- #36
PeterDonis
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Killtech said:as it is the same set everything formulated on one can be translated to an equivalent on the other via the identity bijection
This is the "identity" for the topological space ##\mathbb{R}^2##. But it is not the identity for the manifold "Euclidean plane", because this transformation changes the geometry from the Euclidean plane to something else. So you need to be very clear about exactly what a given transformation does and does not keep the same. At least part of your confusion appears to come from failing to do this.
Killtech said:how is the geometry of space measured?
By the distances between points. In the case of the manifold ##\mathbb{R}^2##, specifying the distance according to your chosen metric between all pairs of points fixes the geometry. And your transformation above does not preserve this: if we take two points A and B, the distance between them according to the Euclidean metric is not the same as the distance between them according to your alternate metric. So again, your transformation changes the geometry.
If you are really concerned about keeping things independent of your choice of units, you can rephrase all of the above in terms of ratios of distances between different pairs of points: for example, take two pairs of points, (A, B) and (C, D), and look at the ratio of the distances between them according to the two different metrics. Everything I said still applies: if any of these ratios change, you have changed the geometry.
Killtech said:using the Euclidean metric the distance between ##d_2(\begin{pmatrix}0 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ 2\end{pmatrix})## is twice as long as ##d_2(\begin{pmatrix}0 \\ 0\end{pmatrix}, \begin{pmatrix}\sqrt 0.5 \\ \sqrt 0.5\end{pmatrix})## and this ratio has no unit anymore.
Yes.
Killtech said:But using the ##d_\infty## metric the same ratio changes to ##2 \sqrt 2##.
Yes. And this is an example of what I said above, that your transformation changes the geometry.
I'm not sure I can make sense of your specific comments about metrology; I think you have confused yourself by failing to pay proper attention to the points I've made above. But I'll try to illustrate what I've said with an example, since you mention SI units: the definition of the SI second in terms of the radiation emitted by a specific hyperfine transition in Cesium is the equivalent of picking a particular pair of points (A, B) in spacetime and calling the spacetime distance between them (which in this case is a time, since the two points are timelike separated) the "standard" distance, and expressing all other spacetime distances as ratios with that standard distance. The SI definition of the meter then just extends this to spacelike distances as well as timelike distances, by fixing the speed of light--which is really the chosen ratio of "space distance" units to "time distance" units--to a particular value. But we could choose different definitions of the second and the meter (and we previously did), without changing any ratios of spacetime distances, i.e., without changing the spacetime geometry. All we would change is which particular spacetime distances we called the "standard" ones.