What Would it Take to Replace the Big Bang?

[luisenrique-arch]

I’d like to share this video from the See the Pattern channel. I think the video does a very good job of explaining what challenges an alternative cosmological theory would have to overcome and what it would need to explain.

https://youtu.be/NKoWo8xGT6A?is=V2XV4ICm1_T86DPt

What do you think?

EDIT: I started reading through the comments under the video and... Jesus Christ, there's so much crackpottery down there.

[Francois-Zinserling]

Anyone who has watched a few cowboy movies will know that people have been 'hanged from them thar trees' for a lot less evidence, and once the lynch mob's mind is made up, it's going to take some shooting to stop it.

It's an incredibly well-done video highlighting the plight of the ΛCDM crowd (and ours) and what it will take to overcome this. Anyone might come with an alternative for the CMB, or dark matter, or dark energy, or redshift, or .... but unless they have a convincing set that overthrows a good chunk of standard model, it's going to just get a sideways glance.

In this way he is absolutely right. We cannot just show an alternate model that 'doesn't require dark matter' if it cannot intuitively lead to solutions for most or all of the rest. It is clear that a new model is needed, but also clear that this model has not been found yet.

[HanDeBruijn]

@Francois-Zinserling writes:

Anyone might come with an alternative for the CMB [ .. ]

Here is the work ACG has done on the CMB. And this has been my best shot at the CMB so far.

$$ u_{CMB} = \frac{4\sigma}{c}\Theta^4 = \frac{W.\mbox{VMT}.e^{-W}}{32\pi.w}\frac{(Hc)^2}{G} $$

Apart from details, the idea is that the CMB energy density at our position is divided by the maximal intrinsic redshift of the energy density at the rim of the Observable universe, which is given by $1+z_{max}=e^W$. This would lead to a enlargement of the temperature $\Theta$ itself by an factor of $e^{W/4} = (1+z)^{1/4}$. However, according to the video by Gareth this factor must be $(1+z)$ instead. Theoretically, the latter is in concordance with Wien's displacement law, as follows.

$$ \lambda_{peak} = \frac{b}{\Theta} \quad \Longrightarrow \quad \frac{\Theta_O}{\Theta_0} = \frac{\lambda_0}{\lambda_O} \quad \Longrightarrow \quad \Theta_O = (1+z_{max})\Theta_0 \approx e^{3.383634283}\times 2.726\mbox{ }K \approx 80.356\mbox{ }K $$

where $\Theta_0$ is the CMB temperature here and now and $\Theta_O$ is the CMB temperature at the rim of the Observable universe. It all means that my best shot at the CMB is plain wrong. Not unexpectedly, because the formula has been found by matching and stretching, while knowing the CMB temperature at our place. But anyway, it's useful to have an estimate for $\Theta_O$ now. (Does everybody see the difference between $0$ and $O$ ?)

It is seen that our maximal redshift as well as our maximal temperature is inside the range of the data displayed in the two screenshots. And because of the logarithmic scale for Redshift in the second picture the temperature rises there as an exponential.

[Francois-Zinserling]

I do regret saying this:

also clear that this model has not been found yet

because there are many sterling works out there, including from this group. I think the point I was trying to make is that the mountain before us cannot be scaled by one person on his own. Case in point the volume of 'evidence' the video shows for the Standard Model.
Let me not be the one to discourage anyone - if anything - let us work together more.

@HanDeBruijn it is my opinion that the 'rim of the observable universe' is just one telescope away from being extended. JWST has probably shown that already and there will likely be better ones soon. This means the 'maximum' x-value on that last graph above will become larger and the y-value must go exponentially higher. Must it go to infinite? - no this is un-physics.

But it is likely there is a cut-off maximum. Then, my opinion - this CMB/Redshift graph is a reverse of a typical $$Ce^{-ax}+B$$ curve where B is the remaining CMB at our measuring point (or a subtle variation there-of), with 'x=0' at the observable boundary, and C is a scaling factor C=1 below). Uhm, well, ok $$e^{ax}+B $$ to the 'left' from us at 0 but we are not the origin, we are the remote measuring point of a distant event. Rem 'B' assumed a constant 2.7K in the S.M.

Crude graph from desmos.com of $$e^{-ax}+B$$ (arbitrary scale!)

The real graph should then look like this, where we are at a point on 'x' where still CMB>0. $$e^{-ax}+B$$ (B=0)

The immediate counter-argument might be that someone 'to the left of us' should read CCMB=0, but if they use the same measuring devices as us, their graph will contain a different chunk of universe but will look exactly like ours - not zero. With the same instruments their 'x' should match ours and their CMB will also appear to be 2.7K.

[HanDeBruijn]

@Francois-Zinserling writes:

it is my opinion that the 'rim of the observable universe' is just one telescope away from being extended.

The 'rim of the observable universe' has a very special (and perhaps rather counter-intuitive) meaning in the Variable Mass Theory (VM). It is e.g. the same for any observer, and its distance is

$$ R_O = W.\frac{c}{H} $$

where $W=3.383634283$ , $c=$ lightspeed in vacuum , $H=$ Hubble parameter .

I am not going to repeat the whole story at this place.

As for the rest of your argument: I just can't make sense of it. So let's repeat.

It seems to me that the labeling of the temperature axis is somewhat sloppy. I would expect a subdivision in the decimal system, which is likely not (?) the case. (Is it a subdivision in units of 5K ? Then where is the 2.7 K at the origin ? )
But let us concentrate on the axis labeled "Redshift". I think it is not really the redshift but (one plus the redshift) instead: $z \rightarrow (1+z)$. And the real variable at that place is not even $(1+z)$ but $s=\log(1+z)$, which is the definition of a logarithmic scale $s$. Now look at the formula for CMB temperature.

$$ \large s=\log(1+z) \quad \Longrightarrow \quad 1+z = 10^s $$

$$ \large \Theta(z) = (1+z)\Theta_0 = \Theta_0 10^s \quad \mbox{with} \quad _0 = \mbox{here and now} $$

Meaning that the linear relationship for $\Theta$ in $z$ looks like an exponential at the logarithmic scale $s$ .
For $z=0$ we simply have the CMB temperature at our place: $\Theta(0) = \Theta_0 10^s = \Theta_0 10^{\log(1+0)} = \Theta_0 10^0=\Theta_0$.

[Francois-Zinserling]

Meaning that the linear relationship for Θ in z looks like an exponential at the logarithmic scale s

I think this is a good reminder of what the graph actually represents.
Meaning my 'it looks like a reverse exponential decay' is a result of the scale, and my whole point thus moot. 🤣

[HanDeBruijn]

@Francois-Zinserling .
Yesterday it looked better than it looks now.

Your whole point may be moot.
But my scale is not good :-(

Update

The picture according to the above would be something like this

which is similar but not quite what's in the video.

[RedshiftDrift]

@luisenrique-arch Unfortunately "See The Pattern" and "DemystifySci" attract a lot of crackpots. It's a bit better in their Patreon channels, but even there the majority don't even understand 10% of what's being discussed in the videos.

The video is good, but the reality is that incomplete cosmologies are not publishable because the 'mob' has decided that alternatives are wrong.

@Francois-Zinserling where can we go to get "some shooting" done? Mainstream cosmologists are too 'chicken' to even risk their skin in a western-type duel/debate!

Here's my comment on See the Pattern/Patreon:

Consider this example: The status of each constraint for a particular mechanism applied to cosmology is

1. Redshift = explained,
2. Distance Ladder = explained,
3. Element Abundance = incomplete,
4. CMB = incomplete,
5. CMB Dipole = explained,
6. CMB Anisotropies = will trivially follow from CMB,
7. CMB Temp & Redshift = will trivially follow from CMB,
8. Large Scale Structure = will follow from previous constraints,
9. Galaxy Evolution = incomplete,
10. Time-domain Behaviour = explained.
    To complete the study, specialized knowledge is needed in: quantum optics, nuclear physics, relativistic scattering, plasma physics, and nonlinear dynamics. No single person is specialized in all these fields, so a team is needed. But building such a team is impossible because no specialist wants to invest their time in a study they don't understand.

We know what it would take to replace the Big Bang, but we don't know how realize this replacement!

[HanDeBruijn]

@RedshiftDrift and @ExpEarth .

From Can the CMB temperature be less than 2.7 K ? this picture:

And another picture from this reference:

The measurements of the CMB temperature in diffuse interstellar medium of the Milky-Way and high redshift galaxies based on excitation of C_I fine-structure and H₂ rotational levels

So what is it ? Is $\Theta(z) = (1+z)\Theta(0)$ indeed linear in the cosmological redshift ? Or is the CMB temperature an absolute constant ?

[ExpEarth]

You most likely have it right with the first answer, T(z) = T0(1 + z) . Gareth had this relation as #4 in his list of things alternative models have to explain. He concedes that there could be errors in the techniques, as @RedshiftDrift has also argued, but at this point the burden is on others show where the measurements are wrong. The relation was used to draw my figure above. It totally fits with the shell universe model, where we have the radiation from the hot shell being redshifted as it goes to the centre. There's a corresponding blueshift in radiation going outwards towards the shell. For an observer at the shell, the centre of the universe would look ver bright. The relation seems to somehow fit your model too, but I can't say for sure, as I still don't understand your model.

[HanDeBruijn]

@ExpEarth .

as I still don't understand your model.

Don't bother, Matt, because as concerning the CMB I do no longer understand it myself too :-(
Note. With Variable Mass (VM), $T$ is systematically replaced by $\Theta$ for temperature, because $T$ has been reserved for Orbital time.