# What does the Kretschmann scalar really tell us about the geometry of spacetime? [duplicate]

This question already has an answer here:

The Kretschmann scalar is one of the measures of spacetime curvature. For flat (Minkowski) spacetime it is zero. The dimensions of the Kretschmann scalar are $[L]^{-4}$. What does that physically signify about the geometry of spacetime?

## marked as duplicate by Kyle Kanos, stafusa, Jon Custer, David Z♦Oct 27 '17 at 20:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

If we restrict ourselves to vacuum solutions then the Kretschmann scalar has a nice simple interpretation as the strength of the local tidal forces.

This happens because for a vacuum solution the Ricci tensor is zero so the Kretschmann scalar depends only on the Weyl tensor, and this tells about the tidal forces acting. The Kretschmann scalar is effectively proportional the square of the tidal force. For the Schwarzschild solution this gives us:

$$F_\text{tidal} \propto \sqrt{\frac{48M^2}{r^6}} \propto \frac{1}{r^3}$$

which we immediately recognise as the Newtonian expression for the tidal force.

For non-vacuum solutions I don't think there is a nice simple interpretation. The Kretschmann scalar is then related to both volume changes and shape changes of a test volume element.