a-cosmology-group/acg#240

@ExpEarth . >the Hubble luminosity. Basically, all objects are emitting energy in relation to their gravitational potential energy, $U$.

Our main reference is about the Gravitational Self Energy Field, as has been mentioned in another reply. Near the end of that page you will find the proper equations. What you call gravitational potential energy should be the same as our self energy. But your formula is missing a factor $2$ in the denominator. Relativity Reexamined is a reference confirming this. Let $G=$ gravitational constant, $M=$ mass of the object, $R=$ radius of the object, then:

$$ U \equiv - \frac{GM^2}{R} \quad (2) \quad \mbox{must be} \quad U \equiv - \frac{GM^2}{2R} \quad (7.13) $$

Furthermore, let $c=$ lightspeed, $r=$ radius. Then we can rewrite Newton’s law of gravitation as follows (see main reference).

$$ F = GM\left[1-GM/(2 c^2 R)\right]/r^2 + G^2 M^2/(2c^2)/r^3 $$

The term $\left[1-GM/(2 c^2 R)\right]$ is lowering the “normal” gravitational force. This could be in concordance with >This energy causes repulsive forces within systems, leading to a local expansion.

The term with $r^3$ causes a perihelion shift. However, as discussed with Pierre, there is a discrepancy with observation. >On this topic, you may have seen all the recent evidence for planets and moons having more internal heat than can be explained.

The (improved) Hubble luminosity is, whith $H_0=$ the Hubble constant:

$$ L_H = \frac{GM^2}{2R}H_0 $$

For the Earth we have the following numbers

G := 6.67408*10^(-11);
Mpc := 3.08567758*10^22;
H_0 := 73.4; # Riess
H_0 := H_0*1000/Mpc;
M := 5.972*10^24;
R := 6.3781*10^6;
L_H := G*M^2/(2*R)*H_0;
                                             15
                       L_H := 0.4438693952 10

Continuing with the Stefan-Boltzmann law:

sigma := 5.670374419*10^(-8);
Theta := evalf((L_H/(4*Pi*R^2*sigma))^(1/4));

                         Theta := 62.55510221

for the excess temperature of the earth: 63 K.

Edit

Same calculation for the sun

M := 1.989*10^30;
R := 696265000;
L_H := G*M^2/(2*R)*H_0;
sigma := 5.670374419*10^(-8);
Theta := evalf((L_H/(4*Pi*R^2*sigma))^(1/4));

                         Theta := 1068.951005

That excess temperature of 1069 K is a sensible part of the sun’s temperature at the surface (5800 K) but it cannot explain for the corona.

@HanDeBruijn

Thanks for your input, Han.

What you call gravitational potential energy should be the same as our self energy. But your formula is missing a factor of 2 in the denominator.

I will check that out in Relativity Reexamined . The factor of 2 in the denominator I assume applies to relativistic objects like black holes. In those situations it could actually help my model.

For the Earth, it is generally expressed as U = 0.6GM^2/R , which is not far from your value actually. Your calculation verifies that the energy the Earth would radiate away is rather small and its effective surface temperature would be sub-freezing if that was the only source of heat. The earth’s surface temperature is dominated by the Sun’s radiation. To test the model, we actually need to look at objects that are far away from the Sun and where other sources of heat can be eliminated. These might include icy moons or planets, provided that tidal forces are absent. For the Earth, expansion would be a possible sink for this energy. Most of the energy released would go into increasing the Earth’s radius and thus its gravitational potential energy.

@ExpEarth . >The factor of 2 in the denominator I assume applies to relativistic objects like black holes.

No Matt, that factor 2 applies to purely Newtonian objects. Please read the suggested section in the book.

The more I think about your Hubble luminosity, the more I find that it’s true, but it also appears more and more trivial. The reason is that (not only) in the Variable Mass Theory the Hubble parameter $H$ is related to Age. Most of the time, the “age of the universe” is put equal to $\approx 1/H$, but in the VMT it can be e.g. the Age of the Earth as well. Fortunately, due to the fourth power law, the exact choice of $A=1/H$ has not much influence on the outcomes for excess temperatures. Obtained results may therefore be expected to remain reasonably accurate. Whatever, quite in general we have

$$ \frac{dU}{dt} = L_H = -U.H = -U/A \quad \Longrightarrow \quad U(t) = U_0.e^{-t/A} $$

which is precisely the well-known formula for an Exponential decay, where $A=$ mean lifetime. Maybe you can add Hubble luminosity to the list of phenomena in Natural sciences at this Wikipedia page :-)

@HanDeBruijn

While I’m happy that you are starting to like the Hubble luminosity, I am not happy that you constantly invoke these exponentially changing processes. All the evidence suggests that the universe carries on in a more or less constant and unchanging manner. The “age of the universe”, 1/H , is not really an age but the period of the cosmic cycle. All the processes of creation and destruction are governed by this cycle. Some examples:

Star formation rates, SFR SFR/unit volume = n H where n is the average stellar number density

Star death rates, SDR SDR/unit volume = n H

rate of hydrogen fusion d rho H(atomic) / dt = - rho H(atomic) H

This is why we don’t generally see stars older than 1/H . I wrote a small paper on this in 1998. http://redshift.vif.com/JournalFiles/Pre2001/V05NO3PDF/v05n3EDW.pdf It needs updating. The numbers still hold true and maybe even more so. I’d be interested to hear suggestions on what other rates could be looked at to check this. My general impression is that people here only look at the Hubble constant in relation to the decay of light. It’s really much bigger than that.

I agree that 1/H is not the “age of the universe”.

I disagree that it is exactly “the period of the cosmic cycle”, because there are cosmic cycles of many kinds and their periods might be k * 1/H, for multiple k (k probably greater than 1 and less than 100 (a guess)).

we don’t generally see stars older than 1/H

I recall there were many papers (of the 1990s) that identified stars ~ 15 Gy and older in the Milky Way, but then their authors were coerced by Big Bang cosmologists to “revise” those ages to under 13.8 Gy.

What is a cosmic cycle?

@ExpEarth .. >The “age of the universe”, 1/H , is not really an age but the period of the cosmic cycle.

I think that the cosmos is not cyclic. And proper understanding what I’ve written reveals that $1/H$ indeed is not really an age. I invite you to figure out what really the meaning is of $A$ in my formulation of an exponential decay. Your last example, named rate of hydrogen fusion contains a differential equation. And the solution is again .. a decay!

$$ \frac{d\rho_{H(atomic)}}{dt} = -\rho_{H(atomic)}H \quad \Longrightarrow \quad \rho_{H(atomic)}(t) = \rho_{H(atomic)}(0) \mbox{ } e^{-Ht} $$

I am not happy that you constantly invoke these exponentially changing processes.

That may be so Matt, but I am not the one to be blamed. Mathematical truth has no mercy, not for me too.

people here only look at the Hubble constant in relation to the decay of light.

But I am not a Tired Light supporter, you know.

@sahil5d . >there are cosmic cycles of many kinds and their periods might be k * 1/H, for multiple k

See comment and replies in LCDM cosmological coincidences. Quote from a relevant reply: >Also take notice of the fact that I’m no longer saying that $H=1/A$ or $H=2/A$ but instead that the Hubble parameter is inversely proportional to Age, with an arbitrary factor $n$, not even necessarily an integer: $H=n/A$.

Or $A=n * 1/H$ which is almost the same as what you are saying.

@mikehelland . >What is a cosmic cycle?

Good question !

@sahil5d

Yes, there can be other cycles, but the one that would be the governor in the observable universe has the Hubble constant H as the principal cycle parameter and 1/H as the cycle period. In my work, I find that black holes (or equivalent term) each have their own Hubble ‘constant’. Their periods are much shorter. Can you think of some other cycles that we can look at for comparison?

I recall there were many papers (of the 1990s) that identified stars ~ 15 Gy and older in the Milky Way, but then their authors were coerced by Big Bang cosmologists to "revise" those ages to under 13.8 Gy.

Yes, there are some stars with greater ages but many too that have lesser ages. The Hubble cycle would refer to the average age of stars per stellar unit mass.

@mikehelland

The easiest way to think of it is that the universe is a sort of forest. The trees (stars) have an average age and average lifetime. These numbers applying to individual trees are not what is governing the cycle though. Imagine you had the whole forest under a glass cover so gases cannot get in or go out. It’s the metabolic rate at which the trees consume and emit gases which actually controls the cycle rate. In a steady state situation, it is the rate at which trees absorb CO2, relative to the total quantity of CO2 present, which would fix the cycle parameter.

@HanDeBruijn

I think that the cosmos is not cyclic. .... Your last example, named rate of hydrogen fusion contains a differential equation. And the solution is again .. a decay!

It’s the same situation as we discussed before regarding the Hubble luminosity. There is no exponential decay! For every H atom that is fused into He, there is an H atom being formed somewhere else, most likely in supermassive black holes (AGNs/quasars).

But I am not a Tired Light supporter, you know.

This could be the root of all your difficulties!

@ExpEarth . There is no discussion about the solution of a differential equation, Matt. Period. And supermassive black holes are not a panacea for all inconsistencies that pop into people’s heads. Don’t get me wrong: I understand the grand idea and might even agree with it. What is IMO lacking is a mathematical formulation that is sufficiently accurate.

The Hagedorn temperature has been tested in labs, and it does create quark-gluon plasma from heavy ions:

https://en.wikipedia.org/wiki/Hagedorn_temperature

Quasars have been measured to exceed this temperature, so it seems pretty certain quasars can boil any element into something that must reconstitute as hydrogen.

@mikehelland . Quote from the Wikipedia page: This quark matter may have been detected [ .. ] (bold font by me). And there is a relevant reply in thread 200 .

@HanDeBruijn

There is no discussion about the solution of a differential equation, Matt. Period.

In this discussion of cycles, we are going in circles. One last try. Suppose you start off with a universe of stars and they all burn the fuel they have and then eventually die. If you look at just those stars alone, you could write your exponential decay law for them. But elsewhere in the universe there are new stars being formed, where that hydrogen fuel has been regenerated at the same rate as it was being consumed in stars. So, on the cosmic scale, there is no change .

@ExpEarth . >If you look at just those stars alone, you could write your exponential decay law for them.

Ah, now we are getting somewhere! That’s is precisely what I am saying. Worse, it is precisely what you are saying all the time too, with your graphs and your formulas. Look at his one: What are the labels “Earth”, “Jupiter”, “WD”, “NS”, “SMBH” for? It means: If you look at just those objects alone. If you include all the rest, then of course there is conservation of energy and the equation for that is simply $dU/dt=0$ or $U=$ constant.

So, on the cosmic scale, there is no change.

You are quite right. And I couldn’t agree more. But your prose is different from your mathematics.

With this graph it’s a different story. Unlike stellar fusion, we can put all the masses and mass systems in the universe on that straight line and they will emit energy at that same rate continuously, unless their radius or mass changes. They are all giving off energy in proportion to their internal gravitational energy. That energy in turn goes into gravity (in my model) and regenerates the CMB. We can write dU/dt = UH because we can also write dU/dt = - UH. So on the whole, yes, U is constant.

May I just say to ACG members generally that if there is anyone interested in doing further research on the Hubble luminosity, I am looking for coauthors on this. I am not good at statistical analysis, as perhaps you already detected. I consider this the most important topic in astrophysics and cosmology and yet it has barely been touched! There would be a ton of new data now. Anyone interested can contact me at: matt.edwards@utoronto.ca

I’m struggling a bit to understand what “Hubble luminosity” means. When I google it, I see mostly references to Hubble’s law, and your papers.

The luminosity of the sun is L☉, and it has that value whether we observe it from Earth, from Pluto, from Andromeda, or from 100 Gpc. An object’s luminosity is what it is, regardless of the distance from which it’s viewed.

Can you elaborate a bit more on what you mean by that?

We can write $dU/dt = UH$ because we can also write $dU/dt = - UH$. So on the whole, yes, $U$ is constant.

With $dU/dt = UH$ and $U=Mc^2$ this could mean that Matt’s Hubble Luminosity is the (missing) counterpart of (my version of) the Variable Mass Theory. There is a section Hubble Parameter at my website for the math. As I’ve said, we’re getting somewhere ..

Sure. The ‘Hubble luminosity’ term was perhaps not wisely chosen, as it leads to Google search results like you found. Other names are welcome.

It is the simplest thing you can imagine. Just take a body’s internal gravitational potential energy (eg, for the Earth, U = - 0.6 GM^2/R) and multiply that by the Hubble constant, or L = -UH . This simple relationship can explain the luminosities of diverse classes of objects, as can be seen in the graph a few posts earlier. It does not explain stellar luminosities well, as fusion processes release much more energy. It only works well when other sources and sinks of internal heating (fusion, tidal, radioactivity) are negligible. For the Earth, I have suggested that a lot of this energy goes into tectonic processes rather than emitted radiation. On the cosmic scale, I link a reverse process of energy absorption to the Hubble redshift and gravity

If you were to replace H with any other value, say 1 inverse second, wouldn’t a similar line appear, just with a different slope?

You would change the value for the Hubble luminosity that way, but it would no longer be equal to the bolometric luminosity of these objects. The point of the graph is to show they are within an order of magnitude of each other over a huge rate of values.

We shall continue our calculations the way we did for the Hubble Luminosity of the Earth. But our goal is a lot more ambitious now: to calculate for the whole Observable universe. All according to (our version of) the Variable Mass Theory (VMT) of course. Previous calculations can be summarized as $U_G + U_{CMB} = 0$, with the Stefan-Boltzmann law for the radiation energy density of the CMB.

$$ U_G = - \frac{G M_O^2}{2R_O}.W.\mbox{VMT} .e^{-W}\quad ; \quad U_{CMB} = 4\pi.wW \left(\frac{c}{H}\right)^3 \frac{4\sigma}{c}\Theta^4 $$

Mass and radius of the Observable universe are given by

$$ M_O = \frac{W}{2}\frac{c^3}{G.H} \quad \mbox{;} \quad R_O = W\frac{c}{H} $$

Gravitational constant $G$, lightspeed $c$, Hubble parameter $H$ are recognized. Other constants ( w , W , VMT ) are purely mathematical.

Another approach - the one employed by @ExpEarth, albeit with a @HanDeBruijn twist - is different from what I’ve done so far and can be summarized as $\mbox{ } 4\pi R_O^2.\sigma\Theta^4 + U_G.e^{-W}.H = 0$ . In words: [ CMB power times area + Hubble luminosity of redshifted gravitational self-energy = 0 ] at the rim of the observable universe. A calculation shall further clarify the meaning of the maths.

W := 3.383634283; # pure math
w := (2-exp(-W)*(W^2+2*W+2))/W; # LATER
G := 6.67408*10^(-11);
Mpc := 3.08567758*10^22;
H := 70.5; # Mike Helland value
H := H*1000/Mpc;
c := 299792458;
VMT := int(exp(-x)/x^2,x=W..infinity);
M_O := W/2*c^3/(G*H);
R_O := W*c/H;
U_G := G*M_O^2/(2*R_O)*W*VMT; # self-energy
U_G := U_G*exp(-W)*H; # times redshift and Hubble Luminosity
sigma := 5.670374419*10^(-8);
Theta := evalf((U_G/(4*Pi*R_O^2*sigma))^(1/4));

         Theta := 2.230197141

This outcome is close to the one with radiation energy density, as previously calculated. It is possible to establish a simple pure mathematical connection between the two models:

$$ \frac{\Theta_{Han}}{\Theta_{Matt}} = \left(\frac{W}{4w}\right)^{1/4} \approx 1.214821788 $$

This will look like Greek to most of ACG, Han. I will try to discuss within the confines of my model. I don’t see your model clearly enough (remember, I am not mathematical like you. I prefer words!)

I also use both methods to determine the CMB temperature in my model (see section 4 of my preprint in discussion #204 ). Here is how I would express things simply in my model. The observable universe arose as a black hole-like structure. During the formation process, the gravitational energy released with the infalling matter got trapped inside the exterior ‘shell’. (I have since modified my model to have a much thicker shell). When you do this and assume the energy ends up as CMB radiation, you get a CMB temperature of 27 K. Now this CMB radiation is constantly being redshifted and regenerated. The CMB redshift causes gravity. The rate of energy loss is u x H . CMB regeneration comes from the Hubble luminosity, -U x H . Since u is about equal to -U , as you say, with our above assumption the two rates are equal. With the Hubble luminosity coming mostly from the outer shell, the temperature of the shell is a bit higher, at 29 K. My model thus predicts 29 K as a maximum CMB temperature. I think you assume like everyone else here that the CMB is constant at 2.7 K. How you got the numbers to square with 2.7 K is not clear to me. I would need to understand your model better.

Your link is to a “value-added catalog”. It takes time to produce those. I don’t think the DESI team is hiding anything, just taking their time to produce the extra data required for a comparable data set. Hopefully we get something like SDSS’s quasar catalogs from DESI. They don’t simply fall out of the raw data though. It takes some effort.

I disagree, it takes time to produce a paper and analysises but the times it takes is roughly independent of the sample size of the dataset, otherwise if it has taken them 1 year to produce this paper and that their next one contains tens as much data will it take them ten years? No they have not manually inspected their 400000 stellar spectras. My conjecture is that the team behind what is one of the most important DESI survey (the only milky way DESI survey) is not a member of the official DESI consortium for inept bureaucratic reasons. A tentative indication of this, is that their paper is not affiliated with the official DESI arxiv account https://arxiv.org/search/astro-ph?searchtype=author&query=DESI+Collaboration The consequence of this conjecture is that we would otherwise be already able to rule out either MOND or DM with near certainty.

With all due respect, I think you’re looking at the project as more of monolith than what it is in reality.

The “I” in DESI stands for instrument. The DESI team works on the instrument.

What you think is important (Milky Way Survey) is not what I think is important (quasar survey).

The people that work on the instrument should be focused on making the instrument and the data that comes directly out of it.

Other teams take that data and compliment them with more analysis (which are often model dependent), and we get these value added catalogs. It’s kind of like blaming the Linux kernel developers for not updating the icons on the desktop. It’s not really how that works.

Personally, I wouldn’t want model dependent analysis at the “kernel” level.