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Dark matter is an inferred gravitating component whose mass is not accounted for by the observed luminous material. Galaxy rotation curves, motions in galaxy clusters and gravitational lensing probe its gravitational contribution through different measurements.^[ref]

For illustration, circular motion in a spherical mass distribution obeys $v_c^2=GM(<r)/r$. A roughly constant orbital speed over increasing radius therefore requires enclosed mass to keep increasing in this model. A realistic inference also models visible matter and the system's geometry; a rotation curve alone is not a direct identification of a new particle.

Dark-matter direct detection searches for interactions in terrestrial targets, while indirect detection of dark matter looks for secondary products. These tests address microscopic models rather than replacing the astronomical gravitational evidence.

Launching at the correct circular orbital speed but at in-plane angle $\alpha$ from the tangent leaves specific orbital energy $-GM/(2R)$, hence semi-major axis $R$. The specific angular momentum becomes $\sqrt{GMR}\cos \alpha$, so $j^2=GMa(1-e^2)$ gives $e=|\sin \alpha |$. The actual periapsis and apoapsis are $R(1\mp e)$, equally far from the intended circular orbit. The maximum geometric radial separation is therefore $Re$. This is not an equal-time separation between two moving satellites.

Write $\mu =GM$ and let $v_c=\sqrt{\mu /r_0}$ be the original circular orbital speed. An impulse leaves the orbital angular momentum unchanged if it is radial: its angular impulse $\mathbf{r}\times \Delta \mathbf{p}$ is zero. The tangential velocity remains $v_c$, and an added radial component $w$ givesThus kinetic energy increases even though angular momentum does not. For a Radial kick from a circular Kepler orbit, the new specific orbital energy and squared specific angular momentum areThe orbital eccentricity is $e=\sqrt{1+2\epsilon j^2/{\mu }^2}=\sqrt{\beta -1}$, while the semi-latus rectum is $p=j^2/\mu =r_0$. HenceFor $\beta =2$ the trajectory is a parabolic Kepler orbit and has no finite furthest radius; its periapsis is $r_0/2$. For $\beta >2$ it is a hyperbolic orbit, with the same formula for $r_{\min }$ but unbounded radius:One must not use the negative denominator in the bound-orbit expression as a negative physical apocentre. On a departing unbound branch the conic's periapsis may already lie in the past; the future minimum from the kick event is then $r_0$. These are ideal conic orbits; a periapsis inside a real planet would cause impact before the ideal continuation.

photon orbit and the Newtonian limit. Write $r_g=GM/c^2$ and $e=E/(m{c}^2)$. For a timelike circular Schwarzschild geodesic,The denominator tends to zero at $r=3r_g$, so the energy per unit rest mass diverges as a massive circular orbit approaches that radius. The corresponding local orbital speed measured by a static observer satisfies $v_{\mathrm{loc}}^2/c^2=r_g/(r-2r_g)$ and tends to one. This suggests a null circular orbit, but one must not simply insert $m=0$ into a massive-particle formula. For a photon, the radial effective potential is proportional to $(1-2r_g/r)/r^2$; its derivative vanishes at $r=3r_g$ and its second derivative there is negative. Thus the photon sphere is at $3GM/c^2$, and its circular null orbits are unstable.

At $r\gg r_g$, expansion givesThe second term is the total Newtonian circular-orbit energy, kinetic plus potential; the relativistic energy also includes rest energy. Here $E$ is the conserved energy referred to a stationary observer at infinity, not the local kinetic energy.

The stable inner edge. At fixed angular momentum, write $\ell =L/(mc)$. The timelike radial geodesic equation can be writtenThe circular condition $V_{\mathrm{eff}}^{\prime }=0$ gives ${\ell }^2=r_g{r}^2/(r-3r_g)$. Substituting into the second derivative taken at fixed $\ell$ givesTimelike circular orbits exist for $r>3r_g$; they are stable against radial perturbations for $r>6r_g$ and unstable for $3r_g<r<6r_g$. ThereforeThe boundary orbit is marginally stable. Differentiating $e(r)$ also gives a minimum there, consistently with this fixed-angular-momentum stability calculation.

accretion efficiency and black-hole spin. Matter supplied slowly from far away initially brings rest energy $d{m}_0{c}^2$. A radiatively efficient, zero-inner-torque thin disk releases the binding energy down to the innermost stable circular orbit; the remaining energy $e_{\mathrm{ISCO}}d{m}_0{c}^2$ is carried into the hole during the plunge. ThusHere $\eta$ denotes the efficiency called $ϵ$ in this question. The ideal relation assumes negligible additional dissipation during plunge and neglects photon capture and advection. A hot radiatively inefficient flow need not have this efficiency.

For a Kerr black hole, an equatorial prograde disk has a smaller innermost stable circular orbit and greater binding-energy efficiency than the Schwarzschild value; a retrograde disk has the opposite trend. The formal extremal prograde limits are $r_{\mathrm{ISCO}}\to r_g$ and $\eta \to 1-1/\sqrt{3}\simeq 0.423$. These are ideal limiting values, not a universal realized efficiency. photon capture and the accretion geometry affect the achievable black-hole spin and escaping radiation.

radiation force and the limiting luminosity. For isotropic luminosity $L$, the outward radiation momentum flux at radius $r$ is $L/(4\pi r^{2}c)$. In fully ionized hydrogen, Thomson scattering acts mainly on the electron, while electrostatic coupling makes the electron and proton move together. Neglecting the electron mass, one electron–proton pair experiencesEquating them yieldsThis Eddington luminosity uses electron-scattering opacity, spherical or isotropic force balance and negligible relativistic corrections at the force-comparison radius. Composition, other opacity, anisotropy and photon trapping can change the relevant limit.

Exponential growth and doubling. Distinguish rest-mass supply ${\dot{m}}_0$ from black-hole mass growth $\dot{M}$:At the Eddington limit,Using the printed scattering cross section gives $t_E\simeq 4.50\times 10^8\mathrm{yr}$. The Salpeter time $t_S$ is an e-folding time; the Eddington-limited doubling time isfor the Schwarzschild thin-disk efficiency. If one approximates $\dot{M}\simeq {\dot{m}}_0$, the factor $1/(1-\eta )$ is dropped and the estimate is about $1.79\times 10^7\mathrm{yr}$. The retained-energy version is the mass-conserving expression.

The efficiency bound. A growth factor $10^8$ requires $N=\log (10^8)=18.4207$ e-folds. Eddington-limited accretion is the fastest growth allowed by the adopted luminosity bound at fixed efficiency, so growth within a time $\Delta t$ requiresThis efficiency upper bound from Eddington-limited growth is an upper bound: higher efficiency radiates more of each supplied rest mass, and the luminosity ceiling therefore allows less mass to enter per unit time. A duty cycle below unity or sub-Eddington luminosity tightens it. The approximate convention $\dot{M}={\dot{m}}_0$ would instead give $\eta \lesssim 0.121$; both figures refer to the same effect, with different treatment of the radiated mass deficit. Mergers, super-Eddington growth or a different seed history would change the premises.

Coherent prograde accretion spins up the hole, moves the inner edge inward and increases $\eta$. For fixed $L/L_{\mathrm{Edd}}$ the mass-growth e-folding time then grows, so the logarithmic growth curve becomes less steep; one cannot simply retain the initial Schwarzschild efficiency throughout. Episodic misaligned accretion can have a different black-hole spin history. In general,

energy available to affect the galaxy. Let $M_g$ denote galaxy mass, keeping it distinct from black-hole mass. At constant efficiency, negligible seed mass and no merger contribution, the emitted energy in building $M_{\mathrm{BH}}=0.002M_g$ isFor $v=300\mathrm{km}{\mathrm{s}}^{-1}$ and the nonspinning efficiency this is approximately $121$. The common small-efficiency approximation gives $2000\eta \simeq 114$, and a fiducial $\eta =0.1$ would give about $222$ with retained mass included. These differences do not alter the conclusion: the available accretion energy substantially exceeds the stated galactic binding-energy estimate.

Only a coupling fraction of order one percent would then be energetically sufficient to affect the gas strongly. radiation, winds or jets can heat gas, drive outflows and suppress further cooling and star formation, providing active-galactic-nucleus feedback and possible self-regulation of black-hole growth. energy availability alone does not prove that the whole galaxy is expelled: most radiation can escape, jets can be collimated, heated gas can cool, and collisionless stars couple differently from gas.

Use the $+---$ metric signature and abbreviate $f(r)=1-2\mu /r$, with $\mu =GM/c^2$ a length. Dots initially mean differentiation with respect to an affine parameter $\lambda$. The geodesic Lagrangian in the equatorial plane isThe planarity of geodesics in spherical symmetry permits this plane to be chosen. More explicitly, the full angular Euler-Lagrange equation is $\ddot{\theta }+2\dot{r}\dot{\theta }/r-\sin \theta \cos \theta {\dot{\varphi }}^2=0$, so initial data $\theta =\pi /2$, $\dot{\theta }=0$ preserve it.

Both $t$ and $\varphi$ are cyclic coordinates. Their Euler-Lagrange equations give conserved conjugate momenta $p_t=c^{2}f\dot{t}$ and $p_{\varphi }=-r^2\dot{\varphi }$. Defining constants $k,h$ therefore givesThe corresponding second-order geodesic equations areand variation of $r$ givesThe geodesic Lagrangian has no explicit $\lambda$ dependence. Its conserved Lagrangian energy function is ${\dot{x}}^{\alpha }\partial \mathcal{L}/\partial {\dot{x}}^{\alpha }-\mathcal{L}=\mathcal{L}$, since it is quadratic in the tangent components. Thus twice this function is a constant:This is the constant squared tangent norm of an affinely parametrized geodesic.

For a massive particle choose $\lambda =\tau$, its proper time, and obtain $\kappa =c^2$. For a photon, the tangent is null and $\kappa =0$. A general rescaling of a massive particle's affine parameter rescales a positive $\kappa$; the numerical value $c^2$ requires the proper time choice. With that choice, $h=L_z/m$ is signed specific angular momentum, of units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$, and $k=E/(m{c}^2)$ is dimensionless conserved specific Killing energy. The time coordinate is normalized to proper time at infinity, so this $E$ is the particle's conserved energy referred to infinity. A photon's constants scale with the normalization of its null tangent; it has no rest frame or proper time parametrization.

Substitution of $\dot{t}=k/f$ and $\dot{\varphi }=h/r^2$ into the norm gives the general radial first integralFor a massive particle this becomesAdopting the usual half-normalization for the effective potential for timelike Schwarzschild geodesics, defineAll terms have units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$, or energy per unit mass. The first two terms are the Newtonian gravitational and centrifugal contributions; the last is the relativistic correction.

A circular orbit requires both $\dot{r}=0$ and $V_{\mathrm{eff}}^{\prime }(R)=0$. Differentiating at fixed $h$ givesThe radial first integral also givesFor the future-directed circular orbit at $R=4\mu$, these constants are $|h|=4\mu c$, $k=1$, and $f=1/2$. ThereforeSince Schwarzschild time agrees with the clock time of a static observer at infinity, the intended infinity-referred tangential orbital speed isThis uses the circumferential areal radius $R$ with the clock at infinity; it is not a local measurement made at the orbit. A local static observer's clock instead has $d{\tau }_{\mathrm{stat}}=\sqrt{f}dt$ and measures $v_{\mathrm{local}}=v_{\infty }/\sqrt{f}=\boxed{c/\sqrt{2}}$. These two operationally different speeds explain the significance of the phrase “at infinity”.

The effective-potential stability of a circular orbit follows fromConsequently the $R=4\mu$ orbit exists but is unstable to radial perturbations. The original plot below shows its position at a maximum, for $h=4\mu c$.

For a weak-field elliptic orbit, the relativistic correction makes the direction of closest approach advance after each radial cycle. The derivations under i and ii below giveper orbit to first order, where $a_{\mathrm{orb}}$ is the semi-major axis and $e$ the orbital eccentricity. For Mercury, the additional Schwarzschild perihelion precession is approximately 43 arcseconds per century, over and above the much larger Newtonian precession produced primarily by other planets. It is the unexplained residual after those effects have been included that provides the classic quantitative test of general relativity; one should not identify the whole observed precession with this relativistic correction.^[ref]

The labels i and ii in the original paper introduce optional orbit-equation aids, rather than separate mandatory subquestions. They are retained here for derivations of both aids and of the perihelion result.

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.

Assume a circular orbit and an orbital inclination $i\simeq 90^{\circ }$, as appropriate for a simple eclipsing binary estimate. At quadrature the two radial velocities are opposite and equal to the orbital speeds. The more massive component has the smaller amplitude. With the centre of mass as origin,Thus $a_1=0.1103\mathrm{AU}$, $a_2=0.2206\mathrm{AU}$ and Kepler's third law givesThe orbital plane contains the observer's line of sight; at quadrature the two velocities point towards and away from that observer, as labelled in the original schematic below. For an elliptic orbit, the measured amplitudes instead give $a\sin i=P(K_1+K_2)\sqrt{1-e^2}/(2\pi )$; the numerical answer requires the circular assumption.

Assume a circular orbit and an orbital inclination $i\simeq 90^{\circ }$, as appropriate for a simple eclipsing binary estimate. At quadrature the two radial velocities are opposite and equal to the orbital speeds. The more massive component has the smaller amplitude. With the centre of mass as origin,Thus $a_1=0.1103\mathrm{AU}$, $a_2=0.2206\mathrm{AU}$ and Kepler's third law givesThe orbital plane contains the observer's line of sight; at quadrature the two velocities point towards and away from that observer, as labelled in the original schematic below. For an elliptic orbit, the measured amplitudes instead give $a\sin i=P(K_1+K_2)\sqrt{1-e^2}/(2\pi )$; the numerical answer requires the circular assumption.

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 $n+1$.