# I Going from Galilei to Minkowski

1. Oct 29, 2017

### Harry83

Hi, I'm starting a double degree in math and physics and am still a relativity newbie. I'm a bit stuck figuring out what exactly means to drop absolute time and simultaneity when making the transition from Galilean spacetime to Minkowski spacetime. Judging from the purely mathematical perspective it looks to me more accurate to say that for timelike paths the change is from a unique preferred frame corresponding to aunique simultaneity slicing in the Galilean case to a whole family of equivalently preferred frames in special relativity corresponding to an infinity of possible simultaneity slicings determining the possible three velocities < c at each point while the conformal structure determined by the indefinite signature allows to keep the relativity principle. Is this about right?

2. Oct 29, 2017

### Staff: Mentor

This is not a correct description. The simultaneity slicing in the Galilean case is unique because the Galilean transformation leaves the time coordinate unchanged; it only affects the space coordinates. But it does not define a preferred frame, because all of the inertial frames related by Galilean transformations are equivalent with respect to Newton's Laws (just as all of the inertial frames related by Lorentz transformations are equivalent with respect to the laws of special relativity).

3. Oct 29, 2017

### robphy

This may be a little abstract....
But since you mentioned math....

Can you write the Galilean transformation as a matrix?
Can you find its eigenvectors and eigenvalues?

In the Lorentz transformations, the corresponding problem leads to different results. In both cases however, there are no timelike eigenvectors (hence no preferred frames).

My preferred motivation is geometrical/physical.
How does one define circles in Spacetime geometry? Use a wristwatch.

4. Oct 29, 2017

### Harry83

Thanks. You're quite right. There sure exists a whole family of equivalent inertial frames also in Galilean spacetime. I wanted to convey the idea of an absolute velocity and I mixed it up with the idea of preferred frame.
So basically is it fair to say that the main difference between that Galilean and SR inertial frames is that the former have a fixed orientation between the time and the space component while the latter don't?
And about absolute simultaneity versus the relativity of simultaneity, is this SR idea that rather than abandoning the simultaneity slicings they are just not unique anymore?

5. Oct 29, 2017

### Harry83

Thanks. I think I clarified above about the preferred frame so it's a non-issue, but I find it interesting the comment about no timelike eigenvectors.

6. Oct 29, 2017

### Staff: Mentor

Not really; it would be better to say that the whole concept of relating the "orientation" of the time axis to the spatial components makes no sense in the Galilean context, because there is no spacetime metric so there is no way to compute an "angle" between the time axis and a spatial vector.

7. Oct 29, 2017

### Staff: Mentor

No, it's more than that, because in SR there is a spacetime metric, whereas in Galilean physics there isn't. See my previous post.

8. Oct 29, 2017

### robphy

Note that there is also no "angle" (arc length along a unit circle) between a time axis and a spatial vector in special relativity (unless you want to use complex angles). In both Galilean and special relativity, one can use a metric / symmetric bilinear form (degenerate in the Galilean case) to define orthogonality between such vectors.

Whether or not one calls the Galilean structures "metrical", there is a substantial geometrical formalism that tries to mimic metrical geometry. The Galilean Spacetime is one of the nine Cayley-Klein geometries in the plane. The other eight are Euclidean, hyperbolic, elliptic, Minkowski, deSitter and anti-deSitter and the Galilean limits of the anti/deSitter Spacetimes.

To explore the affine Cayley-Klein geometries (where the parallel postulate holds), check out my visualization:
https://www.desmos.com/calculator/ti58l2sair

Play with the E slider.
Note how the tangents to the unit circles for the E=0 case coincide... this is absolute simultaneity.

Last edited: Oct 29, 2017
9. Oct 29, 2017

### Staff: Mentor

The word "angle" was probably not the best choice of words. The point is that in SR, you can take an inner product between any two vectors, including the case where one is timelike and the other is spacelike. You can't do that in the Galilean case; you can only take inner products between spatial vectors.

10. Oct 29, 2017

### robphy

The degenerate metric in Galilean Spacetime geometry looks like a projection operator.
The inner product of any timelike vector and any spacelike vector is zero in the Galilean case.

https://en.wikipedia.org/wiki/Newton–Cartan_theory#Classical_spacetimes
Trautman's "Comparison of Newtonian and Relativistic Theories of SpaceTime": http://www.fuw.edu.pl/~amt/CompofNewt.pdf
Ehlers' frame theory in "Examples of Newtonian limits of relativistic spacetimes": http://pubman.mpdl.mpg.de/pubman/item/escidoc:153004:1/component/escidoc:153003/328699.pdf

Last edited: Oct 29, 2017
11. Oct 29, 2017

### Staff: Mentor

I'm not sure that's true. I understand that the two metrics satisfy an orthogonality condition, but that does not mean you can take a timelike vector and a spacelike vector and take their inner product, because to do that you would need a single metric that acts on both, and you don't have that: one metric acts only on timelike vectors and the other acts only on spacelike vectors.

12. Oct 29, 2017

### robphy

In an affine space (like Euclidean, Galilean, and Minkowski), we have an underlying vector space.
We can add and scalar multiply.... we can add timelike vectors to spacelike vectors, etc... indeed, there is no sense of timelike or spacelike without a metric-like structure.

A metric is a type (0,2) tensor field, as added structure to get a spacetime geometry.
One can add a future timelike vector and a past timelike vector to obtain a spacelike vector.
Due to linearity, the metric can't be restricted to only act on vectors of certain types.

Thus, a metric tensor takes two vectors (whatever their type [timelike, spacelike, etc]) and returns a real number (their inner product)... which could be positive, negative, or zero.
There is no problem if the inner product of any timelike vector with any spacelike vector gives zero
(in fact, that's why we have absolute simultaneity in the Galilean case)

.... that's part of the degenerate nature of the geometry... and that's why additional structures are needed (for example, a spatial metric to give non-trivial information for pairs of spatial-vectors since the temporal metric gives zero for such pairs.)

13. Oct 29, 2017

### Staff: Mentor

A geometry, yes. Not necessarily a spacetime geometry.

A metric tensor takes two vectors and returns a real number. But in the case under discussion (Galilean "spacetime"), we have two metric tensors. One only takes spacelike vectors as input; the other only takes timelike vectors as input. So there is no way to get an inner product between a timelike and a spacelike vector in this case.

The sources you link to do not seem to me to claim at any point that there is an inner product between timelike and spacelike vectors in the Galilean case. They only say what I said above.

14. Oct 30, 2017

### vanhees71

Galilei spacetime is not a pseudo-Euclidean 4D affine space but a bundle of 3D Euclidean spaces along the time axis as fiber. That's why there's "absolute time" and "absolute space" in Newtonian physics. It doesn't make sense to define time- and space-like vectors. The direction of time has to be imposed (i.e., the "time axis" is by definition oriented), i.e., you postulate a causality structure ("arrow of time").

Minkowski space is a pseudo-Euclidean 4D affine space. Also here you have to impose orientedness and postulate a causality structure. This is consistently possible because the signature of Minkowski space's fundamental bilinear form is (1,3) (or (3,1) depending on your preferred choice of convention). The physical (classical) symmetry group is the proper orthochronous Poincare group, i.e., the part of the full Poincare group that's continuously connected to the identity operation. Indeed, Nature precisely realizes this symmetry group and not any larger subgroup of the (quantum version of the) Poincare group since the weak interaction breaks C, P, and CP (and thus also T).

15. Oct 30, 2017

### Staff: Mentor

The papers that @robphy linked to use that terminology, though it might be somewhat of an abuse of terminology compared to the relativistic case.

16. Oct 30, 2017

### robphy

There aren't two kinds of indices (one for spacelike and one for timelike).... so a tensor can act on a vector of any type.
As I said earlier, "Due to linearity, the metric can't be restricted to only act on vectors of certain types."

Now, it may be that a certain tensor applied to certain types of vector yield possibly uninteresting quantities (like zero).
But that doesn't stop or restrict applying the tensor to various vectors.

17. Oct 30, 2017

### robphy

Maybe this is all an argument about terminology.
In the end, whatever you want to call things, there are structures (some necessarily degenerate) that try to mimic the situation in a Lorentzian-signature spacetime.

By an affine space, I mean a vector space in which parallelism applies.
The Galilean spacetime has that vector space structure and satisfies the parallel postulate.

Note that a position-vs-time graph (from PHY 101) is a Galilean spacetime diagram.
The position-vs-time graph has a non-euclidean geometry.
Sloped lines are inertial worldlines, and vertical lines are planes of simultaneity.
I can give you a geometric construction suggesting that those sloped lines are orthogonal to those vertical lines.

Define the Galilean circle with a wristwatch... the curve constant interval equal to 1 from the origin.
(That's a hyperbola in Minkowski spacetime.)
Segments from the origin to the circle are unit radii.
Given the circle, the tangents to the circle are orthogonal to the radius.
You can see the limiting cases by playing with the E-slider https://www.desmos.com/calculator/ti58l2sair .
In the E=0 (1+1)-Galilean case, all tangents are collinear (unlike in the Euclidean and Minkowski case).
This collinearity reflects absolute simultaneity.

A major reference for this point of view is Yaglom's
https://archive.org/details/ASimpleNon-euclideanGeometryAndItsPhysicalBasis

https://www.elsevier.com/books/math...f-20th-century-physics/emch/978-0-444-87585-3

Cayley-Klein Geometry is a well-established topic
http://aip.scitation.org/doi/10.1063/1.1664490
http://nyjm.albany.edu/j/2006/12-8v.pdf
https://arxiv.org/abs/0707.2869
http://www.springer.com/us/book/9783642172854 (Richter-Gebert)
https://arxiv.org/abs/physics/9702030

One of my research projects is applying this to
a unified framework for kinematics and dynamics in Galilean and Special Relativity,
that could be used in an undergraduate physics class.

.

Last edited: Oct 31, 2017
18. Oct 30, 2017

### Staff: Mentor

There are upper and lower indices, which are distinct. Both of the papers you linked to carefully use only upper indices for the spatial metric and only lower indices for the temporal one. That indicates that the two operate on distinct domains. The Trautman paper uses the term "form" for something with lower indices, and "vector" for something with upper indices.

Also, on p. 417, Trautman says that the spatial metric "may be used to define the square of any form and of any spacelike vector, but not of timelike vectors". That indicates to me that there is also no way of forming an inner product between a spacelike and a timelike vector.

I'll have to take a look at the other references you give in post #17.

19. Oct 31, 2017

### robphy

The nonzero form is defined on Trautman-p416 $t_{a}=(\partial_a t)$.
Then, a vector $v^a$ such that $v^at_a=0$ is spacelike [and so must be tangent to the hypersurfaces of constant-t... since $v^a$ doesn't pierce the hypersurface (a la MTW's "bongs of a bell")];
if $v^at_a \neq 0$, then $v^a$ is timelike.
In short, if a vector has a nonzero t-component, it's timelike...otherwise it's spacelike.
[Application.. a future-directed unit-timelike vector (of the form $\left( \begin{array}{c} 1 \\ V \end{array}\right)$ can be 4-velocity of an observer traveling with velocity V... this vector, of course, is tangent to the worldline of the observer. In Galilean physics, V can take any finite value. We excluded infinite values... because that would correspond to being tangent to the constant-t hypersurface.. Let's not allow our observers to do that. (The unit magnitude is determined by the temporal metric.) Note the Galilean time-dilation factor between two observers is 1:
$\left( \begin{array}{cc} 1 & V \end{array}\right)\left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right)\left( \begin{array}{c} 1 \\ W\end{array}\right)=1$]

A temporal metric can be defined as $t_{ab}=t_at_b=(\partial_a t)(\partial_b t)$ (see Ehlers-p-A121-near Eq 3a)
Then for spacelike vector $X^a$ and timelike vector $U^a$,
$$X^at_{ab}U^b=(X^at_{a})t_{b}U^b=0$$ can be interpreted by saying that they are orthogonal with respect to this temporal metric...
i.e. their inner product via this temporal metric is zero.
(Of course, it's also zero if $U^a$ is spacelike.
In other words, the temporal metric annihilates spacelike vectors.)

For me, the temporal metric is primary
(since it shows up as a limiting case of the Minkowski spacetime metric, in my formulation.
My unified metric is $M = \left( \begin{array}{cc} 1 & 0 \\ 0 & -\epsilon^2 \end{array}\right)$ ).

Last edited: Oct 31, 2017
20. Oct 31, 2017

### haushofer

How about the inverse metric? See e.g. Dautcourt's "the newtonian limit..."