I Can time be another basis vector under Galilean relativity?

[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 $\tilde v$.

 

[PeterDonis]

I never said that $\tilde{v}$ is "the spatial part of a 4-vector".

You then go on to quote yourself saying exactly that phrase: "the spatial part of a 4-vector". You say it about $\gamma \tilde{v}$, not $\tilde{v}$, but that's merely a quibble, and it's still wrong anyway.

No, $\gamma \tilde{v}$ is not "the spatial part of a 4-vector". It's another 4-vector, since multiplying a 4-vector by an invariant scalar gives another 4-vector. This 4-vector is orthogonal to the 4-velocity of particle 2, which means, as I said, that it's a spacelike 4-vector. But, as I also said, that doesn't mean it's "the spatial part of a 4-vector". It's not. As I said, and you agreed with me saying it since it's in the quote from me that you said you completely agreed with, a spacelike 4-vector is not the same as "the spatial part of a 4-vector". $\gamma \tilde{v}$ is a spacelike 4-vector.

 

[robphy]

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.

I do mean tilde means 4-vector,
as has been the notation in handwritten calculations (sometimes with under-tilde's for covectors) since the arrowhead is typically interpreted as the usual 3-dimensional vectors.

The "correspondence" I refer to is using the same letter, symbol, kernel $v$ for my 4-vector construction
to suggest that it has something to do the 3-d quantity we are more likely familiar with.

• It's like a reboot of a movie or tv-show... I'm hoping to bank on using the earlier intuition, but in the context of a 4-vector calculation. It's along the lines of why some use $t^a$ as 4-velocities.
• In references, like Wald (which I used in my PF Insight), we use a similar correspondence
  

  For an observer moving with 4-velocity $v^a$, the quantity
  $$E_a = F_{ab}V ^b\qquad\qquad\mbox{ (4.2.21)}$$ is interpreted as the electric field measured by that observer,

  This $E^a$ is a 4-vector...
  but with suitable correspondences not made explicit here,
  this corresponds to the 3-D electric field vector $\vec E$ seen by an observer.
  So, I am using this method of presentation since that's what Wald did in his class.

---

introducing the rapidity doesn't really gain us anything conceptually.

I think it does because the
temporal component is $\cosh\theta$
spatial component is $\sinh\theta$
(as components of a unit-vector [similar to Euclidean geometry and trigonometry])
and the ratio is $\frac{\sinh\theta}{\cosh\theta}=\tanh\theta$ [similar to a "slope" in Euclidean geometry].
This to again support that
the spatial component is $\gamma \tilde v$ (with component $\sinh\theta$) [not $\tilde v$].

 

[PeterDonis]

I do mean tilde means 4-vector

Ok, good.

I think it does because the
temporal component is $\cosh\theta$
spatial component is $\sinh\theta$
(as components of a unit-vector [similar to Euclidean geometry and trigonometry])
and the ratio is $\frac{\sinh\theta}{\cosh\theta}=\tanh\theta$ [similar to a "slope" in Euclidean geometry].

But none of these are invariants, and they invite confusion between components of 4-vectors in particular frames and actual 4-vectors. Such as:

This to again support that
the spatial component is $\gamma \tilde v$ (with component $\sinh\theta$) [not $\tilde v$].

No, $\gamma \tilde{v}$ is not "the spatial component". It is a spacelike 4-vector. A 4-vector is not the same as a component (or subset of components) of a 4-vector.

 

[robphy]

No, $\gamma \tilde v$ is not "the spatial part of a 4-vector". It's another 4-vector, since multiplying a 4-vector by an invariant scalar gives another 4-vector. This 4-vector is orthogonal to the 4-velocity of particle 2, which means, as I said, that it's a spacelike 4-vector. But, as I also said, that doesn't mean it's "the spatial part of a 4-vector". It's not.

So, this sounds like the issue of
"Given $\vec W=W_x \hat x + W_y \hat y$,
what do you want to call $W_x$
vs
what you want to call $W_x \hat x$?"

 

[PeterDonis]

So, this sounds like the issue of
"Given $\vec W=W_x \hat x + W_y \hat y$,
what do you want to call $W_x$
vs
what you want to call $W_x \hat x$?"

No, it's not, because $W_x$ in general is not an invariant.

Here we are talking about a case where you have a 4-vector, $A^a$ (the 4-velocity of particle 1), that can be expressed as the sum of two other 4-vectors, $\gamma_{AB} \hat{u_2}$ and $\gamma_{AB} V_{AB} \hat{u_{2 \bot}}$, the second of which you are calling $\gamma_{AB} \tilde{v}$. But this sum is an invariant. The coefficients in the sum are not picked out by choosing any particular frame and looking at components in that frame; they are picked out by the invariant dot product of the two 4-velocities $A^a$ and $B^a$ and the condition of orthogonality to $B^a = \hat{u_2}$. These are all invariants.

 

[robphy]

So, this sounds like the issue of
"Given $\vec W=W_x \hat x + W_y \hat y$,
what do you want to call $W_x$
vs
what you want to call $W_x \hat x$?"

No, it's not, because $W_x$ in general is not an invariant.

Here we are talking about a case where you have a 4-vector, $A^a$ (the 4-velocity of particle 1), that can be expressed as the sum of two other 4-vectors, $\gamma_{AB} \hat{u_2}$ and $\gamma_{AB} V_{AB} \hat{u_{2 \bot}}$, the second of which you are calling $\gamma_{AB} \tilde{v}$. But this sum is an invariant. The coefficients in the sum are not picked out by choosing any particular frame and looking at components in that frame; they are picked out by the invariant dot product of the two 4-velocities $A^a$ and $B^a$ and the condition of orthogonality to $B^a = \hat{u_2}$. These are all invariants.

Ok then,
this sounds like the issue of
"Given $\vec W=W_{\|} \hat X + W_{\bot} \hat X_\bot$,
where $\hat X$ and $\hat X_{\bot}$ are unit-vectors (not necessarily along the $\hat x$-axis) with $\hat X \cdot \hat X_{\bot}=0$,
and $W_{\|}=\vec W\cdot \hat X$
what do you want to call $\left( \vec W\cdot \hat X \right)$
vs
what you want to call $\left( \vec W\cdot \hat X \right) \hat X$?"

 

[PeterDonis]

Ok then,
this sounds like the issue of
"Given $\vec W=W_{\|} \hat X + W_{\bot} \hat X_\bot$,
where $\hat X$ and $\hat X_{\bot}$ are unit-vectors (not necessarily along the $\hat x$-axis) with $\hat X \cdot \hat X_{\bot}=0$,
and $W_{\|}=\vec W\cdot \hat X$
what do you want to call $\left( \vec W\cdot \hat X \right)$
vs
what you want to call $\left( \vec W\cdot \hat X \right) \hat X$?"

I'm not sure what issue there is. Assuming we are now working in Euclidean 3-space, $( \vec{W} \cdot \vec{X} )$ is a scalar and $( \vec{W} \cdot \vec{X} ) \hat{X}$ is a 3-vector. Neither one is a "component" or "part" of a 3-vector.

Similarly, in the spacetime example, $\gamma_{AB}$ and $V_{AB}$ are scalars and $\gamma_{AB} \hat{u_2}$ and $\gamma_{AB} V_{AB} \hat{u_{2 \bot}}$ are 4-vectors. None of them are a "component" or "spatial part" of a 4-vector.

 

[robphy]

Similarly, in the spacetime example, $\gamma_{AB}$ and $V_{AB}$ are scalars and $\gamma_{AB} \hat{u_2}$ and $\gamma_{AB} V_{AB} \hat{u_{2 \bot}}$ are 4-vectors. None of them are a "component" or "spatial part" of a 4-vector.

"temporal part" (or terms like it) is shorthand for "component of $A^a$ parallel to $B^a$"
"spatial part" (or terms like it) is shorthand for "component of $A^a$ orthogonal to $B^a$"
as in below, quoting myself

• 
  $$h_{ab}=g_{ab}-B_a B_b$$
  [snip]
• Thus, we have $$W^b=W^a g_{a}{}^b=
  \underbrace{W^a B_a B^b}_{\mbox{ parallel to $B^a$}}+
  \underbrace{W^a h_{a}{}^b}_{\mbox{orthogonal to $B^a$}}$$
  [snip]
  $$\begin{align*}
  A^b=A^a g_{a}{}^b
  &=A^a B_a B^b + A^a h_{a}{}^b \\
  &=\underbrace{ \gamma\ B^b }_{\mbox{temporal part}}
  + \underbrace{A^a h_{a}{}^b}_{\mbox{spatial part}} \\
  \end{align*}$$

Or from your post

If we look at your decomposition of the 4-velocity Aa of particle 1 in terms of the unit vectors related to particle 2, we have, using my notation γAB for clarity:

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

Maybe there is a better description.
What would you call each term of the right hand side
in my decomposition (or your decomposition) of $A^a$ with respect to $B^a$?

 

[PeterDonis]

What would you call each term of the right hand side
in my decomposition (or your decomposition) of $A^a$ with respect to $B^a$?

The projections of $A^a$ parallel and orthogonal to $B^a$.

 

[Saw]

Please feel free to leave these questions aside, while you discuss your more advanced clarifications.

the spatial-velocity is the spatial-component of the unit-tangent-vector, divided by the temporal component of that unit-tangent vector

I was asking if you were identifying the time basis vector with Galilean spatial velocity because in SR you identify it with 4-velocity and you said that you did not need 4-velocity at Galilean level... So... what is the Galilean time basis vector in your view? Simply the spacetime displacement vector of Bob from event A to B or Alice from A to C? Or maybe we come back to Galilean 4-velocity?

TLDR: yes… but relative-spatial velocity vectors are spacelike.

I find it weird to see the term "spacelike" applied to a rate... To me saying that events 1 and 2 are separated by a spacelike distance meant that there can be no causal influence between them (and that they are simultaneous at least in one frame). What does it mean that a relative velocity is spacelike?

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).

… 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.

No problem with speaking about 4-velocity. I mean, I think that I understand the concept. It means, for every tick of Bob's wristwatch, how much time has elapsed and how much distance Bob has traversed; in Bob's frame, where he is at rest, distance traversed is 0 and time elapsed is that very same tick; in other inertial frames, distance will be whatever and the time elapsed will be the time dilation factor (gamma)...

In SR, the magnitude of this 4-velocity vector is c, i.e. 1 in geometric units. And (if you measure time in length units), would it be dimensionless? In that case I would find this also weird, because I would expect the time basis vector to have units of time or rather of c * t, not to be dimensionless. In my understanding a unit vector is numerically unitary, but it is one unit of the relevant magnitude. Certainly, it shows a direction but has one unit of that direction, whatever it is (eg: it is 1 meter in the X or the Y or the Z direction or 1 s in the T direction or rather 1 meter in the c*T direction)... Otherwise, it may be "tangent to" the time basis vector, sharing its direction, but I would not define it as "the" time basis vector.

 

[PeterDonis]

In SR, the magnitude of this 4-velocity vector is c, i.e. 1 in geometric units. And (if you measure time in length units), would it be dimensionless?

It is dimensionless in geometric units, which is the same as measuring time in length units, or more precisely as using the same units for length and time.

I would expect the time basis vector to have units of time or rather of c * t

You are conflating two different physical interpretations of what you are calling "the time basis vector". This is unfortunately something that is glossed over in many SR texts: the mathematical object that we call "Minkowski spacetime" can be interpreted physically two different ways: as a spacetime, i.e., a 4-dimensional manifold in which points correspond to spacetime events; or as a vector space of spacetime vectors at a point (more technically called the "tangent space" at that point), where "points" in this vector space correspond to vectors with different magnitudes and spacetime directions.

These two interpretations correspond to different units. If you are using the spacetime interpretation, the basis vectors can be thought of as having units of time or length--which, if you are using geometric units, are the same units. But if you are using the tangent space interpretation, the basis vectors have units of speed, which in geometric units is dimensionless.

 

[robphy]

Similarly, in the spacetime example, $\gamma_{AB}$ and $V_{AB}$ are scalars and $\gamma_{AB} \hat{u_2}$ and $\gamma_{AB} V_{AB} \hat{u_{2 \bot}}$ are 4-vectors. None of them are a "component" or "spatial part" of a 4-vector.

Maybe there is a better description.
What would you call each term of the right hand side
in my decomposition (or your decomposition) of $A^a$ with respect to $B^a$?

The projections of $A^a$ parallel and orthogonal to $B^a$.

I would agree and also call it the projections parallel and orthogonal to $B^a$.

However, since my job is about physics,
I also try to give a physical description that we are modeling with that mathematical interpretation.

---

So, although we may disagree,
in my opinion, I think "temporal" and "spatial part" (or terms akin to them) are also appropriate.
These authors appear to think such terms are also appropriate. (Bolding mine.)

• 2.3 The four-velocity, p. 41
  In our four-geometry we define the four-velocity $\vec U$ to be a vector tangent to the world line of the particle, and of such a length that it stretches one unit of time in that particle’s frame.
  ...Thus we could also use as our definition of the four-velocity of a uniformly moving particle that it is the vector $\vec e_0$ in its inertial rest frame. The word ‘velocity’ is justified by the fact that the spatial components of $\vec U$ are closely related to the particle’s ordinary velocity $v$, which is called the three-velocity.

• http://home.uchicago.edu/~geroch/Course Notes
  7. The Geometry of World-Lines, p.16
  In other words, “pure spatial displacements” from p (according to this observer) are
  represented by vectors at p which are orthogonal to the four-velocity of the observer.
  ...He first decomposes $\eta^a$ into the sum of a multiple of his four-velocity
  and a vector orthogonal to his four-velocity:
  $\eta^a = −\xi^a(\eta^m \xi_m) + (\eta^a + \xi^a (\eta^m \xi_m)) \qquad(7)$
  The first term represents what he would call the “temporal displacement” between
  p and q. ... The second term in (7) represents what he would call the “spatial displacement” between p and q. (This term is orthogonal to his four-velocity, and so represents, for him, a spatial
  displacement.)
  
  13. Electromagnetic Fields: Decomposition by an Observer, p.30
  To resolve $F_{ab}$ into “spatial tensors” for our observer, we project the indices of $F_{ab}$ parallel and orthogonal to $\xi^a$.
  ...Note that the determination of an electric field from the electromagnetic field involves a choice of an observer (i.e., of his four-velocity). The remaining piece of $F_{ab}$ is $F_{mn} h^m{}_a h^n{}_b$, a spatial, antisymmetric tensor.
  
  14. Maxwell’s equation, p.32 $$\nabla^{[a}F^{bc]}=0\qquad(25)$$ $$\nabla_b F^{ab}=J^a\qquad(26)$$...We also decompose our derivative operator: let a dot denote $\xi^a \nabla_a$ (the time derivative), and let $D_a = h_a{}^b \nabla_b$ (the spatial derivative). Thus, we now have a special symbol for the spatial and temporal parts (with respect to our field of observers, i.e., with respect to $\xi^a$) of every quantity appearing in (25) and (26).

• (Ch5.1-p96)
  Namely, the vector $G^{ab} u_b$ (as well as $T^{ab}u_b$) cannot have a spatial component, or isotropy would be violated.
  
  (Ch 10-p253)
  The remaining three components of Maxwell's equations do contain second time derivatives of the spatial components of $A_a$...
  
  (p.259)
  In the electromagnetic case, the identity (10.2.6) implied that if the constraint (10.2.3) is satisf‌ied initially and the spatial components of Maxwell’s equations are satisf‌ied everywhere, ... A completely analogous result applies in general relativity . As a consequence of the Bianchi identity,... if the constraints (10.2.28) and (10.2.30) are satisf‌ied initially and the spatial components of Einstein’s equation are satisf‌ied everywhere ,

• (Ch 3.1-p.73)
  A fully geometric equation will involve the test particle's energy-momentum 4-vector, p, not just the spatial part p as measured in a specific Lorentz frame...

• (p.190)
  Since the energy of a photon is related to its frequency by $E = \hbar\omega$,
  $$\hbar \omega = -p\cdot u_{obs},\qquad (9.12)$$ giving the frequency measured by an observer with four-velocity $u_{obs}$. The spatial components $u^i_{obs}$ of the four-velocity are zero for a stationary observer. The time component $u^t_{obs}(r)$ of a stationary observer...

• (p.110)
  Since the quantities on the LHSs of eqns (6.6) and (6.7) are the spatial and temporal
  components of the 4-vector...

 

[robphy]

I was asking if you were identifying the time basis vector with Galilean spatial velocity because in SR you identify it with 4-velocity

and with the unit tangent vector in Euclidean geometry.
It points in the "direction forward along the worldline curve".
Note: Tangency to a curve is independent of a choice of metric.

and you said that you did not need 4-velocity at Galilean level...
So... what is the Galilean time basis vector in your view?

I never said that.
My position is that a worldline has a future unit tangent-vector and this is true in Galilean and Special Relativity.
On a position-vs-time diagram,
an inertial object at rest has its t-axis along the future tangent-vector of its worldline.
So, by the Relativity Principle, any inertial object has its t-axis along the future tangent-vector of its worldline.

Simply the spacetime displacement vector of Bob from event A to B or Alice from A to C?

Yes, but more.

Or maybe we come back to Galilean 4-velocity?

Again, my position is that a worldline has a future unit tangent-vector and this is true in Galilean and Special Relativity.

In the relativity literature, this future unit tangent-vector is called the 4-velocity.
In Special Relativity, the future unit tangent-vector along the worldline is the Minkowski 4-velocity.
In Galilean Relativity, the future unit tangent-vector along the worldline is the Galilean 4-velocity.
In both cases, they are future tangent vectors along the worldline.
What makes them different is that

• the Minkowski metric defines the unit tangent vector
  so that the tips of the 4-velocities from an event lie on a hyperbola/hyperboloid.
• the Galilean metric defines the unit tangent vector
  so that the tips of the 4-velocities from an event lie on a hyperplane.
• just as the Euclidean metric defines the unit tangent vector
  so that the tips of the unit vector from a point lie on a circle/sphere.

---

Here's a well-known paper on Galilean/Newtonian spacetime structure:
"Four-dimensional formulations of Newtonian mechanics and their relation to the special and the general theory of relativity" by Peter Havas (Review of Modern Physics 1964), p.943

We can thus define a Newtonian four-velocity and four-acceleration
by the same Eqs. (31) as before.
The spatial parts of these contravariant vectors are the usual three-velocity v...

 

[Saw]

You are conflating two different physical interpretations of what you are calling "the time basis vector". This is unfortunately something that is glossed over in many SR texts: the mathematical object that we call "Minkowski spacetime" can be interpreted physically two different ways: as a spacetime, i.e., a 4-dimensional manifold in which points correspond to spacetime events; or as a vector space of spacetime vectors at a point (more technically called the "tangent space" at that point), where "points" in this vector space correspond to vectors with different magnitudes and spacetime directions.

These two interpretations correspond to different units. If you are using the spacetime interpretation, the basis vectors can be thought of as having units of time or length--which, if you are using geometric units, are the same units. But if you are using the tangent space interpretation, the basis vectors have units of speed, which in geometric units is dimensionless.

Thanks, this is enlightening, as it does explain the units issue. Can we thus say that we have two options to define the "time basis vector" in SR: (i) one in spacetime (the "object space"?), where the time basis vector would be a 4-displacement vector with time component = 1 unit of time or rather c* time (e.g.: 1 light second) and spatial components = 0 of space (e.g. again 0 light second) and (ii) another in "tangent space" where the time basis vector is the 4-velocity with first component being 1 (dimensionless) and rest of components = 0 (also dimensionless)?

(I am assuming a world with no acceleration where the tangent is always parallel to the worldline. Maybe the introduction of acceleration makes the 4-velocity approach preferable?)

On another note, is that tangent space really a "vector space"?

Again, my position is that a worldline has a future unit tangent-vector and this is true in Galilean and Special Relativity.

In the relativity literature, this future unit tangent-vector is called the 4-velocity.
In Special Relativity, the future unit tangent-vector along the worldline is the Minkowski 4-velocity.
In Galilean Relativity, the future unit tangent-vector along the worldline is the Galilean 4-velocity.
In both cases, they are future tangent vectors along the worldline.
What makes them different is that

• the Minkowski metric defines the unit tangent vector
  so that the tips of the 4-velocities from an event lie on a hyperbola/hyperboloid.
• the Galilean metric defines the unit tangent vector
  so that the tips of the 4-velocities from an event lie on a hyperplane.
• just as the Euclidean metric defines the unit tangent vector
  so that the tips of the unit vector from a point lie on a circle/sphere.

Well, at last I think that I understood you! The 4-velocity approach to the definition of the time basis vector was indeed confusing me, but I believe that the clarification by PeterDonis sorts it out. No matter if the answer to my question (see above) is "the 2 approaches are valid" or "only 4-velocity approach is", your position is now clear to me.

You are making a comparison between Minkowskian, Galilean and Euclidean by looking at worldlines after 1 unit of the (say vertical) axis. For me, at Galilean level, time is a scalar and time axis does not overlap the worldline, it remains stuck where it was as a common axis for all observers. For you, the Galilean time axis does accompany the worldline as the tangent line to it and its unit is the 4-velocity, of which there would also exist a Galilean version, albeit with a numerical common time component. I acknowledge the beauty of the analogy, it is just that I would see concepts like Galilean 4-velocity as a "place-holder". i.e. something that does not really exist in Galilean relativity but is holding a place for another thing that will appear in SR, when you actually need to move forward to the latter.

You have already answered that you don't share my view, due to "this and that", but we interrupted that discussion because I was not understanding your construction. Now that it seems that I do maybe we can return to those issues, like the dictates of the principle of relativity, the meaning of the eigenvector or how you handle in your Galilean spacetime the need to have homogeneous units.

 

[vanhees71]

Of course you can in a manifestly covariant way define something like a relative four-velocity, but the standard description is about a non-covariant three-vector, i.e., you have to boost to the rest frame of one particle.

Let $(u^{\mu})=\gamma_u (1,\vec{\beta}_u)$ and $(w^{\mu})=\gamma_2 (1,\beta_2,0,0)$ four-velocities of the two massive particles. For simplicity I've assumed that the "lab frame" is oriented such that particle 2 moves in 1-direction. To define the relative velocity in the standard HEP and relativistic-transport-theory sense you have to boost to the rest frame of (say) the 2nd particle.
$$u^{\prime \mu}=\gamma_u \begin{pmatrix}\gamma_w (1-\beta_u^1 \beta_w) \\ \gamma_w (\beta_u^1-\beta_w) \\ \beta_u^2 \\ \beta_u^3 \end{pmatrix}.$$
As shown in Sects. 1.6 and 1.7

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

the relative velocity can be written as
$$\vec{\beta}_{\text{rel}}=\frac{1}{u^{\prime 0}} \vec{u}'=\frac{1}{1-\vec{\beta}_u \cdot \vec{b}_w} \left (\vec{\beta}_u-\vec{\beta}_w +\frac{\gamma_u}{\gamma_u+1} \vec{\beta}_u \times (\vec{\beta}_u \times \vec{\beta}_w) \right ).$$
It's magnitude is
$$|\vec{\beta}_{\text{rel}}= \frac{1}{1-\vec{\beta}_u \cdot \vec{\beta}_v} \sqrt{(\vec{\beta}_u-\vec{\beta}_w)^2 - (\vec{\beta}_u \times \vec{\beta}_w)^2}.$$