but I don't believe we have such in our universe....Things like planets/stars etc have both.....a uniform gravitational field is what Einstein imagined for his "equivalence" principle...say from an infinite plane of mass...but mass in our unviverse is like point masses, hence tidal effects.

In case you are interested, a specific cases of this point is the metric of a uniform spherical thin shell: http://arxiv.org/abs/gr-qc/0008065 (Eqn 1 only, rest of article not relevant).

A good rule of thumb might be this (please someone bright verify this):

If the have a metric (like the http://en.wikipedia.org/wiki/Schwarzschild_metric" [Broken]), and you look at the derivates of the temporal component in regards to spatial coordinates then:
1st derivate is related to gravity
2nd derivate is related to curvature & tidal forces

Obviously one can find metrics, where at certain space coordinates only one of the derivates is zero.

This isn't the rule of thumb, and I don't think it helps with the exceptions either.

This is the rule of thumb that gravity=curvature=geodesic deviation=tidal forces. See A.T.'s post for exceptions to the rule of thumb.

The usual way of motivating it is to compare the "local" and "non-local" lift experiments. In Carroll's words, we distinguish between experiments done in "small enough regions of spacetime" versus experiments "in a very big box". It is the non-local, "in a very big box" experiments that show up curvature and geodesic deviation of "true" gravity. http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll4.html

Mathematically this corresponds to curvature and geodesic deviation being characterised by second derivatives, in the sense that higher order derivatives are more non-local than lower order derivatives (I think this is not too cheating a way to think about it, even though strictly speaking, all derivatives are local, since their limits at a point are well-defined). The relevant equations are Eq. 3.21 where the Christoffel symbol is defined in terms of the metric; Eq 3.67 where the curvature is defined in terms of derivatives of Christoffel symbols; and Eq 3.113 which gives the tidal forces in terms of the curvature. http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll3.html

That's true as long as gtt acts as the gravitational potential. Generally, you can extract curvature from the metric components, and "gavity" is not even well defined.

Newtonian gravity itself can be reformulated "essentially exactly" as curved spacetime (Newton-Cartan theory). You can find more detail in Malament's http://arxiv.org/abs/gr-qc/0506065. I don't know whether "essentially exactly" is exactly enough that totally uniform gravity is possible in Newton-Cartan theory.

Newtonian gravity is an excellent approximate description of certain regimes of GR. In those regimes, one can choose coordinates such that Newtonian gravity is the curvature of time. In GR, there is no such thing as totally uniform gravity. There are situations in which gravity is present, and large regions of spacetime are flat, but these flat regions are not infinite in size, and there must be curvature somewhere.

"Uniform gravity" is a theoretical special case of Newtonian gravity, where the Newtonian field is constant for each point of space. It exist only approximately in small regions of space.

You cannot have only one dimension of a manifold intrinsically curved, so you cannot have "just curved time". But you can have a metric, where the distances along the time dimension vary depending on the space coordinate. This does not imply intrinsic curvature yet, but it implies gravity (and gravitational time dilation). http://www.physics.ucla.edu/demoweb...lence_and_general_relativity/curved_time.gif" Note that the cone doesn't have intrinsic curvature, an therefore the title is put in quotation marks.