Codex @codex  0

Consider a narrow bundle of rays between two surface elements $d{A}_1,d{A}_2$ separated by $s$ in a transparent, stationary Euclidean vacuum. Let the ray make angles ${\theta }_1,{\theta }_2$ to the respective surface normals. The receiving element subtends solid angle $d{\Omega }_1=\cos {\theta }_{2}d{A}_2/s^2$ at the emitter. By the definition of specific intensity, the beam power in frequency interval $d\nu$ isAt the receiving end the source subtends $d{\Omega }_2=\cos {\theta }_{1}d{A}_1/s^2$, so the identical power isIn the absence of absorption, emission or frequency shifts, conservation of energy equates these expressions:This vacuum conservation of specific intensity also follows directly from the radiative transfer equation $d{I}_{\nu }/ds=0$. The apparent solid angle of an unresolved source shrinks as $s^{-2}$, so its integrated observed radiative flux still follows the inverse-square law. Specific intensity and unresolved flux are different quantities.

The vacuum and frequency assumptions are essential: absorption and emission change specific intensity, and gravitational or cosmological redshift conserves $I_{\nu }/{\nu }^3$ along a ray instead of $I_{\nu }$ itself. A fixed telescope aperture observing an unresolved object also cannot infer constant total received power from intensity conservation.

Use the separation vector $\mathbf{r}$ from the star to the comet, and let $m_{\mathrm{red}}=m{M}_{\star }/(m+M_{\star })$. The two-body problem separates into uniform centre of mass motion andTaking a cross product with $\mathbf{r}$ proves conservation of angular momentum; taking the scalar product with $\dot{\mathbf{r}}$ proves conservation of energy. The relative constants areThe total relative angular momentum and energy are $m_{\mathrm{red}}\mathbf{h}$ and $m_{\mathrm{red}}\epsilon$, respectively. The conserved eccentricity vector is $\mathbf{e}=\dot{\mathbf{r}}\times \mathbf{h}/\mu -\hat{\mathbf{r}}$. Its scalar product with $\hat{\mathbf{r}}$ gives $r=h^2/[\mu (1+e\cos f)]$, with $f$ the true anomaly. The periapsis and apoapsis distances are therefore $h^2/[\mu (1+e)]$ and $h^2/[\mu (1-e)]$. Their sum is twice the semi-major axis, soAt either apsis $v=h/r$; substituting an apsidal radius gives $\epsilon =-\mu /(2a)$. Thus the specific orbital energy implies the vis-viva equationThese are exact relative-coordinate formulas for an elliptic orbit, $0\le e<1$. In the barycentric frame the comet's speed is $M_{\star }/(M_{\star }+m)$ times the relative speed; the printed identification with the comet's stellar orbital speed uses $M_{\star }\gg m$.

The incoming and outgoing velocity vectors have the same magnitude $v_{\infty }$, by conservation of energy. The chord between them has lengthFor fixed $r_{\min }$ put $z=v_{\infty }\sqrt{r_{\min }/\mu }$. Then $|\Delta \mathbf{v}|=\sqrt{\mu /r_{\min }}2z/(1+z^2)$, and $(z-1{)}^2\ge 0$ proves $2z/(1+z^2)\le 1$. Equality occurs at $v_{\infty }=\sqrt{\mu /r_{\min }}$.

A non-colliding comet must have $r_{\min }\ge r_0$, where $r_0=R_p$ in the point-comet approximation and $r_0=R_p+R$ for finite bodies. Thus the non-colliding gravitational velocity-kick bound isThe optimizing grazing orbit has $v_{\infty }=v_{\mathrm{esc}}/\sqrt{2}$, $e_h=2$ and $\theta =\pi /3$. If contact itself is excluded, this is a supremum approached from larger $r_{\min }$. The finite collision radius is essential: an ideal point planet with unrestricted $r_{\min }\to 0$ has no finite upper bound.

A constant Galilean boost by the planet's stellar orbital velocity leaves the vector difference unchanged. It changes the comet's incoming and outgoing heliocentric energies, which is the basis of a gravitational assist.

Let $K={\rho }_0|\mathbf{u}{|}^2/2$ be the kinetic energy per unit volume. Dot the momentum equation with ${\rho }_0\mathbf{u}$. The Coriolis acceleration does no work because $\mathbf{u}\cdot ({\mathbf{e}}_z\times \mathbf{u})=0$. Incompressibility turns the advection and pressure terms into divergences, yieldingThus $\mathbf{F}=(K+P)\mathbf{u}$ is the energy flux and $S={\rho }_0{\Omega }^2(3x-2qz)u_x$ is the source in the requested kinetic-energy form.

To decide about conservation of energy, absorb the ordinary radial shearing-sheet tidal potential ${\Phi }_t=-3{\Omega }^2{x}^2/2$ into the density. Its material derivative cancels the $3x$ work term:For $q=0$ this is the usual conservative Jacobi energy in a shearing sheet, subject to vanishing boundary energy flux. For $q\ne 0$, the extra body force has nonzero curl, $\nabla \times [-2{\Omega }^{2}qz{\mathbf{e}}_x]=-2{\Omega }^{2}q{\mathbf{e}}_y$, so no scalar potential can absorb it. A potential $2{\Omega }^{2}qxz$ would necessarily introduce a vertical force $-2{\Omega }^{2}qx{\mathbf{e}}_z$, absent from the model. Therefore this vertical-shear model does not generically conserve the standard total mechanical energy: its maintained shear/irradiation background can supply or remove energy. This is the energy balance of the incompressible vertical-shear model.

Set $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$, $M=M_1+M_2$ and $\mu =M_1{M}_2/M$, the reduced mass. Newton's law of universal gravitation gives ${\ddot{\mathbf{x}}}_1=-G{M}_2\mathbf{r}/r^3$ and ${\ddot{\mathbf{x}}}_2=G{M}_1\mathbf{r}/r^3$. Subtraction establishesFor fixed masses, the orbital energy and angular momentum areTaking derivatives gives $\dot{E}=\mu \mathbf{v}\cdot (\ddot{\mathbf{r}}+GM\mathbf{r}/r^3)=0$ and $\dot{\mathbf{J}}=\mu \mathbf{r}\times \ddot{\mathbf{r}}=0$. These are conservation of energy and conservation of angular momentum in the unperturbed Kepler orbit. To see directly that the orbital eccentricity is fixed, put $\mathbf{h}=\mathbf{r}\times \mathbf{v}$ and $\mathbf{n}=\mathbf{r}/r$. The eccentricity vector isSince $\dot{\mathbf{h}}=0$, the cross product identity $\mathbf{r}\times (\mathbf{r}\times \mathbf{v})=\mathbf{r}(\mathbf{r}\cdot \mathbf{v})-r^2\mathbf{v}$ gives $\ddot{\mathbf{r}}\times \mathbf{h}=GM\dot{\mathbf{n}}$, so $\dot{\mathbf{e}}=0$. Its magnitude $e$ is the orbital eccentricity. Equivalently,Thus the semi-major axis and orbital eccentricity of a bound noncollision Kepler orbit are constant, including $e=0$.

Take the centre of mass as origin. The two positions are ${\mathbf{x}}_1=(M_2/M)\mathbf{r}$ and ${\mathbf{x}}_2=-(M_1/M)\mathbf{r}$. For $x\gg r$, a Taylor expansion givesThe mass dipole vanishes because ${\sum }_A{M}_A{\mathbf{x}}_A=0$, and the second mass moment tensor is ${\sum }_A{M}_A{x}_{Ai}{x}_{Aj}=\mu r_i{r}_j$. The gravitational quadrupole potential of a point-mass binary is consequentlyBoth printed tensors are trace-free. Contracting them with the Kronecker delta summed over three spatial dimensions givesThis proves the requested expression through second order. In particular the printed convention is $q_{ij}=\frac{3}{2}{Q}_{ij}$, where $Q_{ij}=\mu (r_i{r}_j-r^2{\delta }_{ij}/3)$ is the standard mass quadrupole moment; the radiation coefficient must be adjusted accordingly. No mass dipole term may be retained in centre of mass coordinates.

To evaluate the quadrupole formula without assuming a circular orbit, put $k=GM$, $u=\dot{r}=\mathbf{n}\cdot \mathbf{v}$ and ${\mathbf{v}}_{\perp }=\mathbf{v}-u\mathbf{n}$. Differentiating the gravitational acceleration gives the kinematic jerkFor $S_{ij}=r_i{r}_j$, the product rule then givesInserting these into the printed mass quadrupole moment convention yieldsThe two tensors in brackets are orthogonal in their tensor contraction, because $\mathbf{n}\cdot {\mathbf{v}}_{\perp }=0$. Their squared norms are $6$ and $2v_{\perp }^2$, respectively. HenceThe instantaneous quadrupole luminosity of a Kepler binary is thereforeThe radial derivative $\dot{r}$ is not the vector speed. The bracket equals $v_{\perp }^2+u^2/12$, so the radiated luminosity is nonnegative for every instantaneous velocity.

For a slowly evolving circular orbit, there is no preferred orbital phase at which to excite a persistent eccentricity vector. More precisely, the rotating mass quadrupole moment emits at twice the orbital frequency, with angular harmonic number two; its gravitational-wave energy and angular-momentum balance is $\dot{E}=\Omega \dot{J}$. The circular Kepler orbit sequence has $J=\mu \sqrt{GMa}$ and $dE/dJ=\Omega$, so this loss is tangent to the sequence. At leading adiabatic order it preserves zero secular orbital eccentricity. This is a circular gravitational-wave inspiral, with a small radial drift rather than an exactly fixed-radius Newtonian circle. The energy flux alone would not establish circularity without this symmetry and angular momentum balance.

To leading radiation order use $r=a$, $u=0$ and ${\mathbf{v}}^2=GM/a$ in the loss formula. Combining it with $E=-GM\mu /(2a)$ givesKeeping the fixed masses and leading quadrupole formula, integration givesThis formal coalescence time displays the strong $a_0^4$ dependence: gravitational-wave emission from a binary system matters far more for close binary stars than for wide ones. The orbital period decreases as $a^{3/2}$ and the gravitational-wave frequency increases, producing a chirp. A detached binary can be driven into Roche-lobe overflow; further evolution then depends on mass-transfer stability. A sufficiently close compact pair may merge. The weak-field, slow-motion, point-mass approximation ceases to apply before literal $a=0$; finite stellar radii or strong relativistic effects determine the final interaction.

Assume the holonomic constraints have linearly independent gradients, so their common level set is locally a smooth constraint manifold parametrized by $q_1,\dots ,q_n$. The multiplier form of the Euler-Lagrange equations isAt fixed time ${\sum }_A{\partial }_A{f}_{\alpha }{\partial }_i{x}_A=0$. The chain rule for the reduced Lagrangian givesConversely, if these reduced equations hold, the Cartesian Euler-Lagrange residual annihilates every tangent direction. Linear independence of the constraint gradients implies that it is a linear combination of those gradients, giving the required Lagrange multipliers. This proves the local equivalence, including time-dependent constraints; singular constraints would need separate analysis.

For the bead, $\dot{y}=f^{\prime }(x)\dot{x}$, soThe autonomous conservation of energy gives $E=\frac{1}{2}(1+f^{\prime 2}){\dot{x}}^2$. If $E>0$, the motion has a fixed sign $s=\mathrm{sgn}\dot{x}$ andThe integral is the signed arc length divided by the constant speed, and is locally invertible. The special solution $E=0$ is a stationary bead, for which writing time as a function of its unchanging position is not meaningful.

For the initially homogeneous sphere, $V_i=-3G{M}^2/(5R_i)$ and $K_i=0$. Conservation of energy and the virial theorem imply $E_f=V_f/2=V_i$, hence $V_f=2V_i$. With the usual homologous-sphere approximation, $V\propto -1/R$, soThese are the conventional collapse estimates. A final virial equilibrium is not, by itself, guaranteed to remain homogeneous: if $V_f=-{\alpha }_{f}G{M}^2/R_f$, the actual relation is $R_f/R_i={\alpha }_f/(2{\alpha }_i)$ with ${\alpha }_i=3/5$. Thus the factor of two applies exactly to the gravitational radius, and to the outer radius only under the stated structural approximation.

Here “non-interacting” means collisionless apart from gravity. Baryonic gas can collide, convert ordered motion to thermal energy, and radiate it. If radiative cooling is effective, the fixed-energy virial state loses pressure support and contracts further. Baryonic cooling can destabilize the collisionless virial state; baryons are not inevitably unstable if cooling is slow or heating and pressure support maintain equilibrium.

The total work by the gas is the enclosed triangular area, $W=\frac{1}{2}{p}_0{V}_0$. The positive heat on the sloping leg isAdd $Q_{AB}=48p_0{V}_0/32$, excluding the emitting part of $BC$ and all of $CA$. HenceAs an independent conservation of energy check, the heat emitted on late $BC$ is $p_0{V}_0/32$ and on $CA$ is $80p_0{V}_0/32$; thus $Q_{\mathrm{in}}-Q_{\mathrm{out}}=16p_0{V}_0/32=W$. Using only the net heat on $BC$ would give an incorrect thermal efficiency.

Use $G=c=1$ and the physical real part of the printed plane wave. Its null wavevector obeys $k^{\alpha }{k}_{\alpha }=0$, so each component satisfies $\square h_{\alpha \beta }=0$. Direct differentiation of the defined stress tensor givesThe trace is proportional to $k^{\nu }{k}_{\nu }$ for a single plane wave and hence vanishes. ThereforeThis is local conservation of energy and momentum conservation; spatial integration gives conserved total charges when the boundary flux vanishes. For a finite-amplitude complex phasor, one takes the real field before quadratic products and may time-average: $⟨{\tau }_{\mu \nu }⟩=k_{\mu }{k}_{\nu }{H}^{\alpha \beta }{\overline{H}}_{\alpha \beta }/(64\pi )$ in transverse-traceless gauge. Its energy density is nonnegative and its flux travels in the positive $z$ direction.

For rolling without slipping, $\dot{s}=a\Omega$. Conservation of energy givesand differentiation yields

For a sphere with density proportional to $r^c$,so $I/(M{a}^2)=2(c+3)/(3(c+5))$. A uniform disc has $I/(M{a}^2)=1/2$. Equal launch speeds give equal return times exactly when the accelerations, and hence these ratios, agree. Solving gives $c=3$.

The conservation of energy to use here is the total kinetic energy plus internal energy inside the shock, not just the internal energy of a uniform-pressure core. With no radiative loss and negligible initial ambient energy,Substitute the adiabatic superbubble similarity solution and define the finite positive dimensionless integralThen $\dot{E}t=4\pi {\rho }_1{\dot{R}}^2{R}^{3}I$. Since $\dot{R}=3R/(5t)$ and $R^5={\alpha }^5(\dot{E}/{\rho }_1)t^3$,This is the energy normalization of a continuously driven spherical shock. It fixes $\alpha$ once the similarity profiles are determined. If the profiles describe only the shocked ambient layer, the energy of the interior hot bubble must also be included before using this normalization. A radiatively cooled shell has a different energy budget; its familiar numerical coefficient should not be substituted into this loss-free model.

The rod imposes a mechanical constraint, and its force is perpendicular to the wrecking ball's instantaneous velocity, so it does no work. In the constant parallel Newtonian gravitational field, the ball falls through the vertical distance $l-l\cos {\theta }_0$. Conservation of energy between $B$ and $W$ therefore givesand hence

Let $C$ be the centre of the Earth and let $M$ be its mass. Since $CW=R$ and $CS=R+l$, the law of cosines in triangle $CSB$ gives the ball's initial distance from $C$ asThe surface value of the Newtonian gravitational field satisfies $g=GM/R^2$. Using the Newtonian gravitational potential energy $-GMm/r$, conservation of energy now givesTherefore

In a steady state, mass conservation makes $\mathcal{F}$ independent of radius. Taking the accretion rate $\dot{M}>0$ for inward flow gives $\mathcal{F}=-\dot{M}$. The conservation of angular momentum equation then makes $-\dot{M}h+\mathcal{G}$ constant. The zero-torque inner boundary condition fixes this constant to $-\dot{M}{h}_{\mathrm{in}}$, soThe viscous torque in an accretion disk is $\mathcal{G}=-2\pi r^3\bar{\nu }\Sigma {\Omega }^{\prime }=2\pi qh\bar{\nu }\Sigma$. Combining these expressions gives the steady accretion disk with arbitrary rotation lawFor $x=r/R_g$, the Paczyński-Wiita circular orbit formulas giveConsequentlyIn particular, $f(6)=0$ and $f\to 1$ far from the black hole, recovering the outer Keplerian accretion disk.

Inside the innermost stable circular orbit, the gas enters the plunging region of a black-hole accretion disk. Its rapid inward motion leaves little time for stresses to exchange angular momentum, motivating the zero-torque inner boundary condition. This is a thin-disc approximation; a strong magnetic stress could change it.

Matter supplied from very large radius has negligible specific orbital energy, while matter crossing the inner edge carries ${\epsilon }_{\mathrm{in}}=-\eta c^2$. With zero inner torque, no energy is supplied by a stress at that edge. If the heat released outside it escapes by radiative transfer, rather than being lost through inward advection, the integrated conservation of energy balance gives

The shell equation is conservative because its radial acceleration is the negative derivative of the specific potential energyMultiplying the equation of motion by $\dot{r}$ givesHenceThis is energy per unit shell mass. The total energy of a uniform sphere requires integrating the energies of its shells.

For a sphere of physical radius $R$, the shell theorem gives $M(r)=M(r/R{)}^3$ and $dM=3M{r}^{2}dr/R^3$. Assemble the sphere shell by shell to obtain its Newtonian gravitational potential energy:The cosmological constant contributes a prescribed one-body potential $-\Lambda r^2/6$, rather than a pair interaction, so there is no additional factor $1/2$:Evaluating at the turnaround radius givesAs in the question, $\Lambda$ here includes the factor $c^2$ that would multiply the geometrical cosmological constant in SI units.

For the final equilibrium apply the scalar virial theorem. The gravitational virial is $W_G$, whereas the cosmological constant force contributesThus, neglecting a surface-pressure term,At fixed $R$, a positive cosmological constant reduces the kinetic energy needed for virial equilibrium because its outward acceleration partially offsets self-gravity.

For the uniform-sphere collapse with a cosmological constant, assume a uniform final sphere of the same mass, with no energy loss or escaping matter. PutThe sphere has no bulk kinetic energy at turnaround, so $E_{\mathrm{ta}}=-A-B$. The final virial theorem givesSince ${\rho }_{\mathrm{ta}}=3M/(4\pi r_{\mathrm{ta}}^3)$,Conservation of energy therefore givesSelect the collapsing branch continuous from $x=1/2$ at $\eta =0$, rather than an unrelated root of the cubic equation.

For $|\eta |\ll 1$, the root lies close to $1/2$. Linearize $x^3$ there:Substitution into the cubic equation gives $(\eta /2-2)x+1-\eta /2\simeq 0$, whenceThe rational expression is an approximation, not an exact solution of the cubic equation. In fact both it and the exact collapsing root agree through order ${\eta }^2$; their first difference is of order ${\eta }^3$.

With zero cosmological constant, spherical collapse gives $r_f=r_{\mathrm{ta}}/2$. For small positive $\Lambda$, the same sphere with the same turnaround radius reaches a smaller final radius. This does not mean that repulsion promotes collapse: as the sphere contracts, $W_{\Lambda }$ becomes less negative, so some released gravitational binding energy pays the energetic cost of moving inward against the repulsive force. Combined with the modified virial theorem, energy conservation requires a more compact endpoint. Positive $\Lambda$ can prevent turnaround altogether; an inward acceleration at $r_{\mathrm{ta}}$ requiresThe perturbative radius formula applies well inside this limit, for spheres that actually turn around. A small negative cosmological constant instead gives $r_f/r_{\mathrm{ta}}>1/2$ at fixed turnaround data.

Ignoring gravitational focusing, a galaxy sweeps a cylinder of volume $\pi R^2{v}_{\mathrm{rel}}T$. Multiplying this by the number density $N$ of target galaxies and averaging the relative speed gives the mean encounter countFor independent encounters modeled by a Poisson process, the probability of at least one galaxy merger is $1-e^{-\mu }$. The printed expression $P=\mu$ is its rare-encounter approximation $\mu \ll 1$; it is not an exact probability for arbitrary $T$.

For an illustrative present-day field population take an effective merger impact parameter $R=20\mathrm{kpc}=0.02\mathrm{Mpc}$, $N=10^{-2}{\mathrm{Mpc}}^{-3}$, and $⟨v_{\mathrm{rel}}⟩=200\mathrm{km}{\mathrm{s}}^{-1}$. These are assumed order-of-magnitude inputs, not precise observational measurements. Taking a Hubble time of $T\simeq 1.4\times 10^{10}\mathrm{yr}\simeq 4.4\times 10^{17}\mathrm{s}$, the given distance conversion yieldsand henceThus the geometric field estimate is of order $10^{-5}$ to $10^{-4}$ per Hubble time. It scales as $R^{2}N⟨v_{\mathrm{rel}}⟩T$ and is very sensitive to environment and the adopted effective merger radius. Enhanced density in groups, gravitational focusing, and the evolution of the galaxy population are all omitted; this number is not a prediction of the full cosmological merger fraction.

Now use the stipulated rapid, distant encounter. The consistent rectilinear trajectory isWith the stated impact parameter along $x$ and velocity along $z$, this trajectory lies in the $xz$-plane. The original PDF's reference to the $xy$-plane is a typo; the local TeX also corrupts the impact-parameter direction. The original PDF fixes that direction as $x$.

Use the positive potential $G{m}_p/|\mathbf{R}-\mathbf{r}|$, whose gradient is the attractive acceleration under the question's sign convention. Its multipole expansion isThe first term has no force; the linear term accelerates the entire galaxy and disappears in the frame following its centre of mass. The leading internal tidal potential is thereforeThis is a quadrupole approximation valid for $r\ll p$, rather than an exact equality for every $r<p$. Its acceleration tidal tensor acts on $\mathbf{r}$ asIn the impulse approximation, each stellar position is held fixed during the flyby and the velocity kick is the integral of this acceleration. Put $s=vt$; terms odd in $s$ integrate to zero, whileThe integrated diagonal coefficients of the tidal tensor are consequently $(2,-2,0)/(v{p}^2)$, givingThe encounter stretches the galaxy along the impact parameter, compresses it along $y$, and gives no net leading kick along the flyby direction.

Denote the pre-encounter stellar velocity by $\mathbf{u}$, to distinguish it from the relative flyby speed $v$. The instantaneous change of specific kinetic energy isUncorrelated kicks with zero mean cross term imply $⟨\mathbf{u}\cdot \Delta \mathbf{v}⟩=0$, so the phase-averaged heating at a specified position isFor an individual star the cross term need not vanish: the formula is an ensemble or orbital-phase average. Integrating over a spherical galaxy of mass $m_g$, symmetry gives $⟨x^2⟩=⟨y^2⟩=⟨z^2⟩=⟨r^2⟩/3$. Therefore the total tidal heating isFor two identical galaxies, each receives this heating with $m_p=m_g$. Adding both contributions gives

In the centre of mass frame, equal masses approach with speeds $v/2$ if $v$ is their relative speed at infinity. The initial orbital energy iswhere $m_g/2$ is the reduced mass. The orbital potential energy vanishes at infinite separation. Within the weak-deflection approximation this asymptotic speed is also the nearly constant speed used in the flyby calculation.

Conservation of energy transfers the positive internal heating out of the relative orbit. Tidal capture of galaxies occurs in this model if $E_{\mathrm{orb}}-\Delta E_{\mathrm{int}}<0$, which givesThe equality is marginal capture. A bound pair still needs subsequent evolution to coalesce, so the result is an approximate capture criterion used here as a merger criterion. Its right-hand side has dimensions of length times speed, as required.