# Homogeneous gravitational field and the geodesic deviation

In General Relativity (GR), we have the geodesic deviation equation (GDE)

$$\tag{1}\frac{D^2\xi^{\alpha}}{d\tau^2}=R^{\alpha}_{\beta\gamma\delta}\frac{dx^{\beta}}{d\tau}\xi^{\gamma}\frac{dx^{\delta}}{d\tau},$$

see e.g. Wikipedia or MTW.

Visually, this deviation can be imagined, to observe the motion of two test particles in the presence of a spherically symmetric mass.In the case of a homogeneous gravitational field, such as a geodesic deviation should not be, because acceleration due to gravity equal at every point.

Clearly, such a field can be realized in a uniformly accelerated frame of reference. In this case, all components of the curvature tensor will be zero, and the equation (1) correctly states that the deviation will not be surviving.

But if a homogeneous field will be created by infinite homogeneous flat layer, in this case, the components of the curvature tensor are non-zero, then by (1) will be a deviation. It turns out that such fields, even though they are homogeneous, can be discerned. I think this situation is paradoxical. Is there an explanation?

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The curvature tensor is the same in all coordinate systems. If you have, say, the Schwarzchild metric, $R_{abc}^{d}$ will transform with the coordinate system and be nonzero in all reference frames. In particular $R_{abcd}R^{abcd}$ will take on the same value at every point no matter what reference frame you choose, accelerated, freely falling, or no.