Time of last scattering - what was it like

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Hello everyone

I've a question about the "time of the last scattering". My understanding is that this took place about 380,000 years after the big bang, and occured when the universe cooled sufficiently to allow protons and electrons to form neutral hydrogens atoms. As a result, light was able to move freely throughout the universe.

My question is, what would this have looked like to someone who was alive at the time who could see light in the frequencies which were emitted? Prior to the l;ast scattering, would they have been able to see a neutral fog, and would there have been an instanteous flash of light across the cosmos as the scattering occured? Or would it all have been less dramatic than this?

As always, please pardon the 'beginners' nature of this question.

Many thanks

Chris
 

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  • #2
marcus
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hot gas, orange-ish light, not an abrupt change

3000 kelvin is the temperature at the surface of a main-sequence star somewhat less massive and hence cooler than the sun

it is not totally opaque, not a sharp barrier. You can see some miles into the star but ultimately your line of sight is blocked by the socalled "optical density" of hydrogen gas at 3000 K.

There is a quantity called the "mean free path" that can be calculated if you know the density and the temperature. The temperature determines the percent ionization.
Then the density determines the abundance of ionized charged particles, which are the main light-scatterers.

I have to go, but it is interesting to think about this. Will get back later.

Back now, notice that expansion has a double effect, it reduces the gas density and it also recduces the temperature which lowers the percent ionization.

there is a ( I would call it) simple but beautiful subtlety here: you don't have to wait for complete transparency for the photons flying free to start.
You only have to wait until the mean free path or line of sight is long enough so that
by the time the photon has gone that far the universe will have expanded some more, enough to give it a new lease on life.
And on and on.
To put it very roughly, you only have to wait until the expected distance a photon can travel is increasing at the rate of one lightyear per year. Then it is the era or epoch of the last scattering because a whole bunch of photons can get started and many of them will make it and become our CMB, our ancient light, 13.7 billion years later.
But it doesn't happen abruptly, in a flash, the way you were asking about
 
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  • #3
Chalnoth
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Hello everyone

I've a question about the "time of the last scattering". My understanding is that this took place about 380,000 years after the big bang, and occured when the universe cooled sufficiently to allow protons and electrons to form neutral hydrogens atoms. As a result, light was able to move freely throughout the universe.

My question is, what would this have looked like to someone who was alive at the time who could see light in the frequencies which were emitted? Prior to the l;ast scattering, would they have been able to see a neutral fog, and would there have been an instanteous flash of light across the cosmos as the scattering occured? Or would it all have been less dramatic than this?

As always, please pardon the 'beginners' nature of this question.

Many thanks

Chris
Well, at the transition, you wouldn't actually be able to tell the difference. You would suddenly start seeing photons from much further away, but without very sensitive instruments, you wouldn't be able to tell.

The reason why is because at that time, the universe was uniform to one part in one hundred thousand. Yes, it is the case that as the plasma turned into a gas, you'd be able to see further and further, but because everything was so extremely uniform, it wouldn't matter whether a photon had come from one centimeter or a light year away: it would still be the same temperature (to within 0.001%).
 
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Chronos
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Much like today, you would be uniformly surrounded by the plasma in every direction. It would be pretty confusing. It would appear to be an impenetrable, bright fog. The optical depth would be virtually zero until it cooled sufficiently to allow more distant photons to reach you.
 
  • #5
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Thanks! so quite literally then a dense bright orange fog which eventually clears.


I suspect that even here we would still have the same questions regarding the size of the complete universe rather than what is to become the observable universe, and also what is happening at some boundary assuming space isnt truely infinite or doesnt roll back in on itself through curved space time?
 
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marcus
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Thanks! so quite literally then a dense bright orange fog which eventually clears.
...
I wouldn't say "literally". the image of fog clearing is more figurative/poetical.

the process of clearing presumably went on over 10s and 100s of years

If you could go back to that time the space around you would not look foggy.

Only a small percentage of H atoms ionize at 3000 K. In effect you would be in a very sparse gas of neutral hydrogen which would seem nearly transparent to you.
Visibility would be good, like on a clear day in earth atmosphere (which is not perfectly transparent either.)

There is a characteristic color of things at 3000K, I don't know how to describe it. How does it look? Like the light from a 100 watt lightbulb I guess. You might call it white but then when you compare it with more perfect daylight white you realize it is slightly orange-ish. That light surrounds you.

If you are going to stick around for a while, you must be in a temporary body made of tungsten, like the flilament of a lightbulb that glows at 3000 K. The light is intense. In that environment your body will eventually get as hot as the filament of a lightbulb.

The density of the hydrogen gas around you is much less dense than the earth's atmosphere at sealevel. We would think of it as a near vacuum.

Only about a quarter of a billion H atoms per cubic meter.

We can be more accurate. Today the ordinary matter in the U is equiv to about 0.2 hydrogen atoms per cubic meter. Back then distances were shorter by a factor of 1100, so cube that to get the factor for volumes. it will be around 1.33 billion. So multiply by 0.2 and get 266 million.

Have to go, back later.
Back now...

So in terms of density, mass per cubic meter. The hot hydrogen around you during that era when the U was slowly becoming transparent, when distances were less by a factor of 1100....
The sealevel atmosphere around you now is about one kilogram per cubic meter, and that gas was, during that era, roughly a billion billion times less dense.
1018 times less dense than the air around you right now. The Usa custom is to call that number a "quintillion".

I guess the words that come to my mind when I think of that is "vacuum furnace".
High vacuum inside, but hot as the tungsten filament of a lightbulb. Just barely ionized, just a very small fraction of the atoms affected, at that temp.

For people who like to calculate, the ordinary matter density of the U today is about
3.788 x 10-28 kg/m3 keeping some unwanted precision to round off later

and that means back at redshift 1100 era the ordinary matter density would have been

5.04 x 10-19 kg/m3

that's why I said on the order of a quintillionth of the density of sealevel air.
 
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I (think!) it was Brian Greene who said something I found interesting (and sort of related to this) but I can't remember exactly what it was. Anyone know what I'm talking about, and could they elaborate a bit?

Basically he was talking about if you were able to observe the universe at a certain, very early time the temperature which would be staggering would actually not be harmful at all. I cannot remember what prevents it from being harmful though. But he remarked that an abundance photons would actually kill you. It may actually have been Brian Clegg too, hum not sure.
 
  • #8
marcus
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the gas temperature wouldn't hurt you, the light temperature would. You can call it the temperature of the radiant heat, or the "abundance of photons". their energy and their numerical abundance. whatever you want. It is the temperature of the surrounding light.

The reason the gas temperature wouldn't hurt you, e.g. by contact, conduction to your protective suit etc., is that the gas is too rarified. It is like a vacuum. Vacuums do not conduct heat.

But light has a temperature. Being in thermal radiation of 3000K is like being in a furnace. Your body comes to equilibrium with the surrounding photons, you heat up to 3000K, and you are toast.

That is why I suggested a temporary tungsten body.

Don't know what Brian G or C were saying, probably something with the same basic physics message as this.
 
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  • #9
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Yes, I believe that is exactly it - thank you. :) This brings me back up to speed and now I can interactively ask what I was wondering when I originally read that. Okay so bear with me I don't have any science background, the gas composing the universe at that time is rarified, or very very low pressure? Almost like a vacuum? How similar is this to low density? I don't think this is entirely correct, but what I am picturing is very very little gas per volume (low density) causing low pressure, and thus the gas which is still hot, is essentially not plentiful enough to have any effect on you? But my understanding is that the matter density was very high back then.

Whereas the photons would be bombarding you, and would be very hot. Does this suggest that at an earlier phase when the universe was incredibly hot, it was only the photons that carried this heat? Thermal cavity radiation is black body radiation, no? I will definitely research it further, as I have almost no understanding of it but think it would greatly clarify this for me. Is it very loosely speaking related to the measure of how energetic a photon is (it's wavelength) equating to it's temperature? Would that mean that these photons that'd be cooking you would be gamma radiation? In which case I just mistakenly thought visible light when reading light which I know I shouldn't do.

Or lastly does it mean that they may indeed be within the visible light spectrum, however - again the 'abundance' of them would cause you to receive much higher energy/heat (enough to kill you) compared with say if you receive visible light spectrum photons on Earth today as you are not receiving even a fraction of the amount.
 
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Oh wow, disregard several parts of my last post. I missed your most recent post (before the one where you responded to me) and saw that it answered my question about, for instance yes it does imply low density.

Can you elaborate slightly on "Just barely ionized, just a very small fraction of the atoms affected, at that temp." I have trouble following this bit.
 
  • #11
marcus
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Can you elaborate slightly on "Just barely ionized, just a very small fraction of the atoms affected, at that temp." I have trouble following this bit.
I will elaborate slightly and maybe someone else can link us to a table that gives the exact percentage ionization of hydrogen gas at 3000 K.

For an intuitive feel, I just google "ionization potential"
There is an ionization potential table that says an H atom needs this amount of energy to ionize: 13.6 eV.

http://www.standnes.no/chemix/periodictable/ionization-potential-table.htm [Broken]

Then you need to be able to convert temperature into energy (electronVolts). Very crudely it is a factor of 10000---that is 10 thousand. So very roughly 3000 K corresponds to 0.3 eV.

So the intuitive idea is that energy is distributed to the atoms in a kind of BELL-CURVE, a sort of lopsided not-quite bell-curve. And most of the atoms get only about 0.3 eV, but a small minority way out in the high energy "tail" of the bell-curve get a big kick of energy now and then and they ionize.

A good thing to do is look at the black body spectrum of the thermal glow at 3000K. The thermal radiation spectrum plotted by photon energy. Most of the photons in the thermal light will have energy around 0.3 eV, but there will be a high energy "tail" and a few photons will have energy of 13.6 eV or more, and they will be able to ionize the H atoms that they hit.

Maybe you can google "black body spectrum" or "thermal radiation" and get a picture of the Planck curve. The lopsided bellshape.
 
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  • #12
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Something I have never really been able to grasp is, if I make a very, VERY large isolated tank and fill it with about a quarter of a billion H atoms per cubic meter in random brownian motion. What is the process by which the gas will clump together, or accumulate, to form a gas ball, eventually becoming large enough to form a star? Wouldnt the gas always want to even out to equal pressure?
 
  • #13
Chalnoth
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Something I have never really been able to grasp is, if I make a very, VERY large isolated tank and fill it with about a quarter of a billion H atoms per cubic meter in random brownian motion. What is the process by which the gas will clump together, or accumulate, to form a gas ball, eventually becoming large enough to form a star? Wouldnt the gas always want to even out to equal pressure?
Well, if you had a closed tank, it wouldn't actually work. You need the tank to be open.

That said, there are two effects at work here. The first is the one that normal thermodynamic calculations don't take into account: gravity. While you can do normal thermodynamics in the presence of an external gravitational field, that approach doesn't work when you take into account the gravitational forces between all of the atoms in the gas. This purely attractive forces throws a massive wrench into the whole system, so much so that nobody knows how to do the entropy properly in that case.

However, fluid dynamics still works just fine in the presence of self gravity. But in the presence of self gravity, you need something else to happen as well: if you imagine one single atom in orbit around some center of mass of the collection, it will remain in that exact same orbit perpetually, unless it loses energy somehow. In practice, the way this works is that gas clouds typically have temperatures much, much larger than the background, so that they emit energy in the form of thermal radiation. As the gas cloud emits energy, it collapses in on itself. This inward collapse, in turn, increases the temperature (self-gravitating systems have negative heat capacity: an outflow of heat leads to an increase in temperature), which makes the system radiate more, which makes it collapse inward even faster.

For a star, this process continues until nuclear fusion is ignited in the core of the star, which causes a tremendous input of additional heat, which stops the collapse.
 
  • #14
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Thanks Chalnoth!
 

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