What causes the sharp if irregular boundary line in the "Cosmic Cliffs" JWST Carina image?

Question

In this recently released JWST image: we can see a relatively sharp, if irregular, boundary line between a region appearing mostly bluish and a region appearing mostly reddish or orange (understood that these are assigned, not natural colors). What is the process that produces such a distinct boundary, or at least a sharp enough transition that it can appear to us that way in an image? If there is a material flow, which way is it moving? What do we understand about the composition or other properties of the material on either side of the boundary?

Answer 1

The "cliff" marks the boundary between the lower, dusty, neutral gas, and the upper ionized region. The ionization is caused by the O and B stars, i.e. the most massive, hottest ones of the stars seen in the top.

The ionized region basically forms a so-called Strömgren sphere, which was calculated analytically by Strömgren (1939) for the spherical region around a single star surrounded by neutral gas.

A battle between ionizing UV radiation and recombining gas

The ionizing radiation carves its way through the neutral gas. Because the cross section $\sigma_\mathrm{H}$ of neutral hydrogen — the main constituent of the gas — is rather high, in dense gas the UV radiation cannot travel very far before encountering a hydrogen atom and ionization it, i.e. splitting it into a proton and an electron. On the other hand, if the gas is dense, it doesn't take long for a proton to meet an electron and recombine.

Hence, the gas around the hot stars will in general be fully ionized out to a certain distance — in the spherical case called the Strömgren radius $R_\mathrm{S}$ — after which it will be neutral. At this distance, you run out of UV photons, so to speak.

The size of the ionized region

The size of the ionized region is given by the balance between the amount of ionizing photons emitted per unit time $\color{red}{Q(H^0)}$, and the rate at which the protons and electrons recombine. This latter rate is, in turn, given by their densities $\color{blue}{n_\mathrm{p}}$ and $\color{blue}{n_\mathrm{e}}$, where in a fully ionized region $\color{blue}{n_\mathrm{p} \simeq n_\mathrm{e}}$, and a quantum mechanical, temperature-dependent factor $\color{green}{\alpha_B(T)}$, where $T$ is the temperature of the gas: $$ \begin{array}{rcl} [\mathrm{volume}] & = & \frac{[\mathrm{ionization\,rate}]}{[\mathrm{recombination\,rate}]} \\ \frac{4\pi}{3}R_\mathrm{S}^3 & = & \frac{\color{red}{Q(H^0)}}{\color{blue}{n_\mathrm{p}n_\mathrm{e}} \color{green}{\alpha_B}}\\ R_\mathrm{S} & = & \left( \frac{3}{4\pi} \frac{\color{red}{Q(H^0)}}{\color{blue}{n_\mathrm{H}^2} \color{green}{\alpha_B}} \right)^{1/3}, \end{array} $$ where in the last step I've called the densities "the number density of hydrogen", $\color{blue}{n_\mathrm{H}}$.

For instance, an O6.5V star emits $\color{red}{Q(H^0)}=6.6\times10^{48}$ photons per second (Panagia 1973). Typical densities of molecular clouds are $\color{blue}{n_\mathrm{H}}=10^{2\text{–}3}\,\mathrm{cm}^{-3}$. The temperatures needed for star formation are $T\sim10^2\,\mathrm{K}$, but once a star has formed, the surroundings will be heated to some $8000\text{–}20\,000\,\mathrm{K}$.

With the approximation from Dijkstra (2017) $$ \color{green}{\alpha_B(T)} = 2.3\times10^{-13}\left(\frac{T}{10^4\mathrm{K}}\right)^{-0.7}\,\mathrm{cm}^3\,\mathrm{s}^{-1} $$ we can write the characteristic size of the ionized bubble as $$ R_\mathrm{S} \simeq 10\,\mathrm{lightyears} \times\color{red}{\left(\frac{Q(\mathrm{H}^0)}{10^{50}\,\mathrm{s}^{-1}}\right)^{1/3}} \color{blue}{\left(\frac{n_\mathrm{H}}{300\,\mathrm{cm}^{-3}}\right)^{-2/3}} \color{green}{\left(\frac{T}{10^4\,\mathrm{K}}\right)^{0.23}}. $$

The size of the boundary

On the other hand, the mean free path $\ell$ of a UV photon is very small: $$ \ell = \frac{1}{\color{blue}{n_\mathrm{H}}\langle \color{magenta}{\sigma_\mathrm{H}} \rangle}, $$ where $\langle\color{magenta}{\sigma_\mathrm{H}}\rangle$ is the average cross section for photons in the UV range. The cross section has a strong wavelength-dependence: $$ \color{magenta}{\sigma_\mathrm{H}}(\lambda) = 6.33\times10^{-18}\left(\frac{\lambda}{912\,\mathrm{{Å}}}\right)^3, $$ and its average hence depends somewhat on how many photons you have at each wavelength (i.e. the spectrum), but e.g. for a $T=4\times10^4\,\mathrm{K}$, you have $\langle\color{magenta}{\sigma_\mathrm{H}}\rangle\simeq3\times10^{-19}\,\mathrm{cm}^2$.

With the same typical densities as above, we can write the the transition boundary layer as $$ \ell \sim 0.01\,\mathrm{lightyears} \, \color{blue}{\left(\frac{n_\mathrm{H}}{300\,\mathrm{cm}^{-3}}\right)^{-1}}. $$

In other words, the reason for the sharp transition is that the typical size of the ionized, bluish region is of the order of 1000 times larger than the transition to the neutral, brownish region. Farther away from the ionizing source, the gas is "self-shielded".

The reason for the irregular boundary

If you have a small region of enhanced density, this will be more efficient at withstanding the ionizing radiation and shielding the neutral gas. In this case you get a "finger" of neutral gas extending into the ionized region, as seen in the upper left part of the image.

This answer on the Pillars Of Creation discusses the irregular shape in more detail.

Note that the above calculations ignore dust and metals, which also absorb light. In the ionized region, dust is exposed to the UV and may be destroyed through sputtering due to collisions with ions, sublimation or evaporation, or even explosions due to the UV radiation.

In the image below, I edited in Webb's image in an larger (2º×2º) view of the full Carina Nebula where you can better see how the rather spherical HII region has carved its way out to the surrounding interstellar medium.

^{Full view of the Carina Nebula. Credit Harel Boren & NASA/ESA/CSA/STScI (but montaged by me).}