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We can represent Newton's free fall as a curved trajectory in 4-dimensional spacetime. Time is on the ordinate and space on the abscissa.

By adding an initial speed to the object, we can reproduce Newton's curves in a 4-dimensional Euclidean space-time.

If we also curve the spatial trajectory due to the initial velocity in the same way, we can reproduce the advance of the perihelion and the doubling of the curvature of light.

We can therefore represent the world line of an object as consisting of a time component and a space component. This world line bends in a gravitational field both in its time component and in its space component.

It all seems natural. So why do physicists use the Minkowski metric since it seems useless here ?

Here is an illustration:

https://i.stack.imgur.com/zmeTI.png

which comes from this question :

Why does the speed of an object affect its path if gravity is warped spacetime?

It seems that in a 4D Euclidean space and with some minor correction about the curvature, Newton's law of gravitation is the same as Einstein's. Why bother with the Minkowski metric ?

Note : A representation without Minkowski's space-time and with absolute time corresponds to Lorentz's aether theory (LET). The Euclidean 4D representation of gravitation is therefore a representation in agreement with LET. Thus LET is compatible with general relativity, provided it is assumed that the aether, like matter and energy, obeys gravitation.

externo
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    If you use a Euclidean metric you'll find the speed of light isn't the same for all observers, which is something of a show stopper for relativity. – John Rennie Dec 03 '22 at 16:35
  • Why should the speed of light always be the same? In Lorentz's theory the speed of light is not isotropic. So I wonder if Lorentz's theory wouldn't lead to this Euclidean representation of gravitation. It seems much simpler to me. – externo Dec 03 '22 at 16:39
  • I wonder if this geometry could not explain the equivalence between Rindler's coordinates (acceleration) and curvature (gravitation) – externo Dec 03 '22 at 17:04
  • @JohnRennie Is light invariant to observers in relative frames or light is invariant in relative frames itself. It seems that transformations are for observer version. Both are not true simultaneously, then why they say that c+c=c. – Neil Libertine Dec 03 '22 at 18:33
  • Minkowski metric is to transform time of one frame to multiple of light, and space is already with relative speed. Purpose is to keep speed of light constant for observers which is already constant in classical relativity. – Neil Libertine Dec 03 '22 at 18:40

1 Answers1

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we can reproduce Newton's curves in a 4-dimensional Euclidean space-time.

Actually, a 4D Euclidean spacetime is not even a theoretical possibility, regardless of any evidence. The issue is that for a metric to qualify as a spacetime metric requires that it produce measurements of both distances and durations. Distances and durations are qualitatively different since one is measured with rulers and the other is measured with clocks. With a 4D Euclidean metric there would be nothing to indicate if a specific spacetime measurement was to be made with a clock or with a ruler. So it fails even in theory to represent a possible spacetime metric.

What is instead possible is to use a pair of degenerate metrics, one for distances and one for durations. This is the approach that was investigated by Cartan in the Newton-Cartan formulation of gravity. You could consider the degenerate distance metric to be a 3D Euclidean metric on space, but not a 4D Euclidean metric on spacetime.

It all seems natural. So why do physicists use the Minkowski metric since it seems useless here ?

The useful thing about the Minkowski metric is that in one single metric it captures the fact that spacetime includes both distances and durations. These are qualitatively distinct measures of spacetime involving distinct classes of experimental devices. So it is the simplest structure that can model spacetime from a pure theory standpoint.

Beyond that, of course, is the experimental evidence. The Minkowski metric naturally leads to an invariant speed. The Newton Cartan approach does not. So the experimental evidence supports the Minkowski metric approach and rejects the two degenerate metrics approach.

It seems that in a 4D Euclidean space and with some minor correction about the curvature, Newton's law of gravitation is the same as Einstein's. Why bother with the Minkowski metric ?

Incidentally, the Minkowski metric is only local in gravity. Similarly, in Newton Cartan gravity the Euclidean metric is only local. Spacetime is curved in the presence of tidal gravity in both cases. So either way you cannot keep a flat metric in the presence of tidal gravity.

Dale
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  • All measurements should be made with rulers. The fourth dimension would be just another spatial dimension, and would coincide with time only for objects at rest. 1 second = 300000000 meters only for motionless clocks. Time counted in seconds would not be a physical dimension, as in classical physics – externo Dec 03 '22 at 18:52
  • @externo said “All measurements should be made with rulers”. Then it is not spacetime, it is just space. – Dale Dec 03 '22 at 18:58
  • Ok, so that would be a 4-dimensional space. – externo Dec 03 '22 at 18:59
  • In 4-dimensional space like described in "Riemann manifolds dual to static spacetimes", https://arxiv.org/abs/2004.10505, by Carolina Figueiredo, José Natário, the coordinate time is affine parameter of a 4-dimensional trajectory with infinitesimal length element $dl=c dt$. The infinitesimal length element of the fourth dimension would be then connected to proper time $ds=c d\tau$. – JanG Dec 03 '22 at 19:18
  • However, in such "spacetime" all matter travel along trajectories with velocity $c$, hence the distance in the fourth dimension cannot be measured by ruler. – JanG Dec 03 '22 at 19:23
  • @Jan Gogolin: "Applying Epstein's correspondence to this spacetime leads to the following Riemannian metric: dt² = dτ² + dx² + dy² + dz²" I think that's it. I didn't know such a study existed. In fact, it is general relativity in accordance with Lorentz's theory. So Lorentz's theory is quite capable of dealing with gravitation. – externo Dec 03 '22 at 20:02
  • @JanGogolin fyi, that journal is rather low quality. It ranks 35 out of 55 in its category http://www.eigenfactor.org/projects/journalRank/rankings.php?search=BU&year=2014&searchby=isicat&orderby=articleinfluence This concept has many inherent problems. First, for this thread the main one is that it is not a spacetime. The elements of an Epstein diagram are not events. (Not important to this thread but I object to calling it a dual space, it is not a dual space) – Dale Dec 03 '22 at 21:28
  • @Dale : This representation is consistent with my point of view. The question of knowing what time is is secondary. From Lorentz's point of view there was only one time, the coordinate time, and the proper times of moving objects were false times. This corresponds to this approach of Epstein. Absolute or coordinate time can be modeled as a 4th spatial dimension. – externo Dec 03 '22 at 22:58
  • @externo this may be “consistent with [your] point of view”, but the dual space is not spacetime. Epstein diagrams are not even isomorphic to spacetime. You can discuss this off topic tangent with Jan if you wish, I have no interest in it – Dale Dec 03 '22 at 23:31
  • @JanGogolin it looks like you have provoked what is likely to be a useless argument with that off topic reference. I will leave you and externo to it. I have no interest in joining. – Dale Dec 03 '22 at 23:34
  • @Dale it was not my intent. My feeling about that paper was mixed but it was peer reviewed so I have thought it would be OK to mentioned it. – JanG Dec 04 '22 at 07:15
  • @externo I am afraid Dale is right if he says Lorentz theory is not general relativity, hence the theory of gravitation. If you have other opinion write down the corresponding equations and try to publish it. – JanG Dec 04 '22 at 07:25