What kind of coordinate change is needed to make gravity disappear?

Question

I understand that the Christoffel symbols associated with the metric will vanish locally once you perform the appropiate change of coordinates. These new coordinates correspond to an observer in free-fall.

Now, if this transformation is dependent on the velocity and acceleration of the free-falling observer, then any velocity should work, as long as the acceleration of this new observer works to "cancel out" the acceleration due to gravity.

But then, acceleration has only three parameters, so how are they enough to make all the Christoffel symbols go to zero? Aren't there like 40 of them? I have considered also rotation, but that only brings like three extra components for the angular velocity, which I'm not completely certain that gravity can produce.

Plus, if the metric is to also be converted into the Minkowski matrix, that's 6 extra components that must go to zero. Does not look good

Answer 1

Perhaps a simpler example of differential geometry helps. We can choose coordinates $\theta,\phi$ for the surface of a sphere, so that $\theta$ is the polar angle: zero at the 'north pole' and $\pi$ at the 'south pole'. $\phi$ is the azimuthal angle, varing from 0 at some arbitrary meridian to $2\pi$.

It can be proved that the only non zero Christoffel symbols are:

$\Gamma^{\phi}_{\theta \phi} = ctg(\theta)$ and $\Gamma^{\theta}_{\phi \phi} = -sin(θ)cos(θ)$

But for any point $(\theta , \phi)$ it is always possible to change coordinates, choosing poles so that the point is now at the equator. And follow a geodesic over this (redefined) equator. In that case $\theta = \frac{\pi}{2} \implies cos(\theta) = 0$, and both symbols vanish.

Note that the Riemann tensor, that involves second derivatives of the metric, doesn't vanish with that change of coordinates.

Answer 2

K.Pull asked: But then, acceleration has only three parameters, so how are they enough to make all the Christoffel symbols go to zero?

One needs to keep in mind that the Christophel symbols are defined in spacetime. As such, one needs to consider the acceleration also as something that has four components. The following might appear a lengthy response but it might be worth a read.

For 3D spatial motion, we have the trajectory as (where $X$ is a chart-map a.k.a. coordinate system) $$X: \mathbb{R} \rightarrow \mathbb{R}^{3}$$ $$X: t \rightarrow \left(x^{1}(t), x^{2}(t), x^{3}(t)\right)$$ Now, for an accelerating particle, the equation of motion would be ($\alpha$ runs from (1,3)) $$m \ddot{X}^{\alpha}(t)=F^{\alpha}(X(t)) \implies \ddot{X}^{\alpha}(t)=f^{\alpha}(X(t)) \tag{1}$$ where in the second half, I only used the force per unit mass (so mass has been removed, so to speak).

As it turns out, Eq. (1) cannot be seen as an autoparallel (uniform, straight-line motion) in the 3D space. One needs to promote this to a 4D spacetime version. One can do that by making the following change $$X: \mathbb{R} \rightarrow \mathbb{R}^{4}$$ $$X: t \rightarrow \left(t, x^{1}(t), x^{2}(t), x^{3}(t)\right)$$

This still results in the same equation of motion, i.e., $$ \ddot{X}^{\alpha}(t)=f^{\alpha}(X(t))$$

but now can be written in a different form, namely $$ \ddot{X}^{\alpha}(t)-f^{\alpha}(X(t)) \dot{X}^{0}(t) \dot{X}^{0}(t)=0 \tag{2}$$ where $\dot{X}^{0}(t) = 1$.

The equation in spacetime can now be written as $$\ddot{X}^{a}+\Gamma_{b c}^{a} \dot{X}^{b} \dot{X}^{c}=0$$ but note that due to Eq. (2), only $\Gamma_{00}^{a} = f^{\alpha}(X(t)) \neq 0$.

This is the geodesic equation where only $\Gamma_{00}^{a} \neq 0$. Now if you choose a frame of reference where even $f^{\alpha}(X(t)) = 0$ (so there is no force), then you can see that even $\Gamma_{00}^{a} = 0$ implying that $\Gamma_{bc}^{a} = 0$. For more, see this nice lecture.