The $g_{00}$ Component of the Metric Tensor

With the transformation coefficients finally evaluated, we can proceed to compute the metric tensor in the quasi-inertial frame. For the 00-component, we have to second order

$$g_{00}={\partial X^{\mu} \over \partial x^{0}} {\partial X^{\nu} \over \partial x^{0}}G_{\mu \nu}.$$ (37)

$$=-\left[1+{2\Phi_0 \over c^2}+{2\Phi_{,j}x^j \over c^2} \right]\l... ...+ {V^2 \over 2 c^2} + {{\bf A}\cdot{\bf r}\over c^2} \right)^2+{V^2 \over c^2}.$$ (38)

Expanding the various products, keeping only terms of $O(2)$, the only terms that survive are:

$$g_{00}=-1-{2\Phi_{,j}x^j \over c^2}-{2 {\bf A}\cdot{\bf r}\over c^2}.$$ (39)

But if the acceleration is separated into a part due to free fall and a non-gravitational part, we have

$$A^j=-\Phi_{,j} + A_{NG}^j.$$ (40)

This component of the metric tensor reduces to the non-gravitational part only:

$$g_{00}=-1-{2 {\bf A}_{NG} \cdot {\bf r}\over c^2}.$$ (41)

There will in general be quadratic terms in the local coordinates $x^i$, but these cannot give rise to any torques on the gyroscope.

The surviving term could be thought of as a local potential due to an induced gravitational field strength brought on by the acceleration:

$$\Phi={\bf A}_{NG} \cdot {\bf r}$$ (42)

which is in complete accord with Einstein's principle of Equivalence.