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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.

It is significant precisely because it is the simplest possible object that can't be set exactly equal to Minkowski space at a point by a choice of coordinates (unlike the metric tensor and the Christoffel symbols), so your paradox isn't a paradox at all.

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I read an article (in Russian) that said that the gravitational field of an infinite homogeneous flat layer - is homogeneous, and the components of the curvature of the field is not equal to zero. This solution, as it says, with some change of coordinates, reduces to the solution of Taub –  Sergio Jan 20 '13 at 20:57
    
@Sergio: Homogenous source does not necessarily imply the same metric as uniform acceleration in flat space. –  Jerry Schirmer Jan 20 '13 at 21:06
    
But what about special case of homogenous source - infinite homogeneous flat layer? In the Newtonian case, the field of such a source is homogeneous. It looks like it will be the same in GR. –  Sergio Jan 20 '13 at 21:11
    
@Sergio: a homogenous field is not a zero field. And it is not equilvalent to uniform acceleration. Curvature can change topology, amongst other things. –  Jerry Schirmer Jan 20 '13 at 21:13
    
What do You mean by the term "zero field"? Do you know any solution in vacuum for the case of infinite uniform massive layer? –  Sergio Jan 20 '13 at 21:21
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