# Find local orthogonal coordinates based on the global metric and geodesics

Consider a Euclidean space $X={\Bbb R}^2$ equipped with a known global $(1,1)$ pseudo-Riemannian metric in non-Cartesian coordinates. I need to define a local orthogonal Cartesian coordinate system of an observer located at a point $P$. When the observer is in a free flight moving along a timelike geodesic, his momentary time axis is tangent to this geodesic. How can I find his space axis? I am interested in the local limit where the geodesics are asymptotically straight. I would appreciate any insight.

When coordinates are involved, you can think of the metric $g$ as a matrix: for two vectors $v, w$, the inner-product is $$g^{ij}v_iw_j = v_ig^{ij}w_j = \langle v, gw\rangle$$ where $\langle\,\,,\,\rangle$ is the ordinary Euclidean metric on $\Bbb R^n$. This is true whether $g$ is Riemannian or pseudo-Rimannian.
So in your case, just multiply the timelike vector by $g$, then find the Euclidean orthogonal to the result to get the spacelike vector.
• No, $g$ need not be diagonal. $$g = \begin{bmatrix} g^{11} & g^{12}\\g^{21} & g^{22} \end{bmatrix}$$ The metric defines an inner product $$g\langle v, w\rangle = \sum_{ij} g^{ij}v_iw_j$$ As is normal in differential geometry, I used the Einstein summation convention to drop the sum signs. But note that $\sum_j g^{ij} w_j$ gives the $i$th component of the vector $u := gw$. and so the whole sum is $\sum_i v_iu_i$, which is just the normal Euclidean inner product. And note that the space itself is neither hyperbolic nor Euclidean. It is the inner products that define these characteristics. – Paul Sinclair Dec 17 '17 at 4:23
• So you don't find the "Euclidean orthogonal in a hyperbolic space". You find the Euclidean orthogonal, period. For two dimensions, there is only one such orthogonal, up to a multiple: $(x, y) \perp (y, -x)$. But you apply it to $gw$, after the affects of the hyperbolic inner product $g$ have been applied. – Paul Sinclair Dec 17 '17 at 4:27