The cosmological principle states that:

The spatial distribution of matter and energy in the universe is homogeneous and isotropic when viewed in a large enough scale.

I find this notion problematic. Consider a box with a homogeneous and isotropic distribution of gas in rest. when a boost is introduced, the gas is contracted (Lorentz-Fitzgerald contraction) along the axis of movement, and the distribution of gas in the box is clearly not isotropic anymore (it is more dense along the axis of movement, meaning there is a preferred direction).

This suggests the cosmological principle is only true for the "rest frame" of the universe- but this implies there is a certain "preferred frame", relative to which an observer can always know how fast it is moving!

Moreover, this implies the Friedman-Lemaître-Robertson-Walker metric (that assumes the cosmological principle) is only correct for the "universal rest frame".

This goes against the invariance principle for inertial systems, that lies at the very heart of relativity.

How can these problems be settled?

  • $\begingroup$ There is a preferred frame, and preferred coordinates (up to spatial rotations). So what? GR says nothing against that. $\endgroup$
    – Ryan Unger
    Jun 24, 2017 at 21:10
  • $\begingroup$ @ocelo7 It seems to me that some kind of notion of a system that is universally preferred can be used (I think?) as "absolute" (to some extent) space and time, according to which all observers can synchronize (in Minkowsky, de-Sitter or anti-de-Sitter spaces), which sounds (perhaps wrongly) as if it should not exist according to special relativity. $\endgroup$
    – A. Ok
    Jun 24, 2017 at 21:21
  • $\begingroup$ Epistemologically, GR and SR are different beasts. In GR one has a fluid (consisting of galaxies and whatnot) whose flow lines can be used to define a cosmic time. The bigger issue with the cosmological principle is justifying that the universe is actually homogeneous and isotropic. It sure doesn't look so from our perspective... $\endgroup$
    – Ryan Unger
    Jun 24, 2017 at 21:26
  • $\begingroup$ With the boost, isn't it more dense everywhere in the box and so it's still a homogenous and isotropic distribution? $\endgroup$
    – user253751
    Jun 25, 2017 at 3:42
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    $\begingroup$ Related: physics.stackexchange.com/q/128198/2451 and links therein. $\endgroup$
    – Qmechanic
    Jun 25, 2017 at 6:47

2 Answers 2


Indeed, if spacetime is homogeneous and isotropic it means the spatial sections are maximally symmetric. Homogeneity means there are 3 Killing vector fields, for which you can pick the x, y and z coordinates. But if you choose other coordinates like radial and angular, it is still symmetric. In fact, spacetime is also isotropic which means rotationally invariant also.

Once you take those 6 symmetries there are not many choices in 3D (the spatial sections) for the geometry of those spaces. The choices are flat, spherical or hyperbolic. Whichever coordinate systems one picks in those spaces (and there are various good choices) the space is symmetric the same way. The rest is simply a time dependence which gives us the scale factor.

See Carrol's lecture notes at https://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll8.html

The coordinate system which takes the time orthogonal to the spatial slices everywhere is then a preferred frame. It is the comoving coordinate system.

Comoving means that if you are at rest in that coordinate system then you are moving along with the large scale flow of matter in the universe. It is in that frame that everything looks homogeneous and isotropic, in the large. In that frame the cosmic microwave background (CMB) looks isotropic.

Like, Carroll states, please note that spacetime is not static, there is a time dependence. Those comoving spatial slices expand and get further away from each other – a galaxy at a certain comoving coordinate with respect to another will get further away from each other because the scale factor a(t) is not constant. Their distance apart will grow proportionately to a(t).

So, yes, there is a preferred frame, it is the one where the symmetries are explicit. Without those symmetries, we would not know how to solve the cosmological equations.

Also note that galaxies and even us can be NOT at rest in the comoving frame. We are not moving exactly along with the flow of the universe. We and our solar system and galaxy etc have peculiar velocities which account for a somewhat higher or lower concentration of mass in our astrophysical near environment. Our peculiar velocity with respect to comoving is about 370 km/s (which includes the solar system around the galaxy, the galaxy and our local cluster – but look the number up, I am going by memory). Hundreds of km/s are not uncommon.

In fact, when we want to see the CMB and measure any anisotropies, we have to first subtract our peculiar velocities as it will have a directional preference.

  • $\begingroup$ Thank you! I still find it strange that we can say our velocity with respect to a universal rest frame (it feels too much like absolute space and time), but I keep getting this answer so the problem is probably with my intuition. $\endgroup$
    – A. Ok
    Jun 24, 2017 at 22:34
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    $\begingroup$ @A.Ok It's not so much a universal frame in which laws are different as a uniquely convenient frame in which to do calculations. To take a simple analogy, the way you'd describe nearby vehicles is based on the rest frame of the ground underneath you, not because it's special in the physics that results but because other frames would provide less insight. $\endgroup$
    – J.G.
    Jun 24, 2017 at 22:47

The notion of symmetries in general relativity is independant of the system of coordinates used, that is why Killing vectors are used. When we say that a spacetime is homogeneous, it means that it has three spacelike Killing vector fields, which will remain true no matter the coordinates used.

There will then of course be a "priviledged" observer for which the metric components will appear to not depend on any spatial coordinates.


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