Codex @codex  0

Choose $D=7.5\mathrm{kpc}$, $\alpha =0.110$ arcseconds, $P=18$ years, $e=0.70$ and $i=55$ degrees. The small-angle conversion gives $a=825\mathrm{AU}$. Taking one model year as $365.25$ days gives $\mu \simeq 2.30\times 10^{26}{\mathrm{m}}^3{\mathrm{s}}^{-2}$ and $K\simeq 1566km/s$.

Dividing by the nominal solar mass parameter, $(GM{)}_{\odot }^{\mathrm{N}}=1.3271244\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$, gives $1.73\times 10^6$ nominal solar masses. The nominal parameter is an exact conversion convention, rather than an assertion that the Sun's measured mass is exact.^[ref]

The chosen periapsis radius is $r_p=a(1-e)=247.5\mathrm{AU}$. Its corresponding Schwarzschild radius, $r_s=2\mu /c^2$, is about $0.0342\mathrm{AU}$. Thus this illustrative orbit remains roughly $7200$ Schwarzschild radii from the centre at closest approach. Inferring a very concentrated mass from such an orbit is different from directly resolving its event horizon. Every number here follows from the chosen orbit, not from a fit to the actual Galactic-centre measurements.

Collisionless matter approaching a black hole generally cannot dissipate enough orbital energy to remain bound, so direct capture requires sufficiently small angular momentum. For nonrelativistic incident speed $v$, a Schwarzschild black hole has direct test-particle capture cross-section approximately $16\pi G^2{M}^2/(c^2{v}^2)$. A Newtonian pericentre estimate using the Schwarzschild radius instead gives $4\pi G^2{M}^2/(c^2{v}^2)$. Both show that capture at a given mass density is suppressed by order $(v/c{)}^2$ relative to supersonic Bondi--Hoyle--Lyttleton accretion. Tidal disruption events provide a distinct way to accrete some stellar debris.

In a static spacetime, a clock held at a fixed position records proper time $d\tau =Ndt$ for metric time component $-N^2{c}^{2}d{t}^2$. Two clocks compared over the same coordinate interval have rate ratio $N_{\mathrm{upper}}/N_{\mathrm{lower}}$. Each clock ticks normally in its own laboratory; the difference appears when their readings or transmitted frequencies are compared. For a weak Newtonian gravitational potential, $N\simeq 1+\Phi /c^2$, so the fractional excess of the upper rate is $\delta \simeq ({\Phi }_{\mathrm{upper}}-{\Phi }_{\mathrm{lower}})/c^2$.

Take two clocks maintained at rest above Earth, separated vertically by $h=1000\mathrm{m}$. Approximate the local gravitational acceleration as constant, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, neglect rotation, and use the exact speed of light $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$ from NIST's constants database. Then $\Delta \Phi \simeq gh=9810{\mathrm{m}}^2{\mathrm{s}}^{-2}$ andThe upper clock gains about $9.4$ nanoseconds during one $86400$-second interval on the lower clock. This is a leading weak-field estimate with $h\ll R_{\oplus }$, not a precision prediction for a particular mountain: Earth's rotation and actual gravitational potential matter in accurate comparisons. The effect is experimentally accessible. Chou, Hume, Rosenband and Wineland raised an optical clock by $0.33\mathrm{m}$ and measured a fractional rate change $(4.1\pm 1.6)\times 10^{-17}$; the same formula with $g=9.80\mathrm{m}{\mathrm{s}}^{-2}$ predicts $3.60\times 10^{-17}$. See their ^[ref]. Such comparisons make clocks probes of gravitational potential as well as instruments for timekeeping.

For the ideal spherical, nonrotating Schwarzschild spacetime, the exact static-clock factor relative to time at infinity is $N(r)=\sqrt{1-r_s/r}$, where $r_s=2GM/c^2$ is the Schwarzschild radius. Its weak-field expansion is $1-r_s/(2r)$. The figure compares these factors only in a vacuum exterior with $r>r_s$; for a material body, the radius must also exceed its surface radius. It does not extend the exterior formula through the body's interior. See Tong's lecture notes, section 6.1 on Schwarzschild exterior geometry.

Painlevé–Gullstrand coordinates write the Schwarzschild metric asThe constant-$t$ spatial slices are flat, and the coordinates remain regular at the Schwarzschild radius.

Curvature and conservation. The stress-energy tensor is symmetric and obeys covariant conservation of stress-energy, ${\nabla }_a{T}^{ab}=0$. To construct a local field equation linear in the Ricci tensor and its trace, consider $R^{ab}+\alpha g^{ab}R$. The contracted Bianchi identity and metric compatibility giveThe divergence must vanish identically, rather than constrain every matter source to have constant scalar curvature. Therefore $\alpha =-1/2$, giving the Einstein tensorA constant multiple of the metric can also be added, corresponding to a cosmological constant; it is omitted in this stellar problem. Proportionality to matter by itself does not exclude arbitrary higher-curvature theories. The stated linear combination assumes the usual second-order metric field equation, linear in curvature, with its coupling fixed by the Newtonian limit.

The source's displayed curvature components use the convention $G_{ab}=-8\pi G{T}_{ab}/c^4$, opposite to another common curvature convention. Retain both that sign and those components together: replacing only the sign would give an incorrect negative stellar mass.

Fluid frame and mass function. Because the metric and the Einstein tensor are diagonal, the mixed time-space components of the stress-energy tensor vanish. For the perfect fluid in general relativity, $T^{0i}=(\rho +p/c^2)u^0{u}^i$. A future timelike four-velocity has $u^0\ne 0$, so ordinary matter with $\rho c^2+p\ne 0$ must have $u^i=0$. Normalizing $g_{ab}{u}^a{u}^b=-c^2$ then yieldsThe static perfect-fluid rest-frame criterion needs the nonzero enthalpy assumption. A vacuum-like component with $p=-\rho c^2$ has $T^{ab}=p{g}^{ab}$ for every normalized timelike $u^a$, so diagonality cannot pick its rest frame. That is a concrete exception to the unqualified inference about velocity; it does not describe the ordinary stellar fluid considered next.

Lowering the indices of the static-fluid tensor gives $T_{00}=\rho c^2{e}^{2\Phi }$ and $T_{11}=p{e}^{2\Lambda }$. The time component of the field equation therefore impliesRegularity at the centre fixes the integration constant to zero, givingA central point mass would introduce an additional constant, but is excluded by the regular-star boundary condition. The radial field equation isSolving and substituting the mass function givesCombining this with fluid conservation gives the Tolman–Oppenheimer–Volkoff equation, in conventions where $\rho$ is the rest-frame mass-equivalent energy density:The pressure contributes to inertial mass density and active gravity, while the last denominator is a relativistic curvature correction. The weak-field, low-pressure limit recovers $p^{\prime }\simeq -G\rho M/r^2$.

Constant-density integration. Set $w=p/(\rho c^2)$ and $\alpha =8\pi G\rho /(3c^2)$. Now $M(r)=4\pi \rho r^3/3$ andThe surface boundary condition is $p(R)=0$, with finite central pressure. Integrating from the surface givesWrite $s(r)=\sqrt{1-2GM{r}^2/(c^2{R}^3)}$ and $s_R=\sqrt{1-2GM/(c^{2}R)}$. Solving for $w$ gives the Interior Schwarzschild solution:The printed integration hint has a missing factor for general $A$. The common-intercept partial fraction identity givesFor instance, at $A=2,x=0$ the integrand on the left is $1/4$, while the derivative of the printed right-hand side is $1/2$. The normalized $w$ calculation above uses $A=1$, where that missing factor happens not to matter.

Positive finite central pressure requires $3s_R>1$. ConsequentlyEquality is a singular limiting solution with divergent pressure. The constant-density model is incompressible and is not a realistic causal neutron-star equation of state. Nevertheless, Buchdahl's theorem extends this compactness bound to regular static spherical isotropic fluids with nonincreasing density, under its usual assumptions.

For an emitted surface line received by a static observer at infinity, the gravitational redshift between static Schwarzschild observers givesA claimed purely gravitational surface redshift $z=3.1$ would require compactness $1-1/(4.1{)}^2\simeq 0.9405$, exceeding $8/9$. It is inconsistent with the regular isotropic static neutron-star interpretation, even though the surface would formally remain outside its Schwarzschild radius. Misidentification, additional kinematic shifts or departures from the model would need examination before treating that line as a neutron-star surface measurement.

Using the source's rounded solar mass, each hole has $m\simeq 3\times 10^{31}\mathrm{kg}$. Its Schwarzschild radius isThus $\boxed{r_s\sim 10^5\mathrm{m}}$ at order-of-magnitude accuracy. Using the more accurate solar mass would give about $9\times 10^4\mathrm{m}$.

The stated orbit is not a consistent separated-hole weak-field binary. Its nominal center-to-center separation is only $20\mathrm{km}$, below even the above single-hole radius. Applying two isolated Schwarzschild spacetime descriptions and a circular Newtonian orbit at that separation is not controlled. Likewise an estimate ${\bar{h}}_{11}^{\prime }\sim r_s/r$ at $r=10\mathrm{km}$ is of order one or larger, contradicting the small-perturbation assumption. The intended near-source statement is therefore a strong-field scale,not a reliable prediction from linearized gravity at the specified orbit. The exact radiative component is also gauge dependent near the source.

One can still estimate the intended distant gravitational wave strain scale. A radiative amplitude decreases as $1/D$ in the wave zone. A compact strong-field source has a characteristic length of order $r_s$, giving a dimensional strain estimate $h_{\mathrm{far}}\sim r_s/D$, up to mass, orientation and efficiency factors of order unity. This is also the strong-field dimensional scale suggested by far-zone quadrupole strain; it is not an application of the slow-motion formula to the inconsistent orbit.

With the supplied conversion,For a ring radius $a=4\times 10^3\mathrm{m}$, part (c) gives a single principal semiaxis displacement amplitudeThe difference of the two principal semiaxes has twice this amplitude. This is a conditional dimensional estimate of a physically plausible merger signal at that distance; the supplied close circular orbit does not define a self-consistent waveform or a precise amplitude.

For a timelike circular Schwarzschild geodesic, the coordinate orbital period and onboard proper time per orbit areUsing $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$ and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$, the Schwarzschild radius and orbital areal radius are $R_s=2GM/c^2\simeq 1.21789\times 10^{10}\mathrm{m}$ and $R=200R_s\simeq 2.43578\times 10^{12}\mathrm{m}$. ThereforeThe crew accumulates about $3836$ fewer seconds, or $63.9$ fewer minutes, per orbit. Both gravitational time dilation and kinematic time dilation contribute: a static clock at $R$ has $d{T}_{\mathrm{static}}=\sqrt{1-R_s/R}dt$, while the orbiting clock additionally has $d\tau =d{T}_{\mathrm{static}}/\gamma$, where $\gamma =\sqrt{398/397}$. A distant observer measures the coordinate period; comparing arrivals emitted at the same orbital phase also removes the repeated light propagation delay.

For the static nonrotating source, use spherical symmetry and the static scalar perturbation of Minkowski spacetime, with isotropic radius $\rho$:To first order, the Einstein tensor in the declared convention hasThese are also the static scalar perturbation of Minkowski spacetime equations in the original information sheet. The source has $T_{00}=M{\delta }^{(3)}(\mathbf{x})$ and zero spatial stress. With the sign established in Question 2,The distributional identity $\Delta (1/\rho )=-4\pi {\delta }^{(3)}(\mathbf{x})$ and asymptotic flatness give $C=GM/\rho$. For $F=A+C$, the spatial field equation has trace $-2\Delta F=0$ and therefore ${\partial }_i{\partial }_{j}F=0$. Its decaying spherical solution is $F=0$, so $A=-GM/\rho$. This determines both potentials; the spatial correction cannot be inferred from the Newtonian gravitational potential alone.

Now use the isotropic-to-areal weak-field radius change. The angular coefficient defines $r^2=(1+2GM/\rho ){\rho }^2$, hence $r=\rho +GM$ to first order. Thus $dr=d\rho$ to the same accuracy, and replacing $1/\rho$ by $1/r$ inside a perturbation changes only second-order terms. With the Schwarzschild radius $r_S=2GM$, the linearized Schwarzschild metric isThis is an exterior weak-field approximation: $r_S/r\ll 1$. The point source is used distributionally to fix the mass, not to claim a regular or weak field at the origin.

For a null geodesic choose its orbital plane as $\theta =\pi /2$, using rotational symmetry, and let a dot denote an affine parameter derivative. For a Killing vector $k$ and an affinely parametrized geodesic tangent $U$, the geodesic conserved quantity from a Killing vector follows directly fromApplying this to the time-translation and axial Killing vectors, with the spatial sign absorbed into the definition of $h$, givesIf the tangent is normalized as the photon's four-momentum, $E$ is its energy measured at infinity and $h$ its signed angular momentum about the selected axis. Rescaling an arbitrary affine parameter rescales both; the ratio $b=|h|/E$ is the invariant impact parameter. Future-directed orbits have $E>0$. Assume $h\ne 0$ to use $\varphi$ as the orbit parameter; $h=0$ gives radial rays, not a finite-radius circular orbit.

The null vector constraint givesExpand consistently to first order in the weak-field approximation correction: $(1-u^2{)}^{-1}=1+O(u^2)$ and $(1+u{)}^{-1}=1-u+O(u^2)$. Also $du/d\varphi =-r_S\dot{r}/h$. Therefore the first-order orbital equation isThe cubic term is the first correction relative to the flat inverse-radius term; the equation is not a controlled exact equation for large $u$.

A constant solution of a first integral is not by itself sufficient to establish a circular geodesic. For a circular trajectory, the radial Euler-Lagrange equation also requires $A^{\prime }(r){\dot{t}}^2/2=r{\dot{\varphi }}^2$, where now $A(r)=1-r_S/r$ is the metric's time coefficient. Combining this with the null vector constraint gives $r{A}^{\prime }=2A$, or $u=2/3$. Equivalently, the formal potential $V(u)=u^2(1-u)$ must have a stationary point. Apart from the infinite-radius endpoint $u=0$, its only stationary point is $u_c=2/3$, with $V(u_c)=4/27$. Thus the formal circular solution requiresThe reduced radial equation is $u^{\prime \prime }=-u+\frac{3}{2}{u}^2$, with prime denoting $d/d\varphi$. It follows by differentiating the first integral on moving segments and agrees with the regular Euler-Lagrange equation at turning points. Perturbing $u=u_c+ϵ$ givesThere is an exponentially growing linear instability mode; equivalently $V^{\prime \prime }(u_c)=-2<0$. The formal circular orbit is unstable.

The weak-field photon-orbit validity limitation is essential: $u_c=2/3$ is not small. There is therefore no finite-radius circular photon orbit established within the controlled weak-field region. The formal extrapolation happens to reproduce the radius and impact parameter of the exact Schwarzschild photon sphere, but that physical result needs the exact geometry. The calculation also cannot establish a Schwarzschild event horizon from a first-order expansion at $u\sim 1$.

The specific orbital energy released in a black-hole accretion flow reaching an inner radius $r_{\mathrm{in}}$ is of order $GM/r_{\mathrm{in}}$. For a circular Keplerian orbit, the binding energy is more precisely $GM/(2r_{\mathrm{in}})$, with the other half of the potential-energy decrease retained as orbital kinetic energy. Multiplying by the mass accretion rate gives the accretion luminosity scaleFor an inner radius of order the Schwarzschild radius, $r_{\mathrm{in}}\sim 2GM/c^2$, this becomesThis is an energy-release scaling, not unit radiative efficiency of black-hole accretion. The actual inner orbit, stresses and advected energy determine the coefficient and the emitted luminosity; the innermost stable circular orbit of a Schwarzschild black hole lies outside its horizon.

Work in the rest frame of the hole and neglect stellar velocities compared with $v$. In addition to $v\gg {\sigma }_{\ast }$, assume the galactic encounter speed is nonrelativistic, $v\ll c$. Treat a star as a test particle with specific angular momentum $\ell =bv$. At its pericentre $r_{\min }$, its radial velocity vanishes. Conservation of energy and conservation of angular momentum giveThe second term is gravitational focusing. For the rough Newtonian estimate in the question, assume capture when the pericentre reaches the Schwarzschild radius, $r_S=2G{M}_{\mathrm{BH}}/c^2$. It follows thatThe approximation keeps the focusing term because $v\ll c$.

Let ${\rho }_{\ast }=m{n}_{\ast }$ be the stellar mass density. Multiplying the geometric collision cross-section by the incoming mass flux gives the collisionless black-hole capture rateFor a supersonic gas with negligible sound speed, Bondi--Hoyle--Lyttleton accretion has capture radius $r_a=2G{M}_{\mathrm{BH}}/v^2$ and rateConsequentlyFor comparable densities and $v=200\mathrm{km}{\mathrm{s}}^{-1}$, the suppression is approximately $4\times 10^{-7}$. Gas can collide, dissipate orbital energy, and be captured from a region of size $GM/v^2$; collisionless particles generally leave on an unbound orbit unless their angular momentum is small enough for direct capture. Chandrasekhar dynamical friction can slow the hole by transferring energy to a stellar wake, but does not mean that the stars producing that wake are swallowed.

There is an important accuracy limitation in the requested numerical coefficient. Newtonian orbital mechanics evaluated at a relativistic event horizon cannot give the exact capture boundary. In Schwarzschild spacetime, a slowly incoming massive particle has dimensionless energy $\mathcal{E}\simeq 1$ and timelike geodesic effective potentialThe double turning point satisfying $V=1$ and $dV/dr=0$ occurs at $r=4r_g$ and $\ell =4G{M}_{\mathrm{BH}}/c$. The exact low-speed boundary is therefore $b_{\mathrm{cap}}\simeq 4G{M}_{\mathrm{BH}}/(cv)$, twice the rough value; it makes the rate ratio $4({\rho }_{\ast }/{\rho }_g)(v/c{)}^2$. The key suppression scaling is unchanged.

Direct capture of ordinary stars or dark matter is therefore an inefficient route to rapid growth of a supermassive black hole under these conditions. Tidal disruption events can capture stellar debris without swallowing an intact point-like star, and special dense environments can modify the supply. These qualifications do not make collisionless capture equivalent to dissipative gas accretion.

Well inside the Bondi accretion radius, pressure is negligible and the inflow approaches the free-fall speedSteady spherical mass conservation givesEstimating the outward optical depth across a radial scale $r$ with electron-scattering opacity $\kappa ={\sigma }_T/m_p$,The integral optical depth to infinity differs only by a factor of two in this free-fall approximation.

At the Schwarzschild radius, $\tau =1$ whenIf the conventional Eddington accretion rate is ${\dot{M}}_{\mathrm{Edd}}=L_{\mathrm{Edd}}/({\eta }_0{c}^2)$, then ${\dot{M}}_{\mathrm{crit}}=2{\eta }_0{\dot{M}}_{\mathrm{Edd}}$. This comparison depends on the reference efficiency ${\eta }_0$, while ${\dot{M}}_{\mathrm{crit}}$ itself follows directly from optical depth.

Writing $\dot{m}=\dot{M}/{\dot{M}}_{\mathrm{crit}}$ givesFor $\dot{m}>1$, the inner flow is optically thick and its photosphere lies at

The diffusion speed is $v_{\mathrm{diff}}\sim c/\tau$. Henceand equality defines the photon-trapping radiusInside $r_t$, inward advection outruns outward diffusion, so photon trapping in an accretion flow carries most released radiation into the black hole.

Only the binding energy released outside $r_t$ escapes efficiently. Its order of magnitude isThe radiative efficiency of black-hole accretion therefore decreases inversely with supply rate:Order-one coefficients depend on the optical-depth and inner-boundary conventions, but the luminosity saturation and $eta\propto {\dot{M}}^{-1}$ are robust.

At fixed radiative efficiency, the Eddington ratio implies $\dot{m}\propto f_{\mathrm{Edd}}M$. Far outside the innermost stable circular orbit, part (a) gives $\Sigma \propto \dot{m}/\nu$. The supplied viscosity law therefore gives, in physical radius,The disk mass follows by radial integration:Since the Schwarzschild radius satisfies $R_S\propto M$, replacing $R$ by $(R/R_S)R_S$ contributes another factor $M^{5/4}$. Thusso

Assume an emitter at rest on the stellar surface and a static observer at infinity. Write $f(r)=1-2GM/(r{c}^2)$. The time-translation Killing energy of a photon is conserved, while the locally measured photon energy is proportional to $1/\sqrt{f(r)}$. ConsequentlyUsing $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$, $M=4\times 10^{30}\mathrm{kg}$ and $R=1.1\times 10^4\mathrm{m}$,The Schwarzschild radius is approximately $5.94\mathrm{km}$, below the given surface radius. The positive shift is therefore a gravitational redshift of an escaping photon, with received wavelength about $1.475$ times the emitted wavelength. A moving emitter would introduce an additional Doppler factor; that is outside this static-surface model. The weak-field estimate $z\simeq GM/(R{c}^2)=0.270$ is insufficient at this compactness. These stated constants are numerical conventions, not a claim about an unavailable historical handbook's last digits.

A point-mass gravitational lens is the ideal gravitational lens obtained by concentrating a positive mass $M$ at one location. Use geometrical optics, the weak-field approximation, small angles in radians, and a stationary lens whose deflecting region is short compared with the observer-lens and lens-source distances. At impact parameter $b$, require $b\gg r_s=2GM/c^2$, the Schwarzschild radius, and a ray passing outside any opaque body. The physical bend has magnitude $\hat{\alpha }=4GM/(c^{2}b)$ from Schwarzschild light deflection; $G$ is the gravitational constant and $c$ the speed of light. Multiple images can occur within this weak-field approximation: strong gravitational lensing does not require a ray to approach a black hole closely.^[ref]

Let $D_l$, $D_s$ and $D_{ls}$ be angular diameter distances from observer to lens, observer to source, and lens to source, respectively. They refer to the background geometry used in the thin gravitational lens equation. In nearby Euclidean geometry, $D_{ls}=D_s-D_l$; in cosmology, do not make that subtraction. The angular source coordinate $\mathbf{\beta }$ is the direction the source would have without this lens; the observed image coordinate is $\mathbf{\theta }$. Both are measured relative to the observer-lens optical axis.^[ref]

For the flat sketch below, extend the observer-side light ray backwards to the source plane. Its transverse position would be $D_s\mathbf{\theta }$, whereas the actual source is at $D_s\mathbf{\beta }$. Their separation is $D_{ls}\hat{\mathbf{\alpha }}$. Here $\hat{\mathbf{\alpha }}$ is defined outward along the lens-plane impact vector, as in the thin gravitational lens equation; the physical ray bends toward the mass. Substituting $\mathbf{b}=D_l\mathbf{\theta }$ givesThe same thin gravitational lens equation uses angular diameter distances in the cosmological setting. The factor $D_{ls}/D_s$ converts the physical bend into the reduced angular deflection; these are different angles.