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Because the CMB is everywhere and is isotropic, if an object would have a certain velocity, it could have a pressure differential produced by the CMB which would produce drag till it would stop with respect to the CMB.

However, wouldn't this mean that there is a 'universal' reference frame created by the CMB? Wouldn't this be going against special relativity assumptions?

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    $\begingroup$ The CMB as a special rest frame is discussed here: physics.stackexchange.com/q/25928 Interactions between a fast-moving object and the CMB are described here: physics.stackexchange.com/a/703204/27732 $\endgroup$
    – Andrew
    Commented May 7, 2022 at 23:04
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    $\begingroup$ physics.stackexchange.com/q/693284 $\endgroup$
    – ProfRob
    Commented May 8, 2022 at 7:32
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    $\begingroup$ Another note: before Einstein, physicists were looking for the absolute rest frame (aether) which would be the only frame in which the speed of light would equal $c$. Because they believed in absolute concepts of spatial distance and time, it would be impossible to observe the speed of light at $c$ in any frame but one. Obviously in our universe, any inertial frame will observe light to travel at $c$, and this is regardless of global features of the universe like the CMB. $\endgroup$
    – RC_23
    Commented May 8, 2022 at 19:58

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The CMB does in fact produce a preferential reference frame. Even without pressure, the preferential frame would be the one that equalizes the red and blue shift in all directions. For example Earth's motion around the Sun and the Sun's motion around the galaxy can be extracted from the red and blue shift in CMB data.

This does not contradict relativity, though, because the equations of physics are still valid and take the same form in any reference frame, even ones moving with respect to the CMB. Also, comparing your speed to the CMB requires looking far away, and relativity is based on the idea that your local neighborhood behaves the same regardless of your state of (inertial) motion.

In the same way, when analyzing motion on Earth, it usually makes sense to define our coordinate system so one axis aligns with gravity (i.e. "up and down"), because gravity is one of the main forces at work, and it reduces the number of sines and cosines needed in the equations. But that doesn't mean you could not obtain equally valid answers in a coordinate system with any orientation. And that "up and down" oriented coord system may not be the most natural choice in other situations, like a long way across the globe, or in the middle of space.

EDIT: In terms of "pressure" from the CMB, I think (?) you are referring to the radiation pressure $P$ of the distant source on a moving perfect reflector, as a function of velocity $v$, which Einstein in 1905 derived to be:

$$P=\frac{1}{4\pi}A^2\frac {(1-\frac{v}{c})}{(1+\frac{v}{c})}$$

where $A$ is the EM field amplitude. So this pressure does depend on velocity.

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  • $\begingroup$ I'm trying to get the SI units right. By $A^2$, do you mean $\epsilon_0E^2$, which has units of pressure? Because if you meant $A$ as in $E=-\nabla\phi-\dot{A}$, I'm not sure what factor you'd include of dimension $\mathsf{T}^{-2}$. $\endgroup$
    – J.G.
    Commented May 8, 2022 at 6:19
  • $\begingroup$ Good point. I tricked you by using the formula from Einstein's original paper (On the Electrodynamics of Moving Bodies, 1905, Sec 8), and he uses CGS or Gaussian units. $A$ is the $E$ or $B$ field, and in that system $\epsilon_0$ is dimensionless. So you are absolutely correct, in SI it would be $\epsilon_0E^2$ $\endgroup$
    – RC_23
    Commented May 8, 2022 at 6:54
  • $\begingroup$ Well, you didn't trick me, because I correctly inferred which field $A$ meant & the need for an $\epsilon_0$ factor. $\endgroup$
    – J.G.
    Commented May 8, 2022 at 7:27
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    $\begingroup$ Not sure what $P$ is but it isn't the net force per unit area on a moving body in the CMB, since it is non-zero when $v=0$., Perhaps it is the spatially averaged radiation pressure? $\endgroup$
    – ProfRob
    Commented May 8, 2022 at 9:27
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    $\begingroup$ You're correct that this does not correspond to the CMB which is in all directions. I just used it as an example to clarify if I understood what type of "pressure" the OP was referring to, e.g. not something associated with the intergalactic medium, etc. $\endgroup$
    – RC_23
    Commented May 8, 2022 at 17:55
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Indeed the existence of the CMB constitutes a reference of velocity.

The thing is: the CMB is not unique in that regard.

As a thought experiment: what is still possible in a Universe that is observationally the same as our Universe, but without CMB? With otherwise full access to any astronomical observations, what is stil possible?

The redshift observations of surrounding galaxies are a rich resource for data mining. Astronomers infer from the distributions of redshifts not only that the Universe is expanding, but also the velocity of our Galaxy with respect to surrounding galaxies. That is how astronomers have found that our Galaxy and the Andromeda galaxy have a velocity towards each other (and it can be projected at what point in the distant future there will be a merger process of the two galaxies.)

So: the overall distribution of galaxies in the Universe provides a reference of velocity of our galaxy relative to all of the other galaxies in the Universe together. This reference of velocity isn't directly accessible of course, the data mining process involves averaging over sufficiently large populations of galaxies, but it's doable.


Relativistic physics makes an assertion about what is still accessible to measurement when the measurement proces is local. The restriction to local measurement serves to highlight a particular property of the physics taking place.

For a spaceship floating in space there is no local measurement that finds the velocity vector of the spaceship with respect to some reference. (By contrast, when you are on a sailing ship in actual water you can at all times measure your velocity with respect to the water adjacent to the ship.)

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The other answers are already good, but I'll take a stab at an extremely short answer, that hopefully cuts at the heart of your confusion:

Yes, the CMB creates a preferred frame, but so does any other physical object, like the apple on my desk — there is a unique frame in which the apple on my desk isn't moving. The CMB is just another physical object (thing made of particles) in our universe.

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There's a distinction between phenomena and physical laws. Phenomena are what actually exists in the universe. Physical laws are what would happen given particular circumstances. Physical laws are the "If A, then B" rules, while phenomena are whether you actually have A or B.

For instance, if you have 100° C water at standard pressure that is boiling, then "you have water", "it's at 100° C", "it's at standard pressure", "it's boiling" are all phenomena. The general pattern of "When you have water at standard pressure, and you raise its temperature to 100° C, it boils" is a reflection of physical laws. If you go to another planet with no water, or where the temperature is always significantly above or significantly below 100° C, then you won't see water boiling, but that doesn't mean the physical laws are any different. "If A, then B" is true even if A and B don't happen to be present.

Relativity just says that you have the same physical laws in every inertial reference frame. It doesn't say that you have the same phenomena. The physical laws by themselves don't make any reference frame special, but the universe can. Relativity doesn't prohibit distinguished reference frames, as long as they're being distinguished by their interactions with what's in the universe, rather than the physical laws of the universe. For instance, a reference frame at rest with respect to the Earth can be distinguished from a reference frame traveling at 100 mph towards the Earth in that people at rest with respect to the latter tend to die rather quickly. This is due to the physical object of the Earth, rather than the physical laws being different in the reference frames.

CMB is a phenomenon, and this phenomenon appears different in different reference frames. The physical laws that govern how it interacts with matter, however, are the same in all inertial reference frames.

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  • $\begingroup$ This answer reminds me of an article by George Ellis on philosophy of cosmology. He used the term "broken symmetry", borrowed from particle physics, to make a similar point $\endgroup$ Commented May 16, 2022 at 2:05
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One thing to note is that the CMB looks different from different locations (and at different times).

So while it provides a local frame of reference for every point in space-time, that frame is not constant over larger amounts of space, and it's likely that observers elsewhere (millions of light years away) observe it differently.

So things are still relative.

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