<p>The electromagnetic energy density is dominated by the cosmic microwave background and the optical/IR background. <a href="https://physics.stackexchange.com/questions/196366/does-the-number-density-of-photons-n-gamma-approx-108-mathrm-m-3-refer/196373#196373]">This Physics SE answer</a>, 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.</p>

<p><em>A plot of <span class="math-container">$\nu I_\nu$</span> (proportional to energy density per logarithmic frequency interval), taken from <a href="https://arxiv.org/abs/1802.03694" rel="noreferrer">Hill et al. (2018)</a></em></p>

<p><a href="https://i.sstatic.net/G8uCd.png" rel="noreferrer"><img src="https://i.sstatic.net/G8uCd.png" alt="enter image description here"></a></p>

<p>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} $$ </p>

<p>The integral of this over all frequencies is
<span class="math-container">$$ u = \frac{8\pi^5 (k_B T)^4}{15 (hc)^3} = 4.2 \times 10^{-14}\ \ {\rm J\ m}^{-3} $$</span></p>

<p>In terms of power, we just multiply by <span class="math-container">$c$</span> to get <span class="math-container">$1.3\times 10^{-5}$</span> Wm<span class="math-container">$^{-2}$</span>.</p>

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

<p>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 <a href="http://www.ing.iac.es/Astronomy/observing/conditions/skybr/skybr.html" rel="noreferrer">Benn &amp; Ellison 2007</a>). Noting that the zero point for the V-band magnitude scale is <span class="math-container">$3.6 \times 10^{-12}$</span> Wm<span class="math-container">$^{-2}$</span> per angstrom for <span class="math-container">$V=0$</span> and taking the optical band as 3000 angstroms, and noting there are <span class="math-container">$5.34\times 10^{11}$</span> square arcseconds in the sky, then a point in space will receive <span class="math-container">$10^{-5}$</span> Wm<span class="math-container">$^{-2}$</span> (or perhaps about half this, because a significant fraction of the dark sky brightness is caused by airglow).</p>

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

<p>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.</p>

<p>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 <span class="math-container">$10^7$</span> stars at 10th magnitude. Both numbers are higher by orders of magnitude than the actual numbers of Galactic stars at these brightnesses.</p>

CC BY-SA 4.0
2019-05-15T21:19:20.380Z
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.
Ingolifs
CC BY-SA 4.0
2019-05-15T21:42:13.530Z
"How much energy, as light and other frequencies" @Ingolifs
ProfRob
CC BY-SA 4.0
2019-05-15T22:18:32.720Z
@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.
SF.
CC BY-SA 4.0
2019-05-15T22:24:40.463Z
(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)
SF.
CC BY-SA 4.0
2019-05-16T08:55:47.253Z
@sf Revised. The CMB is not as totally dominant as I thought from first looking at the plot.
ProfRob
CC BY-SA 4.0
2019-05-16T09:41:47.353Z
Ten microwatts per square meter in visible range... I can work with that.
SF.
CC BY-SA 4.0
2019-05-16T11:11:32.940Z
@SF still thinking about it. There may be a factor of a few over-estimate in the optical number due to geometric factors.
ProfRob
CC BY-SA 4.0
2019-05-16T11:51:30.380Z
@RobJeffries: Sunlight reaching Earth surface is about 500W/m^2 and it's not blinding unless you look directly at the sun. We can see the Milky Way in a dark sky. So, that's a factor of 8-9 orders of magnitude.difference fitting into our eyes' perception; seems already far, so I don't think there's much room for further downgrading radiant intensity of the night sky.
SF.
CC BY-SA 4.0
2019-05-16T13:21:32.367Z
@RobJeffries: Un-accepting then and awaiting a second opinion :)
SF.
CC BY-SA 4.0
2022-02-03T08:05:38.017Z
[Why don't question-askers accept answers?](https://astronomy.meta.stackexchange.com/q/773/7982) occasionally because the answer author is "[still thinking about it.](https://astronomy.stackexchange.com/questions/21039/what-is-the-radiant-intensity-of-the-celestial-sphere?rq=1#comment58169_31882)" :-)
uhoh
CC BY-SA 4.0
2022-02-03T08:16:04.243Z
@uhoh  and a small edit was made. I'm not begging.
ProfRob
CC BY-SA 4.0
2022-02-03T08:22:21.253Z
@ProfRob just wanna let you know I'm still thinking about you! Your answers are always appreciated. Happy lunar new year!
uhoh
CC BY-SA 4.0
2019-05-15T19:28:26.320Z
2019-05-16T13:57:08.183Z
2019-05-16T13:57:08.183Z
ProfRob
<p>Specifically, excluding the Sun and the Moon:<br/>
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?</p>

<p>Say, we've developed a magical solar battery that has 100% efficiency, can absorb all electromagnetic frequencies, is spherical, fully omnidirectional and has <span class="math-container">$1 m^2$</span> 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?</p>

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

CC BY-SA 3.0
2017-05-17T13:34:00.223Z
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.
StephenG - Help Ukraine
CC BY-SA 3.0
2017-05-17T13:45:23.350Z
@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).
SF.
CC BY-SA 3.0
2017-05-19T06:39:40.953Z
@barrycanter: I do know magnitude of individual stars but their angular diameter is minuscule, meanwhile nebulae and the milky way have a pretty low brightness but output their light over much wider area. I'd have to compute weighted average of magnitude of every object n the sky (weighted by angular diameter), and I don't even have a decent database with that data, only some tables for most notable ones.
SF.
CC BY-SA 3.0
2017-05-19T06:44:59.717Z
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.
SF.
CC BY-SA 4.0
2019-05-14T04:48:52.217Z
small edit to bump to active, see [this comment](https://space.stackexchange.com/questions/36094/are-there-parts-of-space-in-which-no-light-would-make-it-to-an-observer#comment114801_36097)
uhoh
CC BY-SA 4.0
2019-05-15T10:06:04.893Z
This answer from the physics site might somewhat answer your question: [https://physics.stackexchange.com/questions/196366/does-the-number-density-of-photons-n-gamma-approx-108-mathrm-m-3-refer/196373#196373] (Answer)
Agerhell
<luminosity><radiation>
What is the radiant intensity of the celestial sphere?
SF.
# What is the radiant intensity of the celestial sphere?

(Asked by: username_0 on 2017-05-17)

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).

**Tags:** luminosity, radiation

> **Comments:**
> **username_1**: 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.  
> **username_0**: @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).  
> **username_0**: @barrycanter: I do know magnitude of individual stars but their angular diameter is minuscule, meanwhile nebulae and the milky way have a pretty low brightness but output their light over much wider area. I'd have to compute weighted average of magnitude of every object n the sky (weighted by angular diameter), and I don't even have a decent database with that data, only some tables for most notable ones.  
> **username_0**: 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.  
> **username_2**: small edit to bump to active, see [this comment](https://space.stackexchange.com/questions/36094/are-there-parts-of-space-in-which-no-light-would-make-it-to-an-observer#comment114801_36097)  
> **username_3**: This answer from the physics site might somewhat answer your question: [https://physics.stackexchange.com/questions/196366/does-the-number-density-of-photons-n-gamma-approx-108-mathrm-m-3-refer/196373#196373] (Answer)  

---

## 1 answers

### Answer by username_4 on 2019-05-15. Score: 10

The electromagnetic energy density is dominated by the cosmic microwave background and the optical/IR background. [This Physics SE answer](https://physics.stackexchange.com/questions/196366/does-the-number-density-of-photons-n-gamma-approx-108-mathrm-m-3-refer/196373#196373]), 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)](https://arxiv.org/abs/1802.03694)*

[![enter image description here](https://i.sstatic.net/G8uCd.png)](https://i.sstatic.net/G8uCd.png)

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](http://www.ing.iac.es/Astronomy/observing/conditions/skybr/skybr.html)). 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.

> **Comments:**
> **username_5**: 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.  
> **username_4**: "How much energy, as light and other frequencies" @username_5  
> **username_0**: @username_5: 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.  
> **username_0**: (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)  
> **username_4**: @sf Revised. The CMB is not as totally dominant as I thought from first looking at the plot.  
> **username_0**: Ten microwatts per square meter in visible range... I can work with that.  
> **username_4**: @SF still thinking about it. There may be a factor of a few over-estimate in the optical number due to geometric factors.  
> **username_0**: @RobJeffries: Sunlight reaching Earth surface is about 500W/m^2 and it's not blinding unless you look directly at the sun. We can see the Milky Way in a dark sky. So, that's a factor of 8-9 orders of magnitude.difference fitting into our eyes' perception; seems already far, so I don't think there's much room for further downgrading radiant intensity of the night sky.  
> **username_0**: @RobJeffries: Un-accepting then and awaiting a second opinion :)  
> **username_2**: [Why don't question-askers accept answers?](https://astronomy.meta.stackexchange.com/q/773/7982) occasionally because the answer author is "[still thinking about it.](https://astronomy.stackexchange.com/questions/21039/what-is-the-radiant-intensity-of-the-celestial-sphere?rq=1#comment58169_31882)" :-)  
> **username_4**: @username_2  and a small edit was made. I'm not begging.  
> **username_2**: @username_4 just wanna let you know I'm still thinking about you! Your answers are always appreciated. Happy lunar new year!  

---

