(1-|\mathbf v|^2)=1$ identically, so $U\cdot U=-1$ (i.e. $-c^2$ in units with $c$) for any $\mathbf v$. The 4-velocity is a unit timelike vector by construction — differentiating $x^\mu$ by the invariant $\tau$ guarantees it. For $\mathbf v=(0.6,0.2,0)$ ($\gamma\approx1.291$) this is mdot(four_velocity([0.6,0.2,0.0]), four_velocity([0.6,0.2,0.0]))→$-1$, independent of speed.

At $\beta=0.6$, $\gamma=1/\sqrt{1-0.36}=1/0.8=1.25$. The 4-momentum is $p^\mu=mU^\mu=(\gamma m,\ \gamma m\mathbf v)$, so

The mass shell then follows from $U\cdot U=-1$ (P3): $p\cdot p=m^2(U\cdot U)=-m^2$, i.e.

So $\sqrt{-p\cdot p}=2$ in every frame, reproducing four_momentum(2.0, [0.6,0,0])→[2.5, 1.5, 0, 0] and invariant_mass→2.0.

Each leg of B's trip spans coordinate time $\Delta t=5$ and displacement $\Delta x=x$, so its proper time is the Minkowski arc length $\sqrt{(\Delta t)^2-(\Delta x)^2}=\sqrt{25-x^2}$. The two symmetric legs give

with equality only at $x=0$: the bent worldline is always shorter (the reversed triangle inequality), so the travelling twin returns younger. Setting $\tau_B=\tfrac12\tau_A=5$ gives $\sqrt{25-x^2}=2.5\Rightarrow x^2=18.75\Rightarrow x=\sqrt{75}/2\approx4.33$ (so $\beta=x/5\approx0.866$, $\gamma=2$). Then proper_time([[0,0,0,0],[5,4.33,0,0],[10,0,0,0]])$\approx5.0$, half of A's $10$.

Take any timelike worldline from $A$ to $B$ and insert an intermediate event $C$ off the straight line. The reversed triangle inequality (P5), applied leg by leg, gives

with equality only when $C$ lies on the straight worldline — every kink (an acceleration) strictly shortens the elapsed proper time. Iterating over all detours, the inertial path is the unique maximum of $\tau=\int d\tau$. This is the principle of maximal aging: a free particle extremizes its proper time, the seed of the geodesic principle in RE-12. Concretely, any bent worldline handed to proper_time returns less than the straight $A\to B$ value of $10$.

Kinetic energy is $T=(\gamma-1)m$, and "$T=n$ times the rest energy $m$" means $(\gamma-1)m=nm$, hence

For $n=1$ the kinetic energy equals the rest energy, so $\gamma=2$ and $v=\sqrt{1-\tfrac14}=\sqrt3/2\approx0.866\,c$. (As $n\to\infty$, $\gamma\to\infty$ but $v\to c$: more kinetic energy buys ever less speed.) Solving kinetic_energy(1, [v,0,0]) = 1 indeed returns $v\approx0.866$, with $\gamma=2$.

With $E=\gamma m$ and $\mathbf p=\gamma m\mathbf v$,

since $\gamma^2(1-|\mathbf v|^2)=1$ at every speed; restoring units, $E^2-|\mathbf p|^2c^2=m^2c^4$. This is just $p\cdot p=-m^2$ in mostly-plus signs. Taking $m\to0$ at fixed $E$ forces $|\mathbf p|=E$: a massless particle is null ($p\cdot p=0$) and moves at $c$. Hence energy(m,v)² − Σ momentum(m,v)²$=m^2$ for any $(m,v)$, and interval2(photon_four_momentum(E, n̂))$=0$.

Each photon carries a null 4-momentum $p_{1,2}=(E,\pm E,0,0)$, so the pair's total is

The 3-momenta cancel while the energies add, so the system is massive ($M=2E$) though each part is massless — mass belongs to the total 4-momentum, not to individual particles. A single photon instead has $P\cdot P=-E^2+E^2=0$, so $M=0$, and it can never decay into massive products: 4-momentum conservation cannot turn one null vector into a timelike one. Thus system_invariant_mass([photon_four_momentum(E,[1,0,0]), photon_four_momentum(E,[-1,0,0])])$=2E$.

At $\beta=0.8$, $\gamma=1/\sqrt{1-0.64}=1/0.6=5/3$. The lumps carry opposite momenta, so the total 4-momentum is purely temporal:

With no net 3-momentum the blob is born at rest, and its rest mass is the system invariant mass

The lost kinetic energy $2(\gamma-1)m$ did not disappear — it became rest mass ($E=mc^2$). This is inelastic_stick(1, [0.8,0,0], 1, [-0.8,0,0])→(3.333, [0,0,0]).

The invariant $s=-P\cdot P$ is frame-independent. In the lab (beam energy $E_b$, momentum $p_b$, on a target $m_p$ at rest), using $E_b^2-p_b^2=m_p^2$,

At threshold the four final particles ($M=4m_p$) sit at rest in the centre-of-momentum frame, so $s=M^2=16m_p^2$. Equating,

The beam needs $6m_pc^2\approx5.6$ GeV of kinetic energy — the 1955 Bevatron antiproton threshold. So threshold_kinetic_energy(1, 1, 4)$=6$.

(a) The pion at rest has $P^\mu=(m_\pi,\mathbf 0)$. Momentum conservation sends the two photons back-to-back with $|\mathbf p_1|=|\mathbf p_2|$, and since each is null, $E_i=|\mathbf p_i|$; energy conservation $E_1+E_2=m_\pi$ with $E_1=E_2$ then gives

(b) Compton scattering shifts the wavelength by $\Delta\lambda=\dfrac{\hbar}{m_ec}(1-\cos\theta)$; on backscatter $\theta=\pi$,

two Compton wavelengths — the maximum possible shift, versus zero in the forward direction. In natural units this is compton_shift(math.pi, 1.0)$=2.0$ against compton_shift(0,1)$=0$.

For a source receding along the line of sight the longitudinal factor is

The frequency halves, and since $\lambda\propto1/f$ the wavelength doubles. The approaching (time-reversed) case is the reciprocal — the Bondi factor $k=\sqrt{(1+\beta)/(1-\beta)}=\sqrt{1.6/0.4}=2$ — so the receding shift is exactly $1/k=0.5$. Hence doppler_longitudinal(0.6)→0.5, bondi_k(0.6)→2.0, and their product is 1.

At the abeam instant ($\theta=90^\circ$) there is no line-of-sight velocity, so classical theory predicts no shift. The relativistic factor at $\theta=\pi/2$ retains only the time-dilation piece:

It is a pure redshift: the moving source's clock runs slow by $\gamma$, so its frequency arrives lowered even with zero radial motion. This transverse effect (Ives–Stilwell, 1938) has no classical analogue. Numerically doppler_transverse(0.5)→0.86603, equal to doppler_general(0.5, π/2).

With $\beta=0.6$, $\gamma=1/\sqrt{1-0.36}=1.25$ and $\cos60^\circ=\tfrac12$. The angular Doppler factor is

and the aberration angle obeys

Boosting the null wavevector $k^\mu=(1,\tfrac12,\tfrac{\sqrt3}{2},0)$ with the RE-03 $\Lambda(0.6)$ reproduces both — Doppler is its time component, aberration its spatial direction — and keeps $k'\cdot k'=0$. So wavevector_frequency(k)/wavevector_frequency(kp)→1.142857, equal to doppler_general(0.6, 60°), and math.degrees(wavevector_angle(kp))→98.2132, the aberration angle, with _mink2(kp)$\approx0$.

For a broadside ray $\theta=90^\circ$ ($\cos\theta=0$) the aberration law gives

For tiny $\beta$, $\arcsin\beta\approx\beta$, so the swing is $\theta'-\tfrac{\pi}{2}\approx\beta=9.93\times10^{-5}$ rad. Converting, $\beta\cdot206265\approx20.5$ arcsec — Bradley's 1727 constant of aberration. The apparent star shifts toward the apex (Earth's instantaneous direction of motion), tracing a small ellipse over the year. Indeed math.degrees(aberration(math.pi/2, 9.93e-5) - math.pi/2)*3600→20.49 arcseconds.

A source isotropic in its rest frame emits half its photons into the rest-frame forward hemisphere ($\theta_{\rm rest}<90^\circ$). Its boundary ray $\theta_{\rm rest}=90^\circ$ aberrates (rest$\to$lab is a $-\beta$ boost) to the lab cone half-angle $\theta_c$ with $\cos\theta_c=\beta$:

As $\beta\to1$ this collapses: $\theta_c\approx\sqrt{2(1-\beta)}=\sqrt{0.02}=0.141$ rad, a forward pencil of order $1/\gamma$. Half the light is beamed into this tight cone — the relativistic headlight effect that makes jets look one-sided. So headlight_halfangle(0.99)→0.14155 rad ($8.110^\circ$), equal to aberration(math.pi/2, -0.99), with $\cos\theta_c\to0.99$.

Length contraction and a photograph answer different questions. Contraction is what simultaneous lab rulers measure: the sphere's extent along $\mathbf v$ is genuinely $1/\gamma$ shorter. A camera instead records photons that arrive together but left at different times — light from the receding far side departed earlier, when the sphere was elsewhere, so that side becomes visible. Combined with aberration of the outline rays — which transform by the same $\cos\theta'=(\cos\theta-\beta)/(1-\beta\cos\theta)$ used throughout this module — the net optical effect is an apparent rotation (Penrose–Terrell, 1959), not a squash, and a sphere's silhouette stays exactly circular. No contradiction: the measured contraction (~RE-04) and the photographed rotation are both right, answering the distinct questions "what do simultaneous rulers read?" versus "what does one lens capture?". Conceptual — no code; see notes.md §6.

Direct evaluation gives $V^\mu W_\mu = 2\cdot1 + 5\cdot0 + (-1)\cdot2 + 3\cdot(-2) = -6$. Under a boost the contravariant $V$ gains a factor $\Lambda$ and the covariant $W$ the inverse-transpose, and the two collapse:

The covariant law is defined as the inverse-transpose precisely so this product is frame-independent - the $\Lambda$ and $\Lambda^{-1}$ annihilate into $\delta^\rho_\nu$. Hence the boosted contraction returns the same value, matching sum(transform_vector(L,V)[i]*transform_covector(L,W)[i]) $=-6$.

Start from the defining Lorentz condition $\Lambda^{\mathsf T}\eta\Lambda=\eta$. Left-multiply by $\eta$ and use $\eta^2=I$, then right-multiply by $\Lambda^{-1}$:

Now form the rank-2 transform of $\eta$ and insert $\eta\eta=I$ at the right:

In index form $\Lambda^\mu{}_\alpha\Lambda^\nu{}_\beta\,\eta^{\alpha\beta}=\eta^{\mu\nu}$: the metric has the same components in every frame. This is exactly transform_tensor2(L, ETA) equalling ETA to machine precision (the demo reports a maximum difference $\sim 2\times10^{-16}$).

Apply the rank-2 law to $\bar T^{\nu\mu}$ and rename the dummy indices $\alpha\leftrightarrow\beta$:

If $T^{\beta\alpha}=+T^{\alpha\beta}$ this equals $+\bar T^{\mu\nu}$; if $T^{\beta\alpha}=-T^{\alpha\beta}$ it equals $-\bar T^{\mu\nu}$. So a boost sends symmetric $\to$ symmetric and antisymmetric $\to$ antisymmetric - symmetry type is a Lorentz-invariant label, which is why the split $T=T^{(\mu\nu)}+T^{[\mu\nu]}$ is frame-independent. This is what transform_tensor2(L, symmetric_part(T)) (stays symmetric) and transform_tensor2(L, antisymmetric_part(T)) (stays antisymmetric) confirm.

(a) Rename $\mu\leftrightarrow\nu$ and use the two symmetries $S_{\nu\mu}=S_{\mu\nu}$, $A^{\nu\mu}=-A^{\mu\nu}$:

so the cross-contraction equals its own negative and must vanish. (b) The trace ties off both indices with the invariant metric, so by P2 ($\eta_{\mu\nu}\Lambda^\mu{}_\alpha\Lambda^\nu{}_\beta=\eta_{\alpha\beta}$):

a Lorentz scalar. These confirm double_contract(symmetric_part(T), antisymmetric_part(T)) $=0$ and trace(transform_tensor2(L, T)) $=$ trace(T) (the demo holds it at $30.0$ before and after the boost).

(a) With $x\cdot x=\eta_{\alpha\beta}x^\alpha x^\beta$ and $\partial x^\alpha/\partial x^\nu=\delta^\alpha_\nu$, the product rule gives

a lower-index vector whose time slot carries the metric's sign, $\partial_0(x\cdot x)=2x_0=-2x^0$. (b) For a single quadratic only the matching second derivative survives, $\Box(x^0)^2=\eta^{00}\cdot2=-2$ and $\Box(x^i)^2=\eta^{ii}\cdot2=+2$. Since $x\cdot x$ carries a minus on its time term, $\Box(x\cdot x)=-(-2)+2+2+2=8=2\,\eta^{\mu\nu}\eta_{\mu\nu}$. These match four_gradient(...) $=[-2,4,-2,1]=2\cdot$lower(x) and dalembertian(...) returning $-2,\,+2,\,8$.

With $k$ constant, each derivative of $f=\cos(k_\mu x^\mu)$ brings down a factor of $k$:

Contracting with $\eta^{\mu\nu}$ to build the d'Alembertian,

so $\Box f=0$ iff $k\cdot k=0$ (a null wavevector). For $k^\mu=(1,1,0,0)$, $k\cdot k=-1+1=0$, hence $\Box f=0$ - the seed of $\Box A^\mu=-\mu_0 J^\mu$ (EM-18) and of $E^2=\mathbf p^2$ for massless quanta. This matches mdot([1,1,0,0],[1,1,0,0]) $=0$ and dalembertian $\approx 0$.

The inverse metric is $g^{\theta\theta}=1$, $g^{\phi\phi}=1/\sin^2\theta$, and the only nonzero derivative is $\partial_\theta g_{\phi\phi}=2\sin\theta\cos\theta$. Substituting into $\Gamma^k_{ij}=\tfrac12 g^{kl}(\partial_i g_{jl}+\partial_j g_{il}-\partial_l g_{ij})$,

The lower pair is symmetric so $\Gamma^\phi_{\phi\theta}=\Gamma^\phi_{\theta\phi}$, and any component needing a derivative other than $\partial_\theta g_{\phi\phi}$ is zero. At $\theta=1$: $-\sin1\cos1=-0.4546$ and $\cot1=0.6421$, matching christoffel(...) with Gam[0][1][1] $\approx-0.4546$ and Gam[1][0][1] $\approx0.6421$.

With $g_{rr}=1$, $g_{\phi\phi}=r^2$, the lone nonzero derivative is $\partial_r g_{\phi\phi}=2r$ and $g^{\phi\phi}=1/r^2$. The same Christoffel formula gives

all others zero. They are nonzero even though the plane is flat: they merely record how the polar basis $(\hat r,\hat\phi)$ swings as you move - a property of the coordinates, not the space. Genuine curvature is the derivative of $\Gamma$ (the Riemann tensor), which vanishes here (RE-11). At $r=1.5$: $\Gamma^r_{\phi\phi}=-1.5$ and $\Gamma^\phi_{r\phi}=1/1.5=0.667$, matching christoffel(...) $\to$ Gam[0][1][1] $\approx-1.5$, Gam[1][0][1] $\approx0.667$.

Lower the connection index, $\Gamma_{kij}\equiv g_{kl}\Gamma^l_{ij}=\tfrac12(\partial_i g_{jk}+\partial_j g_{ik}-\partial_k g_{ij})$, so that

Adding the two lowered symbols, the cross terms cancel and

so $\nabla_i g_{jk}=0$ identically. Metric-compatibility means $\tfrac{d}{d\lambda}\langle V,W\rangle=\langle\nabla V,W\rangle+\langle V,\nabla W\rangle=0$ along a transport, so lengths and angles are preserved and $\nabla$ commutes with raising/lowering. Confirmed: metric_compatibility(...) on the sphere and on the plane-polar metric both return $\approx0$ (the demo gives $0$ to machine precision).

For a curve at fixed $\theta=\theta_0$ swept in $\phi$ at rate $\omega$, the tangent is $\dot x=(0,\omega)$, so in $\ddot x^k=-\Gamma^k_{ij}\dot x^i\dot x^j$ only the $(\phi,\phi)$ term survives:

while $\ddot x^\phi=-\Gamma^\phi_{\phi\phi}\omega^2=0$. This vanishes only when $\sin\theta_0\cos\theta_0=0$, i.e. at the equator $\theta_0=\pi/2$ (or the poles): the equator is a geodesic, every other latitude is not - you must supply $\ddot x^\theta$ ("steer") to hold it. Confirmed: geodesic_acceleration(g,[3.14159/2,0],[0,1]) $\approx[0,0]$ versus geodesic_acceleration(g,[1.0,0],[0,1]) $\approx[0.4546,0]$.

By Gauss-Bonnet the holonomy from transport round the loop equals the curvature integrated over the enclosed cap, $\Delta\alpha=\iint K\,dA$; on the unit sphere $K=1$, so it is simply the area:

For $\theta_0=1.0$, $\Delta\theta=\Delta\phi=0.3$ this is $(\cos1.0-\cos1.3)\cdot0.3=(0.5403-0.2675)\cdot0.3\approx0.0818$ rad. Transporting [1,0] around the rectangle and reading the rotation in the orthonormal frame gives $\approx0.0818$ (the demo measures $0.0818$); reversing the loop flips the sign, and a zero-area out-and-back path leaves the vector unchanged.

Under a coordinate change the connection picks up an inhomogeneous piece,

whose second term is not tensorial - so $\Gamma$ is not a tensor. At any chosen point $p$ one can pick coordinates (geodesic-normal / locally inertial) that make that term cancel $\Gamma(p)$, giving $\Gamma'^k_{ij}(p)=0$. What cannot be removed is $\partial\Gamma$ - the Riemann tensor, which transforms homogeneously. Physically a free-faller transforms gravity away at their location (the connection) but never the tidal field (the curvature): gravity is curvature, not a force on a fixed background. The flat-$\eta$ tests show the limiting case - when $g$ is constant every $\Gamma=0$, the locally inertial frame realised everywhere at once.

By the equivalence principle the climbing photon is equivalent to one received by an observer accelerating upward at $g$; in the flight time $t=h/c$ the receiver gains speed $v=gt=gh/c$ away from the source, so the first-order Doppler shift is

With $g=9.80665\ \text{m/s}^2$ and $h=22.5\ \text{m}$, $\Delta f/f=-9.80665\cdot22.5/(2.998\times10^8)^2=-2.45\times10^{-15}$. It is negative - a redshift - because the photon climbs out, spending energy against the effective field and dropping in frequency. This matches pound_rebka() $=2.455\times10^{-15}$ (magnitude) and redshift_uniform_field(9.80665, 22.5) $=-2.455\times10^{-15}$.

In the rocket, light emitted at the floor reaches the ceiling height $h$ after $t=h/c$; in that interval the ceiling has gained speed $v=at=ah/c$ away from the emission event, so the Doppler shift is $\Delta f/f=-v/c=-ah/c^2$ - this is accelerated_frame_redshift(a,h). A photon rising $h$ in a uniform field $g$ climbs a potential $\Delta\Phi=gh$, giving $\Delta f/f=-\Delta\Phi/c^2=-gh/c^2$ - this is redshift_uniform_field(g,h). Setting $g=a$,

The two are numerically identical for every $a,h$, so no measurement of the photon shift inside the sealed cabin can distinguish acceleration $a$ from a field $g=a$ - the equivalence principle. Confirmed by accelerated_frame_redshift(a, h) $==$ redshift_uniform_field(a, h) (test_equivalence_accelerated_equals_field).

An ideal clock at potential $\Phi$ ticks at proper rate $d\tau/dt\approx1+\Phi/c^2$ (weak field). The fractional amount by which the lower clock falls behind the upper one is then

A photon sent from the lower to the upper clock climbs $\Delta\Phi=\Phi_{\text{upper}}-\Phi_{\text{lower}}>0$, so its redshift is $\Delta f/f=-\Delta\Phi/c^2$ - the same number. Redshift and time dilation are one phenomenon: the light looks reddened only because the deeper clock genuinely runs slow. Confirmed: grav_time_dilation(phi_lo, phi_up) - 1 $==$ grav_redshift(phi_up - phi_lo). For GPS the ground clock (lower $\Phi$) runs slow, so the orbiting clock gains time.

A ray crossing the width $L$ takes $t=L/c$; in that time the floor accelerates up by the transverse drop, and the exit angle is the transverse speed over $c$:

so $\text{drop}=\tfrac12 L\theta$. By the equivalence principle light must bend the same way in a real field $g$. But this argument curves only the time part of the metric; the equal contribution from spatial curvature (RE-11), invisible to the EP, doubles the result - turning Einstein's 1911 figure $0.87''$ into the measured $1.75''$ for starlight grazing the Sun. Confirmed: light_deflection_elevator(g, L) returns (drop, angle) with $\text{drop}=\tfrac12 L\cdot\text{angle}$ (demo: $5.46\times10^{-15}\ \text{m}$ and $1.09\times10^{-15}\ \text{rad}$).

An observer with constant proper acceleration $a$ traces the hyperbola $x^2-(ct)^2=(c^2/a)^2$. A light pulse starting a distance $d$ directly behind can keep pace only out to the break-even separation

For $a=g=9.80665\ \text{m/s}^2$, $d=(2.998\times10^8)^2/9.80665=9.17\times10^{15}\ \text{m}$, and dividing by $9.461\times10^{15}\ \text{m}$ per light-year gives $0.97$ light-years. As $a\to0$, $d\to\infty$ - an inertial observer accelerates not at all and has no horizon. Confirmed: rindler_horizon(9.80665) $=9.165\times10^{15}$, rindler_horizon(0.0) $=$ inf, and rindler_horizon(a)*a $=c^2$.

The local field at radius $r$ is $g(r)=GM/r^2$, stronger on the near mass than the far one. After free fall cancels the common $GM/r^2$, the residual relative acceleration is the field gradient times the separation:

a stretch along $r$. It $\to0$ as $dr\to0$ (so a single point can be made locally inertial) but is nonzero for any finite $dr$, scales as $1/r^3$, and depends on no choice of frame - it is curvature (geodesic deviation, RE-11). For Earth, two masses $1\ \text{m}$ apart give $a_{\text{tidal}}\approx3.1\times10^{-6}\ \text{m/s}^2$, while the local $GM/r^2\approx9.8\ \text{m/s}^2$ that free fall removes is millions of times larger. Confirmed: tidal_acceleration(M, r, dr) scales as $1/r^3$ and $\propto dr$, with tidal_acceleration(M, r, 0) $=0$.

The Eotvos ratio compares the fall rates of two bodies,

If gravitational and inertial mass coincide then $a=(m_{\text{grav}}/m_{\text{inert}})\,g$ is identical for every body, so the weak equivalence principle demands $\eta=0$ for all material pairs. A torsion-balance bound $\eta<10^{-13}$ therefore says any composition-dependent "fifth force" must be weaker than gravity by at least that factor - excluded at the $10^{-13}$ level (Will Table 2.2). Confirmed: eotvos_parameter(a, a) $=0.0$ and eotvos_parameter(1.0, 1.0+1e-13) $\approx1.0\times10^{-13}$.

The round 2-sphere $g=\mathrm{diag}(a^2,\,a^2\sin^2\theta)$ has nonzero Christoffel symbols $\Gamma^\theta{}_{\phi\phi}=-\sin\theta\cos\theta$ and $\Gamma^\phi{}_{\theta\phi}=\cot\theta$. In the notes' convention the single independent Riemann component is

Contracting gives $R_{\theta\theta}=1$, $R_{\phi\phi}=\sin^2\theta$, so the scalar curvature is

both constant. For $a=2$: $R=2/4=0.5$, $K=0.25$ — matching ricci_scalar(sphere_metric(2.0), [1.0, 0.5]) → 0.5 and gaussian_curvature_2d → 0.25.

Schwarzschild is a vacuum solution, $R_{\mu\nu}=0$, so every scalar built from Ricci — including $R$ and $R_{\mu\nu}R^{\mu\nu}$ — vanishes identically. The full Riemann tensor does not vanish, and the lowest invariant that sees it is the Kretschmann scalar

At the horizon $r=2M$ this is $K=48M^2/(2M)^6=3/(4M^4)$, perfectly finite — and since $K$ is a scalar, no coordinate change can make the curvature blow up there. The trouble the metric shows at $r=2M$ lives only in the chart $(t,r)$: a coordinate singularity. At $r=0$, by contrast, $K\to\infty$, an invariant divergence no chart can remove — the true physical singularity. Numerically kretschmann(schwarzschild_metric(1), [0,r,1,0.9]) returns $1.17\times10^{-2}\approx48/4^6$ at $r=4$ while ricci(...) $\approx0$: Ricci-flat yet genuinely curved.

The symmetry-reduced count is $N(n)=n^2(n^2-1)/12$. Plugging in,

The Ricci tensor is symmetric, so it carries $n(n+1)/2$ numbers: $3$ in 2-D, $6$ in 3-D, $10$ in 4-D. Comparing the two counts: in 2-D Riemann's single component is the Gaussian curvature, fixed by $R$ alone; in 3-D Ricci's $6$ exactly matches Riemann's $6$, so Ricci determines Riemann completely — no free field. In 4-D Riemann has $20$ but Ricci only $10$, leaving

components untouched by Ricci — the Weyl tensor, the trace-free part of curvature that survives in vacuum ($R_{\mu\nu}=0$ yet $R_{\mu\nu\rho\sigma}\neq0$, as in Schwarzschild) and propagates as gravitational waves (RE-16).

On the unit 2-sphere $g=\mathrm{diag}(1,\sin^2\theta)$ the one independent Riemann component is $R^\theta{}_{\phi\theta\phi}=\sin^2\theta$ (P1). The fiducial geodesic runs in $\phi$, $u=(u^\theta,u^\phi)=(0,1)$, with neighbour separation $\xi=(1,0)$ in $\theta$. The deviation equation keeps only the term with $b=d=\phi$, $c=\theta$:

Its sign is the physics: $A\cdot\xi=g_{\theta\theta}A^\theta\xi^\theta=-\sin^2\theta<0$, so the neighbour accelerates back toward the fiducial worldline — positive curvature focuses geodesics (the geometric engine of the singularity theorems). At $\theta=1$ this is $A^\theta=-\sin^2(1)\approx-0.708$, matching geodesic_deviation(sphere_metric(1), [1.0,0], [0,1], [1,0]) $\to[-0.708,\,0]$.

Contract the Einstein tensor with the inverse metric, using $g^{\mu\nu}R_{\mu\nu}=R$ and $g^{\mu\nu}g_{\mu\nu}=\delta^\mu{}_\mu=n$:

So the trace flips sign relative to $R$ above $n=2$, and vanishes exactly at $n=2$ (where $G_{\mu\nu}\equiv0$). On the unit 3-sphere ($n=3$, maximally symmetric with $R=6/a^2=6$),

the value test_einstein_trace_identity confirms: ricci_scalar $\approx6$ on _three_sphere(1.0) and the contracted Einstein tensor trace $\approx-3$.

Local energy–momentum conservation, $\nabla_\mu T^{\mu\nu}=0$, is non-negotiable — it is the curved-space promotion of $\partial_\mu T^{\mu\nu}=0$. So whatever curvature tensor sits on the left of the field equation must be identically divergence-free; otherwise setting it $\propto T^{\mu\nu}$ would impose an extra, unphysical constraint on the matter. The contracted Bianchi identity hands exactly one such object for free:

Neither $R^{\mu\nu}$ nor $R\,g^{\mu\nu}$ is separately conserved — only the combination $G^{\mu\nu}$ is, because $\nabla_\mu R^{\mu\nu}=\tfrac12\nabla^\nu R$. Hence the law reads $G_{\mu\nu}=8\pi T_{\mu\nu}$: picking $R_{\mu\nu}\propto T_{\mu\nu}$ instead would force $\nabla^\nu R=0$ (constant scalar curvature), far too rigid for a general source. With $G$ on the left, $\nabla_\mu T^{\mu\nu}=0$ is automatic — the structural reason the Einstein equation of RE-13 takes this form.

The geodesic equation is $\ddot x^\mu + \Gamma^\mu{}_{\nu\rho}\,\dot x^\nu\dot x^\rho = 0$. For the constant Minkowski metric $\eta_{\mu\nu}=\mathrm{diag}(-1,1,1,1)$ every derivative $\partial_\lambda\eta_{\mu\nu}=0$, so each Christoffel symbol $\Gamma^\mu{}_{\nu\rho}=\tfrac12\eta^{\mu\lambda}(\partial_\nu\eta_{\lambda\rho}+\partial_\rho\eta_{\lambda\nu}-\partial_\lambda\eta_{\nu\rho})=0$ vanishes. The equation collapses to $\ddot x^\mu=0$, which integrates twice to

a straight line traced at constant 4-velocity. This must hold because the equivalence principle says a freely falling frame is locally inertial — there special relativity applies and a free particle moves uniformly, so GR has to reduce to $\ddot x^\mu=0$ in that limit. Hence geodesic_rhs(minkowski_metric(), x, v) returns [0,0,0,0] for any x,v.

With $L=\tfrac12 a^2(\dot\theta^2+\sin^2\theta\,\dot\varphi^2)$, the Euler–Lagrange equation for $\theta$ uses $\partial L/\partial\dot\theta=a^2\dot\theta$ and $\partial L/\partial\theta=a^2\sin\theta\cos\theta\,\dot\varphi^2$:

Matching $\ddot\theta+\Gamma^\theta{}_{\varphi\varphi}\dot\varphi^2=0$ gives $\Gamma^\theta{}_{\varphi\varphi}=-\sin\theta\cos\theta$. The $\varphi$-equation $\frac{d}{d\lambda}(a^2\sin^2\theta\,\dot\varphi)=0$ expands to $\ddot\varphi+2\cot\theta\,\dot\theta\dot\varphi=0$, so $\Gamma^\varphi{}_{\theta\varphi}=\cot\theta$. At $\theta=1,\dot\varphi=1$ the acceleration is $\ddot\theta=\sin1\cos1\approx0.45465$ — exactly what geodesic_rhs(sphere_metric(1), [1.0,0], [0,1])[0] and euler_lagrange_gives_christoffel both return.

From P2 the $\theta$-equation of a geodesic is $\ddot\theta=\sin\theta\cos\theta\,\dot\varphi^2$. On the equator $\theta=\pi/2$, $\cos(\pi/2)=0$, so the right side vanishes: $\ddot\theta=0$, consistent with holding $\theta\equiv\pi/2$ (where $\dot\theta=\ddot\theta=0$) — the equator solves the geodesic equation and is a great circle. On a latitude $\theta=\theta_0\ne\pi/2$, staying put forces $\dot\theta=\ddot\theta=0$, yet the equation demands

a contradiction: maintaining that circle needs a non-geodesic (external) force, so it is not free fall. Hence is_geodesic(...) returns True on the equator and False on the latitude.

Proper time is $\tau=\int ds=\int\sqrt{\lvert -dt^2+dx^2\rvert}$ (mostly-plus, timelike branch). The straight worldline keeps $\Delta x=0$, so $\tau=\sqrt{\Delta t^2}=\sqrt{10^2}=10$. The tent is two legs, each with $\Delta t=5,\ \Delta x=3$, giving $ds^2=-25+9=-16$ and $\sqrt{\lvert-16\rvert}=4$ per leg:

The lone minus sign reverses the triangle inequality, so the inertial (straight) twin ages most — maximal aging — while the bent twin returns younger ($8<10$). This matches action_length(...) = 10.0 (straight) vs 8.0 (tent).

For the Killing vector $\xi=\partial_t=(1,0,0,0)$ the conserved projection is $\xi\cdot p=g_{\mu\nu}\xi^\mu\dot x^\nu=g_{tt}\dot t=-(1-2M/r)\dot t\equiv-E$. With $M=1,\ r=10,\ \dot t=1.2$:

Likewise $\xi=\partial_\varphi$ gives $L=g_{\varphi\varphi}\dot\varphi=r^2\sin^2\theta\,\dot\varphi$. Each is constant along any geodesic because $\frac{dQ}{d\lambda}=(\nabla_\mu\xi_\nu)\dot x^\mu\dot x^\nu+\xi_\nu\,\dot x^\mu\nabla_\mu\dot x^\nu=0$ (first term: the symmetric $\dot x^\mu\dot x^\nu$ contracted with the antisymmetric Killing tensor $\nabla_{(\mu}\xi_{\nu)}=0$; second term: the geodesic equation). The code reports $\xi\cdot p=g_{tt}\dot t=-0.96=-E$, matching killing_conserved(...) = −0.96.

Let $N=g_{\mu\nu}\dot x^\mu\dot x^\nu=\dot x_\nu\dot x^\nu$ be the tangent norm. Differentiating along the curve and using metric compatibility ($\nabla_\mu g_{\alpha\beta}=0$),

by the geodesic equation $\dot x^\mu\nabla_\mu\dot x^\nu=0$. So $N$ is constant: a null tangent ($N=0$) stays null and a timelike tangent ($N<0$ in mostly-plus) stays timelike — the causal character is frozen. A geodesic can therefore never convert a photon (null) into matter (timelike). This is exactly why a Schwarzschild null ray keeps $2L\approx0$ from $r=10\to14$ and a sphere great circle keeps $2L=g_{\mu\nu}\dot x^\mu\dot x^\nu$ constant.

With $T_{\mu\nu}=0$ the field equation is $G_{\mu\nu}=0$. Tracing with $g^{\mu\nu}$ in $n=4$ gives $g^{\mu\nu}G_{\mu\nu}=(1-\tfrac n2)R=-R=0$, so $R=0$ and the equation collapses to the plain vacuum condition

The Schwarzschild metric $g=\mathrm{diag}\!\big(-(1-\tfrac{2M}{r}),(1-\tfrac{2M}{r})^{-1},r^2,r^2\sin^2\theta\big)$ has $R_{\mu\nu}=0$ for $r>2M$ (Birkhoff: the unique spherically-symmetric vacuum). It is not flat — the Kretschmann scalar $K=R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}=48M^2/r^6\neq0$ (RE-11), so the trace-free Weyl/tidal curvature survives where Ricci vanishes. Numerically field_equation_residual returns $\max|G_{\mu\nu}|\approx7\times10^{-7}\approx0$ at $r=4,6,10$ (finite-difference noise), confirming the vacuum equation.

Contract $T_{\mu\nu}=(\rho+p)u_\mu u_\nu+p\,g_{\mu\nu}$ with $g^{\mu\nu}$:

The first contraction is the norm $g^{\mu\nu}u_\mu u_\nu=u\cdot u=-1$; the second is $g^{\mu\nu}g_{\mu\nu}=\delta^\mu{}_\mu=n=4$. Hence

Dust $p=0\Rightarrow$ trace $=-\rho$; radiation $p=\rho/3\Rightarrow-\rho+3\cdot\tfrac\rho3=0$ (the conformal/EM source is traceless). The result is a scalar, so it is identical in every frame and on every background — only $u\cdot u=-1$ entered. For $\rho=2.5,\,p=0.4$ the code returns trace $=-1.3=-\rho+3p$ even with the boosted $u$, and $0$ for radiation.

Trace $G_{\mu\nu}=8\pi T_{\mu\nu}$ with $g^{\mu\nu}$. In $n=4$, $g^{\mu\nu}G_{\mu\nu}=(1-\tfrac n2)R=-R$, so with $T\equiv g^{\mu\nu}T_{\mu\nu}$,

Now substitute this back into $R_{\mu\nu}-\tfrac12 R\,g_{\mu\nu}=8\pi T_{\mu\nu}$:

The two forms are the same equation, the trace merely moved from the geometry side onto the source. This trace-reversed form is the convenient one in vacuum: its left side is plain Ricci, so $T_{\mu\nu}=0$ gives $R_{\mu\nu}=0$ in one line, with no need to first argue $R=0$. The round trip is exact — trace_reversed_ricci(metric, T, x), then rebuilding $G=R-\tfrac12(g^{ab}R_{ab})g$, returns $8\pi T$ to machine precision (test_trace_reversed_equivalent_to_einstein).

Geodesics → Newton. For slow motion in a static field the geodesic equation reduces to $\ddot x^i=-\Gamma^i{}_{00}$, with $\Gamma^i{}_{00}=-\tfrac12 g^{ij}\partial_j g_{00}\approx-\tfrac12\partial_i g_{00}$. With $g_{00}=-(1+2\Phi)$,

i.e. Newton's $\ddot{\mathbf x}=-\nabla\Phi$ — this defines $\Phi$ as the gravitational potential. Field equation → Poisson. Linearising $G_{00}$ for the isotropic metric $g_{00}=-(1+2\Phi),\ g_{ij}=(1-2\Phi)\delta_{ij}$ gives $G_{00}\simeq2\nabla^2\Phi$ (the spatial perturbation is essential — $g_{00}$ alone yields only $\nabla^2\Phi$). The $00$ equation $G_{00}=\kappa T_{00}=\kappa\rho$ is then

which is Poisson's $\nabla^2\Phi=4\pi\rho$ only if $\kappa=8\pi$. Numerically newtonian_poisson_residual returns $G_{00}-2\nabla^2\Phi\approx-1\times10^{-5}$ against a signal $\approx-3\times10^{-3}$ (~0.4%), confirming $G_{00}=2\nabla^2\Phi$.

With the $\Lambda$ term the vacuum ($T_{\mu\nu}=0$) equation is $G_{\mu\nu}+\Lambda g_{\mu\nu}=0$, i.e. $R_{\mu\nu}-\tfrac12 R\,g_{\mu\nu}+\Lambda g_{\mu\nu}=0$. Trace it in $n=4$ ($g^{\mu\nu}G_{\mu\nu}=-R$, $g^{\mu\nu}g_{\mu\nu}=4$):

Substituting back, $R_{\mu\nu}=\tfrac12 R\,g_{\mu\nu}-\Lambda g_{\mu\nu}=(2\Lambda-\Lambda)g_{\mu\nu}=\Lambda g_{\mu\nu}$ — a maximally-symmetric Einstein space, curvature alike in every direction. de Sitter realises this with $R_{\mu\nu}=(3/L^2)g_{\mu\nu}$, so matching $\Lambda g_{\mu\nu}$ to $(3/L^2)g_{\mu\nu}$ fixes

It is the positive-curvature $\Lambda$-vacuum, with a cosmic horizon at $r=L$. The code agrees: field_equation_residual(de_sitter_metric(10), 0, x, Lambda=3/10**2) $\approx6\times10^{-7}$, while Lambda=0 leaves a residual $>10^{-3}$ — the wrong $\Lambda$ is genuinely not a solution.

Conservation $\nabla_\mu T^{\mu\nu}=0$ forces the curvature tensor on the left of the field equation to be identically divergence-free. The candidates built from $g$ and its first two derivatives are $R^{\mu\nu}$, $R\,g^{\mu\nu}$, and $g^{\mu\nu}$ itself. The contracted Bianchi identity (RE-11 §3) says

so $R^{\mu\nu}$ alone is not conserved unless $R$ is constant; the unique divergence-free combination is $G^{\mu\nu}=R^{\mu\nu}-\tfrac12 R\,g^{\mu\nu}$, and since $\nabla_\mu g^{\mu\nu}=0$ one may freely add any $\Lambda g^{\mu\nu}$. Hence the left side must be $G_{\mu\nu}$ (up to $\Lambda g_{\mu\nu}$), giving $G_{\mu\nu}+\Lambda g_{\mu\nu}=8\pi T_{\mu\nu}$. Choosing $R_{\mu\nu}\propto T_{\mu\nu}$ instead would impose $\nabla^\nu R=0$ — constant scalar curvature everywhere, far too rigid for a general source — whereas with $G$ the conservation law $\nabla_\mu T^{\mu\nu}=0$ is automatic. This is the structural argument the whole module rests on (notes §2).

In vacuum $R_{\mu\nu}=0$, so the Ricci scalar and every invariant built from $R_{\mu\nu}$ vanishes; the first surviving curvature scalar is the Kretschmann invariant

At $r=0$, $K\to\infty$: tidal forces diverge and no coordinate change removes it — a real curvature singularity, where geodesics end. At the horizon,

which is finite, so spacetime there is perfectly regular — only the $(t,r)$ chart fails (like a map's pole). Hence $r=2M$ is a coordinate singularity (the one-way event horizon) and $r=0$ the physical one. For $M=1$, $K(2M)=3/4$, matching kretschmann_formula(2, 1) → 0.75.

Expand $V(r)=(1-2M/r)(1+L^2/r^2)=1-\dfrac{2M}{r}+\dfrac{L^2}{r^2}-\dfrac{2ML^2}{r^3}$ and differentiate:

(after multiplying by $r^4/2$). The roots $r_\pm=\big[L^2\pm\sqrt{L^4-12M^2L^2}\big]/(2M)$ are real only when $L^4\ge12M^2L^2$, i.e. $L\ge 2\sqrt3\,M$. At equality the discriminant vanishes and the stable minimum ($r_+$) and unstable maximum ($r_-$) merge at

So $r_\text{ISCO}=6M$ — inside it no stable circular orbit exists. This is why isco(1) → 6, isco_from_potential(1) → 6.000, and circular_orbit_radius(2*3**0.5, 1) → 6, while $L$ a hair below $2\sqrt3\,M$ returns None.

For light $V_\gamma(r)=(1-2M/r)L^2/r^2=L^2\big(r^{-2}-2Mr^{-3}\big)$, so

independent of $L$, which factors out entirely. The nature is a maximum: $V_\gamma''=L^2(6r^{-4}-24Mr^{-5})$ evaluated at $r=3M$ is $-6L^2/(243M^4)<0$, so the circular photon orbit is unstable — a light ray there can circle but the slightest nudge sends it spiralling in or out. Hence $r_\text{photon}=3M$, matching photon_sphere(1) → 3 and photon_sphere_from_potential(1) → 3.000.

The semi-latus rectum is $a(1-e^2)$, so the per-orbit advance is $\Delta\varphi=6\pi M/[a(1-e^2)]$. With $M=M_\odot=1476.6\,$m, $a=5.79\times10^{10}\,$m, $e=0.20563$,

Mercury completes $100\times365.25/87.9691\approx 415.2$ orbits per century; converting radians with $1\,\text{rad}=206264.8''$,

This is the famous $43''$ anomaly Einstein recovered. Setting $M\to0$ removes the GR $-2ML^2/r^3$ term and $\Delta\varphi\to0$ — Newtonian ellipses close. The check returns 42.98.

The full Schwarzschild deflection is $\Delta\varphi=4M/b$. Grazing the Sun, $b=R_\odot$:

The equivalence-principle "falling photon" of RE-10 gives only $2M/b=0.88''$: that is the contribution of gravitational time dilation ($g_{tt}$) alone. The spatial curvature $g_{rr}=1/f=(1-2M/r)^{-1}$ — which a uniformly accelerating elevator cannot mimic — supplies an equal second half, doubling the bend to $4M/b$. Hence light_deflection(M_SUN, R_SUN)ARCSEC_PER_RAD → 1.751, exactly 2 (2*M_SUN/R_SUN).

Write $1+z=\sqrt{(1-2M/r_\text{obs})/(1-2M/r_\text{emit})}$. (a) For $r\gg M$ expand each root: $\sqrt{1-2M/r_\text{obs}}\approx1-M/r_\text{obs}$ and $(1-2M/r_\text{emit})^{-1/2}\approx1+M/r_\text{emit}$, so

RE-10's Pound–Rebka $gh/c^2$; for $(10^6,10^9,1)$ this is $10^{-6}-10^{-9}=9.99\times10^{-7}$, the returned $z$. (b) As $r_\text{emit}\to2M$ the denominator $1-2M/r_\text{emit}\to0^+$ while the numerator stays finite, so $1+z\to\infty$: light from just above the horizon is infinitely redshifted and the emitter appears to freeze and fade to black. At $r_\text{emit}=2.0001M$, gravitational_redshift(2.0001, 1e9, 1) ≈ 140 and climbs without bound.

For radial infall from rest at $r_0$ the conserved energy is $E=\sqrt{1-2M/r_0}$, and the radial equation gives $(dr/d\tau)^2=E^2-(1-2M/r)=\dfrac{2M}{r}-\dfrac{2M}{r_0}$ — finite and nonzero at $r=2M$. Integrating $d\tau=dr/\sqrt{2M/r-2M/r_0}$ converges right through the horizon down to $r=0$ in finite proper time $\sim\pi M(r_0/2M)^{3/2}$. But the bookkeeper's clock obeys $dt/d\tau=E/f$ with $f=1-2M/r$, so

Both statements hold because $t$ is a bad coordinate at $r=2M$ (the coordinate singularity of P1): it diverges even though the infaller's proper time — the physical clock — is finite. The distant image therefore asymptotes to the horizon, redshifting to black (P6).

Wavelengths stretch with the scale factor, so the observed-to-emitted ratio is $1+z = a_{\rm obs}/a_{\rm emit}$. With $a_0=1$ today,

For $z=3$: $1+z=4$, so every comoving length — and the universe — has grown by a factor $\mathbf{4}$, and $a_{\rm emit}=1/4=0.25$. The check inverts this: redshift returns $z=a_{\rm obs}/a_{\rm emit}-1$, so redshift(0.25)$=1/0.25-1=3.0$ and the canonical half-size universe redshift(0.5)$=1/0.5-1=1.0$.

Hand the flat FLRW metric to the curvature engine and its time–time Einstein component comes back as

The source is a comoving perfect fluid, whose energy density is $T_{00}=\rho$ (with $g_{00}=-1$, $u_0=-1$, so $T_{00}=(\rho+p)u_0^2+p\,g_{00}=\rho$). The $00$ field equation $G_{00}=8\pi T_{00}$ is therefore

the first Friedmann equation — the coefficient $3$ is not assumed but read off $G_{00}$. The finite-difference curvature confirms it: for de Sitter $a=e^{Ht}$ ($H=1$, $k=0$) G00_from_flrw $\to3.00\approx3H^2$, and for matter $a=t^{2/3}$ at $t=1$ it gives $1.34\approx3(2/3)^2=4/3$.

For $p=w\rho$ the fluid equation $\dot\rho=-3H(\rho+p)$ becomes $\dot\rho/\rho=-3(1+w)\,\dot a/a$, which integrates to

Feed this into flat Friedmann I, $(\dot a/a)^2=\tfrac{8\pi}{3}\rho\propto a^{-3(1+w)}$. Trying $a\propto t^q$ gives $\dot a/a=q/t\propto a^{-3(1+w)/2}\propto t^{-3q(1+w)/2}$, so matching powers of $t$ requires $-1=-3q(1+w)/2$, i.e.

Radiation $w=\tfrac13$: $\rho\propto a^{-4}$, $a\propto t^{1/2}$; matter $w=0$: $\rho\propto a^{-3}$, $a\propto t^{2/3}$; the $\Lambda$ case $w=-1$ is special ($1+w=0$), giving $\rho=$ const and exponential $a\propto e^{Ht}$. The density scalings are exactly era_density: era_density(2.0, 1/3)$=2^{-4}$, era_density(2.0, 0)$=2^{-3}$, era_density(2.0,−1)$=1$, and each $a(t)$ zeroes both Friedmann residuals (test_three_eras_satisfy_both_friedmann).

Friedmann I with $\Lambda=0$ is $H^2=\tfrac{8\pi}{3}\rho-\tfrac{k}{a^2}$. Divide through by $H^2$ and use $\rho_c=3H^2/8\pi$, $\Omega=\rho/\rho_c=8\pi\rho/3H^2$:

Since $a^2H^2>0$, the sign of $\Omega-1$ is exactly the sign of $k$:

The density is its own geometer — weigh the universe and you know whether space curls back on itself. The check sets $\rho=\rho_c$: omega(critical_density(H), H) $\to1.0$ (flat), while the on-shell closed/open densities of test_omega_trichotomy_matches_curvature_sign give $\Omega>1$ and $\Omega<1$ respectively.

In FLRW a comoving perfect fluid has $u^\mu=(1,0,0,0)$ and $T^\mu{}_\nu=\mathrm{diag}(-\rho,p,p,p)$. The $\nu=0$ component of $\nabla_\mu T^{\mu\nu}=0$ is

where the last term supplies the pressure piece $3Hp$, giving $\dot\rho=-3H(\rho+p)$. It is not an independent law: the contracted Bianchi identity $\nabla_\mu G^{\mu\nu}=0$ (RE-11 §3) makes $\nabla_\mu T^{\mu\nu}=0$ automatic, so once the two Friedmann equations (from $G_{00}$ and $G_{ij}$) hold, this one follows — a consistency relation. Its meaning is the first law $dE=-p\,dV$ for a comoving patch $V\propto a^3$: with $E=\rho V$ and $\dot V=3HV$, $\dot E=-p\dot V$ reads $\dot\rho V+\rho(3HV)=-p(3HV)$, i.e. $\dot\rho=-3H(\rho+p)$. Numerically fluid_equation_residual vanishes to machine precision for matter $\rho\propto a^{-3}$ and radiation $\rho\propto a^{-4}$ (test_fluid_equation_matter_and_radiation).

With $\Lambda=0$ the acceleration equation is $\ddot a/a=-\tfrac{4\pi}{3}(\rho+3p)$. Since $a>0$ and, for ordinary matter, $\rho>0$, the expansion accelerates iff the bracket is negative:

So pressure gravitates — it is $\rho+3p$, not $\rho$, that sources deceleration — and only a sufficiently negative pressure overturns the pull. Dark energy with $w=-1$ ($p=-\rho$) clears the bar: then $\rho+3p=-2\rho$, and using flat Friedmann I ($H^2=\tfrac{8\pi}{3}\rho$),

a constant — so $a\propto e^{Ht}$, de Sitter. The check confirms it: friedmann_2_residual(e^{H}, H²e^{H}, 3H²/8π, −3H²/8π) $\approx0$, de Sitter solving Friedmann II with $p=-\rho$ (test_de_sitter_satisfies_both_friedmann_two_ways).

Contract $\bar h_{\mu\nu}=h_{\mu\nu}-\tfrac12\eta_{\mu\nu}h$ with $\eta^{\mu\nu}$, using $\eta^{\mu\nu}\eta_{\mu\nu}=\delta^\mu_\mu=4$:

Feeding this into the reverse map recovers the original, so trace-reversal is an involution:

It is the natural variable because in Lorenz gauge $\partial^\mu\bar h_{\mu\nu}=0$ the linearized Einstein tensor collapses to $G_{\mu\nu}=-\tfrac12\Box\bar h_{\mu\nu}$, turning $G_{\mu\nu}=8\pi T_{\mu\nu}$ into one decoupled wave equation per component, $\Box\bar h_{\mu\nu}=-16\pi T_{\mu\nu}$ (notes §1). No code check — structural.

Acting with $\Box=\eta^{\mu\nu}\partial_\mu\partial_\nu$ on $\bar h_{\mu\nu}=\varepsilon_{\mu\nu}\cos(k_\lambda x^\lambda)$, each $\partial_\mu$ pulls down a factor $k_\mu$:

So $\Box\bar h_{\mu\nu}=-(k_\lambda k^\lambda)\bar h_{\mu\nu}$ vanishes for all $x$ only if $k_\lambda k^\lambda=0$. With $k^\mu=(\omega,\mathbf k)$ in mostly-plus signature, $k_\lambda k^\lambda=-\omega^2+|\mathbf k|^2=0$, i.e. $\omega=|\mathbf k|$ and phase speed $\omega/|\mathbf k|=1=c$. The code's residual coefficient is exactly $\omega^2-k^2$: wave_equation_residual(0.1,0.05,dispersion_omega(1.0),1.0) $=0$ on-shell, while $\omega=1.3,\,k=1$ gives $\approx0.069\neq0$ (off the light cone).

A symmetric $h_{\mu\nu}$ has $10$ independent components. Lorenz gauge $\partial^\mu\bar h_{\mu\nu}=0$ imposes $4$ linear conditions ($\to6$); the residual freedom — any $\xi^\mu$ with $\Box\xi^\mu=0$ preserves Lorenz gauge — supplies $4$ more ($\to2$). For $k^\mu=\omega(1,0,0,1)$, the condition $h_{0\mu}=0$ together with transversality $k^\mu h_{\mu\nu}=0$ kills the time and $z$ rows/columns, and tracelessness $h^i{}_i=0$ forces $h_{xx}=-h_{yy}$, leaving

Thus $10-4-4=2$ physical polarizations. is_transverse_traceless(tt_wave(h_+,h_×,ω,t,z)) returns True, with spatial trace $h_{xx}+h_{yy}+h_{zz}=0$ exactly.

At phase $0$ ($\cos=1$) the geodesic-deviation shift is $\delta\xi^i=\tfrac12 h^i{}_j\xi^j$. Pure $h_+$ has $h^x{}_x=h_+,\,h^y{}_y=-h_+$, so $(1,0)\to(1+\tfrac12 h_+,\,0)$ and $(0,1)\to(0,\,1-\tfrac12 h_+)$ — $x$ stretched, $y$ squeezed; for $h_+=0.2$ these are $(1.1,0)$ and $(0,0.9)$. Pure $h_\times$ has $h^x{}_y=h^y{}_x=h_\times$, so the diagonal $(1,1)/\sqrt2$ scales by $1+\tfrac12 h_\times$ and $(1,-1)/\sqrt2$ by $1-\tfrac12 h_\times$. The area factor is

zero at linear order because $h$ is traceless. ring_response(0.2,0,0,[(1,0),(0,1)])$\to[(1.1,0),(0,0.9)]$ and area_change(0.2,0,0)$\approx-0.01=-\tfrac14(0.2)^2$ — second order.

Far-field $\bar h_{ij}$ is built from time derivatives of the moments of $T^{00}$. The monopole $\int T^{00}d^3y=M$ is the total mass — conserved, so $\ddot M=0$: no monopole radiation. The dipole $\int y_i T^{00}d^3y$ is the center of mass; its first derivative is the total momentum $P_i$, conserved, so its second derivative vanishes: no dipole radiation. Radiation therefore starts at the quadrupole $Q_{ij}=\int y_iy_jT^{00}d^3y$. A spherically symmetric shell has isotropic $Q_{ij}\propto\delta_{ij}$, whose trace-free part $\bar Q_{ij}=Q_{ij}-\tfrac13\delta_{ij}Q$ vanishes. reduced_quadrupole of the symmetric octahedron gives $\max|\bar Q_{ij}|\approx1.3\times10^{-15}$ (luminosity $\approx6.7\times10^{-31}\approx0$), whereas a bar along $x$ gives $\bar Q_{xx}=+5.33,\ \bar Q_{yy}=\bar Q_{zz}=-2.67\neq0$ and radiates.

A binary's mass quadrupole is invariant under a half-turn $\varphi\to\varphi+\pi$, so it returns to the same shape twice per orbit: $f_{\rm GW}=2f_{\rm orb}$. Kepler's third law for total mass $M$ at separation $r$ gives $\omega_{\rm orb}^2=M/r^3$, hence $f_{\rm orb}=\tfrac1{2\pi}\sqrt{M/r^3}$ and $f_{\rm GW}=\tfrac1\pi\sqrt{M/r^3}$. Equating the quadrupole luminosity to the orbital energy loss yields $\dot f=\tfrac{96}{5}\pi^{8/3}\mathcal M_c^{5/3}f^{11/3}$, in which the masses enter only through the chirp mass $\mathcal M_c=(m_1m_2)^{3/5}/(m_1+m_2)^{1/5}$ — so a measured $(f,\dot f)$ determines $\mathcal M_c$ alone. For $36+29\,M_\odot$, chirp_mass(36,29)$\approx28.10\,M_\odot$; the equal-mass check chirp_mass(30,30)$\approx26.12=30/2^{1/5}$, and gw_frequency(36,29,r)$=2\cdot$orbital_frequency(36,29,r).

Both points follow from conservation laws. Electromagnetism has charges of both signs, so the charge dipole $\int\rho\,\mathbf x\,d^3x$ can oscillate freely and radiates at order $1/c^3$. Gravity has only positive mass — no negative gravitational charge — and the mass dipole $\int\rho\,\mathbf x\,d^3x=M\,\mathbf x_{\rm cm}$ has time derivative equal to the total momentum $\mathbf P$, which is conserved; its second derivative therefore vanishes and the dipole cannot source radiation. Radiation is pushed up to the quadrupole, costing a further $1/c^2$ for an overall $1/c^5$. Since $G/c^5\approx10^{-53}$ in SI units, even $M_\odot$-scale compact binaries produce strains $h\sim10^{-21}$, which is why detection demands km-scale interferometers. No code check — structural.

The horizons are the roots of $\Delta=r^2-2Mr+a^2=0$. The quadratic formula gives

real iff $M^2-a^2\ge0$, i.e. $a\le M$ ($a=M$ extremal, $a>M$ a naked ring singularity). For $M=1,a=0.6$: $\sqrt{1-0.36}=0.8$, so $r_\pm=1\pm0.8=(0.2,\,1.8)$. The static limit is $g_{tt}=-(1-2Mr/\Sigma)=0\Rightarrow\Sigma=r^2+a^2\cos^2\theta=2Mr$; at the equator $\cos\theta=0$ gives $r^2=2Mr$, so $r_E=2M=2>r_+$ — the gap is the ergoregion. These match kerr_horizons(1,0.6)$=(0.2,1.8)$, kerr_ergosphere(1,0.6,\pi/2)$=2.0$, and kerr_horizons(1,1.5) raising ValueError since $M^2-a^2<0$.

Kerr was built to solve the vacuum equations $R_{\mu\nu}=0$ (equivalently $G_{\mu\nu}=0$, since $R=0$ in vacuum). The code never assumes this — it checks it: handing $g_{\mu\nu}$ to RE-11's finite-difference ricci returns the actual Ricci tensor, so verify_vacuum $=\max_{\mu\nu}|R_{\mu\nu}|$ is a componentwise validator — one mistyped term (say the cross term $g_{t\phi}$) would leave a non-zero residual and fail the gate. Setting $a=0$ collapses $\Sigma=r^2+a^2\cos^2\theta\to r^2$, $\Delta=r^2-2Mr+a^2\to r^2-2Mr$, and the frame-dragging term $g_{t\phi}=-2Mar\sin^2\theta/\Sigma\to0$, leaving

exactly Schwarzschild. Numerically verify_vacuum(kerr_metric(1,0.5),[0,8,1.0,0.7])$\approx9.9\times10^{-7}\approx0$ (pure finite-difference noise), and kerr_metric(1,0) matches curvature.schwarzschild_metric(1) componentwise.

Trace Einstein's equation $G_{\mu\nu}=8\pi T_{\mu\nu}$: in 4-D $g^{\mu\nu}G_{\mu\nu}=-R$, so $R=-8\pi T^{\mu}{}_{\mu}$. The electromagnetic stress-energy $T^{\rm EM}_{\mu\nu}\propto F_{\mu\alpha}F_{\nu}{}^{\alpha}-\tfrac14 g_{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}$ is trace-free in four dimensions, $T^{\mu}{}_{\mu}=0$, hence

even though $T_{\mu\nu}\neq0$. So the Ricci scalar vanishes while individual components survive: since $R=0$ the mixed Ricci equals the mixed Einstein tensor, $R^{\mu}{}_{\nu}=G^{\mu}{}_{\nu}$, giving $G^{t}{}_{t}=G^{r}{}_{r}=-Q^2/r^4$ and $G^{\theta}{}_{\theta}=G^{\phi}{}_{\phi}=+Q^2/r^4$; the covariant components are then $O(Q^2/r^2)$, e.g. $R_{\theta\theta}=r^2R^{\theta}{}_{\theta}=Q^2/r^2$. At $r=3,\,Q=0.9$ this is $0.81/9=0.09$ — exactly verify_vacuum(reissner_nordstrom_metric(1,0.9),[0,3,1,0.7])$\approx0.09\neq0$ — while curvature.ricci_scalar(...)$\approx10^{-7}\approx0$; sending $Q\to0$ takes both to zero (Schwarzschild, Ricci-flat again).

With a cosmological constant the vacuum equation reads $G_{\mu\nu}+\Lambda g_{\mu\nu}=0$. Contract with $g^{\mu\nu}$, using $g^{\mu\nu}G_{\mu\nu}=-R$ and $g^{\mu\nu}g_{\mu\nu}=4$ in 4-D:

Substituting back, $R_{\mu\nu}=\tfrac12 R\,g_{\mu\nu}-\Lambda g_{\mu\nu}=(2\Lambda-\Lambda)g_{\mu\nu}=\Lambda g_{\mu\nu}$ — constant curvature, the Ricci tensor proportional to the metric. The cosmological horizon sits where the metric function degenerates, $f=1-\Lambda r^2/3=0$, i.e. $r=\sqrt{3/\Lambda}$. For $\Lambda=0.03$ everything matches: verify_einstein_lambda(de_sitter_metric(0.03),[0,3,1,0.7],0.03)$\approx6\times10^{-7}\approx0$, and curvature.ricci_scalar(de_sitter_metric(0.03),[0,3,1,0.7])$\approx0.12=4\cdot0.03$, confirming $R=4\Lambda$.

Foliate spacetime on $t=x^0=\text{const}$. The intrinsic (spatial) metric is just the lower block $\gamma_{ij}=g_{ij}$; the lapse and shift carry the time information via

The lapse is read from the inverse metric because $g^{tt}=-1/\alpha^2$. For inertial Minkowski $g=\mathrm{diag}(-1,1,1,1)$: $g^{tt}=-1\Rightarrow\alpha=1$, no off-diagonal $g_{0i}\Rightarrow\beta^i=0$, and $\gamma=\delta_{ij}$ — flat space sliced flatly. For any static diagonal metric with $g_{tt}=-f$ the inverse gives $g^{tt}=-1/f$, so $\alpha=\sqrt{f}$ and again $\beta=0$. These reproduce adm_decompose(curvature.minkowski_metric(),[0,1,1,0.5])$=(1.0,[0,0,0],I_3)$ and the Schwarzschild lapse adm_decompose(curvature.schwarzschild_metric(1),[0,8,1,0.7])[0]$\approx\sqrt{1-2/8}=0.866$.

The Hamiltonian constraint is $R^{(3)}+K^2-K_{ij}K^{ij}=16\pi\rho$. A moment of time symmetry means the slice is instantaneously at rest, $K_{ij}=0$ (so $K=0$), and the Schwarzschild exterior is vacuum, $\rho=0$; every term but the first drops:

The slice must be scalar-flat. That is not the same as flat: the induced 3-metric has $\gamma_{rr}=1/(1-2M/r)\neq1$ (the Flamm paraboloid), with non-zero 3-Riemann and 3-Ricci — only the particular contraction $R^{(3)}=\gamma^{ij}R^{(3)}_{ij}$ cancels. The code confirms both at once: hamiltonian_constraint_flat(curvature.schwarzschild_metric(1),[0,6,1,0.7])$\approx2.5\times10^{-8}\approx0$ (so $R^{(3)}=0$), while spatial_slice_metric(...)([6,1,0.7])[0][0]$=1/(1-1/3)=1.5\neq1$ — curved, yet scalar-flat, exactly the identity a numerical-relativity code enforces on its initial data.