The Space-Time Components of the Metric Tensor

Having disposed of all but the gravitomagnetic components of the metric tensor, we now need to calculate:

$$g_{0k}={\partial X^{\mu} \over \partial x^0} {\partial X^{\nu} \over \partial x^k}G_{\mu \nu}.$$ (46)

Writing this out, keeping in mind that the calculation needs to be done to $O(3)$, we have

$$g_{0i}=-(1+{2 \Phi \over c^2}){\partial X^0 \over \partial x^0} {... ...\delta_{kl} {\partial X^k \over \partial x^0}{\partial X^k \over \partial x^i}.$$ (47)

The transformation coefficients needed are given in a previous section. The calculation needs to be carried to $O(3)$, but only linear terms in the local variables $x^k$ are needed, since the object of the calculation is to find the rate of precession of a gyroscope placed at the origin. In computing this precession rate, Christoffel Symbols will be needed but these involve first derivatives. Any quadratic terms, such as tidal terms, would vanish when evaluated at the origin.

Also, an expansion of the potential $\Phi$ that occurs is needed to linear order in small distances away from the origin of quasi-inertial coordinates. This means that we make a Taylor expansion:

$$\Phi=\Phi_0+\Phi_{,k}x^k.$$ (48)

When all these expresssions are put into Eq. (49), above, and the expression is expanded, there are a very great many terms which cancel out. We shall demonstrate this with a couple of simple examples.

Cancellation of $\Phi_{0,0}$ terms. The time derivative terms are already of $O(3)$. We get one term multiplying $G_{00}$ which introduces a minus sign, and one such term multiplying the $\delta_{ij}$ factor in $G_{ij}$ from the transformation coefficient

$${\partial X^i \over \partial x^0}.$$ (49)

The result is:

$$-{\gamma x^i\Phi_{0,0} \over c^3}+-{\gamma x^i\Phi_{0,0} \over c^3}=0.$$ (50)

Thus, no such time derivative terms survive.

Cancellation of terms depending only on velocity. Also, all terms that depend only on the relative velocity $V^k$ cancel out. These terms can be identified by setting all acceleration terms or potential terms equal to zero in Eq. (). This would give the following expression:

$$(-1)(1+{V^2 \over 2 c^2 }){V^i \over c}(1+{V^2 \over 2 c^2 })+ \delta_{kl}{V^k \over c}\left[\delta_i^l+{V^l V^i \over 2 c^2} \right]$$ (51)

$$=-{V^i \over c}(1+{V^2 \over c^2 })+(1+{V^2 \over 2 c^2 }){V^i \over c}(1+{V^2 \over 2 c^2 })=0.$$ (52)

The terms that remain in $g_{0i}$ are linear in $x^k$, linear in $V^k$, and linear in the gradients of the gravitational field. The result after a great deal of cancellation, is:

$$g_{0i}={\gamma+1 \over c^3}\left[{\bf V} \cdot {\bf r} \Phi_{,i}-... ...+{1 \over 2 c^3}\left[{\bf V} \cdot {\bf r}A^i- {\bf A} \cdot {\bf r}V^i\right]$$ (53)

Then if the acceleration is separated into that due to the gravitational potential, plus a non-gravitational part, as in Eq. (40), we finally get:

$$g_{0i} = {\gamma+{1 \over 2} \over c^3}\left[ {\bf V} \cdot {\bf ... ... c^3}\left[{\bf V} \cdot {\bf r}A_{NG}^i- {\bf A}_{NG} \cdot {\bf r}V^i \right]$$ (54)

The important terms here are the first terms, proportional to $\gamma+1/2$. The next term would be zero for a circular orbit. The last term we shall set aside for the moment and return to it in a later section.