Spacetime interval and basic properties of light

[Dale]

It looks like magic that ds is invariant. I have the feeling however that the invariance of ds is no coincidence and should follow in a logical way from the general thoughts behind the whole idea.

If you start with the two postulates of relativity, then you can derive the Lorentz transform. Once you have the Lorentz transform you can easily prove the invariance of the spacetime interval.

I just prefer to take a different approach: Just start with the spacetime interval and examine the consequences. You have to start with something, so I like starting with the one formula that you can do the most with. Because you started with one thing, if you find anything that doesn’t match experiments then you know exactly what went wrong, there is only one place to look.

 

[robphy]

good explanation of the concept of invariance. This helps. It looks like magic that ds is invariant. I have the feeling however that the invariance of ds is no coincidence and should follow in a logical way from the general thoughts behind the whole idea.

Here's how I think of the structure (as inspired by the various relativity courses I took and the books I consult).

Euclidean geometry, position-vs-time graphs (with the Newtonian/Galilean structure), and the Minkowski spacetime diagram all have the structure of an affine space (for simplicity, make it a "vector space").

With a vector space, we can add and scalar-multiply.
We can compare the relative sizes of parallel vectors.
But we can't assign a length or magnitude to a vector.

We need additional structure: a metric (a "circle"), some kind of structure [a tensor] that assigns a square-magnitude to a vector.
We can represent this by a https://en.wikipedia.org/wiki/Quadric .
Choose one to represent the tips of all vectors from the origin to be declared as "unit vectors".
The equation of this figure effectively chooses the form of the "invariant" quantity.

(This is similar to the notion of doing
https://en.wikipedia.org/wiki/Straightedge_and_compass_construction
or https://en.wikipedia.org/wiki/Poncelet–Steiner_theorem
rather than straightedge-only constructions.)

(The metric "circle" also defines when two vectors are orthogonal,
as well as provide a measure of angles between radius vectors,
and maybe angles between vectors tangent to the "circles").

If you ask for transformations that preserve that quadric (i.e. keep the magnitudes of vectors unchanged),
you have determined the "boosts" or "rotations".
(related: https://en.wikipedia.org/wiki/Erlangen_program
https://encyclopediaofmath.org/wiki/Erlangen_program )

Example: Modify the E-slider to choose the metric figure
https://www.desmos.com/calculator/kv8szi3ic8

For special relativity, rather than choose a distinguished hyperbola,
I think I can choose a distinguished parallelogram (what will become a "light-clock diamond").

 

[PeterDonis]

All one needs is one timelike vector.

Why? If you're going to consider boosts, you have to have multiple timelike vectors since boosts map timelike vectors to different timelike vectors.

A position-vs-time graph (with its Newtonian/Galilean structure), a Minkowski spacetime diagram, and the usual Euclidean plane
all have the structure of an affine space.

Affine space structure is not enough, is it? Boosts are transformations on a vector space, not on an affine space.

 

[robphy]

All one needs is one timelike vector.

Why? If you're going to consider boosts, you have to have multiple timelike vectors since boosts map timelike vectors to different timelike vectors.

Maybe my phrasing is incomplete.
I'll repeat my revision from #105:

For clarity, maybe I should rephrase:​

a "spacelike" vector is defined as a vector that is orthogonal to some timelike vector.​

(Geometrically, fix a point-event...​

a "timelike" vector from that point-event is a radius vector to a "circle" centered at that point-event ;​

a "spacelike" vector from that point-event is a vector parallel to the tangent to a "circle" centered at that point-event)​

​

In the revised definition,
the spacelike vectors from the origin (which are parallel to tangent-vectors to the metric "circle" centered at the origin)

• in special relativity, point outside of the light-cone (and can regarded as radius vectors to another hyperboloid)
• in Newtonian/Galilean spacetime, point tangent to the absolute-simultaneity hyperplane

as seen in https://www.desmos.com/calculator/kv8szi3ic8 (vary the E-slider).

 

[PeterDonis]

a "spacelike" vector is defined as a vector that is orthogonal to some timelike vector.

Yes, this part I understand.

(Geometrically, fix a point-event...a "timelike" vector from that point-event is a radius vector to a "circle" centered at that point-event ;a "spacelike" vector from that point-event is a vector parallel to the tangent to a "circle" centered at that point-event)

This part I don't understand.

 

[robphy]

(Geometrically, fix a point-event...​

a "timelike" vector from that point-event is a radius vector to a "circle" centered at that point-event ;​

a "spacelike" vector from that point-event is a vector parallel to the tangent to a "circle" centered at that point-event)​

This part I don't understand.

It's the idea that orthogonality is defined by generalizing the idea
that: at a point of a circle, the tangent is orthogonal to the radius.

So, once we have determined a vector is spatial according to some inertial observer,
then it is spacelike (it points outside of the light cone).

(From https://www.physicsforums.com/threads/minkowski-normal-v-euclidean-normal.742230/post-4683143 ...)

As Minkowski suggests (
https://en.wikisource.org/wiki/Translation:Space_and_Time (near [9] )
https://en.wikisource.org/wiki/Page:De_Raum_Zeit_Minkowski_018.jpg ),
where he uses "normal" instead of "orthogonal" or "perpendicular"

Let us decompose a vector drawn from O towards x, y, z, t into its four components x, y, z, t. If the directions of the two vectors are respectively the directions of the radius vector OR of O at one of the surfaces $\pm F=1,$ and additionally a tangent RS at the point R of the relevant surface, then the vectors shall be called normal to each other. Accordingly
$$c^{2}tt_{1}-xx_{1}-yy_{1}-zz_{1}=0,$$ which is the condition that the vectors with the components x, y, z, t and $\displaystyle \left(x_{1}\ y_{1}\ z_{1}\ t_{1}\right)$ are normal to each other.

This construction extends to the Newtonian/Galilean case where their "spacelike vectors" are tangent to its "Galilean circle". In this case, the "Galilean unit circle" and "t=1" hyperplane of simultaneity are on the same hyperplane. (The eigenvectors of a Galilean boost (the Galilean-null vectors) also lie on this hyperplane.)

 

[PeterDonis]

Yes, this part I understand.This part I don't understand.

Perhaps it would help if I described my own understanding of how the Galilean case works. Galilean "spacetime" is a stack of Euclidean 3-spaces, each labeled with a real number $t$ that is its "absolute time". There are two classes of vectors in this spacetime: vectors which are tangent to one of the Euclidean 3-spaces, and vectors which are not. One could label the first class as "spacelike" and the second as "timelike", but to me this labeling is somewhat misleading since there is no spacetime metric for Galilean spacetime so there is no correspondence with the classification of vectors by the sign of their squared norm that we have in Minkowski spacetime.

As for orthogonality, since there is no spacetime metric, if we are going to define orthogonality at all, the only way to do it would seem to be to say that every spacelike vector is orthogonal to every timelike vector. It is also possible, of course, to have spacelike vectors that are orthogonal to other spacelike vectors. This means that the statement that every spacelike vector is orthogonal to some timelike vector, while true, is not really useful for this case. We already have a better way of picking out which vectors are spacelike--they are the ones tangent to one of the Euclidean 3-spaces, as above--so we don't need orthogonality to pick out spacelike vectors, and since any vector that is not spacelike is timelike, we don't need orthogonality to pick out timelike vectors either.

 

[PeterDonis]

It's the idea that orthogonality is defined by generalizing the idea
that: at a point of circle, the tangent is orthogonal to the radius.

But how do you define a "circle" in Galilean spacetime? There is no metric, so the usual definition of a circle as the set of points that are all at the same distance from the center doesn't work.

Also, if we are talking about boosts, we are talking about vectors, not geometric figures, and the whole idea of a "circle" doesn't really make sense for vectors.

 

[Hornbein]

at least what I see is that the subject of this topic where the - sign comes from seems to be an already assumed to be known starting point in your equation (1.2.1) at the first page. so based on that it cannot help me I assume.

The minus sign is there because our 3+1D universe is hyperbolic. x² - y² = C. It only seems strange because we are used to circles, as in x² + y² = C.

With a circle if |x| grows larger then |y| shrinks. With a hyperbola if |x| grows then so does |y|. We never can complete a rotation in a hyperbolic space. Instead we get closer and closer to an asymptote but never get there. This is more or less why we can't exceed the speed of light.

Traditionally we hear of relativistic length contraction and time dilation, but I find that confusing. With a hyperbola if |x| shrinks then so does |y|. So in my opinion shrinking length goes with shorter time. If the other guy's clock slows down then his time is less, not more. It's exactly the same thing, I just find the traditional label weird. It obscures the hyperbolic nature of the beast. Fortunately there is nothing to prevent me from using the concept of time contraction in the privacy of my brain.

By the way, t² - x² - y² - z² = d² is equally valid and is used by some authors. It doesn't matter where the minus sign is.

 

[robphy]

But how do you define a "circle" in Galilean spacetime? There is no metric, so the usual definition of a circle as the set of points that are all at the same distance from the center doesn't work.

Also, if we are talking about boosts, we are talking about vectors, not geometric figures, and the whole idea of a "circle" doesn't really make sense for vectors.

From my draft at https://www.aapt.org/doorway/Posters/SalgadoPoster/Salgado-TrilogyArticle.pdf
(as part of https://www.aapt.org/doorway/Posters/SalgadoPoster/Salgado-GRposter.pdf
https://www.aapt.org/doorway/Posters/SalgadoPoster/SalgadoPoster.htm
http://www.aapt.org/doorway/TGRU/ )

• as Euclid defines a circle in the plane as
  "the locus of points (say) t=1 km from a point [along radial paths from the point, using odometers measuring t]"
• Minkowski might define the Minkowski-circle (a hyperbola) on a spacetime diagram as
  "the locus of events (say) t=1 s from an event [along inertial worldlines, using wristwatches]".
• in hindsight, let's pretend that Galileo might define the Galilean-circle (a t=1 line) on a position-vs-time diagram as
  "the locus of events (say) t=1 s from an event [along inertial worldlines, using wristwatches]"... just like Minkowski.

as seen in https://www.desmos.com/calculator/kv8szi3ic8

• Vary the $E$-slider: E=-1 for Euclid, E=+1 for Minkowski, E=0 for Galilean.
  (Observe the unit "circles", the radial lines, and the tangent lines.)
• Then, for fixed-E, vary the $v_2$-slider to trace out the "circles".
• Open the "Boost" folder, vary the $v_{LAB}$-slider to see the "circle" get preserved by the boost.
  (That is, the boosted radial unit-vectors remain on the "circle".)

The metric for these affine Cayley-Klein geometries is
$$ds^2=dt^2-Edy^2.$$
(Yes, the Galilean metric is degenerate... one needs another quadratic form for the spacelike vectors,
as is well known. See the papers from Trautman, Ehlers, etc...
See also https://www.physicsforums.com/threads/why-is-minkowski-spacetime-non-euclidean.1016402/post-6647305 and links therein.

While "degeneracy" sounds bad,​

note that from a projective-geometry point of view, Euclidean geometry has degenerate features,​

but there they manage to get along and develop Euclidean geometry.)​

The idea of my approach is to carry out all constructions and calculations keeping track of the $E$,
and to do so in a unified way as much as possible.
A construction which seems difficult to be unified is demoted in favor of one that can be unified.
GOAL: Same storyline, different $E$.

(So, I want to define "timelike", "spacelike", and "null" in a unified way from these structures.)​

The boosts for the above are ( from https://www.physicsforums.com/threads/why-is-minkowski-spacetime-non-euclidean.1016402/post-6648907 )

$$\left( \begin{array}{c} t' \\ y' \end{array} \right)
=
\left(
\begin{array}{cc}
\frac{1}{\sqrt{1-E\beta^2}} & \frac{E\beta}{\sqrt{1-E\beta^2}}
\\
\frac{\beta}{\sqrt{1-E\beta^2}} &
\frac{1}{\sqrt{1-E\beta^2}} & \end{array}
\right)\\
\left( \begin{array}{c} t \\ y \end{array} \right)$$

I learned of these 2-D Cayley-Klein geometries (which include these 9 geometries:
Euclidean, Hyperbolic, Elliptical/Spherical, Minkowski, Galilean, deSitter and anti-deSitter and their Galilean limits) from
"A Simple Non-Euclidean Geometry and Its Physical Basis: An Elementary Account of Galilean Geometry and the Galilean Principle of Relativity" by I.M. Yaglom
https://www.amazon.com/dp/0387903321/?tag=pfamazon01-20
and, more recently,
Richter-Gebert's "Perspectives on Projective Geometry: A Guided Tour Through Real and Complex Geometry" https://www.amazon.com/dp/3642172857/?tag=pfamazon01-20 .

To see the unified metric for all 9 geometries, look at page 3 of my poster.)​

So, I am carrying out the constructions and calculations,
especially as it applies to physics (PHY101, Special Relativity, and hopefully the deSitter spaces).

 

[PeterDonis]

"the locus of events (say) t=1 s from an event [along inertial worldlines, using wristwatches]"

But this isn't a circle; it's not even a continuous curve. It's two disjoint Euclidean 3-spaces (one 1 second to the future, the other one second to the past), or two disjoint lines if we eliminate 2 spatial dimensions.

 

[PeterDonis]

my approach

This probably needs to be in a separate thread since it seems well off topic for this one.

 

[robphy]

"the locus of events (say) t=1 s from an event [along inertial worldlines, using wristwatches]"

But this isn't a circle; it's not even a continuous curve. It's two disjoint Euclidean 3-spaces (one 1 second to the future, the other one second to the past), or two disjoint lines if we eliminate 2 spatial dimensions.

In the preceding line, I did specify "(a t=1 line)"... not $t^2=1$.
For completeness,
call it the "future Galilean circle" (t=1) as one would get from future-directed inertial observers from an event.
Similarly, for Minkowski spacetime,
we would refer to the "future Minkowski circle" ($\tau=1$) as one would get from future-directed inertial observers from an event.

...and before it comes up... consider only transformations that preserve the direction of time.

 

[PeterDonis]

For completeness,
call it the "future Galilean circle" (t=1) as one would get from future-directed inertial observers from an event.
Similarly, for Minkowski spacetime,
we would refer to the "future Minkowski circle" () as one would get from future-directed inertial observers from an event.

Ah, ok.

 

[malawi_glenn]

ds is no coincidence and should follow in a logical way from the general thoughts behind the whole idea.

Math is logic. It is not coincidence, it follows from the Lorentz-transformations. The postulate was that $(ct)^2-x^2=0$ for light-like separations in all inertial frames. Then if you want to have a transformation that is linear, only depends on $v$ and reduces to Galilean transformation, the result $(ct)^2 +x^2$ is invariant pops out. It is a prediction of the theory. The language of physics is math which is a logical system. What most people mean when they invoke the word "logic" is usually something else than mathematical logic, like "intuition" etc

Remember that I wrote that it is nothing special with t and x,y,z here. As long has you have quantity that transforms as t and quantities that transforms as x,y,z you have the same form of the invariant.
It comes from the geometry which these transformations induce.

Consider the Euclidean space $\mathbb{R}^3$, rotations and translations preserve the Euclidean distance (norm) $x^2 + y^2 + z^2$, thus it is an invariant under those transformations. Would you look for a "logical" reason why $x^2 + y^2 + z^2$ is invariant under those transformations? Well you could argue "its a sphere with fixed radius, of course it will have the same radius if you rotate or translate it

What would the corresponding "sphere" in this non-Euclidean space of $t$ and $x$ be? (we drop the y and z now for simplicity).
What "shape" would be the same in both $S$ and in $\tilde S$? It would not be circles i.e. $(ct)^2 +x^2 = \text{constant}$. It would be hyperbolas $(ct)^2 -x^2 = \text{constant}$
Points lying on the same hyperbola in this space, which we often call Minkowski space, has the same vaule of $(ct)^2 -x^2$ regardless in which reference system it is calculated.

It is something that is true in general and is nice to work out. For a given transformation in a given space. What are the invariants?

Euclidean geometry, but if I understand that well it means not curved

Euclidean geometry you measure distances with Pythagoras theorem. In Newtonian physics, time is not a coordinate, it is a parameter. Therefore the concept of "space-time" is not well defined.

 

[robphy]

In Newtonian physics, time is not a coordinate, it is a parameter. Therefore the concept of "space-time" is not well defined.

That's not true.
That's what #115 is about.
"Space-time" actually starts with Galilean physics (actually Aristotelian Physics, which lacks the principle of relativity).
Look at

• Geroch (1978), General Relativity from A to B (Ch 2 The Aristotelian View; Ch 3 The Galilean View) https://www.amazon.com/dp/0226288641/?tag=pfamazon01-20
• Penrose (2004), The Road to Reality (Ch 17.1 The spacetime of Aristotelian physics; Ch 17.2 Spacetime for Galilean relativity ) https://www.amazon.com/dp/0679776311/?tag=pfamazon01-20
• Trautman (1964), "Comparison of Newtonian and Relativistic Theories of Space-Time", http://trautman.fuw.edu.pl/publications/Papers-in-pdf/22.pdf
• Cartan (1923) - https://en.wikipedia.org/wiki/Newton–Cartan_theory

 

[malawi_glenn]

That's not true.
That's what #115 is about.

I did not read it in its entirely since it is kinda off-topic here (as @PeterDonis mentioned).
My current conception of Galilean space-time is that it is just a bunch of disjoint R3-sets.

Could you please write down the invariant space-time interval then?

 

[robphy]

I did not read it in its entirely since it is kinda off-topic here (as @PeterDonis mentioned).
My current conception of Galilean space-time is that it is just a bunch of disjoint R3-sets.
View attachment 305725
Could you please write down the invariant space-time interval then?

Yes, but according to the geometrically-oriented relativists* I referenced,
that has a spacetime structure.
(These listed relativists aren't end-users of relativity;
they developed many of the geometric tools we use to study relativity.)

My posts #115 and #122 were in response to
misconceptions and mischaracterizations of the spacetime structures in the Galilean case.

Here are the line-elements (infinitesimal square-intervals}:
\begin{align*}
ds_{Euc}^2 &= dt^2 +dy^2\\
ds_{Gal}^2 &= dt^2\\
ds_{Min}^2 &= dt^2 -dy^2
\end{align*}
$$ds_{unified}^2 = dt^2 - E dy^2$$ as in #115.

Geroch (p. 47):

Penrose (p. 387):

Newton-Cartan:

In Newton–Cartan theory, one starts with a smooth four-dimensional manifold ${\displaystyle M}$ and defines two (degenerate) metrics. ...A temporal metric ${\displaystyle t_{ab}}$ with signature ${\displaystyle (1,0,0,0)}$, used to assign temporal lengths to vectors on ${\displaystyle M}$ and a spatial metric ${\displaystyle h^{ab}}$ with signature ${\displaystyle (0,1,1,1)}$. One also requires that these two metrics satisfy a transversality (or "orthogonality") condition, ${\displaystyle h^{ab}t_{bc}=0}$. Thus, one defines a classical spacetime as an ordered quadruple ${\displaystyle (M,t_{ab},h^{ab},\nabla )}$...
...
One might say that a classical spacetime is the analog of a relativistic spacetime ${\displaystyle (M,g_{ab})}$, where ${\displaystyle g_{ab}}$ is a smooth Lorentzian metric on the manifold ${\displaystyle M}$.

TL;DR The position-vs-time graph from PHY101 has a Galilean spacetime structure... and the concept is well-defined [but not well-known].

 

[vanhees71]

Why? If you're going to consider boosts, you have to have multiple timelike vectors since boosts map timelike vectors to different timelike vectors.Affine space structure is not enough, is it? Boosts are transformations on a vector space, not on an affine space.

An affine space is a set of "points" plus a (real) vector space with some axioms connecting them. All boils down to the standard way you introduce vectors in 2D or 3D Euclidean geometry, i.e., vectors are equivalence classes of ordered pairs of points, $\vec{v}=\overrightarrow{AB}$ (i.e., the two points are connected by a directed straight line), and parallely displaced pairs of other points lead to the same vector by definition.

Now in Euclidean geometry the vector space has a scalar product with the corresponding induced metric, i.e., a positive definite "fundamental bilinear form".

The "only" difference for Minkowski space is that it is a 4D affine space where the fundamental bilinear form of the vector space has signature (1,3) (or equivalently (3,1)), i.e., it's a Lorentzian affine manifold. This difference, however, is very important for physics since it admits to established the known "causality structure" and thus makes it a useful spacetime model, which is not the case for a Euclidean affine manifold. The symmetries of the Lorentzian affine manifold build the (proper orthochronous) Poincare group.

 

[HansH]

Math is logic. It is not coincidence

Consider the Euclidean space $\mathbb{R}^3$, rotations and translations preserve the Euclidean distance (norm) $x^2 + y^2 + z^2$, thus it is an invariant under those transformations. Would you look for a "logical" reason why $x^2 + y^2 + z^2$ is invariant under those transformations? Well you could argue "its a sphere with fixed radius, of course it will have the same radius if you rotate or translate it

What would the corresponding "sphere" in this non-Euclidean space of $t$ and $x$ be? (we drop the y and z now for simplicity).
What "shape" would be the same in both $S$ and in $\tilde S$? It would not be circles i.e. $(ct)^2 +x^2 = \text{constant}$. It would be hyperbolas $(ct)^2 -x^2 = \text{constant}$
Points lying on the same hyperbola in this space, which we often call Minkowski space, has the same vaule of $(ct)^2 -x^2$ regardless in which reference system it is calculated.

It is something that is true in general and is nice to work out. For a given transformation in a given space. What are the invariants?

with 'coincidence' I mean the fact that there is a good reason but I am not able (yet) to see that in my head.

your input about invariance reminds me at a part of mathematics I learned in my pre-universal education about transformations in 3D space and indeed als calculating shapes that kept the same under a transformation. so for the Lorenz transformation you indicate this is a parabola. so I have some work to do to check that by myself I assume. Then I hope I can say: yes this is logical, because then I also can see it in my head.

 

[Ibix]

parabola

Hyperbola: $c^2t^2-x^2=\mathrm{const}$. To get a parabola either the $t$ or $x$ term would need not to be squared.

 

[HansH]

Hyperbola: $c^2t^2-x^2=\mathrm{const}$. To get a parabola either the $t$ or $x$ term would need not to be squared.

yes I actually meant hyerbola.

 

[Sagittarius A-Star]

It looks like magic that ds is invariant. I have the feeling however that the invariance of ds is no coincidence and should follow in a logical way from the general thoughts behind the whole idea.

I think that is possible. The Lorentz-Transformation can be derived from Galileo's principle of relativity (the laws of physics are the same in all inertial reference frames) and from assuming $t' \neq t$, which is the opposite of Newton's assumption of an "absolute time". You can find such a derivation here:
https://www.physicsforums.com/threa...rom-commutative-velocity-composition.1017275/

SR postulate 2 (the vacuum speed of light is the same in all inertial reference frames) is not needed then.

The invariance of $ds^2$ can then be derived with a Lorentz transformation of it's coordinate intervals.

Edit: A bit more is needed. See posting #132.

 

[malawi_glenn]

I think that is possible. The Lorentz-Transformation can be derived from Galileo's principle of relativity (the laws of physics are the same in all inertial reference frames) and from assuming t′≠t, which is the opposite of Newton's assumption of an "absolute time".

Isn't the assumption of the velocity compoisition also needed? Assuming t′≠t is not enough?

 

[Sagittarius A-Star]

Isn't the assumption of the velocity compoisition also needed? Assuming t′≠t is not enough?

The velocity composition is calculated from $dx'/dt'$ (=transformation of a velocity). It's commutative property follows from mathematical group postulates. That they apply physically, is implicitly included in SR postulate 1 (principle of relativity). SR postulate 2 is not needed.

Edit:
If "no absolute time exists" is assumed, then the commutative property of the velocity composition is not implicitly included in SR postulate 1. A bit more is needed. See posting #132.

 

[PeterDonis]

the invariance of ds is no coincidence

Of course not. $ds$ is an invariant because it represents an actual physical observable: the time elapsed on a clock that follows the timelike worldline of which $ds$ is a segment.

 

[vanhees71]

I think that is possible. The Lorentz-Transformation can be derived from Galileo's principle of relativity (the laws of physics are the same in all inertial reference frames) and from assuming $t' \neq t$, which is the opposite of Newton's assumption of an "absolute time". You can find such a derivation here:
https://www.physicsforums.com/threa...rom-commutative-velocity-composition.1017275/

SR postulate 2 (the vacuum speed of light is the same in all inertial reference frames) is not needed then.

The invariance of $ds^2$ can then be derived with a Lorentz transformation of it's coordinate intervals.

You need a bit more:

-the validity of the special principle of relativity (existence of (global) inertial frames)
-space for any inertial observer (i.e., any observer at rest wrt. an arbitrary inertial frame) is a Euclidean 3D affine manifold (implying the corresponding symmetry group ISO(3), i.e., homogeneity and isotropy of affine Eulcidean space)
-for any inertial observer time is homogeneous (invariance of the laws of physics under time translations)
-transformations from one inertial frame to another build a Lie group

This identifies the possible symmetry groups of space as either the Gailei or the Poincare group. Which one is the correct one must be decided by experiment. The answer is obvious :-).

 

[cianfa72]

Yes. And even though they will disagree about the different times and different points, they will all agree that it is a sphere of radius $c\Delta t$. That is what the second postulate means.

Note that the *same* spacetime events that describe the propagating light flash will be distributed over a sphere according to each inertial observer.

 

[cianfa72]

By the way: the very fact that that surface turns out to be a sphere basically boils down to just the isotropic property of the space, right ? In other words, strictly speaking, homogeneity property of the space is not involved, I believe.

 

[Dale]

Homogeneity is also needed. Imagine that everywhere the local speed of light were isotropic, but there were a small patch where the speed of light were slower. Then light comes would get a “dent” as they pass through that patch and would no longer be spheres.

 

[cianfa72]

everywhere the local speed of light were isotropic, but there were a small patch where the speed of light were slower.

Ah ok, your point is that in a small path such that the local one-way speed of light were slower *but* isotropic then those surfaces would not longer be spheres.

 

[Ibix]

Homogeneity is also needed. Imagine that everywhere the local speed of light were isotropic, but there were a small patch where the speed of light were slower.

Then observers on the surface of the patch would see an anisotropic speed of light, no?

After a quick check in Carroll's GR notes, apparently a space that is isotropic everywhere is also homogeneous.

 

[Dale]

Then observers on the surface of the patch would see an anisotropic speed of light, no?

Hmm, yes, that is right. I had missed that.

a space that is isotropic everywhere is also homogeneous.

Yes.

Ok, so I think isotropy without homogeneity is spherical symmetry about one center.

 

[Ibix]

Ok, so I think isotropy without homogeneity is spherical symmetry about one center.

Yes. What Carroll says is

Note that there is no necessary relationship between homogeneity and isotropy; a manifold can be homogeneous but nowhere isotropic (such as $\mathbb{R}×\mathbb{S}^2$ in the usual metric), or it can be isotropic around a point without being homogeneous (such as a cone, which is isotropic around its vertex but certainly not homogeneous). On the other hand, if a space is isotropic everywhere then it is homogeneous. (Likewise if it is isotropic around one point and also homogeneous, it will be isotropic around every point.)

 

[cianfa72]

a manifold can be homogeneous but nowhere isotropic (such as $\mathbb R \times \mathbb S^2$ in the usual metric), or it can be isotropic around a point without being homogeneous (such as a cone, which is isotropic around its vertex but certainly not homogeneous).

I believe we can visualize $\mathbb R \times \mathbb S^2$ topology as a spherical 'onion' with an hole in its center. Each spherical shell parametrized by $r \in \mathbb R$ is homeomorphic to $S^2$.

From my understanding it is actually its natural metric (i.e. its usual metric) that turns it in an homogeneus but isotropic manifold.