# Wiki definition of Fermi coordinates

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• kent davidge
In summary: I don’t know how to do that in math notation.Umm, f(x) = x is zero at x=0, what is its derivative there? F(x,y) =xy is zero along either axis, but what about the partial derivatives on the axes? It’s really that simple an...
As far as it goes, yes, it looks OK, although it is phrased in terms of Riemannian geometry, i.e., locally Euclidean, instead of locally Minkowski.

Note that some textbooks (such as MTW) use the term "Fermi normal coordinates" in a broader sense, as coordinates adapted to a timelike curve, which does not necessarily have to be a geodesic. In the non-geodesic case, the definition has to be broadened somewhat, since the metric will not be exactly Minkowski on the curve and not all of the Christoffel symbols will vanish on the curve. [Edit: corrected.]

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kent davidge
I wonder if wikipedia better called it Riemann coordinates instead. Is not that the definition of Riemann coordinates?

kent davidge said:
Is not that the definition of Riemann coordinates?

Riemann normal coordinates are not quite the same. Riemann normal coordinates are centered on a particular point; Fermi normal coordinates are centered on a particular curve. They do have very similar properties, however.

kent davidge
PeterDonis said:
As far as it goes, yes, it looks OK, although it is phrased in terms of Riemannian geometry, i.e., locally Euclidean, instead of locally Minkowski.

Note that some textbooks (such as MTW) use the term "Fermi normal coordinates" in a broader sense, as coordinates adapted to a timelike curve, which does not necessarily have to be a geodesic. In the non-geodesic case, the definition has to be broadened somewhat, since the metric will not be exactly Minkowski on the curve and not all of the Christoffel symbols will vanish on the curve.
Actually, the metric on origin curve will still be Minkowski, even though the connection coefficients will not vanish on this curve. Even if such coordinates are further extended to allow frame rotation (also done in MTW), the metric exactly on the origin world line remains Minkowski.

PAllen said:
Actually, the metric on origin curve will still be Minkowski, even though the connection coefficients will not vanish on this curve. Even if such coordinates are further extended to allow frame rotation (also done in MTW), the metric exactly on the origin world line remains Minkowski.
can you tell me what page is this discussed in mtw?

kent davidge said:
can you tell me what page is this discussed in mtw?
pp. 327 - 332, in my edition.

kent davidge
PAllen said:
the metric on origin curve will still be Minkowski

Oops, yes, you're right. I've fixed the post.

PAllen said:
pp. 327 - 332, in my edition.
I've been looking. Are you sure that the coordinates they construct there are Fermi coordinates?
I'm asking you that because I'm having a hard time matching what they do with the definitions of Fermi coordinates found across the web.

kent davidge said:
I've been looking. Are you sure that the coordinates they construct there are Fermi coordinates?
I'm asking you that because I'm having a hard time matching what they do with the definitions of Fermi coordinates found across the web.
It is a generalization of Fermi coordinates to allow both rotation around and proper acceleration of the origin world line. If you restrict to the case where the basis is Fermi-Walker transported, and there is no proper acceleration, you end up with traditional Fermi coordinates. But the literature has never been consistent about how restrictive the definition of Fermi coordinates should be, and there are authors who have called the full construction presented in MTW as Fermi Normal Coordinates. Note that at the top of p. 332 they show the reduction to the restrictive definition.

See the following for just one example of an author who calls the complete MTW construction Fermi Normal Coordinates:

https://pdfs.semanticscholar.org/2b9c/8907fa085e26ce87a7d690398f6f00883d7f.pdf

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kent davidge
PAllen said:
the metric on origin curve will still be Minkowski, even though the connection coefficients will not vanish on this curve
How is it possible that the metric remains Minkowskian when its derivatives don't vanish along the curve?

kent davidge said:
How is it possible that the metric remains Minkowskian when its derivatives don't vanish along the curve?

If the curve is a geodesic, then, heuristically, the derivatives pointing in the direction of the curve will all be zero, and the metric will remain Minkowski along the curve.

If the curve is not a geodesic, then, heuristically, the derivatives pointing along the direction of the curve will be just right to compensate for the path curvature of the curve, so that the metric will remain Minkowski along the curve.

cianfa72 and kent davidge
kent davidge said:
How is it possible that the metric remains Minkowskian when its derivatives don't vanish along the curve?

If at some point the metric is diag(-1,1,1,1) then the metric is Minkowskii at that point. There is no requirement for the derivatives to vanish.

cianfa72 and kent davidge
kent davidge said:
How is it possible that the metric remains Minkowskian when its derivatives don't vanish along the curve?
Umm, f(x) = x is zero at x=0, what is its derivative there? F(x,y) =xy is zero along either axis, but what about the partial derivatives on the axes? It’s really that simple an idea.

kent davidge
Sorry coming back to this topic once more. I was reading again the wiki page, and they say "for small t" the coordinates take the form (t,0,0,0).

Is it really only for small t? It doesn't make much sense.

kent davidge said:
they say "for small t" the coordinates take the form (t,0,0,0).

More precisely, it says that the coordinates ##(t, 0, 0, 0)## describe the geodesic near p. However, you are right that for Fermi normal coordinates centered on a geodesic, that is not just for small ##t##, it's for all ##t##, i.e., the geodesic on which the coordinates are centered is described by ##(t, 0, 0, 0)## everywhere.

For Fermi normal coordinates that are centered on a curve ##\gamma## that is not a geodesic, a geodesic passing through the point p, which will have coordinates ##(0, 0, 0, 0)##, and tangent to the (non-geodesic) curve ##\gamma## there, will only be approximated by ##(t, 0, 0, 0)## near p; but for this case the geodesic is not the curve the coordinates are centered on.

kent davidge
I got it.

However that's not what they seem to be describing there. Rather, they are describing a geodesic on which the coordinates are centered. Then wouldn't it be a mistake of them to say that it's only for small t?

kent davidge said:
that's not what they seem to be describing there

Meaning, they are not describing the case for which the curve on which the Fermi normal coordinates are centered is not a geodesic? Yes, that's right, the Wikipedia article is not describing that case.

kent davidge said:
they are describing a geodesic on which the coordinates are centered

Yes.

kent davidge said:
wouldn't it be a mistake of them to say that it's only for small t?

Yes. That's what I said in post #16.

kent davidge

## 1. What are Fermi coordinates?

Fermi coordinates are a set of coordinates used to describe the position and movement of objects in a curved space, such as in general relativity. They are named after physicist Enrico Fermi.

## 2. How are Fermi coordinates different from other coordinate systems?

Fermi coordinates are unique because they are defined in terms of the observer's reference frame. This means that they can be used to describe the position and movement of objects relative to an observer, rather than a fixed point in space.

## 3. What is the significance of Fermi coordinates in physics?

Fermi coordinates are important in the field of general relativity, as they allow us to describe the behavior of objects in a curved space. They are also used in other areas of physics, such as in the study of black holes and gravitational waves.

## 4. How are Fermi coordinates calculated?

To calculate Fermi coordinates, we first need to define a reference frame or observer. Then, we use mathematical equations to transform the coordinates of an object in a curved space to its coordinates in the observer's reference frame.

## 5. Can Fermi coordinates be used in any type of curved space?

Yes, Fermi coordinates can be used in any type of curved space, as long as we have a defined reference frame or observer. They are not limited to any specific type of space or geometry.

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