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 $k_{B}T\ll h\nu$.

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.

A blackbody is an ideal body that absorbs all incident electromagnetic radiation, at every frequency and direction. Its absorptivity is one, so Kirchhoff's law of thermal radiation makes its equilibrium emissivity one as well. At a specified temperature its thermal spectrum is universal: it depends on temperature, not material or shape. This does not mean it looks black when hot; it can glow brightly while remaining a perfect absorber.

A small hole in a nearly isothermal opaque cavity approximates a blackbody, because radiation entering the hole undergoes many interactions before it could escape. In thermal equilibrium the cavity contains an approximately isotropic Planck spectrum. The emitted radiance follows Planck's law, its integrated outward radiant exitance follows the Stefan–Boltzmann law, and its wavelength peak follows Wien's displacement law. A real surface can have wavelength- and direction-dependent emissivity below this ideal limit.

Three physically distinct contributions are unseen dark matter, hot diffuse intracluster gas omitted from the summed stellar mass, and faint stars or stellar remnants missed by transferring the Milky Way mass-to-light ratio to the cluster. The first two are actual additional components; the third changes the inferred stellar budget. Departures from virial equilibrium, velocity anisotropy or extra nonthermal pressure can also bias the dynamical or X-ray estimates, although they need not remove the discrepancy.

Assume an isotropic, stationary spherical galaxy cluster. Its three-dimensional velocity dispersion obeys $⟨v^2⟩=3{\sigma }_{\mathrm{los}}^2$. For a uniform sphere, the virial theorem gives $3M{\sigma }_{\mathrm{los}}^2=3G{M}^2/(5R)$, henceA different density profile changes the coefficient, so $10^{15}{M}_{\odot }$ is the meaningful scale. For isothermal X-ray gas in hydrostatic equilibrium,Using $\mu =0.6$ and a representative slope $-d\ln {\rho }_g/d\ln r=2$–$3$ at $1\mathrm{Mpc}$ gives $M_X\simeq (1.3$–$1.9)\times 10^{15}{M}_{\odot }$. The temperature and extent alone cannot fix this coefficient without a density profile. The total luminosity is $5\times 10^{12}{L}_{\odot }$, so the previously inferred mass-to-light ratio gives

Treat the gravitational field as approximately spherical. The circular orbit condition givesIf every star had a representative mass $m$ and obeyed $L/L_{\odot }=(m/M_{\odot }{)}^3$, its mass-to-light ratio would be $(m/M_{\odot }{)}^{-2}$, giving $\boxed{m\simeq 0.47M_{\odot }}$. This is a one-mass estimate, not an actual average inferred uniquely from the data: a galactic stellar mass function, stellar remnants, dust and nonstellar mass alter the relation. The flattening of the galactic disk also changes the spherical conversion by an order-one factor.

The Milky Way has a thin stellar galactic disk, radius roughly $15$–$20\mathrm{kpc}$ and scale height about $0.3\mathrm{kpc}$, containing gas, dust, young and old stars, spiral arms and most current star formation. Its stellar mass is a few $10^{10}{M}_{\odot }$. A thicker, older galactic disk has scale height around $1\mathrm{kpc}$ and stellar mass of order $10^{10}{M}_{\odot }$, with little cold gas.

The central galactic bulge and bar extend a few thousand parsecs and contain about $10^{10}{M}_{\odot }$, largely older stars. The central supermassive black hole has mass about $4\times 10^6{M}_{\odot }$ and is negligible in the total galactic budget. A diffuse old metal-poor stellar halo, extending tens to over $100\mathrm{kpc}$, contains globular clusters and roughly $10^9{M}_{\odot }$ in stars. An extended dark matter halo reaches roughly $200\mathrm{kpc}$ and dominates the total mass, of order $10^{12}{M}_{\odot }$. These are approximate component scales; the gaseous galactic halo also extends beyond the stellar galactic disk.

The right panel of the figure in solution has luminosity increasing upwards and effective temperature decreasing to the right. An old globular cluster contains a low-mass main sequence, a main-sequence turnoff, subgiants, a red giant branch, a horizontal branch, an Asymptotic giant branch and a faint hot white dwarf sequence. Its high-mass main sequence is absent because those stars have evolved.

If a fraction $f_c=0.1$ of the star participates in hydrogen burning, with initial hydrogen mass fraction $X=0.7$, the available nuclear binding energy is $0.007f_{c}XM{c}^2$. Assuming a roughly constant luminosity and coeval stars,Taking the quoted core fraction as hydrogen fuel itself, $f_{c}X=0.1$, gives $1.98\times 10^{10}\mathrm{yr}$. Stating the fuel convention matters: burning the whole stellar mass overestimates the age by about an order of magnitude. Evolving luminosity and detailed stellar structure are needed for an accurate age.

A free-fall timescale estimate follows from $R/t^2\sim GM/R^2$:Falling to a central point mass gives the more precise factor $\pi /(2\sqrt{2})$, or $1.77\times 10^3\mathrm{s}$; a distributed star changes an order-one coefficient. For any spherical mass distribution with radius $R$,The inequality is strict for an extended star; equality is approached by a thin shell at $R$. For an monatomic gas the virial theorem is $2U+\Omega =0$, so half the gravitational potential energy released by contraction heats the star and half can be radiated. The Kelvin-Helmholtz cooling time is therefore of orderA uniform-density model gives $3G{M}^2/(10RL)\simeq 9.4\times 10^6\mathrm{yr}$. The given bound fixes a scale, rather than an exact contraction history; both estimates are far shorter than the Sun's main-sequence lifetime.

A thin spherical shell has outward pressure force $-4\pi r^2(dP/dr)dr$ and inward gravitational force $GM(r)4\pi r^2\rho (r)dr/r^2$. Hydrostatic equilibrium requiresA uniform-density solar model gives $M(r)=M_{\odot }(r/R_{\odot }{)}^3$, $\bar{\rho }=3M_{\odot }/(4\pi R_{\odot }^3)$, and after integration from $R_{\odot }$ to zero,Thus the appropriate orders of magnitude are $10^{14}$–$10^{16}\mathrm{Pa}$ and $10^7\mathrm{K}$. The uniform-density approximation underestimates the actual central pressure because the Sun is centrally concentrated. Radiation pressure is neglected in applying the ideal gas law.

Adopt the approximate mass-luminosity relation $L/L_{\odot }\simeq (M/M_{\odot }{)}^3$ and equal available fuel fractions. The main-sequence lifetime then scales as $M/L$, givingUsing an exponent $3.5$ instead gives about $5.4\times 10^7$ and $3.1\times 10^8$ years; these are model-dependent estimates. At $r=3r_s=6GM/c^2$ the Newtonian gravitational potential energy released from infinity is $GMm/r=m{c}^2/6$. Half remains as kinetic energy in a circular orbit, so the radiatable binding energy is $GMm/(2r)$:Using the whole potential energy loss would give $1/6$, which does not subtract the final kinetic energy. A fully relativistic calculation for a nonrotating black hole gives $1-\sqrt{8/9}\simeq 5.7\%$. When the surviving star exhausts hydrogen burning, it expands off the main sequence and can fill its Roche lobe. Binary mass transfer to the black hole can form an accretion disk and an X-ray binary; unstable binary mass transfer may instead produce a common envelope. This is conditional on the binary star remaining bound after the first remnant forms. The supplied black hole endpoint is an assumption, not a universal outcome for an isolated $8M_{\odot }$ star.

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.

Kepler's third law for two point masses in a Newtonian gravitational field isHere $a$ is the semimajor axis of their relative orbit, $P$ the orbital period and $G$ the gravitational constant. Equivalently $(P/\mathrm{yr}{)}^2=(a/\mathrm{AU}{)}^3/[(M_1+M_2)/M_{\odot }]$, to the accuracy with which these astronomical units normalize the solar mass.

Shrink the positive outer contour to the branch cut $[-1,1]$. On its upper bank the chosen branch of a multivalued function has $\sqrt{x^2-1}=i\sqrt{1-x^2}$ and on the lower bank it has the negative of this value. The upper bank runs from $1$ to $-1$, and the lower bank runs from $-1$ to $1$. The endpoint circles contribute $O(\sqrt{ϵ})$ and vanish. HenceThe change of variables $x=\sin t$ independently reduces the result to the integral of ${\sin }^{4}t$ over $[-\pi /2,\pi /2]$.

Use the branch cut $[-1,1]$ and the branch of a multivalued function normalized by $\sqrt{z^2-1}\sim z$. The Laurent series from the preceding solution has $z^{-1}$ coefficient $3/8$. Consequently every sufficiently large positively oriented circle givesAlthough $f$ grows cubically, its other Laurent series terms have zero contour integral. Choosing the opposite branch of a multivalued function reverses the sign; neither the branch of a multivalued function nor the orientation can be omitted from the answer.

The square root produces branch points at $z=\pm 1$, with local behaviour proportional to $(z\mp 1{)}^{-1/2}$. One branch cut can join them along $[-1,1]$. Another choice uses the two rays $(-\infty ,-1]\cup [1,\infty )$. Each choice prevents a loop from enclosing just one branch point within a single-valued domain. For the finite-segment choice fix $\sqrt{z^2-1}\sim z$ at infinity. Infinity is then a pole of order three, not a branch point, since

Put $z=e^{i\theta }$, so $d\theta =dz/(iz)$, $\cos 2\theta =(z^2+z^{-2})/2$ and ${\cos }^{2}3\theta =(z^6+2+z^{-6})/4$. The contour integral isThe enclosed poles are $0$ and $\pm 1/\sqrt{2}$. For the rational function displayed without the prefactor $1/(4i)$, the residue at zero is $-21/8$ and the two other residues sum to $27/8$. Therefore the residue theorem givesA useful independent check is the Fourier series $(5-4\cos \varphi {)}^{-1}=\frac{1}{3}[1+2{\sum }_{k\ge 1}{2}^{-k}\cos k\varphi ]$; only its constant and third Fourier coefficients survive.

Cauchy's integral theorem states that a holomorphic function on a simply connected open region has zero contour integral around every closed piecewise smooth contour in that region. More generally the same holds for a contour homologous to zero in its domain of holomorphy. Having no singularity on the curve alone is insufficient: $1/z$ has a nonzero contour integral around the unit circle.

If $f(z)=\sum a_j{z}^j$, then $f(-z)=\sum (-1{)}^j{a}_j{z}^j$ and its residue at zero is $-a_{-1}$. The circle $|z-1|=2$ encloses zero and no other finite singularity. Thus the residue theorem givesfor positive orientation.

For a positively oriented circle of any radius $R>0$, the residue theorem givesOnly the $z^{-1}$ coefficient in the Laurent series contributes. Reversing the orientation changes the sign.