Saha equation (experiment verification?)

  • Thread starter ChrisVer
  • Start date
In summary: I don't know.In summary, the equation for a procedure a^+ + e \leftrightarrow a + \gamma is: \frac{1-X}{X^2} = \eta \frac{g_a}{g_{a+}g_e} \frac{4\sqrt{2} \zeta (3)}{\sqrt{\pi}} \Big( \frac{T}{m_e} \Big)^{3/2} \exp \Big[ \frac{E_{binding}(a)}{T} \Big] where X = \frac{n_{a+}}{n_B} and \eta =
  • #1
ChrisVer
Gold Member
3,378
464
Hi,
The Saha equation for a procedure [itex] a^+ + e \leftrightarrow a + \gamma [/itex] is:
[itex] \frac{1-X}{X^2} = \eta \frac{g_a}{g_{a+}g_e} \frac{4\sqrt{2} \zeta (3)}{\sqrt{\pi}} \Big( \frac{T}{m_e} \Big)^{3/2} \exp \Big[ \frac{E_{binding}(a)}{T} \Big] [/itex]

where [itex]X = \frac{n_{a+}}{n_B}[/itex] and [itex]\eta = \frac{n_B}{n_\gamma} [/itex]

I worked with this equation for [itex]\eta = 5.5 \times 10^{-10}[/itex] and for [itex]a=H[/itex] and [itex]a=^4He[/itex]. The results I got for [itex]\frac{1-X}{X^2}[/itex] are given the figure picture.
3.jpg

The above Helium line was taken by the assumption that baryonic matter consists only out of Helium at recombination time...

From the log(2) we can see the recombination temperature for both these cases.

I have one question though. Today the temperature is [itex]T \approx 2.725~K[/itex]. So the above line today is going to give me the fractional ionization [itex] X [/itex] for today, right? Is it possible to calculate this fraction though experimentally?
If not, what are the experimental informations you can take from such a diagram?
Also without knowing the value today, how can people determine either the recombination temperature or the fraction of [itex]\eta = n_B/n_\gamma[/itex]?
 
Space news on Phys.org
  • #2
ChrisVer said:
So the above line today is going to give me the fractional ionization X X for today, right?
Only if the universe is in thermal equilibrium. Do you expect this?
ChrisVer said:
If not, what are the experimental informations you can take from such a diagram?
The ionization fraction in the early universe.
ChrisVer said:
Also without knowing the value today, how can people determine either the recombination temperature or the fraction of ##\eta = n_B/n_\gamma##?
You have the densities now and can extrapolate that backwards to the early universe.
 
  • #3
So suppose that there are two hypothesis and you want to check them:
1. the baryon number during recombination consisted mainly of He
2. The baryon number during recombination consisted mainly of H
By the plot above, and measurements you can do today, can you verify either the Hyp#1 or Hyp#2?
 
  • #4
The ratio H/He today is nearly the same as during recombination, so it is easy to get that ratio.
To see which type was type did recombination later (=what produced the CMB), you can look at the plot in post 1.
 
  • Like
Likes ChrisVer
  • #5
So in fact, if baryonic number consisted mainly of Helium, the universe would have been transparent to the photons at higher temperatures (=earlier) and thus the radiation temperature from CMB would have been larger today? :)

What I am trying to see is in this:
http://www.helsinki.fi/~hkurkisu/cosmology/Cosmo6.pdf
page 71 in his numbering (or 11 in scrolling)
lowest paragraph
where he states that 90% should have been protons (hydrogen) and the rest Helium..
 
  • #6
The temperature at the (last) recombination would have been higher, but it would have happened earlier, so the universe would still have had time to cool down afterwards. I guess it would depend on the details like the actual helium density (do you convert all hydrogen to helium, or do you just remove hydrogen?).

ChrisVer said:
where he states that 90% should have been protons (hydrogen) and the rest Helium..
Yes, that's still what we see. 90% hydrogen atoms and 10% helium atoms correspond to roughly 75% hydrogen by mass.
 
  • #7
The one plot [for Hydrogen] was taken by considering what they state in the link I sent: there was no Helium (Helium neglected)...
The second plot [for Helium] was taken by considering again that there was only Helium (nothing else consisted the baryonic number).
So I guess I what I did was to remove the Hydrogen in the calculations.
I may also be wrong by trying to figure anything out of that graph...

mfb said:
The temperature at the (last) recombination would have been higher, but it would have happened earlier, so the universe would still have had time to cool down afterwards.

But if the universe became transparent at the recombination time, and that is the time we see in CMB, if it had became earlier we would see its "earlier" stage and not the later one (which occurred at Hydrogen). And so it would appear hotter.
 
  • #8
Hmm, those assumptions will influence η. You also get some inequality between number of helium atoms and number of electrons.
ChrisVer said:
But if the universe became transparent at the recombination time, and that is the time we see in CMB, if it had became earlier we would see its "earlier" stage and not the later one (which occurred at Hydrogen).
Right.
ChrisVer said:
And so it would appear hotter.
Why? It would have cooled down more. There is no reason to assume it would appear hotter.
 
  • Like
Likes ChrisVer
  • #9
mfb said:
Why? It would have cooled down more. There is no reason to assume it would appear hotter.

Oh yes, you are right... so CMB would appear cooler.
 
  • #10
Why cooler?

I'm not sure if it would change anything at all. The thermal energy was dominated by photons anyway, the recombination just changed that number by a tiny amount.
 
  • #11
hmm I guess I haven't really understood the mechanisms of temperature evolution then. But I think I solved my problem with it and you are right...the photon temperature almost always (with some small changes) scaled as 1/R even if CMB was produced by He or by H...

I have one further question, if you are able to help...
If the photons were too many combared to baryons and matter, as the [itex]\eta[/itex] parameter tells us, then why were they trapped into the primodial plasma? I know that plasma works as a trap or perfect reflector (by Compton or Ionization) for photons, but is that the case even if the photon number is over-dominating all the rest constituents?
The [itex]\eta \sim 10^{-9}[/itex] means that to 1 proton you have 1 billion photons... and the number of electrons=number of protons (neutrality) so to 1 proton and 1 electron you have 500 million photons... would all the 500million photons interact with these 2 particles in such a rate that they would get trapped? [or an expression for the interaction rate]
 
Last edited:
  • #12
What do you mean with trapped? There was plasma everywhere, how could you escape something that is everywhere?
 

FAQ: Saha equation (experiment verification?)

1. What is the Saha equation?

The Saha equation is a mathematical formula that relates the ionization state of an element to its temperature and pressure. It was developed by Indian physicist Meghnad Saha in 1920 and is commonly used in astrophysics and plasma physics.

2. How is the Saha equation derived?

The Saha equation is derived from the principles of statistical mechanics and quantum mechanics. It takes into account the ionization energy of an element, as well as the temperature and pressure of the gas in which it is present.

3. What is the significance of the Saha equation in astrophysics?

The Saha equation is essential in understanding the physical processes that occur in stars. It helps to determine the ionization state of elements in stellar atmospheres and plays a key role in predicting the spectra of stars.

4. How is the Saha equation experimentally verified?

The Saha equation has been experimentally verified through spectroscopic observations of stars, including the Sun. By analyzing the spectra of stars, scientists can determine the ionization state of different elements and compare it to the predicted values from the Saha equation.

5. Are there any limitations to the Saha equation?

While the Saha equation is a useful tool in astrophysics, it does have its limitations. It assumes that the gas is in thermal equilibrium and that the atoms and ions are in ideal conditions. It also does not take into account any interactions between particles, which can affect the accuracy of the results.

Similar threads

Replies
0
Views
560
Replies
1
Views
919
Replies
1
Views
999
Replies
9
Views
2K
Replies
22
Views
2K
Replies
8
Views
2K
Replies
4
Views
2K
Back
Top