# I Determination of locally inertial coordinates

1. Apr 3, 2016

### mgal95

Hello,

I am studying on my own from Weinberg's Gravitation and Cosmology and I cannot understand how he derives a solution (pg. 72). I did not know where else to post this thread since it is not homework exercise.

He takes a coordinate system $\xi^a$ "in which the equation of motion of a particle moving freely under the influence of purely gravitational forces is that of a straight line in space-time, that is:
$$\frac{d^2\xi^a}{d\tau^2}=0$$
Next he takes another coordinate system $x^\mu$ which could be "what we will". He does some elementary differentiations, defines the affine connection
$$\Gamma^\lambda_{\mu\nu}\equiv\frac{\partial x^\lambda}{\partial \xi^a}\frac{\partial^2\xi^a}{\partial x^\mu\partial x^\nu}$$
and finally gets the equation
$$\frac{\partial^2\xi^a}{\partial x^\mu\partial x^\nu}=\Gamma^\lambda_{\mu\nu}\frac{\partial \xi^a}{\partial x^\lambda}$$
He states (without giving any other argument) that "The values of the metric tensor and the affine connection at a point $X$ in an arbitrary coordinate system $x^\mu$ provide enough information to determine the locally inertial coordinates $\xi^a(x)$ in a neighbourhood of $X$. The solution (to the above equation) is:

$$\xi^a(x)=\alpha^a+b^a_\mu(x^\mu-X^\mu)+\frac{1}{2}b^a_\lambda\Gamma^\lambda_{\mu\nu}(x^\mu-X^\mu)(x^\nu-X^\nu)+...$$

where $\alpha^a=\xi^a(X)$ and $b^a_\lambda=\frac{\partial\xi^a(X)}{\partial X^\lambda}$ [...] Thus given the affine connection and the metric tensor at $X$, the locally inertial coordinates are determined to order $(x-X)^2$"
Where does this solution come from?

2. Apr 4, 2016

### andrewkirk

The proof is quite long and some authors of physics texts take shortcuts that skate over important points.

For a rigorous proof, a mathematics text is better. For instance in John Lee's 'Riemannian Manifolds' he rigorously proves the existence of 'Normal Coordinates', which are the same thing. Unfortunately though, proofs in a text like that still need adapting to the GR case because there the manifold is pseudo-Riemannian rather than Riemannian.

The general idea of the proof in physics texts is about the number of degrees of freedom we have to construct a solution - hence the reference above to 'provide enough information'. What is being sought is a change of basis matrix $\Lambda$ which, being 4x4, has 16 free variables we can choose. That partially matches the 10 variables that are specified by the metric, with the other six allowing rotations of the frame (ie there are multiple solutions, being rotations of one another). Then there are 40 free variables to choose in the components of the first derivatives of $\Lambda$ wrt each of the four dimensions in the new coord system (after allowing for symmetries). They match up against the 40 constraints imposed by requiring that all first partial derivatives of the metric in the new coordinate system are zero at the point in question.

So in short, there are 56 equations and 62 variables, allowing for a solution space of dimensionality 6, and the solutions in the space are all just rotations of one another.

3. Apr 4, 2016

### Staff: Mentor

A small point: "rotations" here should include boosts, i.e., "spacetime rotations" as well as "spatial rotations".