I know that dark matter is called dark because it interacts very weakly/doesn't interact electromagnetically. Is the same for dark energy, or the name is due to something else? In particular, can dark energy interact electromagnetically?

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    $\begingroup$ both are named "dark" because we don't know what it is. $\endgroup$ Aug 16, 2017 at 17:09
  • $\begingroup$ I understand that if Dark Energy would interact (also?) electromagnetically then at least charged particles would have some other distribution, maybe more smooth.... Because we know charged particles in the universe are not smoothly distributed then dark energy doesn't interact electromagnetically. Is it correct? $\endgroup$
    – Saladino
    Aug 16, 2017 at 17:36
  • $\begingroup$ The dark energy is thought to be uniform across the universe and very weak so it is unknown whether it interacts at all with anything. It is simply proposed as a solution to the accelerated expansion of the universe. $\endgroup$ Aug 16, 2017 at 17:45
  • $\begingroup$ Ok, practically, with our technology, it doesn't interact.... $\endgroup$
    – Saladino
    Aug 16, 2017 at 23:33
  • $\begingroup$ Some advice on English: 1) Avoid starting sentences with "or", "but" or "and", 2) Capitalize the first word of each sentence, proper nouns, and the word "I". Do not capitalize other random nouns. $\endgroup$
    – DanielSank
    Dec 21, 2017 at 17:10

1 Answer 1


There are two main reasons why dark energy is called dark. The first one has to do with the fact that all detections of its effects are indirect. For example, we use the spectra of Type Ia supernovae as standard candles to verify the hypothesis that further galaxies are receding from us at increasing speeds.

This accounts for the observational side.

Theoretically, dark energy is taken to emerge from the cosmological constant $\Lambda$ (This is why the standard model of cosmology is called $\Lambda$CDM). In this cosmology review, we see how dark energy comes from the Friedman metric.

$$ ds^2 = -dt^2 + a(t)[dr^2 +r^2d\Omega^2] $$

The cosmological field equations are

$$ \Big(\frac{\dot{a}}{a}\Big)^2 = \frac{8\pi G}{3}\rho - \frac{k}{a^2} $$


$$ \frac{\ddot{a}}{a} = -\frac{4\pi G}{3}(\rho + 3p) $$

which from conservation of the energy-momentum tensor give a third equation in a system

$$ \dot{\rho} + 3H(\rho + p) = 0 $$

Because we have three equations with three unknowns ($a$, $\rho$, and $ p$), we basically have to take a guess. The usual move is supposing that $p/\rho = w$ where $w = \{1/3, 0, 1, -1\}$ The first three values of $w$ account for different epochs of cosmological evolution: radiation-dominated, dust dominated, and stiff fluid. The last value accounts for the density of dark energy.

Setting $w=-1$ not only gives an expanding universe but results in an equation of state with a negative pressure. Because this does not occur in other areas of physics, it is conceptually unclear what that means, but it is ultimately taken to be an effect of dark energy. Do note, however, that if this value is neither radiation-dominated, dust dominated or a stiff fluid it cannot emit light. It's dark.


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