Why Is Minkowski Spacetime Non-Euclidean?

[PeroK]

is analogous to describing a plane as a sphere of infinite radius

You can only push this sort of flaky mathematics so far before you end up with a contradiction.

 

[robphy]

You can only push this sort of flaky mathematics so far before you end up with a contradiction.

Descriptions are for intuition.
Equations are needed to support the intuition… and provide the context behind the description.
Descriptions alone (without equations) can lead to all sorts of misconceptions.

 

[PeroK]

Descriptions are for intuition.
Equations are needed to support the intuition… and provide the context behind the description.
Descriptions alone (without equations) can lead to all sorts of misconceptions.

The sphere and the plane are topologically distinct. The sphere is locally homeomorphic to the plane. Which is a better description than saying the plane is an infinite sphere- which I suggest is mathematically meaningless.

 

[robphy]

The sphere and the plane are topologically distinct. The sphere is locally homeomorphic to the plane. Which is a better description than saying the plane is an infinite sphere- which I suggest is mathematically meaningless.

Like I said, descriptions need context.
They can’t stand on their own.

Here are some equations in the Desmos below.
$E=1/(c_{max})^2$ in the following.
E=1 is Minkowski, E=0 is Galilean.
Call the terms whatever you want.
This construction is consistent with the physics. Feel free to suggest what may be better descriptions.
https://www.desmos.com/calculator/kv8szi3ic8

 

[Orodruin]

is analogous to describing a plane as a sphere of infinite radius [as one does when describing plane waves].

I think I formally disagree here too. The plane wave approximation is a good approximation far from the source. It is not a spherical wave. (Or vice versa)

A plane is indeed the limit (pointwise) of a set of spheres with increasing radius. It is not a sphere of infinite radius.

The entire mathematical concept of infinity is a tricky one and us physicists often use the nomenclature a bit haphazardly. If you assign infinity as the speed of something you open up the door to questions such as what $\infty + 1$ or $2\infty$ are.

But to the Galilean spacetime, the point is that there is no bound on the speed of information transfer.

 

[Sagittarius A-Star]

E=1 is Minkowski, E=0 is Galilean.
Call the terms whatever you want.
This construction is consistent with the physics. Feel free to suggest what may be better descriptions.

Also Einstein used a sloppy formulation. I understand, what he meant:

The Galilei transformation can be obtained from the Lorentz transformation by substituting an infinitely large value for the velocity of light $c$ in the latter transformation.

Source:
https://en.wikisource.org/wiki/Rela...art_I#Section_11_-_The_Lorentz_Transformation

I think, mathematicians would use a formulation with limit at infinity.
$\lim_{c \rightarrow \infty}{(\gamma (x-vt))} = x-vt, \ \ \ \ \$ $\lim_{c \rightarrow \infty}{(\gamma (t-vx/c^2))} = t$

Edit:

Source 1: Calculation of ...
Limit[(x - v t)/Sqrt[1 - v^2/c^2], c -> Infinity]
https://www.wolframalpha.com/input?i=Limit[(x+-+v+t)/Sqrt[1+-+v^2/c^2],+c+->+Infinity]

Source 2: Calculation of ...
Limit[(t - (v x)/c^2)/Sqrt[1 - v^2/c^2], c -> Infinity]
https://www.wolframalpha.com/input?i=Limit[(t+-+(v+x)/c^2)/Sqrt[1+-+v^2/c^2],+c+->+Infinity]

 

[robphy]

I think, mathematicians would use a formulation with limit at infinity.
$\lim_{c \rightarrow \infty}{(\gamma (x-vt))} = x-vt, \ \ \ \ \$ $\lim_{c \rightarrow \infty}{(\gamma (t-vx/c^2))} = t$

For clarity (and to minimize having to worry about limits), I formulate the idea as
$$
\left( \begin{array}{c} t' \\ \frac{x'}{c_{light}} \end{array} \right)
=
\left(
\begin{array}{cc}
\frac{1}{\sqrt{1-E\beta^2}} & \frac{E\beta}{\sqrt{1-E\beta^2}}
\\
\frac{\beta}{\sqrt{1-E\beta^2}} &
\frac{1}{\sqrt{1-E\beta^2}} & \end{array}
\right)\\
\left( \begin{array}{c} t \\ \frac{x}{c_{light}} \end{array} \right)
$$
where $\beta=v/c_{light}$, where $c_{light}=3\times10^8\ \mbox{m/s}$ [playing the role of a conversion constant].
For $E=0$ (galilean) or $E=+1$ (minkowskian) , one could think of $E$ as if it were $\left(\frac{c_{light}}{c_{max}}\right)^2$.
For $E=-1$, we essentially have a Euclidean rotation matrix.

These equations (based on Appendix A of Yaglom's book from https://www.physicsforums.com/threads/why-is-minkowski-spacetime-non-euclidean.1016402/post-6647305 ) are used in the "formulae" section of my
spacetime diagrammer https://www.desmos.com/calculator/kv8szi3ic8 ,
which visualizes the metric in the Euclidean, Galilean, and Minkowski cases.

(See also: Richter-Gebert's Perspectives on Projective Geometry: A Guided Tour Through Real and Complex Geometry https://www.amazon.com/dp/3642172857/?tag=pfamazon01-20 )

 

[PeroK]

Also Einstein used a sloppy formulation. I understand, what he meant:

Source:
https://en.wikisource.org/wiki/Rela...art_I#Section_11_-_The_Lorentz_Transformation

I think, mathematicians would use a formulation with limit at infinity.
$\lim_{c \rightarrow \infty}{(\gamma (x-vt))} = x-vt, \ \ \ \ \$ $\lim_{c \rightarrow \infty}{(\gamma (t-vx/c^2))} = t$

No matter how large you make $c$ I can choose a large $v$ and we see there is actually no physical difference in the geometry depending on the value of $c$.

In that sense the limit as $c \to \infty$ is not the Galilean geometry.

This is why mathematical analysis is needed so that we know which limits are valid and which are not.

 

[PeroK]

PS the simple and mathematically valid statement is that in Galilean relativity there is no invariant speed. Introducing an invariant speed of "infinity", IMO, is an unnecessary artifice.

 

[Sagittarius A-Star]

In that sense the limit as $c \to \infty$ is not the Galilean geometry.

In that sense, of course!

The "limit-equations" in posting #41 are valid, if they are regarded as purely mathematical equations, without any reference to physics.

For example, no dependency between the mathematical quantities "c" and "v" is assumed, besides the dependencies stated in those equations. You can't write down the GT and assume at the same time, that actual physics, like described in postulate 2 of SR, is valid.

 

[vanhees71]

Isn't the fact that the spacetime metric is not Euclidean due to the fact that there is a maximum velocity in our universe?
maximum velocity $\implies$ everyone agrees on its value $\implies$ derivation of spacetime interval ($ds^2=-dt^2+dx^2+dy^2+dz^2$)
Is this reasoning wrong? Doesn't the fact that spacetime is not Euclidean derive from the fact that there is a maximum velocity?

In a way that's right. You can derive the Lorentz transformation from the assumption that the special principle of relativity holds and assuming that for inertial observers space is Euclidean and time is homogeneous. Using these assumptions you get either the Galilei transformation and from this you can reconstruct Newtonian spacetime (having no limiting speed) or Minkowski spacetime (having a limiting speed). The Euclidean metric also occurs formally in the solutions, but then you want the transformations between inertial frames to form a group, and when using the Euclidean-metric solution then you'd not be able to set up a causality structure. That's possible in Minkowski spacetime, because for time- or light-like separated events the part of the Lorentz group that is continuously connected with the identity, the proper orthochronous Lorentz group $\mathrm{SO}(3,1)$ (in your sign convention for the metric) the time ordering is invariant, and that's why only time- or light-like separated events can be causally connected. For more details, see. e.g.,

V. Berzi and V. Gorini, Reciprocity Principle and the Lorentz Transformations, Jour. Math.
Phys. 10, 1518 (1969), https://doi.org/10.1063/1.1665000

 

[Dale]

I am interested to see, what individual people take as a good explanation respectively as a good answer to the question. I am interested to see what people take as a meaningful explanation. Or what they think is necessary for a meaningful explanation.

For me individually, spacetime is clearly non-Euclidean since spacetime "distances" are measured with both clocks and rulers. I don't know how the distinction could possibly be more blatant.

 

[Dale]

Why, may I ask, do you prefer this sign convention over the other one?

I usually use the convention $ds^2=-c^2 dt^2+dx^2+dy^2+dz^2$ and $c^2 d\tau^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2$, so I can easily switch between the timelike and spacelike conventions.

 

[robphy]

V. Berzi and V. Gorini, Reciprocity Principle and the Lorentz Transformations, Jour. Math.
Phys. 10, 1518 (1969), https://doi.org/10.1063/1.1665000

The $K$ in that paper corresponds to the $E$ in my post #42
in the sense $E=sign(K)$:

cases (pg 1522)
["$K$ is a universal constant having the dimensions of an inverse-square velocity"]
(my $E$ is dimensionless because I have $c_{light}^2=(3\times 10^8\mbox( m/s))^2$ (instead of $1$) in the numerator):

• (A) Berzi K>0 (my E=+1) : Minkowski [eq 40]
• (B) Berzi K=0 (my E=0) : Galilean [eq 41]
• (C) Berzi K<0 (my E=-1): Euclidean [eq 42]

 

[PAllen]

No matter how large you make $c$ I can choose a large $v$ and we see there is actually no physical difference in the geometry depending on the value of $c$.

In that sense the limit as $c \to \infty$ is not the Galilean geometry.

This is why mathematical analysis is needed so that we know which limits are valid and which are not.

Well, while this is true as stated, there is a sense in which it is meaningful to talk about such a limit. Given any physical set up (distances in terms of reference objects, speeds define from these plus a notion of ideal clock), then as c is taken to approach infinity, the error of any observation compared to Galilean approaches zero. This is important because having speeds very low compared to c does not make relativistic effects arbitrarily small. No matter how small the speeds compared to c, if the system is large enough, relativistic effects are also large. Both speeds and distances must be 'small' compared to c or ct for relativistic effects to be insignificant.

 

[PeroK]

Well, while this is true as stated, there is a sense in which it is meaningful to talk about such a limit. Given any physical set up (distances in terms of reference objects, speeds define from these plus a notion of ideal clock), then as c is taken to approach infinity, the error of any observation compared to Galilean approaches zero. This is important because having speeds very low compared to c does not make relativistic effects arbitrarily small. No matter how small the speeds compared to c, if the system is large enough, relativistic effects are also large. Both speeds and distances must be 'small' compared to c or ct for relativistic effects to be insignificant.

Why invent something (an infinite or near infinite $c$) that is simply unnecessary?

 

[Sagittarius A-Star]

Why invent something (an infinite or near infinite $c$) that is simply unnecessary?

In the "limit-equations" in posting #41, c is not set to infinite, nor to a finite value "near infinite", as can be see from the definition of limit at infinity.

 

[PeroK]

In the "limit-equations" in posting #41, c is not set to infinite, nor to a finite value "near infinite", as can be see from the definition of limit at infinity.

In post #41 you reach the sort of contradiction I am talking about. You show that, if we choose a fixed $v$, then in the limit as $c \to \infty$, we have $\gamma \to 1$.

But, as $c \to \infty$ it is not the case that for all $v$ we have $\gamma(v) \to 1$. Because, any $v < c$ is allowed.

And, in fact, the whole idea of setting $c =1$ shows that as $c \to \infty$ the physics does not gradually change depending on how large we make $c$.

In those terms Galilean relativity is not a continuous limit of Special Relativity as $c \to \infty$.

 

[Sagittarius A-Star]

In those terms Galilean relativity is not a continuous limit of Special Relativity as $c \to \infty$.

That is what you wrote already in #43 and I replied to in #45.

 

[PeroK]

That is what you wrote already in #43 and I replied to in #45.

I don't agree with post #45. The conclusion that if you increase the invariant speed $c$ then you approach Galilean relativity in some limiting sense is false. The limit argument doesn't work, IMO.

You only get Galilean relativity by plugging $c = \infty$ directly into the equations. That works, but seems an unnecessary, unphysical artifice.

 

[Sagittarius A-Star]

The conclusion that if you increase the invariant speed $c$ then you approach Galilean relativity in some limiting sense ...

I didn't write or conclude this in #45.

You only get Galilean relativity by plugging $c = \infty$ directly into the equations.

That is the sloppy formulation from Einstein, I wanted to avoid.

 

[PeroK]

That is the sloppy formulation from Einstein, I wanted to avoid.

You can't avoid it. That's my point. The limit argument fails because the Lorentz Transformation entails transforming between any IRFs. And, however large you make $c$ you have all $v < c$ to consider.

 

[Sagittarius A-Star]

You can't avoid it. That's my point. The limit argument fails because the Lorentz Transformation entails transforming between any IRFs. And, however large you make $c$ you have all $v < c$ to consider.

That is again, what you wrote already in #43 and I replied to in #45:

... The "limit-equations" in posting #41 are valid, if they are regarded as purely mathematical equations, without any reference to physics.

For example, no dependency between the mathematical quantities "c" and "v" is assumed, besides the dependencies stated in those equations. You can't write down the GT and assume at the same time, that actual physics, like described in postulate 2 of SR, is valid.

 

[PeroK]

That is again, what you wrote already in #43 and I replied to in #45:

But, there is a mathematical dependency between $v$ and $c$. Relativity is about the group of Lorentz transformations. Not just a transformation for a given $v$.

What is valid is that, for a fixed $c$, as $v \to 0$ then the LT does tend to the Galilean transformation. That's why Galilean relativity is valid for $v << c$.

But SR definitely does not tend to Galilean relativity as $c \to \infty$.

 

[vanhees71]

The Galilean limit of the Poincare group is indeed not trivial. This becomes particularly clear in quantum theory: While the Poincare group has not non-trivial central charges, and you just have to deal with the unitary irreps, the Galilei group has a central charge, which physically is the mass, and you have to deal with ray representations (or the unitary representations of the corresponding central extension of the "classical" Galilei group).

 

[Sagittarius A-Star]

What is valid is that, for a fixed $c$, as $v \to 0$ then the LT does tend to the Galilean transformation.

If you mean $\lim_{v \rightarrow 0}{}$, then you get only a special case of the GT

$\lim_{v \rightarrow 0}{(\gamma (x-vt))} = x-0 \cdot t = x, \ \ \ \ \$ $\lim_{v \rightarrow 0}{(\gamma (t-vx/c^2))} = t$

 

[PeroK]

If you mean $\lim_{v \rightarrow 0}{}$, then you get only a special case of the GT

$\lim_{v \rightarrow 0}{(\gamma (x-vt))} = x-0 \cdot t = x, \ \ \ \ \$ $\lim_{v \rightarrow 0}{(\gamma (t-vx/c^2))} = t$

You're missing the point. If $v$ is small, then $\gamma \approx 1$. And you can study a subset of SR where $v$ is small and you have approximately the Galilean transformation.

If $c$ is large, then not all gamma factors are approximately $1$. You don't have an approximation to Galilean relativity.

 

[Sagittarius A-Star]

If $v$ is small, then $\gamma \approx 1$. And you can study a subset of SR where $v$ is small and you have approximately the Galilean transformation.

That's correct, except if the $x$ in $\gamma (t-vx/c^2)$ is very large, as @PAllen mentioned in #50.

 

[Dale]

as c is taken to approach infinity, the error of any observation compared to Galilean approaches zero

This is really the scientifically important fact. For SR to be a valid theory the differences between it and Galilean relativity must be smaller than the experimental uncertainty in any domain where Galilean relativity has been experimentally validated. You can express that requirement with several different limits, but the important fact is that in any of those limits the difference must approach zero.