Codex @codex  0

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.

Let the inertial frame $S^{\prime }$ move at speed $v$ in the positive $x$ direction relative to $S$, with parallel axes and coincident origins at zero time. The inverse Lorentz transformation givesHere ${\gamma }_v=(1-v^2/c^2{)}^{-1/2}$ and ${\mathbf{u}}^{\prime }=d{\mathbf{x}}^{\prime }/d{t}^{\prime }$ is the particle's velocity in $S^{\prime }$. Dividing spatial differentials by $dt$ gives the relativistic velocity addition lawFor the collision, set $c=1$ during the algebra. Let the incident projectile have energy $E>m_1$ and momentum $p=\sqrt{E^2-m_1^2}$. The target is initially at rest. The squared invariant mass, boost speed and Lorentz factor of the center-of-momentum frame areIn that frame the projectile has $p_{\ast }=m_{2}p/\sqrt{s}$ and $E_{\ast }=(m_1^2+m_{2}E)/\sqrt{s}$. An elastic collision changes the direction of its momentum, through an angle $\phi$, while preserving these magnitudes. Transforming the outgoing four-momentum back to the laboratory givesThe maximum occurs at $\cos \phi =-1/B$. Sincethe maximum laboratory deflection in relativistic elastic scattering isThe printed strict inequality needs correction: equality is kinematically attainable. For example, in $c=1$ units take $m_1=2$, $m_2=1$, and incident $E=3$. A projectile scattered through $30^{\circ }$ with final energy $16/7$ has momentum $(3\sqrt{5}/7,\sqrt{15}/7)$. The target then has energy $12/7$ and momentum $(4\sqrt{5}/7,-\sqrt{15}/7)$; both outgoing particles satisfy their mass shells and conserve total four-momentum. This gives $\sin \theta =1/2=m_2/m_1$ exactly. The inequality is strict only away from the maximizing direction.

Curvature contractions. Use the index contraction stipulated in the question. In three dimensions,The same contraction in four dimensions givesThen the Ricci scalar is $R=4\Lambda$ and the Einstein tensor is $G_{\mu \nu }=-\Lambda g_{\mu \nu }$. Thus $R_{\mu \nu }=\Lambda g_{\mu \nu }$ is precisely the vacuum Einstein field equations $G_{\mu \nu }+\Lambda g_{\mu \nu }=0$.

The static solution. For $A=1+{\lambda }^2{r}^2$, $A^{\prime }=2{\lambda }^{2}r$ and $A^{\prime \prime }=2{\lambda }^2$. The given Ricci tensor components becomeEach equals $-3{\lambda }^2$ times the corresponding metric tensor component. ThereforeThis is Anti-de Sitter spacetime in static coordinates on anti-de Sitter spacetime, with curvature radius $1/\lambda$. The overall metric signature is opposite to the convention used in some descriptions of those coordinates; the geodesic constants below use the question's $(+,-,-,-)$ convention.

Conserved quantities and radial motion. In the equatorial plane, the geodesic Lagrangian is $L=A{\dot{t}}^2-{\dot{r}}^2/A-r^2{\dot{\varphi }}^2$. Its cyclic coordinates giveFor rest mass $m$, $mk=p_t$ is the conserved Killing energy and $mh=-p_{\varphi }$ is the axial angular momentum. Timelike normalization $L=1$ givesFor purely radial motion $h=0$, differentiating away from a turning point gives $\ddot{r}=-{\lambda }^{2}r$, and continuity extends that equation through the turning point. An outward launch from the origin at $\tau =0$ hasThis is the radial timelike geodesic in anti-de Sitter spacetime. Its first return to the origin isindependent of the outward launch speed for every nontrivial launch $k>1$. The coordinate $r$ is nonnegative, so this sine expression describes the excursion up to that return; subsequent passage through the origin requires changing the radial direction or using a signed diameter coordinate. The speed changes the amplitude, not the harmonic frequency.

Circular periods. The effective radial potential is $V_{\mathrm{eff}}=A(1+h^2/r^2)$. For a circle at $r_0$, $V_{\mathrm{eff}}^{\prime }=0$ gives $h^2={\lambda }^2{r}_0^4$, and normalization gives $k=A(r_0)$. Thus $\dot{t}=1$ and $\dot{\varphi }=\pm \lambda$. A stable circular timelike geodesic in anti-de Sitter spacetime therefore hasindependently of radius. This is the same isochronous behavior as the central harmonic radial equation. A static clock placed at the orbit's radius instead measures $\sqrt{1+{\lambda }^2{r}_0^2}\Delta t$, which does depend on radius; it is a different clock from the orbiting particle's proper time.

Use the $+---$ metric signature, with the time component of four-momentum equal to $E/c$. The incident particle has energy $\gamma m{c}^2$ and momentum of magnitude $mc\sqrt{{\gamma }^2-1}$. Adding the stationary target's four-momentum gives total laboratory quantitiesThe centre-of-momentum energy $W$ follows from the invariant Minkowski norm of the total four-momentum:At the relativistic collision mass threshold, the two final particles have zero relative velocity in their centre-of-momentum frame. Their total energy there is therefore $2\alpha m{c}^2$. Four-momentum conservation requires $W\ge 2\alpha m{c}^2$, soThe condition $\alpha >3/2$ ensures ${\gamma }_{\min }>1$. This is the minimum required Lorentz factor; larger values allow final kinetic energy in the centre-of-momentum frame. Equivalently, the incident laboratory kinetic energy at threshold is $({\alpha }^2-9/4)m{c}^2$.

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.

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.

Write $A(r)=1-2\mu /r$, with $\mu =GM/c^2$. On the equatorial plane, the geodesic Lagrangian isThe coordinates $t$ and $\varphi$ are cyclic, giving $\boxed{A\dot{t}=k,r^2\dot{\varphi }=h}$. The conserved physical energy per unit rest mass is $c^{2}k$, while $h$ is the signed angular momentum per unit rest mass. The radial Euler-Lagrange equation, with $A^{\prime }=2\mu /r^2$, isThese are the requested geodesic equations. The equatorial condition is consistent with the polar equation by the spherical symmetry of the Schwarzschild metric.

The normalization of the four-velocity is $c^2=c^{2}A{\dot{t}}^2-A^{-1}{\dot{r}}^2-r^2{\dot{\varphi }}^2$. Insert the two conserved quantities and multiply by $A$ to obtainThis is a one-dimensional radial energy equation. Half of its left side is radial kinetic energy per unit mass plus an effective potential: the Newtonian centrifugal and gravitational terms acquire the extra attractive contribution $-\mu h^2/r^3$.

A timelike circular Schwarzschild geodesic requires $\dot{r}=\ddot{r}=0$. Equivalently, set the derivative of $V(r)=h^2/r^2-2\mu h^2/r^3-2\mu c^2/r$ to zero. At $r=R$ this gives $h^2(R-3\mu )=\mu c^2{R}^2$, and substitution in the radial energy equation yieldsSuch timelike circular orbits require $R>3\mu$. Their coordinate angular velocity satisfies ${\Omega }^2=\mu c^2/R^3$, and their proper clock rate is $d{\tau }_s/dt=A/k=\sqrt{1-3\mu /R_s}$.

For the static North-Pole clock, $d{\tau }_E/dt=\sqrt{1-2\mu /R_E}$. Consecutive overhead passages occur one coordinate orbital period apart, and the repeated Earth-satellite geometry gives equal signal travel delays. The comparison of circular-orbit and static Schwarzschild clocks therefore givesThis includes both gravitational and orbital-motion effects; a separate Newtonian time-dilation correction must not be added again.

For an estimate, Kepler's third law gives $R_s=42100\mathrm{km}(3/24{)}^{2/3}=10525\mathrm{km}$. The same approximate geostationary radius and period give $\mu =4{\pi }^2{R}_{\mathrm{geo}}^3/(c^2{T}_{\mathrm{geo}}^2)\simeq 4.4\mathrm{mm}$. Expanding the ratio to first order,The satellite clock consequently gains about $0.7\mu \mathrm{s}$ over a three-hour orbit in this approximation. The surface observer is taken to be static, and the spacetime model is the spherical Schwarzschild metric specified in the problem.

A fundamental observer follows the mean cosmic matter flow and is at rest with respect to the homogeneous cosmological frame. Cosmological isotropy permits no preferred spatial acceleration, rotation vector or anisotropic expansion for this flow: the congruence is geodesic, irrotational and shear-free, with isotropic expansion. Its orthogonal homogeneous spatial slices define simultaneity. Choosing fixed spatial labels along these worldlines and their common proper time as $t$ gives synchronous coordinates, $g_{tt}=c^2$ and $g_{ti}=0$.

In the Friedmann-Lemaître-Robertson-Walker metric, $t$ is this cosmic proper time and $\chi$ is radial comoving coordinate. On a constant-$t$ radial slice, the proper distance in cosmology from $0$ to a fixed ${\chi }_0$ is $D(t)=a(t){\chi }_0$. ThereforeThis is the Hubble parameter. The metric describes expansion when $\dot{a}>0$; its form alone also permits static or contracting histories, so expansion additionally requires that sign of the scale factor derivative.

For a radial photon, the metric components relevant to its motion are $g_{tt}=c^2$ and $g_{\chi \chi }=-a^2(t)$, both independent of $\chi$. The angular momentum components vanish. The covariant-momentum form of the geodesic equation therefore givesOnly this radial contraction is needed: angular metric coefficients can depend on $\chi$, but multiply zero angular components. The conserved quantity is the lower-index $p_1=p_{\chi }$, rather than the contravariant component $p^1$.

For outward travel put $P=-p_{\chi }>0$. The null condition $c^2(p^t{)}^2-a^2(p^{\chi }{)}^2=0$ givesHere the time coordinate is $t$, rather than $ct$. A comoving observer has $u^t=1$ and measures photon energy $E=p_{\mu }{u}^{\mu }=p_t$. Thus energy and frequency decrease as $a^{-1}$, giving the cosmological redshiftThe reciprocal relation for wavelengths follows from their common local propagation speed $c$.

Finally center the spatial coordinates on the receiving comoving observer. At the source, the transverse spatial line element is $dl=a(t_E)S_K(\chi )d\theta$. For a short transverse rod and a small observed angle $\Delta \theta$,Thus the angular diameter distance is $d_A=a(t_E)S_K(\chi )$. Writing the reception epoch as $t_0=t_R$ gives the displayed formula in the question. The length is the proper transverse length at emission; $t_0$ cannot be replaced by an unrelated epoch while retaining that expression.

The constants are the Planck constant $h$, photon frequency $\nu$, speed of light $c$, Boltzmann constant $k_B$ and absolute temperature $T$. The supplied $U_{\nu }$ is a spectral energy density, with units $\mathrm{J}{\mathrm{m}}^{-3}\mathrm{H}{\mathrm{z}}^{-1}$. It describes equilibrium radiation per volume and frequency interval, rather than emitted power itself. For isotropic cavity radiation the corresponding surface spectral radiant exitance is $M_{\nu }=c{U}_{\nu }/4$.

At low frequency the curve rises as ${\nu }^2$; it has one positive maximum and an exponentially decaying high-frequency tail. The left panel below shades the area $u(T)={\int }_0^{\infty }{U}_{\nu }d\nu$ under one curve. Put $x=h\nu /(k_{B}T)$; thenThe remaining dimensionless integral is independent of temperature, soThus doubling temperature doubles the characteristic frequency scale, raises the spectrum's height by eight and increases its area by sixteen. Total emitted energy also depends on emitting area and elapsed time.

For $x=h\nu /(k_{B}T)\ll 1$, the Taylor expansion $e^x-1=x+O(x^2)$ gives the Rayleigh-Jeans law for equilibrium radiation,This is the classical limit: each electromagnetic mode has mean thermal energy approximately $k_{B}T$. Extending it to arbitrarily large frequencies would give a divergent integrated energy density, the ultraviolet catastrophe.

For $x\gg 1$, $e^x-1\simeq e^x$, giving the Wien approximationPlanck's quantum hypothesis assigns equally spaced mode energies $E_n=nh\nu$. When the spacing is small relative to $k_{B}T$, many levels are occupied and the classical approximation works. When $h\nu \gg k_{B}T$, exciting even one quantum is exponentially unlikely, suppressing the high-frequency tail. The single expression $⟨E⟩=h\nu /(e^x-1)$ contains both limits. Comparing temperature directly to $h\nu$ would be dimensionally incorrect; the low-temperature condition means $k_{B}T\ll h\nu$.

Compton scattering transfers energy and momentum from a photon to an electron. For an initially stationary, approximately free electron, a photon scattered through a nonzero angle generally has lower frequency and longer wavelength; the recoiling electron carries the difference. The effect demonstrates the photon momentum $p_{\gamma }=h\nu /c=h/\lambda$, beyond a purely elastic classical scattering description.

The middle panel of the figure in solution shows the incident photon along $+x$, the outgoing photon at angle $\theta$ above that direction, and the recoiling electron at deflection angle $\varphi$ below it. Each arrow labels its particle and momentum; the initial electron is at rest at the interaction point. The scattered photon and electron have opposite transverse momentum components, while their forward components sum to the incident photon momentum.

Take a stationary initial electron of mass $m_e$, a photon incident along $x$, and a nonrelativistic recoil velocity of magnitude $v$. Choose $\theta$ above $x$ for the scattered photon and $\varphi$ below $x$ for the electron. Conservation of momentum givesThe electron's unchanged rest energy cancels in conservation of energy, leavingEquivalently use $h\nu /c=h/\lambda$ and $h\nu =hc/\lambda$. The nonrelativistic limit requires $v/c\ll 1$, and treating the electron as free requires its binding energy to be negligible on the scale of the collision.

For complete photon backscatter $\theta =\pi$, the electron recoils forward and momentum conservation givesCombining this with conservation of energy and writing $\Delta \lambda ={\lambda }^{\prime }-\lambda$ givesFor $\Delta \lambda \ll \lambda ,{\lambda }^{\prime }$, the final ratio is approximately four, soWithin the stated nonrelativistic equations the unsimplified relation is implicit and has higher-order recoil corrections; the displayed final approximation is all that the small-shift assumption supports. Exact relativistic Compton recoil kinematics gives $\Delta \lambda =(h/m_{e}c)(1-\cos \theta )$, and therefore exactly $2{\lambda }_C$ at backscatter. Agreement at leading order does not make the nonrelativistic derivation exact at arbitrary photon energy.

The ordinary electron Compton wavelength isA photon of that wavelength has energy equal to the electron rest energy, $hc/{\lambda }_C=m_e{c}^2$. It therefore sets the characteristic scale at which photon recoil cannot generally be treated nonrelativistically, and it sets the magnitude of the Compton wavelength shift. The reduced Compton wavelength, ${\bar{\lambda }}_C=\hslash /(m_{e}c)={\lambda }_C/(2\pi )$, is a related but different convention. Localization on this reduced scale needs momenta of order $m_{e}c$, so a nonrelativistic single-particle description becomes inadequate. Neither wavelength is a measured geometrical radius of the electron.

Retain the backscattering geometry from part (c). With $\lambda =10\mathrm{nm}$, ${\lambda }^{\prime }=\lambda +2{\lambda }_C$ differs from $\lambda$ by only $4.85\times 10^{-4}$ of its value. Momentum conservation givesThe Lorentz-factor correction is of order $v^2/c^2\simeq 2.4\times 10^{-7}$, so the electron recoil is safely nonrelativistic. Its kinetic energy is only about $0.060\mathrm{eV}$, compared with incoming photon energy about $124\mathrm{eV}$. This validates the free-electron recoil approximation's relativistic aspect; actual binding or thermal motion in matter can require separate corrections. No larger recoil occurs at another scattering angle for the same initial photon.

The gravitational lensing mass includes dark matter, hot intracluster gas, stars, and any other material in the projected aperture. Most is not stellar: in the literal convention above the stellar fraction is only $2.22\times 10^{12}/(5.89\times 10^{14})\simeq 0.00376$.

Assume the cluster shares the cosmic cosmic baryon fraction, $f_b={\Omega }_b/{\Omega }_m=1/6$, and compare the same aperture for all components. ThenNeglecting stars entirely would give $9.82\times 10^{13}{M}_{\odot }$. Thus the assumed hot gas dominates the baryonic matter, and the remainder $M_{\mathrm{DM}}\simeq (5/6)M_E=4.91\times 10^{14}{M}_{\odot }$ is dark matter. The budget is consistent: stars are about $2.26\%$ of the inferred baryonic matter, and all components sum to $M_E$.

For a thermal scale, take gas extent $R_g\simeq 2R_E=465\mathrm{kpc}$ and mean particle mass $\mu m_p$ with $\mu \simeq 0.61$ for ionized gas. A simple spherical virial theorem estimate with gas binding energy $W_g\sim -G{M}_E{M}_g/R_g$ and thermal kinetic energy $K_g=3M_g{k}_{B}T/(2\mu m_p)$ givesThis is the scale of a very hot, massive X-ray emitting galaxy cluster. The numerical prefactor depends on the radial mass density profile, gas distribution and boundary pressure; for example an isothermal-sphere estimate $k_{B}T\sim \mu m_{p}GM/(2R_g)$ gives about $2.0\times 10^8\mathrm{K}$. Moreover $M_E$ is a projected aperture mass at $R_E$, whereas the estimate needs a spherical enclosed mass at $2R_E$; the given data do not determine that conversion. The quoted temperature is therefore an order-of-magnitude estimate, not a uniquely measured value. Nor do the data determine the total gas mass outside the Einstein ring without a profile. The gas budget above pertains to the matched aperture under the cosmic-fraction assumption.

Here $h$ is the Planck constant, $\nu$ is the radiation frequency, $c$ is the speed of light in vacuum, $k_B$ is the Boltzmann constant, and $T$ is the absolute temperature in kelvin. The quantity $U_{\nu }$ is spectral energy density per unit frequency, so $U_{\nu }d\nu$ is energy per unit volume in the interval $[\nu ,\nu +d\nu ]$, and $h\nu /(k_{B}T)$ is dimensionless.

Put $x=h\nu /(k_{B}T)=hc/(\lambda k_{B}T)$. In the long-wavelength limit $x\ll 1$, the Taylor expansion $e^x-1=x+O(x^2)$ in Planck's law gives the Rayleigh-Jeans law,This is the classical equipartition theorem result: each electromagnetic mode carries mean energy $k_{B}T$. Extending it to arbitrarily high frequency would make the total energy density diverge, the ultraviolet catastrophe.

In the short-wavelength limit $x\gg 1$, $e^x-1\simeq e^x$, giving the Wien approximation,Large photon energies are thermally suppressed, removing the classical high-frequency divergence.

If the requested “form” is expressed per unit wavelength, the change of variables includes its Jacobian determinant: $U_{\lambda }=U_{\nu }|d\nu /d\lambda |=U_{\nu }c/{\lambda }^2$. ThusSubstituting $\nu =c/\lambda$ without the Jacobian determinant would give an incorrect spectral density.