# Is spacetime-curvature relative?

Velocity is relative, which means kinetic energy is. Since, according to general relativity, energy bends spacetime around it, wouldn't this mean observers moving in different inertial frames measure different values of curvature?

I think this is slightly different from this question: Is space curvature relative?

It depends on what you mean by “curvature”.

The most complete description of curvature is in terms of something called the Riemann curvature tensor, $$R_{\mu\nu\lambda\kappa}$$. In four dimensions, it has 256 components (only 20 of which are independent), and the value of these components is different in different reference frames. There is a transformation rule for how this tensor transforms between frames. In other words, the components of the Riemann curvature tensor are relative, i.e. observer-dependent.

However, from this tensor you can construct invariants such as the Ricci scalar $$R$$ which have the same value in all reference frames. These curvature invariants are absolute, i.e., observer-independent.

This is similar to how energy and momentum are relative — they are components of a four-vector which transforms between frames — but a particular combination of them, mass, is absolute.

• Taking into account symmetry the 256 reduce to 55. Still impressive. – my2cts Jan 7 at 20:39
• @my2cts No. For confirmation that 20 is correct, please see mathworld.wolfram.com/RiemannTensor.html. – G. Smith Jan 7 at 20:45
• Could you go into more detail on how we can combine energy and momentum to get mass? Also, isn't mass also relativistic? – gardenhead Jan 7 at 21:05
• In units where $c=1$, the invariant mass (which is sometimes called "rest mass") is $m=\sqrt{E^2-\mathbf{p}^2}$. (With the $c$'s, this is $m=\sqrt{E^2-\mathbf{p}^2 c^2}/c^2$.) The mass is the "length" of the energy-momentum vector $(E,\mathbf{p})$ in Minkowski space. Relativistic mass is just $E/c^2$ and is widely considered an obsolete concept. It is either proportional to the energy, or, in units with $c=1$, equal to the energy. – G. Smith Jan 7 at 21:15
• @my2cts If the downvote is from you, and is because you mistakenly think the number of independent components is 55, it would be polite for you to acknowledge your mistake and reverse your downvote. – G. Smith Jan 7 at 21:21

No, it is absolute. The presence of curvature (non vanishing Riemann tensor) is mathematically equivalent to the existence of geodesical deviation of timelike geodesics. This physically means that there is a relative acceleration of free falling bodies. This fact is absolute, it does not depend on any reference frame as it is a relative phenomenon.

Locally, yes. Globally no.

The Equivalence principle tells you that you can always find a local Inertial reference frame, which means that given a point in spacetime you can always find neighborhood of this point where spacetime is flat, and special relativity holds.

Globally you can't do that because spacetime is, mathematically speaking, a Riemannian manifold, and its curvature is an intrinsic property.