<h3>The Lyman-limit cross section</h3>
<p>The Lyman limit is not a narrow line, like it is for electronic transitions. There is a minimum energy needed to ionize a hydrogen atom — 13.6 eV, corresponding to a wavelength of 912 Å — but higher energies are not a big problem, because — as Jon Custer comments — the excess energy instead goes into kinetic energy of the particles.</p>
<p>If the energy of the photon becomes too large, however, it will eventually be unable to ionize the atom. In fact, the probability of ionization — the cross section — decreases with the cube of the wavelength, i.e. <span class="math-container">$\phi(\lambda) \propto \lambda^3$</span>. In the figure below I've sketched the absorption cross section where you can see that significant absorption actually extends all the way to (a few) 100 Å.</p>
<p><a href="https://i.sstatic.net/sDCAi.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/sDCAi.png" alt="sigLL" /></a></p>
<h3>The intrinsic galaxy spectrum</h3>
<p>But normal galaxies don't emit much light at such high energies. So the reason that the spectra of Lyman-break galaxies &quot;stay&quot; black blueward of the 912 Å Lyman limit is that, at wavelengths shorter than what may be effectively absorbed, there simply isn't much light emitted in the first place.</p>
<p>However, if you consider instead a galaxy with an <a href="https://en.wikipedia.org/wiki/Active_galactic_nucleus" rel="nofollow noreferrer">active galactic nucleus</a>, the emission spectrum extends all the way to soft X-rays (and beyond). For these galaxies you do indeed see the transmitted intensity begin to rise again at short wavelengths.</p>
<h3>High-redshift galaxies</h3>
<p>In my original answer I considered only galaxies at high redhifts (because those are the ones I'm used to thinking of). Here, another effect comes into play which tends to erase even very short wavelengths:</p>
<p>As light travels away from the galaxy, it is redshifted by the expansion of the Universe. Hence, after a little while light that had a too short wavelength will be redshifted closer to the Lyman limit, and if there happens to be neutral hydrogen around at this point (which invariably there is in the early Universe), then it will be absorbed after all.</p>
<h3>The Lyman alpha forest</h3>
<p>This effect it more pronounced for Lyman alpha photons, i.e. the bound-bound transition from the ground state to the first excited state. For this line, only photons with energies very close to the &quot;correct&quot; energy (10.2 eV, or 1216 Å) are able to excite the hydrogen. But expansion eventually redshifts the more energetic part of the spectrum to 1216 Å, so every time there's a hydrogen cloud, this cloud causes a narrow dip in the spectrum.</p>
<p>For intermediate redshifts (<span class="math-container">$z\sim 1\text{–}6$</span>), you will see a bunch of these lines, comprising the so-called &quot;Lyman alpha forest&quot;. At high redshifts (<span class="math-container">$z\gtrsim6$</span>), the Universe is so neutral that all these lines overlap, and no light is transmitted, resulting in the &quot;Gunn-Peterson trough&quot;. At low redshifts (<span class="math-container">$z\lesssim1$</span>), most of the spectrum is transmitted.</p>
<p>This evolution is seen in the figure below:</p>
<p><a href="https://i.sstatic.net/pJfwv.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/pJfwv.jpg" alt="LAF" /></a></p>
<p><sup>Credit: <a href="https://blackholeswithrevelations.wordpress.com/2013/09/19/stfcastro-lyman-alpha-forest/" rel="nofollow noreferrer">Ross McLure</a>.</sup></p>

CC BY-SA 4.0
2022-12-21T18:36:15.943Z
Thanks @pela, so the "cross-section" in your graph means the electron's orbital cross section? And this would correspond to it's energy? Or is this not what that means?
Rich McDaniel
CC BY-SA 4.0
2022-12-21T20:58:46.577Z
@RichMcDaniel Not really, actually: At the atomic level the concept of "cross section" is not that… _classical_. The cross section depends less on the physical size of the scattering agent — indeed _physical size_ loses its meaning — and more on quantum mechanical factors. You can think of the cross section as a "geometrical" way to express the _probability_ of a photon interacting with the particle ("geometrical" in the sense that we express it in units of area).
pela
CC BY-SA 4.0
2022-12-21T20:58:56.583Z
For some processes (e.g. scattering on electrons), the cross section is (almost) independent of the nature of the photon (wavelength, polarization, direction), but for most processes (e.g. electronic transitions between an atom's energy levels), the cross section is highly dependent on wavelength. This is very different from a regular "size" of an object. And in the case of ionization (which is the case in your question) there is a minimum energy required, so for larger wavelengths, the cross section drops to zero, even though you still have an associated "orbital cross section".
pela
CC BY-SA 4.0
2022-12-21T21:12:52.430Z
For instance, at the Lyman limit (where $\lambda=912\,\mathrm{Å}$), the cross section is $\sigma_0=6.3\times10^{-18}\,\mathrm{cm}^2$ (the maximum in my graph above). What this means is that, if you have a density of $n=1$ atom per $\mathrm{cm}^3$, then the mean free path $\ell=1/n\sigma_0$ of a Lyman limit photon is $1.6\times10^{17}\,\mathrm{cm}$, or 0.17 lightyears. But for a photon with $\lambda\simeq500\,\mathrm{Å}$, the cross section is ~6 times smaller, so the mean free path is 1 lightyear, even though you have exactly the same atoms.
pela
CC BY-SA 4.0
2022-12-22T19:13:31.893Z
That makes sense enough given that classical concepts lose meaning at atomic scales. Thanks.
Rich McDaniel
CC BY-SA 4.0
2022-12-22T19:20:16.593Z
Actually I need another clarification: So the Lyman break at the source (not redshifted) would pick up again at around 20-40 nm?   So even though the emission spectrum of a non-AGN galaxy significantly attenuates smaller than UV wavelengths, the remaining balack body radiation would make it out of the emitting galaxy?
Rich McDaniel
CC BY-SA 4.0
2022-12-23T11:03:33.880Z
@RichMcDaniel In principle yes, but a "normal" galaxy doesn't have many ways to produce such high-energy photons. Even the very hottest "O stars", which exist in few numbers, emit only a small fraction below 50 nm. And because these stars are short-lived, they also tend to be still enshrouded in the gaseous cloud from which they were born, so their photons experience a higher opacity than lower-mass stars.
pela
CC BY-SA 4.0
2022-12-23T11:06:53.613Z
In general, most ionizing photons (Lyman continuum photons, or LyC) are absorbed close to the stars, and are "converted" to Lyman series photons (roughly 2/3 of every LyC photon is converted to a Lyman alpha photon). The so-called LyC-leakers are observed not because of the tiny fraction that makes it through the enshrouding gas, but more likely because stellar feedback has cleared paths throughout the interstellar and circumgalactic medium. And still, they typically leak <10%.
pela
CC BY-SA 4.0
2022-12-23T18:54:37.133Z
thanks @pela! I guess I hadn't realized just how little radiation beyond ultraviolet is actually produced.
Rich McDaniel
CC BY-SA 4.0
2022-12-24T12:49:45.813Z
@RichMcDaniel You're welcome :)
pela
CC BY-SA 4.0
2022-12-17T00:37:34.777Z
2022-12-19T16:00:38.070Z
2022-12-19T16:00:38.070Z
pela
<p>From <a href="https://en.wikipedia.org/wiki/Lyman-break_galaxy" rel="nofollow noreferrer">this description</a> of Lyman-break galaxies, I don't understand how:</p>
<blockquote>
<p>...radiation at higher energies than the Lyman limit at 912 Å is almost completely absorbed by neutral gas around star-forming regions of galaxies. In the rest frame of the emitting galaxy, the emitted spectrum is bright at wavelengths longer than 912 Å, but very dim or imperceptible at shorter wavelengths—this is known as a &quot;dropout&quot;, or &quot;break&quot;.</p>
</blockquote>
<p>But the <a href="https://en.wikipedia.org/wiki/Lyman_series" rel="nofollow noreferrer">wikipedia page</a> for the Lyman series states that the highest limit of radiation absorbed or emitted by neutral hydrogen is 91.2 nm:</p>
<blockquote>
<p>the Lyman series is a ... series of transitions and resulting ... emission lines of the hydrogen atom as an electron goes from n ≥ 2 to n = 1 ... The greater the difference in the principal quantum numbers, the higher the energy of the electromagnetic emission.</p>
</blockquote>
<p>It then states that there is an asymptotic limit to this energy as the difference between transition levels approaches infinity:</p>
<blockquote>
<p>There are infinitely many spectral lines, but they become very dense as they approach n = ∞ (the Lyman limit)... &quot;91.1753 nm&quot;</p>
</blockquote>
<p>I don't understand how neutral hydrogen can absorb light from a photon emitted with wavelength shorter than 912 Angstroms if this wavelength is precisely the highest energy photon hydrogen can absorb.</p>
<p>So my question is: How can there exist a wide drop-out in the spectrum of galaxies at wavelengths shorter than about 91.2 nm, if the highest-energy electromagnetic radiation a hydrogen atom can emit or absorb (at n = ∞) is 91.1753 nm?</p>
<p>How is the hydrogen interacting with photons of higher-energies than this?</p>
<p>It must be a conceptual issue I'm not understanding. Should I not view absorption energies as conceptually the same as emission energies?</p>

CC BY-SA 4.0
2022-12-16T22:18:22.217Z
The “limit” takes an electron to be free of the (now) ion. The electron can, of course, absorb more energy and then has kinetic energy flying away from the ex-atom. The free electron levels are a continuum to an infinity.
Jon Custer
CC BY-SA 4.0
2022-12-20T08:21:31.420Z
Hi Rich, I initially gave an incompletely and perhaps misleading answer, but updated now. Hopefully that explains well, otherwise feel free to ask :)
pela
<galaxy><spectroscopy><redshift><james-webb-space-telescope><hydrogen>
What accounts for a Lyman-break for all wavelengths shorter than 91.2nm if the Lyman limit is the highest energy photon that neutral hydrogen absorbs?
Rich McDaniel
# What accounts for a Lyman-break for all wavelengths shorter than 91.2nm if the Lyman limit is the highest energy photon that neutral hydrogen absorbs?

(Asked by: username_0 on 2022-12-16)

From [this description](https://en.wikipedia.org/wiki/Lyman-break_galaxy) of Lyman-break galaxies, I don't understand how:

> ...radiation at higher energies than the Lyman limit at 912 Å is almost completely absorbed by neutral gas around star-forming regions of galaxies. In the rest frame of the emitting galaxy, the emitted spectrum is bright at wavelengths longer than 912 Å, but very dim or imperceptible at shorter wavelengths—this is known as a "dropout", or "break".

But the [wikipedia page](https://en.wikipedia.org/wiki/Lyman_series) for the Lyman series states that the highest limit of radiation absorbed or emitted by neutral hydrogen is 91.2 nm:

> the Lyman series is a ... series of transitions and resulting ... emission lines of the hydrogen atom as an electron goes from n ≥ 2 to n = 1 ... The greater the difference in the principal quantum numbers, the higher the energy of the electromagnetic emission.

It then states that there is an asymptotic limit to this energy as the difference between transition levels approaches infinity:

> There are infinitely many spectral lines, but they become very dense as they approach n = ∞ (the Lyman limit)... "91.1753 nm"

I don't understand how neutral hydrogen can absorb light from a photon emitted with wavelength shorter than 912 Angstroms if this wavelength is precisely the highest energy photon hydrogen can absorb.

So my question is: How can there exist a wide drop-out in the spectrum of galaxies at wavelengths shorter than about 91.2 nm, if the highest-energy electromagnetic radiation a hydrogen atom can emit or absorb (at n = ∞) is 91.1753 nm?

How is the hydrogen interacting with photons of higher-energies than this?

It must be a conceptual issue I'm not understanding. Should I not view absorption energies as conceptually the same as emission energies?

**Tags:** galaxy, spectroscopy, redshift, james-webb-space-telescope, hydrogen

> **Comments:**
> **username_1**: The “limit” takes an electron to be free of the (now) ion. The electron can, of course, absorb more energy and then has kinetic energy flying away from the ex-atom. The free electron levels are a continuum to an infinity.  
> **username_2**: Hi username_0, I initially gave an incompletely and perhaps misleading answer, but updated now. Hopefully that explains well, otherwise feel free to ask :)  

---

## 1 answers

### Answer by username_2 on 2022-12-17. Score: 4 | Accepted answer

### The Lyman-limit cross section

The Lyman limit is not a narrow line, like it is for electronic transitions. There is a minimum energy needed to ionize a hydrogen atom — 13.6 eV, corresponding to a wavelength of 912 Å — but higher energies are not a big problem, because — as username_1 comments — the excess energy instead goes into kinetic energy of the particles.

If the energy of the photon becomes too large, however, it will eventually be unable to ionize the atom. In fact, the probability of ionization — the cross section — decreases with the cube of the wavelength, i.e. $\phi(\lambda) \propto \lambda^3$. In the figure below I've sketched the absorption cross section where you can see that significant absorption actually extends all the way to (a few) 100 Å.

[![sigLL](https://i.sstatic.net/sDCAi.png)](https://i.sstatic.net/sDCAi.png)

### The intrinsic galaxy spectrum

But normal galaxies don't emit much light at such high energies. So the reason that the spectra of Lyman-break galaxies "stay" black blueward of the 912 Å Lyman limit is that, at wavelengths shorter than what may be effectively absorbed, there simply isn't much light emitted in the first place.

However, if you consider instead a galaxy with an [active galactic nucleus](https://en.wikipedia.org/wiki/Active_galactic_nucleus), the emission spectrum extends all the way to soft X-rays (and beyond). For these galaxies you do indeed see the transmitted intensity begin to rise again at short wavelengths.

### High-redshift galaxies

In my original answer I considered only galaxies at high redhifts (because those are the ones I'm used to thinking of). Here, another effect comes into play which tends to erase even very short wavelengths:

As light travels away from the galaxy, it is redshifted by the expansion of the Universe. Hence, after a little while light that had a too short wavelength will be redshifted closer to the Lyman limit, and if there happens to be neutral hydrogen around at this point (which invariably there is in the early Universe), then it will be absorbed after all.

### The Lyman alpha forest

This effect it more pronounced for Lyman alpha photons, i.e. the bound-bound transition from the ground state to the first excited state. For this line, only photons with energies very close to the "correct" energy (10.2 eV, or 1216 Å) are able to excite the hydrogen. But expansion eventually redshifts the more energetic part of the spectrum to 1216 Å, so every time there's a hydrogen cloud, this cloud causes a narrow dip in the spectrum.

For intermediate redshifts ($z\sim 1\text{–}6$), you will see a bunch of these lines, comprising the so-called "Lyman alpha forest". At high redshifts ($z\gtrsim6$), the Universe is so neutral that all these lines overlap, and no light is transmitted, resulting in the "Gunn-Peterson trough". At low redshifts ($z\lesssim1$), most of the spectrum is transmitted.

This evolution is seen in the figure below:

[![LAF](https://i.sstatic.net/pJfwv.jpg)](https://i.sstatic.net/pJfwv.jpg)

Credit: [Ross McLure](https://blackholeswithrevelations.wordpress.com/2013/09/19/stfcastro-lyman-alpha-forest/).

> **Comments:**
> **username_0**: Thanks @username_2, so the "cross-section" in your graph means the electron's orbital cross section? And this would correspond to it's energy? Or is this not what that means?  
> **username_2**: @username_0 Not really, actually: At the atomic level the concept of "cross section" is not that… _classical_. The cross section depends less on the physical size of the scattering agent — indeed _physical size_ loses its meaning — and more on quantum mechanical factors. You can think of the cross section as a "geometrical" way to express the _probability_ of a photon interacting with the particle ("geometrical" in the sense that we express it in units of area).  
> **username_2**: For some processes (e.g. scattering on electrons), the cross section is (almost) independent of the nature of the photon (wavelength, polarization, direction), but for most processes (e.g. electronic transitions between an atom's energy levels), the cross section is highly dependent on wavelength. This is very different from a regular "size" of an object. And in the case of ionization (which is the case in your question) there is a minimum energy required, so for larger wavelengths, the cross section drops to zero, even though you still have an associated "orbital cross section".  
> **username_2**: For instance, at the Lyman limit (where $\lambda=912\,\mathrm{Å}$), the cross section is $\sigma_0=6.3\times10^{-18}\,\mathrm{cm}^2$ (the maximum in my graph above). What this means is that, if you have a density of $n=1$ atom per $\mathrm{cm}^3$, then the mean free path $\ell=1/n\sigma_0$ of a Lyman limit photon is $1.6\times10^{17}\,\mathrm{cm}$, or 0.17 lightyears. But for a photon with $\lambda\simeq500\,\mathrm{Å}$, the cross section is ~6 times smaller, so the mean free path is 1 lightyear, even though you have exactly the same atoms.  
> **username_0**: That makes sense enough given that classical concepts lose meaning at atomic scales. Thanks.  
> **username_0**: Actually I need another clarification: So the Lyman break at the source (not redshifted) would pick up again at around 20-40 nm?   So even though the emission spectrum of a non-AGN galaxy significantly attenuates smaller than UV wavelengths, the remaining balack body radiation would make it out of the emitting galaxy?  
> **username_2**: @username_0 In principle yes, but a "normal" galaxy doesn't have many ways to produce such high-energy photons. Even the very hottest "O stars", which exist in few numbers, emit only a small fraction below 50 nm. And because these stars are short-lived, they also tend to be still enshrouded in the gaseous cloud from which they were born, so their photons experience a higher opacity than lower-mass stars.  
> **username_2**: In general, most ionizing photons (Lyman continuum photons, or LyC) are absorbed close to the stars, and are "converted" to Lyman series photons (roughly 2/3 of every LyC photon is converted to a Lyman alpha photon). The so-called LyC-leakers are observed not because of the tiny fraction that makes it through the enshrouding gas, but more likely because stellar feedback has cleared paths throughout the interstellar and circumgalactic medium. And still, they typically leak <10%.  
> **username_0**: thanks @username_2! I guess I hadn't realized just how little radiation beyond ultraviolet is actually produced.  
> **username_2**: @username_0 You're welcome :)  

---

