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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.

In the stipulated constant-speed model, the Earth-frame speed is $v=D/t=3.844\times 10^8/(76\times 3600)\simeq 1.405\times 10^3$ m/s. The Earth and Moon are stationary endpoints in this model, so their Earth-frame separation $D$ is the proper length. The spacecraft-frame separation, measured simultaneously in that frame, is $D^{\prime }=D/\gamma =D\sqrt{1-v^2/c^2}$. ThusThe spacecraft-frame distance is shorter by about 4.2 mm. This compares simultaneous endpoint measurements in two frames, not the coordinate difference between a launch event and an arrival event. The small-speed length contraction estimate also avoids numerical cancellation when evaluating the tiny contraction.

For a particle of rest mass $m_r$, $E=\gamma m_r{c}^2$ and $p=\gamma m_{r}v$, with $\gamma =(1-v^2/c^2{)}^{-1/2}$. The low-speed expansion is $E=m_r{c}^2+\frac{1}{2}{m}_r{v}^2+O(m_r{v}^4/c^2)$, separating invariant rest energy from the nonrelativistic kinetic energy term.

Conserve the total laboratory energy and momentum at coalescence: $E_{\mathrm{tot}}=4m{c}^2$ and $p_{\mathrm{tot}}=2.4mc$. Since $V=c^2{p}_{\mathrm{tot}}/E_{\mathrm{tot}}$, the combined particle moves at $0.6c$. Its rest mass is $M=\sqrt{4^2-2.4^2}m=3.2m$. The incident particle is not of mass $m$: its stated energy and momentum imply $m_{\mathrm{incident}}=\sqrt{3^2-2.4^2}m=1.8m$. ThusThe extra $0.4m{c}^2$ is internal rest energy retained by the composite after a perfectly inelastic collision; total energy and four-momentum are conserved, while the sum of individual rest masses need not be.

The composite's Lorentz factor is $1.25$. For a rest-frame lifetime $\tau$, the corresponding laboratory time is $1.25\tau$ and its travel distance isIf $\tau$ denotes a mean radioactive lifetime, this is the mean flight distance, not a fixed distance for every decay. The separate frame diagrams below use the common positive collision axis.

A donor star can feed a compact companion through Roche-lobe overflow near the inner Lagrange point, or through capture of a stellar wind. The transferred gas usually carries angular momentum; when its circularization radius exceeds the accretor's effective inner radius it forms an accretion disc. Outward angular momentum transport permits inward mass flow, while dissipation converts gravitational binding energy into heat and radiation. For a compact object with a surface $L\sim GM\dot{M}/R$; for a black hole write $L=\eta \dot{M}{c}^2$, with efficiency determined by the flow and inner boundary. Optically thick disc regions emit a sum of thermal radiation spectrum components; a hot optically thin accretion corona can inverse-Compton scatter softer photons into a harder X-ray component. A black hole has no material surface on which to radiate a stellar boundary layer.

An explicit example is the wind-fed black-hole binary Cygnus X-1. Its massive stellar companion supplies gas, and the accretion flow produces strong X-ray emission.^[ref] Compact-object evidence comes from orbital dynamics rather than X-ray brightness alone. The single-lined binary mass function constrains the unseen mass; inclination, donor properties and distance refine it. Jerome A. Orosz and collaborators combined optical photometry, radial velocities, rotation and a distance constraint to obtain a black-hole mass of $14.8\pm 1.0M_{\odot }$ in their 2011 analysis. The inference depends on the stellar and orbital model, and that historical value should not be confused with a latest measurement.^[ref]

A positive cosmological constant contributes a constant vacuum energy density and negative pressure. In the time-unit convention of Question 2,It therefore causes accelerating expansion once its negative-pressure effect outweighs the decelerating components. Its density stays constant while matter dilutes, so a small early contribution can dominate late. For an isolated positive-vacuum component the scale factor grows exponentially. The cosmological constant problem is the mismatch between the small inferred vacuum density and generic quantum-field vacuum-energy expectations, rather than a failure of the background solution itself.

The key direct expansion evidence came from Type Ia supernova distance-redshift measurements. After light-curve and colour standardization, distant events were fainter than expected in simple decelerating matter-only models, indicating larger luminosity distances. Riess and collaborators' 1998 analysis and Perlmutter and collaborators' independent 42-supernova study favoured an accelerating low-matter-density cosmology with a positive vacuum contribution. Dust, source evolution, selection effects, photometric calibration and peculiar velocities must be controlled: faintness alone is not a model-independent measurement of $\Lambda$.^[ref][ref]

cosmic microwave background acoustic geometry and independent matter-density constraints help break the supernova parameter degeneracy: nearly flat geometry with substantially less than critical matter requires another component within the FLRW model. large-scale structure of the universe and distance measurements provide further consistency tests. The combined evidence supports a dark-energy contribution compatible with a constant; accelerating expansion by itself does not prove exactly constant vacuum energy or rule out every alternative gravity model.

A quasar is a luminous active galactic nucleus powered by gas accretion onto a supermassive black hole. The hot disc produces much of the optical/ultraviolet continuum; surrounding gas gives broad and narrow emission lines, while some systems have relativistic jets. A small emitting region is inferred from rapid variability, $R\lesssim c\Delta t_{\mathrm{obs}}/(1+z)$ when beaming is absent. Sustaining a high luminosity from such a compact region favours accretion over ordinary stellar power. For approximately spherical, ionized hydrogen with electron-scattering opacity,Comparing luminosity with this Eddington scale constrains mass only after assumptions about anisotropy, opacity and accretion state. The emitted power obeys $L=\eta \dot{M}{c}^2$; high binding-energy efficiency makes growth of a massive compact accretor energetically plausible.

Broad-line reverberation mapping estimates a gas radius from a continuum-to-line time lag, $R\sim c\tau$, and combines it with line velocity width to infer $M\sim f_{\mathrm{vir}}R{v}^2/G$. Geometry, inclination and nongravitational gas motions enter the factor $f_{\mathrm{vir}}$. Independent evidence for compact massive nuclei comes from resolved stellar or gas orbits in nearby galaxies. Andrea M. Ghez and collaborators' Galactic-centre stellar-orbit study fitted a common central potential and distance, demonstrating a compact dark mass of order four million solar masses; the Galactic centre is a low-luminosity example rather than a quasar.^[ref]

quasars are an accretion phase of massive nuclear black holes; luminosity, variability and dynamical evidence constrain different aspects of that interpretation. A finite-resolution dark-mass measurement by itself is not a direct image of an event horizon.

A galaxy cluster deflects light from background galaxies according to its projected gravitational field. In a thin gravitational lens approximation,The distances are angular-diameter distances; only mass in the appropriate foreground/background geometry lenses the chosen source this way. Near critical curves, strong gravitational lensing produces highly magnified arcs and multiple images of the same source. Farther out, weak gravitational lensing produces small coherent shear distortions, inferred statistically because individual galaxies have unknown intrinsic ellipticities. The measured reduced shear is $g=\gamma /(1-\kappa )$. Transparent lensing obeys surface-brightness conservation in gravitational lensing at fixed source redshift while changing solid angle and integrated flux.

Strong-lensing image positions and source redshifts constrain the central projected mass; weak shear extends the map to larger radii. Multiple source planes help constrain geometry and degeneracies, but the mass-sheet degeneracy illustrates why image shapes alone do not uniquely fix absolute mass. Point-spread-function anisotropy, selection, shape calibration, cluster-member contamination, source redshift errors and unrelated line-of-sight structure require controls. Lensing does not require hydrostatic equilibrium, making it valuable for disturbed clusters, although interpretation still depends on the gravitational model.

An instructive observation is the merging Bullet Cluster: weak-lensing mass concentrations approximately follow the collisionless galaxies while the dominant visible baryonic plasma, traced by X-rays, is displaced. Clowe and collaborators combined ground-based and Hubble images for lensing reconstruction and compared them with the gas distribution, providing evidence for mass beyond the observed baryons.^[ref]

Take $\tau$ as particle proper time, dots as $d/d\tau$, and $f(r)=1-2\mu /r$ with $\mu =GM/c^2$. In the equatorial plane the geodesic Lagrangian isThe cyclic coordinates give $d(c^{2}f\dot{t})/d\tau =0$ and $d(-r^2\dot{\varphi })/d\tau =0$. Set the resulting constants to $c^{2}k$ and $-h$ respectively. The explicit second-order equations areThe polar equation is satisfied by $\theta =\pi /2$, $\dot{\theta }=0$, consistently with spherical symmetry. The proper-time normalization $d{s}^2=c^{2}d{\tau }^2$ supplies the third first integral:The normalization is $2\mathcal{L}=c^2$; it is distinct from choosing an arbitrary affine tangent's norm.

Stationarity and axial symmetry make the covariant time and azimuthal momenta constant. For physical momentum and coordinate $t$ in seconds, $P_t=m{c}^{2}f\dot{t}=m{c}^{2}k$ and $P_{\varphi }=-m{r}^2\dot{\varphi }=-mh$. Thus $m{c}^{2}k$ is the conserved energy normalized at infinity and $mh$ is the signed axial angular momentum. This is Killing energy with positive time signature; a convention with negative timelike norm uses the opposite contraction sign. The supplied covariant-momentum form of the geodesic equation makes these cyclic constants immediate, but its parametrization must be understood: for $p^{\mu }=d{x}^{\mu }/d\lambda$ it is exactly as written. For physical $P^{\mu }=md{x}^{\mu }/d\tau$, proper-time differentiation instead gives $d{P}_{\mu }/d\tau =({\partial }_{\mu }{g}_{\nu \sigma })P^{\nu }{P}^{\sigma }/(2m)$; equivalently, the displayed mass-free form uses $\lambda =\tau /m$. This affine momentum normalization does not change either conservation law.

Multiply the normalization equation by $f$ and substitute the two cyclic integrals. Rearrangement gives the radial energy equationFor radial marginal escape $h=0$ and the future-directed constant is $k=1$, because the speed at infinity vanishes. Choose the outward root:Hence the marginal escape geodesic in Schwarzschild spacetime hasThe constant $a$ has units of length to the power $3/2$ divided by time. This radius is an areal coordinate, not accumulated local proper distance.

A Schwarzschild static observer at exterior radius $r$ uses local radial rulers $d\ell =dr/\sqrt{f}$ and a local clock $dT=\sqrt{f}dt$. Therefore the local radial speed of a Schwarzschild geodesic isFor the marginal trajectory $k=1$ this givesIts equality to $\dot{r}$ is particular to $k=1$; the local definition, rather than that accidental equality, is the required observer measurement. The Schwarzschild-coordinate velocity is instead $dr/dt=f\sqrt{2GM/r}$.

For nonzero asymptotic speed $0\le u<c$, the same energy constant is $k={\gamma }_{\infty }=(1-u^2/c^2{)}^{-1/2}$. Combining the radial equation with the local-ruler relation gives radial Schwarzschild escape with asymptotic speedThe positive root describes the outward particle, at radii on its trajectory with $r\ge r_0$. Independently, a static observer has four-velocity $U=(1/\sqrt{f},0,0,0)$ in these coordinates, giving relative Lorentz factor $g(U,u_{\mathrm{particle}})/c^2=\sqrt{f}\dot{t}=k/\sqrt{f}$ and the same speed. As $r\to \infty$ it tends to $u$; for $u=0$ it gives the preceding result. Its limiting value approaches $c$ as $r\downarrow 2GM/c^2$, but that is a limit through exterior observers: no static timelike observer exists on or inside the horizon.

For five points, partition the unit square into four squares of side $1/2$. Assign boundary points to just one neighbouring piece. By the pigeonhole principle, two of the points lie in one piece, whose diameter is its diagonal, $\sqrt{(1/2{)}^2+(1/2{)}^2}=\sqrt{2}/2$.

For eight points, use seven smaller-diameter pieces rather than eight rectangles. Start with the convex pentagonTake its four reflected copies at the corners of the square. The remaining pieces are the left triangle with vertices $(0,1/4),(1/4,1/2),(0,3/4)$, its right-hand reflection, and the central diamond with vertices $(1/4,1/2),(1/2,1/4),(3/4,1/2),(1/2,3/4)$. These seven polygons cover the square with disjoint interiors. To verify coverage directly, for $0\le y\le 1/4$ the two bottom pentagons split the horizontal section at $x=1/2$. For $1/4\le y\le 1/2$, the five successive sections have endpoints $0,y-1/4,3/4-y,y+1/4,5/4-y,1$ and belong respectively to the left triangle, bottom-left pentagon, diamond, bottom-right pentagon and right triangle. Reflection gives the top half.

The diameter of a finite convex hull is the greatest distance between its vertices: if two points are convex combinations of vertices, their difference is a convex combination of vertex differences, and the triangle inequality bounds its norm by the largest such distance. Checking the five vertices of $P$ gives diameter $\sqrt{(1/2{)}^2+(1/4{)}^2}=\sqrt{5}/4$; reflections preserve that value. Each remaining triangle and the diamond has diameter $1/2$, which is smaller. After assigning shared boundaries to one piece, eight points in these seven pieces force a pair in a common piece. Hence some pair is at distance at most $\sqrt{5}/4$.

Both arguments are instances of a pigeonhole distance bound from a partition.

Suppose a subset of a metric space is covered by $m$ sets, each of diameter at most $D$. Assign each point to one covering set containing it, choosing a single set at shared boundaries. Among $m+1$ points, the pigeonhole principle puts two in a common set, and their distance is at most $D$. The pieces need not be equal or congruent. This freedom permits better distance bounds than dividing a region only into equal rectangles.

Suppose a group acts by isometries on a metric space $X$, and for points $a,b$ in different orbits one has ${\inf }_{g}d(a,gb)>0$. Then the orbit space with its quotient topology is Hausdorff. Indeed, choose $\epsilon$ less than one third of this positive infimum and saturate the open radius-$\epsilon$ balls about $a$ and $b$ by the action. These saturated sets are open because every acting isometry is a homeomorphism. Their translates cannot meet: their centres stay at least the given positive distance apart. Their disjoint open images in the quotient separate the two orbit points, since the preimage of each image is its saturated set. This applies to the glide reflection $(x,y)\mapsto (x+1,-y)$ on a closed strip: the distance from one point to the translates of another tends to infinity as the integer translation grows, so two distinct orbits have a positive minimum distance.

A small longitudinal electric field drives conductor current in a lossy coaxial line. Its cross product with the azimuthal magnetic field sends energy radially into both conductors, where it becomes heat; forward power decreases with distance.

For the minor circular arc cut off by a chord of length $0\le L\le 2R$ on a circle of radius $R>0$, the sagitta is the maximum perpendicular distance from the chord to that arc:The perpendicular bisector of the chord passes through its midpoint and the center. The Pythagorean theorem gives the center-to-chord distance $\sqrt{R^2-(L/2{)}^2}$; subtracting it from $R$ proves the formula. The major circular arc instead gives $R+\sqrt{R^2-L^2/4}$, so choosing the arc matters.

For $L/R\ll 1$, a Taylor expansion gives $s=L^2/(8R)+O(L^4/R^3)$. In a uniform magnetic field perpendicular to a charged track's motion, $R=p/(|q|B)$, so the sagitta measures the momentum $p$.

Let $K$ be a nonempty compact convex set in a Banach space and $T:K\to K$ a continuous map. For every $\epsilon >0$, there is a finite-dimensional linear subspace $L_{\epsilon }$ and a continuous map $T_{\epsilon }:K\to K\cap L_{\epsilon }$ that approximates $T$ uniformly to within $\epsilon$ in norm.

Choose $y_1,\dots ,y_N\in T(K)$ whose open radius-$\epsilon$ balls cover the compact image $T(K)$. Set $w_i(x)=\max \{0,\epsilon -\Vert Tx-y_i\Vert \}$ andThe denominator is positive. This convex combination belongs to $K\cap \mathrm{span}\{y_1,\dots ,y_N\}$ and depends continuously on $x$. Every contributing $y_i$ has distance less than $\epsilon$ from $Tx$, so the triangle inequality gives $\Vert T_{\epsilon }(x)-Tx\Vert <\epsilon$.

If the continuous dual space $X^{\prime }$ of a normed vector space is norm separable, then $X$ is separable. Choose a dense sequence $f_n$ in the dual unit ball and unit-ball points $x_n$ with $|f_n(x_n)|\ge \Vert f_n\Vert /2$. If their closed linear span were proper, the Hahn-Banach distance formula would give a norm-one functional $f$ vanishing on it. An $f_n$ within $1/4$ of $f$ would simultaneously satisfy $|f_n(x_n)|<1/4$ and $|f_n(x_n)|>3/8$, a contradiction. Rational linear combinations give a countable dense subset of $X$.

The converse fails by duality of l1 and l infinity: ${\ell }^1$ is separable, whereas ${\ell }^{\infty }$ contains uncountably many zero-one sequences at mutual distance one.

For ideal clocks in an inertial frame separated by fixed distance $L$, transporting a clock at small speed $u$ produces lag $\delta =(L/u)(1-\sqrt{1-u^2/c^2})\sim Lu/(2c^2)$. Hence the limit agrees with Einstein synchronisation, assuming the clock hypothesis. A relabelling $t_{\kappa }=t+\kappa x/c$ retains an endpoint offset $\kappa L/c$, so the result does not remove the choice of distant-time convention.

A sphere-plane intersection is a circle, a point, or the empty set. For a sphere of radius $a>0$ whose centre has distance $d$ from the plane, the circular case has radius $\sqrt{a^2-d^2}$ and centre at the orthogonal projection of the sphere centre onto the plane. The Pythagorean theorem proves the radius formula. The cases are respectively $d<a$, $d=a$ and $d>a$; a plane through the centre cuts a great circle.

A point mass constrained to a fixed distance $r$ from a pivot and free to move in three dimensions under uniform gravitational acceleration. With the vertical axis downward and polar angle $\theta$ measured from it, the Lagrangian is the expression in the title. The azimuth $\varphi$ is a cyclic coordinate, so $m{r}^2{\sin }^2\theta \dot{\varphi }$ is conserved angular momentum about the vertical axis. A rigid massless rod enforces the constraint without the loss-of-tension limitation of a string.

On an open set where $u>0$, put $v=\sqrt{u}$. The chain rule gives $\nabla u=2v\nabla v$, so the quadratic eikonal equation is equivalent to $|\nabla v|=1$. For constant positive data $v=v_0$ on a smooth initial curve, the tangential derivative vanishes and the two allowed normal derivatives are ${\partial }_{n}v=\pm 1$. Locally the two solutions are $v=v_0\pm d$, where $d$ is oriented distance along the normal rays; hence $u=(v_0\pm d{)}^2$ wherever the chosen $v$ stays positive and the normal-coordinate map remains invertible.

There are two separate obstructions to using this initial-data chart. A characteristic caustic occurs when normal rays focus and the projection Jacobian determinant vanishes. Alternatively $v$ can reach zero while the geometric rays remain separated: the square-root transformation then ceases to apply and the original Charpit's equations have a degenerate zero-gradient state. A continuation of $u$ beyond either obstruction needs separate justification; squaring an algebraic expression alone does not preserve a prescribed positive square-root branch.

Let a simple closed regular planar curve be invariant under point reflection through a center and remain strictly on one side of a coplanar axis of rotation, at distance $d$ from its center. If $L$ is its arc length, its surface of revolution has surface area $2\pi dL$. To prove this, choose coordinates with the center at the origin and the axis of rotation at $y=-d$. The surface area formula is $2\pi \oint (d+y)ds$. Pair each point with its reflected point: the arc length element $ds$ is unchanged and $y$ changes sign. Thus $\oint yds=0$, proving the result. The one-side condition prevents the surface from passing through the axis of rotation or requiring an absolute value with changing sign.