<p>As a small addendum to Rob Jeffries' excellent answer, I'll note that you probably can't get a scenario of "H-beta, but no H-alpha" for emission lines, either. The emission lines are the reverse of the absorption lines, so to have H-beta emission without H-alpha, you'd have to have $n = 4$ to $n = 2$ transitions <em>without</em> also having $n = 3$ to $n = 2$ transitions. But even if you could somehow magically arrange to have all your H atoms in the $n = 4$ level, some of them would transition to $n = 3$, and then to $n = 2$, so you really can't avoid H-alpha emission.</p>

<p>Furthermore, dust extinction is worse at shorter wavelengths, so dust extinction could in principle give you the appearance of H-alpha emission without H-beta emission, but not the reverse.</p>

CC BY-SA 3.0
2015-07-07T20:57:03.010Z
2015-07-07T20:57:03.010Z
Peter Erwin
<p>In a normal stellar atmosphere, where the temperature decreases with height, I think this is not possible.</p>

<p>H-alpha absorption arises from the $n=2$ to $n=3$ transition; H-beta from the $n=2$ to $n=4$ transition. Thus both transitions are governed by the number of atoms in the $n=2$ lower level.</p>

<p>The absorption coefficient for a transition can be written in proportionality terms by (in local thermodynamic equilibrium)
$$ \alpha_\nu \propto  \nu B_{lu} n_l ( 1 - \exp[-h\nu/kT]),$$
where $B_{ul}$ is the Einstein absorption coefficient, $n_l$ is the population of the lower energy level and $\nu$ is the photon frequency corresponding to the transition.</p>

<p>Using the well-known relationship between the Einstein emission and absorption coefficients and values for the emission coefficients found <a href="http://www.physics.byu.edu/faculty/christensen/Physics%20612/FTI/H%20Einstein%20Coefficients.htm" rel="nofollow">here</a>, we can estimate a ratio of absorption coefficients for a given temperature.</p>

<p>$$ \frac{\alpha_{H\beta}}{\alpha_{H\alpha}} = \frac{g_{n=4}}{g_{n=3}} \frac{A_{H\beta}}{A_{H\alpha}} \left(\frac{\nu_{\alpha}}{\nu_{\beta}}\right)^2 \left(\frac{1 - \exp(-h\nu_{\beta}/kT)}{1-\exp(-h\nu_{\alpha}/kT}\right),$$
where $A_X$ are the Einstein emission coefficients and the statistical weights $g_n$ are given by $2n^2$.</p>

<p>Thus if we take $\nu_\alpha = 4.57\times10^{14}$ Hs, $\nu_{\beta}=6.17\times10^{14}$ Hz, $A_{H\alpha} \simeq 10^{8}$ s$^{-1}$, $A_{H\beta} \simeq 3\times 10^{7}$ s$^{-1}$, then
$$\frac{\alpha_{H\beta}}{\alpha_{H\alpha}} = 0.29 \left( \frac{1 - \exp(-29642/T)}{1 - \exp(-21956/T)}\right)$$</p>

<p>So when $T$ (in Kelvin) is small, the ratio is about 0.3. When $T$ becomes large then the ratio increases, but above $T \sim 12,000$ K, all the Hydrogen is ionised.</p>

<p>Thus in thermodynamic equilibrium is it very difficult to manufacture a circumstance where the optical depth in the H$\beta$ line is larger than that in the H$\alpha$ line.</p>

<p>[I'd be grateful if someone can tell me whether the ratio I calculated above is correct, since I could not find it in any reference and need to work it out from scratch].</p>

CC BY-SA 3.0
2015-07-07T09:54:41.573Z
scratch? That is awesome job!
questionhang
CC BY-SA 3.0
2015-07-07T12:43:58.203Z
I revised the title. It is not necessarily a stellar spectrum.
questionhang
CC BY-SA 3.0
2015-07-07T09:48:07.483Z
2015-07-07T09:48:07.483Z
ProfRob
<p>Generally there may be some line ratio between the two lines.</p>

<p>If  H$\alpha$ is strong, it is quite possible we can see H$_\beta$ too.</p>

<p>Is it possible we can see H$_\beta$, but can not see H$_\alpha$? In what kind of environments can it happen?</p>

<star><spectra><h-alpha>
about H$_\alpha$ and H$_\beta$ lines in astronomical objects' spectra
questionhang
# about H$_\alpha$ and H$_\beta$ lines in astronomical objects' spectra

(Asked by: username_0 on 2015-07-07)

Generally there may be some line ratio between the two lines.

If H$\alpha$ is strong, it is quite possible we can see H$\_\beta$ too.

Is it possible we can see H$\_\beta$, but can not see H$\_\alpha$? In what kind of environments can it happen?

**Tags:** star, spectra, h-alpha


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## 2 answers

### Answer by username_1 on 2015-07-07. Score: 2

As a small addendum to Rob Jeffries' excellent answer, I'll note that you probably can't get a scenario of "H-beta, but no H-alpha" for emission lines, either. The emission lines are the reverse of the absorption lines, so to have H-beta emission without H-alpha, you'd have to have $n = 4$ to $n = 2$ transitions *without* also having $n = 3$ to $n = 2$ transitions. But even if you could somehow magically arrange to have all your H atoms in the $n = 4$ level, some of them would transition to $n = 3$, and then to $n = 2$, so you really can't avoid H-alpha emission.

Furthermore, dust extinction is worse at shorter wavelengths, so dust extinction could in principle give you the appearance of H-alpha emission without H-beta emission, but not the reverse.


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### Answer by username_2 on 2015-07-07. Score: 2

In a normal stellar atmosphere, where the temperature decreases with height, I think this is not possible.

H-alpha absorption arises from the $n=2$ to $n=3$ transition; H-beta from the $n=2$ to $n=4$ transition. Thus both transitions are governed by the number of atoms in the $n=2$ lower level.

The absorption coefficient for a transition can be written in proportionality terms by (in local thermodynamic equilibrium)
$$ \alpha\_\nu \propto \nu B\_{lu} n\_l ( 1 - \exp[-h\nu/kT]),$$
where $B\_{ul}$ is the Einstein absorption coefficient, $n\_l$ is the population of the lower energy level and $\nu$ is the photon frequency corresponding to the transition.

Using the well-known relationship between the Einstein emission and absorption coefficients and values for the emission coefficients found [here](http://www.physics.byu.edu/faculty/christensen/Physics%20612/FTI/H%20Einstein%20Coefficients.htm), we can estimate a ratio of absorption coefficients for a given temperature.

$$ \frac{\alpha\_{H\beta}}{\alpha\_{H\alpha}} = \frac{g\_{n=4}}{g\_{n=3}} \frac{A\_{H\beta}}{A\_{H\alpha}} \left(\frac{\nu\_{\alpha}}{\nu\_{\beta}}\right)^2 \left(\frac{1 - \exp(-h\nu\_{\beta}/kT)}{1-\exp(-h\nu\_{\alpha}/kT}\right),$$
where $A\_X$ are the Einstein emission coefficients and the statistical weights $g\_n$ are given by $2n^2$.

Thus if we take $\nu\_\alpha = 4.57\times10^{14}$ Hs, $\nu\_{\beta}=6.17\times10^{14}$ Hz, $A\_{H\alpha} \simeq 10^{8}$ s$^{-1}$, $A\_{H\beta} \simeq 3\times 10^{7}$ s$^{-1}$, then
$$\frac{\alpha\_{H\beta}}{\alpha\_{H\alpha}} = 0.29 \left( \frac{1 - \exp(-29642/T)}{1 - \exp(-21956/T)}\right)$$

So when $T$ (in Kelvin) is small, the ratio is about 0.3. When $T$ becomes large then the ratio increases, but above $T \sim 12,000$ K, all the Hydrogen is ionised.

Thus in thermodynamic equilibrium is it very difficult to manufacture a circumstance where the optical depth in the H$\beta$ line is larger than that in the H$\alpha$ line.

[I'd be grateful if someone can tell me whether the ratio I calculated above is correct, since I could not find it in any reference and need to work it out from scratch].

> **Comments:**
> **username_0**: scratch? That is awesome job!  
> **username_0**: I revised the title. It is not necessarily a stellar spectrum.  

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