I Can time be another basis vector under Galilean relativity?

[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
$$|\vec{\beta}_{\text{rel}}= \frac{1}{1-\vec{\beta}_u \cdot \vec{\beta}_v} \sqrt{(\vec{\beta}_u-\vec{\beta}_w)^2 - (\vec{\beta}_u \times \vec{\beta}_w)^2}.$$

 

[PeterDonis]

These authors appear to think such terms are also appropriate.

I don't think those authors are using the terms the way you were trying to use them.

Schutz appears to be using "spatial components" to mean the components $U_x$, $U_y$, $U_z$ of the 4-velocity $U$ in some particular frame. You're trying to use terms like "component" to refer to a complete 4-vector, which is what I'm objecting to.

Geroch doesn't use the word "component" or "part" or similar words, except right at the end of the quote you give (and I would object to his usage there on the grounds that it's not consistent with his usage elsewhere in that quote). He uses the word "displacement", which suggests a complete 4-vector, not a piece of one, which is of course consistent with the equation he gives, which is basically the same one we have written here, expressing the 4-velocity of particle 1 in terms of the sum of two 4-vectors: one parallel to the 4-velocity of particle 2 and one orthogonal to it.

Wald uses "components" the same way Schutz does; see above.

MTW uses "spatial part" to refer to a 3-vector, not a 4-vector.

Hartle and Rindler appear to be using "components" the same way Schutz does; see above.

 

[PeterDonis]

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.

One other difference is that in the Galilean case there is no spacetime metric; the "Galilean metric" you refer to is the temporal metric, and there is a separate spatial metric. So there is no well-defined concept of "orthogonality" between timelike vectors and spacelike vectors (and of course there are no null vectors).

 

[PeterDonis]

is that tangent space really a "vector space"?

Yes, definitely. You can check for yourself that it satisfies all of the vector space axioms.

 

[PeterDonis]

I am assuming a world with no acceleration where the tangent is always parallel to the worldline.

A tangent vector is always parallel to the worldline at the point where it is tangent.

Proper acceleration means path curvature, i.e., the tangent vector is not parallel transported along the worldline.

 

[PeterDonis]

For me, at Galilean level, time is a scalar

That is true, but you have to carefully consider what that means. Time being a scalar means that there is an invariant number $t$ at every event, which doesn't change when you change inertial frames. But you can still pick out a "time axis" for each inertial frame by simply using the worldline of its spatial origin.

and time axis does not overlap the worldline

The "time axis" must always overlap some worldline, because some worldline will always be the time axis (the worldline of the spatial origin of whichever frame you have chosen).

If you make a Galilean boost from one inertial frame to another, the planes of simultaneity stay the same, but, heuristically, they "slide" horizontally relative to one another, so that some different worldline is now vertical and becomes the time axis of the new frame. This worldline will be the worldline of the spatial origin of the new frame.

I would see concepts like Galilean 4-velocity as a "place-holder"

No, they're not. They're perfectly well-defined in Galilean relativity. They don't work the same as they do in SR, but that doesn't mean they're not well-defined and valid.

 

[Saw]

"tangent space" where the time basis vector is the 4-velocity with first component being 1 (dimensionless) and rest of components = 0 (also dimensionless)?
(…j
is that tangent space really a "vector space"?

Yes, definitely. You can check for yourself that it satisfies all of the vector space axioms.

Tangent spaces in general, yes. But this particular tangent space is supposed to be made of 4-velocity vectors, right? And if I look at the first axiom of a vector space and try to add one 4-velocity vector to another, I have doubts. Can you do that?

 

[Saw]

The "time axis" must always overlap some worldline, because some worldline will always be the time axis (the worldline of the spatial origin of whichever frame you have chosen).

If you make a Galilean boost from one inertial frame to another, the planes of simultaneity stay the same, but, heuristically, they "slide" horizontally relative to one another, so that some different worldline is now vertical and becomes the time axis of the new frame. This worldline will be the worldline of the spatial origin of the new frame..

If you told me that you can paint it there, overlapping the worldlines, well, the paper accepts anything. But if you affirm that this has a “heuristic“ sense, which one is it? Is it that Bob’s wristwatch is at rest with Bob, so it is always spatially located by the origin of his frame? But that is not “enabling us to learn or discover“ anything about Galilean time, which is Newton’s absolute time and is the same regardless the motion of the clock that registers it. What other heuristic sense do you have in mind?

 

[Saw]

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.

No, they're not. They're perfectly well-defined in Galilean relativity. They don't work the same as they do in SR, but that doesn't mean they're not well-defined and valid.

Well-defined? Yes, Galilean 4-velocity is 3-velocity to which one adds a coordinate time over proper time component which is 1 in all reference frames, since in this context both things coincide. What is the added value of this concept, i.e. its practical use? If it were none, it would fit in the notion of “place-holder”, i.e. an idea for which in this context there is no practical use but may have one once that you move into another territory where coordinate time and proper time do differ.

 

[PeterDonis]

Can you do that?

Sure, just add their components as you would with any pair of vectors.

this particular tangent space is supposed to be made of 4-velocity vectors, right?

Not just 4-velocity vectors, no. All vectors at a point. A vector does not have to have unit norm.

 

[PeterDonis]

Galilean 4-velocity is 3-velocity to which one adds a coordinate time over proper time component which is 1 in all reference frames

If you allow the 3-velocity to change if you change frames, according to the Galilean boost equations, yes, this is one way of defining Galilean 4-velocity.

What is the added value of this concept, i.e. its practical use?

What is the practical use of any mathematical definition? It depends on who is using it and what they are using it for.

 

[PeterDonis]

if you affirm that this has a “heuristic“ sense

All I meant by "heuristic" is that I was giving an ordinary language description, not the actual math. Don't read too much into it.

 

[Saw]

Sure, just add their components as you would with any pair of vectors.

I looked at Wikipedia entry for 4-velocity and it does say (at a note) that a tangent space is a vector space (like the entry about tangent space itself), but the thing is that it also says this with regard to 4-velocity: "addition of two four-velocities does not yield a four-velocity: the space of four-velocities is not itself a vector space."

How does that match with 4-velocity being the time basis vector, given that what you do with basis vectors is precisely scaling them?

Not just 4-velocity vectors, no. All vectors at a point. A vector does not have to have unit norm.

I guess you refer to other derivatives, like 4-force or 4-acceleration... But focusing again on 4-velocity, which is our object of interest because it i said to be at least one option for the time basis vector, I would come back to the issue mentioned before: a basis vector, in a normalized basis, must be unitary but then its role is being scaled by a certain coefficient...

Furthermore, the fact that it is unitary does not mean that it is not a unit of something, i.e. a second, a meter or whatever, how is that consistent with being dimensionless?

To sum up, I have difficulties to see the role of the 4-velocity as the time basis vector. But since this concerns SR and not in particular the Galilean version of the 4-velocity, should I maybe start another thread? Before that, is there any source I should read to know more about the matter?

 

[vanhees71]

In special relavity the four-velocity is by definition
$$u^{\mu}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau},$$
which implies the constraint
$$u_{\mu} u^{\mu}=c^2,$$
i.e., the four-velocity vectors are "on shell" and as such don't form a vector space.

As I repeatedly said, I don't think that in Galilean spacetime a four-vector formalism makes too much sense, although you can of course formulate everything also in an extended Hamilton formalism with an additional arbitrary parameter and then describing the motion of a particle by $t(\lambda)$ and $\vec{x}(\lambda)$.

 

[haushofer]

In special relavity the four-velocity is by definition
$$u^{\mu}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau},$$
which implies the constraint
$$u_{\mu} u^{\mu}=c^2,$$
i.e., the four-velocity vectors are "on shell" and as such don't form a vector space.

As I repeatedly said, I don't think that in Galilean spacetime a four-vector formalism makes too much sense, although you can of course formulate everything also in an extended Hamilton formalism with an additional arbitrary parameter and then describing the motion of a particle by $t(\lambda)$ and $\vec{x}(\lambda)$.

See for this Hamilton formalism of a non-relativistic point particle e.g. chapter 3 of my thesis "Newton-Cartan gravity revisited",

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwiO677A5YP9AhX4gP0HHQumC8kQFnoECBMQAQ&url=https://research.rug.nl/files/34926446/Complete_thesis.pdf&usg=AOvVaw3xeIMItlns2gjYN4ANv5Bd

On page 78 you find the general-covariant action of a point particle in curved (Newtonian) spacetime, eqn. 5.14. There you also see the role of the gauge field belonging to the central extension: it couples to the 4-velocity in the same way as the electromagnetic 4-potential couples to a relativistic particle. Basically this extra coupling is needed if you invoke local Galilei-boosts as a symmetry.

 

[PeterDonis]

the thing is that it also says this with regard to 4-velocity

That's because, as I said, the tangent space is the space of all vectors at a point, not just the space of all unit timelike vectors (which is the space of 4-velocities).

How does that match with 4-velocity being the time basis vector, given that what you do with basis vectors is precisely scaling them?

I have no idea what you mean here. Basis vectors do not have to be unit vectors; in fact, if you choose a null basis vector it obviously can't be a unit vector. An orthonormal basis in spacetime is a basis consisting of one timelike and three spacelike unit vectors, all mutually orthogonal, but that is not the only possible kind of basis. In any case, there is no requirement that basis vectors form a vector space all by themselves; the only requirement is that every vector in the vector space must be a linear combination of basis vectors.

I guess you refer to other derivatives, like 4-force or 4-acceleration

No. I'm just using the simple obvious definition of a vector space.

Take any timelike unit vector and multiply it by any real number. That gives you another timelike vector that is in the tangent space.

Take any spacelike unit vector and multiply it by any real number. That gives you another spacelike vector that is also in the tangent space.

Take any null vector. That is also a vector in the tangent space.

Take any linear combination of the above with real coefficients. That is also a vector in the tangent space.

 

[PeterDonis]

a basis vector, in a normalized basis, must be unitary but then its role is being scaled by a certain coefficient

Scaled how? Where? What are you talking about?

the fact that it is unitary does not mean that it is not a unit of something, i.e. a second, a meter or whatever, how is that consistent with being dimensionless?

A vector with dimensionless units still has a norm, which is a real number.

I have difficulties to see the role of the 4-velocity as the time basis vector.

So far all you have shown is that you have difficulties understanding how vector spaces work.

 

[Saw]

it also says this with regard to 4-velocity: "addition of two four-velocities does not yield a four-velocity: the space of four-velocities is not itself a vector space."

How does that match with 4-velocity being the time basis vector, given that what you do with basis vectors is precisely scaling them?

I have no idea what you mean here.

There may be a misunderstanding here. The fact that the 4-velocities do not form by themselves a vector space is another issue. My "that" refers to the part in red: if you cannot add two four-velocities it means that you cannot multiply one four-velocity by the real number two, i.e. "scale" it by two.

How does that match with 4-velocity being the time basis vector, given that what you do with basis vectors is precisely scaling them?

(...) a basis vector, in a normalized basis, must be unitary but then its role is being scaled by a certain coefficient...

Scaled how? Where? What are you talking about?

I don't think that my use of the verb "scale" is so abnormal or out of place here.

A basis (a set of basis vectors) serves to span a vector space. This means that you can describe any member of the vector space in terms of those basis vectors, i.e. as a linear combination of the latter. The word "linear" is there to denote that you simply scale each basis vector by a scalar, without higher powers. The "combination" part means that, once you have scaled the basis vectors, you add them up. "Scalar" is a generic term that stands for either real or complex numbers, although, as you said, here (SR) we use real numbers (after the idea of scaling time with the imaginary number i has been dropped). A synonym of scalar would be here the term that you also used, coefficients. For the avoidance of doubt, when I say that we scale the time basis vector, what I mean is that we measure with a clock when events 1 and 2 take place and the difference between the two recordings (say, 2 seconds) is what we scale the basis vector by, so as to say that the time component of the spacetime interval separating events 1 and 2 is 2 seconds (or rather c * 2 s).

For completeness, I am aware that we also talk about a scalar as the outcome of a dot product, which is invariant. But that is not really a different meaning. Actually, the dot product of the above-mentioned spacetime interval with itself gives off a scalar because what you do that way is extracting one of those coefficients or scalars accompanying the basis vectors, as it can be readily seen in the example where the two events are measured by the same clock and the outcome of the dot product is the clock's reading (the proper time).

So far all you have shown is that you have difficulties understanding how vector spaces work.

I did not try to show anything, I only asked questions and I think that I have proved that the questions were not without a reason. There will probably be an answer that will prove my doubts groundless, but asking the questions was not unreasonable.

In the light of that, before going ahead with other issues, I would appreciate if you could withdraw that comment, which IMHO is not constructive!

 

[PeterDonis]

The fact that the 4-velocities do not form by themselves a vector space is another issue.

No, it's not an issue. There is no requirement that the 4-velocities all by themselves form a vector space.

if you cannot add two four-velocities

You can; you just don't get another 4-velocity. As above, that is not an issue; there is no requirement that adding two 4-velocities must give another 4-velocity.

it means that you cannot multiply one four-velocity by the real number two, i.e. "scale" it by two.

Wrong. Addition of vectors in a vector space is a different operation from scalar multiplication. There is no relationship between the two.

Again, you do not appear to understand the basics of how vector spaces work. You need that understanding for the subject under discussion.

I don't think that my use of the verb "scale" is so abnormal or out of place here.

It's not a question of your usage (although your usage is not standard). It's a question of you having a proper understanding of the concepts we are discussing. You do not appear to have such an understanding.

I did not try to show anything

I didn't say you did. Your posts show that you don't have a good understanding of vector spaces even though you weren't trying to show that at all.

There will probably be an answer that will prove my doubts groundless

An answer to what? So far you have not asked a cogent question. Everything you have asked is based on an incorrect understanding of the subject matter. The only answer to that is that you need to fix your understanding.

I would appreciate if you could withdraw that comment, which IMHO is not constructive!

Your opinion is mistaken. I am giving you the constructive feedback of telling you that you don't understand a key concept for this discussion, namely vector spaces. Even if I were to withdraw that comment, which I won't, it would not change the fact that you have an incorrect understanding and that is preventing you from even being able to frame a cogent question.

 

[Saw]

Wrong. Addition of vectors in a vector space is a different operation from scalar multiplication. There is no relationship between the two.

This proves that you don't understand the question.

Addition of two vectors of the same magnitude but pointing in different directions is obviously not the same as scaling any of the two vectors by the real number 2.

Addition of two vectors of the same magnitude and being colinear obviously yields the same result as scaling any of them by 2.

Therefore, if you ban the possibility of addition of two vectors, you ban the possibility of adding colinear vectors and therefore you also seem to ban the possibility of scaling them. And if not, you have not mentioned any reason why not.

there is no requirement that adding two 4-velocities must give another 4-velocity.

There seems to be this requirement if 4-velocity is to act as a time basis vector, which is what we are talking about, since what you do with basis vectors, as I explained (without being contradicted), is scaling them, in which case the result of this scaling is obviously the same entity as the non-scaled one. And if not, you have not mentioned any reason why not. If you don't have any reason, say it. If you have it, say it as well, because it is improper for you as a mentor to keep aces up in your sleeve. We are here to learn, not to play childish ego games.

The rest of your post...

Again, you do not appear to understand the basics of how vector spaces work. You need that understanding for the subject under discussion.It's not a question of your usage (although your usage is not standard). It's a question of you having a proper understanding of the concepts we are discussing. You do not appear to have such an understanding.I didn't say you did. Your posts show that you don't have a good understanding of vector spaces even though you weren't trying to show that at all.An answer to what? So far you have not asked a cogent question. Everything you have asked is based on an incorrect understanding of the subject matter. The only answer to that is that you need to fix your understanding.Your opinion is mistaken. I am giving you the constructive feedback of telling you that you don't understand a key concept for this discussion, namely vector spaces. Even if I were to withdraw that comment, which I won't, it would not change the fact that you have an incorrect understanding and that is preventing you from even being able to frame a cogent question.

... is a surprisingly lengthy and repetitive way of reiterating the ad hominem attacks.

If you don't feel like mentoring me, please just leave the discussion, but don't prevent others from doing it by closing a new thread of mine, which is probably what you are paving the way for with your aggressive comments.

Who knows? There may be someone of a different opinion. You don't need to be always right, which seems to be your main goal and activity. For example, not so long ago, you replied to me:

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.

No, they're not. They're perfectly well-defined in Galilean relativity. They don't work the same as they do in SR, but that doesn't mean they're not well-defined and valid.

But recently someone commented:

As I repeatedly said, I don't think that in Galilean spacetime a four-vector formalism makes too much sense

I would kindly request you to engage in a discussion about that, if you wish, and let others help me with my own queries, if they feel like that.

Should you prefer to keep sabotaging me, then I would like to report the situation and ask for an impartial analysis as to whether my questions are "worthy of a polite answer" or not. The analysis should not be, though, whether my questions are "cogent". You know, "cogent" means "persuasive" and is an adjective that is usually applied to "arguments". But, let me insist, I am not trying to persuade anybody about anything, I am only asking questions as a student. And it would be ridiculous that PhysicsForums became known, from now on, thanks to you, as the place where you can only obtain guidance if you convince your mentor from the very beginning that you don't need any guidance, because your question is "cogent"!

 

[PeterDonis]

Addition of two vectors of the same magnitude but pointing in different directions is obviously not the same as scaling any of the two vectors by the real number 2.

Agreed. I have never said otherwise.

Addition of two vectors of the same magnitude and being colinear obviously yields the same result as scaling any of them by 2.

Agreed. I have never said otherwise.

Therefore, if you ban the possibility of addition of two vectors

Nobody has done any such thing. You are the one who keeps making an issue of the fact that adding two 4-velocity vectors gives a vector that is not a unit vector. By making an issue of that non-issue, you are showing that you do not understand the subject under discussion. Yes, it is a fact that adding two unit vectors gives a vector that is not a unit vector. No, that is not a problem, because unit vectors by themselves don't form a vector space. Which is also not a problem, because nobody needs them to be a vector space. You do not appear to understand this.

what you do with basis vectors, as I explained (without being contradicted), is scaling them, in which case the result of this scaling is obviously the same entity as the non-scaled one

I have no idea where you are getting this from. Multiplying a unit vector by the scalar $2$ gives a vector whose norm is $2$, not a unit vector. Yes, that's correct. No, it's not a problem. See above.

 

[PeterDonis]

We are here to learn, not to play childish ego games.

a surprisingly lengthy and repetitive way of reiterating the ad hominem attacks.

Should you prefer to keep sabotaging me

These comments are uncalled for and further indicate that you do not understand the points I have been trying to make. They are also likely to get you a warning if you continue along these lines.

I would like to report the situation

You are always free to use the Report button.

 

[PeterDonis]

As I repeatedly said, I don't think that in Galilean spacetime a four-vector formalism makes too much sense

I would kindly request you to engage in a discussion about that

A discussion about that is fine if you frame a question based on it.

The subthread you are complaining about started with your incorrect assertion in post #126 that the tangent space is only composed of 4-velocity vectors (i.e., unit timelike vectors). It's not. Instead of realizing that it's not and rethinking your position in the light of that new knowledge, you have continued to make further erroneous claims, which I have been correcting (see my post #140 for my latest corrections).

Bear in mind that this thread is not just for you, even though you're the OP. People come to PF all the time to read threads, which means when incorrect statements are made in PF threads, we try to correct them.

 

[Saw]

You are the one who keeps making an issue of the fact that adding two 4-velocity vectors gives a vector that is not a unit vector.

I don't have a problem with that. How could I? It is obvious that, if you scale a unit vector, you get that vector scaled by whatever scalar you have multiplied it by, i.e. not a unit vector anymore.

However, the statement by Wikipedia is the following: "addition of two four-velocities does not yield a four-velocity". That seems to be a problem because it implies that the addition gives off a different object (something that is not a 4-velocity vector), not the same object (a 4-velocity vector) with a coefficient other than 1.

It seems that your ace in the sleeve is that a 4-velocity's magnitude is always 1. You don't need to keep it, I am saying it right now. But if so, what is then the object that yields the addition of two colinear 4-velocities or in general the scaling thereof by some coefficient? What is its name and what kind of object is it?

 

[PeterDonis]

the statement by Wikipedia is the following: "addition of two four-velocities does not yield a four-velocity".

Yes, that's correct. Adding two unit vectors does not give another unit vector. That is what Wikipedia is saying, though of course it's Wikipedia and you should not expect it to say things rigorously.

That seems to be a problem because it implies that the addition gives off a different object

No, it doesn't. You are adding two vectors and getting another vector.

It seems that your ace in the sleeve is that a 4-velocity's magnitude is always 1

It's not an "ace in the sleeve", it's part of the definition of a 4-velocity. It's a unit timelike vector tangent to an object's worldline.

what is then the object that yields the addition of two colinear 4-velocities or in general the scaling thereof by some coefficient? What is its name and what kind of object is it?

A vector. I have already said this multiple times. As above, you are adding two vectors and getting another vector. Or you are multiplying a vector by some scalar and getting another vector. This is vector spaces 101. If you seriously don't understand how this works, please take some time to learn it from a textbook or similar source on the basics of vector spaces.

Your issue seems to be that "4-velocity" or "unit vector" is somehow some "different" type of object from other vectors. As far as vector addition and scalar multiplication is concerned, it's not. It's just a vector. The fact of having unit norm is useful for some other purposes, but it has no significance as far as the basics of vector spaces are concerned.

 

[Saw]

The subthread you are complaining about started with your incorrect assertion in post #126 that the tangent space is only composed of 4-velocity vectors (i.e., unit timelike vectors). It's not. Instead of realizing that it's not and rethinking your position in the light of that new knowledge, you have continued to make further erroneous claims, which I have been correcting (see my post #140 for my latest corrections).

False. As you can see below, I received the information that the tangent space is composed of things other than 4-velocity, I accepted it as good knowledge, and it simply happens that I returned to the question that interested me, because it is related to the object of the thread: where is the time basis vector in Galilean relativity as compared to SR, which is what we are discussing now.

I guess you refer to other derivatives, like 4-force or 4-acceleration... But focusing again on 4-velocity, which is our object of interest because it i said to be at least one option for the time basis vector, I would come back to the issue mentioned before...

I am just trying to understand something that looks weird to me, that the 4-velocity is the time basis vector and I do think that others may have this legitimate lack of understanding. And I do find puzzling this twisted way of mentoring where the objective seems to be, not answering things, but convincing the asker that she should not ask. And I guess there may be many askers who have suffered these methods and may be thinking the same. And this is a shame for PhysicsForums and it should be stopped because it is a most valuable place that does not deserve to gain this ugly image.

 

[PeterDonis]

I am just trying to understand something that looks weird to me, that the 4-velocity is the time basis vector

The 4-velocity is not necessarily the time basis vector. A particular object's 4-velocity can be the time basis vector, if you choose to work in that object's rest frame and you want to use unit vectors as your basis vectors (which is usually a good choice but is not a requirement).

I still do not understand what the issue is. In the previous post of yours that you quoted from, you cut off your quote; here is the rest of that paragraph of your post:

a basis vector, in a normalized basis, must be unitary but then its role is being scaled by a certain coefficient...

I don't see what the issue is here. Yes, an arbitrary vector in a vector space can be written, given a choice of basis, as a linear combination of the basis vectors, i.e., with each basis vector multiplied by some scalar coefficient. As I said before, this is vector spaces 101. I cannot understand what the problem is.

(Btw, "unitary" does not mean the same thing as "unit vector", which is what basis vectors will be in a normalized basis. "Unitary" is a term that applies to operators and means something quite different.)

this twisted way of mentoring where the objective seems to be, not answering things, but convincing the asker that she should not ask

As I have already said, if people make incorrect statements here, we try to correct them. You have made a number of them in this thread, which I have corrected. All of that is a separate issue from what your actual question is (as above, I still don't understand what it is) and getting that question answered.

 

[berkeman]

Thread is closed temporarily for a Mentor discussion...