Jon's Big Theory The universe is an anti de-sitter space within an all but equal and opposite de sitter space.
The zero point energy density of the vacuum is masked by the zero point curvature of space-time.

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Time and The Fourth Dimension.

Stephen Hawking and Leonard Mlodinow state in The Grand Design (page 172)
In the early universe -- when the universe was small enough to be governed by both general ralativity and quantum theory -- there were effectively four dimensions of space and none of time.
...
Suppose the begining of the universe was like the South Pole of the earth, with degrees of latitude playing the role of time. As one moves north, the circles of constant latitude, representing the size of the universe, would expand. The universe would start as a point at the South Pole, but the South Pole is much like any other point. To ask what happened before the begining of the universe would become a meaningless question, because there is nothing south of the South Pole. In this picture space-time has no boundary -- the same laws of nature hold at the South Pole as in other places.

The earth has a south pole and everywhere else is northwards.
The universe has an Early Pole and everywhen else is later.
Taking t as a measure of pure time and u as a measure of the fouth dimension.then, in natural units,
${\displaystyle u^{2}=t^{2}-1}$
We are assuming that the fourth dimension is minimally non-uniform and therefore the fourth dimension incorporates a single unit of space.
When time and space are added together the square of the result is the difference of the two squares.

The graph shows the relationship between the fourth dimension and pure time.
There are positive and negative values of u and there are positive and negative values of t.
However, while we can move from negative u to positive u there are no routes from negative t to positive t.
Time has a beggining, even though the fourth dimension does not.
This satisfies the no-boundary condition

Now consider this. Time begins at t=1. "One what?" one may ask. "One fundamental unit of time." comes the reply. One Planck Unit? Well if Planck Units are fundamental then yes. But when is the Placnk Era? Between t=0 and t=1. We have not only avoided a singularity at the beggining of time, we have avoided the Planck Era in its entiretly.




If we look at the universe with all the dimensions of space seen "edge-on" then we will simply see a line representing the fourth dimension.
A universe in two halves. Not the big bounce.





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Derivatives of u and t

${\displaystyle from\ \ \ u^{2}=t^{2}-1\ \ \ it\ follows\ that}$
${\displaystyle u=\pm \left(t^{2}-1\right)^{1/2}}$
${\displaystyle {\frac {du}{dt}}=\pm {\tfrac {1}{2}}\left(t^{2}-1\right)^{-1/2}\left(2t\right)=\pm {\frac {t}{u}}}$
${\displaystyle \left({\frac {du}{dt}}\right)^{2}={\frac {t^{2}}{u^{2}}}}$
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Change Of Co-ordinates

The general formula for a change of co-ordinate of a tensor F_t is

${\displaystyle F_{t}=F_{x}{\dfrac {\partial x}{\partial t}}+F_{y}{\dfrac {\partial y}{\partial t}}+F_{z}{\dfrac {\partial z}{\partial t}}+F_{u}{\dfrac {\partial u}{\partial t}}}$
since t is only dependant on u, and is fully dependant on u, this simplifies to
${\displaystyle F_{t}=F_{u}{\dfrac {du}{dt}}}$
likewise for ${\displaystyle F_{tt}}$
${\displaystyle F_{tt}=F_{uu}\left({\dfrac {du}{dt}}\right)^{2}}$
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Einstein's field equations

Einstein's field equations may be written
${\displaystyle G_{\mu \nu }+\Lambda g_{\mu \nu }=\kappa T_{\mu \nu }}$
alternatively, using zero point values we get
${\displaystyle G_{\mu \nu }+G_{\mu \nu }^{^{_{ZP}}}=\kappa T_{\mu \nu }+\kappa T_{\mu \nu }^{^{_{ZP}}}}$
Subtracting the second equation from the first we get
${\displaystyle \Lambda g_{\mu \nu }-G_{\mu \nu }^{^{_{ZP}}}=-\kappa T_{\mu \nu }^{^{_{ZP}}}}$
Thus ahowing the cosmological constant to be the difference of the two zero point values.

If we assume that the curvature components are in the fourth dimension and that the energy components are in proper time then

${\displaystyle G_{uu}+G_{uu}^{^{_{ZP}}}=\kappa T_{tt}+\kappa T_{tt}^{^{_{ZP}}}}$

incorporating the cosmological constant

${\displaystyle \Lambda g_{uu}-G_{uu}^{^{_{ZP}}}=-\kappa T_{uu}^{^{_{ZP}}}}$

and adding together

${\displaystyle \Lambda g_{uu}-G_{uu}^{^{_{ZP}}}+G_{uu}+G_{uu}^{^{_{ZP}}}=\kappa T_{tt}+\kappa T_{tt}^{^{_{ZP}}}-\kappa T_{uu}^{^{_{ZP}}}}$

${\displaystyle G_{uu}+\Lambda g_{uu}=\kappa T_{tt}+\kappa T_{tt}^{^{_{ZP}}}-\kappa T_{uu}^{^{_{ZP}}}}$

${\displaystyle G_{uu}\left({\frac {du}{dt}}\right)^{2}+\Lambda g_{uu}\left({\frac {du}{dt}}\right)^{2}=\kappa T_{tt}\left({\frac {du}{dt}}\right)^{2}+\kappa T_{tt}^{^{_{ZP}}}\left({\frac {du}{dt}}\right)^{2}-\kappa T_{uu}^{^{_{ZP}}}\left({\frac {du}{dt}}\right)^{2}}$

${\displaystyle G_{tt}+\Lambda g_{tt}=\kappa T_{tt}{\frac {t^{2}}{u^{2}}}+\kappa T_{tt}^{^{_{ZP}}}{\frac {t^{2}}{u^{2}}}-\kappa T_{tt}^{^{_{ZP}}}}$

${\displaystyle G_{tt}+\Lambda g_{tt}=\kappa T_{tt}{\frac {t^{2}}{u^{2}}}+\kappa T_{tt}^{^{_{ZP}}}\left({\frac {t^{2}}{u^{2}}}-1\right)}$

${\displaystyle G_{tt}+\Lambda g_{tt}=\kappa T_{tt}{\frac {t^{2}}{u^{2}}}+{\frac {\kappa }{u^{2}}}T_{tt}^{^{_{ZP}}}}$


So the vacuum solution is
${\displaystyle G_{tt}+\Lambda g_{tt}={\frac {\kappa }{u^{2}}}T_{tt}^{^{_{ZP}}}}$

Assuming the cosmological constant to be negligable, we derive the Friedman equation for the vacuum.

${\displaystyle G_{uu}=3{\frac {\left(da/du\right)^{2}+kc^{2}}{a^{2}}}}$
${\displaystyle G_{uu}\left({\frac {du}{dt}}\right)^{2}=3{\frac {\left(da/du\right)^{2}+kc^{2}}{a^{2}}}\left(du/dt\right)^{2}}$
${\displaystyle G_{tt}=3{\frac {\left(da/dt\right)^{2}}{a^{2}}}+3{\frac {kc^{2}}{a^{2}}}\left({\frac {du}{dt}}\right)^{2}=3{\frac {{\dot {a}}^{2}}{a^{2}}}+3{\frac {kc^{2}}{a^{2}}}\left({\frac {du}{dt}}\right)^{2}}$
${\displaystyle G_{tt}={\frac {\kappa }{u^{2}}}T_{tt}^{^{_{ZP}}}}$
${\displaystyle 3{\frac {{\dot {a}}^{2}}{a^{2}}}+3{\frac {kc^{2}}{a^{2}}}\left({\frac {du}{dt}}\right)^{2}={\frac {\kappa }{u^{2}}}T_{tt}^{^{_{ZP}}}}$
for an intrinsic curvature of zero
${\displaystyle 3{\frac {{\dot {a}}^{2}}{a^{2}}}={\frac {\kappa }{u^{2}}}T_{tt}^{^{_{ZP}}}}$

Assuming that in fundamental units ${\displaystyle \ T_{tt}^{^{_{ZP}}}=1}$

${\displaystyle 3{\frac {{\dot {a}}^{2}}{a^{2}}}={\frac {\kappa }{u^{2}}}}$

${\displaystyle {\frac {{\dot {a}}^{2}}{a^{2}}}={\frac {\kappa }{3}}{\frac {1}{u^{2}}}}$

${\displaystyle {\frac {\dot {a}}{a}}={\sqrt {\frac {\kappa }{3}}}{\frac {1}{u}}}$

${\displaystyle \int {\frac {\dot {a}}{a}}dt={\sqrt {\frac {\kappa }{3}}}\int {\frac {1}{u}}dt}$

${\displaystyle ln\left(a\right)={\sqrt {\frac {\kappa }{3}}}\ ln\left(t+u\right)\ +\ constant\ of\ integration}$

${\displaystyle a\propto \left(t+u\right)^{\sqrt {\kappa /3}}}$

${\displaystyle a\propto \left(t+{\sqrt {t^{2}-1}}\right)^{\sqrt {\kappa /3}}}$

So there we have it. Inflation, slow roll, with a graceful exit, straight from the vacuum; providing that ${\displaystyle \kappa }$ in natural units is more than three.
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What is the value of ${\displaystyle \kappa }$ in natural units?

Jon's Big Theory

The universe is an anti de-sitter space within an all but equal and opposite de sitter space. The zero point energy density of the vacuum is masked by the zero point curvature of space-time.
The universe has three uniform dimensions of space and a minimally non-uniform dimension of time.
Taking ${\displaystyle u}$ as the co-ordinate of the fourth dimension and ${\displaystyle t}$ as a measure of pure time then, in natural units, ${\displaystyle u}$ is defined by
${\displaystyle u^{2}=t^{2}-1}$
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Derivatives of u and t

from ${\displaystyle \ \ \ u^{2}=t^{2}-1\ \ \ }$ it follows that
${\displaystyle u=\pm \left(t^{2}-1\right)^{1/2}}$
${\displaystyle {\frac {du}{dt}}=\pm {\tfrac {1}{2}}\left(t^{2}-1\right)^{-1/2}\left(2t\right)=\pm {\frac {t}{u}}}$
${\displaystyle \left({\frac {du}{dt}}\right)^{2}={\frac {t^{2}}{u^{2}}}}$
_______________________________________________________________________________________________

Change Of Co-ordinates

The general formula for a change of co-ordinate of a tensor F_t is

${\displaystyle F_{t}=F_{x}{\dfrac {\partial x}{\partial t}}+F_{y}{\dfrac {\partial y}{\partial t}}+F_{z}{\dfrac {\partial z}{\partial t}}+F_{u}{\dfrac {\partial u}{\partial t}}}$
since t is only dependant on u, and is fully dependant on u, this simplifies to
${\displaystyle F_{t}=F_{u}{\dfrac {du}{dt}}}$
likewise for ${\displaystyle F_{tt}}$
${\displaystyle F_{tt}=F_{uu}\left({\dfrac {du}{dt}}\right)^{2}}$
_______________________________________________________________________________________________

Einstein's field equations

Einstein's field equations may be written
${\displaystyle G_{\mu \nu }+\Lambda g_{\mu \nu }=\kappa T_{\mu \nu }}$
alternatively, using zero point values we get
${\displaystyle G_{\mu \nu }+G_{\mu \nu }^{^{_{ZP}}}=\kappa T_{\mu \nu }+\kappa T_{\mu \nu }^{^{_{ZP}}}}$
Subtracting the second equation from the first we get
${\displaystyle \Lambda g_{\mu \nu }-G_{\mu \nu }^{^{_{ZP}}}=-\kappa T_{\mu \nu }^{^{_{ZP}}}}$
Thus ahowing the cosmological constant to be the difference of the two zero point values.

If we assume that the curvature components are in the fourth dimension and that the energy components are in proper time then

${\displaystyle G_{uu}+G_{uu}^{^{_{ZP}}}=\kappa T_{tt}+\kappa T_{tt}^{^{_{ZP}}}}$

incorporating the cosmological constant

${\displaystyle \Lambda g_{uu}-G_{uu}^{^{_{ZP}}}=-\kappa T_{uu}^{^{_{ZP}}}}$

and adding together

${\displaystyle \Lambda g_{uu}-G_{uu}^{^{_{ZP}}}+G_{uu}+G_{uu}^{^{_{ZP}}}=\kappa T_{tt}+\kappa T_{tt}^{^{_{ZP}}}-\kappa T_{uu}^{^{_{ZP}}}}$

${\displaystyle G_{uu}+\Lambda g_{uu}=\kappa T_{tt}+\kappa T_{tt}^{^{_{ZP}}}-\kappa T_{uu}^{^{_{ZP}}}}$

${\displaystyle G_{uu}\left({\frac {du}{dt}}\right)^{2}+\Lambda g_{uu}\left({\frac {du}{dt}}\right)^{2}=\kappa T_{tt}\left({\frac {du}{dt}}\right)^{2}+\kappa T_{tt}^{^{_{ZP}}}\left({\frac {du}{dt}}\right)^{2}-\kappa T_{uu}^{^{_{ZP}}}\left({\frac {du}{dt}}\right)^{2}}$

${\displaystyle G_{tt}+\Lambda g_{tt}=\kappa T_{tt}{\frac {t^{2}}{u^{2}}}+\kappa T_{tt}^{^{_{ZP}}}{\frac {t^{2}}{u^{2}}}-\kappa T_{tt}^{^{_{ZP}}}}$

${\displaystyle G_{tt}+\Lambda g_{tt}=\kappa T_{tt}{\frac {t^{2}}{u^{2}}}+\kappa T_{tt}^{^{_{ZP}}}\left({\frac {t^{2}}{u^{2}}}-1\right)}$

${\displaystyle G_{tt}+\Lambda g_{tt}=\kappa T_{tt}{\frac {t^{2}}{u^{2}}}+{\frac {\kappa }{u^{2}}}T_{tt}^{^{_{ZP}}}}$


So the vacuum solution is
${\displaystyle G_{tt}+\Lambda g_{tt}={\frac {\kappa }{u^{2}}}T_{tt}^{^{_{ZP}}}}$

Assuming the cosmological constant to be negligable, we derive the Friedman equation for the vacuum.

${\displaystyle G_{uu}=3{\frac {\left(da/du\right)^{2}+kc^{2}}{a^{2}}}}$
${\displaystyle G_{uu}\left({\frac {du}{dt}}\right)^{2}=3{\frac {\left(da/du\right)^{2}+kc^{2}}{a^{2}}}\left(du/dt\right)^{2}}$
${\displaystyle G_{tt}=3{\frac {\left(da/dt\right)^{2}}{a^{2}}}+3{\frac {kc^{2}}{a^{2}}}\left({\frac {du}{dt}}\right)^{2}=3{\frac {{\dot {a}}^{2}}{a^{2}}}+3{\frac {kc^{2}}{a^{2}}}\left({\frac {du}{dt}}\right)^{2}}$
${\displaystyle G_{tt}={\frac {\kappa }{u^{2}}}T_{tt}^{^{_{ZP}}}}$
${\displaystyle 3{\frac {{\dot {a}}^{2}}{a^{2}}}+3{\frac {kc^{2}}{a^{2}}}\left({\frac {du}{dt}}\right)^{2}={\frac {\kappa }{u^{2}}}T_{tt}^{^{_{ZP}}}}$
for an intrinsic curvature of zero
${\displaystyle 3{\frac {{\dot {a}}^{2}}{a^{2}}}={\frac {\kappa }{u^{2}}}T_{tt}^{^{_{ZP}}}}$

Assuming that in fundamental units ${\displaystyle \ T_{tt}^{^{_{ZP}}}=1}$

${\displaystyle 3{\frac {{\dot {a}}^{2}}{a^{2}}}={\frac {\kappa }{u^{2}}}}$

${\displaystyle {\frac {{\dot {a}}^{2}}{a^{2}}}={\frac {\kappa }{3}}{\frac {1}{u^{2}}}}$

${\displaystyle {\frac {\dot {a}}{a}}={\sqrt {\frac {\kappa }{3}}}{\frac {1}{u}}}$

${\displaystyle \int {\frac {\dot {a}}{a}}dt={\sqrt {\frac {\kappa }{3}}}\int {\frac {1}{u}}dt}$

${\displaystyle ln\left(a\right)={\sqrt {\frac {\kappa }{3}}}\ ln\left(t+u\right)\ +\ constant\ of\ integration}$

${\displaystyle a\propto \left(t+u\right)^{\sqrt {\kappa /3}}}$

${\displaystyle a\propto \left(t+{\sqrt {t^{2}-1}}\right)^{\sqrt {\kappa /3}}}$

So there we have it. Inflation, slow roll, with a graceful exit, straight from the vacuum; providing that ${\displaystyle \kappa }$ in natural units is more than three.
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What is the value of ${\displaystyle \kappa }$ in natural units?

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Time Has A Begining, The Fourth Dimension Does Not.

The universe has three uniform dimensions of space and a minimally non-uniform dimension of time.
Taking ${\displaystyle u}$ as the co-ordinate of the fourth dimension and ${\displaystyle t}$ as a measure of cosmological time i.e. proper time for all things cosmological, then in natural units, ${\displaystyle u}$ is defined by
${\displaystyle u^{2}=t^{2}-1}$
If we look at the universe with all the dimensions of space "edge-on" then we will simply see a line representing the fourth dimension.
Here we see a line representing the fourth dimension with all the dimensions of space seen "edge on".
The line is continuous, it does not have a beggining or an end.
The graph shows the relationship between the fourth dimension and pure time.
Time does have a beggining.
At the south pole every direction is north.
At u=0 every direction is forwards in time.
When space and time are added together the square of the result is equal to the difference of the squares of the parts. If there is more time added than space then the result is time. If there is more space added than time then the result is space. If the amount of time is equal to the amount of space then the result is Null, neither time nor space. Note 1 second is equivalent to 299 792 458 metres. The fourth dimension is a dimension of time at all points except u=0, where it is Null, and proper time changes direction.
A universe in two halves. Not the big bounce.