I am having some doubts on myself regarding the above concepts in General Relativity.

First, I want to point out how I understand them so far.

A male observer follows a timelike worldline ($\gamma$) in spacetime (because he must have a proper time). He has a frame for himself.

A coordinate is a sets of numbers the observer uses to describe the spacetime in his frame (which is another way to say the spacetime in his view).

The locally flat coordinate of an observer at a time ($s\in\gamma$) is the coordinate (of his frame, of course) in which he sees the metric tensor at a neighborhood of his position be the flat metric (Christoffel symbols vanish):

$$g_{\mu\nu}(s)=\eta_{\mu\nu}$$ $$\Gamma_{\mu\nu}^\rho(s)=0$$

This coordinate depends on and is used naturally by the observer.

Now a locally inertial frame is a frame of any freely falling observer, or any observer following a geodesic ($l$). He may use or may not use the locally flat coordinate of himself. But he has a very special coordinate which is locally flat at every point is his worldline: $$\forall s\in l:$$ $$g_{\mu\nu}(s)=\eta_{\mu\nu}$$ $$\Gamma_{\mu\nu}^\rho(s)=0$$

Do I have any misunderstanding or wrong use of terminology?

Now there should be an freely falling observer $A$ (with his special coordinate) and his wordline crosses the wordline of another (not freely falling) observer $B$. And at the cross point can I believe that the two coordinate (of two frames) may be chosen to be locally identical (or equal) (that is, there exists a linear transformation locally transform one to other)?

  • $\begingroup$ I believe that the derivatives of the metric must also vanish in the locally inertial frame, which is not true of the locally flat frame. $\endgroup$
    – levitopher
    Commented Jan 23, 2015 at 17:29
  • $\begingroup$ There is no locally flat frame in my terminology. There is only locally flat COORDINATE of a frame, and this frame can be a locally inertial frame or not. $\endgroup$ Commented Jan 23, 2015 at 17:32
  • $\begingroup$ That should be the same thing. You can choose the coordinates such that the metric, in those coordinates, is flat. I don't think these two choices can be made simultaneously because the derivatives are not necessarily zero in the locally flat case. $\endgroup$
    – levitopher
    Commented Jan 23, 2015 at 17:35
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    $\begingroup$ Followup question: What happens if the observer is female instead? ;) $\endgroup$
    – Christoph
    Commented Jan 23, 2015 at 18:35
  • 1
    $\begingroup$ There may be an argument for that in some specific circumstance, but since $\Gamma = A+B-C$, where $A$, $B$, and $C$ are first-order derivatives of the metric, $\Gamma=0$ does not imply $A=B=C=0$. For instance, $A=x$, $B=x$, $C=2x$. $\endgroup$
    – levitopher
    Commented Jan 24, 2015 at 19:42

4 Answers 4


In the most general case described by general relativity, it is not possible to find a $neighbourhood$ covered by coordinates $x^\mu$ such that $g^{\mu\nu} = \eta^{\mu\nu}$ in all U. If it were so, you have a zero Riemann tensor, hence the space-time would be flat in all U. You may have space-times with such flat pieces (I think there is no problem in gluing this piece with non-flat pieces, but I may be wrong), but is not the most general case and not what is meant when we say that space-time is locally flat.

What we mean is that the tangent space, in any point is the Minkowski space-time.

This mean that, for any point p, you can find a basis for the tangent space at p (and associated "exponential" coordinates) so that the metric is diag(-,+,+,+) in these coordinates at this point p and the connession coefficients vanish at this point (not in a neighbourhood!)

You can think of these coordinates as those of a inertial observer. Note that there exists several possible coordinates, which are related by a Lorentz transform at the tangent space, and are associated to different observers.

In what sense you can think of these coordinates as those of a inertial observer? In the sense that as long as you are covering a sufficiently small neighbourhood of p, whose dimension will be "smaller the larger the Riemann tensor is at p", you may describe everything happening here as if you were in special relativity. One above all, the geodesics are of the form $d/dt^2 x(\tau) = 0$ and do not accelerate with respect to each other. Of course, actually they do, but these effects are small if you consider small neighbourhood of p and small Riemann at p.

Analogously, Earth is flat at a point in the sense that you can "confuse" the flat tangent space with the actual neighbourhood because the differences are difficult to detect if you zoom enough.


In light of our clarifying discussions, I believe the answer is yes.

I found a nice section on Fermi Normal coordinates here (Section 9):


This seems to be what you mean by "Locally inertial coordinates" - the tetrad is orthonormal with one direction along the curve, and the others along spacelike curves orthogonal to the curve.

Since you can define Fermi normal coordinates anywhere on a timelike geodesic, define them on the intersection of two geodesics. These define a flat metric, so there's no reason why you couldn't pick that metric to be the tetrad for the other observer at the same point.


If the observer is not in free-fall, the metric-tensor $g_{\mu,\nu}(s)$ at the observer's position, expressed in local coordinates around the observer, will not be $\eta_{\mu,\nu}$. Your first assumption about the path $(\gamma)$ is wrong.

I guess what you are aiming at is the notion of the space of coordinates around a point, which is indeed a flat space (since it is (pseudo-)euclidean). This space however serves only to introduce coordinates in an open subset of your manifold by a mapping that is homeomorphic to an open subset of that (pseudo)-euclidean space. This means that the open subset of your manifold is quite the same (up to deformations, that is curvature!!!) as the (pseudo-)euclidean subset.


I think the best way to think of it is as follows.(It's not too different from what everyone has said, but may be put into better perspective).

Choosing a frame of reference is a completely different job than setting up of coordinates. To observe an event in spacetime you must belong to some frame of reference(or equivalently, you create a frame of reference say S, where $\frac{d \vec{r}_{you}}{dt}$ from the frame S is always 0).Notice that I have not defined any coordinates yet. I will lay down coordinates next only to explain the motion of other bodies wrt my frame.

Its clear that coordinates can be defined only after you have chosen your frame of reference.Whenever we do spacetime coordinate transformation say, $x^\mu\rightarrow x^{\mu}{'}$, then we are certainly changing frames.If however we are doing some transformation without any t appearing in the transformation equations, it's change of coordinates.

Next, from all that I have read or come across, the meanings of the two key terms are as follows:

  1. Locally inertial/flat coordinates: Cartesian/Euclidean coordinates laid about some point X in general curved space.
  2. Locally inertial frames: Frames that admit use of Locally inertial/flat coordinates as one of the choices

Hope this helps.


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