a-cosmology-group/acg#184

Thanks for the response.

I get that you may disagree with the assumptions, but that’s the problem I’m interested in.

I’ll update the original post with the steps I’ve broken it into so far.

@mikehelland What are your assumptions?

• Which function $z(d)$ gives the redshift vs distance $d$ from the source? (We know $z(0) = 0$ and $z(R) = \infty$, but what’s the function in between?)

Since the function $f(d)$ is probably not easily integrable in spherical coordinates, why not just integrate numerically?

• Are there galaxies outside the radius R? (“the amount of energy leaving the volume… should tell us the amount of energy entering the volume” seems to indicate that you assume a uniform distribution of galaxies everywhere, correct?)

If the distribution of galaxies is uniform everywhere, then the power leaving the spherical shell of radius $R$ is the same as the power that enters the sphere: the net power transfer is zero.

P.S. “Watts” is power = energy-per-unit-time, but the question is about energy; did you want to say “power leaving the observable region”? (You can edit the question b.t.w.)

Which function gives the redshift vs distance from the source? (We know and , but what’s the function in between?)

r = z/(1+z)

But I recommend we don’t even use z. z is prejudiced toward increasing wavelengths.

$$1+z = \frac{\lambda_{obs}}{\lambda_{emit}}$$

When wavelength increases, z is positive. When z goes to infinity, so does wavelength.

When wavelength decreases, z is negative. When z goes to -1, wavelength goes to 0.

Redshift has an infinite range, and blueshift has a finite range (0 to -1).

We should be concerned with a reduction of frequency and/or energy. For that, let’s quantify blueshift as a positive number instead, so:

$$1+b = \frac{1}{1+z} = \frac{f_{obs}}{f_{emit}}$$

Now redshifts are when b is negative, and they can only go to negative -1. The range of redshifts is now finite.

Now r = -b. So:

$$1-r = \frac{1}{1+z} = \frac{f_{obs}}{f_{emit}}$$

$$b = \frac{1}{1+z} - 1 = -\frac{z}{1+z}$$

the net power transfer is zero.

That’s what I’m counting on though. The amount of energy leaving (joules per second , so watts, right?) should be equal the amount coming in.

But the amount coming in will “die” out before it reaches the center. So there’s a specific amount of energy entering our observable region, but never making it to us. The idea is to connect that to a temperature somehow. If that temperature is set to 2.7 K, then that should be able to determine R.

I Think that HanDiBrujin’s response to your question is correct

@HanDeBruijn

Isn’t that the same as a uniform mass density distribution ?

I don’t think so, but it’s a good question and might help explain what I’m doing.

If we say there are N galaxies per 1 cubic Gigaparsec in a volume V, and their average luminosity is L,

If we can find how might light is hitting the surface area of that volume from the inside, that should tell us how much light is hitting the surface area of that volume from the outside.

So, flux might be what the problem is calling for.

In any case, then we should have a relation to V, N, L, and hopefully from the flux, some kind of temperature.

V, N, L, and the temperature can all be measured. The temperature is 2.7 K.

V should have a radius of about 14 billion light years.

What’s N and L?

This formula of yours is found in thread 160.

$$ d_T = \frac{c}{H_0} \log (1+z) \quad \Longrightarrow \quad 1+z = e^{H_0/c.d_T} $$

When combined with my previous posting, does it mean that

$$ 1+z = e^{H_0/c.R_0} = e^{H_0/c.\sqrt{\mbox{mh}}.c/H_0} = e^{\sqrt{\mbox{mh}}} \quad \mbox{ ?? } $$

Because in that case we have a limited range of maximal available redshifts:

$$ z = e^{\sqrt{3/8\ \vee\ 1/2 \ \vee\ 3/4 \ \vee\ 1 \ \vee\ 2}} - 1 \lt 3.12 $$

for a galaxy nearby the horizon of your Observable universe. This outcome is contradictory to observations. I guess you don’t want to give up your Time Dilated Past hypothesis ?

HanDeBrujin is correct

@HanDeBruijn

In your second equation you equate $d_T$ to $R_0$.

That’s incorrect. $d_T$ goes from 0 to infinity, it is $-ct$ where $t$ has time dilation. $t = -14$ billion years. Take time dilation out of -14 billion years, you get:

$$\tau = \frac{1}{H} (e^{Ht} - 1) = \frac{1}{H} (e^{-1} - 1) = -8.85~ \text{billion years}$$

And this equates to $z=1.718$, so it’s not any type of special time or place.

(Assuming $1/H = 14$ billion years.)

Its not easy for a novice to calculate the amount of energy leaving the observable region ofthe universe

What answer did you get and how did you get it?

@mikehelland . >That’s incorrect.

How so? That time $t$ is your Measured time, right? Strictly speaking that is incorrect. What is really Measured - or better: estimated - are distances, via the Cosmic distance ladder. So-called Cosmic time on the other hand is very much (Big Bang) model dependent (so I don’t trust it). In an eternal and infinite universe - your Observable universe must be a part of it - Measured time is calculated from Measured distance, simply by $-t = r/c$ where $r=$ distance and $c=$ speed of light. I cannot imagine that time undilated past $\tau$ is relevant in this context :-(Perhaps that’s a shortcoming of my imagination).

@mikehelland . Maybe interesting for you too in thread 182 one of the CMB pioneering papers.

@HanDeBruijn

Not quite.

Imagine a universe where distances are static and time is invariant. $r = -ct$. Everything easy there.

Except, when we observe a supernova for example, we observe it to be time dilated. If we say the time between peak brightness and when its dimming rate drops is 25 days, that includes the effects of time dilation. The time we measure is stretched. That’s why I call $t$ “measured” time.

So suppose a universe with static distance, but time is stretched, depending on how far away it is and/or how long ago it happened. Now $-ct > r$. Because time stretched, but distance didn’t.

If we take the time dilation out of $t$ we get Minkowski time $\tau$.

So when $-ct = R$, big R for radius of the observable region, that’s just when the time dilated time is -14 billion years, which corresponds to an untime dilated time of -8.8 billion years, therefore, a $r$ of 8.8 billion light years and a $z$ of 1.7. Nothing remarkable about that time or place.

The horizon is when $-c\tau = r = R$ and $z$ is infinity.

@mikehelland .

The horizon is when $-c\tau = r = R$

Yes of course. Within the framework as currently presented little else is possible. Let’s elaborate a bit.

$$ 1-H_0\tau = 1+H_0.r/c = 1+z = e^{-H_0.t} \quad \Longrightarrow \quad z = H_0/c.R $$

which is the very classical (linear) law as originally formulated by Hubble. Independent of this, the rest of my musings are still going strong, if I’m not wrong:

$$ R = R_O = \sqrt{(mh)}.\frac{c}{H_0} \quad \Longrightarrow \quad z = \sqrt{\frac{3}{8} \vee\frac{1}{2}\vee\frac{3}{4}\vee 1\vee 2} \le 1.414 $$

a value for $z$ which is far away from >$r = R$ and $z$ is infinity.

And how can $z$ be infinite while $R$ is finite in $z=H_0/c.R$ ?

I’m not sure what $\sqrt{mh}$ is, or where your last formula comes from.

In my model:

$$r = \frac{c}{H} \frac{z}{1+z} $$

But my model also has nothing to do with the topic of this thread.

@mikehelland .

The numerical variable $mh$ is connected with the different cosmological models that I have considered. But you can take $\mbox{mh}=1$ as the value that that suits you best - I think so - giving $R=c/H \approx 13.3 \mbox{ } ly$ . >In my model:

$$ r(z) = \frac{c}{H} \frac{z}{1+z} $$

Then indeed you get as the minimum and maximum for $r(z)$ :

$$ r(0) = 0 \quad \mbox{and} \quad r(\infty) = \frac{c}{H} = R $$

Ingenious. But quite recently you have told us a different story, resulting in

$$ r=\frac{c}{H}z \quad \Longrightarrow \quad r(0) = 0 \quad \mbox{and} \quad r(\infty) = \infty $$

It follows from what you have said with $-c\tau=r$. >But my model also has nothing to do with the topic of this thread.

Oops ???

In response to @bgreenman the same remark is uttered and:

The redshift-distance modulus made by the Pantheon+SH0ES is exquisite for any purposes.

Do you mean this one?

$$ d_L = \frac{c}{H_0} (1+z) \log (1+z) \quad \Longrightarrow \quad r = d_L \approx \frac{c}{H} (1+z) z \ne \frac{c}{H} \frac{z}{1+z} $$

One cannot juggle with formulas at will.

Just a note @HanDeBruijn, from a “curve fitting to SNe data” point of view the logarithmic distance is better fitted with a square-root in the function, as in

$$ d_L = \frac{c}{H_0} (1+z) \log (1+z) \quad \Longrightarrow \quad r = d_L \simeq \frac{c}{H_0} \sqrt{1+z} \ z $$

for redshft $z \lesssim 7$.

@mikehelland the data fails to fit the very different function $\frac{c}{H_0} \frac{z}{1+z}$ already for $z \ll 1$, so that can’t be right.

Redshift drift, you are correct, that’s not the fit for the function.

None of the replies (except your original one) seem to address the topic of this discussion. Han keeps confusing different aspects of my model, randomly, out of context. Here’s an accurate summary (though it doesn’t really belong in this topic):

• $H$ - constant with dimension inverse time

• $t$ - Measured time, has time dilation

$$-\infty < t \leq 0$$

• $\tau$ - Minkowski time, time dilation removed

$$-H^{-1} \leq \tau \leq 0$$

$$\tau = H^{-1} (e^{Ht} - 1)$$

• $r$ - Physical distance

$$r = -c\tau = \frac{c}{H} \frac{z}{1+z}$$

• $-ct$ - the speed of light x measured time. This value isn’t particularly physically meaningful. In standard cosmology, it would be the “light travel time distance.” In my model, it’s just the “time-dilated light travel time distance” which doesn’t actually tell us much about anything, but is used to find a luminosity distance.

$$-ct = -\frac{c}{H} \log (H\tau+1) = \frac{c}{H} \log (1+z)$$

• $d_L$ - luminosity distance. This is a hypothetical distance that essentially says, imagine a universe without expansion or time dilation, or tired light or variable mass. How far away would something be in a plain vanilla spacetime, to look like it does in our spacetime?

$$d_L = -ct(1+z) = \frac{c}{H} (1+z) \log (1+z)$$

That’s a decent fit for the SNe data, at H=70.4 km/s/Mpc.

Here’s an accurate summary

• $t$ - measured time

One might think that measured distance then follows as $r=-ct$ . But no: >- $\tau$ - Minkowski time

By sheer logic, it must be then that $r=-c\tau$ and so $r=c/H.z$ . But no: >- $r$ - Physical distance

$$ r = -c\tau = \frac{c}{H} \frac{z}{1+z} $$

Han keeps confusing different aspects of my model

Yes, I’m completely at lost. And wish you the best. I give up.

“By sheer logic” involves some assumptions that I’m not making, apparently.

@HanDeBruijn , I’ve thought about what you said, and I think it’s just a case of “measured time” being a bad name, and unintended implications of it.

In my model (which again, isn’t relevant here so maybe we can discuss this in another thread) time is dilated, and distances are static. t, x, y, z.

When you take the time dilation out of time, you get a new system of coordinates, $\tau$, x, y, z, but only time has changed. These coordinates match Minkowski spacetime.

To differentiate the two times, I used Minkowski time, which seems like a good name, and “measured” time as the other one.

The space coordinates are the same in both systems. Calling $-ct$ “measured distance” makes sense, name-wise. It should be made clear though that the x, y, and z coordinates in the system with measured time are not affected by time dilation, and are not stretching with time. Otherwise you’d be describing a metric with a scale factor, ie, an expanding universe.

In any case, only Euclidean coordinates are required for the problem I’ve stated in this topic.

@mikehelland . Can’t resist ..

Maybe a proper forum for presenting your problem (without any cosmological details please) is Mathematics Stack Exchange. I’ve found a few Questions there that seem to be resemblant to yours:

• Expected overlap of n circles of equal area randomly placed inside a circle of larger area
• How find integral of circle or sphere with another circle(sphere) inside?
• Finding area inside circle and outside another circle

@mikehelland is $R = \frac{c}{H}$? In any case, the numerical integration seems the simplest approach.

The redshift-distance modulus made by the Pantheon+SH0ES is exquisite for any purposes.

https://github.com/PantheonPlusSH0ES/DataRelease/tree/main/Pantheon%2B_Data/1_DATA

Except I don’t see how that’s at all relevant to the questions posed by this discussion topic.

Interesting

Agree

I don’t know

@HanDeBruijn

Similar.

Mine is pretty basic, and I found a way to simplify it and I think get more accurate results.

const mly2mpc = 1 / 3.261564
const mpc2km = 3.08e19
const gly2m = 1000 * mly2mpc * mpc2km * 1000

var Lsol = 3.846e26
var suns = 2e11 // 200 billion suns in a galaxy

// these three values should produce a temperature T
var R = 14
var N = 50000
var L = Lsol * suns

//Stefan-Boltzman constant
const sbc = 5.67e-8 

// for each box
box.flux = (N * L * (1 - r/R) ** 2) / (4 * pi * (r * gly2m) **2)

Basically around N=50000 gives the CMB temp so 50 thousand Milky Way-ish sources per Gly^3 seems to work out.

HanDeBrujin answer is correct