why this egality is exact please ?

[stephaneww]

[math]\Large{ M_{\text{U Hubble at }t_H} = \sum_{1}^{t_H / t_p}    \frac{ m_{Pl}}{2}} [/math]

 

I have an idea but I need confirmation

thanks in advance

Edited October 19, 2021 by stephaneww

[mathematic]

It would help if you defined symbols.

[stephaneww]

Hello

 

M_{U Hubble at tH} : mass of universe at Hubble radius

t_H : Hubble time

t_P : Planck time

m_Pl : Planck mass

Edited October 19, 2021 by stephaneww

[joigus]

It doesn't make sense to me, which doesn't mean it doesn't make sense. What does it mean to add half Planck's mass from one to the ratio Hubble's time/Planck's time?

I can't make heads or tails of it.

Where did you get it from?

 

Mmmmm. I think I have an intuition of what you're trying to do there... Think Hubble-to-Planck units.

Also, what's the purpose?

I'll get back to you. In the meantime, perhaps someone can answer.

[stephaneww]

36 minutes ago, joigus said:

It doesn't make sense to me, which doesn't mean it doesn't make sense. What does it mean to add half Planck's mass from one to the ratio Hubble's time/Planck's time?

I can't make heads or tails of it.

Where did you get it from?

 

 

The formula follows from figure 2 of this article arxiv with a correction on one of the data.

If you test the formula with 2 or 3 values, it matches with exatitude each time after the recombination.

36 minutes ago, joigus said:

Mmmmm. I think I have an intuition of what you're trying to do there... Think Hubble-to-Planck units.

Also, what's the purpose?

I'll get back to you. In the meantime, perhaps someone can answer.

It is indeed a micro macro link,as it is proposed in the arxiv document figure 1.

It is a formula that expresses the growth of the observable universe and why its energy is not conserved.

Edited October 20, 2021 by stephaneww

[stephaneww]

oops i forget this in your edit,sorry

2 hours ago, joigus said:

Also, what's the purpose?

 

replace the "empty set" at the begining in the figure 2, by m_p/2 in a unit of Planck sphere volume (for t₁ = t_p)

at t = 2 t_p you begin the figure 1 with two units of Planck sphere volume

 

the formula explains how, in the cosmological standard model, we go from 1 nucleon at Planck time to about 10⁸⁰ nucleons today at hubble radius.

Edited October 20, 2021 by stephaneww

[studiot]

Quote

why this egality is exact please ?

I don't know enough cosmology to know what the equation is about,

But it is important to realise that the equals sign ( = ) is used by many for two different meanings, viz equality and identity.

 

Which do you mean ?

[stephaneww]

Hello

I am not sure to understand what you put under the term identity.

if I understand correctly the notion of identity, it can be each point = unit of volume of Planck sphere with its associated mass (mp/2) per unit of time (t_p) in the figures quoted above

note that each point can also be a bit of information = nucleon ( identity here)

I am affirmative for the equality of the formula on the other hand.

the notion of summation can be simplified with the help of this calculator for those who do not know this notion

the summation becomes t_H/t_p x m_p/2

 

we only need the Hubble time value (see on wikipedia) chosen to make the calculation.

 

Edited October 20, 2021 by stephaneww

[swansont]

17 hours ago, stephaneww said:

The formula follows from figure 2 of this article arxiv with a correction on one of the data.

That’s a pretty big leap to say it “follows” from that, considering the article makes no mention of any Planck units, or Hubble. Can you show a derivation? Preferably starting from some verifiable equation or identity.

[stephaneww]

33 minutes ago, swansont said:

That’s a pretty big leap to say it “follows” from that, considering the article makes no mention of any Planck units, or Hubble. Can you show a derivation? Preferably starting from some verifiable equation or identity.

Hello

I can do it for the mass of the observable universe

I added a personal hypothesis for the Hubble radius.

It's here :

https://vixra.org/abs/2110.0108

sorry I don't have an entry on arxiv and for the observable universe it's a démonstration , not a spéculation

 

Edited October 20, 2021 by stephaneww

[studiot]

5 hours ago, stephaneww said:

Hello

I am not sure to understand what you put under the term identity.

if I understand correctly the notion of identity, it can be each point = unit of volume of Planck sphere with its associated mass (mp/2) per unit of time (t_p) in the figures quoted above

note that each point can also be a bit of information = nucleon ( identity here)

I am affirmative for the equality of the formula on the other hand.

the notion of summation can be simplified with the help of this calculator for those who do not know this notion

the summation becomes t_H/t_p x m_p/2

 

we only need the Hubble time value (see on wikipedia) chosen to make the calculation.

 

 

Let us say that you have a simple equation

f(x) = 3x -7

The equals sign here is an identity.

This means it is true for any and all values of x.

Or that f(x) and 3x + 7 are identical, and may be substituted for each other in any application.

 

Now let us say that you have an equation

3x-7 = 0

There is only one value of x which satisfies this equation and any other value you assign to x does not.

That is just an equality for x = 7/3

 

Does this help ?

[stephaneww]

I think so

I think I reversed the 2 notions in my previous posts if I understood correctly

thanks

 

hum isn't are two identity ? I'm not sure

Edited October 20, 2021 by stephaneww

[studiot]

51 minutes ago, stephaneww said:

 

hum isn't are two identity ? I'm not sure

No idea what this is  ?

[stephaneww]

26 minutes ago, studiot said:

No idea what this is  ?

if I well undersetand
the summation sigma should be an identity
bit of information = nucleon an equality
and each point = unit of volume of Planck sphere with its associated mass (mp/2) per unit of time (tp) in the figures
 an equality too

 

if I'm not mistaken, does this help you make sense in case ?

Edited October 20, 2021 by stephaneww

[stephaneww]

Hello

 

to try to be simple and educational :

- the big bang theory tells us that in Plank's time, the mass of the universe is 1 Planck mass

- the figure 1 starts with two units (two circles). To have 1 Planck mass at the beginning, each sphere should weigh 1/2 Planck mass

... hence the 1/2 in the formula of the summation.

- figure 2 starts with an empty set according to the authors. however following the same logic as above it starts with 1/2 Planck mass.

- the summation adds 1 unit of 1/2 mPl per unit of Planck time to the previous summation. this is a formula that explains how we go from a universe that weighs one Planck mass at the origin to a universe that weighs 10^54 kg today

my pdf says the same thing with mass or energy densities

after my toy cosmological model gives also the explanation of the homogeneity of the universe. but each thing in its time

[swansont]

2 minutes ago, stephaneww said:

- the big bang theory tells us that in Plank's time, the mass of the universe is 1 Planck mass

Where does the BB theory tell us this?

[stephaneww]

Sorry, it's in the standard cosmological model. edit so also in the big bang theory

My source is a French researcher of the cnrs who doesn't want to be quoted that I say his name.

he told me :

"it is inevitable that at Planck time, the temperature (in units k_B T) is equal to the Planck mass (multiplied by the speed of light squared), and that the horizon contains at that time only of the order of a single particle. This is simply a consequence of the dimensional analysis feasible with the quantities c, hbar and G."

Edited October 21, 2021 by stephaneww

[swansont]

12 minutes ago, stephaneww said:

Sorry, it's in the standard cosmological model. edit so also the big bang theory

My source is a French researcher of the cnrs who doesn't want to be quoted that I say his name.

he told me :

"it is inevitable that at Planck time, the temperature (in units k_B T) is equal to the Planck mass (multiplied by the speed of light squared), and that the horizon contains at that time only of the order of a single particle. This is simply a consequence of the dimensional analysis feasible with the quantities c, hbar and G."

Doesn't want to be quoted? Isn't this in the literature anywhere? If you can't point to a textbook, or at least a peer-reviewed articles, then it's a dubious claim.

Further, "single particle" is not synonymous with "Planck mass"

[stephaneww]

pause this is going too fast for me. i need to take it easy. please give me some time between each answer so i can edit to clarify or correct my mistakes

Edited October 21, 2021 by stephaneww

[swansont]

The critical mass density of the universe equates roughly to 10^-123 planck masses per cubic planck length, and we're thought to be very close to this number You are claiming this number is 1 (https://en.wikipedia.org/wiki/Planck_units#In_cosmology)

That's a pretty large disparity

3 minutes ago, stephaneww said:

 I admit that I didn't understand it either, I'm just an amateur

All the more reason to be meticulous in citing accepted physics, and distinguishing your own ideas.

[stephaneww]

50 minutes ago, swansont said:

Doesn't want to be quoted? Isn't this in the literature anywhere? If you can't point to a textbook, or at least a peer-reviewed articles, then it's a dubious claim.

 

https://fr.wikipedia.org/wiki/Température_de_l'Univers

 

I only found this reference in French.

but I know that in Planck's unit system 'all values can be set equal to 1' or something close to it.

for a Planck time, you have a Planck mass, in relation with a Planck temperature etc...

 

.

Edited October 21, 2021 by stephaneww

[swansont]

8 minutes ago, stephaneww said:

but I know that in Planck's unit system 'all values can be set equal to 1' or something close to it.

Setting values to 1 is a trick used by theorists who don't care about calculations and don't want to carry extraneous terms in their equations. Its usefulness is lost when you actually want to compare to an actual measured or predicted value.

8 minutes ago, stephaneww said:

it is a density and not a mass

divide by the volume of a Planck sphere, you have exactly the Planck mass

1 planck mass per planck volume is a density. The problem is the critical density is a far, far different value. Why is there a difference?

[stephaneww]

pause this is going too fast for me. i need to take it easy. please give me some time between each answer so i can edit to clarify or correct my mistakes

59 minutes ago, swansont said:

The critical mass density of the universe equates roughly to 10^-123 planck masses per cubic planck length, and we're thought to be very close to this number You are claiming this number is 1 (https://en.wikipedia.org/wiki/Planck_units#In_cosmology)

That's a pretty large disparity

 

this large disparity is the  Cosmological constant problem

this is a whole other topic that I have recently covered in the speculation section and on viXra

 

 

Edited October 21, 2021 by stephaneww

[stephaneww]

let's see if we can do without the disgressions that followed this post :

 

1 hour ago, stephaneww said:

Hello

 

to try to be simple and educational :

- the big bang theory tells us that in Plank's time, the mass of the universe is 1 Planck mass

- the figure 1 starts with two units (two circles). To have 1 Planck mass at the beginning, each sphere should weigh 1/2 Planck mass

... hence the 1/2 in the formula of the summation.

- figure 2 starts with an empty set according to the authors. however following the same logic as above it starts with 1/2 Planck mass.

- the summation adds 1 unit of 1/2 mPl per unit of Planck time to the previous summation. this is a formula that explains how we go from a universe that weighs one Planck mass at the origin to a universe that weighs 10^54 kg today

my pdf says the same thing with mass or energy densities

after my toy cosmological model gives also the explanation of the homogeneity of the universe. but each thing in its time

 

Is this question solved for you?

1 hour ago, swansont said:

Where does the BB theory tell us this?

 

[joigus]

I don't think your formula is correct. What you're essentially doing (if I understand you correctly) is adding up \( \frac{m_{P}}{2} \) a certain number of times,

\[ \frac{m_{P}}{2}+\cdots+\frac{m_{P}}{2} \]

and positing that this amounts to Hubble's mass*. Let's call that \( m_{H} \) for brevity if you will.

How many times are you adding up half a Planck mass? As many as there are Planck times in a Hubble time. So, what you're saying amounts to,

\[ m_{H}=\frac{1}{2}\frac{t_{H}}{t_{P}}m_{P}=\frac{1}{2}t_{H}\sqrt{\frac{\hbar c}{G}}\sqrt{\frac{\hbar G}{c^{5}}}=\frac{1}{2}t_{H}\frac{\hbar}{c^{2}} \]

In other words, the "Hubble mass" is proportional to the Hubble time. For starters, I don't think there's any reasonable definition of a Hubble mass that can be related to the expansion parameter. The amount of mass that's trapped within a Hubble radius is the amount that happens to be there due to the whole expansion history of the universe, and should in no way be assumed proportional to time. Keep in mind that the expansion parameter, in the FLRW models can have in principle any time-dependence that you want to postulate. It can be accelerated, decelerated, oscillating, etc.

*I'm assuming that by "Hubble mass" you mean the mass of the universe within a Hubble radius. What I'm trying to tell you is that the amount of mass in there is nothing to do with the Hubble "mechanism".

 

Edited October 21, 2021 by joigus