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The Einstein tensor and its contractions areTo match the supplied Ricci tensor components, use the Riemann curvature tensor conventionThis convention also fixes the sign used for geodesic deviation above. In vacuum, tracing the Einstein field equations gives $-R+4\Lambda =0$. Substitution therefore yields $R_{\mu \nu }=\Lambda g_{\mu \nu }$.

Differentiate the angular field equation to obtainConsequently $R_{00}=\Lambda A=\Lambda g_{00}$ and $R_{11}=-\Lambda /A=\Lambda g_{11}$. The other angular component follows from $R_{33}={\sin }^2\theta R_{22}$. Thus the single angular equation implies all the displayed vacuum equations wherever this coordinate chart is regular.

Integrating $(rA{)}^{\prime }=1-\Lambda r^2$ gives $A=1+C/r-\Lambda r^2/3$. Match $g_{tt}\simeq 1+2\Phi$ with the Newtonian gravitational potential $\Phi =-GM/r$ in the weak-field overlap region to identify $C=-2GM$. Hence the Schwarzschild-de Sitter spacetime hasNegative $\Lambda$ instead gives the Schwarzschild--anti-de Sitter black hole case. This matching fixes the normalization locally; a spacetime with nonzero $\Lambda$ is not asymptotically flat.

For timelike geodesics in the orbital plane, take proper time as parameter and use $L=(A{\dot{t}}^2-{\dot{r}}^2/A-r^2{\dot{\varphi }}^2)/2$. The cyclic coordinates give $A\dot{t}=k$ and $r^2\dot{\varphi }=h$; the angular canonical momentum is $-h$ in this signature. The four-velocity normalization $2L=1$ then givesHere $k$ is conserved Killing energy per unit rest mass in the chosen time normalization, and $h$ is conserved angular momentum per unit rest mass. There is no general interpretation of $k$ as energy measured at an asymptotically flat infinity when $\Lambda \ne 0$.

For radial motion $h=0$, differentiation gives $2\dot{r}\ddot{r}=-A^{\prime }\dot{r}$. The radial Euler-Lagrange equation extends the resulting expression through a turning point where division by $\dot{r}$ would fail. ThusThis is the proper-time second derivative of the areal radius, not a nonzero four-acceleration of the freely falling particle. Positive $\Lambda$ contributes an outward term; negative $\Lambda$ reinforces inward attraction. Restoring units in the slow-motion, weak-field Newtonian limit givesThe extra term describes the gravitational effect of a cosmological constant, not drag or a contact force. The approximation requires $GM/(r{c}^2)\ll 1$, $|\Lambda |r^2\ll 1$ and speed much less than $c$.

Put $v=r\omega$, $\beta =v/c<1$ and $\gamma =(1-{\beta }^2{)}^{-1/2}$. For a photon emitted in laboratory direction $\mathbf{n}$ toward the stationary receiver, its frequency there isSince the receiver lies outside the circle, two emission points have a line of sight tangent to the circle. At those points the source velocity is directly toward or away from the receiver, so $\mathbf{n}\cdot {\mathbf{v}}_e=\pm v$. Thus the doppler extrema for a circularly moving source areThey do not depend on $d$, provided $d>r$; changing $d$ changes the emission and reception phases of the extrema. Closest and farthest source positions instead have transverse velocity relative to that ray, and are not the extrema.

For the corotating receiver, velocities must be evaluated at the actual emission and reception events. An unobstructed vacuum ray satisfies ${\mathbf{x}}_r-{\mathbf{x}}_e=c\Delta t\mathbf{n}$, while rigid steady rotation gives ${\mathbf{v}}_r-{\mathbf{v}}_e=\mathbf{\omega }\times ({\mathbf{x}}_r-{\mathbf{x}}_e)$. Therefore $\mathbf{n}\cdot {\mathbf{v}}_r=\mathbf{n}\cdot {\mathbf{v}}_e$. The emitter and receiver have equal speed magnitudes, hence equal Lorentz factors. The frequency ratio of corotating clocks is consequentlyThus the rim receiver measures ${\nu }_0$, independently of its angular separation. Different transit times or a Sagnac effect do not by themselves imply a frequency change between these identical-radius clocks. The model treats ${\nu }_0$ as a fixed proper clock frequency and assumes a signal path is available; it does not specify the disc's electromagnetic opacity.

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.

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.

Frequency of the orbiting source. Treat the gas blob as a test particle on a timelike circular Schwarzschild geodesic, with pressure and other nongravitational forces negligible. Treat our very distant location as infinity, so the observer's own gravitational potential is negligible. These are the assumptions that make a unique answer in terms of $R_S/r$ possible.

Put $f=1-R_S/r$ and use a local static observer at the emission point as an intermediate measuring frame. The remote observer lies on the symmetry axis of the orbit. A ray arriving on that axis has zero conserved axial angular momentum, so its local direction is perpendicular to the emitter's azimuthal velocity. Bending within a meridional plane does not alter this orthogonality. The transverse relativistic Doppler effect therefore giveswhere $v$ is the orbital speed measured by that local static observer. The gravitational redshift between static Schwarzschild observers then givesA gravitational circular orbit has $v^2=GM/r+O((GM/r{)}^2/c^2)$ in the weak-field limit, so $v^2/c^2=R_S/(2r)+O((R_S/r{)}^2)$. Expanding both factors,Solving for the emitted frequency givesThe gravitational part is $R_S/(2r)$ and the transverse kinematic time dilation contributes another $R_S/(4r)$; the absence of line-of-sight motion does not remove this second effect.

As an exact check, the local speed on a timelike circular Schwarzschild geodesic satisfies $v^2/c^2=\mu /(r-2\mu )$, where $\mu =GM/c^2=R_S/2$. Therefore the face-on frequency shift of a circular Schwarzschild source iswhose expansion agrees with the boxed answer. If the gas is supported or accelerated by pressure or magnetic stresses rather than following a geodesic, its orbital speed is not fixed by the mass and radius alone; the appropriate first-order result is instead ${\nu }_{\mathrm{em}}\approx {\nu }_{\mathrm{ob}}[1+GM/(r{c}^2)+v^2/(2c^2)]$, with $v$ supplied independently.

The conserved initial four-momentum of a particle with positive rest mass satisfies $P_{\mu }{P}^{\mu }=m^2{c}^2>0$. A single photon instead has null four-momentum, $K_{\mu }{K}^{\mu }=0$. Four-momentum conservation for an isolated one-body decay would require $P^{\mu }=K^{\mu }$, contradicting these Lorentz invariants. Thus decay into a single photon is impossible in every inertial frame.

Equivalently, the incoming energy and momentum obey $E^2-c^2|\mathbf{p}{|}^2=m^2{c}^4$, whereas a single outgoing photon obeys $E=c|\mathbf{p}|$. Changing the laboratory speed cannot change this mismatch of the invariant mass shell.

For a spherical mass distribution with constant circular speed $v_c$,For $v_c=220\mathrm{km}{\mathrm{s}}^{-1}$, this is about $0.0140(r/8\mathrm{kpc}{)}^{-2}{M}_{\odot }\mathrm{p}{\mathrm{c}}^{-3}$. It is the spherical model's total density; it is singular if extrapolated to the very centre and is intended only over the flat-rotation region.

For a sharp outer truncation at $r_{\ast }$, the exterior mass is fixed at $M_{\ast }=v_c^2{r}_{\ast }/G$. Integrating the gravitational work from $r$ to the boundary and then to infinity givesThusThe exterior contribution cannot be dropped: at $r=r_{\ast }$ it correctly gives the point-mass escape speed $\sqrt{2}{v}_c$.

If the fastest measured radial speed is assumed to approximate both the total speed and the local escape speed of a bound star, then at $r=8\mathrm{kpc}$,Without those extra assumptions, the measurement gives only a bound: a bound star's total speed is at least its radial component, and the sample maximum need not reach escape, so $r_{\ast }\gtrsim 19.9\mathrm{kpc}$ within this model. An unbound interloper would invalidate even that bound. This is not a determination of the real Milky Way halo edge.

For the two-body problem, the centre of mass lies a distance $a{M}_J/(M_{\odot }+M_J)=a/1048$ from the Sun. With $a=5.2\mathrm{AU}$,The relative circular-orbit period is $P=2\pi \sqrt{a^3/[G(M_{\odot }+M_J)]}\simeq 11.85$ years. The Sun's orbital speed about that centre of mass is thereforeThis is the space-speed amplitude; the observed radial velocity amplitude also includes the orbital inclination.

Differentiate the approximate dispersion: $v_g=3k{v}_{th}^2/\omega$, while $v_{\varphi }=\omega /k$. Therefore $\boxed{v_g{v}_{\varphi }=3v_{th}^2}$. In the high-phase-speed regime that justified the approximation, the group speed is correspondingly small. Trying to make $v_g$ much larger than $v_{th}$ would put the phase speed into the thermally populated distribution, where resonant Landau damping is strong and the undamped-fluid estimate becomes unreliable. This explains the stated tendency for these electron electrostatic waves, not a universal prohibition on fast electromagnetic waves in plasma.

Eclipses justify an approximately edge-on orbit, and equal velocity amplitudes imply equal masses. The relative speed is 200 km s${}^{-1}$, so $a=(v_1+v_2)P/(2\pi )\simeq 1.50\times 10^{10}$ m, or $\boxed{a\simeq 0.100\mathrm{AU}}$. Substitution in the binary period law gives $M_1+M_2\simeq 4.52M_{\odot }$, hence $\boxed{M_1=M_2\simeq 2.26M_{\odot }}$. Inclination corrections scale the dynamical masses as ${\sin }^{-3}i$ if exact edge-on viewing is relaxed.

Assume a central eclipse and nearly uniform projected speed during it. Let $r$ be the smaller radius and $R$ the larger. First-to-second contact requires travel $2r$, whereas the totality interval requires $2(R-r)$. With relative speed 200 km s${}^{-1}$,The light curve has ingress, flat minimum and egress lasting 1.7, 0.8 and 1.7 hours. At the four contacts the projected center separations are $R+r,R-r,R-r,R+r$. A noncentral chord would require an impact parameter and changes the inferred radii; the data alone do not fix that geometry exactly. For comparable masses in an ordinary coeval system, a component larger than its main sequence companion suggests expansion away from the main sequence, conventionally toward a subgiant/early giant track. That is a conditional evolutionary interpretation: neither effective temperature nor a bolometric flux is supplied, so a unique position on the Hertzsprung-Russell diagram cannot be inferred from these radii alone. Binary interactions and the assumed central chord also matter; these numbers do not establish an ordinary single-star evolutionary model.

Expansion redshifts photon energies and preserves a blackbody spectrum, so a comoving observer at $z=2$ sees $T=T_0(1+z)\simeq 8.18$ K. A one-sided dipole excess of 0.105% means $v/c=0.00105$, hence $\boxed{v\simeq 315\mathrm{km}{\mathrm{s}}^{-1}}$. Motion is toward the hotter region. The peak-to-peak contrast is twice this, so using 0.210% for $v/c$ would double the inferred speed incorrectly.

The low-redshift distance gives modulus about 36.28, so $M_V\simeq -20.28$ and $\boxed{L_V\simeq 10^{10}{L}_{\odot }}$. A stellar mass-to-light ratio of order one to a few would imply only $10^{10}$–several $10^{10}{M}_{\odot }$ in stars, below the dynamical estimate. Gas and a dark component can supply additional mass; the precise stellar estimate requires a population-dependent mass-to-light ratio and extinction correction.

An exponential luminous disk has most of its mass within a few scale lengths. Outside that concentration a mass-traces-light model tends toward a declining, approximately Keplerian speed, not an extended flat curve. Adding a spatially extended mass distribution explains the mismatch more naturally than merely rescaling the same concentrated light profile.

Homogeneity makes the inertial transformation linear. Align origins at $t=t^{\prime }=0$, let the second origin have velocity $v$, and impose isotropy and reciprocity. Then $x^{\prime }=\gamma (v)(x-vt)$, $y^{\prime }=y$, $z^{\prime }=z$. If the same limiting signal speed $c$ has trajectories $x=\pm ct$ in both systems, substituting these two trajectories fixes $t^{\prime }=\gamma (v)(t-vx/c^2)$. Requiring the inverse to have the same form with $-v$ fixes ${\gamma }^2(1-v^2/c^2)=1$. Continuity at rest selects the positive square root:Light enters this derivation through the empirical/in-principle operational identification of the invariant speed and Einstein synchronization. Group, homogeneity and isotropy arguments can instead derive a family with a universal parameter $K$, $t^{\prime }=\gamma (t-Kvx)$ and $\gamma =(1-K{v}^2{)}^{-1/2}$. They do not by themselves determine whether $K=0$ or a positive finite value. Empirical light propagation or other relativistic dynamics identifies $K=1/c^2$ for our world. Consequently a derivation need not assume light as its first premise, but it must acquire the physical invariant-speed structure somewhere rather than extract it from linearity alone.^[ref]

In this argument, homogeneity, isotropy, equivalence of inertial frames and the existence/identification of an invariant speed are physical structural posits. Coordinate origins, axis orientation and clock-unit choices are conventions. Einstein synchronization stipulates equal one-way signal times for outward and return paths; the empirical two-way light behavior and material-clock comparisons are not made true merely by that stipulation. Other synchronization coordinates can describe the same operational observations with a different displayed one-way light speed and transformed formula. Distinguishing those choices from the physical invariant structure is essential to the derivation.

Relativity of simultaneity concerns separated events. Under Einstein-synchronized inertial coordinates, $\Delta t^{\prime }=\gamma (\Delta t-v\Delta x/c^2)$. Two events with $\Delta t=0$ and $\Delta x\ne 0$ are therefore generally nonsimultaneous in a relatively moving frame. The distinction is compatible with invariant coincidence at one event and with invariant temporal order for timelike-separated events. It is spacelike separation that permits differing coordinate order.

Conventionality enters when defining distant clock synchronization. A send-and-return signal experiment measures a round-trip duration, but assigning a time to reflection at the distant clock requires a rule. Einstein's convention assigns equal outward and return times. Alternative admissible synchronizations can reassign the one-way coordinate speeds while preserving the observed two-way timings and local clock readings. These alternatives do not permit changing measured proper time at will or making any two causally ordered events operationally interchangeable.

The physical conclusion and convention claim should therefore be separated. No preferred inertial simultaneity relation is selected by the ordinary Lorentz-invariant structure. Once the standard synchronization is fixed, relativity between inertial frames follows quantitatively. Reparameterizing distant times shows a conventional element in their coordinate representation; it does not eliminate every empirical constraint from relativity. Matter dynamics, causal cones and clock transport still restrict what counts as an equivalent description.

A claim that simultaneity is simply unreal overlooks its well-defined local and frame-relative uses. A claim that it is entirely conventional overlooks the invariant difference between causal and spacelike separation. The defensible account distinguishes those uses rather than expecting one slogan to settle them all.^[ref]

The age difference is the difference between proper times along two worldlines joining the same departure and reunion events. In the inertial Earth frame a traveler who spends total coordinate time $T$ moving at speed magnitude $v$, with negligible-duration turnaround, accumulates $T\sqrt{1-v^2/c^2}$ while the stay-at-home clock accumulates $T$. These are invariant readings at reunion. Acceleration permits the change of traveling inertial frame, but the result is determined by the whole path, not a universal rule that acceleration itself slows clocks.

Relativity of simultaneity is important when explaining the same result using the traveler's successive inertial coordinates. At the turnaround, the traveler changes which distant Earth event is assigned the same time. The simultaneity slice changes, producing a jump in the coordinate time attributed to Earth in the ideal instantaneous limit. On each long inertial leg the traveler can assign a slow rate to the distant Earth clock without contradicting Earth's larger total elapsed time, because those assignments do not cover the reunion comparison with a single simultaneity convention.

The key distinction is between explaining a coordinate-based apparent symmetry and deriving an invariant result. One can calculate both proper-time integrals in one inertial frame without ever using the traveler's simultaneity jump. Conversely, claiming that each twin must simply see the other younger confuses two reciprocal comparisons between different sets of simultaneous distant events. Radar signals and Doppler shifts provide an operational account with finite signal travel times and the same reunion readings.

Thus relativity of simultaneity resolves a specific misleading inference in the paradox, but is not an extra physical mechanism that changes a clock at a distance. The worldline asymmetry and the invariant proper-time calculation suffice for the prediction; simultaneity explains why reciprocal inertial time-dilation statements cannot be pasted together indiscriminately.^[ref]

The geometrical explanation uses $d{\tau }^2=d{t}^2-d{\mathbf{x}}^2/c^2$ along each timelike path. In flat spacetime an inertial path between fixed timelike endpoints locally maximizes proper time, unlike the familiar shortest-path property of Euclidean distance. This is a physical metric property tested by clocks, not an appeal to a diagram alone. The explanation is satisfactory insofar as it connects that metric, specified worldlines and actual clock behavior; it does not require an absolute flow of time or a special clock-damaging action at the turnaround.

At fixed projectile mass and the assumed common quarter-power law, changing diameter from 30 to 75 m increases energy by $(75/30{)}^4$ and speed by $(75/30{)}^2=6.25$. The corresponding primary impact speed is about 15.6 km s${}^{-1}$. Five percent transferred to secondary-forming ejecta gives $v_{ej}\simeq 781$ m s${}^{-1}$. ThereforeThis assumes a 45-degree launch and similar target properties/coupling, as well as the source's fixed mass and velocity-transfer rules. Such a range is a noticeable fraction of the lunar radius, so curvature would produce a correction; the supplied flat-surface formula is an estimate rather than exact lunar orbit dynamics.

The relative orbital speed is $Z{e}^2/(4\pi {ϵ}_0\hslash n)$. Divide the circumference by this speed to findPositronium has twice the hydrogen period at the same principal quantum number. The symbol $n$ here labels principal levels with $n=1$ the ground state; if “nth excited state” is read literally as excitation count, replace $n$ by $n+1$.

Neglect upstream speed and hydraulic losses. If $H$ is the upstream free-surface height above the crest, the specific energy at the crest is $H=y+v^2/(2g)$ and flow per unit width is $Q=yv$. At the critical broad-crested control section, extremizing $Q^2=2g{y}^2(H-y)$ gives $y=2H/3$ and $v^2=gy$. ThereforeFinite approach speed, friction and contraction introduce measured discharge coefficients; the ideal result assumes a long enough crest for the near-hydrostatic critical-flow approximation.

A Pelton wheel splits and turns a free jet through nearly 180 degrees in moving buckets, transferring momentum to the runner. For ideal bucket speed $V$, the absolute exit speed is $2V-u$ and force is $2\rho Q(u-V)$. Power $2\rho QV(u-V)$ is maximized at $V=u/2$, giving $P_{\max }=\rho Q{u}^2/2$. If the same inflow must now pass a slit of height $d$, continuity gives $u=Q/d$. Using the preceding flow rate with $H=y_{\min }$,The head must actually be sufficient to supply this jet speed: $u^2/(2g)$ cannot exceed the available hydraulic drop. An arbitrarily small slit cannot produce infinite power from a fixed-height reservoir; it would reduce discharge or raise water until another outlet, including overtopping, operates. The printed scaling presupposes compatible head and fixed-flow conditions.

The cup's rotation relative to the radial basis has angular speed $\varphi$, so its absolute orientation has speed $\alpha =\Omega +\varphi$. With $\mathcal{R}(\theta )$ the planar rotation matrix,At $\varphi =-\Omega$, $\alpha =0$: the cup translates on a circle while its axes remain parallel to fixed inertial axes. Thus $x=R\cos \Omega t+x_1$, $y=R\sin \Omega t+y_1$. Cancellation of orientation rotation does not remove acceleration of the cup's center.