Coordinate Transformations

The objective here is to calculate the metric tensor in the freely falling frame, so we need the coordinate transformations to quadratic order in $x^k$ in order to calculate the transformation coefficients to linear order. In the previous lecture the following expression for the coordinate transformations was obtained:

$$X^{\mu}(P)=X^{\mu}(P_0)+\Lambda_{(i)}^{\alpha} x^i-{1 \over 2}\Gamma_{\alpha \beta}^{\mu}\Lambda_{(i)}^{\alpha} x^i\Lambda_{(j)}^{\beta} x^j.$$ (24)

First we evaluate the linear sums. This calculation is not too bad. We find:

$$\Lambda_{(i)}^0 x^i={\bf V}\cdot{\bf r}\left[1-{(2+\gamma) \Phi_0 \over c^2}+{V^2 \over 2c^2} \right]$$ (25)

and

$$\Lambda_{(i)}^k x^i =x^k\left[1+{\gamma \Phi_0 \over c^2}\right+V^k{V^2 \over 2 c^2}$$ (26)

The quadratic terms obviously involve a lot of calculation, since the Christoffel symbols of the second kind must be calculation. However, if one keeps track of the fact that the spatial part of the transformation need be calculated only to $O(2)$, and the time part only to $O(3)$, there is a lot of simplification. I'm not going to do all that here, but here is the result for the transformations:

$$X^0=\int_{{\rm path}}^{x^0} K dx^0+{{\bf V}\cdot{\bf r}\over c} \... ...er 2c^2} -{\Phi_{,i}x^i \over c^2} \right]+{r^2 \gamma \Phi_{0,0} \over 2 c^2 }$$ (27)

$$X^k=X_0^k+x^k\left[1+{\gamma \Phi_0 \over c^2} +{\gamma \Phi_{,j}... ...right]+V^k{{\bf V}\cdot{\bf r}\over 2 c^2} -{r^2 \gamma \Phi_{,k} \over 2 c^2 }$$ (28)

where

$$r^2 = x^i x^j \delta_{ij} = (x^1)^2+(x^2)^2+(x^3)^2.$$ (29)

Even though there are in principle an enormous number of terms in the summations, they reduce to only a very few terms involving derivatives of the potentials.

One can check that at $x^k=0$, the $X^k$-coordinates have the values corresponding to following the origin of the local frame. The transformations have built Lorentz transformation, breakdown of synchronization, rescaling of lengths due to the potentials, and the spatial axes do not rotate since if one is considering a gyroscope in orbit, the terms in velocity will have the orbital period but there are no terms in these equations that would make a net rotation of the spatial axes build up.

Now given these transformations, Eqs. (27) and (28), we can regard them as given, no matter where they came from, and calculate the metric tensor in the new reference frame. According to the Principle of General Covariance, any coordinate transformation is allowable-but of course one has to relativity principles in the proper way to interpret the results. So the next job is to calculate the transformation coefficients.