Are the Christoffel symbols all zero in gravity-free space?

Question

I was looking at the geodesic equation, $$\ddot{x}^\mu + \Gamma^\mu{}_{\nu\rho} \dot{x}^\nu \dot{x}^\rho = 0, $$ and thinking about how to identify gravity-free spaces by looking at the Christoffel symbols $\Gamma^\mu{}_{\nu\rho}$. For example, if they were zero, under certain limits I could check what kind of expressions/signs metric derivatives take. Can this be used to show that the space is gravity-free?

Answer 1

No, not necessarily.

Flat space (which is what I assume you mean when you say gravity-free space) is special because it's possible to choose a global coordinate system in which all of the Christoffel symbols vanish. However, it's easy to make a choice of coordinate system (for example, spherical coordinates) for which the Christoffel symbols are generically non-zero, even in the absence of spacetime curvature.

Answer 2

The Christoffel symbols are not tensors, which specifically means that even if there is some coice of coordinates which makes them vanish, there is some other choice of coordinates for which they don't.

In terms of equations of motion, this means that for a given choice of coordinates, even though the space is flat, geodesics might not be given as $r(t) = (at+b, ct+d, et+f, gt+h)$ for real numbers $a, b, c, d, e, f$, even though there is some coordinate system for which they are.

However, the Riemann curvature tensor, given by $$ R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho{}_{\nu\sigma} - \partial_\nu\Gamma^\rho{}_{\mu\sigma} + \Gamma^\rho{}_{\mu\lambda}\Gamma - \Gamma^\lambda{}_{\mu\sigma}\Gamma^\rho{}_{\nu\lambda} $$ is a tensor, so if there is a choice of coordinates for which it vanishes, then it vanishes in all coordinate systems.

Answer 3

No. General Relativity makes no difference between gravitational and inertial ('fictitious') forces, and in accelerated frames of references, the Christoffel symbols will be non-zero even in spacetimes we would think of as free of gravity.

Answer 4

Can this be used to show that the space is gravity-free?

In contrast to the other answers here, I will start by saying that the answer to this question is "yes". However, the answer to the converse question (which you unfortunately put in the title), is emphatically "no". Consider for example regular Euclidean space with spherical coordinates. The Christoffel symbols are not zero, but the space is obviously flat. On the other hand, if the Christoffel symbols are zero, the space under consideration must be flat.

Answer 5

There are two ways to think about it and which of them is more appropriate depends on what you define as "gravity".

1. When the symbols are zero, the equations of motion are manifestly special-relativistic. In other words, when the symbols are zero, the coordinate velocities of a projectile will remain constant with time for a free particle. If we mean by the absence of gravity that the coordinate velocities of a projectile remain constant with time (i.e. we identify the absence of gravity with the inertial character coordinate set-up) then the symbols being zero does imply the absence of gravity.

2. When the symbols are non-zero at a point, it is always possible to find a coordinate transformation that makes the symbol zero at that point and vice-versa. This makes the above definition of gravity highly coordinate dependent. That is to say that the definition that identifies the absence of gravity with the inertial character of the coordinates will conclude that even if the underlying spacetime is the same, there exists gravity in one frame and there doesn't exist gravity in another frame. Thus, we define a different and a more useful notion of gravity by identifying gravity with the curvature of spacetime. The curvature of spacetime is a coordinate independent fact. Physically, if two geodesics that are parallel at one point converge or diverge as they progress then the curvature is said to be non-zero and vice-versa. All the physical bodies or energy densities that change the topology of the spacetime and makes it non-Minkowskian, always introduce the curvature in the spacetime. It is for this reason that identifying gravity with the curvature is more useful. Because then the absence or the presence of gravity will tell us whether your underlying spacetime is Minkowskian or not. With this definition of gravity, the vanishing symbols do not mean that gravity is absent. It is the Reimann curvature tensor that would have to vanish to mean that the gravity is absent.