I Spacetime interval and basic properties of light

[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.

 

[Ibix]

I think Carroll means the surface of a hyper-cylinder (or whatever the name for it is), where every $\mathbb{S}^2$ is identical and stacked one on top of another, not ever-smaller spheres nested one inside the other, which is what I think you mean. Every point is identical (diffeomorphic to any other), but not every direction.

 

[cianfa72]

I think Carroll means the surface of a hyper-cylinder (or whatever the name for it is), where every $\mathbb{S}^2$ is identical and stacked one on top of another, not ever-smaller spheres nested one inside the other, which is what I think you mean. Every point is identical (diffeomorphic to any other), but not every direction.

Yes, if we take in account (as required) the metric. However --from just a topological point of view-- I believe the 'infinite spherical onion with internal hole' might make sense though.

Ah ok, on the hypercylinder surface the homogeneity is 'encoded' by the diffeomorphism between every pair of points on it.