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Spacetime manifolds, as well as their simplified form, twodimensional spacetime diagrams, are always Lorentzian (Lorentzian metric $ds^2 = dt^2 - dx^2$). Normally, the Lorentzian metric is hidden because spacetime is represented on a Euclidean support such as a sheet of paper, and we can measure the corresponding Euclidean distance ($ds^2 = dt^2 + dx^2$) even though it is completely meaningless. For instance, the Lorentzian metric in the diagram is $4$, but the measurement of the worldline yields the meaningless distance of $5.83$.

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Is there a way to get rid of this superfluous Euclidean metric of the support of observation (such as a sheet of paper), or is the Euclidean metric a necessary accessory of the Lorentzian spacetime manifold?

Moonraker
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    Are you aware of the derivation of Lorentz transformations from a metric background and relativity? – FlatterMann Oct 03 '22 at 09:15
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    You are drawing the diagram on a piece of paper that has two spacelike dimensions i.e. you are drawing a timelike axis on a spacelike dimension. But unless you have access to Lorentzian paper I don't see how you can get around this. You just have to remember not to take distances on the paper literally. – John Rennie Oct 03 '22 at 10:21
  • Possible duplicates: https://physics.stackexchange.com/q/129187/2451 , https://physics.stackexchange.com/q/729771/2451 and links therein. – Qmechanic Oct 03 '22 at 10:33
  • @John Rennie - Thank you for your answer which helps! I understand it in the sense that currently it is not tried to "get rid" of this superfluous Euclidean metric, and that it is just not taken into account. – Moonraker Oct 03 '22 at 11:44
  • Is there a way to get rid of this superfluous Euclidean metric… I don't get it. Do not draw spacetime diagrams and you are set. There are ways to visualize spacetime other than spacetime diagrams (e.g. animations). – A.V.S. Oct 04 '22 at 06:59
  • @A.V.S. - What kind of animations? How are light rays and light cones represented (with null Lorentzian spacetime interval zero)? – Moonraker Oct 05 '22 at 08:14
  • What kind of animations? I am thinking about something like this (and other gifs from the same page), with the screen representing space and animation frames corresponding to evolution w.r.t. time. How are light rays and light cones… on this kind of animation one could do instantaneous snapshots of wavefronts corresponding to signal emission at particular events. – A.V.S. Oct 06 '22 at 10:50
  • @A.V.S. - Interesting idea, but I would say that this is rather representing R3 space (in time slices). The animation provides the 4th dimension of time, and in your brain (or in the data base of a computer, e.g. in the form of a .csv file) there are formed Euclidean fourdimensional manifolds. The meaningless Euclidean metric is still dominating the hidden Lorentzian metric. – Moonraker Oct 06 '22 at 14:06
  • I am beginning to understand. You seem to believe that collection of points that could be labelled by four real numbers carry an inherent euclidean metric. But this is not true, we only have an inherent manifold topology of $\mathbb{R}^4$, but there is no “god-given” euclidean metric that goes with it, and there could be other topologies on a spacetime. See wiki, – A.V.S. Oct 06 '22 at 16:04
  • @A.V.S. - Very good comment in the heart of my question: On one hand there is no "god-given" metric, on the other hand it seems hard to get rid of the Euclidean metric. If I have a .csv file spacetime manifold with two points 1;1;1;1 and 6;4;1;1, this corresponds approximately to the diagram I sketched. Any person who reads and interprets the file will choose automatically the Euclidean diagram for the description, and the Euclidean metric appears. – Moonraker Oct 06 '22 at 16:58

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Here is the attempt of an answer:

The Lorentzian metric of a spacetime diagram is $ds^2= dt^2 - dx^2$ (c=1), and by this, it is a function of the coordinates dt and dx. The Euclidean metric $ds^2= dt^2 + dx^2$ has the same arguments, and for every Lorentzian spacetime interval we may define a Euclidean interval, simply by replacing the minus sign with a plus sign. That means that the Lorentzian spacetime implies automatically the possibility of the existence of a Euclidean metric, and there is no way to "get rid" of this possibility.

The resulting Euclidean metric is not completely meaningless: It is part of a Euclidean manifold which corresponds to the sheet of paper on which is sketched the spacetime diagram. This Euclidean geometry is not referring to the underlying physical reality (which is described by the Lorentzian metric) but it is describing the "interface" of observation. The same is true for fourdimensional spacetime manifolds, we may always define a Euclidean metric where the minus sign of the Lorentzian metric is substituted with a plus sign. By consequence, we may distinguish two metrics with respect to spacetime: The Euclidean metric of the observer and the Lorentzian metric of the underlying observed universe.

Your comments are welcome.

Moonraker
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