Can exact Lorentzian kinematics emerge from Euclidean space?

[Ans]

TL;DR

Can a theory on 4D Euclidean space produce exact Lorentz transformations and an invariant signal speed while the underlying metric stays Euclidean and no Wick rotation is used?

Can exact Lorentzian kinematics arise from a theory formulated on a four-dimensional Euclidean space without a Wick rotation?

Suppose a theory is formulated on a four-dimensional Euclidean space with a positive-definite metric.

Assume that:

• the metric always remains Euclidean;
• no coordinate is analytically continued (t \to it);
• no real coordinate transformation is claimed to convert the Euclidean metric into a Minkowski metric.

Can such a theory nevertheless contain a physical sector in which the Lorentz transformations hold exactly, with an exact finite invariant signal speed?

I do not mean approximate Lorentz invariance in a low-energy or long-wavelength limit. I mean exact special-relativistic kinematics.

The usual signature argument shows that a positive-definite Euclidean metric cannot be transformed into a Lorentzian metric by a real nonsingular coordinate transformation.

But does it also imply the stronger statement that exact Lorentzian kinematics cannot arise in a theory whose underlying geometric structure remains Euclidean?

In other words, are these two statements equivalent?

1. A Euclidean metric cannot be transformed into a Lorentzian metric.
2. A theory formulated on Euclidean space cannot produce exact Lorentz transformations in its physical sector.

The first statement is standard. Does the second actually follow from it?

 

[Sagittarius A-Star]

Suppose a theory is formulated on a four-dimensional Euclidean space with a positive-definite metric.

Assume that:

• the metric always remains Euclidean;
• no coordinate is analytically continued (t \to it);
• no real coordinate transformation is claimed to convert the Euclidean metric into a Minkowski metric.

Can such a theory nevertheless contain a physical sector in which the Lorentz transformations hold exactly, with an exact finite invariant signal speed?

Given the assumptions in the paper "Nothing but Relativity", including group properties of a transformation (two boosts in one direction are equivalent to one boost of the same form),
https://arxiv.org/abs/physics/0302045
the Lorentz transformation as physically relevant follows only for one of three cases.

Transformartion equation (27) in the article can be derived without the 2nd postulate.
K<0 : can be excluded (causality would be not invariant)
K=0 : makes it the Galilean transformation
K>0 : makes it the Lorentz transformation
An experiment is needed to decide between Galilean- or Lorentz-transformation.

From this with twin paradox-experiments we know, that the Lorentz transformation is physically valid.

With unit system $c=1$, in the $xt$-plane:
LT
$(1)\ x'=\gamma (x-vt)$
$(2)\ t'=\gamma (t-vx)$
Add (1) and (2):
$(3)\ x'+t'=\gamma (x+t-v(x+t))=\gamma(1-v)(x+t)$
Subtract (2) from (1):
$(4)\ x'-t'=\gamma (x-t+v(x-t))=\gamma(1+v)(x-t)$
Multiply (3) and (4):
$(5)\ x'^2-t'^2=\gamma^2(1-v^2)(x^2-t^2)$
$$x'^2-t'^2=x^2-t^2$$
The Lorentz transformation is equivalent to Minkowski metric and not to 4D-Euclidean metric.

 

[PeterDonis]

Can such a theory nevertheless contain a physical sector in which the Lorentz transformations hold exactly, with an exact finite invariant signal speed?

The concept of "signal speed" would not even make sense in a Euclidean theory such as you describe, since such a theory has no notion of "time". In standard GR terminology, all four coordinates on the 4D Euclidean space are spacelike.

 

[PeterDonis]

the Galilean transformation

A Galilean spacetime is not the same thing as a 4D Euclidean space.

 

[Sagittarius A-Star]

A Galilean spacetime is not the same thing as a 4D Euclidean space.

That's true, but I didn't claim that it is the same. With "4D-Euclidean metric" I referred only to the OP.

 

[Ibix]

I think it depends what you mean. You can have Lorentzian dynamics in Lorentz Ether Theory, which has a 3d Euclidean space and a separate time parameter, and the Lorentz behaviour that provides the illusion of a 4d spacetime happens for some unspecified reason.

If you mean a 4d Euclidean space where one direction is timelike, I don't see what would pick out the dimension to call timelike. You can certainly draw such things (it's what a Minkowski diagram is), but there we impose a Minkowski metric on the diagram artificially and just arbitrarily pick a couple of directions to call null.

So my immediate thought is that you can impose Lorentzian dynamics on other geometries, but it has to be put in manually.

 

[Roberto Pavani]

Only if you solve the problem of time, as PeterDonis pointed out, in R4 you have four spatial directions and no time.

Some related work: analogue gravity (Barceló, Liberati & Visser, Analogue Gravity, Living Rev. Relativity 14, 3, 2011), effective Lorentzian metrics emerging from non-relativistic substrates. It's close to the spirit of the question, though not a full answer: it gives effective/emergent Lorentz invariance, not the exact kinematics the OP is asking about.

 

[Sagittarius A-Star]

A Galilean spacetime is not the same thing as a 4D Euclidean space.

Yes. It is the unphysical first case $K=-1$ in posting #2, which is an Euclidean rotation in the $xt$ plane.
$\begin{pmatrix} x'\\ t' \end{pmatrix} = \frac{1}{\sqrt{1-Kv^2}} \begin{pmatrix} 1 & -v\\ -Kv & 1 \end{pmatrix}$

$\begin{pmatrix} x'\\ t' \end{pmatrix} = \frac{1}{\sqrt{1+v^2}} \begin{pmatrix} 1 & -v\\ v & 1 \end{pmatrix} \begin{pmatrix} x\\ t \end{pmatrix}$


$\begin{pmatrix} x'\\ t' \end{pmatrix} = \begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} x\\ t \end{pmatrix}, \qquad \theta=\arctan v$

 

[PeterDonis]

It is the unphysical first case $K=-1$ in posting #2, which is an Euclidean rotation in the $xt$ plane.

In this case $t$ isn't timelike, so there's no justification for even calling it a "time". But I guess one could view that as yet another reason for calling it "unphysical".

 

[PeterDonis]

a 4d Euclidean space where one direction is timelike

There is no such thing. "Timelike" means a squared interval of the opposite sign from spacelike. There is no such thing in a Euclidean space; all squared intervals have the same sign.

 

[Dale]

signal speed

I also don’t know what speed would mean in such a manifold.

The Lorentz transform treats time and space differently. Since in your space there is nothing that distinguishes the different directions any method that ends in the Lorentz transform would not be natural.

 

[Ibix]

There is no such thing. "Timelike" means a squared interval of the opposite sign from spacelike. There is no such thing in a Euclidean space; all squared intervals have the same sign.

That's kind of the point I was trying to make - all the directions are the same, so which one is time?

The only solution I can think of is something like a Minkowski diagram, which is a Euclidean plane where we impose Lorentzian physics by fiat rather than having it follow from the geometry.

 

[PeterDonis]

all the directions are the same, so which one is time?

None of them.

The only solution I can think of is something like a Minkowski diagram, which is a Euclidean plane where we impose Lorentzian physics by fiat rather than having it follow from the geometry.

You can't "impose Lorentzian physics by fiat" in a geometry that has all spacelike intervals. The whole point of Lorentzian physics is that the geometry is Minkowski, not Euclidean. The OP even rules out the dodge of using Wick rotation to construct a kinda sorta correspondence between the two.

 

[Ibix]

You can't "impose Lorentzian physics by fiat" in a geometry that has all spacelike intervals.

I could observe the paths of some particles constrained (for the sake of simplicity) to move in 1d. Then I could plot them on a Minkowski diagram. Then I could cut off the top half of the diagram and pass it to you, and you could then fill in the top half using your knowledge of relativity and (give or take measurement precision etc) it would exactly reproduce the top half of my diagram. That is, the top half of the diagram is predictable from the bottom half given knowledge of relativistic physics. What is that if not relativistic physics imposed on a Euclidean plane?

 

[PeterDonis]

the top half of the diagram is predictable from the bottom half given knowledge of relativistic physics. What is that if not relativistic physics imposed on a Euclidean plane?

The fact that I can draw the diagram of relativistic physics on a Euclidean plane does not mean I am imposing relativistic physics on a Euclidean plane. The diagram is just a diagram of the physics; it's not the actual physics. The map is not the territory.