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Our night sky is filled with stars. On a dark night a significant fraction of the sky is light. This light, we are told, has been in transit for many millions of years. There must therefore be quite a lot of light in transit at any one time.

There are many questions here where we are reminded that light has relativistic mass and therefore has a gravitational effect.

How much light is there in transit at any one time? Perhaps a measure per cubic something would be interesting.

And, of course, what gravitational effect would this light exert? Has it's effect been taken into account when the calculations of the amount of dark matter/energy is made?

Qmechanic
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OldCurmudgeon
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    The title is just too good a straight line to resist: (in your best Groucho voice) "It's not heavy. It's light!" ::rimshot:: – dmckee --- ex-moderator kitten May 05 '12 at 02:24
  • Nice question. What if there was massive amount of photons long ago, and they all flew much further than size of matter? May they still be counted and affect total numbers –  May 29 '12 at 00:53
  • related: http://physics.stackexchange.com/questions/75753/how-much-of-the-universes-mass-is-bound-up-in-photons-traveling-between-stars –  Aug 31 '13 at 00:03

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Actually, on a dark night, the fraction of the sky that is light is pretty negligible. That's what it means to be a dark night ;-)

It's actually not hard to get an estimate of the density of light in the universe. Let's say that "light" includes photons of all wavelengths (not just visible light) for simplicity. A straightforward way to do it is to point a wide-spectrum telescope at the night sky and see how much fast it collects energy. (You have to point it away from the sun and other nearby sources like the galactic disc, because these sources emit a large amount of energy at Earth, which is not representative of the universe as a whole.) This was done with the COBE and WMAP satellites (and more recently Planck, with essentially the same result). They found that, if you eliminate contributions from a few specific nearby sources, the radiation in the universe follows a blackbody spectrum with a temperature of $2.73\text{ K}$. You can calculate the energy density of such a blackbody like this:

$$u = \frac{4\sigma T^4}{c} = 4.2\times 10^{-14}\ \mathrm{\frac{J}{m^3}}$$

This is just four thousandths of a percent of the critical energy density, which is

$$\Omega = \frac{3 c^2 H^2}{8\pi G} = 8.6\times 10^{-10}\ \mathrm{\frac{J}{m^3}}$$

On the other hand, the density of normal matter (atoms) is estimated to be about 4.6% of the critical density - four orders of magnitude higher. So the photon energy density is completely insignificant compared to the density of baryons, dark matter, or dark energy, and that means it has a negligible gravitational effect on the cosmological evolution of the universe.

Of course, in the vicinity of a star, the photon energy density is much higher because of the star's higher blackbody temperature. Let's take the sun, for example. The sun has a surface temperature of $5778\text{ K}$, which means the intensity of radiation it emits is

$$I = \frac{P}{A} = \sigma T^4 = 6.31\times 10^7\ \mathrm{\frac{W}{m^2}}$$

This corresponds to an energy density in the vicinity of the sun's surface of

$$u_\text{rad} = \frac{I}{c} = 0.211\ \mathrm{\frac{J}{m^3}}$$

However, the energy density of the matter that constitutes the sun is

$$u_\text{matter} = \rho_\odot c^2 = 1.2\times 10^{20}\ \mathrm{\frac{J}{m^3}}$$

So again, the gravitational effect of the photons, even on a local scale, is completely negligible.

Incidentally, this would not have been the case in the early universe, when the photons had much higher energy and thus their energy density relative to matter was much higher.

David Z
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  • +1 Informative - Cover the gravitational effect and we're done. – OldCurmudgeon May 06 '12 at 09:35
  • Oh yeah, forgot about.that part... I'll get to it later today. – David Z May 06 '12 at 20:50
  • btw - there's a comment here that evaluates your omega to 0.855E-9 joules per cubic meter. Could you add that for completeness (and balance) please. ... Looking forward to your gravity figures. – OldCurmudgeon May 07 '12 at 21:00
  • Actually on reflection: I didn't entirely forget, since what I was going to say about the gravitational effect was basically that it's negligible on cosmological scales - that's shown by the fact that the universe-wide photon density is so much less than that of matter. But for completeness, I've added a corresponding argument for local scales. – David Z May 08 '12 at 03:39
  • Thank you David. Are you confident that although the density is so small it should be negligible, the fact that it is continuous across vast reaches of space does not affect the calculations? – OldCurmudgeon May 08 '12 at 08:03
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    That's why I calculated a density rather than an amount of energy. By definition it is an average over the entire observable universe. To put it another way, all these cosmological calculations depend on the average density being basically constant across the entire observable universe. – David Z May 08 '12 at 08:17
  • So the relative number of photons is almost negligible in energ-mass when it comes to the dark matter and dynamics of the universe? – Alchimista Oct 22 '20 at 11:50
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Basically all starlight was produced by nuclear fusion. When fusing hydrogen all the way to the top of the binding curve at iron, about 9 MeV per proton is released. This is about 1% of the proton mass-energy, so the total relativistic mass of all the light in the universe can be no more than about 1% of the mass of the matter. This is not a complete answer, but I think it shows that this is most likely negligible when considering cosmological evolution.

  • But surely millions of years worth of light is in transit at any one moment. Doesn't that add up to even a little more than a hill of beans? – OldCurmudgeon May 05 '12 at 00:35
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    Yes, but stars will continue to burn for millions of years, so at any give time only a very tiny fraction of their mass is being converted to light. The 1% bound is for the energy released over the entire stellar lifetime. –  May 05 '12 at 00:47
  • So you're saying that because light is really really small it's actually insignificant. ... are you a politician of some sort? :) – OldCurmudgeon May 05 '12 at 00:51
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    Unfortunately, a politician is a lot more likely to take a tiny thing and act like it's a huge problem. But yes, usually a 1% effect will be insignificant - although of course 1% of the mass-energy of the universe is absurdly large on an everyday scale. Everything is relative. –  May 05 '12 at 00:59
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    In addition to the starlight, there is also the energy density of the CMB "light", but this is still small compared with the energy locked up in the baryons (about a thousandth). This is what people mean when they say our current era is "matter dominated". Of course it wasn't always the case.... – twistor59 May 05 '12 at 07:22
  • I think David Z must have pressed enter at the precise moment (in some frame) when I pressed enter for that last comment. – twistor59 May 05 '12 at 07:25
  • Stars will continue to burn for millions of years, but light emitted by stars that exploded millions of years ago is still reaching us today. I don't disagree that the overall mass of the electromagnetic spectrum is very small compared to matter, but it still must be accounted for even if just as a margin of error. – Drunken Code Monkey Oct 09 '15 at 05:22
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I recall reading recently (do not remember where) that the mass equivalent of the photons in the universe $P_{ou}$ is about 90% due to the cosmic microwave radiation (CMR). However, the best values I found for calculating the mass equivalence of the radiation in the observable universe is based on three variables.

(1) The mass of matter in the observable universe: $M_{ou}$ = $10^{53}$ kgm. The source is

How much energy is in the Universe as photons? .

(2) The value of $\Omega_r$ = 8.24 $10^{-5}$. The source is

http://hyperphysics.phy-astr.gsu.edu/hbase/Astro/denpar.html.

(3) The value of $\Omega_m$ = 0.27. The source is the same as (2).

From these values the calculation of $P_{ou}$ is

$P_{ou}$ = $M_{ou}$ x $\Omega_r$ / $\Omega_m$ = 3.7 $10^{49}$ kgm.

Buzz
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