Construction of Tetrad

Construction of the tetrad of basis vectors for the freely-falling, but non-rotating frame, proceeds much as before. A big difference here is that the spatial members of the tetrad are not generated by parallel propagation along the geodesic. One member of the tetrad is still taken to be tangent to the geodesic $G$, by choosing:

$$\Lambda_{(0)}^{\mu}={dX^{\mu} \over ds }.$$ (4)

We still impose the same orthonormality conditions on the members of the tetrad:

$$\Lambda^{\mu}_{(\alpha)} \Lambda^{\nu}_{(\beta)}G_{\mu \nu}=\eta_{\alpha\beta},$$ (5)

where the metric tensor appearing in Eq. (5) is evaluated on the geodesic itself. We shall denote the value of the potential $\Phi$ on $G$ by $\Phi_0$. The zero-member of the tetrad does satisfy the equation of parallel propagation:

$${d \Lambda_{(0)}^{\mu} \over ds}+ \Gamma_{\alpha \beta}^{\mu} \La... ...^2}+ \Gamma_{\alpha \beta}^{\mu} {dX^{\alpha} \over ds}{dX^{\beta} \over ds}=0.$$ (6)

and I shall assume that somehow this equation has been solved so that the position of the origin of coordinates of the quasi-inertial frame is known as a function of $X^0$. This position will be denoted by:

$$X^k(X^0)=X_0^k.$$ (7)

To the required order of the calculation, the proper time is obtained from the scalar invariant:

$$-(dx^0)^2 = G_{00}(dX^0)^2+G_{ij}dX^i dX^j=-\left[1+{2 \Phi_0 \ov... ...ver (dX^0)^2}+{dX^2 \over (dX^0)^2}+{dX^2 \over (dX^0)^2}\right)\right](dX^0)^2$$ (8)

$$=-\left[1+{2 \Phi_0 \over c^2}-{V^2 \over c^2} \right](dX^0)^2.$$ (9)

Solving this equation for $dX^0$, keeping only terms of $O(2)$, we have

$$dX^0=\left[1-{ \Phi_0 \over c^2} + {V^2 \over 2 c^2}\right]dx^0.$$ (10)

There are no terms in $\gamma$ in this equation because they would be of higher order in the scheme of approximation. The quantity in square brackets occurs so often in what follows that I shall give it an abbreviation:

$$K=\left[1-{ \Phi_0 \over c^2} + {V^2 \over 2 c^2}\right].$$ (11)

Here $V$ is the velocity of the gyroscope as observed in the original reference frame, in which the mass source is at rest. The zero member of the tetrad can be written in terms of $K$ as:

$$\Lambda_{(0)}^{\mu}=\{K,KV^1,KV^2,KV^3\},$$ (12)

correct to $O(3)$. Time in the quasi-inertial frame is defined by a standard atomic clock at the position of the gyroscope. The proper time on this clock is taken as the time variable $x^0$. We can imagine integrating Eq. (10) along the geodesic path to obtain the first part of the transformation to quasi inertial coordinates:

$$X^0=\int_{{\rm path}}^{x^0} K dx^0.$$ (13)

The next part of the calculation is to determine the time components of the spatial members of the tetrad using orthogonality:

$$\Lambda^{\mu}_{(0)} \Lambda^{\mu}_{(i)}G_{\mu \nu}=\eta_{0i}=0.$$ (14)

Thus writing out the sums and keeping in mind that in the orthogonality relation, the metric is evaluated at the origin of the quasi-inertial coordinates, gives

$$-\Lambda_{(0)}^0 \Lambda_{(i)}^0\left[1+{2 \Phi_0 \over c^2} \rig... ...)}^k \Lambda_{(i)}^l \delta_{kl} \left[1-{2 \gamma \Phi_0 \over c^2} \right]=0.$$ (15)

Inserting the values from Eq. (12) gives

$$-K\Lambda_{(i)}^0\left[1+{2 \Phi_0 \over c^2} \right]+K V^k\Lambda_{(i)}^l \delta_{kl} \left[1-{2 \gamma \Phi_0 \over c^2} \right]=0.$$ (16)

K cancels out, and one can solve to the required order of accuracy:

$$\Lambda_{(i)}^0=V^k \Lambda_{(i)}^l \delta_{kl} \left[1-{2( \gamma+1) \Phi_0 \over c^2} \right].$$ (17)

Thus the 0 component of each of the spatial members of the tetrad will be determined when the spatial components themselves are known. We shall choose these latter components to be parallel to the axes of the reference frame we started with, and non-rotating, so I can just write down the solution that works:

$$\Lambda_{(k)}^l=\delta_k^l\left[1+{\gamma \Phi_0 \over c^2}\right]+{V^k V^l \over 2c^2}.$$ (18)

The correction to the Kronecker delta arises because the gravitational potential of the mass source causes a change in the calculation of proper distance at the position of the gyroscope. The velocity terms represent Lorentz contraction. Substituting these components into Eq. (17) gives

$$\Lambda_{(i)}^0=V^i\left[1-{(2+\gamma) \Phi_0 \over c^2 } +{V^2 \over 2c^2}\right].$$ (19)

These results completely determine the non-rotating tetrad. It is straightforward to check that the orthonormality condition is satisfied to $O(3)$. We summarize the orders of the calculation as follows:

$$\Lambda_{(0)}^0: {\rm to O(2)};$$ (20)

$$\Lambda_{(0)}^k: {\rm to O(3)};$$ (21)

$$\Lambda_{(i)}^0: {\rm to O(3)};$$ (22)

$$\Lambda_{(i)}^j: {\rm to O(2)}.$$ (23)