Proving an Equation for a Single Component Flat Universe

[Arman777]

I am trying to prove that for a single component flat universe

$$\frac{dz}{dt_0} = H_0(1 + z) - H_0(1 + z)^{\frac{3 + 3w}{2}}$$

For a single component flat universe,
$q = \frac{2}{3 + 3w}$
$a(t) = (t/t_0)^q$
$t_0 = qH_0^{-1}$
$1 + z = (t_0/t_e)^q$

Now here is my approach,

$$\frac{dz}{dt_0} = qt_0^{q-1}t_e^{-q} = q (t_0/t_e)^qt_0^{-1}$$
$$\frac{dz}{dt_0} = q ( 1 + z) q^{-1}H_0 = H_0 (1 + z)$$

Can someone point it out what am I missing. Like I am trying over 3 days now. Either I suck at math or there's something that I am missing

Edit: This identity can be derived from above identities but I wanted to write
$$t_e = \frac{t_0}{(1 + z)^{3(1+ w)/2}} = \frac{2}{3(1 + w)H_0} \frac{1}{(1+z)^{3(1+ w)/2}}$$

 

[PeterDonis]

I am trying to prove that for a single component flat universe

Your first equation doesn't make sense; $t_0$ is a constant, so $dz / dt_0$ is meaningless.

 

[Arman777]

Your first equation doesn't make sense; $t_0$ is a constant, so $dz / dt_0$ is meaningless.

I know I am also confused too. This is a question in the Cosmology book

I tried to think it as $dz/dt$ and then make $t = t_0$ substitution but that also does not work.

 

[PeterDonis]

This is a question in the Cosmology book

What book?

 

[Arman777]

What book?

Barbara Ryden 2nd Ed

 

[Vanadium 50]

This is is not a cosmology problem.
This is not a calculus problem.
It is not even an Algebra 2 problem.
It is an Algebra 1 problem.

 

[PeterDonis]

Moving to homework forum since this is a problem from a textbook.

 

[Arman777]

This is is not a cosmology problem.
This is not a calculus problem.
It is not even an Algebra 2 problem.
It is an Algebra 1 problem.

Well that's not much helpful. Can you guide us ?

 

[Vanadium 50]

Well that's not much helpful.

Well, that's not much effort.

Since past experience indicates that you're never going to put the effort in, and this thread is going to around and around in circles, the answer is:

3(1+w)/2 < 1
w < -1/3

 

[George Jones]

I am trying to prove that for a single component flat universe

$$\frac{dz}{dt_0} = H_0(1 + z) - H_0(1 + z)^{\frac{3 + 3w}{2}}$$

For a single component flat universe,
$q = \frac{2}{3 + 3w}$
$a(t) = (t/t_0)^q$
$t_0 = qH_0^{-1}$
$1 + z = (t_0/t_e)^q$

Now here is my approach,

$$\frac{dz}{dt_0} = qt_0^{q-1}t_e^{-q} = q (t_0/t_e)^qt_0^{-1}$$
$$\frac{dz}{dt_0} = q ( 1 + z) q^{-1}H_0 = H_0 (1 + z)$$

Can someone point it out what am I missing. Like I am trying over 3 days now. Either I suck at math or there's something that I am missing

You are missing something: $t_e$ is a function of $t_0$, i.e., $t_e = t_e \left( t_0 \right)$. Consequently, when $z$ is differentiated with respect to $t_0$, because of the quotient (or product) rule, you should get a second term.

Imagine that, over millions of years, you watch a distant galaxy through a telescope. As you watch, you see the galaxy progress in time. But what you see in the eyepiece is the galaxy as it was at the time of emission, i.e., the emission time $t_e$ changes as the observer's time $t_0$ changes.

I have derived the required result using this method, but I doubt that what I have done is what Ryden expects. I will look into this.

 

[Orodruin]

Honestly, the question is not very well defined as what is being observed is not well defined. If you consider observing a comoving object, then you need to take the change in $t_e$ with $t_0$ into account. However, if you observe something like a spacelike surface of constant cosmological time (such as the CMB), then clearly $t_e$ would be fixed.

It should be better specified what is intended by ”light source”.

 

[Vanadium 50]

Gentlemen, it is not nearly this complicated. At t0, every element on the right hand side has a value, so dz/dt0 does too. If it has a value, it has a sign.

 

[Arman777]

3(1+w)/2 < 1
w < -1/3

It does not make any sense. We don't know the value of w.

 

[Vanadium 50]

What does the question ask for?

 

[Arman777]

You are missing something: tetet_e is a function of t0t0t_0, i.e., te=te(t0)te=te(t0)t_e = t_e \left( t_0 \right).

If you consider observing a comoving object, then you need to take the change in tetet_e with t0t0t_0 into account.

What kind of dependence are we talking about ? Can you show by using equations ?

but I doubt that what I have done is what Ryden expects

Why so ?

I am kind of lost. Is this dependence can be derived from the equations in the first post or is it something else ?

 

[Arman777]

It should be better specified what is intended by ”light source”.

I guess yeah. And as Peter Donis mentioned, derivative of $z$ w.r.t $t_0$ seems really odd.

 

[Arman777]

What does the question ask for?

We are trying to prove the identity. The question says "Show that ..".

 

[Orodruin]

What does the question ask for?

The question asks you to first show the relation and then to find particular values of w. The OP’s issue is related to the first part, not the second.

 

[George Jones]

I guess yeah. And as Peter Donis mentioned, derivative of $z$ w.r.t $t_0$ seems really odd.

The idea is to let the observation time $t_0$ progress into the future. I do not like this notation, because, for example, it would mean that the the expression $a \left( t \right) = \left( t/t_0 \right)^q$ (valid for fixed $t_0$) would have to change to something like $a \left( t \right) = A t^q$. (Why?) Better to drop the subscipt $0$ from the observation time, i.e., just write the observation time as $t$. Then, $a \left( t \right) = \left( t/t_0 \right)^q$ for a fixed $t_0$ is valid. The subscipt $0$ can be inserted at the end

Take a comoving galaxy as the light source. Taking into account that $t_e$ is a function of observation time $t$, find $dz/dt$. To differentiate $a \left( t_e \right)$, use the chain rule.

 

[Arman777]

I am trying but I am kind of stuck again.

$\frac{dz}{dt_0} = \frac{d}{dt_0}(t_0^q t_e^{-q} - 1)$
So we can write
$\frac{dz}{dt_0} = qt_0^{q-1}t_e^{-q} - qt_0^qt_e^{-q-1} \frac{dt_e}{dt_0}$
So first term is $H_0(1+z)$
$\frac{dz}{dt_0} = H_0(1+z) - qt_0^qt_e^{-q-1} \frac{dt_e}{dt_0}$
In the other side we have again $H_0(1+z)$
$$\frac{dz}{dt_0} = H_0(1+z)[1 - t_e^{-1}t_0 \frac{dt_e}{dt_0}]$$

Is it correct up to here ?

 

[George Jones]

$$\frac{dz}{dt_0} = H_0(1+z)[1 - t_e^{-1} \frac{dt_e}{dt_0}]$$

Is it correct up to here ?

Almost, but I think that you are missing a factor of $t_0$ in the second term in the square brrackets on the right.

 

[Arman777]

q−10

Almost, but I think that you are missing a factor of $t_0$ in the second term in the square brrackets on the right.

Yes you are right.
so we have

$$\frac{dz}{dt_0} = H_0(1+z)[1 - t_e^{-1}t_0 \frac{dt_e}{dt_0}]$$

So it means that $dt_e/dt_0 = 1/(1 + z)$ ?