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There is a genuine inconsistency in this part. For any nonnegative metallicity distribution function, its maximum is at least its mean. A population of positive mean solar metallicity cannot have a maximum equal to half that mean. For example, a uniform abundance distribution on $[0,2Z_{\odot }]$ has mean $Z_{\odot }$ and maximum $2Z_{\odot }$. Thus the printed maximum-to-mean ratio cannot be proved. The following calculation explains the supplied function, gives the consistent distribution, and also treats the literal birth-metallicity interpretation separately.

Let $f(\mu )$ denote the supplied abundance function. Algebraically it isWith $\delta =1-\mu$, a Taylor series of $(1-\delta {)}^{Y+1}$ givesConsequentlyThis is an early-consumption expansion: its fractional correction is $(1-Y)\delta /3+O({\delta }^2)$. Small yield alone does not make the expression linear throughout most of the gas-consumption history. At fixed $0<\mu <1$, the small-$Y$ expansion instead is

In a well-mixed closed-box model of galactic chemical evolution, the gas fraction of a galaxy also labels the permanently locked stellar mass: $d{M}_s=-M_{g,\mathrm{init}}d\mu$. Newly formed stars inherit the current gas-phase metallicity $Z_g(\mu )$. Therefore the mass-weighted mean stellar metallicity satisfies the cumulative and birth stellar metallicities relationApplying the derivative to $f$ givesThus the supplied function has the natural interpretation of a cumulative mean, not an individual star's birth abundance. Direct integration of $1-u^Y$ verifies that it gives exactly $f$. This is the saturating closed-box enrichment law: it corresponds to effective stellar yield $Y(1-Z_g)$, with $Z_g$ measured as a fraction. The usual constant-yield trace-metal model instead gives $Z_g=-Y\log \mu$. Both agree to first order when $Y|\log \mu |\ll 1$; their exact formulae should not be identified outside that regime.

Let the final gas fraction be ${\mu }_f>0$ and ${\delta }_f=1-{\mu }_f$. In the consistent mean-metallicity interpretation, ${\overline{Z}}_{\ast }({\mu }_f)=Z_{\odot }$ while the most recently formed stars have the maximum $z_f=1-{\mu }_f^Y$. At early consumption,so the corrected result isThe assumption here includes ${\delta }_f\ll 1$, equivalently $2Z_{\odot }/Y\ll 1$ in this approximation. Exactly, $z_f=1-{\mu }_f^Y$ and $Z_{\odot }=f({\mu }_f)$ determine the maximum implicitly; it remains larger than the mean, but its ratio to the mean need not equal two after substantial consumption.

For the mass-weighted metallicity distribution function, stars formed while the gas fraction falls by $-d\mu$ have probability $dP=-d\mu /{\delta }_f$. Writing their individual abundance as $z=1-{\mu }^Y$ gives $\mu =(1-z{)}^{1/Y}$ and the change of variablesThe density is zero outside this interval. Its cumulative probability is $[1-(1-z{)}^{1/Y}]/(1-{\mu }_f)$, so it integrates to one; integrating $zF(z)$ reproduces $f({\mu }_f)$. A star-count distribution has the same form if the initial mass function and survival selection give a fixed number of observed stars per unit locked mass. Luminosity-weighted samples require different weights.

For trace abundances, $(1-z{)}^{1/Y-1}\simeq e^{-z/Y}$ to leading order at fixed $z/Y$. The corresponding ordinary closed-box metallicity distribution isIn the early-consumption regime $z_f\ll Y$, this becomes approximately uniform, $F\simeq 1/(Y{\delta }_f)$ on $[0,Y{\delta }_f]$. Its mean is half its upper endpoint, again producing the corrected maximum $2Z_{\odot }$.

If the supplied $f(\mu )$ is instead retained literally as the individual birth abundance, set $q=f(\mu )$ rather than $z=1-{\mu }^Y$. The observation-time population mean is then ${\delta }_f^{-1}{\int }_{{\mu }_f}^{1}f(u)du$, not $f({\mu }_f)$. Its exact distribution can be specified parametrically usingPositivity follows because $f$ is the average of the increasing function $1-(1-\delta {)}^Y$. Its support is $0<q<f({\mu }_f)$, and normalization follows directly from $dP=d\delta /{\delta }_f$. At early consumption it is uniform with $F_{\mathrm{literal}}\simeq 2/(Y{\delta }_f)$, maximum $Y{\delta }_f/2$ and mean $Y{\delta }_f/4$. Even this literal alternative gives maximum twice the population mean, never half. The distinction between birth abundance and cumulative mean resolves the calculation; the printed inverse maximum-to-mean ratio is false under either interpretation.

Divide metal enrichment by gas depletion in the closed-box model of galactic chemical evolution:The logarithm appears on integrating from $M_0$ and $Z=0$:With the gas fraction of a galaxy $\mu =M_g/M_0$, this isIt gives $Z=0$ at $\mu =1$ and increasing gas-phase metallicity as gas is consumed, independently of the time dependence of the star formation rate. For this closed initially star-free system, $M_0=M_g+M_s$, so this initial-reservoir fraction equals the present gas-to-total baryonic fraction. That equality fails in an open system.

Let $G$ be the remaining gas mass, $S$ the formed stellar mass, and $G_0$ the initial gas mass; initially $S=0$ and the gas-phase metallicity is zero. Take the stellar yield $y$ and the mass-loading factor $\nu$ to be constant, with homogeneous mixing and no galactic gas inflow. With the specified zero bulk return, mass conservation givesNew metals add $ydS$, while star formation and the galactic outflow remove metals at the existing gas-phase metallicity $Z$. ThusThe leaky-box model of galactic chemical evolution therefore givesFor the usual gas fraction of a galaxy, $\mu =G/(G+S)$, the denominator is the baryonic mass still in the system, which decreases under the galactic outflow. Since $G_0=G+(1+\nu )S$, we have $G_0/G=1+(1+\nu )(1-\mu )/\mu$. In the remaining-mass convention,If instead $\mu$ denotes $G/G_0$ relative to the fixed initial reservoir, the same calculation gives $\boxed{Z=y\ln (1/\mu )/(1+\nu )}$. Stating the denominator matters: these two gas fraction of a galaxy conventions coincide for the closed-box model of galactic chemical evolution, recovered at $\nu =0$, but differ when gas escapes. Neglecting bulk return while retaining a nonzero stellar yield is the intended trace-metal approximation.

Let $\psi (t)$ be the star formation rate, $M$ the remaining gas mass, and $S$ the mass put into stars. With zero returned bulk gas, pristine galactic gas inflow, and no galactic outflow, mass conservation givesA well-mixed gas reservoir loses existing metals at rate $Z\psi$ and gains newly synthesized metals at rate $y\psi$; the incoming gas brings none. ThusFor $0<\nu <1$, eliminate time and integrate:The proportional-infall model of galactic chemical evolution therefore gives, for initially pristine gas,The displayed form requires $Z_0=0$; more generally $Z=y/\nu +(Z_0-y/\nu )(M/M_0{)}^{\nu /(1-\nu )}$. Also $S=(M_0-M)/(1-\nu )$ for an initially star-free reservoir, so $M/M_0$ is not the gas fraction of a galaxy $M/(M+S)$. As a check, the $\nu \to 0$ limit is the closed-box model of galactic chemical evolution relation $Z=y\log (M_0/M)$; at $\nu =1$ the gas mass is constant and one must instead solve $\dot{Z}=(y-Z)\psi /M_0$. Retaining a nonzero stellar yield while setting bulk mass return to zero is the prescribed trace-metal approximation, not a literal claim that stellar ejecta have no mass.

Use the instantaneous recycling approximation with a well-mixed gas reservoir, constant stellar yield $p$, no gas inflow, and an irreversible galactic outflow. Let $M_s$ be the mass locked into long-lived stars and remnants after prompt recycling, and put $\psi (t)={\dot{M}}_s(t)\ge 0$. The yield $p$ is the newly synthesized mass of heavy elements returned to the gas per unit increase of $M_s$. Thus $M_h$ in $Z=M_h/M_g$ denotes the heavy-element mass in the gas, excluding metals already locked into stars.

Assume $M_s(0)=0$, $M_g(0)=M_0$, and $Z(0)=Z_0$; initially pristine gas means $Z_0=0$. Take the outflow rate to be $\alpha \psi$ with constant $\alpha \ge 0$, and assume its metallicity equals the current gas-phase metallicity $Z$. This is the well-mixed leaky-box model of galactic chemical evolution. The parameter $\alpha$ is a mass-loading factor relative to the net rate of locking mass into stars. If a gross star formation rate $\Psi$ and a prompt returned fraction $\mathcal{R}$ are used instead, $\psi =(1-\mathcal{R})\Psi$ and a wind rate $\eta \Psi$ corresponds to $\alpha =\eta /(1-\mathcal{R})$.

Conservation of mass givesand the baryonic mass remaining in the box is $M_g+M_s=M_0-\alpha M_s$. Newly made metals enter the gas at rate $p\psi$, while pre-existing metals are locked into stars at rate $Z\psi$ and leave in the wind at rate $\alpha Z\psi$. HenceUsing $M_h=Z{M}_g$ and the product rule,The cancellation expresses the fact that a well-mixed wind removes gas and metals in the same proportion, while fresh stellar production raises the abundance in the remaining gas.

Divide this equation by ${\dot{M}}_g=-(1+\alpha )\psi$ along the evolving reservoir, or integrate in time through intervals with no star formation. The resulting gas-phase metallicity isEquivalently, its fully explicit dependence on a prescribed net star formation rate isvalid while $M_g(t)>0$. With pre-existing stars, replace $M_s(t)$ in the gas-mass relation by $M_s(t)-M_s(0)$. Without specifying $\psi (t)$ or a gas-consumption law, the model fixes $Z$ as a function of gas mass but does not determine a unique function of time.

For the closed-box model of galactic chemical evolution, set $\alpha =0$. Then $M_g=M_0-M_s$ andAt the same remaining fraction $g=M_g/M_0$ of the initial gas reservoir, the leaky box has the logarithmic coefficient $p/(1+\alpha )$, lower than the true stellar yield by the wind factor. This coefficient is often called an effective yield, but the definition of the gas fraction must be specified.

In particular, the observable gas fraction of a galaxy is often $\mu =M_g/(M_g+M_s)$, using the mass still present, rather than $g$. In our initially star-free model,ThereforeFor $0<\mu <1$ and $\alpha >0$, the leaky-box enrichment is smaller at the same $\mu$: $\ln [1+\alpha (1-\mu )]\le \alpha (1-\mu )<\alpha \ln (1/\mu )$. If effective yield is defined observationally as $(Z-Z_0)/\ln (1/\mu )$, it is consequently not simply the constant $p/(1+\alpha )$ at all $\mu$.

As an explicit time example, choose a linear gas-consumption law $\psi =ϵM_g$, with constant $ϵ>0$. The mass and metallicity equations giveThe closed box with the same $ϵ$ has the same metallicity at a given time, while the leaky box exhausts its gas faster. Thus a comparison at fixed time depends on the chosen star formation rate, even though the comparison at fixed gas fraction above is unambiguous.

Finally, the assumption about wind metallicity is essential. If the expelled gas has abundance $Z_w$ instead, its metal-loss rate is $\alpha Z_w\psi$, and the same bookkeeping yieldsA wind enriched relative to the ambient gas lowers the enrichment rate; the logarithmic solution derived above applies to $Z_w=Z$.