It seems that we are moving relative to the universe at the speed of ~ 600 km/s. This is the speed of our galaxy relative to the cosmic microwave background.

Where does this rest frame come from? Is this special in any way (i.e., an absolute frame)?

EDIT: I think the more important question is "where does the CMB rest frame come from?".


5 Answers 5


I found this answer at Professor Douglas Scott's FAQ page. He researches CMB and cosmology at the University of British Columbia.

How come we can tell what motion we have with respect to the CMB? Doesn't this mean there's an absolute frame of reference?

The theory of special relativity is based on the principle that there are no preferred reference frames. In other words, the whole of Einstein's theory rests on the assumption that physics works the same irrespective of what speed and direction you have. So the fact that there is a frame of reference in which there is no motion through the CMB would appear to violate special relativity!

However, the crucial assumption of Einstein's theory is not that there are no special frames, but that there are no special frames where the laws of physics are different. There clearly is a frame where the CMB is at rest, and so this is, in some sense, the rest frame of the Universe. But for doing any physics experiment, any other frame is as good as this one. So the only difference is that in the CMB rest frame you measure no velocity with respect to the CMB photons, but that does not imply any fundamental difference in the laws of physics.

“Where does it come from?” is also answered:

Where did the photons actually come from?

A very good question. We believe that the very early Universe was very hot and dense. At an early enough time it was so hot, ie there was so much energy around, that pairs of particles and anti-particles were continually being created and annihilated again. This annihilation makes pure energy, which means particles of light - photons. As the Universe expanded and the temperature fell the particles and anti-particles (quarks and the like) annihilated each other for the last time, and the energies were low enough that they couldn't be recreated again. For some reason (that still isn't well understood) the early Universe had about one part in a billion more particles than anti-particles. So when all the anti-particles had annihilated all the particles, that left about a billion photons for every particle of matter. And that's the way the Universe is today!

So the photons that we observe in the cosmic microwave background were created in the first minute or so of the history of the Universe. Subsequently they cooled along with the expansion of the Universe, and eventually they can be observed today with a temperature of about 2.73 Kelvin.


@starwed points out in the comments that there may be some confusion as to whether someone in the rest frame is stationary with respect to the photons in the rest frame. I found a couple more questions on Professor Scott's excellent email FAQ page to clarify the concept.

In your answer to the "How come we can tell what motion we have with respect to the CMB?" question, there is one more point that could be mentioned. In an expanding universe, two distant objects that are each at rest with respect to the CMB will typically be in motion relative to each other, right?

The expansion of the Universe is certainly an inconvenience when it comes to thinking of simple pictures of how things work cosmologically! Normally we get around this by imagining a set of observers who are all expanding from each other uniformly, i.e. they have no "peculiar motions", only the "Hubble expansion" (which is directly related to their distance apart). These observers then define an expanding reference frame. There are many different such frames, all moving with some constant speed relative to each other. But one of them can be picked out explicitly as the one with no CMB dipole pattern on the sky. And that's the absolute (expanding) rest frame!

Assumptions: From most points in the universe, one will measure a CMBR dipole. Thus, one would have to accelerate to attain a frame of reference "at rest" relative to the CMBR. Questions: Having attained that "rest frame", would one not have to accelerate constantly to stay at rest (to counter attraction of all the mass scattered around the universe)? [abridged]

I think the assumption is wrong, and therefore the question doesn't need to be asked.

The fact that there's a CMB dipole (one side of the sky hotter and the other side colder than the average) tells us that we are moving at a certain speed in a certain direction with respect to the "preferred" reference frame (i.e. the one in which there is no observed dipole). To get ourselves into this dipole-free frame we just have to move with a velocity which cancels out the dipole-producing velocity. There's no need to accelerate (accept the rapid acceleration you'd need to do to change velocity of course).

Our local motion (which makes us move relative to the "CMB frame" and hence gives us a dipole to observe) is caused by nearby clusters and superclusters of galaxies pulling us around. It's true that over cosmological timescales these objects are also moving. And so if we wanted to keep ourselves always in the dipole-free frame we'd have to make small adjustments to our velocity as we moved and got pulled around by different objects. But these changes would be on roughly billion year timescales. And so to get into the frame with no CMB dipole basically just requires the following 3 steps: (1) observe today's dipole; (2) move towards the coldest direction at just the right speed to cancel the dipole; and (3) maintain basically that same velocity forever.

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    $\begingroup$ Actually, the answer is to "where the photons come from", and not "where does the rest frame come from"? $\endgroup$
    – Jus12
    Commented Nov 16, 2011 at 18:20
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    $\begingroup$ By definition, the rest frame is the photons. That's the microwave part of the Cosmic Microwave Background rest frame. $\endgroup$
    – ghoppe
    Commented Nov 16, 2011 at 20:12
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    $\begingroup$ Err, that's a bit misleading. The bit above makes it seem that, in the rest frame, the photons are stationary. But that's not true, since they are not all moving in the same direction. The key is that in the natural rest frame, the CMB will be isotropic -- it'll look the same in all directions. But once you move relative to that frame, there's a preferred direction. $\endgroup$
    – starwed
    Commented Nov 5, 2012 at 14:38
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    $\begingroup$ @ghoppe I think what Jus was trying to ask is why is there a frame where the average velocity of those photons is 0? Why wouldn't it be like the Cauchy distribution, which has no expected value, and whose sample averages are the same as the initial distribution? $\endgroup$
    – user76284
    Commented Dec 29, 2018 at 3:34
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    $\begingroup$ @user76284 the Cauchy distribution can only lack an expected value because it's supported on the whole real line, which is unbounded (goes off to infinity). The velocities of photons, essentially, live on a sphere which is bounded, and so their distribution must have an average. $\endgroup$ Commented May 5, 2022 at 18:21

The best theory we have, at present, for addressing this question, is Einstein's General Theory of Relativity. His earlier Special Theory of Relativity postulated the equivalence of all inertial (i.e., non-accelerating) reference frames for formulating the basic laws of nature but ignored gravitational phenomena. the General Theory incorporated gravity, treating it as a manifestation of the curvature of space-time geometry, and used tensor calculus to formulate the laws of nature in a form that was invariant under any change of reference frame or coordinate system whatsoever. It was also the first theory that could model the dynamical evolution of the universe as a whole. The subsequent discovery of both the Big Bang expanding evolution of the universe and the cosmic microwave background radiation (CMB), then provided data which could be used to define a coordinate system relative to which the largest scale features of the universe could be most simply described. This is a so-called co-moving coordinate system, in which the spatial coordinates expand with the separating galactic clusters and the CMB has the most spherically symmetric distribution of intensity with frequency. This coordinate system has been referred to as defining the "rest frame of the universe" and our motion relative to it can be measured by the dipole moment distribution we observe superimposed on the otherwise spherically symmetric CMB. But while the expansion and the CMB are phenomena that look simpler in the co-moving coordinates, the laws that govern the expansion and the CMB, when expressed in tensor form, have the same form in any coordinates. So the co-moving coordinate system, the "rest frame of the universe", is (presently) only special, or "preferred" for the description of the largest scale features of the universe.

  • $\begingroup$ But aren't the largest scales precisely what we're really interested in ? $\endgroup$
    – cumfy
    Commented Apr 25, 2020 at 11:42
  • $\begingroup$ It's dubious that 1915's GR is the best theory for analyzing the situation: 1929's Einstein-Cartan Theory provides an inflationary mechanism, not requiring the hypothetical scalar field characterizing most inflationary models, for explaining spatial & energetic expansion. It's described by Nikodem Poplawski in numerous 2010-2021 papers, whose preprints can be found by his name on the Arxiv website. There are two big problems with his model: It's potentially past- (as well as future-) eternal, which leaves it controversial for many theologians, & it's (reportedly) more complex than GR. $\endgroup$
    – Edouard
    Commented Sep 27, 2021 at 5:22

The cosmic background radiation is a very redshifted image of the last scattering surface. The last scattering surface isn't a single object, it's how the universe looked far enough back in time that it was made entirely of hot opaque plasma. A few moments later, the temperature cooled enough that particles could become electrically neutral and light wasn't immediately absorbed by the next particle over. The universe became transparent and the light you see from the last scattering surface is just now reaching you.

The plasma that comprised the last scattering surface was very hot, and its individual particles had quite a lot of motion in them. The rest frame of the cosmic background radiation is an average of these particles, not some special reference frame unique to physics.

It's also important to note that it isn't just "a ball out there somewhere", but a thing that happened everywhere, including where you're standing! But for where you're standing, that was billions of years ago, and you've moved since then and become a human standing on a planet instead of a plasma. When you look billions of years away, you see the universe as it was billions of years ago. You see the plasma that hasn't had the chance to move yet.

TL;DR: No, it isn't special. It comes from us having moved around since then.


The photons that forms the CMB come from what is called the last scattering surface. This surface is a different one depending on the position in spacetime of the observer. So each observer will have a different reference frame in which the CMB has no dipole term.

  • $\begingroup$ I can agree that the age and relative location of the LSS is dependent on spacetime coordinates of the observer, but it doesn't necessarily follow that this means different observers in their own comoving frame would see the others as having a peculiar velocity. Can you cite references for that or provide further reasoning to back up that claim? $\endgroup$
    – Jim
    Commented Apr 20, 2017 at 14:31
  • $\begingroup$ @Jim The last scattering surface comprises whatever gas molecules were last to reionize along the line of sight. Different observers will see different clouds in the distance undergoing reionization. $\endgroup$ Commented Apr 20, 2017 at 14:44
  • $\begingroup$ (… eh, reionization doesn't quite work like that, but I think the general idea is right: which particular opaque objects comprise the CMB depends on the observer's location.) $\endgroup$ Commented Apr 20, 2017 at 14:51
  • $\begingroup$ I don't disagree with that. But if the different clouds were all about the same peculiar velocity, then observers now should be able to agree on the comoving frame. My curiosity lies in how you concluded that different clouds means different comoving frames $\endgroup$
    – Jim
    Commented Apr 20, 2017 at 15:10
  • $\begingroup$ Jim, I think that to have one unique reference frame and therefore a definition of a cosmic time one need further assumptions. You can read about it at " A first course in general relativity" by Bernard F. Schultz page 320, 321. But these assumptions are made to make easy calculations and one should keep in mind that there are no privilege frame in the Universe as one would expect from general relativity theory. $\endgroup$ Commented Apr 24, 2017 at 6:05

Since decoupling happened when the Universe was only about one thousandth of its present size, and the photons have been travelling uninterrupted since then, they come from a considerable distance away. Indeed, a distance close to the size of the observable Universe. Those we see originate on the surface of a very large sphere centred on our location, shown in Figure 10.2, called the surface of last scattering. Its radius is of order 6000 $h^{- 1}$ Mpc (see Problem to.5). Of course, there is nothing special about this particular surface, except that it happens to be at just the right distance that the photons have reached us by today. There are photons originating at every point, and observers in different parts of the Universe (if there are any!) will see photons originating from different large spheres. of the same radius, centred on their location.

from Introduction to Modern Cosmology by Andrew Liddle. Page 80

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    $\begingroup$ This is a direct copy-paste of Liddle's Introduction to Modern Cosmology. We don't usually mind using references, but you must indicate that it's a quote (you can use the > before the text) and from what source! Otherwise, it's basically plagiarism. $\endgroup$
    – Kyle Kanos
    Commented Apr 21, 2017 at 10:02
  • $\begingroup$ I told just before the reference, the author and the page. $\endgroup$
    – user153604
    Commented Apr 21, 2017 at 10:18
  • $\begingroup$ I just want to understand things. Don't worry I'm not going to argue anymore. $\endgroup$
    – user153604
    Commented Apr 21, 2017 at 10:26

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