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# The end of the quantum vacuum catastrophe ?

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Little side note hint, the corresponding blackbody temperature of the universe will be roughly the inverse of the scale factor. (the above should explain why inverse)

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Sorry for the timeout, I was busy IRL.

You and Marcus. Jorrie did a great job to help people . I'm sure you already know where I go, but you want to make me work

Well,  I move slowly : $a$ is a scaling factor, i.e. a multiplier of an elastic ruler between two combile points (i.e. fixed without expansion)

for example,  between two combile points $A$ and $B$, at $t_0$ the distance is $D_0$, then at $t_1$ the distance between $A$ and $B$ will be $a(t_1)*D$

$S=1/a$ and $z=S-1$ , now it's clear to me.

Quote

s 0.001 this is the CMB surface of last scattering. the a with values greater than 1 is future expansion. (in this case roughly 88 Gyrs into the future lol

But, I don't understand  this quote... surface of the sphere of CMB ??? temperature of the CMB ??? with whitch unit ??? where come "88 Gyrs into the future" ???

and most importantly, I do not see how to derive $a$ to get $\dot{a(t)}$ and $\ddot{a(t)}$

Edited by stephaneww
latex

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edit latex

and most importantly, I do not see how to derive $a$ to get $\dot{a}(t)$ and $\ddot{a}(t)$

Edited by stephaneww

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well, always slowly

I went  to school again.

I had look here : https://en.wikipedia.org/wiki/Derivative , but I don't find what is $\Delta y$ and what is $\Delta x$ for this thread

edit:

if I have the answer for $\dot{a}(t)$, I think I'll find $\ddot{a}(t)$

Edited by stephaneww

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The scale factor time derivitaves are simple ratios.

The scale factor today is 1 you compare the radius at the time you wish to examine and compare it to the radius today. If for example the radius then is 1/2 the radius today. Then $\ddot{a}(t)=0.5$

it is as simple as that. If you have 3 dots then you compare to two dots. If two dots compare to 1 dot. They are just ways to keep track of the sequence of time comparisons

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I'm still very slow, yet I'm pretty sure it's very simple ...

1. $\ddot{a}$ is almost ok. If I just put a ratio on the colone $R(Gly)$. Otherwise, I spent 2 days on the problem without progress ....

2. To agree on the numeric values of the data, I propose to use the values you took for the Cosmological Calculator, at the bottom of page 2 of this thread (With "Number of Steps = 20") . For example I find, $S=3,208, H=7,244*10^{-18}s^{-1}$ and $\epsilon=8,435*10^{-10}$Joules/m^3 for  $T(Gly)=2,9777$, and finally $\epsilon_\Lambda=5,394*10^{-10}$Joules/m^3, which is $\Lambda=1,007*10^{-35}*10^{-35}s^{-2}$

3. On this link: http://www.astronomy.ohio-state.edu/~dhw/A5682/notes4.pdf, I tried to use this formula to no avail: $P=w \epsilon$ with $w=-1$. (page 5 of notes4.pdf) where $( \epsilon+3P)=-2P$. Is $\epsilon$ the critical density ???

4. I do not find the equality Eq. (9) from http://www.scholarpedia.org/article/Cosmological_constant and I can not find the result: For $\frac{\ddot{a}}{a}$, I find values in $10^{-35}$
for the second part of the tie

5. I especially think that my brain is getting too old to learn alone...

Edited by stephaneww
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2 hours ago, stephaneww said:

I'm still very slow, yet I'm pretty sure it's very simple ...

1. a¨ is almost ok. If I just put a ratio on the colone R(Gly) . Otherwise, I spent 2 days on the problem without progress ....

2. To agree on the numeric values of the data, I propose to use the values you took for the Cosmological Calculator, at the bottom of page 2 of this thread (With "Number of Steps = 20") . For example I find, S=3,208,H=7,2441018s1 and ϵ=8,4351010 Joules/m^3 for  T(Gly)=2,9777 , and finally ϵΛ=5,3941010 Joules/m^3, which is Λ=1,00710351035s2

3. On this link: http://www.astronomy.ohio-state.edu/~dhw/A5682/notes4.pdf, I tried to use this formula to no avail: P=wϵ with w=1 . (page 5 of notes4.pdf) where (ϵ+3P)=2P . Is ϵ the critical density ???

4. I do not find the equality Eq. (9) from http://www.scholarpedia.org/article/Cosmological_constant and I can not find the result: For a¨a , I find values in 1035
for the second part of the tie

5. I especially think that my brain is getting too old to learn alone...

Your getting there no worries. No the e for  energy density is not the critical density. The critical density is the density that expansion will halt  its expansion but for a flat universe will take an infinite amount of time. Your numbers are within the right order of magnitude which is good. The confusion your having with e in the equation is related to thinking of it as the critical density. It isn't its the total energy for a spatially flat universe it should be approximately the critical density which you have above.

$\epsilon_t=\epsilon_\lambda+\epsilon_{matter}+\epsilon_{radiation}$ in the notation being used in this paper

So for the individual equations of state you need to look at the mixed states of the fluid equation and use the condition $P<\frac{-\epsilon}{3}$

to derive w=-1 see bottom of that link. The previous steps is matter then radiation equations of state.

Here this procedure has a more thorough procedure to follow

hopefully you will see the connection to the fluid equation without giving the direct answer. As the first link note4 is giving 3 separate examples of deriving the individual equations of state from the fluid equation (specifically the deceleration equation which is the hint for the minus sign. See note if $\epsilon$ an P are both positive the universe decelerates.

Edited by Mordred

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On 09/09/2017 at 6:07 PM, Mordred said:

I'm a little busy this weekend but would like to assist you on your modelling approach, so once I get a chance I will probably provide some further details to assist you in moving forward into your approach.

In particular we will need to test if your approach gives the equation of state w=-1 for the Lambda term under scalar modelling. Though I see nothing above that shouldn't, one cannot assume that lol.

well, I'm not sure, but I don't see how inclued my approach for equation of state w=-1.

Furthermore,

I find $\frac{\ddot{a}}{a}$ at about a numeric value of 1

while, $\frac{4\pi G}{3c^2}*(\epsilon+3P)$ is about a numeric value of $10^{-35}$

so I tend to think that equality is not verified, and therefore my proposal on the vacuum catastrophe is wrong ...

edit : I wanted to say energy and not critical density

Edited by stephaneww

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While I agree that equality of your proposal isn't definitive. I think you may not be grasping the usages of the dimensionless constants for the scale factor nor the w term.

These two constants by themselves are simply dimensionless ratios only. They have no units unto themselves. The units of measure of those formulas cancel under dimensional analysis.

Edited by Mordred

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55 minutes ago, Mordred said:

While I agree that equality of your proposal isn't definitive. I think you may not be grasping the usages of the dimensionless constants for the scale factor nor the w term.

These two constants by themselves are simply dimensionless ratios only. They have no units unto themselves. The units of measure of those formulas cancel under dimensional analysis.

That's perfectly true ...

... And so,  I think that it's not me who can finish that work

Of course, other contributions, in addition to what has been written to confirm or invalidate the first message of the thread, are welcome ...

Edited by stephaneww

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Hello.

Certainly I didn't succed to prove ma first proposition with the equations of state, but using my method for the vacuum catastrophe applied to the Hubble constant, a parallel of the same type can be made from a point of view of quantum mechanics relativistic deterministic

$H_0=67,9 Km/s/Mpc = 2,2005*10^{-18} s^{-1}$, $T(Gly)=13,787*10^9$

$\frac{\hbar*H_0}{l_p^3}=5,4965*10^{52}\text{ Joules/m^3} \text{, }(H_0 \text{ in }s^{-1})$

to follow the method of the vacuum catastrophe used in this tread we raise this volumetric energy density at square :

$(\frac{\hbar*H_0}{l_p^3})^2=3,0211*10^{105}\text{ Joules^2/m^6}$

$\frac{E_p}{l_p^3}=4,6332*10^{113}\text{ Joules/m^3}$

$(\frac{\hbar*H_0}{l_p^3})^2/\frac{E_p}{l_p^3}/8/\pi*3=7,7834*10^{-10}\text{ Joules/m^3}=$

$\text{exactly critical density of energy per volume at Hubble radius (edit : not necessarily at Hubble radius) in }\Lambda \text{CDM model}$

the difference with the vacuum catastrophe is a factor 3

Is it enough to validate the method?

________________________

notes:

Hubble radius based on $T(Gly)=13,787*10^9$ convert in seconds =$4,3509*10^{17}s$ is

$R=T(s)*c=4,3509*10^{17}s*299792458m/s=1,3044*10^{26}m$

$c= \text{light speed in vacuum}$

for calculation of critical densité, I use $H_0=67,9 Km/s/Mpc = 2,2005*10^{-18} s^{-1}$

Edited by stephaneww

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7 hours ago, stephaneww said:

Hello.

Certainly I didn't succed to prove ma first proposition with the equations of state, but using my method for the vacuum catastrophe applied to the Hubble constant, a parallel of the same type can be made from a point of view of quantum mechanics relativistic deterministic

$H_0=67,9 Km/s/Mpc = 2,2005*10^{-18} s^{-1}$, $T(Gly)=13,787*10^9$

$\frac{\hbar*H_0}{l_p^3}=5,4965*10^{52}\text{ Joules/m^3} \text{, }(H_0 \text{ in }s^{-1})$

to follow the method of the catastrophe of vacuum used in this tread we raise this volumetric energy density at square :

$(\frac{\hbar*H_0}{l_p^3})^2=3,0211*10^{105}\text{ Joules^2/m^6}$

$\frac{E_p}{l_p^3}=4,6332*10^{113}\text{ Joules/m^3}$

$(\frac{\hbar*H_0}{l_p^3})^2/\frac{E_p}{l_p^3}/8/\pi*3=7,7834*10^{-10}\text{ Joules/m^3}=$

=$\frac{3*c^2*H_0^2}{8\pi G}=\rho_c$ to use the formula... (with LFRW equations)... it will be clearer

Edited by stephaneww

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Yep looks much better.

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and so Mordred please  :

17 hours ago, stephaneww said:

...

the difference with the vacuum catastrophe is a factor 3

Is it enough to validate the method?

...

is the factor 3 link to a question of pression (states equations with $w$)?

Edited by stephaneww

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oops I thougt I had find something interresting but not finaly, sorry

finaly I let this :

when we make  energy density per volume of the cosmological constant (in Joule / m ^ 3) / Planck force (in Newton) * 8 * pi, I find exactly the numerical value of the cosmological constant in m ^ -2.

can you explain to me why, confirm or correct me please?

Edited by stephaneww

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there is no question finaly : it's an another form of equations I posted before, sorry

Edited by stephaneww

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hello

in my opinion, there is really something auround squares or square roots and 8 pi about the vacuum catastrophe :

$\frac{\hbar^2*\Lambda}{L_p^3* E_p} /8\pi=$

the exact value of volumic energy density of cosmological constant exprimed in Joules/m^3

$\hbar$  reduced Planck constant (Joules*s)

$\Lambda$ cosmological constant (s^-2)

$L_p$ Planck length (meters)

and

$E_p$ Planck energy (Joules)

thank you in advance if you can explain it to me or correct me please.

Edited by stephaneww
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On ‎03‎/‎02‎/‎2018 at 12:08 PM, stephaneww said:

for hello

in my opinion, there is really something auround squares or square roots and 8 pi about the vacuum catastrophe :

2ΛL3pEp/8π=

the exact value of volumic energy density of cosmological constant exprimed in Joules/m^3

reduced Planck constant (Joules*s)

Λ cosmological constant (s^-2)

Lp Planck length (meters)

and

Ep Planck energy (Joules)

thank you in advance if you can explain it to me or correct me please.

oops sorry, this is only another writing of the equation of the vacuum catastrophe = 1/tp^2 * 1/Λ * 8pi.

does this mean tant 1/tp^2 can't be considered as an energy in quantum mechanics as we do for Λ in the relativity? I don't know.

only the begining is perhaps interesting, but I'm not sure of that

Edited by stephaneww

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On ‎05‎/‎02‎/‎2018 at 1:15 AM, stephaneww said:

oops sorry, this is only another writing of the equation of the vacuum catastrophe = 1/tp^2 * 1/Λ * 8pi.

does this mean tant 1/tp^2 can't be considered as an energy in quantum mechanics as we do for Λ in the relativity? I don't know.

only the begining is perhaps interesting, but I'm not sure of that

it's the same  with lp^2 and Λ in m^-2 so we have an area divided by an area, but it's not relevant :

It is only now that I understood that it is very likely the multiplication of an area and the inverse of an area  :

before making a detour by the pressures of a perfect fluid (that of the cosmological constant) and the mechanical definition of a pressure:

$P_{\Lambda}=-\frac{c^4*\Lambda}{8\pi*G}$

$P=\frac{F}{S}$

with
$P$ : pressure
$F$ : force
and
$S$ : area

now by isolating the Force  we have
$F=P*S$

assume that the Planck pressure is the same behavior as that of the cosmological constant, ie that of a perfect fluid, as seen above, a sign near.
in terms of units,
for $P$ is in Pascal (symbol Pa: $kg*m^{-1}*s^{-2}$ that is identical in terms of unity to Joules / m^3 ( in other words, a volume energy density)

the Planck Force $F_p$ = volumetric energy density of Planck * (Planck length)2

it can be seen that the Planck Force =(c4/G) is also present in $P_{\Lambda}$

isolate the Planck Force in this last equality we have: $F_p$= $P_{\Lambda}*2*(4*\pi/\Lambda)$ or in terms of dimensions $\Lambda$ is in $m^{-2}$
so $1/\Lambda$ is in $m^2$.

$4*\pi/\Lambda$ probably due to the fact that in the case of the cosmological constant the surface to be considered is that of a sphere and not a plan like in the quantum version

factor $2$ is already present for relationships in cosmology and Planck dimensions, as in Planck mass flow and Hubble radius mass flow, for example.
or maybe it comes from the fact that it takes 2 surfaces to see the energy of the vacuum via the Casimir effect

$F=P*S$ is probably (within a factor of 2) a formula common to quantum mechanics and cosmology from their volume energy density and their respective area

multiply both values in $m^2$ and $m^{-2}$ would be meaningless, count only the question of the Planck force in both cases, from the volumetric energy densities

Edited by stephaneww

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Um,

many things are badly said ...

to be more clear : Planck force is the common point between the volumetric energy density of the cosmological constant and that of Planck

vacuum catastrophe appears by omitting to multiply by their respective surfaces (except for factor 2)

is it wright please ?

Edited by stephaneww

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do you think that this is enought to accept the model of my first post ? (without equations state )

$\frac{\Lambda_0}{3*H_0^2}=\Omega_{\Lambda_0}=0.69$ source : http://www.cnrs.fr/publications/imagesdelaphysique/couv-PDF/IdP2008/03-Bernardeau.pdf (page 10)

The cosmological constant appears here as a numeric multiple of a pulse expressed in $s^{-1}$ (squared) more precisely that of $(H_0)$ : Hubble constant (SI unit)

thanks in advance for yours answers

CNRS* = C : center, N : national, R : research, S: scientific

Edited by stephaneww

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I propose a simplified version here :

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