Killing vectors in Robertson-Walker mertric

[paweld]

Let's consider the following metric:
$$g=dt^2 - a^2(t)\frac{dx^2+dy^2+dz^2}{1+\frac{x^2+y^2+z^2}{4}}$$
It can be also express in different coordinates as:
$$g=dt^2 - a^2(t)\left( \frac{dr^2}{1-r^2} + r^2(\sin^2(\theta) d\varphi^2+\theta^2) \right)$$

Of course it admits killing vetctors which are generators of rotations:
$$L_z=x\partial_y - y\partial_x, \ldots$$
Can anyone found different killing vector?
This metric describes spatialy homogenous universe so it should have
translation symmetry but it's not apparent in this coordinates.
Why it's hidden?
Thanks for any replies.

 

[George Jones]

Sorry for picking this up so late. Did you mean to omit $k$, i.e., did you mean to consider only the $k = 1$ case? If so, do you know do which Riemannian manifold models each spatial section, and can you guess what the spatial isometry group is?

It might be helpful to consider first the $k=0$ case.

 

[Entropia]

The Robertson-Walker metric is a solution to Einstein's field equations that describes a homogeneous and isotropic universe. As such, it should have translation symmetry, meaning that the metric should remain unchanged under translations in space. However, in the given coordinates, it is not immediately apparent that there is translation symmetry.

To understand why this symmetry is hidden, we need to look at the Killing vectors in this metric. Killing vectors are vector fields that preserve the metric, meaning that they leave the metric unchanged when acted upon. In other words, they represent symmetries of the metric.

In the given metric, we can see that there are three Killing vectors that correspond to rotations in the x, y, and z directions. These are represented by the generators L_x, L_y, and L_z, respectively. However, these are not the only Killing vectors in this metric.

Another Killing vector that is often overlooked is the generator of time translations, which is given by:

L_t=\partial_t

This vector represents the symmetry of the metric under translations in time. In other words, it tells us that the metric remains unchanged at different points in time.

So why is this symmetry hidden? It is because of the choice of coordinates. In the given coordinates, the metric appears to have a time-dependent term, a(t), multiplying the spatial terms. However, if we transform to a different set of coordinates, such as comoving coordinates, where the spatial coordinates are scaled by the expansion factor a(t), we can see that the metric becomes time-independent:

g=dt^2 - dr^2 - r^2(\sin^2(\theta) d\varphi^2+\theta^2)

In these coordinates, it is clear that the metric has translation symmetry, with the time coordinate remaining unchanged under translations.

In summary, the Robertson-Walker metric does have translation symmetry, but it is hidden in the given coordinates due to the choice of coordinates. The existence of the time translation Killing vector shows that the metric remains unchanged at different points in time, and thus has translation symmetry.