stephaneww

3 minutes ago, Mordred said:

No because velocity is a vector. It has a direction.

if the velocity vector is the time arrow (everywhere the same in all directions) it still can't work?

18 minutes ago, Mordred said:

I think the problem your having is how expansion works. There is no Centre of the universe nor does it expand  in any direction from the Centre.  

Also the singularity isn't at any particular direction from us if we were to look into the past. Assuming we can see far enough back in time you would not be able to point at any particular direction and say the BB occurred  in that direction

I knew that and I understand that, no problem, we agree.

Okay, let me try this  :
here .https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1117499  we have also

[math]E_{end}/V_{max}=\frac{1.448∗10^{76}J}{2.179*10^{85}m^3}=5.243*10^{-10}J/m^3[/math] (the Density Energy of cosmological constant, I just realized that.)

[math]5.243*10^{-10}J/m^3.c.\pi=1/1.994.J/m^{-3}.m/s[/math] in another word the energy Density of the cosmological constant at the singularity of the Big Bang multipiled by a velocity [math](c.\pi)[/math] in [math]m/s[/math].

Doesn't that fit that definition ?

Quote

Intensity can be found by taking the energy density (energy per unit volume) at a point in space and multiplying it by the velocity at which the energy is moving. The resulting vector has the units of power divided by area (i.e., surface power density).

souce : https://en.wikipedia.org/wiki/Intensity_(physics)

I must admit that all this is beyond my current knowledge. There remains that perfect equality that I still do not understand.

Um, which post are you talking about, please?

1.https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1117499
2.https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1117557

both ?

Do you remember how to write the energy density of the cosmological constant expressed with Planck's force. Is it impossible to deal with that?

There's no way to say that Planck's field =[math] \phi_p=1/2 .h[/math] is involved ?

Finally, there is no way to say that [math]\Lambda[/math] in [math]m^{-2}[/math] is the division by the surface ?

Quote

Intensity can be found by taking the energy density (energy per unit volume) at a point in space and multiplying it by the velocity at which the energy is moving. The resulting vector has the units of power divided by area (i.e., surface power density).

souce : https://en.wikipedia.org/wiki/Intensity_(physics)

I wouldn't sleep so dumb tonight:
[math]J/m^3 . m/s[/math],  is also the unit of the surface power density : [math]kg/s^3[/math] or  [math] W/m^2 [/math]

I let you do the dimensional analysis that goes well with [math]1/2, \hbar,t_p, \Lambda,8 \pi[/math] based on the first Wikipedia link in the article about the surface power density (electromagnetic radiation)

Then follow the link Particle model and quantum theory, 

With the factor 1/2, you should find the value [math]1/1,994 kg/s^3[/math] 

We should have, with the values from the first post on page 4, and second post page 5 :

[math]\frac{1}{2}\frac{h}{t^2}\frac{\Lambda}{8\pi}=0.50144. kg/s^3=1/1.994.kg/s^3[/math]

But as I have very little knowledge of quantum mechanics, and nothing in electromagnetism, I don't know how to interpret this result

A little help is welcom, thanks in advance

there was still one mistake left at the end of first post page 4 😓

please read : ( [math] \Omega_\Lambda, at H(p,minimum)[/math] missed )

[math]E_{tot,end}.c.\pi=1.446*10^{76}.c.\pi=1.364*10^{85}J.m/s[/math]

and :

[math]1.364*10^{85}/V_{max}=1.3364*10^{85}/2.719*10^{85}=1/1.994 \frac{J}{m^3}m/s[/math], in another word a energy density that displaces.

 hoping I didn't leave any more mistakes to be left behind...

Um, again a mistake at the top of  page 4, sorry :

On 9/4/2019 at 10:40 AM, stephaneww said:

...

So we will have:
- [math]D=6.118∗10^{-113}J/m^3[/math]

...

[math]D=6.118∗10^{-133}J/m^3[/math], -133 is the good exponant 

the next value was correct :

[math]=6.073*10^{-31}s^{-1}[/math]

What do you think of this paper that as gone through a peer review process please  ?

http://journalofcosmology.com/JOC26/General%20Relativity%20and%20Effects%20in%20Time%20as%20Causation%20JofC.pdf

On 9/9/2019 at 6:11 AM, stephaneww said:

- [math]V_{max}.ρ_c,at H(p,minimum).c.π=1.364∗10^{85}J.m/s[/math]

-  [math]V_{max}/1.364∗10^{85}=2.719∗10^{85}/1.364∗10^{85}=1/1.994.sJ^{-1}m^{-2}[/math] 

[math]V_{max}/1.364∗10^{85}=2.719∗10^{85}/1.364∗10^{85}=1/1.994.J^{-1}m^2.s[/math] 

are the good units if I don't make a new mistake. 

but apart from the numerical value = 1/2, I don't know how to interpret it

the other values and units are fair.

1 hour ago, swansont said:

I don’t understand why you think this matters.

I think I didn't understand the initial question then.

1 hour ago, swansont said:

Does it change the fact that each velocity exist for only a single value of time?

Yes, of course, that doesn't change anything.

1 hour ago, swansont said:

Velocity is a vector, so you should have stopped after saying it has each velocity for precisely 0 seconds.

 

edit: and now I see Strange has made this same point.

you need an acceleration to get the speed: before v=0.

you can set t=0 at the time of the acceleration event but the time exists before v=0 and after v returns to 0

2 hours ago, swansont said:

"Time never stops" is not a proper response to the question of whether a duration of zero is possible.

Yes for minutes of one hour for example: the zero evolves from 0 to 1 for 1 minute, the same for hours of the day.

2 hours ago, swansont said:

Answer the kinematics question from above:
If I toss a ball upward, it has a velocity in the upward direction. Then it slows down under gravity, and will have a velocity in the downward direction.

What is the duration over which it has zero velocity?

The reference material is the earth for this measurement. Time continues to pass. When the speed is zero, the answer is the same as above

On 9/4/2019 at 10:40 AM, stephaneww said:

In addition, we have:
- Vmax.ρc,at H(p,minimum).c.π=1.364∗10^85J.m/s
and 
- [math]V_{max}/1.364∗10^{85}=2.719∗10^{85}/1.364∗10^{85}=1/1.994.s.J^{−1}m^{−1}[/math]  

Um, I believe there are errors on the physical dimensions.

I'll let you fix it until I have a moment to time it.

On 8/26/2019 at 1:24 AM, Farid said:

Hi everyone,

I wanted to create this thread because of the duration of time itself. Time cannot have a duration of more than zero seconds. Time having a duration of more than zero seconds means that when time is one second, an amount of time that is the duration of that time passes and time is still one second, which is impossible. 

 

 

Hi,

No and the answer is very simple: after 60 seconds a hand that counts the seconds on a dial watch goes through 0 and starts running again for a new cycle. you add 1 to the number of cycles of the seconds. time never stops.

edit  :

for the universe what exists at point 0 is the singularity of the Big Bang

now that the model is complete, I allow anyone interested to write this thread according to the scientific rules for a free consultation publication on arXiv. indeed it is not in my competence....

Well, we have:

On 4/27/2019 at 7:03 PM, stephaneww said:

Fp : Planck force

1 the energy density by volume of the quantum vacuum as A=Fp/lp^2 = 4.633*10^113 Joules/m^3 (formula derived from dimensional analysis in Planck units)

2 the energy density by volume of the vacuum of the cosmological constant as B= Fp* Lambda /8/ pi = 5.354*10^-10 Joules/m^3 

the ratio between the two being the number in factor of 10^122 undimensioned

the value of the adimensionless factor, X, is more precisely : A/B=X=8.654*10¹²², X is called " The cosmological constant problem ".

and :

21 hours ago, stephaneww said:

Moreover, with the geometric mean we had

[math]C=\sqrt{AB}[/math]

or

[math]C^2=AB[/math]
or 
[math]1/B=AC^{-2}[/math]

important notes : B page 1, is different of the formula of [math]B=\hbar \Lambda^2 c /(8 \pi)^2[/math], page 2

B page 1  [math]=C[/math] (The energy density of the cosmological constant) page 2

so

[math]1/B=[/math]

[math]\frac{(8*\pi)^2}{\hbar \Lambda^2 c}=\frac{F_p}{l_p^2}\frac{(8\pi)^2}{F_p^2 \Lambda^2}[/math]

so

[math]\frac{1}{\hbar  c}=\frac{1}{l_p^2}\frac{1}{F_p}[/math]

so

[math]\frac{F_p}{\hbar  c}=\frac{1}{l_p^2}[/math]

so

[math]l_p^2=\frac{\hbar  c}{ F_p}[/math] with [math]F_p=\frac{c^4}{G}[/math] cf wikipedia

so

 [math]l_p^2=\frac{\hbar G}{ c^3}[/math] with [math]t_p^2=\frac{l_p^2}{ c^2}[/math] cf french wikipedia

so

[math]\frac{l_p^2}{ c^2}=\frac{\hbar G}{ c^5}[/math]

so 

[math]t_p=\sqrt{\frac{\hbar G}{ c^5} }[/math] i.e  the expected formula for the beginning of a new cycle

hey hey hey, no, it's very simple with Planck's force and we find exactly Planck's time at the origin of the cycles.

I need to rest. The demonstration is coming soon.

8 hours ago, stephaneww said:

outch, I hope to do it more simply with Planck's force.

um, it seems to be a dead end....

outch, I hope to do it more simply with Planck's force.

Thanks for the note.

I just wanted to try to get back to planck time. I'm still going to try

Okay, but I prefer zero point energy with [math]\epsilon=\frac{1}{2}\frac{h}{t_p}=9.78*10^8J[/math] (with p for "Planck")?

Here's why:

By reducing it to an energy density we have:
[math]\epsilon_p / l_p^3=1/2 \frac{\hbar}{t_p}/l_p^3 = 2.316*10^{133} J/m^3[/math]

and the energy density of the quantum vacuum
[math]A=m_p c^2/l_p^3=4.63* 10^{133}=[/math] 

[math] 2 \epsilon_p / l_p^3[/math] 

In other words, with the circle we have the occupation of the part of the -y axis that completes the +y part. But you're right, I have to take into account the time arrow. i.e. go back to a sinusoidal

Moreover, with the geometric mean we had:
[math]C=\sqrt{AB}[/math]
or 
[math]C^2=AB[/math]
or 
[math]1/B=AC^{-2}[/math]

Okay, I think I understand.

Preliminary question:

Do we also have zero point energy [math]\epsilon=\frac{1}{2} \frac{h}{t_p}=9.78*10^8J[/math] (with p for "Planck")?

The symmetric in this representation would be in the negative part of the Y axis????
So imaginary numbers but I don't understand the need for them 

For the zero-point energy I think I may have a solution that would fit with this geometric representation.
I just need to recover it and put it in latex format.

47 minutes ago, Mordred said:

I really don't know how you plan on applying time as a sinusoidal. 

I changed my mind after that: I apply time like a circle. so 3d space would be euclian, but not the space time

see :

19 hours ago, stephaneww said:

 

I think it is better to say that time describes a circle of center M  and that the 3D space is euclian

"Something like a Yin and yang…" 

What do you think of this approach, please?

47 minutes ago, Mordred said:

Think of it as the average value.

So this is the point M on wikipedia, correct ?

Do I still have to recover the zero-point Energy ?

14 hours ago, stephaneww said:

14 hours ago, stephaneww said:

  16 hours ago, stephaneww said:

On 9/5/2019 at 1:27 AM, Mordred said:

You would need a mechanism of collapse to get a new BB.

 I have no idea at the moment.

Maybe because the time is sinusoidal

this hypothesis is based on the geometric representation of the geometric mean of Wikipedia.  (time go from C' to A ?)

as a reminder, it is this average that is used to solve the problem of the cosmological constant

but I don't know how to develop this hypothesis more precisely at the moment.

perhaps you can help me ?

13 hours ago, stephaneww said:

Normally the energy density of the cosmological constant should be the point M

I think it is better to say that time describes a circle of center M  and that the 3D space is euclian

"Something like a Yin and yang…" 

What do you think of this approach, please?

6 hours ago, Mordred said:

 in a homogenous and isotropic (uniform distribution the energy density should be the same everywhere) however when you state that under QM your already referring to the mean average. It is no different

I'm not sure I understand. Can you develop it, please? (my translator stumbles upon "mean average" )