Wiki definition of Fermi coordinates

[kent davidge]

Is the wikipedia definition of Fermi coordinates accurate? https://en.wikipedia.org/wiki/Fermi_coordinates

 

[PeterDonis]

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.]

 

[kent davidge]

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

 

[PeterDonis]

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.

 

[PAllen]

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.

 

[kent davidge]

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?

 

[PAllen]

can you tell me what page is this discussed in mtw?

pp. 327 - 332, in my edition.

 

[PeterDonis]

the metric on origin curve will still be Minkowski

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

 

[kent davidge]

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.

 

[PAllen]

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

 

[kent davidge]

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?

 

[PeterDonis]

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.

 

[pervect]

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.

 

[PAllen]

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.

 

[PeterDonis]

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?

 

[PeterDonis]

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.

they are describing a geodesic on which the coordinates are centered

Yes.

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.