# Is space curvature relative?

If I have an object that is of some length $l$ moving at a relativistic velocity $v$ for some reference frame in a 1D universe, then length contraction states that $l=\gamma\times l'$. But at the frame of reference, there is no length contraction, since there $v=0$. The curvature of space should be Euclidean there. However, if I move my frame of reference to the object that is traveling with velocity $v$, the exact opposite is true, with there being no curvature near the primed frame from the primed point of view. What if one of the objects is accelerating? If I stand on a planet, will my geometry of space agree with what someone who's not standing on one sees?

-
suggested edit: the space isn't curved for either reference frame due to their relative velocity. Length contracts according to velocity, space curves according to gravitational fields (which are complimentary to acceleration). Your final sentence, however, still poses a reasonable question. –  Alan Rominger Oct 2 '12 at 4:56

Let's stay with your initial example of length contraction in special relativity. Length is an example of something that is not invariant, that is different observers will measure different lengths. However there is a quantity called proper time, $\tau$, that is an invariant in SR:

$$d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2$$

By this I mean that for any two spacetime points all observers will measure the same proper time between the two points. Note that different observers will measure different values of $dt$, $dx$, etc, but when combined to give $d\tau$ all observers will get the same value. In fact from the invariance of $d\tau$ you can derive the Lorentz transformations and all the weird effects seen in SR.

So the proper time is an example of an invariant, i.e. something that does not depend on the frame of the observer. In fact it's an invariant in general relativity as well, though the formula to calculate the proper time (called the metric) is generally more complicated because it has to take the curvature into account.

The principal invariant in GR is the Riemann curvature tensor, and this is what is generally meant by the term curvature. So the answer to your question is that yes, in GR the curvature is an invariant. But this is not as simple as it sounds. The Riemann tensor is generally represented as a matrix that has 20 independant components. Different observers will measure different values for the components of the matrix, but when combined they always give the same physical quantity. This is similar to the situation in SR where different observers will measure different values for $dt$, $dx$, etc but when combined will get the same value for $d\tau$.

To take your example of standing on the Earyh watching an apple fall. You feel the gravitational force of the earth and see the apple accelerating down. The apple feels no force, and sees the Earth and you accelerating towards it. You and the apple agree on the value of the Riemann curvature tensor, but in your different frames it produces apparently different behaviour.

-
Your answer to general relativity is what I was looking for. My introduction to GR was simply "gravitational and inertial accelerations cannot be told apart," rather than the more formal one you described, which answers my question. –  VF1 Oct 2 '12 at 16:07