I Can time be another basis vector under Galilean relativity?

[Saw]

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.

 

[vanhees71]

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) $(0,1)^{\text{T}}$.

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.

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.

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.

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.

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.

 

[Saw]

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...)

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.

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.

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.

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?).

 

[robphy]

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".

- 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-$\vec w$.)

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.

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.)

 

[Saw]

• 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:

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:

One eigenvector is (according to your notation where you put t as first component and x as second), the one you mention, $(0,x)^{\text{T}}$

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.

• 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...

 

[robphy]

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

{{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.
  

 

[Saw]

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.

 

[robphy]

Thanks indeed for the complete explanation.
...
[snip] ...Is this more or less the meaning?

Yes, I'm glad this is now clear.

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)

(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/ )

Bottom line:

• Observer-Wristwatch-time is a scalar.
• Observer 4-velocity for the observer-time-axis is a vector.

 

[Saw]

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.

 

[vanhees71]

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.

 

[Saw]

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.

Further thoughts on the same subject, in the line of relating the Galilean eigenvector to an infinite speed.

You cannot define it as a speed, just like you don't define it as the speed of light in SR. It must be a displacement vector composed of x and t intervals. If it travels instantaneously, it means that, despite employing t = 0, its x is everywhere. Would it then make sense to define it, not as $(0,1)^\top$, assuming that x = 1, but as $(0,x)^\top$, assuming that x is everywhere? (note that in fact, in the derivation of the eigenvalue, x was the X coordinate of "a certain vector"; we can assume that it was a unitary vector, but that is an assumption; understanding now what it means, I would leave the x).

Also, it is to be noted that in SR the eigenvectors are the lightlike vectors going in each direction, X and -X. Here, instead, we have only one Galilean eigenvector. This may imply that its expression is telling you precisely this: "don't see me as a vector that is somewhere, looking either left or right, but as one that is everywhere at the same time".

In conclusion, this is what I would understand from your development, in terms of eigenvectors / eigenvalues:

eigenvectors:
- just like Minkowskian observers have a finite speed instrument with which they fix lengths and synch clocks (actually 2 instruments, one going in each direction), Galilean observers have also an infinite speed instrument for the same purposes;
- in agreement with its nature of eigenvectors, these instruments remain on their respective lines despite boosts and this requires that they keep the same correlation between x and t, i.e. the same speed, either a finite or infinite one, as the case may be: in the Minkowskian case this has the simple reading of the same correlation between t and x, in the Galilean case what remains invariant is the statement "I am a vector that in no time can reach everyhwere";

eigenvalues: in the Galilean case, the eigenvalue = 1 entails that the scale is maintained and objects keep their lengths and clocks keep their time rates and remain synched, whereas in the Minkowskian case we only manage to keep the invariance of the speed of light at the cost of a dilation and a change of scale, the change of scale being given by Doppler-Bondi factor and being reciprocal, so that each frame sees the other as "out of scale" (and hence length-contracted, time-dilated and de-synched).

Would you agree to this para-phrasing of your points?

I am quite happy with it, as it accommodates the idea of speed as a quality of the eigenvector and the idea of scale (affecting both length and time) as a quality of the eigenvalue.

 

[robphy]

---

Further thoughts on the same subject, in the line of relating the Galilean eigenvector to an infinite speed.

[ snip]

Would you agree to this para-phrasing of your points?

I think you have mostly got it.

However, these passages sound contradictory (word-coloring by me)

Further thoughts on the same subject, in the line of relating the Galilean eigenvector to an infinite speed.

You cannot define it as a speed, just like you don't define it as the speed of light in SR.

... [snip] ...

eigenvectors:
- just like Minkowskian observers have a finite speed instrument with which they fix lengths and synch clocks (actually 2 instruments, one going in each direction), Galilean observers have also an infinite speed instrument...

I would describe an eigenvector of the (1+1)-Lorentz-boost
as having a velocity-magnitude of "the speed of light".
This speed is not attainable by any future-timelike vector,
although a sequence of repeatedly-boosting a future-timelike vector tends to it.

This is in analogy to what I said in #47,
where an eigenvector of the (1+1)-Galilean boost
is not attainable by any future Galilean-timelike vector,
although the sequence of repeatedly-boosting a future Galilean-timelike vector tends to it.
I describe that as an infinite speed (although I don't want to get into discussions about infinity).
At this point, I'm more concerned with what the mathematical structure implies
in terms of describing the physics and less concerned about what the structure is called.
(Since I recently re-watched on YouTube Feynman's The Pleasure of Finding Things Out....
this made me think about the passage about "knowing the name of something".)

By the way, one could describe the finite speed of light eigenvector in the (1+1)-Lorentz boost
by an infinite rapidity, not attainable by a timelike vector, but it is an upper bound.
The Galilean rapidity associated with its eigenvector, not attainable by a Galilean-timelike vector,
is also infinite.

In terms of coordinates, in light-cone $(u,v)$ coordinates,
where frontward-future eigenvector is $(1,0)$ with finite speed,
nonzero timelike-vectors can't have one of these coordinates as zero.
Similarly, in $(t,x)$ coordinates, where the eigenvector is $(0,1)$ with my "inifinte" speed,
nonzero Galilean-timelike-vectors can't have the t-coordinate as zero.

In the business of extending an idea by a specific set of analogies,
we sometimes have to loosen the traditional definition of a word.
At this stage,
I am happy calling the Galilean-eigenvector $(0,1)^\top$ an "infinite" speed
because I can and have used the mathematical construction to make progress toward
formalizing the analogies and connecting them to the physics.
There may be a better term.

But in my current intuition, I think of velocity as a slope.
On occasion, in Euclidean geometry, I may have a vertical segment,
where one has a similar issue with the "slope" of a vertical segment.
How one handles this situation could be how we should handle the description of $(0,1)^\top$.

---

Would it then make sense to define it, not as $(0,1)^⊤$, assuming that x = 1, but as $(0,x)^⊤$, assuming that x is everywhere?

An eigenvector is really a direction, a slope (if I am permitted an extension on the term).
When $(0,1)^⊤$ is an eigenvector $(0,17)^⊤$ is also an eigenvector. And, of course,
they have the same eigenvalue.

---

The discussion of the eigenvector-eigenvalue table and its interpretation from my poster
has been useful. It helps me address issues that may be unclear, which would help
when I present these ideas in a talk, a poster, or a paper.
...but the "eigenvectors and eigenvalues" concept is only the tip of the iceberg
of the Galilean spacetime (which was introduced by the eigenchris video you linked in the OP).

What is more interesting and more useful is
how well it can describe Galilean physics
(as a stepping stone as to how well Minkowski-spacetime geometry
can describe [Special Relativistic-]Spacetime physics).

 

[Saw]

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, (...)

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, (...). 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.

I am not qualified to comment on the rest of your post, but I am happy to see that we are all non the same page in the description of this point (well, that I landed over that page, you were both already there...!): if I am not mistaken in my previous post, the Galilean eigenvector would be this rigid rod that you mention, which can sustain infinite sound velocity and serve for synching clocks and fixing lengths and which, under a boost, would keep its spacetime direction (i.e. its speed would remain infinite) and would be multiplied by an eigenvalue = 1, thus preserving the scale (measurements of time and length remain invariant).

 

[Saw]

the "eigenvectors and eigenvalues" concept is only the tip of the iceberg
of the Galilean spacetime (which was introduced by the eigenchris video you linked in the OP).

Surely it is and we have to come back to the issue of time as a vector vs a scalar, but please let me say a couple of things more on the eigenvector subject, because this is a little like the foundation of the building.

An eigenvector is really a direction, a slope (if I am permitted an extension on the term).
When $(0,1)^⊤$ is an eigenvector $(0,17)^⊤$ is also an eigenvector. And, of course,
they have the same eigenvalue.

That is enlightening. You are telling me that one must not read an eigenvector as if it were a basis vector (you must not interpret that at time = 0, all frames agree that a bird is overflying x = 1 or whatever), but read that all frames agree that the eigenvector's slope is whatever. So it can make sense that the second term is arbitrarily anyone.

these passages sound contradictory (word-coloring by me)

But (apart from the just mentioned correction, which I take) I don't think that my view of the eigenvector contradicts what has just been said.

At this point, I'm more concerned with what the mathematical structure implies in terms of describing the physics and less concerned about what the structure is called.

Yes, and when semantics turns muddled, I suggested before (#14), in order to describe the physics, to look at the operational meaning of concepts.

I am happy calling the Galilean-eigenvector $(0,1)^\top$ an "infinite" speed
because I can and have used the mathematical construction to make progress toward
(...)
There may be a better term.

What I said is that an eigenvector is a "measurement instrument", whose measured values are cross-frame "shared", although what is shared may be more or less: it is essential that there is cross-frame agreement on the ratio between the values, it is contingent that there is also agreement on the scale (which depends on the eigenvalue).

So I think that we converge on the essence: I agree that what is essential is the ratio (here the infinite speed), it is just that I see this as the output of a measurement instrument: it is the rigid rod inspired by vanhees71's comment (see my post #63), one endowed with infinite sound speed.

I would describe an eigenvector of the (1+1)-Lorentz-boost
as having a velocity-magnitude of "the speed of light".
This speed is not attainable by any future-timelike vector,
although a sequence of repeatedly-boosting a future-timelike vector tends to it.

This is in analogy to what I said in #47,
where an eigenvector of the (1+1)-Galilean boost
is not attainable by any future Galilean-timelike vector,
although the sequence of repeatedly-boosting a future Galilean-timelike vector tends to it.
I describe that as an infinite speed (although I don't want to get into discussions about infinity).

A usual difficulty that I have is that you apply the SR distinctions (timelike, spacelike...) to Galilean relativity, where they are not supposed to apply... Isn't the key that in Galilean relativity we assume that all vectors are timelike precisely because we assume that the eigenvector (= shared instrument = a rod endowed with total rigidity and infinite sound speed) can be manufactured, even if it can't?

My take would rather be that Galilean relativity sees this infinite speed as achievable, but one can see that it is wrong.In particular, if we take the slope as run over rise = horizontal over vertical = x over t (i.e. we maintain the rule x over t, to get a speed, although following the relativity convention, we put t as vertical), then we get whatever x over 0, which is an undefined expression. Whenever I spot an undefined, I interpret it as a thing that breaks the rules of math and hence is a physical impossibility, unless interpreted as a limit. So this would be an indication that you need evolution to a better paradigm because the thing that the Galilean transformation hinges on (its eigenvector) turns out to be a physical impossibility.

(Although one should be prudent in this, however, since our mathematical approach has assumptions leading to this step which a Galilean might contest...)

By the way, one could describe the finite speed of light eigenvector in the (1+1)-Lorentz boost
by an infinite rapidity, not attainable by a timelike vector, but it is an upper bound.
The Galilean rapidity associated with its eigenvector, not attainable by a Galilean-timelike vector,
is also infinite.

In terms of coordinates, in light-cone $(u,v)$ coordinates,
where frontward-future eigenvector is $(1,0)$ with finite speed,
nonzero timelike-vectors can't have one of these coordinates as zero.
Similarly, in $(t,x)$ coordinates, where the eigenvector is $(0,1)$ with my "inifinte" speed,
nonzero Galilean-timelike-vectors can't have the t-coordinate as zero.

Just noting that I am too tired and will need time to assimilate this...

 

[vanhees71]

Surely it is and we have to come back to the issue of time as a vector vs a scalar, but please let me say a couple of things more on the eigenvector subject, because this is a little like the foundation of the building.

How often do I have to repeat this! There is no single physical theory, where time is a vector!!! It's a real oriented parameter parametrizing the causal order of things.

In Newtonian mechanics it's just a real number, and at each point in time there's a 3D Euclidean affine space, i.e., Newtonian spacetime is a fiber bundle, at least that's the natural mathematical structure depicted closest the physical meaning of the Newtonian spacetime model, not these artificial "Newtonian four-vectors" we are unfortunately still discussing and which obviously are very confusing for you.

In SR spacetime is described by a 4D Lorentzian (pseudo-Euclidean) affine manifold. Choosing an inertial reference frame can mathematically be defined as choosing a space-time origin (the arbitrary event, we label with $t=0$ and $\vec{x}=0$) and a Lorentzian basis (tetrad). Then the corresponding "coordinate time" is the time-component of the four-vector wrt. this choice of (global) inertial reference frame. Another much more convenient choice of time is a scalar, depending on the physical situation you want to describe. E.g., if you do special-relativistic point-particle mechanics for a single particle (maybe moving in some field, like an electromagnetic field) a convenient choice for the time parameter is the particle's proper time, which is a Lorentz scalar, parametrizing the worldline of the particle.

 

[Saw]

In Newtonian mechanics it's just a real number, and at each point in time there's a 3D Euclidean affine space, i.e., Newtonian spacetime is a fiber bundle, at least that's the natural mathematical structure depicted closest the physical meaning of the Newtonian spacetime model, not these artificial "Newtonian four-vectors" we are unfortunately still discussing and which obviously are very confusing for you.

I have no problem with Newtonian time as a real number. That was the point of the OP: that the video showed a "time basis vector" that seemed incompatible with Galilean relativity and not the good way of introducing SR, as it implied that you will find in the latter what already exists somehow in the former. I must say however that the discussion is being very productive for me.

There is no single physical theory, where time is a vector!!! It's a real oriented parameter parametrizing the causal order of things.

(...)

In SR spacetime is described by a 4D Lorentzian (pseudo-Euclidean) affine manifold. Choosing an inertial reference frame can mathematically be defined as choosing a space-time origin (the arbitrary event, we label with $t=0$ and $\vec{x}=0$) and a Lorentzian basis (tetrad). Then the corresponding "coordinate time" is the time-component of the four-vector wrt. this choice of (global) inertial reference frame. Another much more convenient choice of time is a scalar, depending on the physical situation you want to describe. E.g., if you do special-relativistic point-particle mechanics for a single particle (maybe moving in some field, like an electromagnetic field) a convenient choice for the time parameter is the particle's proper time, which is a Lorentz scalar, parametrizing the worldline of the particle.

Ah, but here you are pointing out that my understanding of SR is wrong. Maybe! The thing is that I have seen a few descriptions of SR in terms of linear algebra, with the LT presented as a change of basis and the basis vectors being (putting t first and x later and assuming just one spatial dimension) a unit spatial vector (0,1) and a unit time vector (1,0), the combination of which (with the specific metric of SR where the dot product has minus sign) gives a spacetime interval as distance between two events.

In fact, if we talk about a four-vector (a 2-vector if we restrict ourselves to T and X), the time-component that you talk about must be a vector as well: a coefficient multiplied by a basis vector. This coefficient will be "coordinate time"... or "proper time", if the interval in question happens to be timelike and the inertial reference frame that we are talking about also happens to be present at both events of the interval. But... if I look at the Euclidean analogy, there may also be situations in which one frame has one single axis linking the two extremes of the distance. For example, you may measure the height of a tree with an X axis paralell to the ground + an Y axis parallel to the tree, whereas my X axis is a ladder leaning on the top of the tree. So your solution comes from the combination of two coordinate values, while mine flows from a single proper value. With regard to another problem, it may be the other way around. But because of this, we don't stop saying that the basis vector of X axis is a vector.

Also note that there is proper length as well, and because of this we don't stop saying that there is a space basis vector.

Another much more convenient choice of time is a scalar, depending on the physical situation you want to describe. E.g., if you do special-relativistic point-particle mechanics for a single particle (maybe moving in some field, like an electromagnetic field) a convenient choice for the time parameter is the particle's proper time, which is a Lorentz scalar, parametrizing the worldline of the particle.

Are you here talking about 4-velocity vectors instead of 4-displacement (i.e. the tangent to the wordline of a particle)?

Well, I do understand that proper time comes here in the denominator, even in SR, as a Lorentz scalar.

But then I would not say that "there are no time basis vectors in SR". I would say that coordinate time (which may sometimes coincide with proper time) is the magnitude of the time basis vector component that you use to compose a 4-displacement vector, although then such vector divided by proper time (coming here as an invariant or frame-independent value) gives you the 4-velocity of a particle.

I must admit, though, that I have difficulties wrapping my head around the concept of 4-velocity and that I may not perceive the importance that you and robphy give it. And I am also prepared to be shown that I am wrong, but please see above the reasons why I am saying what is after all the object of the OP: I do see the possibility of talking about a Minkowskian time basis vector, while I don't see a Galilean time basis vector.

 

[vanhees71]

Ah, but here you are pointing out that my understanding of SR is wrong. Maybe! The thing is that I have seen a few descriptions of SR in terms of linear algebra, with the LT presented as a change of basis and the basis vectors being (putting x first and t later and assuming just one spatial dimension) a unit spatial vector (0,1) and a unit time vector (1,0), the combination of which (with the specific metric of SR where the dot product has minus sign) gives a spacetime interval as distance between two events.

But that's correct, but still time is not a vector but either a component of a four vector given a basis ("coordinate time") or a scalar (like "proper time" of a particle).

In fact, if we talk about a four-vector (a 2-vector if we restrict ourselves to T and X), the time-component that you talk about must be a vector as well: a coefficient multiplied by a basis vector. This coefficient will be "coordinate time"... or "proper time", if the interval in question happens to be timelike and the inertial reference frame that we are talking about also happens to be present at both events of the interval. But... if I look at the Euclidean analogy, there may also be situations in which one frame has one single axis linking the two extremes of the distance. For example, you may measure the height of a tree with an X axis paralell to the ground + an Y axis parallel to the tree, whereas my X axis is a ladder leaning on the top of the tree. So your solution comes from the combination of two coordinate values, while mine flows from a single proper value. With regard to another problem, it may be the other way around. But because of this, we don't stop saying that the basis vector of X axis is a vector.

A component is a component (a number in the field of the vector space, i.e., here in the field of real numbers), not a vector. It's very important to distinguish components from vectors although unfortunately in most textbooks they say something like $x^{\mu}$ is a vector, although this symbol means the components of a vector wrt. a (sometimes no explicitly!) given basis. A lot of confusion can be avoided when one clearly distinguishes vectors (invariant objects) from vector components (transforming by a linear transformation under change of a basis).

Also note that there is proper length as well, and because of this we don't stop saying that there is a space basis vector.

There are time-like and spacelike vectors indeed, but that doesn't mean that time and space are vectors.

Are you here talking about 4-velocity vectors instead of 4-displacement (i.e. the tangent to the wordline of a particle)?

Well, I do understand that proper time comes here in the denominator, even in SR, as a Lorentz scalar.

But then I would not say that "there are no time basis vectors in SR". I would say that coordinate time (which may sometimes coincide with proper time) is the magnitude of the time basis vector component that you use to compose a 4-displacement vector, although then such vector divided by proper time (coming here as an invariant or frame-independent value) gives you the 4-velocity of a particle.

Of course, there are time-like vectors in SR and you can use them also as basis vectors (as one does when choosing a pseudo-Cartesian basis, which is most convenient), but that doesn't imply that time is a vector!

I must admit, though, that I have difficulties wrapping my head around the concept of 4-velocity and that I may not perceive the importance that you and robphy give it. And I am also prepared to be shown that I am wrong, but please see above the reasons why I am saying what is after all the object of the OP: I do see the possibility of talking about a Minkowskian time basis vector, while I don't see a Galilean time basis vector.

Four-velocity is a time-like four-vector along the worldline of a (massive) particle, which obeys $u_{\mu} u^{\mu}=1$, when defining it as $u^{\mu}=\mathrm{d}_{\tau} x^{\mu}/c$.

It's a time-like basis vector, but time itself is not a vector.

 

[robphy]

---

Surely it is and we have to come back to the issue of time as a vector vs a scalar, but please let me say a couple of things more on the eigenvector subject, because this is a little like the foundation of the building.

How often do I have to repeat this! There is no single physical theory, where time is a vector!!! It's a real oriented parameter parametrizing the causal order of things.

@Saw,
from my examples in #58 (clarifying #54 ) ,
in accord with @vanhees71 , wristwatch-time is a scalar, not a vector.
My final summary is

• Observer-Wristwatch-time (the age of the observer along his worldline) is a scalar.
• Along the observer-worldline at a particular wristwatch-time (age),
  the Observer 4-velocity vector (a unit tangent vector to his worldline)
  determines the instantaneous observer-time-axis "forward-into-the-future along the worldline".

like

• Car odometer-reading (the distance traveled along its path) is a scalar.
• Along the car's path at a particular odometer reading (distance traveled),
  the car's direction (a unit tangent vector to his worldline)
  determines the instantaneous direction of "forward along the path" for the car.
• If the car travels in a straight line,
  the odometer reading is the "radius", a scalar,
  a number that tells you how far away you are from the origin.
• The "unit radius vector" is a vector that points in some direction
  that determines the direction "away from an origin O".

I have nothing further to add on this issue.

---

Concerning eigenvectors...

Just noting that I am too tired and will need time to assimilate this...

Take as much elapsed wristwatch-time ( a scalar ) as you need.

 

[Saw]

A component is a component (a number in the field of the vector space, i.e., here in the field of real numbers), not a vector. It's very important to distinguish components from vectors although unfortunately in most textbooks they say something like $x^{\mu}$ is a vector, although this symbol means the components of a vector wrt. a (sometimes no explicitly!) given basis. A lot of confusion can be avoided when one clearly distinguishes vectors (invariant objects) from vector components (transforming by a linear transformation under change of a basis).

The meaning of this is clear to me, since quite a few years ago... :) although thanks for taking the time to specify it!

It seems that we may be using different terminologies (after all English is only my fourth language...). :)

I (of course) assume that a vector is an invariant object, which can be described in different manners, i.e. from different perspectives or, in our context, with different bases.

The (as a minimum linearly independent, ideally orthogonal and unitary) vectors that compose a basis serve to describe (span) a vector space. This description is made by associating a scalar of the corresponding field of the vector space (in our case a real number, in other cases it may be a complex number) to each basis vector.

I have always referred to this scalar as the "coordinate" or "coefficient" scaling the basis vector as per a given basis.

Instead, I have used the term "component" to refer to, as I said in my previous post, the basis vector as scaled by the relevant coefficient. The use of the term seems logical, because, when the time comes for putting together the components (i.e. to compose them), what you compose is not the coefficients but the basis vectors as scaled by the coefficients.

However, clearly above you are using the term "components" as meaning what I called "coefficients". Which is then the right term for what I called "components", i.e. the basis vectors as scaled by scalars?

This is important to know because here for example...

time is not a vector but either a component of a four vector given a basis ("coordinate time")

what do you mean by "component of a four vector given a basis ("coordinate time")"?

First option: you are sticking to your above-mentioned definition, i.e. you are referring to the number scaling the time basis vector. But that number is a scalar, no matter whether it is wrist-watch time, i.e. it has been measured by a clock being present at the two events, or it has been measured by two previously synced distant clocks. Certainly, the latter thing does not make the number in question less of a number. A different thing is that the second thing ("coordinate time"), unlike "proper time", will never be the scalar being spit out by a dot product of a ST vector with itself as the invariant solution to a problem, but that is another issue (see below).

Second option: you use now another meaning and you refer to the time basis vector as scaled by the coefficient. In that case, if you use now "component" with the meaning that I usually ascribe to the word, then I don't know why you say that it is not a vector. A unit basis vector, however scaled, is obviously a vector.

or a scalar (like "proper time" of a particle).

I think that what you mean here is what I hinted at above: you mean that proper time is a scalar in the sense that it is given off by a dot product of a ST vector with itself as the invariant solution to a problem, whereas coordinate time would never be that. No problem with that, if that is what you mean.

Conclusion: in SR,

time itself is not a vector.

sure, I am not saying that it is, because after all "what is time" if not, as Einstein said, what is measured by a clock?

The only thing that I am saying is that in SR there is a time basis vector, whereas in Galilean relativity there isn't any. I am also aware that in SR proper time is a scalar that is isolated by combining the time basis vector with space basis vectors, as scaled in the basis that you are using, and that this is an invariant number with which you solve problems.

What I don't know is what ropphy is saying about SR and Galilean relativity being the same in this respect, do you? (Under Galileo you don't need a time basis vector to combine with the space basis vector and thus learn the wrist-watch reading between two distant events, since here time is universal and all time is wrist-watch time.)

 

[robphy]

The only thing that I am saying is that in SR there is a time basis vector, whereas in Galilean relativity there isn't any. I am also aware that in SR proper time is a scalar that is isolated by combining the time basis vector with space basis vectors, as scaled in the basis that you are using, and that this is an invariant number with which you solve problems.

What I don't know is what ropphy is saying about SR and Galilean relativity being the same in this respect, do you? (Under Galileo you don't need a time basis vector to combine with the space basis vector and thus learn the wrist-watch reading between two distant events, since here time is universal and all time is wrist-watch time.)

A "time basis vector" (what I would call "a 4-velocity vector that is tangent to a worldline")
is needed in both SR and Galilean because (in inertial case) it is used to describe
the spacetime-displacement vector
"from the 'tick-0 on Alice's worldline'-event to the 'tick-1 on Alice's worldline'-event ",
which is different from the spacetime-displacement vector
"from the 'tick-0 on Bob's worldline'-event to the 'tick-1 on Bob's worldline'-event ",

If all one would want to do is label "hyperplanes of simultaneity"
(this hyperplane is "all space at t=0", that hyperplane is "all space at t=1", ...)
and thus only care about the "elapsed time between events"
then I could see why you may be saying that
you need the "time-basis vector" in SR (since the hyperplanes depend on inertial observer)
but
you don't need the "time-basis vector" in the Galilean case (since the hyperplanes are independent of inertial observer).

However, we generally want to do more than just label hyperplanes of simultaneity.
Implicit in what I said above,
we would also want to identify (e.g.)
the set of events that correspond to x=0 (,y=0, z=0) according to Alice, for each reading on Alice's wristwatch,
which is different from
the set of events that correspond to x=0 (,y=0, z=0) according to Bob, for each reading on Bob's wristwatch,
Thus, we also care about the "spatial-separation between events at different times",
(so that we can answer "Is that particle at rest in my frame? and if not, how fast is it traveling in my frame?")
and this depends on observer.

From #58,

Every inertial observer would like to draw her spacetime diagram
with her worldline vertically upwards (in the time-upward convention),
as granted by the relativity principle. This is the diagram for the Galilean case.So, we need the "time basis vector" ("a 4-velocity vector") in both the SR and Galilean case.

 

[Saw]

If all one would want to do is label "hyperplanes of simultaneity"
(this hyperplane is "all space at t=0", that hyperplane is "all space at t=1", ...)
and thus only care about the "elapsed time between events"
then I could see why you may be saying that
you need the "time-basis vector" in SR (since the hyperplanes depend on inertial observer)
but
you don't need the "time-basis vector" in the Galilean case (since the hyperplanes are independent of inertial observer).

Here you seem to be talking about "elapsed time between events" that are "simultaneous", even if separated by distance, so elapsed time between them is 0. You note that, since simultaneity is absolute in Galilean relativity, there is nothing here related to time that is observer-dependent. So no need for a "time-basis vector" for this set of events. I cannot agree more.

However, we generally want to do more than just label hyperplanes of simultaneity.
Implicit in what I said above,
we would also want to identify (e.g.)
the set of events that correspond to x=0 (,y=0, z=0) according to Alice, for each reading on Alice's wristwatch,
which is different from
the set of events that correspond to x=0 (,y=0, z=0) according to Bob, for each reading on Bob's wristwatch,
Thus, we also care about the "spatial-separation between events at different times",
(so that we can answer "Is that particle at rest in my frame? and if not, how fast is it traveling in my frame?")
and this depends on observer.

Here you talk about events with "spatial separation" but happening "at different times" and you say that this "depends on observer". But what "depends on observer"? Let us scrutinize each element of the situation to check if anything related to time is observer-dependent.

- The spatial separation certainly is frame-dependent, but this is the x coordinate, it is not due to the time period elapsed between the two events, which is the same for all observers, no matter if they measure with a wristwatch present at the two events or with distant clocks.
- The speed of any object traveling between the two events is also frame-dependent, but this is an automatic consequence of the previous statement: it is due to the distance traversed being relative, while the time period remains absolute.
- Certainly, if you look at Alice and Bob, as you propose, who are the observers located at the origin of each frame, it is clear that their wristwatches get progressively apart from each other... Does it mean anything in terms of time? No, because the watches keep ticking in perfect sync, as could be checked through the rigid rod with infinite sound speed which we identified as the eigenvector of the Galilean transformation. And if they stopped to do so, we would recalibrate them with that magical rod so that they keep ticking in sync. Obviously, you don't mean that the mere fact that the watches are getting apart from each other makes time observer-dependent, just like if the watches were of different colors, this would not make their time readings color-dependent.
- You say the "set of events" for Alice and Bob (from their respective vantage points, x=0 and x'=0) are "different". "Different" in what sense? If an event is characterized by an x and a t coordinate, the x coordinate is different but the t coordinate is not.

So far I did not spot here, either, anything related to time that is observer dependent.

So, we need the "time basis vector" ("a 4-velocity vector") in both the SR and Galilean case.

4-velocity seems to be a key concept in your argument. Do you mean that, for example, the 4-velocity of a particle traveling between the two events would be different in each frame? But let us open up this concept.

In SR it means 4 observer-dependent coordinates (coordinate time and the 3 spatial coordinates) divided by proper time measured from the frame at rest with the particle, which is an invariant concept. In SR coordinate time and proper time are different, except in the frame of the particle in question.

But in Galilean relativity, time is absolute, so coordinate time is always the same as proper time. Hence the first coordinate is always 1 in any frame. In turn, the three spatial coordinates are observer-dependent, but they are not related to time. Thus the concept is just 3-velocity to which you have added a value related to time, which is invariantly 1. And you conclude that this way you have found a "time-basis vector"?

Honestly, this looks to me contrived: you have artificially mixed, under an ad hoc concept (Galilean 4-velocity), absolute time separation between two events with the relative spatial distance between them, but through this forced cohabitation of one with the other you are not managing to make time observer-dependent in any sense.

Every inertial observer would like to draw her spacetime diagram
with her worldline vertically upwards (in the time-upward convention),
as granted by the relativity principle.

What prevents it? Why do you need a time-basis vector for this? Do you rather mean that if I draw my vertical line perpendicular to my X axis, then yours must be inclined? Sure, but this has nothing to do with the time axis. You don't need to attribute this different inclination to any observer dependency in terms of time. You just have to admit that the different evolution over time of the origins of the frames reflects their progressive spatial separation.

See this image which I posted before, now improved on the basis of our discussion on eigenvectors, where the vertical (absolute time) and horizontal (absolute length) orange axes reflect the common scale arising from eigenvalue = 1.

 

[haushofer]

Yes, but that curvature is not derived from a spacetime metric, since there is no spacetime metric in Galilean spacetime. The curvature is derived from a connection, but that connection has nothing to do with any spacetime metric (since there isn't one).

The connection can be derived from degenerate space and time metrics, of which the "inverses" are related by projective relations. Metric compatibility however doesn't fix the connection completely; the ambiguity is a two-form K. See e.g.

https://arxiv.org/abs/1011.1145

In that paper it is shown how NC theory is derived from the Bargmann algebra, and how the connection is uniquely solved for in terms of the Vielbeine (gauge fields of translations) and gauge field of the central extension.

 

[Saw]

@robphy
Since you know more than me and have been patient enough to keep objecting me, I have made more effort to converge to you.
I understand that you are telling me that one Galilean frame sees that the other’s time-basis vector is dilated, despite measuring the same time, because it measures that same time along more spatial distance. So would that be the “observer-dependency“ that you note “related to time”?
Is that in line with defining the Galilean transformation as a shear mapping, as sometimes it is said?

 

[vanhees71]

The connection can be derived from degenerate space and time metrics, of which the "inverses" are related by projective relations. Metric compatibility however doesn't fix the connection completely; the ambiguity is a two-form K. See e.g.

https://arxiv.org/abs/1011.1145

In that paper it is shown how NC theory is derived from the Bargmann algebra, and how the connection is uniquely solved for in terms of the Vielbeine (gauge fields of translations) and gauge field of the central extension.

I guess one can derive a Newtonian theory of gravitation by making the Galilei symmetry local, as one can derive Einstein-Cartan theory by making the Poincare group local starting from general relativity.

 

[haushofer]

I guess one can derive a Newtonian theory of gravitation by making the Galilei symmetry local, as one can derive Einstein-Cartan theory by making the Poincare group local starting from general relativity.

Yes, but you need the Bargmann algebra, i.e. the central extension. That extension is crucial. Ultimately, the gauge field of that generator will give (after gauge-fixing your coordinates) the Newton potential. Curiously, a nonrelativistic point particle then couples to this gauge field just as a relativistic point particle couples to the electromagnetic 4-vector A_mu. See also

https://arxiv.org/abs/1206.5176

Or, if your Dutch is reasonable,

https://www.google.com/url?sa=t&sou...4QFnoECAwQAQ&usg=AOvVaw3VwJnCJF78arsiOAAfiisA

 

[vanhees71]

I see. So from this point of view there's some classical argument that the true Galilei symmetry is the central extension Bargmann algebra. That's anyway the case for quantum theory.

Although my name suggests otherwise, I don't speak any Dutch. I guess, however, maybe it's possible to have an idea what the paper is about ;-)).

 

[haushofer]

I see. So from this point of view there's some classical argument that the true Galilei symmetry is the central extension Bargmann algebra. That's anyway the case for quantum theory.

Although my name suggests otherwise, I don't speak any Dutch. I guess, however, maybe it's possible to have an idea what the paper is about ;-)).

I'd thought so. I once was in the Scientology department of New York out of curiosity, and said I was Dutch, at which the poor fellow offered me a German ("Deutsch") translation of Diagenesis :P

But yes, indeed. Central extensions are often associated with quantisation. But the fact that the Lagrangian of a non-rel. point particle transforms into a total derivative under boosts necessitates an adjustment of the corresponding Noether charge to be conserved. In the Poisson brackets this introduces an extension in the Poisson bracket

$$
\{Q_B, \ Q_P \} \sim M
$$

motivating the central extension (P stands for spatial translations, B stands for boosts).

 

[robphy]

From #70 (reiterating #58, which is reiterating #54)

A "time basis vector" (what I would call "a 4-velocity vector that is tangent to a worldline")
is needed in both SR and Galilean because (in inertial case) it is used to describe
the spacetime-displacement vector
"from the 'tick-0 on Alice's worldline'-event to the 'tick-1 on Alice's worldline'-event ",
which is different from the spacetime-displacement vector
"from the 'tick-0 on Bob's worldline'-event to the 'tick-1 on Bob's worldline'-event ".

I repeat this because it is a natural thing to do... and to have.
But for some reason, you ignore it.

---

Honestly, this looks to me contrived: you have artificially mixed, under an ad hoc concept (Galilean 4-velocity), absolute time separation between two events with the relative spatial distance between them,

it seems you think I made artificial made this up or forced it to be so.
I am going to argue that the structure is already there, as part of the position-vs-time diagram.

As soon as you write down the Galilean transformations in matrix form,
you are working with Galilean vectors, with time and space components.
That's what the Galilean transformation does...
it maps a Galilean 4-velocity to another Galilean 4-velocity.

But you seem unhappy with that.
So, let me not use the word 4-velocity, since I think it is a distraction at this stage.
let's use instead "unit tangent vector" and think of it as a unit displacement .

So, a Galilean boost transformation maps a unit tangent vector (of an inertial worldline)
to another unit tangent vector.
That's what this diagram shows---making the tilted tangent vector vertical.
(No need to mention or invoke 4-velocity.)

(The fact that the horizontal lines (hyperplanes) are unchanged is irrelevant
to the transformation making the tilted tangent vector vertical.)

---

 

[vanhees71]

As I said, I don't feel "unhappy" with it. In fact I use this formalism with spacetime vectors in Galilean relativity myself, because it's convenient to deal with the Galilei group, but contrary to the spacetime models of relativity there's no further physically relevant structure than the fiber-bundle structure. Time is simply absolute, i.e., a oriented $\mathbb{R}^1$, along which affine euclidean $\mathbb{E}^3$ manifolds are pinned.

 

[robphy]

For me, in my Spacetime Trig approach,
I can prove many kinematical results in special relativity
by direct analogy with Euclidean geometry via
a one-parameter family of Cayley-Klein geometries
from E=+1 for Minkowski to E=-1 Euclidean, where E=0 is Galilean spacetime geometry,
which is the topic of the eigenchris video linked in the OP.

My participation in this thread is my response to the OP,
which questions the usefulness of the Galilean spacetime.

I think the value of the approach is that many of the surprising results of SR (special relativity)
can be shown to have a "Galilean limit" precisely,
and seems to suggest the Galilean-spacetime version
and Euclidean-geometric version of the special relativistic result.
This suggests that the result might not be as "How could that be?"-surprising
if one was already
aware of the Euclidean and Galilean versions...
or one might be able to use Euclidean and (if illuminated) Galilean versions to obtain the SR-result.

It's not just about such analogies in a few situations,
as seen in scattered throughout the literature
as if one chose (possibly because one already knows the answer for this situation)
which term would be subjected to the limit.

Rather, it's a unified approach that dictates which terms are subjected,
due to a parameter in the Cayley-Klein approach (at the level of its projective-geometric foundations).

But, yes, in Galilean geometry in isolation
may be appear uninteresting and not useful.

 

[Saw]

A "time basis vector" (what I would call "a 4-velocity vector that is tangent to a worldline")
is needed in both SR and Galilean because (in inertial case) it is used to describe
the spacetime-displacement vector
"from the 'tick-0 on Alice's worldline'-event to the 'tick-1 on Alice's worldline'-event ",
which is different from the spacetime-displacement vector
"from the 'tick-0 on Bob's worldline'-event to the 'tick-1 on Bob's worldline'-event "

I repeat this because it is a natural thing to do... and to have.
But for some reason, you ignore it.

Sorry if I sound sometimes too radical, it is just that I am struggling with the ideas!

Let us check if I can follow you.

Say event A is when Bob and Alice meet ( t = 0), event B when Bob's wristwatch ticks 1s and event C when Alices' wristwatch also ticks 1s.

A-B is different from A-C.

The description of each set of events is different in each frame:

- In his coordinate system, Bob will paint his T axis as a vertical line (his wordline) and segment A-B as overlapping/tangent to such line, but he will paint Alice's T axis as a diagonal line and segment B-C as overlapping to that diagonal line.
- In her own coordinate system, will do the opposite: she will paint her T axis as a vertical line, the segment A-C also vertical, but Bob's T axis and segment A-B as diagonal.
- The projection from event C to Bob's axis is overlapping with the projection from B to Alice's axes, meaning that they both agree on simultaneity and that the time lapses A-B and A-C are 1 s.

But why are the two pairs of events different? Because the wristwatches are moving wrt to each other.

And why does each frame make a different description of each pair? Because each frame deems itself motionless and attributes that progressive spatial separation to the other frame. So the thing that can be inferred here is that two Galilean frames have different rulers but not that they have different clocks, different time units, time basis vectors, or even time axes...

I understand that it may be difficult to accept that there are no different time basis vectors if there are different time axes, with different inclinations. That is why at a given moment, instead of the above-mentioned description where each frame paints its own T axis, I proposed this drawing, where the T axis is common.

I don't know how you see this. I think that it is consistent with the very helpful guidance that you've given me in understanding the eigenvector of the Galilean transformation. Can the eigenvector, i.e. common instrument that all frames share (the rod with infinite sound speed), also be the "common clock"? You may say: "but that rod only works for fixing absolute length or ensuring absolute simultaneity, it is not a clock, which is another instrument at rest with each frame!". Well, just imagine that the rod is, due to some mechanism, oscillating up and down, and it is everywhere, thus providing all vehicles with a common time. In SR that would not work, because the driving force keeping the rod in motion would have limited propagation speed, but this would not be a problem under Galilean relativity thanks to the admission of an infinite speed.

As soon as you write down the Galilean transformations in matrix form,
you are working with Galilean vectors, with time and space components.
That's what the Galilean transformation does...
it maps a Galilean 4-velocity to another Galilean 4-velocity.

But there may be something that I am missing, since as usual, however, when you bring in the Galilean 4-velocity, I get lost.

I think that I understand the SR 4-velocity as something different from the ST displacement: it is the rate of change of ST displacement wrt to proper time.

In the Galilean context, based on your comment above...

A "time basis vector" (...)
is needed (..) to describe
the spacetime-displacement vector
"from the 'tick-0 on Alice's worldline'-event to the 'tick-1 on Alice's worldline'-event ",
which is different from the spacetime-displacement vector
"from the 'tick-0 on Bob's worldline'-event to the 'tick-1 on Bob's worldline'-event "

...I thought that the time basis that you invoke is simply the "ST" displacement vector A-B or B-C.

Same puzzlement for SR: why do you identify the SR time basis with 4-velocity instead of simply with clock unit readings, either at the origin or with any other clock, as they are supposed to be synced as per the frame's standard?

 

[robphy]

It's too difficult for me to manage multiple issues in each post.
So, I will work on a piece at a time

But why are the two pairs of events different? Because the wristwatches are moving wrt to each other.

And why does each frame make a different description of each pair? Because each frame deems itself motionless and attributes that progressive spatial separation to the other frame. So the thing that can be inferred here is that two Galilean frames have different rulers but not that they have different clocks, different time units, time basis vectors, or even time axes...

They have identically constructed clocks (synchronized once in the past).
And, for any given event, when Alice and Bob use their clocks,
the assign time coordinates which, in the Galilean case, agree.

Since each clock has its own worldline (locating where and where each clock has been),
they each have a unit tangent vector.

(I will describe everything using tangent vectors because
the word "4-velocity" is introducing a distraction, and
is thus an easy target to complain about.
So, I'm sticking with unit tangent vectors, which should be clear and uncontroversial.)

It is useful to find a quantity to describe how different these two tangent vectors are---
something to describe the separation between their directions.
In Euclidean geometry, we use "relative angle" and we use "relative slope"=tan(relative angle).
In Minkowski geometry, we use "relative rapidity" and we use "relative velocity"="relative slope on the diagram"=tanh(relative rapidity).
It turns out (in the literature I referenced earlier, and in my poster) in the Galilean geometry,
we use the relative Galilean-rapidity, and we can calculate a "relative slope",
which is proportional to the rapidity (as opposed to a nonlinear relation because of tan or tanh).

Thus, the "relative spatial velocity" between the moving clocks can be obtained
using the two tangent vectors.
This relative velocity determines the "v" in the Galilean boost to transform from Alice's frame to Bob's frame.
That seems like a useful physically-interesting quantity.

 

[robphy]

I understand that it may be difficult to accept that there are no different time basis vectors if there are different time axes, with different inclinations. That is why at a given moment, instead of the above-mentioned description where each frame paints its own T axis, I proposed this drawing, where the T axis is common.

Allowing each frame to draw their own position-vs-time diagram (in Galilean and Special Relativity)
is akin to allowing each orientation of graph paper to have its own x- and y-axes in Euclidean geometry.
So, I am merely following (as far as I can) what worked well for Euclidean geometry
and in retrospect will see that will also work well for Special Relativity.

That Euclidean geometry grants me the freedom to do geometry in any orientation
is akin to the Principle of Relativity granting me the freedom to study physics in any inertial reference frame.

However, your alternative proposal deviates from this freedom
and, to quote an earlier post of yours directed at the "Galilean 4-velocity"

Honestly, this looks to me contrived: you have artificially mixed, under an ad hoc concept

Your alternative proposal distinguishes one T-axis, which is not in the spirit of the Principle of Relativity.
Further, this alternative proposal won't carry over into Special Relativity.
So, this becomes further "common-sense baggage" that you'll have to unload.

I don't know how you see this. I think that it is consistent with the very helpful guidance that you've given me in understanding the eigenvector of the Galilean transformation. Can the eigenvector, i.e. common instrument that all frames share (the rod with infinite sound speed), also be the "common clock"?

Your interpretation of the eigenvector of the Galilean transformation isn't my interpretation.
As we have seen, there are no eigenvectors of the Galilean boost with nonzero t-components.
So, the "common clock" (interpreted as a worldline with a
tangent vector with a nonzero t-component)
can't have a tangent vector that is an eigenvector of the Galilean boost.

 

[Saw]

It's too difficult for me to manage multiple issues in each post.
So, I will work on a piece at a time

Same for me! The priority is to check if I understand this point of yours, so let us make it clear before anything else.

When I say "this point", I mean the link that you establish between the unit tangent vector, relative velocity btw the two frames (and their clocks) and the time basis vector.

Since each clock has its own worldline (locating where and where each clock has been),
they each have a unit tangent vector.

(I will describe everything using tangent vectors because
the word "4-velocity" is introducing a distraction, and
is thus an easy target to complain about.
So, I'm sticking with unit tangent vectors, which should be clear and uncontroversial.)

I suppose there is a typo and you meant "when and where each clock has been". As to what the unit tangent vector is, I have no problem. I think that I more or less described it in my previous post:

Say event A is when Bob and Alice meet ( t = 0), event B when Bob's wristwatch ticks 1s and event C when Alices' wristwatch also ticks 1s.

A-B is different from A-C.

The description of each set of events is different in each frame:

- In his coordinate system, Bob will paint his T axis as a vertical line (his wordline) and segment A-B as overlapping/tangent to such line, but he will paint Alice's T axis as a diagonal line and segment B-C as overlapping to that diagonal line.
- In her own coordinate system, will do the opposite: she will paint her T axis as a vertical line, the segment A-C also vertical, but Bob's T axis and segment A-B as diagonal.
- The projection from event C to Bob's axis is overlapping with the projection from B to Alice's axes, meaning that they both agree on simultaneity and that the time lapses A-B and A-C are 1 s.

You also want to make an analogy with Euclidean and Minkoswskian geometry, where you introduce the generalized concepts of angle and tangent of this angle, which for me is perfect in itself. A different thing is how to interpret this analogy, but we are not there yet. Let me use this picture from your poster, somehow manipulated, as a reference:

The analogy is spotting that there is an "angle" in the three geometries: the distance between the final point of Bob's unit tangent vector (event B in my notation) and the final point of Alice's (event C) = a circle-like angle (Euclidean), a degenerate circle or straight line-like angle (Galilean) and a hyperbolic angle (Minkowskian).

In all cases, the tangent of the "angle" is the slope" o ratio between the relevant values: a purely spatial ratio = x/y (in the Euclidean case) or a spacetime ratio = x/t. i.e. a velocity (in the Galilean and Minkowskian cases).

Please confirm if I reflected it well.

If so, what I don't understand, as noted in previous posts, is where you place the time basis vector:

- My first interpretation is that you equate it with the "spacetime displacement vector" A-B (Bob) or A-C (Alice), which is relative: in Galilean case, only due to relative distance; in Minkowskian case, in more respects.
- Then you add the concept of "velocity", which is the tangent of the "angle" between the two displacements To be noted: this velocity, which is...

the "v" in the Galilean boost to transform from Alice's frame to Bob's frame

.... ie., the relative velocity between the two watches and between the frames, is invariant, both in the Galilean and in the Minkowskian geometries (for different reasons in each case), so it is a different thing.

Surely, the two concepts are related and they are both useful, but they are different. And then one has to decide what to call the "time basis vector", whether one thing or the other. In my naive understanding, the time basis vector should be the unit of the time axis and this in turn the output of a clock, with which precisely you calculate the velocity of things, either of the other frame or of projectiles moving wrt both frames, as a ratio between distance traversed and time elapsed...

On another note, strictly speaking, your literal wording is always equating time basis with "4-velocity", but we had agreed to leave aside this other concept "for the time being". I let you choose whether to bring it into the discussion or not.

 

[robphy]

In all cases, the tangent of the "angle" is the slope" o ratio between the relevant values: a purely spatial ratio = x/y (in the Euclidean case) or a spacetime ratio = x/t. i.e. a velocity (in the Galilean and Minkowskian cases).

Please confirm if I reflected it well.

Yes, in your rotated-axes convention (following space-running-horizontally, time-running upwards).
In my poster, I keep the time-axis horizontal (as in PHY 101 texts) and refer to the spatial-direction as y
in order to make comparisons with Euclidean geometry and intro (Galilean) physics easier.

(However, it could be argued that familiar "angle" as arc-length-on-unit-circle
isn't as natural as construction as the associated "sector area"-intercepted in unit circle.)

In my naive understanding, the time basis vector should be the unit of the time axis and this in turn the output of a clock

I would agree with this, provided that this is
the unit tangent-vector to the worldline for any clock (whatever its motion and in whatever case (SR or Galilean)).

Even if a symmetry exists (like absolute time)​

that suggests that
each observer can give-up his or her clock​

(as if surrendering a democratic right [the principle of relativity])​

in favor a universal (autocratic absolute timekeeper) Big Ben to tick for the universe,
​

the unit tangent vectors to various worldlines can and should​

still function as [possibly viewed by some as redundant and useless] time-basis vectors.

(In surrendering, the ability to understand relativity in hindered...​

"do you mean we each can really keep our own time with our own watches?" )​

​

---

I am keeping the focus on the unit tangent vector, but that is what the geometrical analogy dictates.
In special relativity, one can interpret that unit tangent vector as the so-called 4-velocity (geometrically, a dimensionless quantity [since it's a unit-vector], despite its name and despite its historical introduction),
where the ratio of its sides [in (1+1)-spacetime] is the dimensionless version of the more-familiar "spatial velocity" v appearing in the boost. But these aspects appear to provide a distraction from the main issue.
So, I have and will continue to avoid it... because I can support my viewpoint that time-axes are associated with the unit tangent-vector of every clock worldline.

 

[vanhees71]

Yes, in your rotated-axes convention (following space-running-horizontally, time-running upwards).
In my poster, I keep the time-axis horizontal (as in PHY 101 texts) and refer to the spatial-direction as y
in order to make comparisons with Euclidean geometry and intro (Galilean) physics easier.

But the Galilean space-time plane is not a Euclidean affine space (neither is the Minkowski space-time-diagram plane)!

 

[robphy]

Yes, in your rotated-axes convention (following space-running-horizontally, time-running upwards).
In my poster, I keep the time-axis horizontal (as in PHY 101 texts) and refer to the spatial-direction as y
in order to make comparisons with Euclidean geometry and intro (Galilean) physics easier.

But the Galilean space-time plane is not a Euclidean affine space (neither is the Minkowski space-time-diagram plane)!

Yes, and that's good...
since these have different metrics, different structures that support or don't support a causal structure.

But the point of analogies
is that you can use
what is common [ about being an affine space (a vector space who has forgotten its origin) ]
and then point out
what is different [ depending on your metric or quadratic form ].

• Two given lines intersecting in the plane do so at the same point in all three cases (Euc, Gal, Mink).
• Two lines that are parallel in Euclidean geometry
  are also parallel in Minkowski spacetime and Galilean spacetime geometries.
  • A vector perpendicular to those lines depends on the metric.
  • How far apart the lines are depends on the metric.
• The operation $\vec A+ \vec B= \vec C$ gives the same resulting vector in all three cases.
  • However, the "magnitude of $\vec C$" depends on the metric.

And metric dependence is directly traceable to where the E-parameter ($\epsilon^2$ in my poster)
appears. So the storyline I use goes something like this

• In Euclidean space, a circle
  (points equidistant from a reference point using odometers that can travel in any direction)
  has the form $t^2+y^2=R^2$ (where the spatial $t$-axis is parallel to the straight-line odometer path and the space $y$-axis is perpendicular to the path
  (where perpendicular is parallel to the tangent to the circle)
  and $R$ is the odometer reading).
• In a PHY 101 position-vs-time graph, a Galilean circle
  (the set of events from a reference event that occur "1 second later" according to wristwatches that were at that event and can travel inertially with any velocity),
  as revealed by experiments done at terrestrial speeds and extrapolated to arbitrarily large speeds,
  has the form $t^2+(0y^2)=R^2$
  (where time $t$-axis is parallel to the inertial wristwatch worldline and the spatial $y$-axis is perpendicular to that worldline
  (where perpendicular is parallel to the tangent to the circle)
  and $R$ is the wristwatch reading).
• In a Minkowski position-vs-time graph (a spacetime diagram), a Minkowski circle
  (the set of events from a reference event that occur "1 second later" according to wristwatches that were at that event and can travel inertially with any velocity)
  as revealed by experiments done at terrestrial speeds and high-energy particle speeds,
  has the form $t^2-y^2=R^2$
  (where time $t$-axis is parallel to the inertial wristwatch worldline and the spatial $y$-axis is perpendicular to that worldline
  (where perpendicular is parallel to the tangent to the circle)
  and $R$ is the wristwatch reading).
• So, a Galilean might have said "a generalize circle has the form
  $t^2+(k) y^2=0$, where $k=1$ for Euclidean and $k=0$ for Galilean.
  And there are many similarities (and differences) between the equations
  and the geometrical constructions.
• But then (newsflash)
  experimental data (from unstable high-speed particles that should have decayed)
  says that for position-vs-time graphs is seems that $k\neq 0$ , in fact $k<0$!
  For some convenient choice of units, a curve-fit gives a value of $k=-1$ ,
  which implies there are asymptotic speeds for wristwatches that are finite.
  This $k$ is $(-E)$ in my past discussions in this thread ( and $-\epsilon^2$ in my poster).
  
  (Again this is along the lines of
  If Maxwell had worked between Ampère and Faraday: An historical fable with a pedagogical moral
  https://aapt.scitation.org/doi/10.1119/1.12239 .)

So, it is a good thing that

the Galilean space-time plane is not a Euclidean affine space (neither is the Minkowski space-time-diagram plane)!

a very good thing.

 

[vanhees71]

Yes, and that's good...
since these have different metrics, different structures that support or don't support a causal structure.

But the point of analogies
is that you can use
what is common [ about being an affine space (a vector space who has forgotten its origin) ]
and then point out
what is different [ depending on your metric or quadratic form ].

But from a didactic point of view there are good and bad analogies. Claiming to have a Euclidean plane although the very point of the diagram is that it's not a Euclidean plane is a bad analogy. That's why I find Minkowski diagrams in principle not a good didactic tool, but I have to teach them, because my high-schhool-teacher students need to understand them to explain them to their highschool students. Unfortunately Minkowski diagrams are still considered as good didactical tools, and you have to spend a lot of time to hammer into the students the fact that they must not read them as a Euclidean plane, and that what's unit ticmarks on two Lorentzian reference frames don't look like the Euclidean unit ticmarks in a Cartesian coordinate system.

Fortunately nobody came up with the idea to use "Galilei diagrams" in Newtonian mechanics, i.e., there you don't need to introduce the also pretty complicated fiber-bundle geometry to visualize the Galilei-boost transformation between inertial frames, because fortunately nobody uses them.

 

[robphy]

But from a didactic point of view there are good and bad analogies. Claiming to have a Euclidean plane although the very point of the diagram is that it's not a Euclidean plane is a bad analogy. That's why I find Minkowski diagrams in principle not a good didactic tool, but I have to teach them, because my high-schhool-teacher students need to understand them to explain them to their highschool students. Unfortunately Minkowski diagrams are still considered as good didactical tools, and you have to spend a lot of time to hammer into the students the fact that they must not read them as a Euclidean plane, and that what's unit ticmarks on two Lorentzian reference frames don't look like the Euclidean unit ticmarks in a Cartesian coordinate system.

Fortunately nobody came up with the idea to use "Galilei diagrams" in Newtonian mechanics, i.e., there you don't need to introduce the also pretty complicated fiber-bundle geometry to visualize the Galilei-boost transformation between inertial frames, because fortunately nobody uses them.

A PHY 101 position-vs-time graph is a Galilean spacetime diagram.

No one has really used the metric structure explicitly,
although we have all used it implicitly.

It seems we have learned how to use PHY 101 diagrams
without reading them as Euclidean planes,
because we have done a lot of practice with them.
(However, I have seen occasional student answers
using the Pythagorean theorem on a PHY 101 position-vs-time graph.)

Like any tool, one has to be taught how to use it... and how not to use it.
In addition, the typical user does not need to know all of the underlying mechanisms
to make use of it.
I know how to read a watch without knowing how it all works.

We don't have to get into the various structures in Euclidean geometry for relatively-simple problems.
Do we say,
to study a block on an incline, we can use SO(2) symmetry to construct a coordinate system parallel and perpendicular to the plane. So, let's first understand what SO(2) is.

Likewise, we don't have to get into many details underlying the Galilean diagram to use it.

However, we might want to drop hints to prepare them for special relativity.
Again, in isolation, the Galilean diagram doesn't seem worth it.
But it may be helpful to interpret what is going on special relativity.

 

[vanhees71]

A PHY 101 position-vs-time graph is a Galilean spacetime diagram.

No, it's a graph, showing the function $x=x(t)$, but it's never used as a Galilean spacetime diagram. Nobody with a clear mind would draw the time coordinate of another inertial observer moving with the constant velocity $v$ against the first one, then construct unit ticmarks at this new $t'$ axis, and they don't have the same Euclidean distance as the unit ticmarks on the original $t$ axis although this complicated construction just takes care of the fact that $t'=t$ and $x'=x-v t$.

No one has really used the metric structure explicitly,
although we have all used it implicitly.

And that's the problem! You are always inclined to look at spacetime diagrams as if they were Euclidean planes, because we are used to do Euclidean geometry by drawing on a sheet of paper.

It seems we have learned how to use PHY 101 diagrams
without reading them as Euclidean planes,
because we have done a lot of practice with them.
(However, I have seen occasional student answers
using the Pythagorean theorem on a PHY 101 position-vs-time graph.)

Ok, to what purpose? I've no clue, where such an idea has a useful application in Newtonian mechanics.

Like any tool, one has to be taught how to use it... and how not to use it.

Indeed, and introducing spacetime diagrams you have to force yourself and your students to not to read it as if it were a geometric construction in a Euclidean plane ;-)), i.e., you have to use considerable time to explain how NOT to use it.

In addition, the typical user does not need to know all of the underlying mechanisms
to make use of it.
I know how to read a watch without knowing how it all works.

We don't have to get into the various structures in Euclidean geometry for relatively-simple problems.
Do we say,
to study a block on an incline, we can use SO(2) symmetry to construct a coordinate system parallel and perpendicular to the plane. So, let's first understand what SO(2) is.

No, there you can use Euclidean geometry, where it is applicable, i.e., in the 3D space of an inertial observer in Newtonian mechanics ;-).

Likewise, we don't have to get into many details underlying the Galilean diagram to use it.

We just don't need such diagrams for anything. In my opinion also Minkowski diagrams are not really needed, since where they can be used the calculations using Lorentz transformations and the Minkowski-space "metric" is much more straight forward and thus less probable to lead to misconceptions.

However, we might want to drop hints to prepare them for special relativity.
Again, in isolation, the Galilean diagram doesn't seem worth it.
But it may be helpful to interpret what is going on special relativity.

Of course, it also depends on each individual person. Maybe some people find Minkowski diagrams helpful. For me they are a pain, because they confuse me always when I want to explain them ;-)).

 

[Saw]

@robphy please let me come back for a moment to my prior question.

I had specifically asked:

where you place the time basis vector:

- My first interpretation is that you equate it with the "spacetime displacement vector" A-B (Bob) or A-C (Alice), which is relative: in Galilean case, only due to relative distance; in Minkowskian case, in more respects.
- Then you add the concept of "velocity", which is the tangent of the "angle" between the two displacements To be noted: this velocity, which is (...) the relative velocity between the two watches and between the frames, is invariant, both in the Galilean and in the Minkowskian geometries (for different reasons in each case), so it is a different thing.

(...) And then one has to decide what to call the "time basis vector", whether one thing or the other.

Your comments are:

the unit tangent vectors to various worldlines can and should​

still function as [possibly viewed by some as redundant and useless] time-basis vectors.​

​

(...)​

​

---

In special relativity, one can interpret that unit tangent vector as the so-called 4-velocity (geometrically, a dimensionless quantity [since it's a unit-vector], despite its name and despite its historical introduction),
where the ratio of its sides [in (1+1)-spacetime] is the dimensionless version of the more-familiar "spatial velocity" v appearing in the boost.

Do you mean that the answer is "somehow both", in the sense that your "unit tangent vector" (spatial velocity in Galilean relativity, 4-velocity in SR) is the unit vector of the spacetime displacement vector of the wristwatch?

 

[vanhees71]

A Galilei spacetime diagram, with two inertial reference frames depicted, looks as follows (I think I've posted it already above):

I've drawn the time-axis to the right. Usually for the analogous Minkowski diagram one points it up, but the math is of course the same. As you clearly see, the time unit tic on the $t'$ axis is NOT the same as that on the $t$ axis when falsely interpreting the space-time diagram as a "Euclidean plane"!

 

[robphy]

Do you mean that the answer is "somehow both", in the sense that your "unit tangent vector" (spatial velocity in Galilean relativity, 4-velocity in SR) is the unit vector of the spacetime displacement vector of the wristwatch?

TLDR: yes… but relative-spatial velocity vectors are spacelike. Unit-tangent vectors to worldlines are timelike.

Details follow:

Worldlines in SR and in Galilean relativity are timelike curves, that is curves with timelike tangents.

“Unit timelike” means a displacement vector that is a radius vector for the “unit circle” in that geometry.
This is the displacement vector that gets you from one tick on an inertial wristwatch’s worldline to its next tick.

The tangent-vector-to-the-worldline is timelike.

In both the SR and Galilean cases,
the spatial-velocity is the spatial-component of the unit-tangent-vector, divided by the temporal component of that unit-tangent vector. (Explicit details below.)

In SR, this is $v=\frac{\Delta x}{\Delta t}=\frac{\sinh\theta}{\cosh\theta}=\tanh\theta$, which is akin to the slope in Euclidean geometry.

In the Galilean case, there is an analogous construction that begins
$v=\frac{\Delta x}{\Delta t}$ and uses the Galilean-analogue of the trig function (as defined by I.M.Yaglom). Whereas the velocity is a nonlinear function of rapidity in SR, and the slope is a nonlinear function of angle in Euclidean geomety, the Galilean-velocity is a linear-function of the Galilean-rapidity [which leads to the additivity of Galilean-velocities].

The spatial-velocity vector is spacelike. The 3-velocity (as a vector in spacetime) is spacelike.Expanding to (3+1)-spacetime, the spatial-velocity vector of particle (with magnitude $\frac{\Delta x}{\Delta t}$ is a spacelike vector (since it is parallel to the measurer’s tangent-line-to-the-unit-“circle”, following Minkowski’s definition from his “Space and Time”). This geometrical description (radius-tangent and timelike-spacelike) works the same in both SR and Galilean… it’s just that different “circles” lead to different tangent-lines.

So, to summarize:
Given two unit-timelike tangent vectors at an event (each tangent to an inertial worldline),
the [relative-]spatial-velocity is a spacelike-vector.

Alice will decompose Bob’s unit-timelike tangent vector into parts parallel and perpendicular to Alice’s unit timelike-tangent. Alice will describe Bob’s spatial velocity with a spacelike-vector that is purely spatial for Alice.

Similarly, Bob will decompose Alice’s unit-timelike tangent vector into parts parallel and perpendicular to Bob’s unit timelike-tangent. Bob will describe Alice’s spatial velocity with a spacelike-vector that is purely-spatial for Bob.

In the literature (using geometric units), the “unit timelike vectors that are tangent to worldlines” are called 4-velocities (although they are unit-less… since it is a unit-vector… we have divided by its magnitude… so the 4-velocity has magnitude 1. It is basically a direction-vector along the time-axis of the worldline it is tangent to).

Thus,
the unit tangent vector to a worldline (often called the 4-velocity) is timelike.
The 3-velocity (as a spacetime vector) describing relative-velocity is spacelike.

Maybe it’s more than you wanted… but I’ve tried to be complete and logical in the presentation… and I delayed (until now) the name “4-velocity” because of the distraction it has caused.

 

[vanhees71]

The relative velocity is the four-velocity vector of one particle in the restframe of the other particle and as such timelike or for "massless particles" lightlike. The usual frame-dependent three-velocity $\vec{v}=\mathrm{d} \vec{x}/\mathrm{d} t$ is a strange object with not so simple transformation properties. It's usually always better to use covariant vector/tensor components and then calculate those non-covariant objects from them.

EDIT: See the correction to this in #96 in response to @PeterDonis 's posting #95.

 

[PeterDonis]

The relative velocity is the four-velocity vector of one particle in the restframe of the other particle

A vector is not "in" any particular frame; it's an invariant geometric object. There is no such thing as "the four-velocity of one particle in the restframe of the other particle".

If you have two particles whose worldlines cross at some event, the dot product of their two 4-velocity vectors at that event gives the $\gamma$ factor $1 / \sqrt{1 - v^2}$, which is the simplest representation of their "relative velocity".

 

[vanhees71]

You are right. I've described it too sloppily. One more try:

To get the relative velocity of particle 1 relative to particle 2 you Lorentz boost to the rest frame of particle 2 (using the four-velocity of particle 1), i.e., you are in the reference frame, where
$$u_1=\begin{pmatrix}\gamma \\ \gamma \vec{v} \end{pmatrix}, \quad u_2=\begin{pmatrix} 1 \\ \vec{0} \end{pmatrix}.$$
Then $\vec{v}$ is the relative velocity of particle one relative to particle 2. That's the standard definition used in HEP when defining invariant cross sections and in relativistic Boltzmann transport theory. For details also see Sect. 1.6 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

and for transport theory

https://itp.uni-frankfurt.de/~hees/publ/kolkata.pdf

 

[robphy]

You are right. I've described it too sloppily. One more try:

To get the relative velocity of particle 1 relative to particle 2 you Lorentz boost to the rest frame of particle 2 (using the four-velocity of particle 1), i.e., you are in the reference frame, where
$$u_1=\begin{pmatrix}\gamma \\ \gamma \vec{v} \end{pmatrix}, \quad u_2=\begin{pmatrix} 1 \\ \vec{0} \end{pmatrix}.$$
Then $\vec{v}$ is the relative velocity of particle one relative to particle 2. That's the standard definition used in HEP

So, $\hat u_1$ and $\hat u_2$ are 4-velocities.
and in $\hat u_2$'s frame, $\hat u_1=\gamma \hat u_2 + \gamma \tilde v$,
where $\tilde v$ is purely-spatial according to $\hat u_2$ (since $g(\hat u_2, \tilde v)=0$).
That is to say,
$\gamma \hat u_2$ is timelike
and
$\gamma \tilde v$ is spacelike
Thus, the relative velocity (the 3-velocity) is the ratio $\frac{\gamma \tilde v}{\gamma}= \tilde v$, which is spacelike.

In terms of rapidity, this is
$\gamma \hat u_2=\cosh\theta\ \hat u_2$
and
$\gamma \tilde v=\sinh\theta\ \hat u_{2\bot}$.
Thus, the relative velocity (the 3-velocity) is the ratio $\tilde v=\displaystyle \frac{\sinh\theta \hat u_{2\bot} }{\cosh\theta}=\tanh\theta\hat u_{2\bot}$, which is spacelike.

This is in accord with my earlier statement

Thus,
the unit tangent vector to a worldline (often called the 4-velocity) is timelike.
The 3-velocity (as a spacetime vector) describing relative-velocity is spacelike.

 

[PeterDonis]

So, $\hat u_1$ and $\hat u_2$ are 4-velocities.
and in $\hat u_2$'s frame, $\hat u_1=\gamma \hat u_2 + \gamma \tilde v$,
where $\tilde v$ is purely-spatial according to $\hat u_2$ (since $g(\hat u_2, \tilde v)=0$).

I'm not sure about this. $\vec{v}$ is a 3-vector, not a 4-vector. So I don't think the vector equation you are writing here is valid, since it mixes 3-vectors and 4-vectors. Also, you can't plug a 3-vector into the 4-d spacetime metric to check for orthogonality.

 

[robphy]

I'm not sure about this. $\vec{v}$ is a 3-vector, not a 4-vector. So I don't think the vector equation you are writing here is valid, since it mixes 3-vectors and 4-vectors. Also, you can't plug a 3-vector into the 4-d spacetime metric to check for orthogonality.

Intuitively, the observer with 4-velocity $B^a$
decomposes a 4-vector $W^a$
into the sum of a 4-vector parallel to $B^a$ and [the rest] a 4-vector metric-orthogonal to $B^a$.
Using $(+,-,-,-)$

• The parallel component is $W^a B_a B^b = (W^a B_a) B^b$
• The orthogonal component is gotten by
  forming the projection operator $h_{ab}=g_{ab}-B_a B_b$ orthogonal to $B^a$,
  then writing $W^a h_{a}{}^b =W^a ( g_a{}^b-B_a B^b)= W^b - (W^aB_a)B^b$.
  Note: $W^a h_{a}{}^b B_b = (W^b - (W^aB_a)B^b)B_b = W^b B_b -(W^aB_a)(1)=0$.
  Thus, $W^a h_{a}{}^b$ is orthogonal to $B^b$.
• Thus, we have $W^b=W^a g_{a}{}^b= \underbrace{W^a B_a B^b}_{\mbox{ parallel to B^a}}+ \underbrace{W^a h_{a}{}^b}_{\mbox{orthogonal to B^a}}$
• I think it can be shown that 4-vectors of the form $W^a h_{a}{}^b$
  span a 3-dimensional space that is orthogonal to $B^a$,
  and can be identified as a " 3-vector according to $B^a$ ".

When $W^a$ is the 4-velocity $A^a$, we have
$$\begin{align*}
A^b=A^a g_{a}{}^b
&=A^a B_a B^b + A^a h_{a}{}^b \\
&=\underbrace{ \gamma\ B^b }_{\mbox{temporal part}}
+ \underbrace{A^a h_{a}{}^b}_{\mbox{spatial part}} \\
&=\cosh\theta_{AB}\ \hat B^b + \sinh\theta_{AB}\ \hat B_{\bot}^b\\
&=\cosh\theta_{AB}\ ( \hat B^b + \tanh\theta_{AB}\ \hat B_{\bot}^b)\\
&=\gamma_{AB} \ ( \hat B^b + V_{AB} \ \hat B_{\bot}^b)
\end{align*}$$ where I have emphasized the unit 4-vectors $\hat B^b$ and $\hat B_{\bot}^b$.

 

[vanhees71]

So, $\hat u_1$ and $\hat u_2$ are 4-velocities.
and in $\hat u_2$'s frame, $\hat u_1=\gamma \hat u_2 + \gamma \tilde v$,
where $\tilde v$ is purely-spatial according to $\hat u_2$ (since $g(\hat u_2, \tilde v)=0$).
That is to say,
$\gamma \hat u_2$ is timelike
and
$\gamma \tilde v$ is spacelike
Thus, the relative velocity (the 3-velocity) is the ratio $\frac{\gamma \tilde v}{\gamma}= \tilde v$, which is spacelike.

In terms of rapidity, this is
$\gamma \hat u_2=\cosh\theta\ \hat u_2$
and
$\gamma \tilde v=\sinh\theta\ \hat u_{2\bot}$.
Thus, the relative velocity (the 3-velocity) is the ratio $\tilde v=\displaystyle \frac{\sinh\theta \hat u_{2\bot} }{\cosh\theta}=\tanh\theta\hat u_{2\bot}$, which is spacelike.

This is in accord with my earlier statement

I don't know, what your symbols mean, but it doesn't look right, because $\vec{v}$ is not the spacial part of a four-vector. It's not manifestly covariant but a "three velocity".