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Use the Euler-Bernoulli beam equation for a uniform slender elastic beam, small deflection and small slope, neglecting axial tension, shear deformation, rotational inertia of a cross-section and damping. Let $E$ be Young's modulus, $I={\int }_A{z}^{2}dA$ the second moment of area about the neutral axis, and $\rho$ the mass per unit length. If density instead denotes mass per volume, replace this $\rho$ by ${\rho }_{\mathrm{bulk}}A$. The bending stiffness is $EI$; the small-slope curvature is $y_{xx}$. The bending and kinetic energies areTwice integrating the variation of $U$ by parts gives the restoring load $-EI{y}_{xxxx}$ per unit length. Newton's second law therefore givesAt a pivot, displacement is fixed but slope is free; zero end bending moment gives $EI{y}_{xx}=0$. This also follows from the natural boundary term $EI{y}_{xx}\delta y_x$ in the variation.

With separation of variables, $y=XT$ givesChoose the time normalization $T(0)=1$. Initial rest removes the sine time term, so $T(t)=\cos \omega t$. The spatial characteristic roots are $\pm ik,\pm k$, givingThe two conditions at $x=0$ give $B+D=0$ and $-B+D=0$, hence $B=D=0$. At $L$, addition and subtraction of $X(L)=X^{\prime \prime }(L)=0$ give $C\sinh kL=0$ and $A\sin kL=0$. Nontrivial positive-frequency modes therefore have $C=0$ and $k_n=n\pi /L$. Thus the simply supported beam normal modes areThe zero-frequency cubic solution also vanishes under these four endpoint conditions, so $n=0$ is not an additional mode.

For the specified release, small slope requires $|a|L\ll 1$. Expand the initial deflection in the Fourier sine series. Its coefficients areWriting ${\omega }_1={\pi }^2\sqrt{EI/\rho }/L^2$, the resulting motion isOnly odd normal modes occur because the initial profile is symmetric about the midpoint, and their amplitudes decrease as $n^{-3}$. The frequencies grow as $n^2$, so the motion is dispersive and is not a single shape multiplied by one cosine. Nevertheless the ideal undamped solution is exactly periodic with period $2\pi /{\omega }_1$. At $t=\pi /(2{\omega }_1)$ all odd-mode cosines vanish, so the bar is instantaneously straight; at $t=\pi /{\omega }_1$ it has the opposite initial deflection. The first mode dominates displacement, with third-mode amplitude only $1/27$ of the first.

There is initial curvature incompatibility at a moment-free beam end: the prescribed parabola has $y_{xx}(x,0)=2a$, whereas a moment-free pivot requires zero curvature at an endpoint. Therefore the data do not define a classical solution with curvature continuous through the two end-time corners. They do define a finite-energy weak solution after release of the forces or end moments that prepared the initial shape. The sine expansion represents that solution: its displacement reproduces the initial parabola, each mode satisfies the released endpoint conditions, and ${\sum }_n{n}^4|A_n{|}^2<\infty$ gives finite bending energy. One must not differentiate the initial series twice at its endpoints and assign the interior parabola's curvature there. This qualification leaves the modal displacement prediction intact.

Faraday's law for the moving loop is $\mathcal{E}=-d{\Phi }_B/dt$, equivalently $\mathcal{E}=\oint (\mathbf{E}+\mathbf{v}\times \mathbf{B})\cdot d\mathbf{\ell }$ for its moving circuit. Lenz's law fixes its orientation: the induced current opposes the change in linked magnetic flux. The minus sign represents this opposition, not opposition to the existing field in every phase of motion.

Unroll the small loop's motion locally into translation to the right at speed $v\simeq b\omega$, where $b$ is the radius marked in the source figure. Take the imposed field into the page. On entry the linked into-page flux increases, so the current is counterclockwise. The leading vertical side carries current upward and feels a leftward magnetic force $iBa$. The in-field parts of the horizontal sides feel equal opposite vertical forces; the trailing side is outside the field. On exit the current is clockwise. The trailing vertical side is now the one in the field and again carries upward current, producing leftward drag; the horizontal forces reverse and still cancel. When the entire loop is within the uniform region, its flux is constant, so $i=0$ and every side force is zero under the stated negligible-self-inductance approximation.

For a loop crossing a magnetic field boundary, overlap changes at rate $av$, henceFor approximately constant speed this gives two negative rectangular force pulses of duration $a/v$, separated by a short zero-force interval while the loop is fully inside. Their time integral isThe quoted impulse is its magnitude; this impulse of magnetic braking opposes the motion. In fact its independence of speed remains exact within this locally translational sharp-boundary model even if speed changes: each crossing contributes $-(B^2{a}^2/R)\int vdt=-(B^2{a}^2/R)\int dx$ over distance $a$.

Write the system's moment of inertia as $\mathcal{I}$. Since $d{K}_{\mathrm{rot}}/dt=\tau \omega =F_{x}b\omega =F_{x}v$, the local energy balance during overlap isThus rotational energy lost is exactly resistive heat in this model. At nearly constant $v$, each pass givesFor a finite speed reduction the exact relation is the integral of $i(t{)}^{2}R$, rather than initial speed times impulse; the equality of work and heating still holds. Fringing fields, loop curvature and finite self-inductance smooth the ideal pulses. A practical user control is to vary the electromagnet current, hence $B$: braking scales as $B^2$.

Take $\zeta$ as signed distance from the neutral axis within the beam cross-section. Bending into a large radius $R$ changes the length of a fibre proportionally to $\zeta /R$, so linear elasticity gives $\sigma =E\zeta /R$, up to the curvature sign convention. The moment magnitude isHere $I$ is the second moment of area about the neutral axis, with units of length to the fourth power, not the mass moment of inertia. A homogeneous cross-section has its neutral axis through the centroid under pure bending.

For the straight pin-ended beam, the compressive dead load produces bending moment $-Fy$ at transverse offset $y$. In the small-slope approximation, curvature is $y^{\prime \prime }$ with the corresponding sign convention, givingThe stated endpoints fix positions rather than slopes: this interpretation is consistent with the illustrated end forces and the later sinusoidal unstressed shape. Truly clamped endpoints would have a different threshold. A nonzero solution $y=A_n\sin (n\pi x/L)$ requires $F_n=n^2{\pi }^{2}EI/L^2$. Thus the first Euler load is $F_c={\pi }^{2}EI/L^2$. Below it the perfect straight state is stable, at it the first mode is marginal in linear theory, and above it the straight state is unstable. The linear equation selects thresholds but cannot set the amplitude or describe a finite postbuckled equilibrium.

For the initially curved compressed beam, elastic stress responds to the change from its stress-free curvature. With $y=y_0+\eta$, the moment is $EI(y^{\prime \prime }-y_0^{\prime \prime })=EI{\eta }^{\prime \prime }$, so force balance givesPut $q=\pi /L$ and seek the branch continuously connected to the unloaded shape. Since $y_0=Y\sin qx$, substitute $\eta =C\sin qx$ to obtain $-EI{q}^{2}C=-F(Y+C)$, orAway from eigenvalues, the homogeneous solution $A\sin (\sqrt{F/(EI)}x)+B\cos (\sqrt{F/(EI)}x)$ vanishes under the two end conditions, so this particular solution is the unique linear equilibrium. For $0\le F<F_c$ and $Y>0$, the maximum additional displacement is ${\eta }_{\max }=YF/(F_c-F)$ and maximum total displacement is $y_{\max }=Y/(1-F/F_c)$, both at midspan.

This sinusoidal imperfection amplification of a column produces deflection at every positive load and strong amplification near $F_c$. The apparent divergence signals loss of the small-slope model before infinite displacement; nonlinear geometry is needed there. The formal continuation above the first threshold is not a stable physical loading branch. Both total and additional displacement are distinguished in the graph.

For a slender elastic beam undergoing small bending, assume plane cross-sections remain plane and normal to the neutral axis. Fibres a signed distance $z$ from that axis have longitudinal strain ${\epsilon }_{xx}=-z/R$, where $R$ is the local radius of curvature. Hooke's law gives stress ${\sigma }_{xx}=-Ez/R$, where $E$ is Young's modulus. Zero resultant axial force places the neutral axis through the section's centroid for a homogeneous beam.

The bending moment is the moment of this stress distribution. Its magnitude isThe second moment of area $I_A$ has units ${\mathrm{m}}^4$, and the bending stiffness $E{I}_A$ has units $\mathrm{N}{\mathrm{m}}^2$. A rectangular section of width $b$ and depth $h$ has $I_A=b{h}^3/12$, so placing more material far from the neutral axis strongly resists bending. This is the purpose of an I-section.

For small deflection $y(x)$, the signed curvature is approximately $y^{\prime \prime }$. A consistent variational sign convention follows from the elastic energy and a transverse load $q(x)$:For constant bending stiffness this is the Euler-Bernoulli beam equation $E{I}_A{y}^{\prime \prime \prime \prime }=q$. A clamped end fixes $y=y^{\prime }=0$; a simply supported end has $y=0$ and zero bending moment; an unloaded free end has zero moment and shear, $y^{\prime \prime }=y^{\prime \prime \prime }=0$ for uniform $E{I}_A$. Endpoint loads replace the corresponding zero-force boundary conditions.

For a cantilever of length $L$ carrying a tip force $F$, choose positive deflection along $F$. The internal bending gives $E{I}_A{y}^{\prime \prime }=F(L-x)$, with $y(0)=y^{\prime }(0)=0$. Two integrations giveThis approximation requires small slopes, elastic strains and a length large compared with section dimensions. Shear deformation, rotary inertia, plastic yielding and large rotations require extensions. Adding inertia gives flexural waves with ${\omega }^2=(E{I}_A/{\rho }_m{A}_s)k^4$ for a uniform beam, where ${\rho }_m{A}_s$ is mass per unit length.

Over the plotted range, approximate the surface tension by $\gamma (T)={\gamma }_0-\beta T$ with constant positive $\beta =-{\gamma }^{\prime }$, and take $C_A=70\mathrm{mJ}{\mathrm{K}}^{-1}=0.070\mathrm{J}{\mathrm{K}}^{-1}$. A reversible adiabat hasIntegrating givesThe arbitrary fixed reference temperature $T_{\ast }$ makes the logarithm dimensionless; changing it only shifts the integration constant. This is the dimensionally precise version of writing $C_A\ln T+\beta A$ as constant. Each term has units of entropy. Between two states,If the graph had substantial curvature, the entropy differential would still hold, but this particular constant-$\beta$ exponential would no longer follow.

For a slender elastic beam undergoing small bending, cross-sections remain approximately plane and perpendicular to the centreline. At distance $z$ from the neutral axis, longitudinal strain is ${ϵ}_{xx}=-z\kappa$, where $\kappa$ is curvature. Linear elasticity gives ${\sigma }_{xx}=E{ϵ}_{xx}$, so the bending moment has magnitudeHere $I$ is the second moment of cross-sectional area, not a mass moment of inertia. For width $b$ and depth $h$ in the bending direction, $I=b{h}^3/12$. The cubic depth dependence explains why a deep beam resists bending more effectively than the same material spread sideways.

For a uniform beam with transverse displacement $y(x)$, $\kappa \simeq y^{\prime \prime }$ and bending energy is $\frac{1}{2}\int EI(y^{\prime \prime }{)}^{2}dx$. Balancing transverse load $q(x)$ yields the Euler-Bernoulli beam equation $EI{y}^{\prime \prime \prime \prime }=q$ under a consistent sign convention. Supports specify values of deflection, slope, moment or shear; the equation alone does not fix the response.

Now apply a compressive end load $P>0$. A small deflection shortens the axial projection by $\frac{1}{2}\int (y^{\prime }{)}^{2}dx$, so the total potential energy relative to the straight configuration isIts stationary equation isFor pinned ends, $y(0)=y(\ell )=0$ and the end moments vanish, so $y^{\prime \prime }(0)=y^{\prime \prime }(\ell )=0$. The candidate shapes $y=A\sin (n\pi x/\ell )$ require $P=EI(n\pi /\ell {)}^2$. The first loss of stability is consequentlyAt smaller $P$, the quadratic energy is positive; beyond this threshold the straight configuration no longer minimizes it. End restraint changes the threshold: clamped–clamped gives $4{\pi }^{2}EI/{\ell }^2$ and clamped–free gives ${\pi }^{2}EI/(4{\ell }^2)$, equivalently an effective-length correction. A column buckles in the direction of smaller flexural rigidity. Euler buckling is a geometric instability within elastic response; yielding or crushing can occur first for a short thick strut. Large deflections, shear deformation and imperfections require refinements of the ideal slender-beam model.

For a slender elastic beam, the Euler-Bernoulli beam equation assumes small strain, small deflection slopes and cross-sections that remain plane and normal to the deflected centreline. A fibre at signed distance $y$ from the neutral axis has longitudinal strain ${ϵ}_{xx}=-y\kappa$, where $\kappa$ is the centreline curvature. Hooke's law gives ${\sigma }_{xx}=-Ey\kappa$, with Young's modulus $E$.

Choose the internal bending-moment sign so that $M=-\int y{\sigma }_{xx}dA$. ThenHere $I_A$ is the second moment of area, with units ${\mathrm{m}}^4$, not the mass moment of inertia of a rotating body. For a uniform material under pure bending, zero axial resultant places the neutral axis through the section's centroid.

For small slopes, $\kappa \simeq w^{\prime \prime }(x)$. Combining the moment-curvature relation with transverse force and moment balance gives, with a consistent load sign convention,For constant stiffness this becomes $E{I}_A{w}^{(4)}=q$. The bending stiffness $E{I}_A$ has units $\mathrm{N}{\mathrm{m}}^2$. Boundary conditions specify displacement and slope at a clamped end, or bending moment and shear force at a free end.

For a cantilever of length $\ell$ carrying a transverse end load $F$, take deflection positive towards the load. The moment magnitude is $M(x)=F(\ell -x)$, and integration with $w(0)=w^{\prime }(0)=0$ givesThe free-end moment vanishes and the tip shear supplies the applied load. For a rectangular section of width $b$ and bending thickness $h$, $I_A=b{h}^3/12$, so increasing thickness strongly increases bending resistance. The model ceases to be reliable for large deflections, plastic strains or short thick beams where shear deformation matters.

Assume a slender homogeneous elastic beam obeying linear elasticity, with plane sections remaining plane and no resultant longitudinal force. A length element of its neutral axis subtending a small angle $d\varphi$ has length $Rd\varphi$. A neighbouring fibre at distance $z$ from that axis has a length change of magnitude $zd\varphi$, giving longitudinal strain $\epsilon =z/R$ with signs chosen consistently for tensile and compressed fibres. By Hooke's law, its longitudinal stress is $\sigma =E\epsilon =Ez/R$.

The torque of these longitudinal forces about the neutral axis is thereforeHere $E$ is Young's modulus and $I$ the second moment of area. The displayed relation uses a signed radius and bending moment; an unsigned geometrical radius is $|R|=EI/|B|$. For consistency with the negative bending moments in the later parts, take the deflection $y$ positive downward and the signed curvature as $1/R\simeq y^{\prime \prime }$. Then the small-slope Euler-Bernoulli beam equation reads $EI{y}^{\prime \prime }=B$. Reversing the vertical coordinate reverses the corresponding sign relation but leaves the physical deflection unchanged.

Take downward deflection $y(x)$ positive, with the support level $y=0$. The rectangular section second moment of area about the horizontal neutral axis isFor small slopes the curvature relation from part (b), with the stated sign convention, isIntegrating twice givesThe support boundary conditions $y(0)=y(L)=0$ give $C_2=0$ and $C_1=q{L}^3/24$. HenceThis is a downward displacement; its upward component is $-y(x)$. As checks, reflection symmetry gives $y^{\prime }(L/2)=0$, and the maximum downward deflection isThe units are length: $\rho g/(E{b}^2)$ has units ${\mathrm{m}}^{-3}$. The width $a$ cancels because both weight and bending resistance scale with it. The small-slope approximation requires $\rho g{L}^3/(E{b}^2)\ll 1$; shear strain is also neglected, as appropriate for a slender elastic beam.

A lipid bilayer is fluid within its plane, so a bare equilibrated bilayer has no static in-plane shear modulus. Stretching its area is nevertheless expensive, whereas bending is much softer. The Helfrich membrane energy expresses this distinction throughwith membrane bending rigidity $\kappa$, mean curvature $H$, Gaussian curvature $K$, spontaneous curvature $C_0$ and membrane tension $\sigma$. Area dilation additionally costs approximately $K_A(\Delta A{)}^2/(2A_0)$. For a closed surface of fixed topology, the Gauss-Bonnet theorem makes the integrated Gaussian-curvature term constant.

These energies control vesicle shapes, budding and the ease of forming membrane tubes. Asymmetric leaflet composition or bound proteins can change the preferred curvature. Real cell membranes can also be coupled to a cytoskeleton, which supplies shear elasticity absent from a free lipid bilayer. Thus a red-cell shape cannot generally be inferred from the bare bilayer model alone.

Two useful measurements probe different terms. Optical tracking of thermal membrane flickering measures fluctuation amplitudes; for small height modes, a normalization with $h_q=A^{-1}\int h{e}^{-i\mathbf{q}\cdot \mathbf{x}}{d}^{2}x$ gives $⟨|h_q{|}^2⟩=k_{B}T/[A(\kappa q^4+\sigma q^2)]$. Micropipette aspiration controls pressure while measuring the membrane drawn into a narrow pipette, linking apparent area change to tension and, at larger strain, area dilation. The observables test the elastic coefficients rather than merely showing a membrane image.

Take overdots in this question to mean differentiation with respect to the orbiting particle's proper time $\tau$. Write $\mu =GM/c^2$, $x=r/\mu$ and $\ell =hc/(GM)$, with $h$ the specific angular momentum. Then the dimensionless timelike effective potential isThe last term is the relativistic correction to the Newtonian centrifugal-plus-gravitational potential. At large $r$ the two agree. Newtonian angular momentum gives an infinite inner centrifugal barrier for nonzero $h$; the relativistic correction lowers that barrier and permits plunge. The exterior problem starts at $r=2\mu$, where $U=-c^2/2$; extrapolating the potential towards $r=0$ is not a static exterior description.

Circular orbits obeyThere are two extrema when $h^2>12(GM/c{)}^2$: an inner maximum representing an unstable circular orbit and an outer minimum representing a stable one. At equality they merge at $r=6\mu$; below it neither exists. More explicitly, evaluate the curvature at a circular orbit:Timelike circular orbits require $r>3\mu$. They are radially stable for $r>6\mu$, so the marginally stable inner limit is $r_{\mathrm{ISCO}}=6GM/c^2$. The Newtonian potential instead has one stable minimum at $r=h^2/(GM)$, without an innermost stable circular orbit.

For the Schwarzschild circular-orbit local speed in the question, $\gamma$ is the Lorentz factor relative to a local Schwarzschild static observer, not $dt/d\tau$. That observer's clock obeys $dT=\sqrt{1-2\mu /r}dt$. Since the particle passes with local speed $v$, $rd\varphi /d\tau =\gamma v$. Circular normalization givesThus $v=c/2$ at the innermost stable circular orbit. Using Schwarzschild time in place of the local clock would give a different speed.

For the conserved particle energy measured at infinity, substitute $h^2$ in the circular energy equation. Setting $y=\mu /r$ givesTake the positive root for a future-directed exterior orbit. It approaches $m{c}^2-GMm/(2r)$ at large radius. The marginally bound circular Schwarzschild orbit solves $k=1$, giving $y(4y-1)=0$ and hence $r_{\mathrm{mb}}=4GM/c^2$ at finite radius. It is unstable; strictly bound circular orbits have $r>4\mu$, with $E<m{c}^2$. The phrase smallest bound orbit here denotes this marginal circular limit, not a prohibition on a bound-energy trajectory plunging below it.

At the innermost stable circular orbit,A radiatively efficient thin disc with a zero-torque inner boundary condition can therefore release about 5.7% of the accreted rest energy before the plunge. energy advection, stress at the inner edge and photon capture can alter the escaping luminosity, so this is a model efficiency rather than a universal emitted fraction.

For the Kerr ISCO spin limits of an equatorial prograde disc around a rapidly rotating Kerr black hole, the innermost stable circular orbit moves inward and its specific energy decreases. In the ideal extremal limit its Boyer-Lindquist radius tends to $\mu$, $E/(m{c}^2)\to 1/\sqrt{3}$, and the corresponding binding-energy efficiency tends to 42.3%. Retrograde rotation has the opposite trend: the extremal limit is $r_{\mathrm{ISCO}}=9\mu$, $E/(m{c}^2)=5/(3\sqrt{3})$, giving about 3.8%. The prograde limiting coordinate radius does not identify a massive orbit with the null horizon generator.^[ref]

Finally $\dot{t}=k/(1-2\mu /r)=1/\sqrt{1-3\mu /r}$. Dividing $\dot{\varphi }=h/r^2$ by this givesIts algebraic form is exactly Newtonian, but $r$ is the Schwarzschild spacetime areal radius and $t$ Schwarzschild time, normalized at infinity. Local clock rates, orbit stability and binding energy remain relativistic. With $G{M}_{\odot }\simeq 1.3271\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$,The marginally bound circular-orbit frequency is about $0.404\mathrm{kHz}$. These Schwarzschild ISCO orbital frequencies and marginally bound frequencies are rotation frequencies per distant stationary observer's time, as intended by using $d\varphi /dt$. A static observer at finite radius $r_o$ instead divides both frequencies by $\sqrt{1-2\mu /r_o}$, leaving their ratio unchanged. Identifying an observed quasi-periodic oscillation with the innermost stable circular orbit orbital frequency is an additional astrophysical assumption.

The weak equivalence principle says that freely falling test bodies with identical initial conditions follow the same trajectory independently of composition, when nongravitational forces and appreciable self-gravity are negligible. In Newtonian language it implies universal gravitational-to-inertial mass ratio. The Einstein equivalence principle additionally requires local Lorentz invariance and local position invariance of nongravitational experiments: a sufficiently small freely falling laboratory realizes special relativity. The strong equivalence principle extends universality to gravitational experiments and self-gravitating bodies.

A falling elevator makes the local content clear. Uniform gravitational acceleration can be removed along a freely falling worldline by a suitable frame, but curvature cannot generally be removed over an extended region. Nearby geodesics have relative accelerationfor the displayed Riemann-sign convention. tidal forces therefore remain detectable even in free fall. Equivalence motivates a metric description and gravitational clock shifts, but does not uniquely select Einstein's field equations.

Tests compare accelerations of different materials, frequencies of clocks at different gravitational potentials, or the response of self-gravitating bodies. lunar laser ranging times pulses reflected by lunar retroreflectors and fits Earth-Moon motion in the Sun's field; a differential free-fall violation would add a characteristic periodic range perturbation. The fit must also model tides, station and reflector geometry and propagation delays, with solar-radiation pressure and reflector thermal effects checked as possible mimics. Williams and collaborators' 2012 analysis is a concrete example of this method, published after this examination.^[ref]

In general relativity, light follows null geodesics. A weak gravitational field outside an isolated spherical mass gives the Schwarzschild light deflectionfor impact parameter $b\gg GM/c^2$. At the solar limb this is approximately $1.75$ arcseconds. A Newtonian corpuscular calculation gives half this value; the full result depends on spatial curvature as well as the gravitational time dilation. The parametrized post-Newtonian formalism predicts $\alpha =2(1+{\gamma }_{\mathrm{PPN}})GM/(b{c}^2)$, with ${\gamma }_{\mathrm{PPN}}=1$ in general relativity.

The Eddington experiment compared photographic star positions near the eclipsed Sun with reference exposures taken without the Sun in that field. Rays bend toward the Sun, but their backward extrapolations place the apparent stellar images farther from it. Plate scale, focus changes, atmospheric refraction and comparison-field alignment must be separated from a radial displacement proportional to $1/b$. The original report discusses the instruments, plate reductions and competing predicted deflections; the test is a quantitative comparison with uncertainty, not a unique proof of a gravitational theory.^[ref]

radio interferometry permits an independent test without an eclipse. A 2005 Very Long Baseline Array experiment observed four compact sources at 43, 23 and 15 GHz. Differential astrometry suppressed common antenna errors, while a fit $\mathbf{p}(\nu )={\mathbf{p}}_0+\mathbf{I}{\nu }^{-2}$ separated refraction in the solar corona from achromatic gravitational deflection. Tropospheric errors and radio-source structure also required calibration. The reported ${\gamma }_{\mathrm{PPN}}=0.9998\pm 0.0003$ was consistent with general relativity.^[ref]

tidal forces measure differences in gravitational acceleration across an extended region. Neighboring freely falling particles each have zero four-acceleration, but their separation need not accelerate uniformly. Let $u^{\mu }$ be the tangent to a reference timelike geodesic and ${\xi }^{\mu }$ its infinitesimal separation from a neighboring one, chosen from an affinely parametrized geodesic congruence. Using the Riemann curvature tensor convention specified in Solution 3,Changing the overall curvature convention changes this displayed sign. The covariant derivatives compare vectors by parallel transport, avoiding a coordinate-dependent subtraction of accelerations.

In the Newtonian limit, expand the force field at two nearby points:For $\Phi =-GM/r$, a local orthonormal frame aligned with the radius givesThere is radial stretching and transverse compression; the trace vanishes outside the mass. Two freely falling test masses and a measurement of their changing separation implement gravity gradiometry. Their common acceleration drops out, while a relative acceleration proportional to separation remains. A local inertial frame removes Christoffel symbols at a point, but cannot remove nonzero curvature over a finite region.

A tidal force is a difference of gravitational acceleration across an extended region. In Newtonian gravity, $g_i=-{\partial }_i\Phi$. Two neighboring freely falling particles separated by ${\xi }^j$ have, to first order,For a point mass, $\Phi =-GM/r$. In a local orthonormal radial/transverse frame,A small freely falling cloud stretches radially and compresses transversely. Outside the source the three eigenvalues sum to zero, consistently with ${\nabla }^2\Phi =0$; within matter the trace is $-4\pi G\rho$. These are accelerations of separations, not the much larger common acceleration of the cloud's centre.

In general relativity, freely falling test particles follow timelike geodesics. A freely falling frame can eliminate the connection at one event, but not the Riemann curvature tensor throughout a finite laboratory. Let $u$ be the geodesic tangent and $\xi$ a connecting vector for neighboring geodesics with $[u,\xi ]=0$. In the explicitly chosen curvature conventionthe geodesic deviation equation isIt follows by commuting ${\nabla }_u{\nabla }_{\xi }u$, using ${\nabla }_{u}u=0$ and zero torsion. In a parallel-propagated orthonormal frame, its spatial components directly describe measured relative acceleration. The weak-field, slow-motion limit recovers the Newtonian Hessian above. Reversing the curvature sign convention reverses the displayed curvature side, without changing the physical stretching. The equivalence principle removes locally uniform gravity, while geodesic deviation measures the residual tidal field. The connection between parallel transport, curvature and neighboring geodesics is developed systematically in the standard geometric construction.^[ref]

The cosmological principle idealizes spatial slices as homogeneous and isotropic. The corresponding Friedmann-Lemaître-Robertson-Walker metric can be writtenHere $t$ is cosmic time, $a(t)$ has dimensions of length in this convention, and $\chi$ is a dimensionless comoving coordinate. The sign of $K$ gives positive, zero or negative spatial curvature. It does not alone determine global topology. A comoving pair at fixed radial coordinates has proper separation scaling with $a(t)$, so the Hubble parameter is $H=\dot{a}/a$, and cosmological redshift satisfies $1+z=a(t_{\mathrm{obs}})/a(t_{\mathrm{em}})$.

For a perfect fluid of mass-equivalent energy density $\rho$ and pressure $p$, Einstein's equations giveThe last equation is local energy conservation for constant $\Lambda$, with vacuum energy kept separate from $\rho ,p$. A component with $p=w\rho c^2$ has $\rho \propto a^{-3(1+w)}$: pressureless matter dilutes as $a^{-3}$ and radiation as $a^{-4}$, the additional radiation factor being photon redshift. With $K=\Lambda =0$, matter domination gives $a\propto t^{2/3}$ and radiation domination gives $a\propto t^{1/2}$. Positive vacuum energy alone gives exponential expansion, $a\propto e^{Ht}$ with $H=c\sqrt{\Lambda /3}$.

Geometry, contents and initial conditions together determine the expansion history. A closed matter-only model recollapses; an open matter-only model expands indefinitely. With a cosmological constant, positive spatial curvature no longer by itself implies recollapse. Present critical density is ${\rho }_c=3H_0^2/(8\pi G)$ and the density parameters obey ${\Omega }_m+{\Omega }_r+{\Omega }_{\Lambda }+{\Omega }_K=1$, where ${\Omega }_K=-K{c}^2/(a_0^2{H}_0^2)$. These models describe the large-scale background, while local structure and inhomogeneities require perturbations or more general metrics.

Ordinary component differences at neighboring points depend on how the local bases vary. The covariant derivative corrects for that variation:Each upper tensor index adds a connection term and each lower index subtracts one. A scalar has ${\nabla }_{a}f={\partial }_{a}f$, while the derivative of a tensor transforms as a tensor with one additional covariant index. The Christoffel symbols themselves are not tensor components; their inhomogeneous transformation cancels that of the ordinary derivative.

For the Levi-Civita connection,These express metric compatibility and zero torsion. Along $x^a(\lambda )$, parallel transport is the initial value problemIt preserves inner products for the Levi-Civita connection. Parallel transport of the tangent itself gives the affinely parametrized geodesic equation, ${\ddot{x}}^a+{\Gamma }_{bc}^a{\dot{x}}^b{\dot{x}}^c=0$.

Transport generally depends on the path. On a round sphere, transport a tangent vector around the triangle bounded by two meridians $90$ degrees apart and the intervening equatorial arc. All three sides are geodesics; the enclosed spherical area is $\pi a^2/2$. The returning vector is rotated by $90$ degrees, with sign set by traversal orientation. Locally this holonomy is controlled by curvature, even though the connection can be made zero at a single point. A vector can therefore have changing coordinate components while remaining parallel, and constant components do not generally mean parallel transport.

This construction compares directions at separate points through a chosen connection, rather than assuming that a curved manifold supplies a path-independent comparison.^[ref]

Local equivalence and tides. In Newtonian mechanics, inertial mass $m_{\mathrm{i}}$ measures the response to a force, whereas passive gravitational mass $m_{\mathrm{g}}$ couples to an imposed Newtonian gravitational field. A freely falling body obeysThe observed composition independence of free fall says that this ratio is universal. With a common choice of units it is one. This is the Newtonian content of the weak equivalence principle for test particles; it does not by itself specify all nongravitational physics.

The Einstein equivalence principle further says that local nongravitational experiments in a sufficiently small freely falling laboratory have the form predicted by special relativity, independently of the laboratory's location and motion. In such a local inertial frame, the metric tensor is locally the Minkowski metric and the Christoffel symbols vanish at the selected event. The strong equivalence principle extends the universality claim to local gravitational experiments and bodies with significant gravitational self-energy. None of these statements removes measurable tidal forces across a finite apparatus.

In an ideal small freely falling lift, a released ball initially floats relative to the cabin, because both undergo the same free fall. Two separated balls, however, need not retain their separation: they sample different parts of the gravitational field. For a point source, linearizing $\mathbf{g}=-GM\mathbf{r}/r^3$ across a small separation $\mathbf{\xi }$ gives the tidal tensor predictionA radial separation stretches with coefficient $2GM/r^3$, whereas a transverse separation is compressed with coefficient $-GM/r^3$. Initially parallel velocities therefore develop relative motion even though the individual bodies experience zero proper acceleration.

The relativistic statement is geodesic deviation. With the Riemann curvature tensor conventiona connecting vector between neighboring freely falling worldlines satisfiesHere $D/d\tau$ is the covariant derivative along the reference geodesic, and $U^{\mu }$ its four-velocity. Curvature can be nonzero where the Christoffel symbols have been made zero. In the weak, static, slow-motion limit this equation reduces to ${\ddot{\xi }}^i=-{\partial }_i{\partial }_j\Phi {\xi }^j$, recovering the tidal tensor above. A finite falling lift can detect gravity through these relative accelerations; local equivalence does not imply a globally uniform field.

Weak-field connection and potential. Take $x^0=ct$ and the $+---$ metric signature. There is an index error in the printed formula for the Christoffel symbols: its first differentiated metric component must be $g_{\nu 0}$, not $g_{\mu 0}$. The correctly contracted expression isThe offending term vanishes under the stipulated time independence, so this correction does not change the intended calculation. Write $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$ and treat both $h$ and its derivatives as first order. To that order $g^{00}=1-h_{00}$, $g^{0i}=O(h)$, and $g^{ij}=-{\delta }^{ij}+O(h)$. A stationary metric tensor givesFor the temporal component the inverse-metric coefficient is already first order, soFor a spatial component the zeroth-order inverse metric contributes:The time component of the approximate geodesic equation says that $dt/d\tau$ is constant to the retained order. In its spatial component, convert the proper time derivatives to coordinate-time derivatives and cancel $(dt/d\tau {)}^2$:Comparison with the Newtonian equation $d^2\mathbf{x}/d{t}^2=-\nabla \Phi$ identifies $h_{00}=2\Phi /c^2+C$. For an isolated source, choose the gravitational potential to tend to zero at infinity and normalize the time coordinate there, giving $C=0$. ThusThe quantity $\Phi$ is the Newtonian gravitational potential, or gravitational potential energy per unit mass, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}={\mathrm{m}}^2{\mathrm{s}}^{-2}$. For a point source, $\Phi =-GM/r$. A static clock has $d\tau =\sqrt{g_{00}}dt\approx (1+\Phi /c^2)dt$, connecting the Newtonian limit with gravitational time dilation.