3
$\begingroup$

What would happen if all dark energy (cosmological constant) instantly converted to radiation due to vacuum metastability event, while otherwise vacuum properties remain the same?

Dark energy density is $10^{-27} kg/m^3$. How will it affect Earth and the Sun?

$\endgroup$
6
$\begingroup$

Convert mass density to energy density, to get $9 \times 10^{-11}\,\mathrm{J/m^3}$ which isn't very much. After all, an adult human is on order of one eighth to one quarter of a cubic meter.

So, assuming a blackbody spectrum so that the radiation is essentially all low energy and does not ionize the immediate temperature rise will be trivial. The planet itself will be similarly subject to a trivial increase in total energy (it's a lot of energy for the whole Earth, but it is spread out over a very large volume and it a pittance compared to the energy already present).

But all of space that is not filled with matter will be radiating in all direction. We need to know how hot that radiation field is.

The total energy density of a blackbody field is $$ u = \frac{4 \sigma T^4}{c} \;,$$ where $\sigma = 5.67 \times 10^{-8} W m^{-2} K^{-4}$. So, solving for temperature we get $$ T = \left( \frac{u c}{4 \sigma} \right)^{0.25} = 19 \,\mathrm{K} \;,$$

Overall result on earth: pretty boring, except that radio astronomers will be vexed.

Very cold objects in space will tend to warm up, but even Kuiper belt objects are a bit warmer than this (circa $50 \,\mathrm{K}$) as far as we know.

Basically a big "Meh!".


If you want it to be exciting make a non-thermal assumption for the radiation field.

$\endgroup$
  • $\begingroup$ Would not all objects warm up, not only very cold, since energy transfer is one-directional? $\endgroup$ – Anixx Nov 7 '16 at 10:20
  • 1
    $\begingroup$ Sure, but the effect is likely to make a bigger difference to cold bodies because (a) they tend to have lower specific heat capacities and (b) the same small increase in temperature represents a bigger fractional change. $\endgroup$ – dmckee Nov 7 '16 at 16:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.