<p>Yes, light gravitates. The gravitational charge is energy. Well, gravity is a spin-2 force, so you really have momentum and stress as well, but they are analogous to a generalization of electric current.</p>

<p>In general, anything that contributes to the <a href="http://en.wikipedia.org/wiki/Stress-energy_tensor" rel="noreferrer">stress-energy tensor</a> will have some gravitational effect, and light does that, having both an energy density and putting a pressure in the direction of propagation.</p>

<blockquote>
  <p>But while that light is travelling through space, its energy is "unavailable" to the rest of the universe. </p>
</blockquote>

<p>Not quite. It still gravitates. However, the radiation-dominated era was before about 50k years after the Big Bang, but it is long past. Today the gravitational effect of radiation is cosmologically negligible. We live in a transition between matter-dominated and dark-energy-dominated eras.</p>

<blockquote>
  <p>Given that all stars on the sky is sending photons that reaches every square centimeter of the earth surface, the amount of pressure should sum up to be quite large.</p>
</blockquote>

<p>The light pressure on any surface is proportional to the light energy density incident on it. Thus we can check this line of reasoning directly by observing that the sky is dark at night.</p>

<p><em>Why</em> it is dark at night is probably deserves its own question (cf. also <a href="http://en.wikipedia.org/wiki/Olbers%27_paradox" rel="noreferrer">Olbers' paradox</a>), but it is pretty clear that it is in fact quite small. To be fair, we should check more than the visible range, but even so the sky is pretty dark. Thus on average, light pressure is very small.</p>

<p>We have the privilege of being close to a star, but even during the day, the light pressure due to the Sun is on the order of micropascals.</p>

<blockquote>
  <p>... there should be energy amounting to the gravity of about 1.7*10^22 suns distributed throughout the universe.</p>
</blockquote>

<p>And this is a tiny amount. As you just said, this is the equivalent of about 0.034% of the total mass of stars in the universe, which is in turn constitute but a fraction of the matter in the universe. So why you are surprised that its effect is negligible? It's literally thousands of times less than the uncertainty in the measurements of the amount of matter in the universe.</p>

CC BY-SA 3.0
2014-01-10T12:20:22.903Z
2014-01-10T12:20:22.903Z
Stan Liou
<p>Light causes gravity while travelling, a clear yes, by <a href="http://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence" rel="nofollow noreferrer">Einstein's famous mass-energy equvalence</a>. (Compare <a href="https://physics.stackexchange.com/questions/6197/do-two-beams-of-light-attract-each-other-in-general-theory-of-relativity">this discussion on StackExchange</a>.)</p>

<p>The gravitational pull of light is negligible to other mass in large scale.
Only a small fraction of mass of a star is transformed into light during its lifetime, and only a small part of the ordinary matter has ever been a star. A fraction of the ordinary (standard model particles) matter consists of neutrinos (neutrinos and electrons are leptons). The baryonic matter consists mainly of hydrogen and some helium (nuclei) formed shortly after the big bang.</p>

<p>A small fraction of mass of a star consists of photons, travalling out of the star. <a href="http://www.astro.uni-bonn.de/~nlanger/siu_web/ssescript/new/chapter4-5.pdf" rel="nofollow noreferrer">This travel can take millions of years</a>.</p>

<p>The effect of light on asteroids is not negligible, but it's not the gravitational pull. It' mainly the <a href="http://en.wikipedia.org/wiki/Yarkovsky%E2%80%93O%27Keefe%E2%80%93Radzievskii%E2%80%93Paddack_effect" rel="nofollow noreferrer">YORP</a> effect.
Dust is also affected by light.</p>

CC BY-SA 3.0
2014-01-07T14:47:16.273Z
So, even though that the majority of light that has ever been emitted by the universes' hundreds of billions of galaxies is still in travel, the effect is negible? In every single coordinate in the universe, a photon is crossing for every single light emitting star with a direct path to it. The amount of light "in travel" is also ever increasing, meaning that the combined energy of all other mass is ever decreasing until the point that the mass becomes part of a black hole. How can scientists be sure that it is negligible?
frodeborli
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2014-01-07T15:12:29.367Z
Take the average background temperature af about 3 K; that's the mean temperature, and therefore the overall electromagnetic radiation equilibrium. Consider the average space at a black radiator (http://en.wikipedia.org/wiki/Planck%27s_law). Take a look at the Stefan-Boltzmann law (http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law): The energy of the total radiation is proprtional to the 4th power of the temperature. Now calculate the mass per volume corresponding to this radiation energy, and compare it with the mean density of the local universe.
Gerald
CC BY-SA 3.0
2014-01-07T15:57:30.717Z
(sorry for the two typos above "of about 3K", "as a black radiator") Decreasing mass doesn't necessarily mean converging towards zero, unless you propose, that every particle will decay eventually into photons. There is at least no experimental evidence for this assumption. Not all mass needs to end in a black hole in a unviverse with accelerated expansion. It just cools down.
Gerald
CC BY-SA 3.0
2014-01-07T17:33:18.673Z
@Gerald: It is useful to remember, though, that back in the days of radiation-dominated universe the gravity pull from the light was seriously important.
Alexey Bobrick
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2014-01-07T17:43:28.893Z
I'm not quite sure, whether we are talking of the same thing. Radiation pressure took an important role (see http://en.wikipedia.org/wiki/Radiation_pressure). I'm not aware of an important role of the gravity pull of light. Can you point to a source, which provides evidence?
Gerald
CC BY-SA 3.0
2014-01-07T18:35:21.070Z
I am not sure that I am following you, but the amount of "emptiness" containing only the sum of photons in travel must be in many orders of magnitude larger than any mass in existence. And I do propose that as long as an object is radiating light, it is losing energy and thus is decreasing its own gravitation.
frodeborli
CC BY-SA 3.0
2014-01-07T18:44:52.917Z
Sorry, I meant volume of emptiness must be hundreds of orders of magnitude larger than any volume inhabited by mass. How can that be negligible on cosmic scale?
frodeborli
CC BY-SA 3.0
2014-01-07T20:13:02.880Z
It's not empty, see http://en.wikipedia.org/wiki/Warm%E2%80%93hot_intergalactic_medium and http://en.wikipedia.org/wiki/Interstellar_medium
Gerald
CC BY-SA 3.0
2014-01-10T12:21:53.160Z
I think the first sentence is exactly backwards, but otherwise generally agree with this approach.
Stan Liou
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2014-01-10T23:03:08.920Z
The effect is tiny in comparison with benching of light due to gravity. A more detailed discussion about the first sentence, see here: http://physics.stackexchange.com/questions/6197/do-two-beams-of-light-attract-each-other-in-general-theory-of-relativity
Gerald
CC BY-SA 3.0
2014-01-11T04:41:18.807Z
What I mean is simply that mass has gravitational effects because it has energy (and a lot of it), which shows up in the $T^{00}$ component of the stress-energy tensor. Instead of explaining gravity trying to explain gravity as effect of mass, which is incorrect anyway, one should instead recognize that it's the energy that the gravitational charge in a way analogous to, say, electric charge.
Stan Liou
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2014-01-12T17:54:51.590Z
Ok, now I understand what you mean. That's agreed.
Gerald
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2014-01-07T12:48:25.377Z
2014-01-10T23:05:18.273Z
2017-04-13T12:40:15.377Z
Gerald
<p>Old question, but I'll address something that hasn't been brought up by the previous answers.</p>

<h2>Photons $\simeq$ CMB photons (to first order)</h2>

<p>As the others has already said: yes, light has energy and hence it gravitates. The bulk of photons that permeate the Universe isn't of stellar origin, though, but is in fact the cosmic microwave background, the energy density of which several orders of magnitude larger than other photons, as seen in the graph from <a href="https://physics.stackexchange.com/a/196373/70207">this answer to "Number density of CMB photons"</a>. In terms of number density, there are 4-500 photons per cm$^3$.</p>

<h2>Space is big and isotropic</h2>

<p>Since CMB photons are isotropically distributed, the ever-so-small radiation pressure is equal in all directions, and hence cancels out. And although we're all the time bombarded by both CMB photons and stellar photons, space is so mind-bogglingly big (<a href="https://en.wikiquote.org/wiki/The_Hitchhiker&#39;s_Guide_to_the_Galaxy#Chapter_8" rel="nofollow noreferrer">D. Adams, 1978</a>) that if you consider a random photon in the Universe, the probability of it hitting anything at all is negligible. Roughly 90% of the CMB photons have traveled for 13.8 billion years without hitting anything; the remaining 10% interacted with the free electrons that were released after reionization, but weren't absorbed, just polarized, and by far most of these interactions took place shortly after reionization; by now, the Universe has simply expanded too much. </p>

<h2>Photons are redshifted</h2>

<p>Although there <em>is</em> energy in photons, and hence they add to gravitation, first of all they're homogeneously distributed in the Universe (and thus pulls equally in all directions), and second their energy density is negligible compared to baryons ("normal matter" like gas, stars, and planets), dark matter, and dark energy. In fact, their relative densities are $\{\rho_\mathrm{bar},\rho_\mathrm{DM},\rho_\mathrm{DE},\rho_\mathrm{phot}\}/\rho_\mathrm{total} = \{0.05,0.27,0.68,10^{-4}\}$. But this was not always the case. As the Universe expands and new space is created, the density of matter decreases as $1/a^3$, where $a$ is the scale factor ("size") of the Universe. The same is true for photons, but since additionally they're redshifted proportionally to $a$, their energy density decreases as $1/a^4$. That means that as you go back in time, the relative contribution of photons to the energy budget increases, and in fact until the Universe was 47,000 years old, its dynamics was dominated by radiation.</p>

CC BY-SA 3.0
2016-02-04T15:24:08.460Z
The biggest a-ha in your answer was that photons are redshifted - which I haven't considered. Just curious: regarding isotropic distribution of photons, how can you be sure about that?
frodeborli
CC BY-SA 3.0
2016-02-04T15:35:36.427Z
@frodeborli: If you look at a map of the CMB, such as [this one](http://www.esa.int/spaceinimages/Images/2013/03/Planck_CMB), you'll see that it's isotropic to one part in ~1e5. Note that on a map like this, two important *an*isotropies have been removed: 1) Because we're inside the Milky Way, there's an extra signal from sources in the Galactic disk, and 2) because we're moving through space at some 500 km/s (in comoving coordinates), the CMB is slightly blueshifted — and hence more energetic — in the direction in which we're moving, and correspondingly redshifted in the opposite direction.
pela
CC BY-SA 3.0
2016-02-04T17:20:03.220Z
Yes, so it appears isotropic in our region of space. But I don't consider this proof that photons are isotropic in their *distribution* throughout space. That very distant star you're looking at *is*, from our perspective, in a universe that is only 47000 years old.
frodeborli
CC BY-SA 3.0
2016-02-04T22:12:54.033Z
And we see those distant old stars in every direction @frodeborli. If you have some complicated theory to explain it, good for you, but the Occam's razor causes scientists prefer the simpler theory of isotropic distribution.
kubanczyk
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2016-02-04T23:01:34.907Z
@kubanczyk “Make things as simple as possible, but not simpler.”. Regardless of that; you can't possibly conclude beyond doubt that photons are evenly distributed throughout space, *based solely* on the fact that we are receiving them somewhat evenly distributed at this tiny planet. There are many photons that we will never receive here, and you don't know where they are heading or how many they are. There might/probably are trillions of super energetic GRBs shooting through space that we will never see; simply seeing them would cause a sterile earth.
frodeborli
CC BY-SA 3.0
2016-02-04T23:22:10.650Z
Science cannot prove *anything*. Science can partially predict future experiences of humans, and it can improve itself in predicting more and more correctly. Any additional propositions (like "super energetic GRBs") are as much useful as they allow to make additional predictions about our future experiences. Saying "I don't know if there is a monster under the bed, so I don't know if it kills me" simply doesn't help me to achieve evolutionary advantage, so I don't say it: Occam's razor. Only if I have experience of hearing heavy breath under my bed, I'll think about 'monster theory'.
kubanczyk
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2016-02-05T08:58:22.480Z
@frodeborli: Claiming that the observed isotropy of the CMB doesn't imply homogeneity would imply that we occupy a special place in the Universe. That thought is so terrifying that you would really have to come up with a sound and falsifiable theory to justify the claim, in order to be taken seriously. Isotropy implying homogeneity is not "oversimplifying things", it's the most natural expectation. But of course it's not a proof, as with everything in physics.
pela
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2016-02-05T14:51:31.527Z
@pela It does not imply homogenity. I don't understand why it even suggests homogenity. If you draw 10^100 infinitely long and fairly thin lines throughout the universe, the odds of any one of those lines intersecting with earth is still tiny. You can't conclude that since nobody is pointing a flashlight at you, there are no flashlights. And you can't see the ray coming from a flashlight, unless it's pointing at you.
frodeborli
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2016-02-05T15:02:10.057Z
@kubanczyk I agree, and I was not seeking for you too claim that science proves anything. I simply don't *understand* how scientist can feel that it's logical to conclude that due to us seeing isotropic background radiation- everybody else must be seeing isotropic background radiation,regardless of where they are. In fact, we know that an object 10 billion light years away from us is, right now as we are looking at it, existing in a brighter/younger universe and it's difficult to know if that is isotropic.Every object must see themself as being in the darkest area of the universe, us included.
frodeborli
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2016-02-05T15:15:08.393Z
@frodeborli: I don't think I understand you. If the Universe looks the same in all directions, then unless we are in a special place, it must _be_ the same in all directions, i.e. it must be homogeneous. See e.g. [this answer on the cosmological principle](http://astronomy.stackexchange.com/questions/13189/on-the-cosmological-principle/13193#13193). By the way, if you draw $10^{100}$ lines in the observable Universe, every cm$^2$ of you, and of the Earth, will be intersected by $10^{42}$ lines
pela
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2016-02-05T15:20:35.577Z
I don't really get your example with the flashlights. It seems to describe a scenario where something _isn't_ observed. If I don't see anybody pointing flashlights at me from any direction, I can't conclude that there aren't any people pointing flashlights in different directions, but I _can_ conclude that there aren't many. But in the CMB case, we see $10^{13}$ CMB going through every cm$^2$ of a detector every second, regardless of the direction we look in. Unless we are in a special place, another observer in another part of the Universe would see the same.
pela
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2016-02-05T16:12:58.787Z
@pela Thank you for calculating that. I was initially going for a much lower number (10^40), but got a bit brave. Still I am surprised. Curious as to how you calculate that. Regarding homogeneousness, due to relativity - the universe is brighter the further you look, *for the object you look at* due to that object being in a younger universe. The object you are looking at will appear to have stronger gravitational forces affecting it. The density of empty space due to radiation is a function of distance from the observer.
frodeborli
CC BY-SA 3.0
2016-02-05T16:26:34.713Z
@pela Another observer in another place (5 billion light years away) in the universe will probably observe a similar density as we do, in 5 billion years. But at that time we will see a darker universe. It must be a gradient, it can't be homogenous. Also, due to the vastness of the universe, if we look south at a distant star 13 billion ly away, and then look north 13 billion ly away, those two stars are very close to each other. Looking a bit further, the stars actually occupy the same space - and that obviously will appear homogenous.
frodeborli
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2016-02-06T00:17:22.343Z
@frodeborli: An order-of-magnitude estimate of the number of lines penetrating a cm$^2$ — the "line flux" $F_\mathrm{lines}$ — is simply your number of lines $N$, divided by the cross section of the Universe, i.e. $F_\mathrm{lines} \sim N / \pi R_\mathrm{Uni}^2 = 10^{100} / \pi (14.4\,\mathrm{Gpc})^2 \sim 10^{42}\,\mathrm{cm}^{-2}$. But I don't understand your statement "the Universe looks brighter, the farther we look". Why would an object be brighter due to the Universe being younger? Anyway, you're right that the radiation density is a function of distance from us, as per discussed above.
pela
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2016-02-06T00:18:10.950Z
I think perhaps I see what you mean. By "the Universe being brighter far away" you mean it was hotter? Don't forget that as we look far away, we also look at the past. Another observer 5 Gly away will, as you say, observe the same as us, but not in 5 Gyr. He will observe the same _right now_. Of course, if _we_ look at that observer _right now_, we will see him as he were billions of years ago (9.5 Gyr ago; not 5 Gyr as you might think, since the Universe is expanding). To see how he looks right now, we will have to wait some 7 Gyr.
pela
CC BY-SA 3.0
2016-02-06T00:18:19.543Z
The gradient you talk about is a gradient in _time_, not space. If you froze the Universe right now and took off in a spaceship, you would have a very boring journey; as you arrive to galaxies that before you departed looked less evolved, you'd find that when you arrive, they'd be just as evolved as the Milky Way (statistically speaking, of course).
pela
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2016-02-06T13:41:01.390Z
@pela That gradient in time is what implies that the universe is brighter for objects far away. As light and gravity have propagated for 10 billion years before reaching us, the source of that light was in a much brighter universe. Since simultaneousness is relative, that object is currently in a bright universe, while we are in a dark universe (and vice versa); thus brightness must be a gradient.
frodeborli
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2016-02-06T14:02:27.420Z
Since most of what we are looking at, at great distances, are very "young objects" existing in a much smaller/denser universe I would expect us to see a somewhat uniform universe. But that does not mean that if we had been elsewhere, and experienced 13.7 billion years, that what we would see there is the same as what we see here, now. Regarding the space ship journey; your story is true *if* the region of space your travel to is as dense as our region of space. If not, that region might be older or younger than our region of space.
frodeborli
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2016-02-06T16:28:43.090Z
At the risk of repeating myself: If you observe an isotropic Universe, and you assume that we don't occupy a special place in it, then yes, another observer at another place would see exactly the same as we do (modulo statistics). There's no way around that. That makes my story true: If you freeze the Universe, then every part of it is 13.8 Gyr old, and every part you visit in this frozen will thus look the same to you when you get there (mod. statistics). The reason distant galaxies look younger is not because they are younger right now, it's only because the light took some time to get here.
pela
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2020-01-12T11:07:49.200Z
@pelo I revisited this thread just now, and I can see that we were speaking past each other. I understand your reasoning, but I was focused on a different vay of thinking about relativity. To me the word NOW also contains spatial coordinates. If we froze the universe NOW then travel to a distant star, then we will reach a young star. The destination will not age as we approach it, if universe was frozen as seen from earth.
frodeborli
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2020-01-12T20:03:00.517Z
@frodeborli Hmm… if I understand you correctly, then that's not true. If I observe a galaxy which is 1 Glyr away, then what I see is a 1 Gyr old picture, i.e. if I could see a clock, that clock would say that Universe is 1**2**.8 Gyr old. But that's only because light took 1 Gyr to reach me. In reality the galaxy _is_ older, and if I freeze space and go there, then once I get there, the clock will say "1**3**.8 Gyr".
pela
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2020-01-13T07:56:57.350Z
@pela Naturally, but I am talking about relativity of simultaneity. If my calendar say it is the 12 o'clock on the 13th of January 2020 at the same time as I see a supernova explode then, *in my frame of reference* - those two events are simultaneous. So when I say that "now", that distant supernova is in a younger universe, then that is a valid view. Saying that it is in a universe that is as old as ours is an unprovable prediction, although I agree with that prediction.
frodeborli
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2020-01-13T08:24:40.170Z
@pela Strictly speaking, you can't be sure that the clock would say 13.8 Gyr when you arrive - because you can't know the relative amount of gravitational time dilation in that region of space, compared to our region of space. *Right now* in your way of thinking, the universe will not be the same age everywhere.
frodeborli
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2020-01-13T11:35:32.763Z
@frodeborli Well, unless the clock I'm looking at resides in some freakish potential well, or sitting on the arm of an alien who likes to fly around at 99% the speed of light, I'm okay with completely neglecting that. I'm referring to comoving time, but of course you can always conjecture some scenario that makes my statement untrue. If that's what you're after, then we are indeed talking past each other.
pela
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2020-01-13T11:37:43.113Z
And yes, in physics all predictions are unprovable. But if you're questioning the standard model of cosmology, then that is a whole other discussion.
pela
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2020-01-13T19:12:47.037Z
@pela Point is; now means now according to a single particular frame of reference. I don't like mixing frames of references when talking about physics. So when I say that the supernova happened 400 years ago, then that is absolutely correct unless you complicate things by selecting a different reference frame. When I say that right now, that distant stellar object exists in a denser and younger universe - it is because I assume we all agree on which frame we are talking about. When you decided to disagree with me, it is because you decided to mix frames.
frodeborli
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2020-01-14T09:43:57.663Z
@frodeborli Okay so _usually_ in astronomy we don't really care what goes on in some galaxy _right now_. We don't know _exactly_ how some distant galaxy has evolved since it emitted the light we see, but we know — in a statistical sense — how it and other equally distant galaxies have evolved. We do, perhaps confusingly, say that "galaxy GN-z11 resides in an environment that is ~1000 times as dense as our local, current Universe", but even though it sounds as we mean "right now", it is understood that in reality, right now it will reside in an environment that is similar to our local Universe.
pela
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2020-01-14T09:44:28.793Z
So, _if_ we nevertheless use the term "right now", we mean according to a comoving observer, which except for small peculiar velocities is the same frame in which we do our observations.
pela
CC BY-SA 3.0
2016-02-04T14:51:26.970Z
2016-02-04T14:51:26.970Z
2017-04-13T12:40:15.377Z
pela
<p>When light is emitted by for example a star, that star loses energy - which causes it to reduce its gravity. Then that energy begins a journey for potentially billions of years, until it reaches some other object.</p>

<p>When that light reaches a surface, such as another star or galaxy, it will give that energy to the destination star in the form of heat. This causes the receiver to increase its energy, in turn restoring a sort of balance. It also causes the receiver to emit a minute amount of more light again, almost like a reflection.</p>

<p>It will also excert pressure on the receiving surface once it reaches its destination, be it a star, a rock or anything else.</p>

<p>But while that light is travelling through space, its energy is "unavailable" to the rest of the universe. Naturally I ask the following question: </p>

<p><strong>Will light cause gravity, while it is traveling?</strong></p>

<p>Every single star emits light in every direction, and will eventually reach every other star in the universe. At any single point in the universe, there must be a continous ray of light coming from every single other star in the universe, that has a direct path to that point. Given that all stars on the sky is sending photons that reaches every square centimeter of the earth surface, the amount of pressure should sum up to be quite large.</p>

<p><strong>Is the amount of pressure really neglible, given that every single atom on any surface is receiving light from every single lightsource on the sky?</strong></p>

<p>Based on a calculation found at <a href="http://solar-center.stanford.edu/FAQ/Qshrink.html" rel="noreferrer">http://solar-center.stanford.edu/FAQ/Qshrink.html</a> the sun will during its lifetime emit 0.034 % of its total mass as energy. Assuming the sun is average, and that there are about 10^24 stars in the universe, and all of these stars on average are half way through their lifetime, there should be energy amounting to the gravity of about 1.7*10^22 suns distributed throughout the universe.</p>

<gravity><universe><light>
How does light affect the universe?
frodeborli
# How does light affect the universe?

(Asked by: username_0 on 2014-01-07)

When light is emitted by for example a star, that star loses energy - which causes it to reduce its gravity. Then that energy begins a journey for potentially billions of years, until it reaches some other object.

When that light reaches a surface, such as another star or galaxy, it will give that energy to the destination star in the form of heat. This causes the receiver to increase its energy, in turn restoring a sort of balance. It also causes the receiver to emit a minute amount of more light again, almost like a reflection.

It will also excert pressure on the receiving surface once it reaches its destination, be it a star, a rock or anything else.

But while that light is travelling through space, its energy is "unavailable" to the rest of the universe. Naturally I ask the following question:

**Will light cause gravity, while it is traveling?**

Every single star emits light in every direction, and will eventually reach every other star in the universe. At any single point in the universe, there must be a continous ray of light coming from every single other star in the universe, that has a direct path to that point. Given that all stars on the sky is sending photons that reaches every square centimeter of the earth surface, the amount of pressure should sum up to be quite large.

**Is the amount of pressure really neglible, given that every single atom on any surface is receiving light from every single lightsource on the sky?**

Based on a calculation found at <http://solar-center.stanford.edu/FAQ/Qshrink.html> the sun will during its lifetime emit 0.034 % of its total mass as energy. Assuming the sun is average, and that there are about 10^24 stars in the universe, and all of these stars on average are half way through their lifetime, there should be energy amounting to the gravity of about 1.7\*10^22 suns distributed throughout the universe.

**Tags:** gravity, universe, light


---

## 3 answers

### Answer by username_4 on 2016-02-04. Score: 5 | Accepted answer

Old question, but I'll address something that hasn't been brought up by the previous answers.

## Photons $\simeq$ CMB photons (to first order)

As the others has already said: yes, light has energy and hence it gravitates. The bulk of photons that permeate the Universe isn't of stellar origin, though, but is in fact the cosmic microwave background, the energy density of which several orders of magnitude larger than other photons, as seen in the graph from [this answer to "Number density of CMB photons"](https://physics.stackexchange.com/a/196373/70207). In terms of number density, there are 4-500 photons per cm$^3$.

## Space is big and isotropic

Since CMB photons are isotropically distributed, the ever-so-small radiation pressure is equal in all directions, and hence cancels out. And although we're all the time bombarded by both CMB photons and stellar photons, space is so mind-bogglingly big ([D. Adams, 1978](https://en.wikiquote.org/wiki/The_Hitchhiker's_Guide_to_the_Galaxy#Chapter_8)) that if you consider a random photon in the Universe, the probability of it hitting anything at all is negligible. Roughly 90% of the CMB photons have traveled for 13.8 billion years without hitting anything; the remaining 10% interacted with the free electrons that were released after reionization, but weren't absorbed, just polarized, and by far most of these interactions took place shortly after reionization; by now, the Universe has simply expanded too much.

## Photons are redshifted

Although there *is* energy in photons, and hence they add to gravitation, first of all they're homogeneously distributed in the Universe (and thus pulls equally in all directions), and second their energy density is negligible compared to baryons ("normal matter" like gas, stars, and planets), dark matter, and dark energy. In fact, their relative densities are $\{\rho\_\mathrm{bar},\rho\_\mathrm{DM},\rho\_\mathrm{DE},\rho\_\mathrm{phot}\}/\rho\_\mathrm{total} = \{0.05,0.27,0.68,10^{-4}\}$. But this was not always the case. As the Universe expands and new space is created, the density of matter decreases as $1/a^3$, where $a$ is the scale factor ("size") of the Universe. The same is true for photons, but since additionally they're redshifted proportionally to $a$, their energy density decreases as $1/a^4$. That means that as you go back in time, the relative contribution of photons to the energy budget increases, and in fact until the Universe was 47,000 years old, its dynamics was dominated by radiation.

> **Comments:**
> **username_0**: The biggest a-ha in your answer was that photons are redshifted - which I haven't considered. Just curious: regarding isotropic distribution of photons, how can you be sure about that?  
> **username_4**: @username_0: If you look at a map of the CMB, such as [this one](http://www.esa.int/spaceinimages/Images/2013/03/Planck_CMB), you'll see that it's isotropic to one part in ~1e5. Note that on a map like this, two important *an*isotropies have been removed: 1) Because we're inside the Milky Way, there's an extra signal from sources in the Galactic disk, and 2) because we're moving through space at some 500 km/s (in comoving coordinates), the CMB is slightly blueshifted — and hence more energetic — in the direction in which we're moving, and correspondingly redshifted in the opposite direction.  
> **username_0**: Yes, so it appears isotropic in our region of space. But I don't consider this proof that photons are isotropic in their *distribution* throughout space. That very distant star you're looking at *is*, from our perspective, in a universe that is only 47000 years old.  
> **username_5**: And we see those distant old stars in every direction @username_0. If you have some complicated theory to explain it, good for you, but the Occam's razor causes scientists prefer the simpler theory of isotropic distribution.  
> **username_0**: @username_5 “Make things as simple as possible, but not simpler.”. Regardless of that; you can't possibly conclude beyond doubt that photons are evenly distributed throughout space, *based solely* on the fact that we are receiving them somewhat evenly distributed at this tiny planet. There are many photons that we will never receive here, and you don't know where they are heading or how many they are. There might/probably are trillions of super energetic GRBs shooting through space that we will never see; simply seeing them would cause a sterile earth.  
> **username_5**: Science cannot prove *anything*. Science can partially predict future experiences of humans, and it can improve itself in predicting more and more correctly. Any additional propositions (like "super energetic GRBs") are as much useful as they allow to make additional predictions about our future experiences. Saying "I don't know if there is a monster under the bed, so I don't know if it kills me" simply doesn't help me to achieve evolutionary advantage, so I don't say it: Occam's razor. Only if I have experience of hearing heavy breath under my bed, I'll think about 'monster theory'.  
> **username_4**: @username_0: Claiming that the observed isotropy of the CMB doesn't imply homogeneity would imply that we occupy a special place in the Universe. That thought is so terrifying that you would really have to come up with a sound and falsifiable theory to justify the claim, in order to be taken seriously. Isotropy implying homogeneity is not "oversimplifying things", it's the most natural expectation. But of course it's not a proof, as with everything in physics.  
> **username_0**: @username_4 It does not imply homogenity. I don't understand why it even suggests homogenity. If you draw 10^100 infinitely long and fairly thin lines throughout the universe, the odds of any one of those lines intersecting with earth is still tiny. You can't conclude that since nobody is pointing a flashlight at you, there are no flashlights. And you can't see the ray coming from a flashlight, unless it's pointing at you.  
> **username_0**: @username_5 I agree, and I was not seeking for you too claim that science proves anything. I simply don't *understand* how scientist can feel that it's logical to conclude that due to us seeing isotropic background radiation- everybody else must be seeing isotropic background radiation,regardless of where they are. In fact, we know that an object 10 billion light years away from us is, right now as we are looking at it, existing in a brighter/younger universe and it's difficult to know if that is isotropic.Every object must see themself as being in the darkest area of the universe, us included.  
> **username_4**: @username_0: I don't think I understand you. If the Universe looks the same in all directions, then unless we are in a special place, it must _be_ the same in all directions, i.e. it must be homogeneous. See e.g. [this answer on the cosmological principle](http://astronomy.stackexchange.com/questions/13189/on-the-cosmological-principle/13193#13193). By the way, if you draw $10^{100}$ lines in the observable Universe, every cm$^2$ of you, and of the Earth, will be intersected by $10^{42}$ lines  
> **username_4**: I don't really get your example with the flashlights. It seems to describe a scenario where something _isn't_ observed. If I don't see anybody pointing flashlights at me from any direction, I can't conclude that there aren't any people pointing flashlights in different directions, but I _can_ conclude that there aren't many. But in the CMB case, we see $10^{13}$ CMB going through every cm$^2$ of a detector every second, regardless of the direction we look in. Unless we are in a special place, another observer in another part of the Universe would see the same.  
> **username_0**: @username_4 Thank you for calculating that. I was initially going for a much lower number (10^40), but got a bit brave. Still I am surprised. Curious as to how you calculate that. Regarding homogeneousness, due to relativity - the universe is brighter the further you look, *for the object you look at* due to that object being in a younger universe. The object you are looking at will appear to have stronger gravitational forces affecting it. The density of empty space due to radiation is a function of distance from the observer.  
> **username_0**: @username_4 Another observer in another place (5 billion light years away) in the universe will probably observe a similar density as we do, in 5 billion years. But at that time we will see a darker universe. It must be a gradient, it can't be homogenous. Also, due to the vastness of the universe, if we look south at a distant star 13 billion ly away, and then look north 13 billion ly away, those two stars are very close to each other. Looking a bit further, the stars actually occupy the same space - and that obviously will appear homogenous.  
> **username_4**: @username_0: An order-of-magnitude estimate of the number of lines penetrating a cm$^2$ — the "line flux" $F_\mathrm{lines}$ — is simply your number of lines $N$, divided by the cross section of the Universe, i.e. $F_\mathrm{lines} \sim N / \pi R_\mathrm{Uni}^2 = 10^{100} / \pi (14.4\,\mathrm{Gpc})^2 \sim 10^{42}\,\mathrm{cm}^{-2}$. But I don't understand your statement "the Universe looks brighter, the farther we look". Why would an object be brighter due to the Universe being younger? Anyway, you're right that the radiation density is a function of distance from us, as per discussed above.  
> **username_4**: I think perhaps I see what you mean. By "the Universe being brighter far away" you mean it was hotter? Don't forget that as we look far away, we also look at the past. Another observer 5 Gly away will, as you say, observe the same as us, but not in 5 Gyr. He will observe the same _right now_. Of course, if _we_ look at that observer _right now_, we will see him as he were billions of years ago (9.5 Gyr ago; not 5 Gyr as you might think, since the Universe is expanding). To see how he looks right now, we will have to wait some 7 Gyr.  
> **username_4**: The gradient you talk about is a gradient in _time_, not space. If you froze the Universe right now and took off in a spaceship, you would have a very boring journey; as you arrive to galaxies that before you departed looked less evolved, you'd find that when you arrive, they'd be just as evolved as the Milky Way (statistically speaking, of course).  
> **username_0**: @username_4 That gradient in time is what implies that the universe is brighter for objects far away. As light and gravity have propagated for 10 billion years before reaching us, the source of that light was in a much brighter universe. Since simultaneousness is relative, that object is currently in a bright universe, while we are in a dark universe (and vice versa); thus brightness must be a gradient.  
> **username_0**: Since most of what we are looking at, at great distances, are very "young objects" existing in a much smaller/denser universe I would expect us to see a somewhat uniform universe. But that does not mean that if we had been elsewhere, and experienced 13.7 billion years, that what we would see there is the same as what we see here, now. Regarding the space ship journey; your story is true *if* the region of space your travel to is as dense as our region of space. If not, that region might be older or younger than our region of space.  
> **username_4**: At the risk of repeating myself: If you observe an isotropic Universe, and you assume that we don't occupy a special place in it, then yes, another observer at another place would see exactly the same as we do (modulo statistics). There's no way around that. That makes my story true: If you freeze the Universe, then every part of it is 13.8 Gyr old, and every part you visit in this frozen will thus look the same to you when you get there (mod. statistics). The reason distant galaxies look younger is not because they are younger right now, it's only because the light took some time to get here.  
> **username_0**: @pelo I revisited this thread just now, and I can see that we were speaking past each other. I understand your reasoning, but I was focused on a different vay of thinking about relativity. To me the word NOW also contains spatial coordinates. If we froze the universe NOW then travel to a distant star, then we will reach a young star. The destination will not age as we approach it, if universe was frozen as seen from earth.  
> **username_4**: @username_0 Hmm… if I understand you correctly, then that's not true. If I observe a galaxy which is 1 Glyr away, then what I see is a 1 Gyr old picture, i.e. if I could see a clock, that clock would say that Universe is 1**2**.8 Gyr old. But that's only because light took 1 Gyr to reach me. In reality the galaxy _is_ older, and if I freeze space and go there, then once I get there, the clock will say "1**3**.8 Gyr".  
> **username_0**: @username_4 Naturally, but I am talking about relativity of simultaneity. If my calendar say it is the 12 o'clock on the 13th of January 2020 at the same time as I see a supernova explode then, *in my frame of reference* - those two events are simultaneous. So when I say that "now", that distant supernova is in a younger universe, then that is a valid view. Saying that it is in a universe that is as old as ours is an unprovable prediction, although I agree with that prediction.  
> **username_0**: @username_4 Strictly speaking, you can't be sure that the clock would say 13.8 Gyr when you arrive - because you can't know the relative amount of gravitational time dilation in that region of space, compared to our region of space. *Right now* in your way of thinking, the universe will not be the same age everywhere.  
> **username_4**: @username_0 Well, unless the clock I'm looking at resides in some freakish potential well, or sitting on the arm of an alien who likes to fly around at 99% the speed of light, I'm okay with completely neglecting that. I'm referring to comoving time, but of course you can always conjecture some scenario that makes my statement untrue. If that's what you're after, then we are indeed talking past each other.  
> **username_4**: And yes, in physics all predictions are unprovable. But if you're questioning the standard model of cosmology, then that is a whole other discussion.  
> **username_0**: @username_4 Point is; now means now according to a single particular frame of reference. I don't like mixing frames of references when talking about physics. So when I say that the supernova happened 400 years ago, then that is absolutely correct unless you complicate things by selecting a different reference frame. When I say that right now, that distant stellar object exists in a denser and younger universe - it is because I assume we all agree on which frame we are talking about. When you decided to disagree with me, it is because you decided to mix frames.  
> **username_4**: @username_0 Okay so _usually_ in astronomy we don't really care what goes on in some galaxy _right now_. We don't know _exactly_ how some distant galaxy has evolved since it emitted the light we see, but we know — in a statistical sense — how it and other equally distant galaxies have evolved. We do, perhaps confusingly, say that "galaxy GN-z11 resides in an environment that is ~1000 times as dense as our local, current Universe", but even though it sounds as we mean "right now", it is understood that in reality, right now it will reside in an environment that is similar to our local Universe.  
> **username_4**: So, _if_ we nevertheless use the term "right now", we mean according to a comoving observer, which except for small peculiar velocities is the same frame in which we do our observations.  

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### Answer by username_1 on 2014-01-10. Score: 6

Yes, light gravitates. The gravitational charge is energy. Well, gravity is a spin-2 force, so you really have momentum and stress as well, but they are analogous to a generalization of electric current.

In general, anything that contributes to the [stress-energy tensor](http://en.wikipedia.org/wiki/Stress-energy_tensor) will have some gravitational effect, and light does that, having both an energy density and putting a pressure in the direction of propagation.

> But while that light is travelling through space, its energy is "unavailable" to the rest of the universe.

Not quite. It still gravitates. However, the radiation-dominated era was before about 50k years after the Big Bang, but it is long past. Today the gravitational effect of radiation is cosmologically negligible. We live in a transition between matter-dominated and dark-energy-dominated eras.

> Given that all stars on the sky is sending photons that reaches every square centimeter of the earth surface, the amount of pressure should sum up to be quite large.

The light pressure on any surface is proportional to the light energy density incident on it. Thus we can check this line of reasoning directly by observing that the sky is dark at night.

*Why* it is dark at night is probably deserves its own question (cf. also [Olbers' paradox](http://en.wikipedia.org/wiki/Olbers%27_paradox)), but it is pretty clear that it is in fact quite small. To be fair, we should check more than the visible range, but even so the sky is pretty dark. Thus on average, light pressure is very small.

We have the privilege of being close to a star, but even during the day, the light pressure due to the Sun is on the order of micropascals.

> ... there should be energy amounting to the gravity of about 1.7\*10^22 suns distributed throughout the universe.

And this is a tiny amount. As you just said, this is the equivalent of about 0.034% of the total mass of stars in the universe, which is in turn constitute but a fraction of the matter in the universe. So why you are surprised that its effect is negligible? It's literally thousands of times less than the uncertainty in the measurements of the amount of matter in the universe.


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### Answer by username_2 on 2014-01-07. Score: 4

Light causes gravity while travelling, a clear yes, by [Einstein's famous mass-energy equvalence](http://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence). (Compare [this discussion on StackExchange](https://physics.stackexchange.com/questions/6197/do-two-beams-of-light-attract-each-other-in-general-theory-of-relativity).)

The gravitational pull of light is negligible to other mass in large scale.
Only a small fraction of mass of a star is transformed into light during its lifetime, and only a small part of the ordinary matter has ever been a star. A fraction of the ordinary (standard model particles) matter consists of neutrinos (neutrinos and electrons are leptons). The baryonic matter consists mainly of hydrogen and some helium (nuclei) formed shortly after the big bang.

A small fraction of mass of a star consists of photons, travalling out of the star. [This travel can take millions of years](http://www.astro.uni-bonn.de/~nlanger/siu_web/ssescript/new/chapter4-5.pdf).

The effect of light on asteroids is not negligible, but it's not the gravitational pull. It' mainly the [YORP](http://en.wikipedia.org/wiki/Yarkovsky%E2%80%93O%27Keefe%E2%80%93Radzievskii%E2%80%93Paddack_effect) effect.
Dust is also affected by light.

> **Comments:**
> **username_0**: So, even though that the majority of light that has ever been emitted by the universes' hundreds of billions of galaxies is still in travel, the effect is negible? In every single coordinate in the universe, a photon is crossing for every single light emitting star with a direct path to it. The amount of light "in travel" is also ever increasing, meaning that the combined energy of all other mass is ever decreasing until the point that the mass becomes part of a black hole. How can scientists be sure that it is negligible?  
> **username_2**: Take the average background temperature af about 3 K; that's the mean temperature, and therefore the overall electromagnetic radiation equilibrium. Consider the average space at a black radiator (http://en.wikipedia.org/wiki/Planck%27s_law). Take a look at the Stefan-Boltzmann law (http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law): The energy of the total radiation is proprtional to the 4th power of the temperature. Now calculate the mass per volume corresponding to this radiation energy, and compare it with the mean density of the local universe.  
> **username_2**: (sorry for the two typos above "of about 3K", "as a black radiator") Decreasing mass doesn't necessarily mean converging towards zero, unless you propose, that every particle will decay eventually into photons. There is at least no experimental evidence for this assumption. Not all mass needs to end in a black hole in a unviverse with accelerated expansion. It just cools down.  
> **username_3**: @username_2: It is useful to remember, though, that back in the days of radiation-dominated universe the gravity pull from the light was seriously important.  
> **username_2**: I'm not quite sure, whether we are talking of the same thing. Radiation pressure took an important role (see http://en.wikipedia.org/wiki/Radiation_pressure). I'm not aware of an important role of the gravity pull of light. Can you point to a source, which provides evidence?  
> **username_0**: I am not sure that I am following you, but the amount of "emptiness" containing only the sum of photons in travel must be in many orders of magnitude larger than any mass in existence. And I do propose that as long as an object is radiating light, it is losing energy and thus is decreasing its own gravitation.  
> **username_0**: Sorry, I meant volume of emptiness must be hundreds of orders of magnitude larger than any volume inhabited by mass. How can that be negligible on cosmic scale?  
> **username_2**: It's not empty, see http://en.wikipedia.org/wiki/Warm%E2%80%93hot_intergalactic_medium and http://en.wikipedia.org/wiki/Interstellar_medium  
> **username_1**: I think the first sentence is exactly backwards, but otherwise generally agree with this approach.  
> **username_2**: The effect is tiny in comparison with benching of light due to gravity. A more detailed discussion about the first sentence, see here: http://physics.stackexchange.com/questions/6197/do-two-beams-of-light-attract-each-other-in-general-theory-of-relativity  
> **username_1**: What I mean is simply that mass has gravitational effects because it has energy (and a lot of it), which shows up in the $T^{00}$ component of the stress-energy tensor. Instead of explaining gravity trying to explain gravity as effect of mass, which is incorrect anyway, one should instead recognize that it's the energy that the gravitational charge in a way analogous to, say, electric charge.  
> **username_2**: Ok, now I understand what you mean. That's agreed.  

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