# Does isotropy imply homogeneity?

This question comes from exercise 27.1 in Gravitation by Misner, Thorne and Wheeler. They required the following:

Use elementary thought experiments to show that isotropy of the universe implies homogeneity.

I know homogeneity as the universe is the same everywhere at a given time, and isotropy is related to direction.

I wonder how the isotropy of the universe implies homogeneity.

• Why would it???? I can put you at the center of a big white ball and the world will look perfectly isotropic, but it will be far from homogeneous. – CuriousOne Jun 27 '15 at 4:48
• As John Rennie points out quite rightly, one needs a stronger condition than just isotropy for one observer to make a non-trivial question out of it. As an experimentalist who doesn't have access to anything but one location that immediately jumps at me as a non-trivial difference between theory and experiment. :-) – CuriousOne Jun 27 '15 at 5:52
• Possible duplicate: physics.stackexchange.com/q/24881/2451 , physics.stackexchange.com/q/47234/2451 and links therein. – Qmechanic Jun 27 '15 at 7:25

When MTW say the universe is isotropic, they mean it is isotropic everywhere i.e. at all points in the universe.

It's easy to construct universes that are isotropic at a single point and not homogeneous, for example CuriousOne's suggestion of a ball with density that is a function of distance from the centre. However this ball is only isotropic if you are at the centre of the ball. If you require the ball to be isotropic everywhere you necessarily require it to be homogeneous.

MTW actually give you the answer (in a technical form) to exercise 27.1 in the paragraph just above the exercise next to the side note:

Isotropy implies fluid world lines orthogonal to homogeneous hypersurfaces

This is closely related to the fact that in a Euclidean space, coordinate translations can be generated by performing two successive rotations around different points, as isotropy is essentially rotation invariance and homogeneity translation invariance. Suppose we have a rotation $$R(\vec{r}_0)$$ respect to $$\vec{r}_0$$ defined through the action on any point $$\vec{r}$$ as

$$R(\vec{r}_0)\vec{r}=\vec{r}_0+R(\vec{r}-\vec{r}_0)$$

Then clearly $$R(\vec{r}_0)\vec{r}_0=\vec{r}_0$$. For simplicity, we simply use $$R$$ to denote a rotation respect to the origin. Then for any point $$\vec{r}$$, two successive rotations around origin and $$\vec{r}_0$$ respectively would give

$$R^{-1}(\vec{r}_0)R\vec{r}=\vec{r}_0+R^{-1}(R\vec{r}-\vec{r}_0)=\vec{r}+(I-R^{-1})\vec{r}_0$$

Then for any translation $$\vec{a}$$, we can choose the coordinate system such that $$\vec{a}=(a,0,0)$$, then set

$$\displaystyle R^{-1}=\left(\begin{array}0 &-1&\\1&&\\&&1\\\end{array}\right)$$

and $$\vec{r}=(a/2,a/2,0)$$, we get

$$\vec{r}+\vec{a}=\vec{r}+(I-R^{-1})\vec{r}_0=\vec{r}_0+R^{-1}(R\vec{r}-\vec{r}_0)=R^{-1}(\vec{r}_0)R\vec{r}$$

Then if the space is invariant under rotations with respect to any point, it will be invariant under translation. In curved spacetimes, instead of global rotations, we need to consider Killing vectors. And similarly, existence of Killing vectors for isotropy at every point implies the existence of Killing vectors for homogenity. For details, see Chapter 13 of Weinberg's extraordinary book, Gravitation and Cosmology.

A few years late here, but I think a clear way of thinking about this is any two points in the universe, A and B, will be connected by a great circle drawn around C. If the universe is isotropic at point C then the points A and B must look the same. This logic can then be extended to any two points in the universe.

This logic clearly relies on that which was pointed out by John Rennie, that the universe must be isotropic everywhere.

• any two points in the universe, A and B, will be connected by a great circle drawn around C There are multiple things that don't make sense to me about this. "Great circle" is terminology used for spherical geometry, but the spatial topology of our universe probably isn't spherical. Or are you talking about the celestial sphere as seen from C? Also, it's not clear to me whether you're imagining C to be some special point (if so, that?) or whether you're missing a quantifier: ... for every point C? there exists a point C? Is C the location of an observer? – user4552 Aug 21 '19 at 13:33
• I wasn't implying any special geometry, just a circle connecting the points. And the way I think about it fits best with "There exists a point C", I wasn't implying any special point in the universe. Thank you for your corrections. – Tom Kite Aug 22 '19 at 14:15