# Extent of coordinate freedom to set metric components along a spacetime path

If we describe spacetime with a Lorentzian manifold, it is always possible to choose a coordinate system such that at any particular point $x^\alpha$, the components of the metric are: $$g_{\mu\nu}(x^\alpha) = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 &-1 & 0 & 0 \\ 0 & 0 &-1 & 0 \\ 0 & 0 & 0 &-1 \end{array} \right)$$

But our freedom is much greater than this. In curved spacetime, the equivalence principle suggests we can choose coordinates such that the metric is of that form for every point along a chosen time-like geodesic. And in flat spacetime, we can see an explicit example that it doesn't even need to be a geodesic, for the Rindler metric has that form on every point of a particular worldline with constant proper acceleration. I have a feeling this is possible for any time-like worldine.

So my question is:

Given a coordinate system and metric for a Lorentzian manifold and a time-like worldline on this spacetime, is it always possible to find a coordinate transformation such that for every point on the world line the components of the metric in this coordinate system are just (1,-1,-1,-1) on the diagonal?

I realize that even for simplified cases (say a geodesic on the Schwarzschild background), such a coordinate system could be incredibly complicated. So if someone creates an incredibly complicated explicit construction, please also show that the solution to the equations you setup exist with some kind of existence proof.

This originally started from trying to find such a local coordinate chart for a free falling observer towards a blackhole, but realized I didn't know a good mathematical way to represent such coordinate freedom to get me started. Eventually I ended up pondering this current question. So even if you can't give a full answer, but can suggest some mathematical tools or where to read up on them, any help would be appreciated.

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Related: the case of null geodesics. – Stan Liou Dec 26 '13 at 9:16

What you want are Fermi geodesic coordinates.

From some initial point $p$ on a timelike geodesic with four-velocity $u$, take the proper time $\tau$ as the time coordinate and select three orthonormal vectors $\{{\mathbf{e}}_\hat{\alpha}\}$ that serve as a basis for the orthogonal complement of the geodesic's tangent vector at $p$. Parallel-transport the spatial vectors along the geodesic by solving the equation $$\nabla_u \mathbf{e}_{\hat\alpha} = 0\text{,}$$ where the metric-compatibility of the Levi-Civita connection ensures that the four basis vectors stay orthonormal along the points of the geodesic.

At every $\tau$, take the $3$-manifold of spacelike geodesics going orthogonally to the given timelike geodesic. On this manifold, we can construct the usual Riemann normal coordinates using our orthonormal spatial vectors: basically, pick a unit vector $\alpha{\mathbf{e}}_\hat{1} + \beta{\mathbf{e}}_\hat{2} +\gamma{\mathbf{e}}_\hat{3}$, send out a spatial geodesic there, and then label each point on it with coordinates $(\tau,s\alpha,s\beta,s\gamma)$, where $s$ is distance along the spatial geodesic.

In the geodesic case, not only does the metric Minkowski form on the points of the geodesic, but the Christoffel symbols also vanish there. In general, we can also use Fermi-Walker transport to consider an accelerated observer that is also rotating. This is described, e.g., in MTW §13.6. This makes the the Christoffel symbols have $\Gamma^{\hat{0}}_{\hat{j}\hat{0}} = \Gamma^{\hat{j}}_{\hat{0}\hat{0}} = a^\hat{j}$, the components of the acceleration four-vector, as well a term corresponding to the rotation of the observer in $\Gamma^{\hat{j}}_{\hat{k}\hat{0}}$.

Things you might also be interested in: tetrads and Gullstrand-Painlevé coordinates for the Schwarzschild spacetime that correspond to the frame field of Lemaître observers freely-falling from rest at infinity.

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A naive question : For each point ($\vec x, x^o$), we are able to find a frame $F_{\vec x, x^0}$ where the metric tensor is diagonal. Now, for a given path $\vec x = \vec f(x^0)$, not necessarily a geodesic, we should be able to find frames $F_{\vec f(x^0), x^0}$, where, for each $x^0$, the metric tensor would be diagonal – Trimok Dec 26 '13 at 18:18
@Trimok: a very good question, since I screwed up describing what acceleration screws up. (: Yes, you're right; the specialness of geodesic (and if we parallel-transport our spatial vectors, nonrotating) observers is that the coordinates are outright locally inertial along the worldline, rather than just Minkowski-form. Thanks. – Stan Liou Dec 26 '13 at 19:17
So the answer is yes, there is enough coordinate freedom to do this for any time-like worldline? If the time-like worldline is a geodesic, I need to read up on "Fermi geodesic coordinates", and if the worldline is not a geodesic I can use "Fermi-Walker transport". Is that a correct large overview? I need to read up a bunch for the details to sink in. Also, can you expand on "outright locally inertial along the worldline, rather than just Minkowski-form"? By outright locally inertial do you mean the metric is 1,-1,-1,-1 diagonal and the first derivatives (in christoffel symbol) are zero? – Queasaurus Dec 26 '13 at 19:34
@Queasaurus: yes, that's right, except it's the first derivatives. The F-W transport is for the general case, and is the same as a parallel transport if acceleration is zero. Be aware that different sources are slightly inconsistent about the name; e.g., MTW calls the geodesic case "Fermi normal coordinates", while others use that name for the general case, probably because Fermi considered only geodesics and later Walker generalized it. – Stan Liou Dec 26 '13 at 19:54