# Determine the mass of baryonic matter based on relativity constants ?

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• 2019-Jan-30, 02:16 AM
Reality Check
Quote:

Originally Posted by stephaneww
This is just a shortcut of language from me

You have been writing that the Planck force is the force in the Planck units. That starts in your first post. All systems of units are equal. Selecting to use one is a matter of convenience, not physics.

Quote:

Originally Posted by stephaneww
explain how you calculate a force with only one mass please ? ...

The Planck force is not a force between any masses, it is a unit of force like a Newton. The Planck force is not calculated, it is derived from the other Planck units. Planck force is the derived unit of force resulting from the definition of the base Planck units for time, length, and mass.

If we did an calculation of the Newtonian gravitational force between 2 units of mass then we have F = G (1 kg) (1 kg)/r^2 or F = G (1 mp) (1 mp) )/r^2, or etc. The force will be related to the unit of mass squared. The distance r is irrelevant. Put it to 1 meter, 1 light-year, a billion light-year and the force is still related to the unit of mass squared.

It is the Newtonian law of gravitation has two masses in it. These can be two equal masses but that is not needed. The Newtonian law of gravitation is not used for cosmology for many reasons. For example, there is no way for the universe to expand as observed using Newtonian gravitation. Specifically in Newtonian gravitation the cosmological constant that you want to include does not exist.

Standard cosmology has the universe filled with homogeneous and isotropic matter. That is the cosmological principle. My first thought was that the well known shell theorem suggests that the net Newtonian force at any point by the surrounding matter would be zero. Maybe it is not. However it is definitely not a force that pushes mass away to create the expanding universe we observe. The Newtonian gravity between the Milky Way and other galaxies will not push them away.

Quote:

Originally Posted by stephaneww
...It's exactly what I do here

Which is what I have been telling you is wrong - you divide the masses by 2 and so ignore half the baryonic mass of the universe to get a wrong value for the Planck force unit of measurement Fp.
The error is better stated in your post just above that equation: "so, in cosmology we need two constant mass and egal". In cosmology there are never 2 masses. There is the total mass of the universe with a homogeneous and isotropic distribution. It is a continuous distribution. Alternately you can think of it as an infinite number of masses. The actual universe has many more than 2 masses in it, e.g. maybe 2 trillion galaxies each with billions of stars. One thing that the cosmological principle does is allow cosmology to have solutions that are relatively easy to calculate.

N.B. We do not know how big the universe is or even if it has a size and so cannot know the mass of the universe. We can estimate the mass inside the observable universe. We know that the universe is much bigger (estimates of 250 times larger to 10^23 times larger to even larger) and so the baryonic mass of the universe is much bigger than that estimate. We also know that baryonic matter is a small part of the mass of the universe (there is also dark matter) and that most of the gravitational effects in GR comes from dark energy. This is why the cosmological parameters you have referred to are densities.
• 2019-Jan-30, 03:12 AM
stephaneww
Quote:

Originally Posted by Reality Check
You have been writing that the Planck force is the force in the Planck units. That starts in your first post. All systems of units are equal. Selecting to use one is a matter of convenience, not physics.

Quote:

Originally Posted by stephaneww
...
edit :The Planck force has something special: I selected because it's related to cosmology and GR (the pound force not)

see here : https://fr.wikipedia.org/wiki/Consta...nergie_du_vide

Quote:

Originally Posted by Reality Check
The Planck force is not a force between any masses, it is a unit of force like a Newton. The Planck force is not calculated, it is derived from the other Planck units. Planck force is the derived unit of force resulting from the definition of the base Planck units for time, length, and mass.

here however you begin with 2 mass and didn't answer my question...
Quote:

Originally Posted by Reality Check
If one did an irrelevant calculation of the Newtonian gravitational force between 2 units of mass then we have F = G (1 kg) (1 kg)/r^2 or F = G (1 mp) (1 mp) )/r^2, or etc. The force will be related to the unit of mass squared.

Quote:

Originally Posted by stephaneww
and what is the value of r to have Fp? ...

Quote:

Originally Posted by Reality Check
If we did an calculation of the Newtonian gravitational force between 2 units of mass then we have F = G (1 kg) (1 kg)/r^2 or F = G (1 mp) (1 mp) )/r^2, or etc. The force will be related to the unit of mass squared. The distance r is irrelevant. Put it to 1 meter, 1 light-year, a billion light-year and the force is still related to the unit of mass squared.

if you want to be coherent you have to use the unit of lenght in each system unit

at this point, I think that my calculations can be qualified as being realized in the Planck unit system and what we can call the "LambdaCDM model cosmological unit system" for the second parts of the messages #1 and #2
• 2019-Jan-30, 03:42 AM
Reality Check
Quote:

Originally Posted by stephaneww

A waste of my time, stephaneww, because that link does not say that Fp is special in cosmology. All it does is use the Plank force unit of measurement once to simplify an equation. There is an equation using G and c. We want to make it simpler by setting G = 1 and c = 1. That is Planck units.

The English Cosmological constant article doe not mention Planck force at all.
• 2019-Jan-30, 03:47 AM
Reality Check
Quote:

Originally Posted by stephaneww
here however you begin with 2 mass and didn't answer my question...

My post was that you are wrong because Planck force is the derived unit of force resulting from the definition of the base Planck units for time, length, and mass., i.e. there are no 2 masses.

I gave an irrelevant (this means "not about your misunderstanding about the Planck force") calculation of 2 masses because that is Newton's law of gravitation. That force depends on the masses multiplied together. If they are equal then it is mass squared.

I answered your question: The distance r is irrelevant. Put it to 1 meter, 1 light-year, a billion light-year and the force is still related to the unit of mass squared.

FYI : Put r = rp and the force is still related to the unit of mass squared.
• 2019-Jan-30, 03:55 AM
stephaneww
Quote:

Originally Posted by Reality Check
A waste of my time, stephaneww, because that link does not say that Fp is special in cosmology. All it does is use the Plank force unit of measurement once to simplify an equation. There is an equation using G and c. We want to make it simpler by setting G = 1 and c = 1. That is Planck units.

The English Cosmological constant article doe not mention Planck force at all.

It isn't important, the relation is true. The English Force Planck mentions the GR...
Quote:

This limit property is valid both for gravitational force and for any other type of force.
• 2019-Jan-30, 04:02 AM
Reality Check
Quote:

Originally Posted by stephaneww
... what we can call the "LambdaCDM model cosmological unit system" for the second parts of the messages #1 and #2

There is no "LambdaCDM model cosmological unit system", stephaneww. Any unit system can be used. An English scientist might use imperial units. A French scientist might use SI units. An Argentinian scientist might use Planck units. An Egyptian scientist might use ancient Egyptian units. A Roman scientist might use ancient Roman units. An MIT scientist might use Smoots. A possible scientist on an alien planet can use their own units. The next day they may all select to use something different.

Cosmology literature tends to be in SI units because that is the standard (it is in the name). If you want we can go through some of the well known papers so that you can see that.
• 2019-Jan-30, 04:03 AM
stephaneww
Quote:

Originally Posted by Reality Check

FYI : Put r = rp and the force is still related to the unit of mass squared.

I'don't understand what is "FYI" (I'm french, what does it means please ?

what is rp ?
• 2019-Jan-30, 04:27 AM
stephaneww
Quote:

Originally Posted by Reality Check
Any unit system can be used. An English scientist might use imperial units. A French scientist might use SI units. An Argentinian scientist might use Planck units. An Egyptian scientist might use ancient Egyptian units. A Roman scientist might use ancient Roman units. An MIT scientist might use Smoots. A possible scientist on an alien planet can use their own units. The next day they may all select to use something different.

It doesn't matter, I have already verify :
Quote:

Originally Posted by stephaneww
Quote:

Originally Posted by Reality Check;
Measurements can be made in any system of units that a person wants to use.

except, if you use another system of units, that you must then multiply or divide to obtain Planck's mass and Planck's length, and/or, the physical notions G, c, h, Lambda, Mpc. by the equivalents of the imperial units that you use to conserve the consistency between the SI and Planck's units in your system of units. You do not change anything at the end. (I have tested for inch and pound)

• 2019-Jan-30, 06:10 AM
tusenfem
Quote:

Originally Posted by stephaneww
I'don't understand what is "FYI" (I'm french, what does it means please ?

what is rp ?

• 2019-Jan-30, 08:05 PM
stephaneww
Quote:

Originally Posted by Reality Check
You have been writing that the Planck force is the force in the Planck units. That starts in your first post. All systems of units are equal. Selecting to use one is a matter of convenience, not physics.

The Planck force is not a force between any masses, it is a unit of force like a Newton. The Planck force is not calculated, it is derived from the other Planck units. Planck force is the derived unit of force resulting from the definition of the base Planck units for time, length, and mass.

There is others ways to calculâte them with G, c and hbar

Quote:

Originally Posted by Reality Check
If we did an calculation of the Newtonian gravitational force between 2 units of mass then we have F = G (1 kg) (1 kg)/r^2 or F = G (1 mp) (1 mp) )/r^2, or etc. The force will be related to the unit of mass squared. The distance r is irrelevant. Put it to 1 meter, 1 light-year, a billion light-year and the force is still related to the unit of mass squared.

except than if you want Fp, r must be equal to lp

Quote:

Originally Posted by Reality Check
It is the Newtonian law of gravitation has two masses in it. These can be two equal masses but that is not needed. The Newtonian law of gravitation is not used for cosmology for many reasons. For example, there is no way for the universe to expand as observed using Newtonian gravitation. Specifically in Newtonian gravitation the cosmological constant that you want to include does not exist.

the Planck lenght doesn't exist too in Newtonian gravitation, however you can use it to calculte Fp

Quote:

Originally Posted by Reality Check
Standard cosmology has the universe filled with homogeneous and isotropic matter. That is the cosmological principle. My first thought was that the well known shell theorem suggests that the net Newtonian force at any point by the surrounding matter would be zero. Maybe it is not....

I think it's not, because in your link of shell theorem, the density of ordinary matter must be constant. It's not the case in the observable universe in expansion

Quote:

Originally Posted by Reality Check
However it is definitely not a force that pushes mass away to create the expanding universe we observe. The Newtonian gravity between the Milky Way and other galaxies will not push them away.

have the same value than Fp, and the pression associates to the energy density of the cosmological constant is negative, so...

Quote:

Originally Posted by Reality Check
Which is what I have been telling you is wrong - you divide the masses by 2 and so ignore half the baryonic mass of the universe to get a wrong value for the Planck force unit of measurement Fp.
The error is better stated in your post just above that equation: "so, in cosmology we need two constant mass and egal". In cosmology there are never 2 masses. There is the total mass of the universe with a homogeneous and isotropic distribution. It is a continuous distribution. Alternately you can think of it as an infinite number of masses. The actual universe has many more than 2 masses in it, e.g. maybe 2 trillion galaxies each with billions of stars. One thing that the cosmological principle does is allow cosmology to have solutions that are relatively easy to calculate.

You forgot that this 2 masses are separed by a radius . This radius exist in Einstein Universe, but I didn't find how to link it with my propositions.

Quote:

Originally Posted by Reality Check
N.B. We do not know how big the universe is or even if it has a size and so cannot know the mass of the universe. We can estimate the mass inside the observable universe. We know that the universe is much bigger (estimates of 250 times larger to 10^23 times larger to even larger) and so the baryonic mass of the universe is much bigger than that estimate. We also know that baryonic matter is a small part of the mass of the universe (there is also dark matter) and that most of the gravitational effects in GR comes from dark energy. This is why the cosmological parameters you have referred to are densities.

If you accept, we can put this limit:

The Newtonian part of CDM is due to gravitational influence being regarded as only within the sphere, not outside the sphere of an expanding universe. As such any mass outside the expanding sphere is not included in the WMAP/PLANCK data as they are limited to the visible universe. Due to the BB model this mass is ignored totally because there is supposed to be nothing existing outside as the BB created everything, even space time.
• 2019-Jan-31, 01:48 AM
Reality Check
Quote:

Originally Posted by stephaneww
There is others ways to calculâte them with G, c and hbar...

This is not the actual errors in your ATM idea, stephaneww, and replies irrelevant to what I wrote.
The Planck force is not an actual force that is calculated. The Planck force is the derived unit of force resulting from the definition of the base Planck units for time, length, and mass.

I wrote that IF we do calculations that actually use Newtonian gravitation (not what you are doing) then we MAY get certain results.

Standard cosmology has the universe filled with homogeneous and isotropic matter. That is the cosmological principle. We test the cosmological principle and see that on cosmological scales matter is homogeneous and isotropic.

Quote:

Originally Posted by stephaneww
You forgot that this 2 masses are separed by a radius ...

I was reading your English again, stephaneww, and coming to the same conclusion. What you mean by "we need to divide by two" is "we need to divide into two".

But now you emphasize the massive flaw in your ATM idea. The cosmological constant is not a distance
Quote:

In cosmology, the cosmological constant (usually denoted by the Greek capital letter lambda: Λ) is the energy density of space, or vacuum energy, that arises in Albert Einstein's field equations of general relativity.
N.B. This article has a strange "true dimension" sentence that seems to say there is a "true" system of units. This is obviously wrong since system of units are equally "true".

The cosmological constant is not in GR as a distance between anything. GR has a convention (a accepted way of doing things) of setting G = 1 and c = 1. This is a way to make the equations look simple. That might be what gives units of length^-2.

Cosmological constant:
In what may be the GR convention, the cosmological constant measured by Planck may be 1.1056 10^-52 m^-2.
In "reduced Planck units", the cosmological constant measured by Planck has no units! It is just a number.
In natural units which include Planck units, the cosmological constant measured by Planck is 4.33×10^−66 eV^2.
• 2019-Jan-31, 02:17 AM
Reality Check
Quote:

Originally Posted by stephaneww
You forgot that this 2 masses are separed by a radius . This radius exist in Einstein Universe, but I didn't find how to link it with my propositions..

My previous post explains that in GR is not any kind of distance. just has the units of length^-2 in a convention (set G = c = 1 to make equations simpler) or a system of units.

But say that we ignore this flaw and let you do that. That radius would be about 10^26 metres or about 10,000,000,000 light years. That has no importance in cosmology.
You are treating the universe as two equal bodies of baryonic matter separated by an enormous distance of about 10,000,000,000 light years (the r in Newton's law of gravitation). This is not the universe we exist in. The evidence is that universe has baryonic and non-baryonic spread throughout it. There is certainly no 10 billion light year gap.

That toy universe cannot expand or accelerate because all you include is Newton's law which gives an attractive force between the 2 bodies. The universe actually "collapses' since over time the bodies will collide. No expansion implies that in that universe the value of = 0 with units of length^-2 or eV^2 or no units or whatever else someone chooses to use.
• 2019-Jan-31, 02:39 AM
Reality Check
Quote:

Originally Posted by stephaneww
If you accept, we can put this limit:

The Newtonian part of CDM is due to gravitational influence being regarded as only within the sphere, not outside the sphere of an expanding universe. As such any mass outside the expanding sphere is not included in the WMAP/PLANCK data as they are limited to the visible universe. Due to the BB model this mass is ignored totally because there is supposed to be nothing existing outside as the BB created everything, even space time.

I cannot accept something that is obviously wrong ( is not any kind of distance, treating as kind of distance leads to a universe that is not ours).

Your "limit" is not good cosmology but may be a result of English as a second language.
Big Bang
Friedmann–Lemaître–Robertson–Walker metric (the GR solution used in cosmology for a "homogeneous, isotropic, expanding (or otherwise, contracting) universe that is path-connected, but not necessarily simply connected").

There is no "outside the sphere of an expanding universe" since a universe is everything that there is. By definition, there is no outside to a universe. The universe is not expanding into an outside.
We are limited to the observable (visible) universe. That allows us to measure the densities that control the expansion of the universe. The mass is not used.
The BB model has mass existing outside of the observable universe. We just cannot observe that mass. The entire universe is filled with mass in the BB model.
The BB model does not say the BB created everything, even space time. We know that as we get closer to t = 0, densities and pressures increase so much that our physics breaks down. This is called the initial singularity. We cannot know what happened at t = 0 in the BB model and so it starts at t > 0 with existing spacetime.
• 2019-Feb-01, 01:03 AM
stephaneww
Quote:

Originally Posted by Shaula
Do you have a theory that led to that equation?

Hello Shaula

I think I finally found one. I have it with the numerical values, I still have to finalize the formulas, check the dimensional analysis and do the translation in English.

It's based on the static universe of Einstein
• 2019-Feb-01, 02:20 AM
Shaula
Quote:

Originally Posted by stephaneww
I think I finally found one. I have it with the numerical values, I still have to finalize the formulas, check the dimensional analysis and do the translation in English.

Hope to see it presented here soon then. Good luck finalising it.
• 2019-Feb-01, 02:28 AM
Reality Check
Quote:

Originally Posted by stephaneww
...It's based on the static universe of Einstein

That is a problem for your ATM idea, stephaneww, because Einstein's static universe was wrong. It is probable that Einstein considered the introduction of the cosmological constant into GR to allow a static universe, "the greatest blunder of his life".
Investigating the legend of Einstein’s “biggest blunder”
Quote:

An Einstein scholar concludes that the famed physicist probably did use those words to describe the cosmological constant.
...
In 1956, the year after Albert Einstein’s death, physicist George Gamow wrote a Scientific American article on the Big Bang model that included a tantalizing historical tidbit: “Einstein remarked to me many years ago that the cosmic repulsion idea was the biggest blunder he had made in his entire life.”1 Gamow repeated the claim in his 1970 autobiography,2 and the story became one of the great legends of 20th-century physics. The story only got juicier with the 1998 discovery that the rate of cosmic expansion is increasing, an observation that cast Einstein’s “biggest blunder” in a new light.
We have known that this was a "blunder" since the discovery that the universe is expanding in 1927 (Georges Lemaître) and 1929 (Edwin Hubble). The Einstein universe
Quote:

Because the Einstein universe soon was recognized to be inherently unstable, it was presently abandoned as a viable model for the universe. It is unstable in the sense that any slight change in either the value of the cosmological constant, the matter density, or the spatial curvature will result in a universe that either expands and accelerates forever or re-collapses to a singularity.

After Einstein renounced his cosmological constant, and embraced the Friedmann-LeMaitre model of an expanding universe,[5] most physicists of the twentieth century assumed that the cosmological constant is zero. If so (absent some other form of dark energy), the expansion of the universe would be decelerating. However, after Saul Perlmutter, Brian P. Schmidt, and Adam G. Riess introduced the theory of an accelerating universe during 1998, a positive cosmological constant has been revived as a simple explanation for dark energy.
Any scientific theory based on a theory that has an overwhelming amount of evidence showing to be wrong, starts with an enormous handicap. The author has to show that all of that evidence is either explained by that base theory or is wrong. That cannot be done in less than 30 days.
• 2019-Feb-01, 03:15 AM
stephaneww
Hum. This seems too simple : I reverse matter density and density of the cosmological constant in the equation of the static universe of Einstein. because it's the density of the cosmological constant witch is constant and not the density of matter

The value of the measured cosmological constant is exactly

edit

is in kg/m^3 to have in m^-2
• 2019-Feb-01, 03:28 AM
Reality Check
Quote:

Originally Posted by stephaneww
Hum. This seems too simple: ...

Too simple to be correct and very wrong in any case, stephaneww, because the universe is expanding. Simply a matter density is not an energy density. They cannot be swapped because they are two different things.

ETA: Even the English in the article does not help: "ρ is the energy density of the matter in the universe". The cosmological constant is energy density related to spacetime, not matter. They are not the same thing.
• 2019-Feb-01, 09:43 PM
slang