# Number Theory/Why is the vaccum space not curvature free?

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Why is the vaccum space not curvature free? is an important question as in general relativity it is assumed that a vaccum space is in fact empty (no matter and energy), but in quantum mechanics there is no such thing as an empty space.

## Why is the vaccum space not curvature free?

By convention in the theory of general relativity we have for a vaccum space:

${\displaystyle \begin{array}{lr} T_{ik} = 0\text{,} & R_{ik} = 0 \end{array}}$

This is of course a simplification as we want to show. Using the Friedmann${\displaystyle -}$equations:

${\displaystyle \begin{array}{ll} \displaystyle\frac{\dot R(t)^2}{R(t)^2} & = \displaystyle\frac{8 \pi G}{3} \rho + \displaystyle\frac{\Lambda c^2}{3} - \displaystyle\frac{\kappa c^2}{R^2} \\ & \\ \displaystyle\frac{\ddot R(t)^2}{R(t)^2} & = - \displaystyle\frac{4 \pi G}{3} \bigg(\rho + \displaystyle\frac{3p}{c^2}\bigg) \displaystyle\frac{\Lambda c^2}{3} \\ \end{array}}$

we see that for the Planck time ${\displaystyle t_P = \sqrt{\frac{\hbar G}{c^5}}}$ the energy density ${\displaystyle W(t) = \frac{d\mathcal{E}}{dV}}$ was maximal, simply due to that the volume ${\displaystyle V(t_P) = \min}$ and (if we assume) ${\displaystyle \mathcal{E} =}$ const. The Friedmann${\displaystyle -}$equations yield:

${\displaystyle W(t_p)> W(t) > W(t_{\infty}) = 0 }$

since an energy flux exsits for ${\displaystyle t > t_P}$, which is given by:

${\displaystyle 0 < \vec j^{\mu} = W \displaystyle\frac{dx^{\mu}}{dt} = W \vec v }$

Hence, the vacuum space is not curvature free. Another way of realizing this is by using thermodynamics. After Planck's law, which inforces the law of Stefan${\displaystyle -}$Boltzmann, we have for the temperature:

${\displaystyle T = \sqrt[4]{\displaystyle\frac{P}{\sigma A}} }$

Due to ${\displaystyle T_{\text{space}} \neq 0}$ this yields:

${\displaystyle \displaystyle\frac{8 \pi G}{c^4} T_{ik} > 0 \Longrightarrow R_{ik} > 0 }$

We point out, that the curvature is so small that for the calculations in the theory of general relativity it makes no difference. The data of course supports this statement since:

${\displaystyle \Omega_{\text{tot}} = 1,00 \pm 0,02}$

or

${\displaystyle \Omega_{\text{tot}} = 1.0005 \pm 0,00065}$

This still means that space is flat.

## Consequences ${\displaystyle R_{ik} > 0}$

• First of all it is important to realize, that the result for ${\displaystyle R_{ik} > 0}$ is derived completely by the theory of general relativity.
• ${\displaystyle R_{ik} > 0}$ ${\displaystyle \Longleftrightarrow}$ ${\displaystyle T_{ik} > 0}$ means, that there is not one volume in the universe (cosmos) that is completely curvature free! Equivalent to it, we can say, that there is not one volume in the universe, where there is no energy, which is obvious due to equation (of the energy diffusion flux):

${\displaystyle \displaystyle\frac{d\mathcal{E}}{dV} \displaystyle\frac{dx^{\mu}}{dt} = W \vec v }$

This fact includes even voids, because the energy density is distributed equally.

• This is of course nothing new to quantum mechanics, where, with the elaboration of quantum electrodynamics, it was established that the vacuum was not empty. This is expressed best by Heisenberg's uncertainty principle:

${\displaystyle \Delta E \Delta t \geq \displaystyle\frac{h}{4\pi} }$