2
$\begingroup$

I know that the term 'dark' is used because a) dark matter does not interact with light b) we know so little about them. Dark matter I guess could be just an unexplained type of particle we don't know about. But what about dark energy?

When we say 'dark energy', are we saying that it's a different concept altogether than energy? Is it saying that there is a third energy type to potential and kinetic? Or is it just 'normal' energy that we cannot explain the origin of? Or do we just not know?

$\endgroup$
2
  • $\begingroup$ Have you tried reading en.wikipedia.org/wiki/Dark_energy This answers the questions you ask. $\endgroup$
    – Eletie
    Apr 6, 2021 at 16:59
  • $\begingroup$ I can see that it answers it, but I'm unable to understand it; I'm not sure the correct SE protocol is for this scenario. $\endgroup$
    – yolo
    Apr 6, 2021 at 17:01

2 Answers 2

0
$\begingroup$

What we know is that the mass-energy density of the universe must be much higher than expected from baryonic matter/dark matter in order to fit the accelerated expansion observed. It could be a particle. It could some gravitational feature we don't understand... The thing is, what we currently think is that the density of dark energy is constant over time. That's pretty crazy by itself, but it's even crazier if you try to figure out how to do that with some kind of particle.

The most reasonable way to get a density that doesn't change over time, is to say that it's a property of space; if you get more space, you get more "space energy". Dark matter is likely to be a particle, since it "clumps" around galaxies. But dark energy seems to be basically everywhere.

$\endgroup$
0
$\begingroup$

When we say 'dark energy', are we saying that it's a different concept altogether than energy?

Yes, it is. Thermodynamics fundamental equation states that when gas volume increases,- then gas internal energy should drop, assuming it has constant entropy:

$$ \mathrm {d} U=-p\,\mathrm {d} V $$

But contrary to expectations, universe does not slows-down in expansion, but even accelerates the expansion with increasing universe volume. This means that in given thermodynamics equation above pressure term is negative, that's why universe has overall increasing "internal energy".

How pressure can be negative ? Pressure can be expressed as vacuum energy per unit volume : $$ p=\frac {E}{V} $$

Substitute Einstein's mass-energy relation (noticing that multiplying both equation sides by $-1$ does not change anything in principle mathematically, however physical meaning will be different) $$ -E=-mc^2$$ So we get for pressure :

$$ p=\frac {-mc^2}{V} $$

This equation can be a bit shortened, due to the fact that it involves vacuum density :

$$\boxed {p= \rho_{\text{vac}} c^2 }$$

Currently vacuum density is estimated to be : $$ \rho _{\text{vac}}=-5.96\times 10^{-27}{\text{ kg/m}}^{3} $$

Negative vacuum density lets for universe to "accumulate" internal energy. It needs some sort of negative mass, which fills all vacuum space. Negative mass gives birth to "negative gravitation", i.e. negative mass is repulsion force, not attraction like gravity. So two units of negative mass will propel from each other apart. That's why "dark energy" is called dark, it can be thought as a synonym to a negative energy.

$\endgroup$
1
  • 1
    $\begingroup$ Notice, that in some references, (wiki, etc) negative pressure is expressed as $p=- \rho_{\text{vac}} c^2$. But I'm against such form, because it implies that ordinary mass can somehow produce negative pressure, while basically it can't. So better is- to get used to negative density, which better recalls the fact that we need negative mass for injecting negative pressure. $\endgroup$ Apr 7, 2021 at 8:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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