Transformation Coefficients

In the tensor transformation of the metric tensor, we need to compute

$$g_{\alpha\beta}={\partial X^{\mu} \over \partial x^{\alpha}} {\partial X^{\nu} \over \partial x^{\beta}}G_{\mu \nu}.$$ (30)

The calculation is carried out near the origin of the quasi-inertial frame, where the transformations apply, and the metric tensor components are evaluated near the origin by a Taylor expansion:

$$\Phi=\Phi_0+\Phi_{,i} x^i$$ (31)

where as always, the repeated latin index means a summation from 1 to 3. The quantity $\Phi_0$ has to be considered a function of time if the gyroscope orbit is not circular. The tranformation coefficients only need to be calculated to linear terms in $x^i$ since we are interested in showing that linear terms in $g_{00}$ and $g_{ij}$ cancel out so that only tidal terms remain. Such tidal terms, quadratic terms in $x^$, cannot affect gyroscope motion when the gyroscope is at the origin, so we will not worry about them. The partial derivatives needed for the transformation coefficients are, to linear order, easy to get from the above.

In calculating these partial derivatives, one will encounter the partial derivative of various components $V^i$ with respect to $x^{\mu}$. These velocity components are however, evaluated at the origin of local coordinates and are not functions of $x^k$. They depend in $X^0$, and through the factor $K$, on $x^0$ (see Eq. (10)). So for example,

$${\partial V^i \over \partial x^0}={dV^i \over dX^0}{dX^0 \over dx^0}={dV^i \over dX^0}K=K{A^i \over c},$$ (32)

where $A^i$ is a component of acceleration (with the correct units) of the gyroscope, as observed in the original frame in which the mass source is at rest. In many cases the correction terms in $K$ can be neglected, as will be seen, since the factors of velocity always already have in them a factor $1/c$.

To give a concrete example,

$${\partial X^0 \over \partial x^0}=K+K{{\bf A}\cdot{\bf r}\over c^... ......]=1-{\Phi_0 \over c^2} + {V^2 \over 2 c^2} + {{\bf A}\cdot{\bf r}\over c^2}.$$ (33)

The calculation is carried out to $O(2)$, and linear in $x^i$, which is all that is necessary. Continuing with this process, we get:

$${\partial X^0 \over \partial x^k}={V^k \over c}\left[1-{(2+\gamma... ...V}\cdot{\bf r}\over c} {\Phi_{,k} \over c^2}+ {\gamma x^k \Phi_{0,0} \over c^2}$$ (34)

$${\partial X^l \over \partial x^0}=V^l K + {x^l\gamma \Phi_{0,0} \over c^2}+{V^l {\bf A}\cdot{\bf r}\over c^3}+{A^l {\bf V}\cdot{\bf r}\over c^3}.$$ (35)

$${\partial X^l \over \partial x^k}=\delta_k^l\left[1+{\gamma \Phi_... ...ver 2 c^2}+{\gamma x^l \Phi_{,k} \over c^2} - {\gamma x^k \Phi_{,l} \over c^2}.$$ (36)