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A curve of growth describes the equivalent width $E(m)=\int (1-e^{-m{\kappa }_{\nu }})d\nu$ as a function of absorber mass path or column density. Its weak-line limit is $E\simeq mS$. Saturation makes growth sublinear: a pure Gaussian core gives $E\sim 2{\Delta }_D\sqrt{\log {\tau }_0}$, while saturated Lorentzian line shape wings give $E\sim 2\sqrt{mS{\Delta }_L}$. The spectral coordinate and coefficient convention must be consistent.

An absorption equivalent width is $W_{\lambda }=\int [1-I_{\lambda }/I_{\lambda }^0]d\lambda$, the wavelength width of a completely black interval that removes the same continuum-normalized area. For a narrow spectral line profile, small optical depth and negligible stimulated emission, the energy-density convention for Einstein coefficients gives $W_{\lambda }=h{B}_{12}{N}_1/{\nu }_{12}$. This follows by replacing $1-e^{-{\tau }_{\nu }}$ by ${\tau }_{\nu }$ and using $|d\lambda /d\nu |=c/{\nu }^2$. At redshift $z$, the observed and rest-frame values satisfy $W_{\mathrm{obs}}=(1+z)W_{\mathrm{rest}}$.

With the spontaneous emission term suppressed, $j_{\nu }=0$. A uniform absorption coefficient over length $\ell$ gives directlyThe normalized specific intensity deficit is unchanged by conversion between wavelength and frequency, since both numerator and denominator acquire the same Jacobian determinant. The Boltzmann distribution and the Einstein coefficients give $n_2{B}_{21}/(n_1{B}_{12})=e^{-h{\nu }_{12}/(k_B{T}_e)}\ll 1$. Consequently, in the optically thin limit,Here $N_1=n_1\ell$ is the column density, and $\nu \simeq {\nu }_{12}$ has been used in slowly varying factors. This linear relation between equivalent width and column density fails once the central optical depth becomes appreciable.

For the unresolved Lyman-alpha line, the lower $1s$ and combined upper $2p$ state degeneracies are $g_1=2$ and $g_2=6$, so the Einstein coefficients giveThe measured equivalent width $W=7.0\times 10^{-12}\mathrm{m}$ then impliesAssume a central transverse line of sight through a uniform filament, so $\ell =d=0.5\mathrm{Mpc}=1.54\times 10^{22}\mathrm{m}$. Nearly all neutral hydrogen is in its ground state at this temperature, henceThe cylinder's length $L$ determines its total volume but does not enter a transverse column density. Geometry must be specified: an axial line of sight has $\ell =L$ and gives $n_{\mathrm{HI}}=4.2\times 10^{-7}{\mathrm{m}}^{-3}$ for the same measured equivalent width. Intermediate chords give intermediate path-dependent estimates. The numbers above treat $W$ as a rest-frame equivalent width; an observed value at redshift $z$ must first be divided by $1+z$.

Let $m$ be the absorber mass path along a ray. If the spectral line profile is independent of position, Beer-Lambert law gives ${\mathfrak{T}}_{\nu }(m)=e^{-m{\kappa }_{\nu }}$. A finite-band mean and the line's equivalent width are respectivelyWhen the band contains all significant absorption, ${\int }_{\mathcal{B}}{\mathfrak{T}}_{\nu }d\nu \simeq \Delta {\nu }_{\mathcal{B}}-E$. An unqualified integral of transmission over the entire infinite line domain diverges because transmission tends to one far from line centre; $E$ is the finite and physically useful line deficit.

This is nonexponential band transmission: each monochromatic transmission is exponential, but their average generally is not. Differentiating with respect to mass path givesMore decisively, define the absorption coefficient averaged over the transmitted spectrum byThen $\boxed{a_{\mathrm{eff}}^{\prime }(m)=-{\mathrm{Var}}_m(\kappa )\le 0}$. The core saturates first and the surviving radiation increasingly samples weakly absorbing wings. The effective attenuation per additional mass therefore decreases. Jensen's inequality also gives $\overline{\mathfrak{T}}\ge e^{-m⟨\kappa ⟩}$, with equality for a spectrally constant coefficient. Thus the band transmission decays more slowly than the exponential with its initial mean attenuation.

The corresponding curve of growth has $E(m)\simeq mS$ for an optically weak line, and sublinear growth after saturation. For the Gaussian profile, with ${\tau }_0=mS/(\sqrt{\pi }{\Delta }_D)\gg 1$, absorption extends to $|\delta \nu |\simeq {\Delta }_D\sqrt{\log {\tau }_0}$, so $E\sim 2{\Delta }_D\sqrt{\log {\tau }_0}$. For a strongly saturated pure Lorentz profile the wings instead give $E\sim 2\sqrt{mS{\Delta }_L}$. These asymptotes apply while the line remains isolated and its important wings remain within the range where the assumed profile and nearly constant Planck function are appropriate. The integrated absorption is not an exponential either: $E^{\prime }>0$ and $E^{\prime \prime }\le 0$.

For the single-line greenhouse deficit, use hydrostatic balance, fixed surface pressure $p_s$, constant gravity $g$, and the dry adiabatThe gas constant and specific heat refer to the background gas. Let ${\mu }_d=\cos \tilde{\theta }>0$ be the supplied effective ray cosine. The vertical optical depth and slant transmittance from level $p$ to space areThis permits pressure-dependent line widths. With local thermodynamic equilibrium, negligible infrared scattering, a black thermal surface, and no incident thermal radiation from space, the formal radiative transfer solution isThe weights add to one. Subtract the transparent-surface radiance and put $B(p)=B_{{\nu }_0}(T(p))$, using the narrow-line approximation only in the line deficit. Define its pressure-dependent equivalent widthApproximating each hemispheric flux by $\pi$ times the effective-ray radiance preserves the black-surface flux. The outgoing longwave flux is thereforeIntegration by parts gives the particularly transparent formBoth boundary terms vanish: $E(0,q)=0$ and $B(p_s)-B(p_s)=0$. The dry adiabat and monotonic Planck function give $dB/dp>0$. Atmospheric emission replaces part of the warm surface radiation by emission from cooler levels, reducing emission to space for a specified $T_s$. The inequality is strict when opacity occurs over a nonzero range with temperature contrast; an isothermal absorbing layer at the surface temperature would not produce such a contrast.

For a spherical planet the mean absorbed instellation is $F_{\ast }=(1-\alpha )L_{⊛}/4$. The required global energy balance isConsequently every equilibrium of this model has $T_s\ge T_{\mathrm{bare}}=[(1-\alpha )L_{⊛}/(4{\sigma }_{\mathrm{SB}}){]}^{1/4}$, with strict warming when $D>0$. At fixed temperature profile, increasing $q$ increases $E$ and hence $D$. This argument does not assume that the greenhouse gas changes the stellar-band albedo, surface pressure, or background thermodynamics; those are held fixed. For sufficiently broad or overlapping saturated wings the full frequency-dependent Planck function must replace the narrow-line approximation.

For a black surface thermally coupled to a dry convective atmosphere, define $E(p,q)=\int [1-{\mathfrak{T}}_{\nu }(p)]d\nu$, the line equivalent width between pressure $p$ and space. If the line is sufficiently narrow to evaluate the Planck function at its centre, the outgoing flux is $\sigma T_s^4-D$. Integration by parts of the formal radiative transfer solution gives the displayed deficit, nonnegative because temperature and the Planck function increase downward. In the weak-line limit with constant integrated strength, $E\simeq qSp/(g{\mu }_d)$ and the leading deficit is independent of normalized line shape.