2
$\begingroup$

Photons produced during inflation, e.g. by annihilation of antimatter and matter, and not somehow reabsorbed would presumably have been highly redshifted by the end of the inflationary epoch and even more so by now.

If they had been produced in sufficient abundance, could the gravity of their collective energy have been sufficient to bunch them and (given their individual extremely low energy) keep them moreorless "marching on the spot" within bunches and unable to radiate as a photon normally would.

(Note that photons can be trapped in suitable materials, so the idea of trapping them isn't absurd per se, even though managing it in a vacuum with only other photons to help maybe is?! )

An ultra-long wavelength would also explain why these photons are not absorbed by even supermassive black holes nor, if such turns out to be the case, collect in the interiors of planets and stars

Also, it would presumably explain why they do not interact with other forms of matter or photons, especially much shorter wavelength photons.

Furthermore, if long-wavelength photons "leak" over time from accumulations of them manifesting as dark matter, they would presumably proceed on their merry way as normal radiating photons and form an increasingly uniform background, as if the dark matter was slowly evaporating. So could that in turn manifest as apparently accelerating dark energy?

Improve this question
$\endgroup$
1
  • $\begingroup$ I think observations broadly rule out the possibility that dark matter is made of stuff with zero rest mass. This is what the word "cold" in "cold dark matter" is getting at. But I'm not sure so this is not an answer. $\endgroup$
    – Andrew Steane
    Commented Jun 22, 2021 at 13:13

2 Answers 2

Reset to default
3
$\begingroup$

No. Structure formations tells us that dark matter is non-relativistic. The technical buzzword for that is "cold" as opposed to relativistic = hot; thus the "C" in our Lambda-CDM cosmological standard model. More precisely, its density scales with like the expansion factor of the universe $a$ like $a^{-3}$ as expected for matter, and not as $a^{-4}$ as expected for radiation.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ the $a^{-4}$ expectation doesn't rely on any specific assumptions on wavelength? typically wavelength is assumed to be comparable or much shorter than mean free path, specially in vacuum $\endgroup$
    – lurscher
    Commented Jun 22, 2021 at 14:01
  • 2
    $\begingroup$ I am tempted to answer @lurscher's comment with a, "no, no specific assumption other than this being relativistic"... the derivation of the Friedman equations is really quite basic GR, not much else going in at all. On the other extreme for example, this $a^{-4}$ also holds for any particle as long as it's relativistic. But, truth be told, sometimes implicit assumptions sneak in somewhere in basic derivations without being made explicit, so I'm opting here for a more careful "uhmmm, actually I'm not sure..." ;) $\endgroup$
    – rfl
    Commented Jun 22, 2021 at 14:29
0
$\begingroup$

In the current cosmological model, the Big Bang photons are trapped in the consecutive plasmas until 280.000 years after the BB , when they no longer interact dominantly with matter and end up as the cosmic microwave background. Microwaves are very low energy photons which means there is no possibility to have gravitational (or other) attractions , that is what decoupling means. Photons are out of the game, and particularly very long wavelength ones, because the longer the wavelength the smaller the energy, and dark mass is all about missing bulk masses.

Improve this answer
$\endgroup$
1
  • $\begingroup$ i was envisaging photons with galactic wavelengths, or even multiples thereof. But I can see that would require them to be almost unimaginably profuse to have any discernable effect let alone a "bulk mass". Also presumably they would have had to survive, either directly or from precursors, from before the electroweak and previous symmetry breaking stages of the Big Bang. So maybe that is also problematic. Oh well, just a passing thought. $\endgroup$
    – John R Ramsden
    Commented Jun 24, 2021 at 15:44

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.