Specifically, excluding the Sun and the Moon:
How much energy, as light and other frequencies reaches a point in space on the average, if you add up all the sources like distant stars, the Milky Way, distant galaxies, nebulae etc.? How much brighter is the space than "perfect blackness" on the average?

Say, we've developed a magical solar battery that has 100% efficiency, can absorb all electromagnetic frequencies, is spherical, fully omnidirectional and has $1 m^2$ of cross-section surface (56cm radius); we sent it out halfway to Proxima Centauri and we measure how much power it produces - how many watts we'd get out of the battery utilizing all the starlight available?

Note: If "all frequencies" is not easily obtainable, the luminous counterpart to radiant values will be okay (only taking visible spectrum into account).

  • $\begingroup$ The power density of all EM radiation at a given point in space is going to depend on the precise distribution of EM sources (i.e. how distant and in what direction). Clearly if you're real close to e.g. the Sun, the power density will be different than if you're a light year away - that $\frac 1 {r^2}$ factor is important. $\endgroup$ – StephenG May 17 '17 at 13:34
  • $\begingroup$ @StephenG: Definitely a light year away or more. I specifically requested excluding the "local sources" - "Halfway to Proxima Centauri," But a less "sci-fi" interpretation would be "while in orbit, during New Moon, with the craft on the night side, point a broad-angle light-meter away from Earth and read it out. How many watts/steradian or lumens/steradian does it read? (multiply by 4pi for whole sky). $\endgroup$ – SF. May 17 '17 at 13:45
  • $\begingroup$ You can work this out for yourself. You know how luminous the Sun is, it's magnitude, and the magnitude of the other stars (you can adjust for the change going halfway to Proxima). You also know magnitude is a logarithmic scale. Exponentiate, add, and that's pretty much it. I can help (contact info in profile) though I won't do it for you. $\endgroup$ – user21 May 19 '17 at 0:37
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    $\begingroup$ BTW, to whoever voted to close this as too broad: "too broad" is if the answer required is too long for a Q&A site. Here the answer is literally a single number, plus the unit. Something like "2.7 Watt/m^2" or "11 lumens" - numbers obviously made up because I don't know the right ones. $\endgroup$ – SF. May 19 '17 at 6:44
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    $\begingroup$ This answer from the physics site might somewhat answer your question: [physics.stackexchange.com/questions/196366/… (Answer) $\endgroup$ – Agerhell May 15 '19 at 10:06

The electromagnetic energy density is dominated by the cosmic microwave background and the optical/IR background. This Physics SE answer, contains the plot below, showing the contribution of energy density at different frequencies. You can integrate under this "by eye" to see that the CMB contribution is the largest followed closely by an optical/IR bump.

A plot of $\nu I_\nu$ (proportional to energy density per logarithmic frequency interval), taken from Hill et al. (2018)

enter image description here

The CMB is approximately isotropic blackbody radiation with a temperature of 2.73 K and therefore a specific intensity of $I_{\nu} = \frac{2h\nu^3}{c^2}\frac{1}{\exp[h \nu/k_B T] -1} $$

The integral of this over all frequencies is $$ u = \frac{8\pi^5 (k_B T)^4}{15 (hc)^3} = 4.2 \times 10^{-14}\ \ {\rm J\ m}^{-3} $$

In terms of power, we just multiply by $c$ to get $1.3\times 10^{-5}$ Wm$^{-2}$.

Note that you can't harness this unless you have a receiver that is colder than the CMB.

You can do a rough calculation for the contribution of optical background light by noting that the dark sky at a good, dark astronomical site is around 22 mag per square arcsecond in the V band. Only about 0.1-0.2 mag of this could be attributed to light pollution at the best astronomical sites (see for example Benn & Ellison 2007). Noting that the zero point for the V-band magnitude scale is $3.6 \times 10^{-12}$ Wm$^{-2}$ per angstrom for $V=0$ and taking the optical band as 3000 angstroms, and noting there are $5.34\times 10^{11}$ square arcseconds in the sky, then a point in space will receive $10^{-5}$ Wm$^{-2}$ (or perhaps about half this, because a significant fraction of the dark sky brightness is caused by airglow).

This roughly agrees with what is shown in the plot. Contributions at mid-infrared and at shorter wavelengths than optical do not contribute appreciably.

It might be thought that discrete sources or the Milky Way might contribute more, but I think that isn't the case. The Milky Way surface brightness is only about a factor of three above the dark sky brightness, but of course occupies a tiny fraction of the total sky.

A way to see that discrete stars don't contribute much is to note that the optical flux from the entire dark sky is equivalent to about 1000 zeroth magnitude stars or $10^7$ stars at 10th magnitude. Both numbers are higher by orders of magnitude than the actual numbers of Galactic stars at these brightnesses.

  • $\begingroup$ Nice answer. I think the OP was mainly interested in light in the visual range, though. If you can find the total visual light flux, and a figure to compare it to (sunlight/moonlight/venuslight flux, for example), I'll award you the bounty. $\endgroup$ – Ingolifs May 15 '19 at 21:19
  • $\begingroup$ "How much energy, as light and other frequencies" @Ingolifs $\endgroup$ – ProfRob May 15 '19 at 21:42
  • $\begingroup$ @Ingolifs: It's a valuable answer and combining it with the one linked from Physics.SE I should be able to get values for more "useful" frequencies. I'll wait a bit with accepting in case someone is willing to expand or correct something, but in general it's quite satisfactory. $\endgroup$ – SF. May 15 '19 at 22:18
  • $\begingroup$ (also, while these photons are way more numerous, they are quite low-energy. I'm not sure if a smaller but more energetic fraction doesn't contribute more,,, I'd need to plug the numbers for that though) $\endgroup$ – SF. May 15 '19 at 22:24
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    $\begingroup$ @sf Revised. The CMB is not as totally dominant as I thought from first looking at the plot. $\endgroup$ – ProfRob May 16 '19 at 8:55

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