Why is it natural to impose the condition that the metric remains unchanged under parallel transport? In Wald’s General Relativity, given a metric tensor $g_{ab}$, and two vectors $v^a,w^b$, the author said it is “natural” to impose the condition that the $g_{ab}v^aw^b$ is invariant under parallel transport, which leads us to “naturally” choose a derivative operator.
But I can’t see why it is “natural”. Can anyone explain this to me?  
 A: First I'll amplify on the point made by Avantgarde in a comment. Let's say you're in outer space, in free-fall conditions, and you have two gyroscopes. You orient the gyroscopes so that their axes are perpendicular. The directions of their axes can be described by three-vectors, but these three-vectors can also be promoted to four-vectors in a natural way, by requiring that they be vectors of simultaneity in your frame, i.e., they represent purely spatial displacements to you. Let these two vectors be $v^a$ and $w^b$. Then the orthogonality of their axes can be expressed as $v^aw_a=0$, which is equivalent to $g_{ab}v^aw^b=0$.
Now as time goes on, and you free-fall along with the gyroscopes, we expect the axes of these two gyroscopes to remain orthogonal. This is really just the equivalence principle at work. The properties of spacetime are always locally equivalent to the properties of Minkowski space. Mathematically, this means that we expect $v^aw_a$ to remain zero. This can be generalized to cases where the vectors are not necessarily spacelike or where the inner product is nonzero.
At a more philosophical/conceptual level, if this kind of inner product were to change under parallel transport, then we would probably attribute the observed change to a change in the metric. But it doesn't really make sense to talk about measuring a change in the metric, because the metric is our only tool for making measurements in the first place. So for example, if the metric were to change by a factor of 2, how would we know? That would be like saying that all objects, including rulers and clocks, changed by a factor of 2, but that would be unobservable because nothing would change relative to anything else.
A: A world line of a test particle is a physical object which, according to the equivalence principle, can be described in a local Lorentz frame as a straight line and in any coordinate frame as a geodesic.  
Local Lorentz frame
$g_{\alpha \beta} = \eta_{\alpha \beta}$ flat spacetime metric
$\partial g_{\alpha \beta} / \partial x^\gamma = 0$
Note: A local Lorentz frame is the closest thing to a global Lorentz frame.
$d^2x^\alpha / d\tau^2 = 0$ straight line
$\tau$ proper time  
Any coordinate frame
$d^2x^\alpha / d\tau^2 + \Gamma^\alpha_{\beta \gamma} (dx^\beta / d\tau) (dx^\gamma / d\tau) = 0$ geodesic
$\tau$ affine parameter  
The consistency of the two representations demands $\Gamma^\alpha_{\beta \gamma} = 0$ in any local Lorentz frame, i.e. requires any local Lorentz frame to be a local inertial frame.  
That means that in a local Lorentz frame the covariant derivative $\nabla_\gamma$ shows:
$\nabla_\gamma g_{\alpha \beta} = \partial g_{\alpha \beta} / \partial x^\gamma - \Gamma^\mu_{\alpha \gamma} g_{\mu \beta} - \Gamma^\mu_{\beta \gamma} g_{\alpha \mu} = 0$
The covariant derivative of the metric vanishes because the partial derivative and the $\Gamma's$ vanish separately. However the equation is tensorial, hence it is valid in any arbitrary coordinate system.  
That is the physical basis why the metric remains unchanged under parallel transport.
