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To quote Wikipedia:

Balmer Jump is caused by electrons being completely ionized directly from the second energy level of a hydrogen atom (bound-free absorption), which creates a continuum absorption at wavelengths shorter than 364.5 nm.

Based on the graphs below we used in class, the non-grey energy distribution of the energy flux is continuously lower than the grey case for wavelengths shorter than $364.5nm$, which according to Wikipedia is due to the bound-free absorptions of electrons in the $n=2$ energy state.

I understand why the energy flux gets lower for those wavelengths but I still cannot grasp why the discontinuity occurs. Is it because for wavelengths higher than $365.4nm$, the bound-free absorption of electrons of $n=3$ starts? And do bound-bound absorptions not occur at all for stars with these discontinuities?

I would be very thankful for any clarification on this subject.

enter image description here

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  • $\begingroup$ I wonder if there is any relationship to Why does titanium oxide around Betelgeuse produce this particular sawtooth-shaped absorption spectrum? $\endgroup$
    – uhoh
    Jan 30, 2023 at 20:46
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    $\begingroup$ Thank you for your comment. Indeed the two spectra do have similar shapes, but reading the comments of the post mentioned it looks like the Betelgeuse's spectrum is due to the TiO behavior as a heavy metal, whereas the Balmer Jump is supposedly due to the bound-free transitions of n=2 Hydrogen electrons. Thank you for the reference though, I will try to dig deeper that way as well. $\endgroup$
    – Hika
    Feb 1, 2023 at 16:43

2 Answers 2

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The Balmer break comes from a combination of two main things: the ability of photons with high enough energies (wavelengths shorter than 364.5 nm) to ionize hydrogen atoms that are in the $n = 2$ energy level, and the inability of any photons with lower energies (wavelengths longer than 364.5 nm) to ionize the atoms. This leads to an abrupt change in absorption of outgoing photons, from high absorption at wavelengths shorter than 364.5 nm, to no absorption at longer wavelengths.

Another important factor: Any photon with a wavelength shorter than 364.5 nm can ionize an $n = 2$ hydrogen atom -- but the probability (due to the cross-section of the photon-atom interaction) goes down as the wavelength gets shorter. That's why you get the sawtooth pattern: very-short-wavelength photons can ionize the atoms, but are unlikely to. As the wavelengths get longer, the probability of ionizing a photon increases, up to the point of 364.5 nm. Once the wavelength is longer than that, ionization is simply impossible, so the absorption goes to zero.

In practice, the total absorption doesn't go to zero, because there are other sources of absorption, including hydrogen atoms in the higher states, like the $n = 3$ state (which has its own "Paschen jump"), and things like H$^{-}$ hydrogen ions. You also have to think about things like how many atoms are in different energy levels. In a very hot atmosphere, most of the hydrogen atoms will be in higher energy levels (or even ionized), so you won't get much of a Balmer break.

do bound-bound absorptions not occur at all for stars with these discontinuities?

They absolutely do occur -- they produce absorption lines, such as those of the Balmer series (which are not shown in the simplified, cartoon spectra in your figure). If you have enough hydrogen atoms in the $n = 2$ state to produce a noticeable Balmer break, then you'll automatically have Balmer absorption lines (H$\alpha$, H$\beta$, etc.) due to longer-wavelength photons being absorbed by the same atoms.

These lines do get closer together as you approach 364.5 nm from the long-wavelength side, which has the effect of making the peak of the sawtooth a bit rounded, but they don't produce the jump -- they make it a little bit more gradual. (Your figures do not show the effects of the absorption lines.)

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  • $\begingroup$ Wow, this makes so much sense to me. Thank you very much for this detailed explanation. $\endgroup$
    – Hika
    Feb 3, 2023 at 5:31
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I don't quite understand your plot (does "grey" mean "blackbody"?), but I think you've stated what's going on pretty clearly (I'm reading Fundamental Astronomy by Karttumen et al to make sure).

The Balmer series starts with n=2 to n=3 at 656 nm, and as the final energy level increases, the lines get closer together (I found this nifty calculator, which gives answers that look right).

$$n_f=5\qquad \lambda=434~nm$$ $$n_f=6\qquad \lambda=410~nm$$ $$n_f=7\qquad \lambda=397~nm$$ $$...$$ $$n_f=50,\qquad \lambda=365.091~nm$$ $$n_f=51,\qquad \lambda =365.068~nm$$

And so forth, until the free state at 364.7 nm. Visually, this looks like some kind of continuum absorption, since the lines are not resolvable. But after that, you don't hit the next absorption line (Lyman) until 121.6 nm (that's n=1 to n=2). So calling that a jump (or discontinuity) is really more about the distribution of the lines - they go from being very close together to being regularly spaced apart again. Presumably we could also define "Lyman jump" at 91.2 nm, and Pachsen jump at 821 nm, etc.

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  • $\begingroup$ This isn't really correct. The "jump" is an increase in absorption at wavelengths shorter than 364.7 nm, due to photoionization of $n = 2$ atoms. The crowding together of individual absorption lines with higher and higher values of $n_{f}$ modifies the shape of the jump immediately longward of 364.7 nm, but that's all. $\endgroup$ Feb 1, 2023 at 20:21
  • $\begingroup$ @PeterErwin: Ah yeah thanks, that's closer to what the OP was asking, that explains his plots. The description in Karttuman et al is more about the crowding of the lines, but yeah, once you can ionize from the n=2 state, you'll get excess absorption because of it. $\endgroup$
    – cduston
    Feb 1, 2023 at 22:06

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