4

In cosmology empty space has an energy density $\rho_{\Lambda}$ of

$$\rho_{\Lambda}=\rho_c \cdot \Omega_{\Lambda}\cdot c^2$$

with $\Omega_{\Lambda}$ beeing the dark energy fraction ($0.683$ according to Planck 2013) and $\rho_c$ beeing todays critical density defined by

$$\rho_c=3 H_0^2/8/\pi/G$$

where $H_0=2.176\cdot 10^{-18}\, \text{s}^{-1}$ is the Hubble constant, and $G=6.674\cdot 10^{-11}\, \text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2}$ Newtons constant. This is in units of $\text{Joule}/\text{m}^3$ or $\text{Pascal}$

$$\rho_{\Lambda}=5.2\cdot 10^{-10} \, \text{kg}\, \text{m}^{-1}\, \text{s}^{-2}$$

Now the universe is expanding, and since the volume increases, so does energy. The rate at which space expands is as mentioned above $2.176\cdot10^{-18}\, \text{m}/\text{m}/\text{s}$ which means that every meter grows by $2.176\cdot10^{-18}$ meters every second.

So one cubic meter, $1\, \text{m}^3$, every second gives birth to

$$\Delta{V} = V_2-V_1=6.528\cdot 10^{-18} \, \text{m}^3$$

Where the volume $V_1$ = $r^3$ with $r=1\, \text{m}$, and $V_2=r\cdot(1+H_0\cdot \Delta{t})$ with $\Delta{t}=1\, \text{s}$

When we multiply the new born volume $\Delta{V}$ with the dark energy density $\rho_{\Lambda}$ and divide it by $\Delta{t}$, we get in units of power, $\text{kg}\,\text{m}^2\,\text{s}^{-3}$, the value of

$$3.394\cdot 10^{-27} \, \text{Watt}$$

Is my interpretation that every cubic meter generates a power of $3.394\cdot 10^{-27} \, \text{Watt}$ correct, or is there a flaw in my considerations?

Yukterez
  • 11,553
  • $\rho_\Lambda$ and $\rho_c$ can't both have units of energy density if you are multiplying one by $c^2$ to get the other... and $\rho_c$ apparently has units of energy density according to your equations... so $\rho_\Lambda$, which you call an energy density, looks like it is not an energy density. – hft Mar 03 '15 at 06:54
  • @hft: Energy and mass are equivalent. So the factor of $c^2$ is just Einsteins $e=m\cdot c^2$. $\rho_{c}$ is the critical density in $kg/m^3$, while dark energy has units of $Joule/m^3$. In relativity this is equivalent by a factor of $c^2$ – Yukterez Mar 03 '15 at 06:59
  • your rho_c has units of kg/(ms^2), which is already an energy density. Not mass density. So you are not saying E=mc^2, you are saying E=Ec^2 – hft Mar 03 '15 at 07:01
  • @hft: http://s1.postimg.org/3sxkj2kbj/Unbenannt.png – Yukterez Mar 03 '15 at 07:04
  • So, you changed the "c" in the first equation with units of "kg/(ms^2)" to a Lamda. Sure, that is fine now. – hft Mar 03 '15 at 07:05
  • Yes, damn latex [: – Yukterez Mar 03 '15 at 07:06
  • Simon, when switching from meters to cubic meters, why don't you cube the numbers? Space expansion is happening in all directions of the 3D volume. – Moonraker Mar 03 '15 at 07:48
  • Nice ( and exact ) calculus ... –  Feb 05 '16 at 00:39
  • @igael It not infinitiesimal exact since I use Δt -> 1 sec in the example above to make it easier to calculate, but there is also a version with the limit going to zero at http://yukterez.net/forum/viewtopic.php?f=3&t=65 In the link it is not the power generated by 1 m³ like here but the volume of a sphere of radius earth..moon, just in case you don't speak german [: The result is almost the same with Δt -> 1 sec oder Δt -> 0 – Yukterez Feb 05 '16 at 04:18
  • the $\Delta V$ was correct , the cubic and square may be ignored and the result seems ok here. In the german page, it is the power for the whole volume $r^3$ and not for 1 $m^3$ else I don't understand the meaning of $L$. The limit for $\Delta t$ -> 0 must be equal to the value for $\Delta t = 1$ –  Feb 05 '16 at 05:05
  • sorry, the german page says merely that you need this big volume for 1 W :) –  Feb 05 '16 at 05:18

1 Answers1

2

is there a flaw in my considerations?

For one thing, the universe is not just empty space.

But, anyways... yeah, if you have a volume of constant energy density and you increase that volume while keeping the energy density constant then... yeah, you increase the energy. It's true.

hft
  • 19,536