Calculate the time of the surface of last scattering

In summary, the Friedmann equations were used to calculate the energy density of the radiation field at the beginning of the universe. The observed 75 25 per cent hydrogen/helium ratio* implied that the temperature of the radiation must have been 10^9 K. After some 300,000 years, its temperature would have fallen to 10^3 K, cool enough for electrons to associate with protons to produce neutral hydrogen atoms.
  • #1
g.lemaitre
267
2

Homework Statement



By extrapolating the Fricdmann equations back in time, they discovered that the energy density of this radiation field must have initially considerably exceeded that of matter (as defined by E = mc^2). Radiation density is the amount of energy in a given volume of space, and it can be expressed as the temperature of the black body emitting the same energy. The observed 75 25 per cent hydrogen/helium ratio* implied that the temperature of the radiation must have been 10^9 K. Electromagnetic radiation readily interacts with electrically charged protons and electrons, so while the fireball was dominated by the radiation field the mean distance traveled by a photon was minuscule. As the Universe expanded, the radiation field would have cooled*. After some 300,000 years, its temperature would have fallen to 10^3 K, cool enough for electrons to associate with protons to produce neutral hydrogen atoms.


Homework Equations


The Attempt at a Solution



I would like to do this calculation. I don't even know where to begin since there are quite a lot of Friedmann equations out there. Which Friedman equation do I use and how do I use it?
 
Physics news on Phys.org
  • #2
I should also mention that I'm trying to figure out how many years elapsed from the BB to the surface of last scattering. I also want to figure out the temperature at certain times.
 
  • #3
This is the standard way: let's denote time today as [itex] t_0 [/itex] and time of last scattering as [itex] t_* [/itex], (and normalize so that [itex]a_0 = 1 [/itex])

[tex] t_0 - t_* = \int_{t_*}^{t_0} dt = \int_{a_*}^{a_0} da \frac{dt}{da} = \int_{a_*}^{1} \frac{da}{a H} [/tex]

Then plug in Hubble parameter from Friedmann equations and solve the integral. So for example, if you had a flat dust universe, then
[tex] H = H_0 a^{-3/2} [/tex]
and
[tex] t_0 - t_* = H_0^{-1} \int_{a_*}^{1} da \sqrt{a} = \frac{2}{3 H_0} (1 - a_*^{3/2}) [/tex]
 
  • #4
I don't know how to normalize. I'm assuming you learn that in QM. Is it easy or should I just wait until I finish QM or is there a youtube video that will show me how to do it easily.
 
  • #5
g.lemaitre said:
I don't know how to normalize. I'm assuming you learn that in QM. Is it easy or should I just wait until I finish QM or is there a youtube video that will show me how to do it easily.

You do it by saying that [itex] a_0 = 1 [/itex]. It is not mathematically very demanding :-) Basically, you are free to choose the value of scale factor on a single moment of time (this corresponds to a choice of units basically) and making the choice above is very reasonable, as it simplifies most formulas.
 
  • #6
Is all I have to do is plug the numbers into here and solve?

[tex] = \frac{2}{3 H_0} (1 - a_*^{3/2}) [/tex][/QUOTE]

I don't think the value for [tex]a_*[/tex] is known. For [tex]H_0 [/tex] I'm just going to use 22.2 (km/s)Mly^-1 which I think is the same as 71 (km/s)Mpc^-1.

Try to use less jargon with me when you're answering my question. My knowledge of math is a bit shaky.
 
  • #7
g.lemaitre said:
I don't think the value for [tex]a_*[/tex] is known.

What makes you think that? What do you know then?
 
  • #8
Well, in every equation you've given me there is at least one unknown, so I can't get the answer
 
  • #9
g.lemaitre said:
Well, in every equation you've given me there is at least one unknown, so I can't get the answer

Of course there is one unknown: this is the thing that determines when last scattering happened. In the original post you gave the temperature for it. I gave a formula which has the scale factor. They are related via [itex] \frac{T_1}{T_2} = \frac{a_2}{a_1} [/itex]
 

Related to Calculate the time of the surface of last scattering

1. How is the time of the surface of last scattering calculated?

The time of the surface of last scattering is calculated by measuring the cosmic microwave background radiation (CMB) and using the distance that the light has traveled to reach us from the moment of recombination, which occurred approximately 380,000 years after the Big Bang.

2. What is the significance of the time of the surface of last scattering?

The time of the surface of last scattering marks the moment in the history of the universe when the first atoms were able to form and light was able to travel freely. This moment is crucial in understanding the evolution of the universe and the formation of structures such as galaxies and stars.

3. How accurate is the calculation of the time of the surface of last scattering?

The calculation of the time of the surface of last scattering is considered to be very accurate, with a margin of error of only a few hundred years. This is due to the precision of modern technology and the abundance of data collected from various sources, such as satellites and ground-based telescopes.

4. Can the time of the surface of last scattering change?

No, the time of the surface of last scattering is a fixed point in the history of the universe and does not change. However, our understanding of this moment may evolve as new data and theories emerge.

5. What insights can we gain from studying the time of the surface of last scattering?

Studying the time of the surface of last scattering allows us to gain a better understanding of the early universe and its evolution. It also provides evidence for the Big Bang theory and helps us understand the formation of the cosmic microwave background radiation, which is a crucial piece of evidence for the Big Bang model.

Similar threads

Replies
1
Views
904
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
864
  • Advanced Physics Homework Help
Replies
1
Views
2K
Replies
1
Views
353
  • Thermodynamics
Replies
5
Views
725
  • Other Physics Topics
2
Replies
56
Views
5K
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
859
Replies
4
Views
1K
Back
Top