I know that galaxies are moving away from us, and so can see that it's intuitive that if space was expanding, then the astronomical observations from Earth would be the same as at all other points in the universe.

But it's that intuition often overlooks details. So, what I would like to know, is whether there is proof that smoothly expanding space in 4 dimensions, accommodates every observer point in this way.

Does anyone know?

  • $\begingroup$ 4 dimensions? What are you assuming is the 4th dimension? $\endgroup$
    – Sponge Bob
    Aug 17, 2015 at 21:45
  • $\begingroup$ Hi Lucy, it's not really intuition that counts, as I have found out by being wrong lots of times, it's the math and the observations. We don't obviously enough, I suppose, have absolute proof that an observer at the edge of our observable universe, sees the universe the way we do, but we would normally work on the principle that he/she/it does. Look up Copernican Principle on Wikipedia. $\endgroup$
    – user81619
    Aug 17, 2015 at 21:55
  • $\begingroup$ Related: physics.stackexchange.com/q/25591/2451 and links therein. $\endgroup$
    – Qmechanic
    Aug 18, 2015 at 9:35

2 Answers 2


You are approaching the question from the wrong end.

The expansion of the universe is described by a particular solution to Einstein's equation called the FLRW metric. To derive this metric we have to make some assumptions, and the key assumptions are that the universe is isotropic and homogeneous i.e. that it is the same everywhere.

So the universe being the same everywhere is an initial assumption that goes into describing how the universe expands. It isn't something that is derived from the way the universe expands. You can of course show that the FLRW metric implies isotropy and homogeneity (as Timaeus does) but that's because those assumptions were built into it from the beginning.

  • $\begingroup$ I thought the question was to check rigorously whether expanding space succeeds at the goal of making a homogeneous isotropic spacetime. And the metric doesn't imply that so I'll go edit my answer. $\endgroup$
    – Timaeus
    Aug 18, 2015 at 14:07
  • $\begingroup$ @Timaeus: your answer is quite correct. The metric is independant of $x$, $y$ and $z$, so it implies homogeneity. My point is that homogeneity was an assumption made when deriving the FLRW metric in the first place, so of course it implies homogeneity. $\endgroup$ Aug 18, 2015 at 14:15
  • $\begingroup$ I thought I was a bit vague about the coordinates being global and all of R4 in particular becsuse if other questions I answered for the OP. Which made me realize I wasn't clear in arguing that it was isotropic and homogeneous. Now I have shown that. Which is what I think they wanted, whether the models are known to be homogeneous and isotropic rather than something that intuitively seems like it and so is our best attempt or our only hope. $\endgroup$
    – Timaeus
    Aug 18, 2015 at 14:40
  • $\begingroup$ Also you can derive the metric, but the metric doesn't make the space isotropic. I'm still not sure the OP asked about isotropy but in case they did want that the local form of metric doesn't give you global isotropy. So you are claiming I showed the FLRW metric implies isotropy, and I give a counter example. The metric alone does not imply isotropy. $\endgroup$
    – Timaeus
    Aug 18, 2015 at 14:43
  • $\begingroup$ Hi thanks John Rennie and Timaeus. I didn't actually know what you just said so that's appreciated. But, surely for robustness you'd still want to see if you get the result from another direction? I was just thinking that due to the geometric aspects, it would be easy to demonstrate that it does happen? $\endgroup$ Aug 19, 2015 at 16:28


You can make a model where you have coordinates $t, x,y,z$ where for any $x,y,z$ the universe looks the same.

The metric ends up looking e.g. like $$ds^2=dt^2-(a(t))^2(dx^2+dy^2+dz^2)$$ and you can move your $x,y,z$ to have any value and everything looks the same (though things do look different for different values of $t$). You end up with the densities of matter and radiation being solely functions of $t$ (which is just a a coordinate it doesn't say how fast clocks tick, that will depend on the path the clocks takes and depends on the full metric above). And then you also need to solve for $a(t)$ as a function of time.

The function $a(t)$ relates coordinate distances to actual metric distances. Being a certain coordinate distance away could end up becoming much closer or much farther away based on how the scale factor $a(t)$ changes.

Now having this metric does mean everything looks the same at every point in space. But it doesn't mean your space looks the same in every direction. This is because there are still many spacetime's that have that metric.

For instance you can consider the space $$\{(t,a,A,y,z):a^2+A^2=1\}.$$ Then $x$ can go around like an angle in the $a,A$ plane. In this space we have the same metric as $\{(t,x,y,z)\}$ so locally everything looks the same but in one of them when we travel around $2\pi$ in the $x$ direction we end up back where we started and that doesn't happen if we move in the $y$ direction or the $z$ direction.

So if you want the universe to look the same at every place and in all directions you have to be more careful than just having one of the local metrics that works for that. But that does show that it is hard to check.

One way to check is to start with $\{(t,x,y,z)\}$ and $d\tau^2=dt^2-(a(t))^2(dx^2+dy^2+dz^2)$ and then do a transformation $x\mapsto X=x+\Delta x,$ $y\mapsto Y= y+\Delta y$ and $z\mapsto Z=z+\Delta z,$ for fixed $\Delta x,$ $\Delta y,$ and $\Delta z.$ Then $\{(t,x,y,z)\}=\mathbb R^4=\{(t,X,Y,Z)\}$ and $dt^2-(a(t))^2(dX^2+dY^2+dZ^2)=dt^2-(a(t))^2(dx^2+dy^2+dz^2)= d\tau^2$ so the translation changed nothing. The objects and functions are the same, just the names and labels which were arbitrary. This truly shows it is the same at every point.

Then do another transformation, this time rotating $(x,y,z)$ to $(x',y',z')$ by a fixed amount and direction and noting that $\{(t,x,y,z)\}=\mathbb R^4=\{(t,x',y',z')\}$ and $d\tau^2=dt^2-(a(t))^2(dx^2+dy^2+dz^2)=dt^2-(a(t))^2(dx'^2+dy'^2+dz'^2)\}.$ This shows it looks the same in all directions.

To be clear, this cannot be done for the non isotropic case I mentioned before: $\{(t,a,A,y,z):a^2+A^2=1\}$ with $d\tau^2=dt^2-(a(t))^2(da^2+dA^2+dy^2+dz^2).$ that is because the rotation doesn't map that set $\{(t,a,A,y,z):a^2+A^2=1\}$ to $\{(t,x,y,z)\}.$ Showing that they are different was established when going around $2\pi$ in one takes you back and that doesn't happen for the other. I was just trying to make it clear that we really have shown that the space looks the same in all directions when we have $\{(t,x,y,z)\}$ and $d\tau^2=dt^2-(a(t))^2(dx^2+dy^2+dz^2).$

  • $\begingroup$ This is brilliant. I'll have to study it to understand. Which I'll do so thanks. $\endgroup$ Aug 19, 2015 at 16:31
  • $\begingroup$ hi - I'm just feeding back on these answers. What I would like to know, is whether it works in the most basic geometrical sense. For example, it can be shown for a region of points with a square ordering (i.e. the native points on a piece of graph paper). $\endgroup$ Mar 20, 2016 at 3:36
  • $\begingroup$ @LucyMeadow I'm not certain how you want to define expanding space in a situation that isn't homogeneous (which is the case usually considered and where the definition is easier). Most of my answer is just saying that you can't determine whether your spacetime is isotropic or non isotropic just from looking at the form of the metric. Part of me wants to delete my answer because my opinion is that the answer to your original question is actually no (though it could depend on how you define expansion). $\endgroup$
    – Timaeus
    Mar 20, 2016 at 4:09

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