Unification of Variable Mass and Tired Light

[HanDeBruijn]

From a reply by @ExpEarth :

We are all just incredibly stubborn in backing our own theories and not anyone else's!

From another reply by @RedshiftDrift :

VM may be compatible with other TL models, but that's not on topic for this STz discussion.

Therefore I've decided to open another discussion - this one - instead of adding further "not on topic" comments to the thread by Louis.
From the same STz thread a far more sensible remark.

This makes the $\exp(-H_0 t)$ property of STz common to that of thermal exchanges, where the temperature varies with time as $\exp(-a t)$ when an object cools down. There is no new physics in STz, this property of heat exchanges was known by Carnot in the 1820s!

That's indeed an important clue !! Variable Mass and Tired Light can thus be unified because they form sort of a cosmic Heat exchanger. A heuristically relevant set of equations from Wikipedia is in A model of a simple heat exchanger, with $u=$ energy density, $t=$ time and $\gamma T$ to be replaced soon.

$$ \begin{cases} \large \frac{du_1}{dt} = \gamma (T_2-T_1) \\ \large \frac{du_2}{dt} = \gamma (T_1-T_2) \end{cases} $$

Heat exchangers have been bed, bath and breakfest for quite some years when I was an engineer.
There is a second piece of necessary information:

And so the total Hubble is the sum of the intrinsic and the tired-light one:

$$ \large H_0 = H_{VM} + H_{TL} $$

A third piece of heuristics is found in a discussion with @ExpEarth about his celebrated Hubble luminosity.
Combining now the above intuitions (with a dimensions check) finally leads to

$$ \begin{cases} \large \frac{du_1}{dt} = H_{TL}(u_2-u_1) \\ \large \frac{du_2}{dt} = H_{VM}(u_1-u_2) \end{cases} $$

The equations are partly solved by addition and subtraction. Energy conservation is recognized with the first.

$$ \frac{d(u_1+u_2)}{dt} = 0 \quad \Longrightarrow \quad u_1+u_2 = A $$

$$ \frac{d(u_1-u_2)}{dt} = -(H_{VM}+H_{TL})(u_1-u_2) \quad \Longrightarrow \quad u_1-u_2 = B e^{-H_0 t} $$

with $H_0 = H_{VM}+H_{TL}$ . The equations are solved completely by further addition and subtraction.

$$ u_1 = \frac{1}{2}\left[A + B e^{-H_0 t}\right] \quad ; \quad u_2 = \frac{1}{2}\left[A - B e^{-H_0 t}\right] $$

Assume that the constants $A$ and $B$ are positive. Let $u_1$ be attached to Tired Light (TL) and let $u_2$ be attached to Variable Mass (VM). Then we can make a preliminary sketch of the (TL)-(VM) energy exchange as a function of cosmic distance.

[RedshiftDrift]

@HanDeBruijn
The two coupled differential equations for $u_1$ and $u_2$ are symmetric with respect to VM and TL. Why do you pick the initial condition $u_1(t=0)>A/2$, $u_2(0)<A/2$ ? $\quad$ This choice gives a positive slope for VM which I would interpret as more energy, or blue-shift. What does the solution mean?
$u_1(0)<A/2$, $u_2(0)>A/2$ is also possible, no?

In writing

the total Hubble is the sum of the intrinsic and the tired-light one: $H_0 = H_{VM} + H_{TL}$

you lump together all tired-light models. But this is incomplete as there are several distinct tired-light models.
To include all these models, your equations for a "multi-element heat exchanger" would be more like

$$ \begin{cases} \large \frac{du_1}{dt} = H_{VM}(u_1 - \sum_{i\ne 1} u_i) \\ \large \frac{du_2}{dt} = H_{STz}(u_2 - \sum_{i\ne 2} u_i) \\ \large \frac{du_3}{dt} = H_{NTL}(u_3 - \sum_{i\ne 3} u_i) \\ \large \frac{du_4}{dt} = H_{PR}(u_4 - \sum_{i\ne 4} u_i) \\ \large \frac{du_5}{dt} = H_{CREIL}(u_5 - \sum_{i\ne 5} u_i) \\ \large \frac{du_6}{dt} = H_{SIM}(u_6 - \sum_{i\ne 6} u_i) \\ \large \frac{du_7}{dt} = H_{PTH}(u_7 - \sum_{i\ne 7} u_i) \\ .\\ .\\ .\\ \large \frac{du_j}{dt} = H_j(u_j - \sum_{i\ne j} u_i) \\ .\\ .\\ .\\ \end{cases} $$

There are about $80 \gtrsim j$ of these distinct TL models¹

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P.S. A countercurrent heat exchanger is more efficient and produces a decrease in the temperature of both liquids as a function of position.

Footnotes

1. VM:Variable Mass, STz:Stimulated Transfer redshift, NTL:New Tired Light, PR:Plasma Redshift, CREIL:Coherent Raman Effect on Incoherent Light, SIM:Scattering in Intergalactic Medium, PTH:Plasma Theory of Hubble. ↩

[HanDeBruijn]

@RedshiftDrift . Confession: before I went to sleep it looked better than it looks now :-( Must go back to the drawing board, I guess.

[RedshiftDrift]

Confession: before I went to sleep it looked better than it looks now
@HanDeBruijn, confession: this happens to me all the time...

But that's better than stubbornly keeping alive a bad idea (like a zombie Big Bang model).

[HanDeBruijn]

There is an error in the above solution procedure for the equations

$$ \begin{cases} \large \frac{du_1}{dt} = H_{TL}(u_2-u_1) \\ \large \frac{du_2}{dt} = H_{VM}(u_1-u_2) \end{cases} $$

The first equation is partly solved by addition, yes, but it went wrong:

$$ \frac{d(u_1+u_2)}{dt} = 0 \quad \mbox{must be} \quad \frac{d\left[(u_1/H_{TL})+(u_2/H_{VM})\right]}{dt} = 0 $$

Resulting in

$$ H_{VM}u_1+H_{TL}u_2 = A \quad \mbox{instead of} \quad u_1+u_2 = A $$

still to be combined ith $H_0 = H_{VM}+H_{TL}$ and

$$ u_1-u_2 = B e^{-H_0 t} $$

Now multiply the latter with $(H_{TL},H_{VM})$ and perform further addition and subtraction with the former.

$$ u_1 = \left[A + B H_{TL} e^{-H_0 t}\right]/H_0 \quad ; \quad u_2 = \left[A - B H_{VM} e^{-H_0 t}\right]/H_0 $$

Written otherwise, with $u_\infty = u_1(\infty) = u_2(\infty) = A/H_0 =(H_{VM}u_1+H_{TL}u_2)/(H_{VM}+H_{TL}) =$ weighted mean, and replacing time $t$ by distance $x=ct$ with $c=$ lightspeed:

$$ u_1(x) = u_\infty + B \frac{H_{TL}}{H_0} e^{-(H_0/c)x} \quad ; \quad u_2(x) = u_\infty - B \frac{H_{VM}}{H_0} e^{-(H_0/c)x} $$

Fortunately, nothing becomes essentially different with the above preliminary sketch.
But I'm not satisfied with the result for Variable Mass. Indeed an exponential function comes out but previous "established" theory suggests that associated energy density goes

$$ \mbox{like} \quad \frac{\rho c^2}{\rho_0 c^2} = e^{-(H_0/c)x} \quad \mbox{and not like} \quad \simeq \left(1-e^{-(H_0/c)x}\right) $$

where $\rho=$ mass density.

Update(s).

It's easy to see that $u_\infty = 0$. Because VM starts with zero energy (density) at $t=-\infty$ and TL ends with zero energy (density) at $t=+\infty$ .

$$ u_{TL}(x) = + B \frac{H_{TL}}{H_0} e^{-(H_0/c)x} \quad ; \quad u_{VM}(x) = - B \frac{H_{VM}}{H_0} e^{-(H_0/c)x} $$

Only some magic is needed now to flip the minus in $(-B)$ into a plus. Unfortunately though: magic is not science.

[ExpEarth]

Hi @HanDeBruijn and @RedshiftDrift. I'm not sure exactly how your respective models got into heat exchangers, but I really like this heat exchanger discussion. I am trying to apply it to my shell universe model also #256 .

In my model, we have the hottest CMB temperatures in the shell and the coldest at the centre of the universe. So to me this looks like the countercurrent heat exchanger. The hot CMB photons flow inward and are cooled along the way by the spacetime filaments it interacts with. The spacetime filaments are a static element, almost like a crystal. The energy of the filaments increases in density towards the shell, since the mass density is greater there, and so their effective temperature increases there also. On the inward trip CMB photons thus continually move into regions of colder spacetime temperature. The energy density of the CMB photons is therefore greater than the energy density of the spacetime filaments in the regions they cross. There is thus a net flow of energy from CMB photons to filaments and this produces gravity in my model. We have H being related to G.

Now the cold reverse flow is just the same CMB photons, now cooled down, moving from the centre of the universe back towards the shell. These photons are heated up again along the way by contact with the very same spacetime filaments. But because the energy density of the filaments is greater now, this now reheats the CMB photons - by the Hubble luminosity.

The heat exchangers shown above do not show the cosmic situation appropriately, if we assume a static universe. There's no flow of energy in or out of the system. It's just recycling of energy.

Anyone seen a heat exchanger like this before? Am trying to diagram it.

[RedshiftDrift]

The "heat exchanger" analogy simply describes energy exchange.

The cosmic system you describe moves energy inward and outward inside a closed volume.

It would be like using a heat exchanger with the two outputs connected to the two inputs (if I understand this correctly? As in diagram below?)

Not exactly the kind of thing engineers would like to see 😉 You will have to make your own diagram!

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"Poor artist's" version of a 1D 'radial' heat exchanger showing the energy flow in the shell universe (image from this site)
The colour coding is misleading because "red" is hot but represents "blue"-shifted photons...!

[ExpEarth]

Thanks for the useful links, Louis. I agree that engineers won't like this heat exchanger universe. It's doing no work - useless!

I will try to get a diagram together.

[ExpEarth]

Look what you started @HanDeBruijn and @RedshiftDrift ! I think I will be modifying my gravity model as a result of this exercise.

Okay, it works something like this. The CMB photons that contact baryons in the shell are scattered toward the interior. The CMB photons are the thin coloured arrows. The recoil from the reemission gives the baryons a push towards the exterior (large white arrows). Similarly, when a reheated CMB photon passes through the universe and hits a shell baryon, it pushes the baryon towards the exterior (white arrow again). So both interactions tend to push the shell towards a larger radius. This also has the effect of pulling the connected filaments of spacetime in that direction.

Now the CMB photons are redshifted as they move inward and blueshifted as they move back towards the shell. In the redshift, they impart some of their momentum to the spacetime filaments, pushing them towards the centre (small black arrows). When the CMB photons head back to the shell, they are reheated by the spacetime filaments in those regions. This causes the filaments to lose a portion of their momentum that was in the direction of the CMB photon, resulting again in a net push towards the centre.

The end result is that the forces on the shell baryons and the spacetime filaments are balanced. In essence, hotter CMB photons lose energy to cooler CMB photons moving in the opposite direction. (On this point, I notice a slight correspondence with your model @RedshiftDrift , where you have wave energy being transferred to a wave moving the opposite way.)

This is a work in progress. Hey, I'm not expecting anyone to love this heat exchanger!

[RedshiftDrift]

Continuing this discussion of @ExpEarth's gravity model here.