Diffusion-limited escape

A diagram showing that hydrogen diffusion in the upper atmosphere is the bottleneck for hydrogen escape on Earth, following from that given in Catling and Kasting (2017), p. 147.[1]

Hydrogen escape on Earth occurs at ~500 km altitude at the exobase (the lower border of the exosphere) where gases are collisionless. Hydrogen atoms at the exobase exceeding the escape velocity escape to space without colliding into another gas particle.

For a hydrogen atom to escape from the exobase, it must first travel upward through the atmosphere from the troposphere. Near ground level, hydrogen in the form of H₂O, H₂, and CH₄ travels upward in the homosphere through turbulent mixing, which dominates up to the homopause. At about 17 km altitude, the cold tropopause (known as the "cold trap") freezes out most of the H₂O vapor that travels through it, preventing the upward mixing of some hydrogen. In the upper homosphere, hydrogen bearing molecules are split by ultraviolet photons leaving only H and H₂ behind. The H and H₂ diffuse upward through the heterosphere to the exobase where they escape the atmosphere by Jeans thermal escape and/or a number of suprathermal mechanisms. On Earth, the rate-limiting step or "bottleneck" for hydrogen escape is diffusion through the heterosphere. Therefore, hydrogen escape on Earth is diffusion-limited.

By considering one dimensional molecular diffusion of H₂ through a heavier background atmosphere, you can derive a formula for the upward diffusion-limited flux of hydrogen (${\displaystyle \Phi _{l}}$):[1][7]

${\displaystyle \Phi _{l}=Cf_{T}(H)}$

${\displaystyle C}$ is a constant for a particular background atmosphere and planet, and ${\displaystyle f_{T}(H)}$is the total hydrogen mixing ratio in all its forms above the tropopause. You can calculate ${\displaystyle f_{T}(H)}$by summing all hydrogen bearing species weighted by the number of hydrogen atoms each species contains:

${\displaystyle f_{T}(H)=f_{H}+2f_{H_{2}}+2f_{H_{2}O}+4f_{CH_{4}}+...}$

For Earth's atmosphere, ${\displaystyle C=2.5\times 10^{13}}$cm⁻² s⁻¹,[7] and, the concentration of hydrogen bearing gases above the tropopause is 1.8 ppmv (parts per million by volume) CH₄, 3 ppmv H₂O, and 0.55 ppmv H₂.[1] Plugging these numbers into the formulas above gives a predicted diffusion-limited hydrogen escape rate of ${\displaystyle \Phi _{l}=4.3\times 10^{8}}$H atoms cm⁻² s⁻¹. This calculated hydrogen flux agrees with measurements of hydrogen escape.[10]

Note that hydrogen is the only gas in Earth's atmosphere that escapes at the diffusion-limit. Helium escape is not diffusion-limited and instead escapes by a suprathermal process known as the polar wind.

Transport of gas molecules in the atmosphere occurs by two mechanisms: molecular and eddy diffusion. Molecular diffusion is the transport of molecules from an area of higher concentration to lower concentration due to thermal motion. Eddy diffusion is the transport of molecules by the turbulent mixing of a gas. The sum of molecular and eddy diffusion fluxes give the total flux of a gas ${\displaystyle i}$ through the atmosphere:

${\displaystyle \Phi _{i}=\Phi _{i}^{mol}+\Phi _{i}^{eddy}}$

The vertical eddy diffusion flux is given by

${\displaystyle \Phi _{i}^{eddy}=-Kn{\frac {df_{i}}{dz}}}$

${\displaystyle K}$ is the eddy diffusion coefficient, ${\displaystyle n}$ is the number density of the atmosphere (molecules cm⁻³), and ${\displaystyle f_{i}}$ is the volume mixing ratio of gas ${\displaystyle i}$. The above formula for eddy diffusion is a simplification for how gases actually mix in the atmosphere. The eddy diffusion coefficient can only be empirically derived from atmospheric tracer studies.[11]

The molecular diffusion flux, on the other hand, can be derived from theory. The general formula for the diffusion of gas 1 relative to gas 2 is given by [12]

${\displaystyle {\vec {v}}_{1}-{\vec {v}}_{2}=-D_{12}\left({\frac {n^{2}}{n_{1}n_{2}}}\nabla \left({\frac {n_{1}}{n}}\right)+{\frac {m_{2}-m_{1}}{m}}\nabla (\ln {P})+\alpha _{T}\nabla (\ln {T})-{\frac {m_{1}m_{2}}{mkT}}({\vec {a}}_{1}-{\vec {a}}_{2})\right)}$

Variable	Definition
${\displaystyle {\vec {v}}_{1}}$, ${\displaystyle {\vec {v}}_{2}}$	velocity of gas 1, 2 (cm s−1)
${\displaystyle w_{1}}$, ${\displaystyle w_{2}}$	vertical velocity of gas 1, 2 (cm s−1)
${\displaystyle D_{12}}$	binary diffusion coefficient (cm2 s−1 molecules−1)
${\displaystyle b_{12}}$	binary diffusion parameter (${\displaystyle 2.6\times 10^{19}}$ cm−1 s−1 for H)[1]
${\displaystyle n_{1}}$and ${\displaystyle n_{2}}$	number densities of gas 1 and 2 (molecules cm−3)
${\displaystyle n}$	${\displaystyle n_{1}+n_{2}}$(molecules cm−3)
${\displaystyle f_{1}}$	mixing ratio of gas 1
${\displaystyle m_{1}}$and ${\displaystyle m_{2}}$	molecular mass of gas 1 and 2 (in kg molecule−1)
${\displaystyle m}$	${\displaystyle (n_{1}m_{1}+n_{1}m_{2})/(n_{1}+n_{2})}$
${\displaystyle k}$	Boltzmann's constant (${\displaystyle 1.38\times 10^{-23}}$J K−1)
${\displaystyle T}$	Temperature (K)
${\displaystyle {\vec {a}}_{1}}$and ${\displaystyle {\vec {a}}_{2}}$	acceleration of gas 1 and 2 from gravity, electric fields, etc. (cm s−2)
${\displaystyle g}$	gravitational acceleration (9.81 m s−2 on Earth)
${\displaystyle \alpha _{T}}$	thermal diffusivity (~-0.25 for H or H2 in air)
${\displaystyle P}$	air pressure (Pa)

Each variable is defined in table on right. The terms on the right hand side of the formula account for diffusion due to molecular concentration, pressure, temperature, and force gradients respectively. The expression above ultimately comes from the Boltzmann transport equation.[13] We can simplify the above equation considerably with several assumptions. We will consider only vertical diffusion, and a neutral gas such that the accelerations are both equal to gravity (${\displaystyle {\vec {a}}_{1}={\vec {a}}_{2}=g}$) so the last term cancels. We are left with

${\displaystyle w_{1}-w_{2}=-D_{12}\left({\frac {n^{2}}{n_{1}n_{2}}}{\frac {d}{dz}}\left({\frac {n_{1}}{n}}\right)+{\frac {m_{2}-m_{1}}{m}}{\frac {d}{dz}}(\ln {P})+\alpha _{T}{\frac {d}{dz}}(\ln {T})\right)}$

We are interested in the diffusion of a lighter molecule (e.g. hydrogen) through a stationary heavier background gas (air). Therefore, we can take velocity of the heavy background gas to be zero: ${\displaystyle w_{2}=0}$. We can also use the chain rule and the hydrostatic equation to rewrite the derivative in the second term.

${\displaystyle {\frac {d}{dz}}\ln {P}={\frac {1}{P}}{\frac {dP}{dz}}={\frac {-mg}{kT}}}$

The chain rule can also be used to simplify the derivative in the third term.

${\displaystyle {\frac {d}{dz}}\ln {T}={\frac {1}{T}}{\frac {dT}{dz}}}$

Making these substitutions gives

${\displaystyle w_{1}=-D_{12}\left({\frac {n^{2}}{n_{1}n_{2}}}{\frac {df_{1}}{dz}}+{\frac {(m_{2}-m_{1})g}{kT}}+{\frac {\alpha _{T}}{T}}{\frac {dT}{dz}}\right)}$

Note, that we have also made the substitution ${\displaystyle n_{1}/n=f_{1}}$. The flux of molecular diffusion is given by

${\displaystyle \Phi _{1}^{mol}=w_{1}n_{1}=-D_{12}n_{1}\left({\frac {n^{2}}{n_{1}n_{2}}}{\frac {df_{1}}{dz}}+{\frac {(m_{2}-m_{1})g}{kT}}+{\frac {\alpha _{T}}{T}}{\frac {dT}{dz}}\right)}$

By adding the molecular diffusion flux and the eddy diffusion flux, we get the total flux of molecule 1 through the background gas

${\displaystyle \Phi _{1}=\Phi _{1}^{mol}+\Phi _{1}^{eddy}=-Kn{\frac {df_{1}}{dz}}-D_{12}n_{1}\left({\frac {n^{2}}{n_{1}n_{2}}}{\frac {df_{1}}{dz}}+{\frac {(m_{2}-m_{1})g}{kT}}+{\frac {\alpha _{T}}{T}}{\frac {dT}{dz}}\right)}$

Temperature gradients are fairly small in the heterosphere, so ${\displaystyle dT/dz\approx 0}$, which leaves us with

${\displaystyle \Phi _{1}=-Kn{\frac {df_{1}}{dz}}-D_{12}n_{1}\left({\frac {n^{2}}{n_{1}n_{2}}}{\frac {df_{1}}{dz}}+{\frac {(m_{2}-m_{1})g}{kT}}\right)}$

The maximum flux of gas 1 occurs when ${\displaystyle df_{1}/dz=0}$. Qualitatively, this is because ${\displaystyle f_{1}}$ must decrease with altitude in order to contribute to the upward flux of gas 1. If ${\displaystyle f_{1}}$ decreases with altitude, then ${\displaystyle n_{1}}$ must decrease rapidly with altitude (recall that ${\displaystyle f_{1}=n_{1}/n}$). Rapidly decreasing ${\displaystyle n_{1}}$ would require rapidly increasing ${\displaystyle w_{1}}$ in order to drive a constant upward flux of gas 1 (recall ${\displaystyle \Phi _{1}=w_{1}n_{1}}$). Rapidly increasing ${\displaystyle w_{1}}$ isn't physically possible. For a mathematical explanation for why ${\displaystyle df_{1}/dz=0}$, see Walker 1977, p. 160.[3] The maximum flux of gas 1 relative to gas 2 (${\displaystyle \Phi _{l}}$, which occurs when ${\displaystyle df_{1}/dz=0}$) is therefore

${\displaystyle \Phi _{l}=D_{12}n_{1}\left({\frac {(m_{2}-m_{1})g}{kT}}\right)}$

Since ${\displaystyle D_{12}=b_{12}/n}$,

${\displaystyle \Phi _{l}={\frac {b_{12}g(m_{2}-m_{1})}{kT}}{\frac {n_{1}}{n}}={\frac {b_{12}g(m_{2}-m_{1})}{kT}}f_{1}}$

or

${\displaystyle \Phi _{l}=Cf_{1}}$

This is the diffusion-limited flux of a molecule. For any particular atmosphere, ${\displaystyle C}$ is a constant. For hydrogen (gas 1) diffusion through air (gas 2) in the heterosphere on Earth ${\displaystyle m_{air}-m_{hydrogen}\approx 4.8\times 10^{-26}}$, ${\displaystyle g=9.81}$m s⁻² ,and ${\displaystyle T\approx 208}$ K. Both H and H₂ diffuse through the heterosphere, so we will use a diffusion parameter that is the weighted sum of H and H₂ number densities at the tropopause.

${\displaystyle b_{12}=b_{H}{\frac {n_{H}}{n_{H}+n_{H2}}}+b_{H2}{\frac {n_{H2}}{n_{H}+n_{H2}}}}$

For ${\displaystyle n_{H}\approx 1.8\times 10^{7}}$ molecules cm⁻³, ${\displaystyle n_{H2}\approx 5.2\times 10^{7}}$ molecules cm⁻³, ${\displaystyle b_{H}\approx 2.73\times 10^{19}}$ cm⁻¹s⁻¹, and ${\displaystyle b_{H2}\approx 1.46\times 10^{19}}$ cm⁻¹s⁻¹, the binary diffusion parameter is ${\displaystyle b_{12}=1.8\times 10^{19}}$. These numbers give ${\displaystyle C=2.9\times 10^{13}}$molecules cm⁻² s⁻¹. In more detailed calculations the constant is ${\displaystyle C=2.5\times 10^{13}}$molecules cm⁻² s⁻¹.[7] The above formula can be used to calculate the diffusion-limited flux of gases other than hydrogen.

Every rocky body in the solar system with a substantial atmosphere, including Earth, Mars, Venus, and Titan, loses hydrogen at the diffusion-limited rate.

For Mars, the constant governing diffusion-limited escape of hydrogen is ${\displaystyle C_{mars}=1.1\times 10^{13}}$ molecules cm⁻² s⁻¹.[1] Spectroscopic measurements of Mars' atmosphere suggest that ${\displaystyle f_{T}(H)=(30\pm 10)\times 10^{-6}}$.[14] Multiplying these numbers together gives the diffusion-limited rate escape of hydrogen:

${\displaystyle \Phi _{l}^{mars}=C_{mars}f_{T}(H)=(3.3\pm 1.1)\times 10^{8}}$ H atoms cm⁻² s⁻¹

Mariner 6 and 7 spacecraft indirectly observed hydrogen escape flux on Mars between ${\displaystyle 1\times 10^{8}}$and ${\displaystyle 2\times 10^{8}}$ H atoms cm⁻² s⁻¹.[15] These observations suggest that Mars' atmosphere is losing hydrogen at roughly the diffusion limited value.

Observations of hydrogen escape on Venus and Titan are also at the diffusion-limit. On Venus, hydrogen escape was measured to be about ${\displaystyle 1.7\times 10^{7}}$ H atoms cm⁻² s⁻¹, while the calculated diffusion limited rate is about ${\displaystyle 3\times 10^{7}}$H atoms cm⁻² s⁻¹, which are in reasonable agreement.[16][1] On Titan, hydrogen escape was measured by the Cassini spacecraft to be ${\displaystyle (2.0\pm 2.1)\times 10^{10}}$ H atoms cm⁻² s⁻¹, and the calculated diffusion-limited rate is ${\displaystyle 3\times 10^{10}}$H atoms cm⁻² s⁻¹.[17][1]