Mass of the observable Universe is known to be $1.50×10^{53}$ kg. Age is approximately known to be 13.7 billion years.The observable Universe is a sphere with diameter of roughly $8.8\times10^{26}$ m.

Mass, Length intervals and Time intervals are relative quantities.

My question is: Are the mass, diameter and age of the Universe frame dependent?


2 Answers 2


Spacetime geometry is, some details aside, very similar to ordinary geometry. Length and time are as relative in spacetime geometry as length is relative in ordinary geometry. And length isn't relative in ordinary geometry. A curve or line segment has only one length. If you have two points on a surface, there may be a most natural path between them (such as a straight line on a plane or a great-circle arc on a sphere) and you can measure the length of that curve. If there is no most natural path then the distance between the points still isn't relative; it's just ill-defined.

In cosmology, spacetime has a certain shape, and the "age of the universe" and "radius of the observable universe" are certain measurements of it, like the waist and inseam of a pair of jeans. They make sense as measurements because spacetime has that shape. These measurements are only defined "relative to the shape of spacetime", but the universe we live in objectively does have that shape. They aren't relative to an observer. If two tailors measure different values for the inseam then at least one of them is wrong.

The large-scale shape of the universe is a bit simpler than a pair of pants; it's somewhat more analogous to this:


The age of the universe is the outseam from the apex to the base. The diameter of the observable universe is part of the "waist", but the analogy breaks down here because the universe is larger than the observable universe. The point, though, is that some paths along this surface make more sense than others, and those are the paths used to define these quantities. You could spiral your measuring tape around the pant leg, but there's no reason to do it.

(The universe is not perfectly symmetrical, since there are local concentrations of mass like the Milky Way, and so these measurements are not precisely defined, but the same is true of the pair of jeans. That there is some fuzziness in the definition doesn't mean it can't be measured at all.)

The mass of the observable universe is the density of the universe times $\frac43 πR^3$, where $R$ is the radius of the observable universe. The density is related to spacetime geometry by the GR field equation, so you could say that this is just another measurement of the shape of spacetime. Another way to look at it is that it's the sum of the rest mass of every object in the universe, though that glosses over some subtleties. What it isn't is the relativistic mass relative to some arbitrarily moving local inertial frame, because that wouldn't make any more geometric sense than the spiral inseam.

  • $\begingroup$ 'where $R$ is the radius of the observable universe'. That implies a flat universe, right? $\endgroup$
    – JanG
    Nov 22, 2023 at 16:36
  • $\begingroup$ @JanG: in a closed universe the whole volume is V=2π²rκ³ where rκ=c/H/√[-Ωκ] is the radius of curvature, which is the distance from you to yourself divided by 2π. $\endgroup$
    – Yukterez
    Nov 22, 2023 at 16:41
  • $\begingroup$ @Yukterez. Yes, then I understand. I would prefer if relativists would use the name 'apparent radius' or something similar instead of the word 'radius' which is irrevocable associated with a flat geometry. $\endgroup$
    – JanG
    Nov 22, 2023 at 16:53
  • 1
    $\begingroup$ @JanG It's correct because the spatial slices are flat. I vaguely recall that there are subtleties when the spatial curvature is nonzero (you shouldn't just multiply density by volume), but that problem doesn't arise in the real world so I ignored it. $\endgroup$
    – benrg
    Nov 22, 2023 at 17:18
  • $\begingroup$ @benrg Thanks for clarification. $\endgroup$
    – JanG
    Nov 22, 2023 at 17:21

Yes, diameter and age are frame-dependent, due to length-contraction and time-dilation. To a "moving" observer, both are reduced, although the diameter depends on which direction you measure in. Not the mass, although this is just a question of terminology, as the modern definition of a particle's "mass" means its rest mass, which does not change with speed.

In more detail, take Friedmann-Lemaitre-Robertson-Walker spacetime as a model of a universe homogeneous and isotropic in space. Use hyperspherical coordinates, in which the metric is $ds^2 = -dt^2 + R(t)^2(d\chi^2 + S_k(\chi)^2d\Omega^2)$. Here $\chi$ is a radial coordinate, $R(t)$ is the scale factor, and the later terms are unimportant in the following. An observer comoving with the matter has 4-velocity $(1,0,0,0)$, hence coordinate time $t$ coincides with their proper time. Hence they say the universe is 13.7 Gyr old, using the scale factor values which cosmologists report. Now an observer moving radially (in our coordinates) with speed $\beta$ relative to the comoving observers has 4-velocity $u^\mu = (\gamma,\beta\gamma/R(t),0,0)$. Here $\gamma = (1-\beta^2)^{-1/2}$ is the Lorentz factor. The first component says $dt/d\tau = \gamma$, so coordinate time $t$ passes more quickly than this observer's proper time. If they maintain the same relative speed over the history of the universe, they measure a time of $13.7/\gamma$ Gyr. But note this trajectory requires acceleration, if $\beta$ and $R'(t)$ are nonzero.

As for length, consider the vector $\xi^\mu = (\beta\gamma,\gamma/R(t),0,0)$. This forms an orthonormal pair with the moving observers: $\langle\mathbf u,\boldsymbol\xi\rangle = 0$ and $\langle\boldsymbol\xi,\boldsymbol\xi\rangle = 1$, and intuitively is like a little ruler. Along this direction, the proper length satisfies $d\chi/ds = \xi^\chi = \gamma/R(t)$ locally, in contrast with the comoving observers which have $d\chi/ds = 1/R(t)$. Hence the radial direction is shortened by a factor of $\gamma$, relative to the comoving observers. If you strung a line of these moving observers across the universe, all moving in the same direction, and (theoretically) combine their local measurements, then the universe diameter would be shortened in this direction. But it is unchanged in any orthogonal direction.


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