I Can time be another basis vector under Galilean relativity?

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The discussion critiques a video that attempts to frame Galilean relativity in terms of spacetime as a vector space, suggesting that this approach obscures rather than clarifies the transition to special relativity (SR). It argues that in Galilean relativity, time and space are fundamentally different, with time not being a homogeneous unit that can be combined with spatial dimensions. The concept of a Galilean spacetime interval is deemed impractical and lacking in utility, as it does not solve any real problems. Additionally, the absence of an invariant speed in Galilean relativity contrasts with SR, where time and space are homogenized through the speed of light. Ultimately, the discussion emphasizes that while mathematical representations may be convenient, they do not capture the distinct nature of time in the Galilean framework.
  • #31
vanhees71 said:
A rotation matrix has (at least) one real eigenvalue 1 and a real eigenvector, giving the direction of the rotation axis.

I was also struck by his statement "none" for eigenvectors of the Euclidean rotation, since there is as you point out an eigenvector constituted by the rotation axis, which (I understand) is the line passing through the fixed point of the rotation, i.e. visually, a line perpendicular to the page where we paint the 2D rotation, right?

But note that in the image that I copied in post #25 from robphy's paper he himself does mention an eigenvector for the Euclidean rotation with coordinates (0,0), corresponding to such perpendicular line?

And another interesting point is what would be the eigenvector of the Galilean boost? In this discussion he mentioned "infinite speed" while in the said paper he mentioned "time". I said that I prefer the second option, although I am not sure if eigenvector would be in this context the accurate term or we are talking about an analog of it.

I think that a parenthesis to clarify this subject would be useful, as it is relevant to the main issue.
 
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  • #32
Saw said:
Sorry, can you please quote which specific part of my digression is of no relevance to learning SR or GR? Just to know what to take into account in the future.
All of the digression in trying to rework Galilean relativity.
 
  • #33
PeroK said:
All of the digression in trying to rework Galilean relativity.
The very object of the thread? You deem it irrelevant to learning SR and GR? Well, the author of the reference video thinks that his reworking of Galilean relativity makes it easier to understand SR and GR. One of the mentors (robphy) is also very keen on the idea and has made extensive study on that. I am myself fond of the abstract idea (who could not be?, it is obvious that you better understand things by contrast with their close cousins), but I am not so sure about those particular reworkings, about their aptness to better teach SR and GR. I think that this opinion is shared by another mentor, vanhees71.
 
  • #34
Saw said:
The very object of the thread?
Yes.
Saw said:
Well, the author of the reference video thinks that his reworking of Galilean relativity makes it easier to understand SR and GR.
He may be right and he may be wrong.
Saw said:
One of the mentors (robphy) is also very keen on the idea and has made extensive study on that.
I didn't say everyone would agree.
 
  • #35
Saw said:
I was also struck by his statement "none" for eigenvectors of the Euclidean rotation, since there is as you point out an eigenvector constituted by the rotation axis, which (I understand) is the line passing through the fixed point of the rotation, i.e. visually, a line perpendicular to the page where we paint the 2D rotation, right?

But note that in the image that I copied in post #25 from robphy's paper he himself does mention an eigenvector for the Euclidean rotation with coordinates (0,0), corresponding to such perpendicular line?

And another interesting point is what would be the eigenvector of the Galilean boost? In this discussion he mentioned "infinite speed" while in the said paper he mentioned "time". I said that I prefer the second option, although I am not sure if eigenvector would be in this context the accurate term or we are talking about an analog of it.

I think that a parenthesis to clarify this subject would be useful, as it is relevant to the main issue.
The null vector never is an eigenvector of any matrix. Because trivially ##\hat{A} \vec{0}=\vec{0}## for any matrix ##\hat{A}##.

As I said, for me these pseudo-4D extension is just a convenient way to realize all Galilei transformations (including Galilei boosts) by ##4 \times 4##-matrix multiplications, but that's all there is to it. It doesn't add any additional structure to the description of Galilean spacetime than is already there when interpreting it as a fiber bundle, which is the natural choice for the Galilei-Newton spacetime.

Concerning the Galilei boost, let's to for 1D motion. Then the "vector" is
$$\vec{x}=\begin{pmatrix} t \\ x \end{pmatrix}$$
and the boost is described by
$$\vec{x}'=\begin{pmatrix} t \\ x-v t \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ v & 1 \end{pmatrix} \begin{pmatrix} t \\ x \end{pmatrix}.$$
This "Jordan matrix" has only one eigenvector ##(0,1)^{\text{T}}## with eigenvalue ##1##, which simply tells you that at time ##t=0## all the points on the ##x## axis are unchanged by the transformation, which is, however trivial anyway.
 
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  • #36
vanhees71 said:
The null vector never is an eigenvector of any matrix. Because trivially ##\hat{A} \vec{0}=\vec{0}## for any matrix ##\hat{A}##.

As I said, for me these pseudo-4D extension is just a convenient way to realize all Galilei transformations (including Galilei boosts) by ##4 \times 4##-matrix multiplications, but that's all there is to it. It doesn't add any additional structure to the description of Galilean spacetime than is already there when interpreting it as a fiber bundle, which is the natural choice for the Galilei-Newton spacetime.

Concerning the Galilei boost, let's to for 1D motion. Then the "vector" is
$$\vec{x}=\begin{pmatrix} t \\ x \end{pmatrix}$$
and the boost is described by
$$\vec{x}'=\begin{pmatrix} t \\ x-v t \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ v & 1 \end{pmatrix} \begin{pmatrix} t \\ x \end{pmatrix}.$$
This "Jordan matrix" has only one eigenvector ##(0,1)^{\text{T}}## with eigenvalue ##1##, which simply tells you that at time ##t=0## all the points on the ##x## axis are unchanged by the transformation, which is, however trivial anyway.
Thanks, for me you are answering the question and would only continue the thread if robphy has more to say.
Just two comments:
- The convenient way to realize the matrix multiplication... does it have to do with "homogeneous coordinates", in the vein of what you can do with a translation? (See my post #12.)
- You may want to put the missing negative sign by the v term in the transformation matrix.
 
  • #37
vanhees71 said:
A rotation matrix has (at least) one real eigenvalue 1 and a real eigenvector, giving the direction of the rotation axis. That's why the Euclidean spacetime model is not appropriate, because it doesn't admit causal ordering or if you want to enforce causal ordering you have to restrict the rotations, but then the transformations from one inertial frame to another one don't form a group.

The context of this discussion (starting from the OP and the link to eigenchris's YouTube video) has always been on a two-dimensional plane,
and when I joined, I brought up the Cayley-Klein approach
with the parameter E to move between
the Euclidean plane, the (1+1)-Galilean diagram, and the (1+1)-Minkowski diagram.

(This "Spacetime Trigonometry" approach is heavily inspired by
A Simple Non-Euclidean Geometry and Its Physical Basis by I.M. Yaglom
https://www.amazon.com/dp/0387903321/?tag=pfamazon01-20
and modern more mathematical (but less physical) reference is
Perspectives on Projective Geometry: A Guided Tour Through Real and Complex Geometry by Jürgen Richter-Gebert
https://www.amazon.com/dp/3642172857/?tag=pfamazon01-20
)

vanhees71 said:
The null vector never is an eigenvector of any matrix. Because trivially ##\hat{A} \vec{0}=\vec{0}## for any matrix ##\hat{A}##.
In relativity, we have null-vectors (nonzero vectors with zero dot-product with itself)
that are not the zero-vector.

Saw said:
One of the mentors (robphy) is also very keen on the idea and has made extensive study on that.
btw.. I'm not a PF mentor.
 
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  • #38
vanhees71 said:
This "Jordan matrix" has only one eigenvector ##(0,1)^{\text{T}}## with eigenvalue ##1##, which simply tells you that at time ##t=0## all the points on the ##x## axis are unchanged by the transformation, which is, however trivial anyway.
This Galilean-boost eigenvector corresponds to the limiting infinite speed (infinite slope) of all Galilean-timelike vectors,
similar to the Lorentz-boost eigenvectors corresponding to limiting [finite] velocity of all Minkowski-timelike vectors.

update: The Galilean boost eigenvector also happens to correspond with the concept of "absolute time" because this eigenvector is along the Galilean-circle. Since all tangent vectors to the Galilean circle are parallel, all Galilean observers agree on those events being simultaneous (since the tangents to the circle for each Galilean-4-velocity are Galilean-orthogonal to that 4-velocity).
This is unlike the situation for tangents to a circle (in 2D Euclidean space) and tangents to a hyperbola (in (1+1)-Minkowski spacetime).

The Galilean eigenvalue of 1 means that all Galilean observers will agree on the magnitude of displacement vectors parallel to ##(0,1)^{\text{T}}##. Physically, all Galilean observers agree on the length of a ruler. This is "absolute length".

Many results in Galilean geometry are trivial... and since much of the physical interpretations are part of our "common sense", it is difficult to generalize to the non-trivial case without guidance.
The point of my approach is that I am trying to suggest that
many things one learns in "common sense" nonrelativistic physics can be reframed geometrically
to suggest (to provide guidance) to the non-intuitive relativistic physics
(as opposed to the familiar refrain
that I have heard that
one might have to start all over from scratch---
or in the words found in the old video game Marble Madness (and likely elsewhere),
"everything you know is wrong").

Admittedly, the approach is not that easy right now,
since it's unfamiliar.
But I have seen enough parallels in the literature
and now see a way to formalize the analogies and
visualize them with (say) desmos ( https://www.desmos.com/calculator/kv8szi3ic8 )...
so that they are not just cherry-picked...
but that they follow from similar constructions in the three geometries.

So, the reframing is to help build on what one knows (but likely from a different viewpoint)
on the way to special relativity
and beyond (e.g. tensors: http://www.opensourcephysics.org/CPC/posters/salgado-talk.pdf
from http://www.opensourcephysics.org/CPC/abstracts_contributed.html ).
 
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  • #39
robphy said:
btw.. I'm not a PF mentor.
But you mentor people, like me, which is what matters!

robphy said:
Many results in Galilean geometry are trivial... and since much of the physical interpretations are part of our "common sense", it is difficult to generalize to the non-trivial case without guidance.
The point of my approach is that I am trying to suggest that
many things one learns in "common sense" nonrelativistic physics can be reframed geometrically
to suggest (to provide guidance) to the non-intuitive relativistic physics
(as opposed to the familiar refrain
that I have heard that
one might have to start all over from scratch---
or in the words found in the old video game Marble Madness (and likely elsewhere),
"everything you know is wrong").

My interest in the eigenvectors issue has temporarily faded, but you make it revive with this excellent comment. I have names for this phenomenon that you mention, like "backward generalization" or "orphan elements". I mean that an analogy is a one-to-one correspondence between the sets of elements of two or several situations, as well as with the generalized set of elements that embraces all sets. And it often happens that one element looks "orphan" since you don't see its correspondence in the most elementary situation. You don't perceive that connection precisely because it is so "common sense" or "trivial" that you take it for granted. But when you find it, this is most rewarding because it links and thus illuminates both the elementary and the more advanced situations, thanks to this "backward generalization".
robphy said:
This Galilean-boost eigenvector corresponds to the limiting infinite speed (infinite slope) of all Galilean-timelike vectors,
similar to the Lorentz-boost eigenvectors corresponding to limiting [finite] velocity of all Minkowski-timelike vectors.

update: The Galilean boost eigenvector also happens to correspond with the concept of "absolute time" because this eigenvector is along the Galilean-circle. Since all tangent vectors to the Galilean circle are parallel, all Galilean observers agree on those events being simultaneous (since the tangents to the circle for each Galilean-4-velocity are Galilean-orthogonal to that 4-velocity).
This is unlike the situation for tangents to a circle (in 2D Euclidean space) and tangents to a hyperbola (in (1+1)-Minkowski spacetime).

The Galilean eigenvalue of 1 means that all Galilean observers will agree on the magnitude of displacement vectors parallel to ##(0,1)^{\text{T}}##. Physically, all Galilean observers agree on the length of a ruler. This is "absolute length".
I understand less the "limiting infinite speed" version of the eigenvector. It is easier to catch the "absolute time" version of the eigenvector and the "absolute length" as eigenvector = 1. If I am not mistaken, this means that the Galilean matrix transformation hinges on the fact that the time vectors remain intact (no change of direction, no change of size), just like the LT of SR hinges on the fact that light vectors experience no change of direction, although they are dilated by Bondi factor... Does this mean that you are assuming measurement of Galilean length with sonar method, as you mentioned in another post?
 
  • #40
Saw said:
But you mentor people, like me, which is what matters!
Thanks for the vote of confidence.
Saw said:
My interest in the eigenvectors issue has temporarily faded, but you make it revive with this excellent comment. I have names for this phenomenon that you mention, like "backward generalization" or "orphan elements". I mean that an analogy is a one-to-one correspondence between the sets of elements of two or several situations, as well as with the generalized set of elements that embraces all sets. And it often happens that one element looks "orphan" since you don't see its correspondence in the most elementary situation. You don't perceive that connection precisely because it is so "common sense" or "trivial" that you take it for granted. But when you find it, this is most rewarding because it links and thus illuminates both the elementary and the more advanced situations, thanks to this "backward generalization".
Interesting term.
An aspect of some parts of physics is to find patterns from initially-apparently-unrelated ideas.
Some successful unifications include "electricity and magnetism", "terrestrial and universal gravitation", etc...
Saw said:
I understand less the "limiting infinite speed" version of the eigenvector. It is easier to catch the "absolute time" version of the eigenvector and the "absolute length" as eigenvector = 1. If I am not mistaken, this means that the Galilean matrix transformation hinges on the fact that the time vectors remain intact (no change of direction, no change of size), just like the LT of SR hinges on the fact that light vectors experience no change of direction, although they are dilated by Bondi factor...
These conclusions are inspired by seeing corresponding results in the Euclidean and Minkowski cases (sources of some guidance to get from the Galilean case to the Special Relativistic case).

Saw said:
Does this mean that you are assuming measurement of Galilean length with sonar method, as you mentioned in another post?
I'm not sure how the radar-measurements of special relativity
will map down to an analogous procedure that is Galilean invariant.



In summary, I think eigenchris' presentation (which I have only glanced at) seems like it is compatible with the Galilean approaches I have been looking at. In some of his other videos, he has done a good job of explaining some ideas that I had difficulty explaining to myself and to others.
So, I think his videos have good value.
...but, yes, it might not be for everyone.
 
  • #41
robphy said:
In summary, I think eigenchris' presentation (which I have only glanced at) seems like it is compatible with the Galilean approaches I have been looking at. In some of his other videos, he has done a good job of explaining some ideas that I had difficulty explaining to myself and to others.
So, I think his videos have good value.
...but, yes, it might not be for everyone.

I totally agree. As I said from the start for me eigenchris' series is excellent and I also learn a lot from it. I understand that others may be indifferent to this attempt of seeking an analogy among the 3 areas (Euclidean, Galilean, Minkowskian), but I find it most useful, as it illuminates the three of them.

The thing is, however, that I see the analogy working differently from what you and eigenchris claim:

- instead of presenting Galilean time as another basis vector almost on the same footing as the 3 spatial basis vectors, I would precisely remark that Galilean time is not such thing; time is here, instead, as you rightly point out BTW, the eigenvector of the matrix transformation and as such it is an invariant thing, which distinctly differentiates it from that poor variant thing that spatial basis vectors are

- and the shift to SR means that light or, well, to be more accurate, two lightlike (and hence null) vectors take the role of time as eigenvectors, at the cost of time jumping on the wagon of variant things to rub elbows with the 3 spatial basis vectors, as a new basis vector, almost on the same footing as its new brothers, although it keeps a difference in that, when the moment for combination arrives (dot product of the spacetime displacement vector with itself) it is preceded by an opposite sign (though even this should be covered by a generalized expression which is blind to this detail).
robphy said:
I'm not sure how the radar-measurements of special relativity
will map down to an analogous procedure that is Galilean invariant.

Neither do I. I mentioned this only because you did mention a measurement with "sonar" in the context of the discussion about how to make time and space homogeneous...

But I would not dwell more on specific aspects in this post, without testing your interest in more discussion. Certainly, even if you don't agree with my different way of visualizing the analogy as stated above, I find the exchange most useful. If you find it of any use, I would suggest two debates:

- Getting deeper into the issue of what eigenvectors mean and how they operate in each of the 3 areas.
- This issue of conversion btw time and space units, where your views are less clear to me. You and eigenchris (I asked him and he is in your line) see no problem in expressing Galilean time measurements in length units by multiplying the former by an arbitrary speed, which will forcefully be variant, unlike c in SR. We did not advance much in this leg and it deserves discussion.

Apparently, PF does not like long discussions, but I bet this would not be a problem here: some of us (strange people) find analogies fruitful; this particular analogy (with Galilean and Euclidean areas) helps (at least us) to better understand SR and the subject about how this analogy works is not yet peaceful among us.

Also, I would not call this speculation! Say a carpenter has been using a saw for ages and one day he decides to discuss with his colleagues why toothed-things cut wood well, we would not ban the poor guy from learning something from that discussion!
 
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  • #42
Saw said:
I understand that others may be indifferent to this attempt of seeking an analogy among the 3 areas (Euclidean, Galilean, Minkowskian), but I find it most useful, as it illuminates the three of them.
IMO, this is illuminating:

http://www2.physics.umd.edu/~yakovenk/teaching/Lorentz.pdf
 
  • #43
robphy said:
This Galilean-boost eigenvector corresponds to the limiting infinite speed (infinite slope) of all Galilean-timelike vectors,
similar to the Lorentz-boost eigenvectors corresponding to limiting [finite] velocity of all Minkowski-timelike vectors.
I don't see this infinite limiting speed here. Of course you can get it by a symmetry analysis using the special principle of relativity, homogeneity of time, and Euclidicity of space for all inertial observers as well as the demand that the symmetry transformations must build a group and that the spacetime reconstructed from these symmetry group must admit a "causality structure".

From this analysis you finally get two space-time models: Galilei-Newton spacetime and Minkowski spacetime. The former can, understood in the right way, be interpreted as a deformation of the Minkowski spacetime in the limit of the limiting speed ##c \rightarrow \infty##.

The apparent posibility of a Euclidean affine manifold as a spacetime must be abandoned, because it doesn't admit a causality structure.
robphy said:
update: The Galilean boost eigenvector also happens to correspond with the concept of "absolute time" because this eigenvector is along the Galilean-circle. Since all tangent vectors to the Galilean circle are parallel, all Galilean observers agree on those events being simultaneous (since the tangents to the circle for each Galilean-4-velocity are Galilean-orthogonal to that 4-velocity).
Which "Gailean circle"? It's of course true that this analysis uses the concept of absolute time (and in a certain sense also an absolute space) in defining, how the Galilei boosts look like.

You can draw Galilean spactime diagrams much analogous to Minowski spacetime diagrams in SR. Guess, why the former aren't even used by physics-didactics people although they are very much obsessed about the latter ones!
robphy said:
This is unlike the situation for tangents to a circle (in 2D Euclidean space) and tangents to a hyperbola (in (1+1)-Minkowski spacetime).

The Galilean eigenvalue of 1 means that all Galilean observers will agree on the magnitude of displacement vectors parallel to ##(0,1)^{\text{T}}##. Physically, all Galilean observers agree on the length of a ruler. This is "absolute length".

Many results in Galilean geometry are trivial... and since much of the physical interpretations are part of our "common sense", it is difficult to generalize to the non-trivial case without guidance.
Galilean geometry is far from trivial. In some sense it's more complicated than Minkowskian geometry, at least concerning the Galilei group vs. the Poincare group when analyzed for their use in QT ;-).
 
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  • #44
vanhees71 said:
Galilean geometry is far from trivial. In some sense it's more complicated than Minkowskian geometry, at least concerning the Galilei group vs. the Poincare group when analyzed for their use in QT ;-).
And this is the problem. How does a student ever learn SR when they are sidetracked onto the Poincare group before they have touched on the Lorentz Transformation?
 
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  • #45
The Lorentz group is a subgroup of the Poincare group. Of course you should study it thoroughly, but without the spactime translations you'll have a hard time to construct the dynamics of relativistic QT.
 
  • #46
vanhees71 said:
The Lorentz group is a subgroup of the Poincare group. Of course you should study it thoroughly, but without the spactime translations you'll have a hard time to construct the dynamics of relativistic QT.
How much Group Theory does the OP know?
 
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  • #47
robphy said:
This Galilean-boost eigenvector corresponds to the limiting infinite speed (infinite slope) of all Galilean-timelike vectors,
similar to the Lorentz-boost eigenvectors corresponding to limiting [finite] velocity of all Minkowski-timelike vectors.

vanhees71 said:
I don't see this infinite limiting speed here. Of course you can get it by a symmetry analysis using the special principle of relativity, homogeneity of time, and Euclidicity of space for all inertial observers as well as the demand that the symmetry transformations must build a group and that the spacetime reconstructed from these symmetry group must admit a "causality structure".

Consider an analogue for a power iteration method for attempting the computation of an eigenvector of a matrix.
Repeatedly applying a Galilean-boost to a Galilean 4-velocity,
\left( \begin{array}{cc}<br /> 1 &amp; 0 \\<br /> v &amp; 1<br /> \end{array} \right)^n<br /> \left( \begin{array}{c}<br /> 1 \\<br /> u<br /> \end{array} \right)<br /> =\left( \begin{array}{c}<br /> 1 \\<br /> nv+u<br /> \end{array} \right)<br />
tends to a Galilean 4-velocity with unbounded speed as n\rightarrow \infty.

vanhees71 said:
You can draw Galilean spactime diagrams much analogous to Minowski spacetime diagrams in SR. Guess, why the former aren't even used by physics-didactics people although they are very much obsessed about the latter ones!
Every introductory physics textbook draws a Galilean spacetime diagram ( a position-vs-time graph ),
although they probably don't know that.

robphy said:
Many results in Galilean geometry are trivial... and since much of the physical interpretations are part of our "common sense", it is difficult to generalize to the non-trivial case without guidance.

vanhees71 said:
Galilean geometry is far from trivial. In some sense it's more complicated than Minkowskian geometry, at least concerning the Galilei group vs. the Poincare group when analyzed for their use in QT ;-).

Yep, I agree.
That's why I said "Many results in Galilean geometry are" and not "Galilean geometry is".
Some aspects (like the degenerate properties of the metrics) introduce complications not seen in the Euclidean and the Minkowski case.
 
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  • #48
robphy said:
Consider an analogue for a power iteration method for attempting the computation of an eigenvector of a matrix.
Applied to a Galilean 4-velocity,
\left( \begin{array}{cc}<br /> 1 &amp; 0 \\<br /> v &amp; 1<br /> \end{array} \right)^n<br /> \left( \begin{array}{c}<br /> 1 \\<br /> u<br /> \end{array} \right)<br /> =\left( \begin{array}{c}<br /> 1 \\<br /> nv+u<br /> \end{array} \right)<br />
tends to a Galilean 4-velocity with unbounded speed as n\rightarrow \infty.Every introductory physics textbook draws a Galilean spacetime diagram ( a position-vs-time graph ),
although they probably don't know that.
That's not a Galilean spacetime diagram depicting two different inertial frames.

I once did this for curiosity. The result is

galilei-diagram.png

I think it's really good for nothing, confusing the simple formulae of the Galilei transformation even more than the Minkowski diagram for the somewhat more complicated formulae of the Lorentz transformation.
 
  • #49
vanhees71 said:
That's not a Galilean spacetime diagram depicting two different inertial frames.

I once did this for curiosity. The result is

View attachment 321007
I think it's really good for nothing, confusing the simple formulae of the Galilei transformation even more than the Minkowski diagram for the somewhat more complicated formulae of the Lorentz transformation.

In a similar attempt, I have drawn this, which tries to convey that the different vertical positions that the origins O and O take at each instant do not constitute, strictly speaking, different time axes. The timeline is a different thing that does not form part of the "basis" just like the lightlike vectors being the eigenvectors in SR.

1674577433313.png
 
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  • #50
vanhees71 said:
That's not a Galilean spacetime diagram depicting two different inertial frames.
But it is. You (and likely many others) just don't see it and maybe don't appreciate it.
Maybe the situation is uninteresting to you.

(For comparisons with standard position-vs-time diagrams and xy-axes in the Euclidean plane,
the time axes below all run horizontally to the right. I use "y" for space in the spacetime diagrams.)


From earlier...
1674575328016.png


Here's a more interesting application:
Rolling without slipping as a superposition of motions in two reference frames.
Though one need not actively invoke or recognize the underlying Galilean geometry,
it is implicitly being used.

1674576918422.png

vanhees71 said:
I once did this for curiosity. The result is

View attachment 321007
I think it's really good for nothing, confusing the simple formulae of the Galilei transformation even more than the Minkowski diagram for the somewhat more complicated formulae of the Lorentz transformation.

I acknowledge your opinion.

Sometimes things (even apparently silly and useless things, at first glance) are done for practice,
in preparation for more complicated operations.
Hideo Miyagi said:
"Paint the fence"

Sometimes one needs to see it in comparison to the other E-alternatives,
like in time dilation and length contraction
(as in my https://www.desmos.com/calculator/kv8szi3ic8 and in my poster linked above https://www.aapt.org/doorway/Posters/SalgadoPoster/Salgado-GRposter.pdf ).

1674575901248.png


1674575943970.png


These diagrams suggest (in hindsight) that Galilean is an intermediate geometry.
With guidance, one can be led to Special Relativity
(as in the Parable that I wrote attached to the end of the poster where
  • a Galilean tries to interpret a position-vs-time graph geometrically
    in the spirit of Euclidean geometry with an experimentally determined circle on that diagram (where arc-lengths of worldlines are measured by wristwatches)
  • then Minkowski (with improved experimentation) determines a more realistic circle on that diagram.
)
 
  • #51
vanhees71 said:
Concerning the Galilei boost, let's to for 1D motion. Then the "vector" is $$\vec{x}=\begin{pmatrix} t \\ x \end{pmatrix}$$ and the boost is described by $$\vec{x}'=\begin{pmatrix} t \\ x-v t \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ v & 1 \end{pmatrix} \begin{pmatrix} t \\ x \end{pmatrix}.$$ This "Jordan matrix" has only one eigenvector ##(0,1)^{\text{T}}## with eigenvalue ##1##, which simply tells you that at time ##t=0## all the points on the ##x## axis are unchanged by the transformation, which is, however trivial anyway.

I have been thinking about this and... are you sure of the last sentence?

First, if we are to be consistent in objecting to the view that time can be another basis vector of Galilean bases, we should not represent it with an x component, whatever it is (even if it is 0). That is why I wondered earlier if we should call it at all an "eigenvector" or rather a scalar, a scalar on which the Galilean transformation hinges, just like other transformations hinge on a vector. (Or, if you wish, we could talk about an eigentensor of rank 0!)

But anyhow, even if consider time as a vector and an eigenvector, what is 1 and what is 0 in ##(0,1)^{\text{T}}##? It seems that if the expression refers to the time unitary vector, then x = 0 and t = 1. So how can you say that the eigenvalue = 1 means that "at time ##t=0## all the points on the ##x## axis are unchanged by the transformation"? Should it not rather mean that t=1s (just like any other t value of the timeline) is not changed in size by the transformation? Instead, the x value is changed by the transformation, obviously, since it becomes x'.

Also note that, if we are serious about Galilean time being this (a scalar), with regard to it the two concepts (eigenvector and eigenvalue) merge: since the number has no direction, the fact that it is unaltered by the transformation, simply means that its magnitude remains unchanged.

All this looks like truisms or trivial remarks, which is a good indication that it is valuable as an example of what I called in post #39 an orphan element of the analogy: the functional equivalent of the SR eigenvector and eigenvalue (lightlike vectors and Doppler factor) is the Galilean time, a simple scalar.

PS: What to do with length? Length l of objects, unlike x value and like t value, also remains unaltered by the Galilean transformation. Two possibilities: try to link it to time, as apparently robphy is doing, or view it as another scalar on which the transformation hinges, probably the latter.
 
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  • #52
Saw said:
I have been thinking about this and... are you sure of the last sentence?
That was the question about the eigenvectors and eigenvalues of
$$\hat{B}=\begin{pmatrix}1 &0 \\ v & 1 \end{pmatrix}.$$
First you need for the eigenvalues [EDIT: forgotten square added]
$$P_B(\lambda)=\mathrm{det} (\hat{B}-\lambda \hat{1}) = (1-\lambda)^2.$$
There's only one eigenvalue ##\lambda=1##. To find the eigenvectors you need
$$\hat{B} \vec{x} = \vec{x} \; \Rightarrow \; \begin{pmatrix}t \\v t+x \end{pmatrix}= \begin{pmatrix}t \\ x \end{pmatrix}.$$
For ##v \neq 0## obviously the 2nd component of this vector equation tells you that the only solution is ##t=0## and ##x \neq 0##, i.e., there's only one eigenvector (up to an arbitrary factor of course) ##(0,1)^{\text{T}}##.
Saw said:
First, if we are to be consistent in objecting to the view that time can be another basis vector of Galilean bases, we should not represent it with an x component, whatever it is (even if it is 0). That is why I wondered earlier if we should call it at all an "eigenvector" or rather a scalar, a scalar on which the Galilean transformation hinges, just like other transformations hinge on a vector. (Or, if you wish, we could talk about an eigentensor of rank 0!)
As I said, I don't see much use for introducing such spacetime vectors for Galilei-Newton spacetime, but it's of course not in any way wrong to do so if you wish. I don't understand what you want to say concerning eigen vectors, scalars, and tensors.
Saw said:
But anyhow, even if consider time as a vector and an eigenvector, what is 1 and what is 0 in ##(0,1)^{\text{T}}##? It seems that if the expression refers to the time unitary vector, then x = 0 and t = 1. So how can you say that the eigenvalue = 1 means that "at time ##t=0## all the points on the ##x## axis are unchanged by the transformation"? Should it not rather mean that t=1s (just like any other t value of the timeline) is not changed in size by the transformation? Instead, the x value is changed by the transformation, obviously, since it becomes x'.
Time is not a vector but a component of the above introduced spacetime vector. For the other questions see the solution of the eigenvalue problem above.
Saw said:
Also note that, if we are serious about Galilean time being this (a scalar), with regard to it the two concepts (eigenvector and eigenvalue) merge: since the number has no direction, the fact that it is unaltered by the transformation, simply means that its magnitude remains unchanged.
Galilean time is a scalar with respect to spatial rotations but, in this formalism with Galilean four-vectors, not with respect to the Galilei boosts, i.e., not with respect to the Galilei group.
Saw said:
All this looks like truisms or trivial remarks, which is a good indication that it is valuable as an example of what I called in post #39 an orphan element of the analogy: the functional equivalent of the SR eigenvector and eigenvalue (lightlike vectors and Doppler factor) is the Galilean time, a simple scalar.
Sure, if it helps you to understand Newtonian and/or SR better, why not use it? I find it that it overcomplicates trivial things and doesn't shed any light on the really interesting properties of the Galilei vs. Poincare group, which in its full physical meaning comes into the light only when analyzing these symmetries in the context with quantum (field) theory.
Saw said:
PS: What to do with length? Length l of objects, unlike x value and like t value, also remains unaltered by the Galilean transformation. Two possibilities: try to link it to time, as apparently robphy is doing, or view it as another scalar on which the transformation hinges, probably the latter.
The length of objects is of course time-independent in Galilean spacetime. In the Galilean four-vector formalism all time components of objects are the same, ##t##, because time is absolute in Galilean spacetime. That's why differences between events have time component ##0##, And the Euclidean length of the usual spatial 3-vectors are thus invariant under all Galilei transformations (including temporal and spatial translations, spatial rotations, and Galilei boosts).

There's no useful 4D metric or pseudo metric for Galilean spacetime, and that's why I think that this formalism is too clumsy to be helpful in understanding the Galilean spacetime geometry.
 
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  • #53
vanhees71 said:
That was the question about the eigenvectors and eigenvalues of
$$\hat{B}=\begin{pmatrix}1 &0 \\ v & 1 \end{pmatrix}.$$
First you need for the eigenvalues
$$P_B(\lambda)=\mathrm{det} (\hat{B}-\lambda \hat{1}) = (1-\lambda).$$
There's only one eigenvalue ##\lambda=1##. To find the eigenvectors you need
$$\hat{B} \vec{x} = \vec{x} \; \Rightarrow \; \begin{pmatrix}t \\v t+x \end{pmatrix}= \begin{pmatrix}t \\ x \end{pmatrix}.$$
For ##v \neq 0## obviously the 2nd component of this vector equation tells you that the only solution is ##t=0## and ##x \neq 0##, i.e., there's only one eigenvector (up to an arbitrary factor of course) ##(0,1)^{\text{T}}##.
Thanks, noted, I follow you now!
But then I have a technical question, since, according to my Linear Algebra notes:
- a square n x n matrix has n eigenvalues, so here we should have 2
- if only one appears, it may be that it is "repeated", meaning that there may be two eigenvectors with the same eigenvalue.
One eigenvector is (according to your notation where you put t as first component and x as second), the one you mention, ##(0,x)^{\text{T}}##, which assumes ##v \neq 0## and ##t=0##, i.e. there is relative motion but the film has not started.
But shouldn't we also consider the other possibility that you dismiss, where there is no relative motion, yes, we are in a static situation, ##v = 0##, but time is passing by, ##t\neq 0##, so the other eigenvector would be ##(t,0)^{\text{T}}##?
The first would account, actually, for absolute length, the second for absolute time.
(I admit that this sounds weird, but that's life with this trivial but complex Galilean relativity...)

vanhees71 said:
There's no useful 4D metric or pseudo metric for Galilean spacetime, and that's why I think that this formalism is too clumsy to be helpful in understanding the Galilean spacetime geometry.
Yes, I agree with that, with the utmost respect for those who defend that formalism. I also subscribe to their wish to find a smooth transition to SR, although IMHO the way to do it didactically, without misleading, should be pointing out that "going 4D" is not what Galilean relativity "somehow" does, but precisely what it does not and what SR does.

vanhees71 said:
As I said, I don't see much use for introducing such spacetime vectors for Galilei-Newton spacetime, but it's of course not in any way wrong to do so if you wish.
But then here you become too lenient... To me, it is simply wrong, for the above-mentioned reason.

vanhees71 said:
Galilean time is a scalar with respect to spatial rotations but, in this formalism with Galilean four-vectors, not with respect to the Galilei boosts, i.e., not with respect to the Galilei group.

So in this Galilean 4D formalism time is treated as a vector with respect to the Galilei boosts, isn't it? It should not be, IMHO.

vanhees71 said:
I don't understand what you want to say concerning eigen vectors, scalars, and tensors.

I am simply saying that in Galilean relativity absolute time and absolute length are scalars, for all purposes. Despite that, by playing with the matrix of the Galilean Transformation, we have observed that there are two (two if you buy my first comment) things that remain invariant when that matrix is applied to them. But I don't like to call those things "eigenvectors", because that would amount to admitting that Galilean time (and length) are vectors. So I call them "eigen" (because they remain unaffected under a transformation) + "tensors" (because scalars are tensors of rank 0, arent't they?).
 
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  • #54
Saw said:
But then I have a technical question, since, according to my Linear Algebra notes:
- a square n x n matrix has n eigenvalues, so here we should have 2
  • A planar rotation is 2x2 matrix has no eigenvectors [since all nonzero vectors are mapped to other vectors not proportional to it] and thus no eigenvalues.
    (This doesn't count: https://www.wolframalpha.com/input?...sqrt(1+v^2)],+[v/sqrt(1+v^2),1/sqrt(1+v^2)]+] )
  • A (1+1)-Lorentz boost is 2x2 matrix that has exactly two eigenvectors (two lightlike vectors) and the corresponding eigenvalue are k and 1/k. (https://www.wolframalpha.com/input?...sqrt(1-v^2)],+[v/sqrt(1-v^2),1/sqrt(1-v^2)]+])
    Note the eigenvectors are independent of v... they are frame-invariant directions.
  • A (1+1)-Galilean boost is 2x2 matrix that has exactly one eigenvector, and its eigenvalue equals 1.
    (https://www.wolframalpha.com/input?i=eigenvectors+of+[++[1,0],+[v,1]+] )
    Note the eigenvector is independent of v... it's a frame-invariant direction.

    [ Wolframalpha does offer a "generalized eigenvector"...
    but I am not venturing in that possible extension. May be useful: https://www2.math.upenn.edu/~moose/240S2013/slides7-31.pdf ]

    The eigenvalue being 1 means that
    the spacetime displacement vector \left( \begin{array}{c} 0 \\ L \end{array} \right)
    (which physically represents a "length" L (as the distance between two parallel worldlines measured as measured by the displacement vector between two events at the same time according to the measurer)
    is mapped to the same vector.
    In short, the measured length of an object is independent of the observer measuring the length.
    This is "absolute length".
Saw said:
- if only one appears, it may be that it is "repeated", meaning that there may be two eigenvectors with the same eigenvalue.
Concerning this bullet point, the 2x2 identity matrix has repeated eigenvalues and has every direction as an eigenvector. However, this is not the case for (1+1)-Galilean boost. There is only one eigenvector and its eigenvalue is 1. (A vector ##\vec w## that is not of the form ##(0,L)^{\text{T}}## is mapped to another vector not-parallel-to-##\vec w##.)

Saw said:
One eigenvector is (according to your notation where you put t as first component and x as second), the one you mention, ##(0,x)^{\text{T}}##, which assumes ##v \neq 0## and ##t=0##, i.e. there is relative motion but the film has not started.
But shouldn't we also consider the other possibility that you dismiss, where there is no relative motion, yes, we are in a static situation, ##v = 0##, but time is passing by, ##t\neq 0##, so the other eigenvector would be ##(t,0)^{\text{T}}##?
No. ##(0,x)^{\text{T}}## does NOT assume ##v \neq 0## and ##t=0##.
##(0,x)^{\text{T}}## is not a Galilean 4-velocity, a unit timelike 4-vector (a vector of the form ##(1,v)^{\text{T}}##).
v is a slope.
There is no implied motion for ##(0,x)^{\text{T}}##. It is a purely-spatial displacement.
Saw said:
So in this Galilean 4D formalism time is treated as a vector with respect to the Galilei boosts, isn't it? It should not be, IMHO.

[snip]

I am simply saying that in Galilean relativity absolute time and absolute length are scalars, for all purposes. Despite that, by playing with the matrix of the Galilean Transformation, we have observed that there are two (two if you buy my first comment) things that remain invariant when that matrix is applied to them.

[space]

But I don't like to call those things "eigenvectors", because that would amount to admitting that Galilean time (and length) are vectors. So I call them "eigen" (because they remain unaffected under a transformation) + "tensors" (because scalars are tensors of rank 0, arent't they?).
  • Time is not a vector in spacetime, not in Galilean relativity and not in Special relativity.
    Time is a scalar, read off by a wristwatch.
    Perhaps "Wristwatch time" (Minkowski's "proper time") is more descriptive.
  • The Galilean "4-velocity", a unit timelike 4-vector (a vector of the form ##(1,v)^{\text{T}}##),
    is the tangent vector to the worldline of an object.
    This vector maps, for example,
    the event "object A's watch (on object A's worldline) reads time tick-1" to
    the event "object A's watch (on object A's worldline) reads time tick-2",
    and similarly for other objects (generally traveling with different speeds) and other tick-and-(tick+1)'s.

    (The "relativity-principle" implies there are no timelike eigenvectors.
    There is no distinguished finite speed for 4-velocities [along timelike worldlines].)
  • "Absolute time" implies that when observers A and B (regardless of their states of motions)
    agree on the elapsed time between two events,
    as if the spacetime diagram can be sliced (foliated) into "parallel planes of events"
    that all observers agree are simultaneous.

In my opinion, various mathematical objects in this approach
nicely match up with the physics that it is trying to model.

My strategy is to let the mathematical model (and its consequences)
guide the interpretation... and let's see where it gets us.
We do this in special relativity with the Minkowski metric to try to explain what we observe.
It seems like a natural question to see what we get starting with a Galilean metric
to explain what we observe [within the context and mindset of a Galilean].

It might be good to transcribe the abstract of a paper I mentioned earlier
by Jammer and Stachel
"If Maxwell had worked between Ampère and Faraday: An historical fable with a pedagogical moral"
American Journal of Physics 48, 5 (1980); https://doi.org/10.1119/1.12239
If one drops the Faraday induction term from Maxwell’s equations, they become exactly Galilei invariant. This suggests that if Maxwell had worked between Ampère and Faraday, he could have developed this Galilei‐invariant electromagnetic theory so that Faraday’s discovery would have confronted physicists with the dilemma: give up the Galileian relativity principle for electromagnetism (ether hypothesis), or modify it (special relativity). This suggests a new pedagogical approach to electromagnetic theory, in which the displacement current and the Galileian relativity principle are introduced before the induction term is discussed.
In short, one uses experiments (done in an alternate sequence) to first deduce
physics in the context of Galilean relativity. Then, when new and improved experiments
are done, one is forced to replace Galilean relativity by Einstein/Minkowskian relativity.
(In the real sequence of events, Galilean relativity is only a passing footnote...
and we are thrust into Einstein's Special Relativity, with little experience with invariance.
Similarly, further experiments will suggest that we need to go to General Relativity.
But, pedagogically, it might be good to learn to walk before we try to run.)
 
  • #55
robphy said:
  • A (1+1)-Lorentz boost is 2x2 matrix that has exactly two eigenvectors (two lightlike vectors) and the corresponding eigenvalue are k and 1/k. (https://www.wolframalpha.com/input?i=eigenvectors+of+[++[1/sqrt(1-v^2),v/sqrt(1-v^2)],+[v/sqrt(1-v^2),1/sqrt(1-v^2)]+])
    Note the eigenvectors are independent of v... they are frame-invariant directions.
  • A (1+1)-Galilean boost is 2x2 matrix that has exactly one eigenvector, and its eigenvalue equals 1.
    (https://www.wolframalpha.com/input?i=eigenvectors+of+[++[1,0],+[v,1]+] )
    Note the eigenvector is independent of v... it's a frame-invariant direction.

    (...)

    The eigenvalue being 1 means that
    the spacetime displacement vector \left( \begin{array}{c} 0 \\ L \end{array} \right)
    (which physically represents a "length" L (as the distance between two parallel worldlines measured as measured by the displacement vector between two events at the same time according to the measurer)
    is mapped to the same vector.
    In short, the measured length of an object is independent of the observer measuring the length.
    This is "absolute length".

Well, yes, it is clear to me that an eigenvector is something that is not kicked-off its line and (if its eigenvalue is 1) keeps the same magnitude despite the transformation, here the Galilean boost (and, in this sense, it is independent of v).

I talked about the option of ##v \neq 0## based on this sentence of vanhees71, which you may clarify with him:
vanhees71 said:
For ##v \neq 0## obviously the 2nd component of this vector equation tells you that the only solution is ##t=0## and ##x \neq 0##, i.e., there's only one eigenvector (up to an arbitrary factor of course) ##(0,1)^{\text{T}}##.
But what is clear is that one must find the eigenvector/s, as vanhees71 mentioned, as a solution to these equations:

\begin{array}{l}<br /> t = t\\<br /> x + vt = x<br /> \end{array}

One hypothesis is t = 0 and then you get (0,x). If you call that (0,L) and associate the eigenvalue 1 to the fact that in the Galilean context there is "absolute length" (lengths are not perceived to contract because of relative motion), that is perfect to me. I also said it here:

Saw said:
One eigenvector is (according to your notation where you put t as first component and x as second), the one you mention, ##(0,x)^{\text{T}}##
Saw said:
The first would account, actually, for absolute length

If you then say that v = 0 is not only possible but what has to be done to solve the equations, this is perfect for me. It is not only consistent with the former case (where nothing changes if it is not only t = 0 but also v = 0), but also allows another solution: just make v = 0 and t = 0 and you get (t,0).

Mathematically this is clearly a possible solution (at least as possible as the previous one) and, from a logical point of view, it seems a must. Otherwise, you have accounted for "absolute length" and not for "absolute time"!

On another note, why should the Galilean matrix be an exception to the rule that a square n x n matrix needs two eigenvectors? Besides, the link for Wolframalpha about this does not work.

Conclusion: thanks for the tip that v can be made equal to 0, now all is in order, but for me with a repeated eigenvalue and 2 eigenvectors, if we abuse of the name, since we are in face of things that are not really vectors.

Now to the question of what they are.

robphy said:
  • Time is not a vector in spacetime, not in Galilean relativity and not in Special relativity.
    Time is a scalar, read off by a wristwatch.
    Perhaps "Wristwatch time" (Minkowski's "proper time") is more descriptive.
Good that you say that time is a scalar in Galilean relativity and not a vector. Note however that we are here because you and eigenchris claim that there exists a Galilean spacetime as vector space, with a separation spacetime vector btw events, composed of scaled basis time and space vectors, like here:

\vec S = \color{red}{x{\vec e_x}} + \color{red}{t{\vec e_t}} = \color{blue}{x{\vec e_x}} + \color{blue}{t{\vec e_t}}

even if you also admit that ultimately, in terms of metric, there is no single spacetime metric and you make the dot product of time interval and spatial distance independently.

But what puzzles me is that you say that time is not a vector in SR. Well, of course, "time is a vector" in SR is sloppy language. What we mean is that in order to calculate the spacetime distance between two events you make a linear combination of a time basis vector with a space basis vector.

And of course, when that spacetime distance is timelike its magnitude is proper time, i.e. the time read by a wristwatch that has been present at both events, but another frame which is in motion with regard to the latter will have a coordinate time interval which can only lead to the same invariant result by combination (subtraction) with a space interval.

Well, I don't need to clumsily explain to you these obvious things that you perfectly know...
 
  • #56
Saw said:
vanhees71 said:
That was the question about the eigenvectors and eigenvalues of
$$\hat{B}=\begin{pmatrix}1 &0 \\ v & 1 \end{pmatrix}.$$
First you need for the eigenvalues [EDIT: forgotten square added]
$$P_B(\lambda)=\mathrm{det} (\hat{B}-\lambda \hat{1}) = (1-\lambda)^2.$$
There's only one eigenvalue ##\lambda=1##. To find the eigenvectors you need
$$\hat{B} \vec{x} = \vec{x} \; \Rightarrow \; \begin{pmatrix}t \\v t+x \end{pmatrix}= \begin{pmatrix}t \\ x \end{pmatrix}.$$
For ##v \neq 0## obviously the 2nd component of this vector equation tells you that the only solution is ##t=0## and ##x \neq 0##, i.e., there's only one eigenvector (up to an arbitrary factor of course) ##(0,1)^{\text{T}}##.

One eigenvector is (according to your notation where you put t as first component and x as second), the one you mention, ##(0,x)^{\text{T}}##, which assumes ##v \neq 0## and ##t=0##, i.e. there is relative motion but the film has not started.
But shouldn't we also consider the other possibility that you dismiss, where there is no relative motion, yes, we are in a static situation, ##v = 0##, but time is passing by, ##t\neq 0##, so the other eigenvector would be ##(t,0)^{\text{T}}##?
The first would account, actually, for absolute length, the second for absolute time.
(I admit that this sounds weird, but that's life with this trivial but complex Galilean relativity...)

Note that
<br /> \left( \begin{array}{cc} 1 &amp; 0 \\ v &amp; 1 \end{array} \right)<br /> \left( \begin{array}{c} a \\ b \end{array} \right)<br /> =\left( \begin{array}{c} a \\ av+b \end{array} \right)<br /> \stackrel{?}{=} <br /> \lambda \left( \begin{array}{c} a \\ b \end{array} \right)<br />
Use
Code:
{{1,0},{v,1}}{{a},{b}}
https://www.wolframalpha.com/input?i={{1,0},{v,1}}{{a},{b}}

To find an eigenvector (i.e., acceptable pairs of (a,b) where at least a or b is nonzero),
we must have ##a=ka## and ##av+b=kb## for some real ##k##.
  • case ##v\neq 0##: implies ##(ka)v+b=kb##, we must have ##k=1## and (to kill off ##kav##) ##a=0##.
    Thus, we have an eigenvector ##(0,b)^{\text{T}}## with ##k=1##.
  • case ##v= 0##: implies we could have ##a\neq 0## and ##b=0## as you say.
    Thus, we have an eigenvector ##(a,0)^{\text{T}}## with ##k=1##, as you say.

    But note that we can also have ##(a,b)^{\text{T}}## with ##k=1##, with ##b\neq 0##.
    So, we also have an eigenvector ##(a,b)^{\text{T}}## with ##k=1##, with ##b\neq 0##.

    In fact, any nonzero vector ##(a,b)^{\text{T}}## ( including ##(0,b)^{\text{T}}## ) is an eigenvector in the case ##v=0##
    because the boost matrix in the ##v=0## case is ##\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)##.
    So, while ##(a,0)^{\text{T}}## is (trivially) an eigenvector so is any other nonzero vector ##(a,b)^{\text{T}}##.
    In other words, there is nothing special about ##(a,0)^{\text{T}}## in case of the zero-boost (the identity).
  • Think physically for a moment.
    If a non-trivial boost had a timelike eigenvector,
    then that implies a "preferred frame of reference"
    (since the timelike eigenvector will select a special 4-velocity vector,
    picking out a special inertial worldline).
    That violates the principle of relativity.
 
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  • #57
robphy said:
there is nothing special about ##(a,0)^{\text{T}}## in case of the zero-boost (the identity).
Thanks indeed for the complete explanation.

There was some confusion about the role of v. Now I think that all matches btw your explanations and vanhees71's. It is confirmed (I had even mentioned it, but then lost the assumption because I thought that you were not assuming it...) that, in order to determine the eigenvectors, we need that the transformation matrix is not the identity matrix because in that case it does nothing and hence not only the eigenvectors, all vectors would remain invariant! So we must start assuming that ##v \neq 0##.

A different thing is that, once the transformation is applied because ##v \neq 0##, it will do its job no matter the ##v##. I guess that that was the message of the bullets in your post #54.

For completeness, I will say that what Wolfram link does is giving you the rank of the matrix, i.e. the number of linearly independent rows and columns that it contains, which here is 1 despite of the fact that the matrix is 2x2. I investigated a little and found that you can have a matrix with a number of eigenvectors higher than its rank, but... that is the identity matrix, which is not apt here...

So we only have one (non-trivial) eigenvector which you express as ##(0,L)^{\text{T}}##. This makes sense to me if visualized as follows: assuming ##t = 0## is not so much placing ourselves at the start of the film, but considering a time interval = 0, that is to say, the film has been frozen at a given instant with a given picture; then the eigenvalue = 1 means that all frames agree that any x interval (which can be the length of any object) has the same value as per their respective measurements. In other words, they all use the same scale, there is "absolute length". This is unlike what happens in SR where the eigenvalue is the Doppler-Bondi factor and so the eigenvectors experience dilation and there are different scales in the Minkowski diagram. Is this more or less the meaning?

But then Galilean relativity also has "absolute time". Clearly, it is a scalar under Galilean relativity (BTW you did not reply to my puzzlement at your statement that it is also a scalar and not a vector in SR). But how do you accommodate this geometrically? Maybe under the trivial null eigenvector (0,0)?

Edit: or maybe as another meaning implicit in the existing single eigenvector? I must admit that I did not yet understand the last comment about thinking it physically and the impossibility of a timelike eigenvector. Keep thinking about it.
 
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  • #58
Saw said:
Thanks indeed for the complete explanation.
...
[snip] ...Is this more or less the meaning?
Yes, I'm glad this is now clear.

Saw said:
But then Galilean relativity also has "absolute time". Clearly, it is a scalar under Galilean relativity (BTW you did not reply to my puzzlement at your statement that it is also a scalar and not a vector in SR). But how do you accommodate this geometrically? Maybe under the trivial null eigenvector (0,0)?

Edit: or maybe as another meaning implicit in the existing single eigenvector? I must admit that I did not yet understand the last comment about thinking it physically and the impossibility of a timelike eigenvector. Keep thinking about it.

The eigenvectors of the boost should primarily be about maximum signal speeds.
So, if I were to redo that poster,
I would say "absolute maximum signal speed is infinite" instead of "absolute time"
for the ##(0,1)^\top## eigenvector of the Galilean boost.
And the eigenvectors represent limiting the possible 4-velocities (future-timelike directions)
in the Galilean case, as the lightlike eigenvectors in (1+1)-Minkowski spacetime.
(Note that the ##(0,1)^\top## eigenvector of the Galilean boost is also a null vector with respect to the temporal-Galilean metric. Together with the discussion below about absolute time,
this means that ##(0,1)^\top## is both null and spacelike in Galilean geometry,
as a result of "opening up the light cones" from the Minkowski case).The notion of "absolute time" is really about
different frames of reference agreeing on what sets of events are simultaneous.


But before I get to absolute time, I repeat my statements in #54
  • Time is not a vector in spacetime.
    Time is a scalar, read off by a wristwatch.
    Perhaps "Wristwatch time" (Minkowski's "proper time") is more descriptive.
    It is analogous to a distance as read off an odometer that is traveled along a path.
    The odometer reading has no direction.
  • The "4-velocity" is a unit timelike 4-vector
    is the unit tangent vector to the worldline of an object.
    This vector maps, for example,
    the event "object A's watch (on object A's worldline) reads time tick-1" to
    the event "object A's watch (on object A's worldline) reads time tick-2",
    and similarly for other objects (generally traveling with different speeds) and other tick-and-(tick+1)'s.
    Physically, we think of this vector as the "time axis of an observer".
A further geometric analogy might help distinguish
the observer's-time-axis (specified by a unit vector)
from the wristwatch time (a scalar).
  • The "unit radius vector" is a vector that points in some direction in the plane away from an origin O.
  • The "radius" is a scalar, a number that tells you how far away you are from that origin.

A set of events are simultaneous for an inertial observer
when that observer assigns to those events
the same value of t (as read off that observer's wristwatch).
Geometrically, those events lie on a hyperplane that is orthogonal to that inertial observer.

Orthogonality is defined by the circle in that geometry (related to the metric and dot-product).
The construction (familiar from Euclidean geometry and also given by Minkowski in his "Space and Time") is to find the tangent line to the "circle" (a hyperbola in (1+1)-Minkowski spacetime)
at the tip of the timelike 4-vector. To construct the inertial observer's spatial axis,
draw a vector parallel to this tangent line through the tail of the 4-velocity vector.

In 2D Euclidean geometry, the tangent lines to circles have different inclinations.
For each possible orientation of the rectangular coordinate system labeled by its x-axis,
the y-axis (the x=0 line) of one system is not parallel to the y-axis of another system.
That is, when one system assigns the same x-coordinate to all points on a line,
another system will generally assign the points on that line different individual x-coordinates.

A similar thing happens in special relativity.
The tangent lines to the unit future-hyperbolas (the unit Minkowskian circles whose events are one tick in the future, as read off by wristwatches on inertial worldlines) from an event O have different inclinations.
The inertial observers indexed by their timelike 4-velocities will have different y-axes (t=0 lines) for each observer.
Thus, two events simultaneous (having the same t-coordinate assignment) in one frame
may not be simultaneous in another.
This is the relativity of simultaneity.

For Galilean relativity, with its Galilean circle ( the "t=1" line),
all tangent lines coincide. So, if an observer says two events are simultaneous,
then all other observers will agree.
This universal agreement is absolute simultaneity...
and is a peculiarity of the E=0 case. Among the E-parametrization of geometries,
the Galilean situation (our common sense) is the exception, not the general rule.

To appreciate this, here is an animation from my spacetime diagrammer ( https://www.desmos.com/calculator/kv8szi3ic8 ),
showing the cases for E varying from -1 (Euclidean) to 0 (Galilean) to +1 (Minkowski)

d8q62.gif

(if the above isn't animated, visit https://i.stack.imgur.com/d8q62.gif
which is taken from my answer at https://physics.stackexchange.com/questions/673969/euclidean-space-to-minkowski-spacetime )Because of this universal agreement of simultaneity,
we can slice up spacetime into universally accepted hyperplanes of simultaneity,
which can be labeled by any observer's wristwatch.
This universal labeling by a wristwatch (a scalar) is absolute time.

The relativity-principle says that there is no preferred wristwatch,
no preferred 4-velocity vector (no timelike eigenvector).

(Note that each choice of 4-velocity will
"bevel the deck" of hyperplanes-of-universal-simultaneity differently,
which effectively assigns the same value of the y-coordinate
to events parallel to the 4-velocity.)
...From my ancient set of webpages on relativity
(now archived at http://visualrelativity.com/LIGHTCONE/ )
gal1-m.gif
Bottom line:
  • Observer-Wristwatch-time is a scalar.
  • Observer 4-velocity for the observer-time-axis is a vector.
 

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  • #59
robphy said:
The eigenvectors of the boost should primarily be about maximum signal speeds.
So, if I were to redo that poster,
I would say "absolute maximum signal speed is infinite" instead of "absolute time"
for the ##(0,1)^\top## eigenvector of the Galilean boost.
And the eigenvectors represent limiting the possible 4-velocities (future-timelike directions)
in the Galilean case, as the lightlike eigenvectors in (1+1)-Minkowski spacetime.
(Note that the ##(0,1)^\top## eigenvector of the Galilean boost is also a null vector with respect to the temporal-Galilean metric. Together with the discussion below about absolute time,
this means that ##(0,1)^\top## is both null and spacelike in Galilean geometry,
as a result of "opening up the light cones" from the Minkowski case).The notion of "absolute time" is really about
different frames of reference agreeing on what sets of events are simultaneous.
Well, it is true that thanks to an infinite signal speed, which would be absolute by definition, you would ensure absolute clock synching and you would be able to check if the objects change their length and the clocks slow down or not due to motion..., which they are not supposed to do in this context. So you propose to visualize the Galilean eigenvector as a sort of "super-lightlike vector"?

More later on the nature of time as scalar vs. vector.
 
  • #60
I still think it's much more worthwile to argue with physics rather than spacetime diagrams. I must admit that I'm myself forced to teach Minkowski diagrams to my high-school teacher students only, because that's the way they have to present it at school (if it's presented at all for that matter :-(). I find them much more difficult to understand than the pretty simple math of four-vectors and four-tensors and the Poincare/Lorentz transformations.

So much more important is a discussion of the physical concepts, and it's a great opportunity to discuss the interdependence between theory and observations and their quantification.

In my opinion one should also emphasize what the spacetime models (i.e., Newton-Galilei, Einstein-Minkowski, and Einstein GR/Einstein-Cartan) in "practical use" (where the Einstein-Cartan manifolds are still rather rarely discussed in lack of feasible observations concerning the torsion in fermionic polarized matter) have in common.

Galilei-Newton and Einstein-Minkowski spacetime have in common the special principle of relativity, i.e., the existence and equivalence of all "inertial frames of reference" together with the Euclidicity of space for all inertial observers. From a physical point of view in Newtonian physics (i.e., classical non-relativistic mechanics) the paradigm are point particles and actions at a distance, such that one can establish the existence of rigid bodies, which enable global clock synchronizing by using rigid rods, which have an infinite sound velocity, i.e., you can synchronize all clocks with instantaneously propagating signals, and thus you can even synchronize all clocks, no matter in which however accelerated motion they are, which ensures that you have an absolute time, even when using non-inertial frames of reference, which are as easy or difficult to realize as inertial frames of refence by just using three non-complanar rigid rods fixed at an origin, which may move in any way against an arbitrary inertial reference frame, i.e., not only the inertial but all observers describe space as a 3D Euclidean affine manifold.

In SR there are no rigid rods and no faster-than-light signal propagation that can be used to synchronize all clocks in arbitrary motion relative to any inertial frame, let alone even non-inertial frames. Clock synchronization in arbitrary (inertial or non-inertial) is a matter of definition or convention, and the most simple one for inertial frames is the use of light signals and light clocks as, e.g., in Einstein's famous paper of 1905. Mathematically this is the choice of pseudo-Cartesian (Lorentzian) basis vectors in Minkowski spacetime, i.e., the possibility to choose a global constant tetrad fieldm which establishes the SR spacetime manifold to be a 4D pseudo-Euclidean affine space with a fundamental form with signature (1,3) or equivalently (3,1), i.e., a 4D Lorentzian affine manifold.

The consequences are pretty severe. E.g., there's no consistent classical model for closed systems of interacting point particles, there's even a "no-go theorem" that such a model can at least not been formulated with the action principle. The successful descriptions of matter, both in the classical and in the quantum realm, are local (quantum) field theories (relativistic hydro/magnetohydro, Boltzmann(-Uehling-Uhlenbeck) transport, to some extent Kadanoff-Baym,...).

Last but not least GR spacetime can be seen as a gauge theory which makes the Poincare symmetry local, i.e., there are no more global inertial reference frames but only local ones, and you can reinterpret this as a pseudo-Riemannian (Lorentzian) differentiable manifold.

As it further turns out when considering matter content with "particles" with spin one has to extend this scheme to a Einstein-Cartan manifold with torsion. However for the usual macroscopic/astronomical observables, where only a semi-classical description of matter (hydro/magneto-hydro, or semi-classical Boltzmann-Uehling-Uhlenbeck transport) and the (classical) electromagnetic field is needed, one is lead back to the Lorentzian torsion-free spacetime model to GR. Whether the prediction of the necessity of torsion in polarized/spinning matter is correct or not is, as far as I know, not yet possible to be observed.
 

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