Equilibrium point of matter and radiation density

  • #1
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Summary:

Radiation and Matter equilibrium point and expansion rate calculation

Main Question or Discussion Point

We want to calculate the ao/a(teq) of the equilibrium point between ρm and ρrm= ρr )

My book solves it this way;

ρm(t) / ρr(t)= a(t) ⇒
⇒ (ρm/ ρr)teq =1 =
= (ρm/ ρr)o * a(teq)/ ao

I don't understand the a(teq)/ ao part. If ρm(t)= ροo3 and ρr(t)= ροo4 then it should be
ρm(t)/ ρr(t) = ρom/ ρor * ao

After that it continues like;
ρmr = Ωmr
and it goes ao/ a(teq)= 1 + zeq =
=(Ωmr)o

what does the zeq mean and how did that come up from the equation?

(2)Also, there's an equilibrium point for dark energy and radiation density as well and a similar problem like above. Does dark energy have a known ω value? Like the ω=1/3 for radiation. I was looking for its value and couldn't find it, so I thought we maybe don't know it because it's dark energy?
 

Answers and Replies

  • #2
Orodruin
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Summary: Radiation and Matter equilibrium point and expansion rate calculation

If ρm(t)= ρο/αo3 and ρr(t)= ρο/αo4
This cannot be true. The LHS depends on t and the RHS does not. Your RHS should contain the scale factor at time t (and assumes that the scale fsctor at the present time is normalised to one).
 
  • #3
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This cannot be true. The LHS depends on t and the RHS does not. Your RHS should contain the scale factor at time t (and assumes that the scale fsctor at the present time is normalised to one).
Ok then ρm(t)= ροmo(t)3 and likewise for the ρr(t). Still doesn't explain why the equation turns this way.
Also what zeq is.
 
  • #4
George Jones
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Ok then ρm(t)= ροmo(t)3 and likewise for the ρr(t). Still doesn't explain why the equation turns this way.
What do you mean by ##a_0 \left(t\right)##? Usually, ##a_0## is defined by ##a_0 = a \left(t_0\right)##, where ##t_0## is the time "now".

Also what zeq is.
##z## is is often called "redshift", and is defined as the relative change factor, i.e., ##z## is defined by

$$z=\frac{\Delta a }{a} = \frac{a_0 - a \left(t\right)}{a \left(t\right)} = \frac{a_0}{a \left(t\right)} - 1.$$

##z_{eq}## is the value of ##z## when ##t = t_{eq}##.
 
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  • #5
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What do you mean by ##a_0 \left(t\right)##? Usually, ##a_0## is defined by ##a_0 = a \left(t_0\right)##, where ##t_0## is the time "now".
Yes, that's what I mean. The a in time now, just writing as ao instead of a(to) because that's the symbol we also use in class
 
  • #6
Orodruin
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Yes, that's what I mean. The a in time now, just writing as ao instead of a(to) because that's the symbol we also use in class
This in no way explains if you have gotten things correctly. The point was that it is not ##a_0## that should go into the equations you wrote down.
 
  • #7
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can I say a(t)= a(teq)/ao ? Doing the calculations this is what is missing for the equation mathematically, but not even sure if that's right to say about the expansion rates.
I feel like I'm missing a property of how the different times in the expansion rate work. Like, are there three different times in here (t, teq, to) or two where t=teq and to.
 
  • #8
PeterDonis
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can I say a(t)= a(teq)/ao ?
That makes no sense.

##a(t)## is a function: it is the scale factor as a function of time.

##a_{teq}## and ##a_0## are numbers: they are the values of the function at particular times, ##t_{eq}## and now.

You can't possibly get back a complete function by taking a ratio of two of its values at particular points.
 
  • #9
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Summary: Radiation and Matter equilibrium point and expansion rate calculation

Also, there's an equilibrium point for dark energy and radiation density as well and a similar problem like above.
Hi,
The equations you posted are derived from the parametrized Friedmann equations. The Friedmann equations are here: Friedmann and the parametrized equations are here: Parametrized .
They are parametrized in terms of the present values of the density parameters of the universe.

So going through the calculations of the second link and also from the following paper:Hogg, we have the following:

* Matter-Radiation Equality = 3372.8252559496009308896965250168609930765452393629
* CMB (decoupling) = 1091.3773947272224484356389171311764619349297783378
* Lambda-Radiation (Reionization) = 7.5799342318325121165504399316399123652696055240218
* Lambda-Matter Equality = 0.44694551856761346726843414251672002564496590529955

The answer is the redshift of those events. At this event (matter-radiation) the universe was 51825 years old, the Hubble constant was 10,422,612 km/sec/Mpc, the matter density was 0.501, radiation density was 0.498

Another Wiki article on the derivation of the parametrized equation here: Hubble Parameter
 
  • #10
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Hi,
The equations you posted are derived from the parametrized Friedmann equations. The Friedmann equations are here: Friedmann and the parametrized equations are here: Parametrized .
They are parametrized in terms of the present values of the density parameters of the universe.

So going through the calculations of the second link and also from the following paper:Hogg, we have the following:

* Matter-Radiation Equality = 3372.8252559496009308896965250168609930765452393629
* CMB (decoupling) = 1091.3773947272224484356389171311764619349297783378
* Lambda-Radiation (Reionization) = 7.5799342318325121165504399316399123652696055240218
* Lambda-Matter Equality = 0.44694551856761346726843414251672002564496590529955

The answer is the redshift of those events. At this event (matter-radiation) the universe was 51825 years old, the Hubble constant was 10,422,612 km/sec/Mpc, the matter density was 0.501, radiation density was 0.498

Another Wiki article on the derivation of the parametrized equation here: Hubble Parameter
Thanks. I found out how the redshift zeq came up from the a, but I still don't understand why
(ρm/ ρr)teq = (ρm/ ρr)o * a(teq)/ ao

Anyone has any idea? What's the issue with the densities and the a(teq)/ ao ?

If we say ρm(t)/ ρr(t)= ρom * ao4 / ρor * ao3 then we get
ρm(t)/ ρr(t)= ρom* ao/ ρor
and not ρm(t)/ ρr(t)= ρom* a(t)/ ρor * ao which is the correct answer
 
  • #11
Orodruin
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If we say ρm(t)/ ρr(t)= ρom * ao4 / ρor * ao3
Again. We do not say this. It is wrong and unreasonable. Stop claiming this.

The answer is already in your first post.
 
  • #12
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To reach equality: we just need to take matter density and divide by radiation density, but for the time or in my calculation case, the redshift to which this is relevant we need to include the scale factor which is related to the hubble constant. a is the function a(t) for time.

Btw, you need to read up on the book's nomenclature.
 
  • #13
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Again. We do not say this. It is wrong and unreasonable. Stop claiming this.

The answer is already in your first post.
You're referring to the fact that I didn't write it like ρr=ao(t)4? I corrected it, so no need to mention it again. I know a is time dependent. Other than pointing it out again, I know the answer is in my op. I just don't understand how it came up.
 
  • #14
Orodruin
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You're referring to the fact that I didn't write it like ρr=ao(t)4?
This is also wrong. In fact, it is a contradiction in terms as pointed out in post #8: ##a_0## is a number defined as ##a_0 = a(t_0)## where ##t_0## is the current time, it does not depend on ##t## and therefore writing ##a_0(t)## is meaningless.
 
  • #15
PeterDonis
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I just don't understand how it came up.
That's because you keep on confusing yourself with wrong notation, and switching from one wrong notation to another into the bargain.

Let's restate what you say your book says:

$$
\frac{\rho_m(t)}{\rho_r(t)} = a(t)
$$

The first thing to note is that, as written, this is incorrect (which makes me think you are incorrectly quoting your book). To get a correct equation, we make use of the following equations for the functions ##\rho_m(t)## and ##\rho_r(t)##:

$$
\rho_m(t) = \frac{\left( \rho_m \right)_0}{a(t)^3}
$$

$$
\rho_r(t) = \frac{\left( \rho_r \right)_0}{a(t)^4}
$$

Taking the ratio of these two functions gives:

$$
\frac{\rho_m(t)}{\rho_r(t)} = a(t) \frac{\left( \rho_m \right)_0}{\left( \rho_r \right)_0}
$$

Notice the extra ratio of the values of the densities "now" on the RHS. Notice also that this equation shows that we are using the common convention where the value of the scale factor "now" is ##1##, i.e., ##a_0 = a(t_0) = 1##.

Now we just look at what happens if we plug ##t_{eq}## into the equation above:

$$
\frac{\left( \rho_m \right)_{eq}}{\left( \rho_r \right)_{eq}} = a_{eq} \frac{\left( \rho_m \right)_0}{\left( \rho_r \right)_0}
$$

If we put back the ##a_0## (which is ##1##, so it doesn't change anything to put it back), we get

$$
\frac{\left( \rho_m \right)_{eq}}{\left( \rho_r \right)_{eq}} = \frac{a_{eq}}{a_0} \frac{\left( \rho_m \right)_0}{\left( \rho_r \right)_0}
$$
 
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  • #16
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That's because you keep on confusing yourself with wrong notation, and switching from one wrong notation to another into the bargain.

Let's restate what you say your book says:

$$
\frac{\rho_m(t)}{\rho_r(t)} = a(t)
$$

The first thing to note is that, as written, this is incorrect (which makes me think you are incorrectly quoting your book). To get a correct equation, we make use of the following equations for the functions ##\rho_m(t)## and ##\rho_r(t)##:

$$
\rho_m(t) = \frac{\left( \rho_m \right)_0}{a(t)^3}
$$

$$
\rho_r(t) = \frac{\left( \rho_r \right)_0}{a(t)^4}
$$

Taking the ratio of these two functions gives:

$$
\frac{\rho_m(t)}{\rho_r(t)} = a(t) \frac{\left( \rho_m \right)_0}{\left( \rho_r \right)_0}
$$

Notice the extra ratio of the values of the densities "now" on the RHS. Notice also that this equation shows that we are using the common convention where the value of the scale factor "now" is ##1##, i.e., ##a_0 = a(t_0) = 1##.

Now we just look at what happens if we plug ##t_{eq}## into the equation above:

$$
\frac{\left( \rho_m \right)_{eq}}{\left( \rho_r \right)_{eq}} = a_{eq} \frac{\left( \rho_m \right)_0}{\left( \rho_r \right)_0}
$$

If we put back the ##a_0## (which is ##1##, so it doesn't change anything to put it back), we get

$$
\frac{\left( \rho_m \right)_{eq}}{\left( \rho_r \right)_{eq}} = \frac{a_{eq}}{a_0} \frac{\left( \rho_m \right)_0}{\left( \rho_r \right)_0}
$$
Thank you very much!
 

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