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.


3 Answers 3


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 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), 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.


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". 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) 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.

  • $\begingroup$ 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? $\endgroup$
    – frodeborli
    Feb 4, 2016 at 15:24
  • $\begingroup$ @frodeborli: If you look at a map of the CMB, such as this one, you'll see that it's isotropic to one part in ~1e5. Note that on a map like this, two important anisotropies 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. $\endgroup$
    – pela
    Feb 4, 2016 at 15:35
  • $\begingroup$ 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. $\endgroup$
    – frodeborli
    Feb 4, 2016 at 17:20
  • $\begingroup$ 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. $\endgroup$
    – kubanczyk
    Feb 4, 2016 at 22:12
  • $\begingroup$ @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. $\endgroup$
    – frodeborli
    Feb 4, 2016 at 23:01

Light causes gravity while travelling, a clear yes, by Einstein's famous mass-energy equvalence. (Compare this discussion on StackExchange.)

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.

The effect of light on asteroids is not negligible, but it's not the gravitational pull. It' mainly the YORP effect. Dust is also affected by light.

  • $\begingroup$ 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? $\endgroup$
    – frodeborli
    Jan 7, 2014 at 14:47
  • 1
    $\begingroup$ 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 (en.wikipedia.org/wiki/Planck%27s_law). Take a look at the Stefan-Boltzmann law (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. $\endgroup$
    – Gerald
    Jan 7, 2014 at 15:12
  • $\begingroup$ (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. $\endgroup$
    – Gerald
    Jan 7, 2014 at 15:57
  • $\begingroup$ @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. $\endgroup$ Jan 7, 2014 at 17:33
  • 1
    $\begingroup$ 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. $\endgroup$
    – Stan Liou
    Jan 11, 2014 at 4:41

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