First do we have anyway to even estimate the mass of the entire observable universe? And then is there any data that shows mass being gained or lost? Would we ever know if someone was playing with the til.

Also I want to be clear that I am not talking about small masses on the outskirts of the "universe" or small discrepancies in measurement or anything of the sort.

Note: I would like to add that maybe we should define the observable universe as NOW (x-date) so that we aren't calculating a moving target.

  • $\begingroup$ Mass is due to higgs boson and I think you are referring to matter and energy of any kinds. $\endgroup$
    – user6760
    Apr 4, 2015 at 6:48
  • $\begingroup$ conservation of mass and energy: lightandmatter.com/html_books/7cp/ch01/ch01.html It's not known whether the universe is a closed system. $\endgroup$ Apr 4, 2015 at 11:44
  • $\begingroup$ Are we able to measure the mass of the whole universe? That puts a wrinkle into the question. My guess would be no because there's no known method and no observation for the creation of new mass that I know of, unless dark energy has mass, in which case, the answer would probobly be yes. $\endgroup$
    – userLTK
    Apr 5, 2015 at 4:19
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    $\begingroup$ @user6760 That's not quite true. Only some mass comes from the Higgs mechanism, namely the mass of the $W^{\pm}$ and $Z$ bosons for the weak force. The proton, however, only gets 1% of its mass from its constituent quarks, and the quark masses are intrinsic, rather than derived from the Higg's mechanism. The rest comes from the quarks' kinetic energy and the strong force binding them. $\endgroup$ Jun 27, 2015 at 0:25
  • $\begingroup$ Somewhere out there is the very edge of the observable universe, and at that edge, there'll be hydrogen atoms completely oblivious to the fact that they're receding from us at near lightspeed. They're just drifting around with the usual assortment of local, random velocities. When two such atoms collide, one might be kicked up to above lightspeed relative to us and disappear from the observable universe. That'd lower the observable mass. It's just a little kinetic thing across a barrier, but we might even pick up a photon from collisional ionization in such an event. $\endgroup$ Jun 30, 2015 at 18:18

4 Answers 4


Yes, the mass of the observable Universe is always increasing.


Even if you're only referring the "ordinary" matter (such as stars, gas, and bicycles) and dark matter, the mass of the observable Universe does increase, not because mass is being created, but because the size of the observable Universe increases. In a billion years from now, we can see stuff that today is too far away for the light to have reached us, so its radius has increased. Since the mass $M$ equals density $\rho_\mathrm{M}$ times volume $V$, $M$ increases.

As called2voyage mentions, we have several ways of measuring the density, and we know it's close to $\rho_\mathrm{M}\simeq 2.7\times10^{-30}\,\mathrm{g}\,\mathrm{cm}^{-3}$ (Planck Collaboration et al. 2020). The radius is $R = 4.4\times10^{28}\,\mathrm{cm}$, so the mass is $$ M = \rho_\mathrm{M} \times V = \rho_\mathrm{M} \times \frac{4\pi}{3}R^3 \simeq 10^{57}\,\mathrm{g}, $$ or $5\times10^{23}M_\odot$ (Solar masses).

Mass increase of matter

Every second, the radius of the observable Universe increases by $dR = c\,dt = 300\,000\,\mathrm{km}$, in addition to the expansion. Here, $c$ is the speed of light, and $dt$ is the time interval that I choose to be 1 second. That means that its mass (currently) increases by $$ \begin{array}{rcl} dM & = & A \times dR \times \rho_\mathrm{M}\\ & = & 4\pi R^2 \times c\,dt \times \rho_\mathrm{M}\\ & \sim & 10^6\,M_\odot\,\text{per second,} \end{array} $$ where $A=4\pi R^2$ is the surface area of the Universe.

Dark energy

However, another factor contributes to the mass increase, namely the so-called dark energy, which is a form of energy attributed to empty space. And since new space is created as the Universe expands, dark energy is being created all the time. Currently, the energy density of dark energy, expressed as mass density through $E=mc^2$, is more than twice that of matter ($\rho_\Lambda \simeq 6\times10^{-30}\,\mathrm{g}\,\mathrm{cm}^{-3}$).

The rate at which the observable Universe grows due to expansion can be calculated from the Hubble law, which says that objects at a distance $d$ from us recedes at a velocity $$ v = H_0 \, d, $$ where $H_0\simeq 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$ is the Hubble constant. Expansion thus makes the edge of the observable Universe recede at $v=H_0 R = 3.2c$ (yes, more than three times the speed of light), in addition to the factor of $1c$ that comes from more light reaching us (as above).

Mass increase of dark energy

Hence, every second the "total" radius of the observable Universe (i.e. expansion + more light) increases by $dR = (3.2c + 1c)\times dt$, such that the increase in mass/energy from dark energy is $$ \begin{array}{rcl} dM & = & A \times dR \times \rho_\Lambda\\ & = & 4\pi R^2 \times (3.2c + 1c)dt \times \rho_\Lambda\\ & \sim & 10^7\,M_\odot\,\text{per second,} \end{array} $$ an order of magnitude more than that of regular/dark matter.

  • $\begingroup$ You used the radius of about 46 billion light years, the current estimate and the answer below, Dr. Jagadheep Pandian used 13.8 billion light years so you came up with answers that were off by about a factor of 30 using the same density. I'm curious which one is correct. My guess is that yours is, but I'm curious to have that verified. $\endgroup$
    – userLTK
    Jul 31, 2015 at 12:38
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    $\begingroup$ @userLTK: Yes, it seems Jagadheep has an error in his description. Although the Universe is 13.8 billion years (Gyr) old, we can see farther than 13.8 billion lightyears (Gly) away, because the Universe has expanded in the meantime. The exact result can be found by integrating (numerically) the Friedmann equation over time, and turns out to be roughly 46.5 Gly. Since incidentally, this result can also be expressed as roughly 14 billion parsec (Gpc), it might also be this that confused Jagadheep. $\endgroup$
    – pela
    Jul 31, 2015 at 14:00
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    $\begingroup$ What a wonderful and interesting answer, but I will never understand metric expansion or how the observable universe can expand by 3 or 4 times c. So I will try to chip away at my ignorance by asking If we watched extremely red-shifted galaxies near the edge of the observable universe for a very long time, how would they change? Would more appear? $\endgroup$
    – uhoh
    Jul 3, 2021 at 0:32

The mass of the observable universe can be derived from its density.

According to Dr. Jagadheep D. Pandian:

The density of matter in the universe can be measured by various means, which are too technical to go into at this point: people measure the density by studying the fluctuations in the Cosmic Microwave Background, superclusters, Big Bang nucleosynthesis, etc.

Using the density and the size of the observable universe, the mass can be derived to be 3 x 1055 g. This figure includes both dark matter and traditional matter.

I imagine that a historical deviation in mass might be detectable if it were significant, but I can't imagine what might cause a deviation that would be detectable on the scale we're talking about.


  • $\begingroup$ As pointed out by userLTK, the author of this source has used a value of the radius of the observable Universe which is approximately a factor of 3 too small. This could either be because he forgot to account for the fact that the Universe is expanding, or because he accidentally used lightyears instead of parsecs (see my comment to userLTK under my own answer). $\endgroup$
    – pela
    Jul 31, 2015 at 14:03
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    $\begingroup$ No, they just used the wrong value for the radius of the observable universe. $\endgroup$
    – ProfRob
    Jul 31, 2015 at 17:58

The apparent density of the universe is inconsistent with available observations. Because of the relatively slow speed of light, observations do not reflect actual distributions.

Galaxies are now being observed in positions and size billions of years ago. While our own Milky Way has a computational time dispersion over millions of years.

Add to this that galaxies are observably moving apart from each other at greater than the speed of light, means that there are galaxies moving away from us at speeds greater than the speed of light and are likely not detected because of their weakly dispersed signature.

The observable universe is likely skewed with sufficient variability in trustworthy evidence, that makes measuring universal density impractical, and untrustworthy.

  • $\begingroup$ The density of the Universe is not determined by counting visible galaxies. And obviously the finite speed of light is taken into account in all calculations. No inconsistencies here. $\endgroup$
    – pela
    Dec 28, 2016 at 23:33
  • $\begingroup$ If we observe galaxies moving apart from each other at greater than the speed of light, then we have galaxies moving apart from us at greater than the speed of light. Also, based upon the separation due to relativity (entropy) and the universe as space/time experiencing dissipation. Based on the speed of light there should be a limit to which we can see the edge of our known universe based on light. As galaxies in our very far field are all travelling faster than the speed of light away from us. Unless some at the far reaches are travelling towards us, which would mean a larger universe $\endgroup$
    – Toni
    Mar 15, 2017 at 16:34
  • $\begingroup$ If the Big Bang is like an explosion, AND Dark Matter acts like a rubber band. Then surrounding systems of galaxies were blown outward, until Dark Matter pulls outer galaxies back in. Which would indicate that potentially there was no Big Bang. That a much larger universe than what we perceive is in perpetual oscillation modes. NOT saying it is, but I personally do not know of a related contradiction. $\endgroup$
    – Toni
    Mar 15, 2017 at 16:42
  • $\begingroup$ math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed. html If gravity propagates at the speed of light, then again the density calculations based on gravimetrics is incredibly skewed. The question being. "What can be evaluated to determine an accurate distribution of mass? Including White Matter, Dark Matter, and other forms contributing to density that we as yet cannot detect?" $\endgroup$
    – Toni
    Mar 15, 2017 at 16:59
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    $\begingroup$ Sorry Toni, but you seem to have several misunderstandings about astronomy. Galaxies recede from each other at a rate proportional to their separation, so for sufficiently large distances, yes, they recede faster than the speed of light. That's no problem for the calculations we do. Galaxies were not "blown outwards" by Big Bang. That's not how Big Bang happened, and galaxies weren't around until several hundred million years later. Dark matter doesn't "act like a rubber band and pull galaxies back"; dark matter acts through gravity, just like ordinary matter, and the two are thoroughly mixed. $\endgroup$
    – pela
    Mar 15, 2017 at 21:14

There remains a question of semantics, which is simply, what is meant by the "observable universe." The fact is, different people are going to mean different things by that very phrase. Indeed, the Wiki on the "observable universe" contradicts itself in its very first paragraph, stating first that "The observable universe is a spherical region of the Universe comprising all matter that may be observed from Earth at the present time, because light and other signals from these objects have had time to reach Earth since the beginning of the cosmological expansion", but then four sentences later it changes its meaning, saying "Every location in the Universe has its own observable universe, which may or may not overlap with the one centered on Earth." So they seem to pick a particular universal age for their meaning, but not necessarily a particular vantage point. But note that in this meaning, there is no way to answer how it changes with time, as it exists only at a single time.

This meaning gives us various choices for how to extend the "observable universe" forward and backward in time. For example, we could take all the stuff in today's "observable universe" from Earth, and ask where that stuff will be in the future, and where it was in the past. Then we can use language like "when the observable universe was the size of a grapefruit," etc., but notice the ambiguity: when applied to the future, like "what will the observable universe be at such-and-such a time," we invariably imagine updating what could be seen by the beings of the day, but when applied to the past, we generally don't imagine there are any beings at all, so we don't update what their observable universe would be, we take our own and just shrink it.

So quite frankly, the term is really something of a mess, and so to answer your question, we would need to clarify which meaning you are taking. Let's assume you mean the "obserable universe" that is constantly updating what hypothetical beings could observe had they existed on Earth at the time, then we do have a time-dependent mass. As the Earth ages, there will be more time for light to reach us, so the observable universe will increase in size, but it will not necessarily increase in mass. Assuming the acceleration continues as expected, the observable universe mass will increase by something like a factor of 2, reach a maximum, and then begin to decrease. Its size will always be increasing with time, but its mass will then be decreasing.

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    $\begingroup$ The term "observable Universe" has a precise and well-understood meaning: it is the events that that are in the conformal causal past of a given comoving observer at a given time. $\endgroup$
    – John Davis
    Dec 28, 2016 at 19:36
  • $\begingroup$ I also forgot to add, providing the Universe is matter-dominated or dark energy dominated then the mass of the observable Universe will always increase. $\endgroup$
    – John Davis
    Dec 28, 2016 at 20:07
  • $\begingroup$ What I'm saying is that your meaning is not what you will find in most places. I'm not claiming that a precise meaning is impossible, I'm saying it's not what is used. $\endgroup$
    – Ken G
    Dec 28, 2016 at 20:13
  • $\begingroup$ Also, why does it need to be the "conformal" causal past-- is not the causal past already a well-defined concept? And should not a non-comoving observer have an observable universe as well? Finally, your claim does not sound correct-- I think it is widely accepted that a dark energy dominated universe has an observable universe mass that falls with time, that's certainly what the Wiki claims. $\endgroup$
    – Ken G
    Dec 28, 2016 at 20:17
  • $\begingroup$ On that last point, it could be I am confusing it with the event horizon, whereby galaxies whose evolution we could in principle monitor over cosmic time fall out of our sphere of influence, and we from theirs. So we could only see their evolution up to a point, but they would appear to slow to a standstill and be deeply redshifted, so be unobservable in practice, but the meaning of observable universe is only about what we could see in principle and if we could see them at any stage of their history. $\endgroup$
    – Ken G
    Dec 28, 2016 at 20:29

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