Can time be another basis vector under Galilean relativity?

[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".

 

[robphy]

[snip]
$\gamma \hat u_2=\cosh\theta\ \hat u_2$
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.

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".

All quantities with $\tilde{\phantom{v}}$ (tilde) are 4-vectors.
All quantities with $^\widehat{\phantom{v}}$ (hat) are unit 4-vectors, with square-norm $1$ for timelike and $-1$ for spacelike.

I did not use the arrowhead anywhere. I never wrote $\vec v$.
There are no explicit 3-vector quantities.
But, in my last post, I am suggesting that any 4-vector constructed with the projection tensor $h_{ab}$ is orthogonal to the $B_a$~observer and can be identified with a 3-vector for the $B_a$~observer.

I agree with you that

$\vec{v}$ is not the spacial part of a four-vector.

and I never said that.
From what I wrote above,
$$\gamma \tilde v=\sinh\theta\ \hat u_{2\bot}$$ is the spatial component (spatial part) of a four-vector...
akin to what you wrote

$$u_1=\begin{pmatrix}\gamma \\ \gamma \vec{v} \end{pmatrix}, \quad u_2=\begin{pmatrix} 1 \\ \vec{0} \end{pmatrix}.$$

Using your notation, $\vec v$ is ratio $\frac{\Delta \vec x}{\Delta t}$ (like a slope)
$$\vec v=\frac{\mbox{spatial part}}{\mbox{temporal part}}=\frac{\gamma \vec v}{\gamma}$$

And, I had formed the same ratio
when I wrote
"Thus, the relative velocity (the 3-velocity) is the ratio $$\tilde v=\displaystyle \frac{\sinh\theta \hat u_{2\bot} }{\cosh\theta}=\tanh\theta\hat u_{2\bot}$$"

So, I think we are saying the same thing... but my notation may be unfamiliar to you.

---

(I used a similar approach in my PF insight
The Electric Field Seen by an Observer: A Relativistic Calculation with Tensors
which tries to follow the approach in Ch 13 of Geroch's General Relativity lecture notes (draft at http://home.uchicago.edu/~geroch/Course Notes ). The spatial-velocity construction is in Ch 7.

I never liked the 3-vector approach and I never liked the differential approach from old relativity texts to develop relativistic formulas. But when Geroch showed these geometrically-motivated 4-vector methods in class, it was an eye-opener.)

 

[PeterDonis]

From what I wrote above,
$$\gamma \tilde v=\sinh\theta\ \hat u_{2\bot}$$ is the spatial component (spatial part) of a four-vector

You're not using your own notation consistently.

The 4-velocity of particle 1, in the rest frame of particle 2, has components $(\gamma, \gamma v_x, \gamma v_y, \gamma v_z)$. You can write this as $(\gamma, \gamma \vec{v}) = \cosh \theta \hat{u_2} + \sinh \theta \hat{u_{2 \bot}}$, but in those equations, while $\hat{u_2}$ and $\hat{u_{2 \bot}}$ are unit 4-vectors, $\vec{v}$ is not. It's a 3-vector defined in particle 2's rest frame. Or, if you like, it's a 3-vector defined in the Euclidean 3-space orthogonal to particle 2's worldline. But none of those things justify putting a tilde over $v$, because none of those things make $v$ a 4-vector.

 

[robphy]

From what I wrote above,
$$\gamma \tilde v=\sinh\theta\ \hat u_{2\bot}$$ is the spatial component (spatial part) of a four-vector...

You're not using your own notation consistently.

The 4-velocity of particle 1, in the rest frame of particle 2, has components $(\gamma, \gamma v_x, \gamma v_y, \gamma v_z)$. You can write this as $(\gamma, \gamma \vec{v}) = \cosh \theta \hat{u_2} + \sinh \theta \hat{u_{2 \bot}}$, but in those equations, while $\hat{u_2}$ and $\hat{u_{2 \bot}}$ are unit 4-vectors, $\vec{v}$ is not. It's a 3-vector defined in particle 2's rest frame. Or, if you like, it's a 3-vector defined in the Euclidean 3-space orthogonal to particle 2's worldline. But none of those things justify putting a tilde over $v$, because none of those things make $v$ a 4-vector.

I repeat that I did not use the arrowhead anywhere (including the passage you quoted),
.... except to refer to other people's notation.
I wrote $\tilde v$ (with a tilde) to suggest a correspondence (as described below) but not an equality.
I am not promoting someone's $\vec v$ (with arrowhead) to a 4-vector.
I am, however, doing a 4-vector calculation to suggest ( without explicit projection mappings )
what could be connected with what people describe as 3-vectors.Generally, in my past posts (here in this thread and elsewhere [like in my PF Insight]),
I rarely (if ever) use the arrowhead for a 4-vector---
for a 4-vector,
I always use a tilde $\ \tilde{\vphantom{v}}\$ or an abstract index (as in $v^a$)...
because I don't want it to be confused with a three-dimensional vector.

I never explicitly wrote $(\gamma, \gamma \vec{v}) = \cosh \theta \hat{u_2} + \sinh \theta \hat{u_{2 \bot}}$ , so that is NOT MY notation.

(While I may want to convey or suggest the physical identification,
I will not write what your wrote
but I am sure that it requires
additional notation about mapping vectors via projections to convey the mathematical identification of,
e.g., the "unit-4-vector in the x-direction" with the "unit 3-vector in the x-direction" although we will likely refer to both as $\hat x$... etc..
So, I avoid doing that and hoping that we can understand the context without having to formulaically write it down.
Along these lines, I try to stick with 4-vector notation as much as possible.
And I think am consistent.
(I could be mistaken... so if you could find an expression in this thread
with an arrowhead $\vec v$ that I use but in not reference to someone else's notation,
I will stand corrected.)
I avoid a 3-vector notation as much as possible, unless I am referring to someone else's notation.
)

When I write

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 $g(\hat u_2, \tilde v)=0$).

The right-hand side is a sum of two 4-vectors $\gamma \hat u_2$ and $\gamma \tilde v$.
So, I use
$\gamma \tilde v$ (with the tilde)
to suggest a connection to (but not explicitly equate [using projection operations])
$\gamma \vec v$ (with the arrow).

If it is so bothersome, I should have just written
$\hat u_2$'s frame, $\hat u_1=\gamma \hat u_2+\gamma \tilde Q$.
I would carry out the same construction to obtain $\tilde Q$
as the ratio of the spatial-component of $\hat u_1$ (according to $\hat u_2$)
and the temporal-component of $\hat u_1$,
and then note that it seems to have the property of the so-called 3-velocity, written as $\vec v$ (with the arrowhead) in @vanhees71 's notation.

To be completely precise, I would probably have to use something like
Jantzen's Spacetime Splitting formalism
(see
"2.1.1 Observer orthogonal decomposition"
"2.1.2 Observer-adapted frames"
"2.1.3 Relative kinematics: algebra"
and onward of http://www34.homepage.villanova.edu/robert.jantzen/gem/gem.pdf#page=27 ).

 

[PeterDonis]

I repeat that I did not use the arrowhead anywhere (including the passage you quoted),

I didn't say you did. I did, to emphasize my point.

I wrote $\tilde v$ (with a tilde) to suggest a correspondence (as described below) but not an equality.

I'm not sure I understand. Earlier, you said:

All quantities with $\tilde{\phantom{v}}$ (tilde) are 4-vectors.

So when you write $\tilde{v}$, I interpret it as a 4-vector, because you told me to. I'm trying to be clear about exactly which 4-vector it is supposed to refer to. But if it's a 4-vector, then it's not "the spatial part of a 4-vector", it's just a 4-vector--with, as I said before, a time component that is zero in one particular frame, the rest frame of particle 2, which means it must be a spacelike 4-vector, as you have said. But "spacelike 4-vector" is not the same as "spatial part of a 4-vector".

If you now say a tilde doesn't mean an actual 4-vector, just some kind of "correspondence", then I no longer understand your notation.

The right-hand side is a sum of two 4-vectors $\gamma \hat u_2$ and $\gamma \tilde v$.

If you define $\gamma$ as the dot product of the two 4-velocities, i.e., $\gamma = g_{ab} A^a B^b$, then $\gamma$ is an invariant (although $\gamma$ then becomes a somewhat confusing notation for this invariant; maybe $\gamma_{AB}$ would be better), so yes, multiplying it by some 4-vector does give another 4-vector. We already know which 4-vector $\hat{u_2}$ is; it's just the 4-velocity of particle 2, i.e., $B^a$ (since 4-velocities are already unit vectors). The question is, which 4-vector is $\tilde{v}$?

If we look at your decomposition of the 4-velocity $A^a$ of particle 1 in terms of the unit vectors related to particle 2, we have, using my notation $\gamma_{AB}$ for clarity:

$$
A^a = \gamma_{AB} \hat{u_2} + \gamma_{AB} V_{AB} \hat{u_{2 \bot}}
$$

where $V_{AB}$ is obtained by solving the equation $\gamma_{AB} = 1 / \sqrt{1 - V_{AB}^2}$. We would also have $\gamma_{AB} = \cosh \theta$ and $\gamma_{AB} V_{AB} = \sinh \theta$, where $\theta$ is the "relative rapidity", and therefore $V_{AB} = \tanh \theta$; but since $\gamma_{AB}$ is the obvious invariant here, the dot product of the two 4-velocities, introducing the rapidity doesn't really gain us anything conceptually.

Given that equation, when you write $\gamma \tilde{v}$, implying that $\tilde{v}$ is a 4-vector, and where $\gamma$ is what I have been calling $\gamma_{AB}$ above, then we must have $\tilde{v} = V_{AB} \hat{u_{2 \bot}}$. In other words, this is a spacelike 4-vector, orthogonal to the 4-velocity of particle 2, that points in the direction of relative motion between particle 1 and particle 2, and has a magnitude equal to $V_{AB}$ as defined above. Calling $V_{AB}$ the "magnitude of the relative velocity" would be reasonable.

 

[robphy]

So when you write $\tilde v$, I interpret it as a 4-vector, because you told me to. I'm trying to be clear about exactly which 4-vector it is supposed to refer to. But if it's a 4-vector, then it's not "the spatial part of a 4-vector", it's just a 4-vector--with, as I said before, a time component that is zero in one particular frame, the rest frame of particle 2, which means it must be a spacelike 4-vector, as you have said. But "spacelike 4-vector" is not the same as "spatial part of a 4-vector".

Yes, I completely agree.

I never said that $\tilde v$ is "the spatial part of a 4-vector". (If anyone said that, it wasn't me.)
What I did say is

From what I wrote above,
$$\gamma \tilde v=\sinh\theta\ \hat u_{2\bot}$$ is the spatial component (spatial part) of a four-vector...

says that $\gamma \tilde v$ is "the spatial part of a 4-vector",
that is, $(\gamma \tilde v)$ is "the spatial part of a 4-vector", not $\tilde v$.