How do you calculate the amount of energy leaving the observable region?

[mikehelland]

Note: Hopefully this problem and its solution is generic enough to be useful to a variety of models, not just one.
Update: the problem has been restated to be as clear as possible, and I've included updates on my attempt
Update Nov26: I think I have some results, reported below

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The problem, cosmologically speaking

Assume a volume that has an average of N light sources per cubic unit, each with an average luminosity of L. Within this volume is a sphere with radius R.

Assume that each light source, for whatever reason, reaches a maximum radius of R, having a power factor (?) of $1 - r/R$ from its source.

Note: this loss is in addition to the inverse square law.
Note2: I changed this to $(1 - r/R)^2$ so there's a loss due to redshift, and a loss due to time dilation.

How much energy from sources inside the sphere gets out? How much energy from outside the sphere gets in? Presumably, they should be equal. Can the energy per second coming in be related to a temperature T? Thus creating a relationship between R, N, L and T?

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The problem, mathematically speaking

I decided to start 2D. I made a main circle $C_0$. I started at the perimeter, lined it with boxes, and worked my way in. Obviously, zero energy is getting out here. This just fills the circle. So now choose any box in the circle, and do the same thing for it, make a bunch of boxes. Some of this will go outside the original circle.

https://mikehelland.github.io/hubbles-law/other/cmbenergy.htm

I then extended it to 3D:

(just one moveable sphere)
https://mikehelland.github.io/hubbles-law/other/cmbenergy3d.htm

(does all possible spheres)
https://mikehelland.github.io/hubbles-law/other/cmbenergy3dall.htm

Using this page I bumped the left slider down to "10". That's numerically integrating the sphere with a dx of 10. R in the model is 140. So if the units are in hundred million light years, 10 is 1 billion light years.

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Results

I think my conclusion is that, assuming N = 1 light source per cubic billion light years, each has a luminosity of L = 1 watt, and the radius is R = 14 billion light years, then:

• 15 megawatts is landing outside the main sphere

I'm not sure what this really tells me. I think what I want to know is the luminosity of the light that intersects the main sphere's surface area. That's:

• 0.2 megawatts

So I guess you divide that by the surface area? And then you increase N and L until you get... I was trying to get to a temperature somehow. Now I've kind of lost track of how I was going to do that.

Temperature

Assuming we can use:

$$\displaystyle T={\sqrt[{4}]{\frac {L}{4\pi R^{2}\sigma }}}$$

const pc2ly = 3.261564
const pc2km = 3.08e19

var L // Luminosity of a box
var Lsol = 3.846e26
var suns = 2e11 // 200 billion suns in a galaxy

var R = 140

// in meters
var Rm = R/10 / pc2ly * pc2km * 1000 * 1000

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

T = (L * Lsol * suns / (4 * pi * sbc * Rm**2))**(1/4)

I get a temp of 0.188 K.

That's assuming 1 galaxy per cubic giga-lightyears, with the luminosity of 200 billion suns.

To find the right N (number of galaxies per Gly^3) and average luminosity of a galaxy L, it seems I'm only off by a factor of 50K somewhere:

(50000 * L * Lsol * suns / (4 * pi * sbc * Rm**2))**(1/4) 

2.8 K

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Any comments or suggestions?

[HanDeBruijn]

Let's say there's a spherical volume of space with a radius of R, with galaxies uniformly distributed within.

Can we call that our Observable universe, with a Cosmological horizon at R ?

Assume that to an observer at the center of the volume, light coming from R has a z of infinity. So, light leaving the center has a z of infinity at R.

Does that mean that there is no preferred (space-time) position for an observer ?
However, I don't think there is a z of infinity; instead a maximal redshift $z=e^W-1\approx 28.5$ may be theoretized

Based on these assumptions, no light from the center will make it outside of the volume.

Not exactly. There is an interesting publication, called Relativistic Escape Velocity using Special Relativity. See thread 153.
Please follow the reasoning in that paper up to Eq. (7). Then I propose to continue as follows.

$$ m_0c^2 - \frac{G M m_0}{R} = m_0c^2\sqrt{1-\frac{v^2}{c^2}} $$

There is a removable singularity for $v=c$ whch is purely mathematical and defnitely not physical.
This makes our Special Relativity analysis essentially different from its GR counterpart and much less problematic.
But using a computer mathematical application as suggested in the paper is not necessary:

$$ \left(1 - \frac{G M}{R c^2}\right)^2 = 1-\frac{v^2}{c^2} $$

$$ v = c\sqrt{1 - \left(1 - \frac{G M}{R c^2}\right)^2} = \sqrt{\frac{2GM}{R}-\left(\frac{GM}{Rc}\right)^2} $$

So the escape velocity according to SR is

$$ v=\sqrt{\frac{2GM'}{R}} \quad \mbox{with} \quad M' = M + \left[-\frac{GM^2}{2Rc^2}\right] $$

The term between square brackets is recognized as the negative mass of the Self Energy Field of a body with radius R and mass M.
Let's specialize this for our model of the Observable universe, the one I've been trying to explain in thread 179.

$$ M'_O = \frac{W}{2}\frac{c^3}{G.H}(1-g) \quad ; \quad g = \frac{W^2}{4}\mbox{VMT} \quad ; \quad R_O = W\frac{c}{H} $$

$$ v = \sqrt{2G\frac{W}{2}\frac{c^3}{G.H}\frac{H}{Wc}(1-g)} \quad \Longrightarrow \quad \Large\boxed{v = c\sqrt{1-g}} \normalsize \approx 0.9671398994 \times c $$

With Special Relativity instead of GR, the escape velocity for our Observable ("black hole") universe is calculated to be slightly less than the speed of light. As a consequence, light can escape from - as well as enter into - that part of the universe. Contradictory to

Based on these assumptions, no light from the center will make it outside of the volume.

[mikehelland]

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.

[RedshiftDrift]

@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.)

[mikehelland]

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:

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

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

[mikehelland]

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.

[bgreenman]

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

[HanDeBruijn]

Let's say there's a spherical volume of space with a radius of R, with galaxies uniformly distributed within.

Isn't that the same as a uniform mass density distribution $\rho_0$ ? No need to do it numerically; you can do it analytically.
I think that a handful of useful ingredients has been provided in thread 127.
(I'm leaving out the Variable Mass Theory in the sequel because nobody seems to be understanding or interested)
If $M_O$ is the mass and $R_O$ is the radius of your (Observable) universe then we have

$$ M_O = \frac{4}{3}\pi R_O^3\rho_0 $$

There are two ways to define the maximal radius of a black hole (universe), the common General Relativistic (Einstein) one and the one derived with Special Relativity and Newtonian Dynamics.

$$ R_E = \frac{2GM_O}{c^2} \quad \mbox{or} \quad R_N = \frac{GM_O}{c^2} $$

And so

$$ R_O = G\frac{(8 \mbox{ or } 4)}{3 c^2}\pi R_O^3\rho_0 \quad \Longrightarrow \quad R_O^2 = \frac{3 c^2}{(8 \mbox{ or } 4)\pi G\rho_0} $$

Repeated for convenience, from that previous thread as mentioned:

$$ H_0 = \sqrt{\chi G \rho_0} \quad \mbox{with} \quad \chi = 8\pi \times \begin{cases} 1/8 & (: \mbox{OI}) \\ 1/6 & (: \mbox{Mach}) \\ 1/3 & (: \Lambda\mbox{CDM}) \end{cases} $$

Combine this with the formula for $R_O$ :

$$ G\rho_0 = \frac{H_0^2}{\chi} \quad \Longrightarrow \quad R_O^2 = \frac{c^2}{H_0^2} \frac{3\chi}{(8 \mbox{ or } 4)\pi} $$

Your radius of the Observable universe is thus proportional to the Hubble length $c/H_0 \approx 13.3 \mbox{ } ly$.

$$ R_O = \sqrt{\frac{3\chi}{(8 \mbox{ or } 4)\pi}} \frac{c}{H_0} =\sqrt{\mbox{mh}}.\frac{c}{H_0} $$

where you can choose from

$$ \mbox{mh} = \left(\frac{3}{8} \vee\frac{1}{2}\vee\frac{3}{4}\vee 1\vee 2\right) $$

It follows anyway that there is a maximum ${\bf R_O \le } \frac{c}{H_0}\sqrt{2} \approx {\bf 18.8 \mbox{ } ly}$ .

[mikehelland]

"By sheer logic" involves some assumptions that I'm not making, apparently.

[mikehelland]

@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.

[HanDeBruijn]

[HanDeBruijn]

@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

[RedshiftDrift]

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

[bgreenman]

More work is needed on measuring distances, HanDeBrujin is right

[mikehelland]

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.

[bgreenman]

Interesting

[HanDeBruijn]

@mikehelland .
Maybe it’s better practice to leave the original text of your question intact and augment it with a clearly marked Update. Because now people’s quotes and accompanying responses have become vacuous.

[bgreenman]

Agree

[HanDeBruijn]

@mikehelland . Your last approaches seem to be be similar to those in comments at Cosmology Slides: asking for reviews.

[bgreenman]

I don't know

[mikehelland]

@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.

[bgreenman]

HanDeBrujin answer is correct