# Lorentz boost -- speed or velocity?

• I
In summary, the Lorentz boost formula for spacetime coordinates involves a factor (t-vx/c^2), where v is the speed and x is the distance. This can be confusing because v and x can also refer to velocity and position, and the direction of the movement can affect the result. However, the boost is the same regardless of whether the objects are moving towards or away from each other. The boost can be expressed as a matrix, which involves projection operators and the Lorentz factor.
Gold Member
TL;DR Summary
The v and x in the factor (t-vx/c^2) in the Lorentz boost: if v is the speed and x is the distance, then it doesn't matter if the movement is away or towards or anything else, the boost will be the same, but if vx is the dot product of velocity and a spatial position vector, then the direction makes a difference So which is it?
The Wikipedia article on Lorentz transformations is a bit confusing by its using speed and velocity almost interchangeably: of course γ (Gamma) stays the same, but (letting c=1) t'=γ(t-vx) , then if this is v⋅x, and x stays the same, then there would be a difference if something were going away from the other at v or going towards each other at -v. I seem to recall that there is no difference, indicating that the scalars are what is meant, but in the Wiki article, there is a section in which they are, so I am obviously overlooking something.

##v## is velocity.

Orodruin said:
vv is velocity.
So does that mean that there is a difference in the boost between if two objects are traveling away or towards each other?

So does that mean that there is a difference in the boost between if two objects are traveling away or towards each other?
Yes. Technically this is akin to the question "do I rotate clockwise or anti-clockwise" in Euclidean geometry.

Orodruin said:
Technically this is akin to the question "do I rotate clockwise or anti-clockwise" in Euclidean geometry.
Thank you. I was also put off by the fact that most explanations of the solution to the twin paradox use a Minkowski diagram (e.g., on the Wiki page for the twin paradox) show the amounts of dilation to be the same for the outward and inward journeys.

Thank you. I was also put off by the fact that most explanations of the solution to the twin paradox use a Minkowski diagram (e.g., on the Wiki page for the twin paradox) show the amounts of dilation to be the same for the outward and inward journeys.
Time dilation depends on ##v^2##, not ##v##, so yes, they are the same.

Orodruin said:
Time dilation depends on v2v^2, not v, so yes, they are the same.
But the v2 only appears in the gamma (γ), but as the entire boost is
t'=γ(t-vx/c2)
then it seems that the boost depends also on the sign of v.

So does that mean that there is a difference in the boost between if two objects are traveling away or towards each other?

A boost is a change of reference frame. There is no sense in which one reference frame is moving "away from" or "towards" another reference frame. The relationship is one of relative velocity.

Likewise, the relative velocity of two objects is not defined by whether they are moving towards each other or not. An object may move towards you, past you and away from you, but its relative velocity remains the same throughout.

The sign of the relative velocity (positive or negative) is determined by whether you take the object to be moving in the positive x-direction or negative x-direction; not whether it's moving towards you or away from you.

But the v2 only appears in the gamma (γ), but as the entire boost is
t'=γ(t-vx/c2)
then it seems that the boost depends also on the sign of v.
The Euclidean analogue of this statement is
essentially saying cosine is an even function of $\phi$ but a rotation depends on the sign of $\phi$.

The time dilation formula is essentially the act of finding the temporal component of a timelike vector. The $\gamma$ can be expressed as $\cosh\theta$ , where $\tanh\theta=v$, ...think adjacent side.

But the v2 only appears in the gamma (γ), but as the entire boost is
t'=γ(t-vx/c2)
PeroK said:
A boost is a change of reference frame.
Responding to @nomadreid by adding specificity to @PeroK ’s statement, a Lorentz boost is a change of inertial reference frame. In the twin paradox, the traveler will either not be inertial, or else will have to switch to a different frame at some point in her journey (at least if she wants to continue to consider herself at rest). Therefore, the Lorentz boost does not cover the case of the traveler (at least, not in any naive sense).

Comparing wristwatches of two different people who start together, travel apart, and end back together is analogous to comparing the lengths of two different curves that connect two points. More than analogous, actually; it’s pretty much the same.

Summary:: The v and x in the factor (t-vx/c^2) in the Lorentz boost: if v is the speed and x is the distance, then it doesn't matter if the movement is away or towards or anything else, the boost will be the same, but if vx is the dot product of velocity and a spatial position vector, then the direction makes a difference So which is it?

The Wikipedia article on Lorentz transformations is a bit confusing by its using speed and velocity almost interchangeably: of course γ (Gamma) stays the same, but (letting c=1) t'=γ(t-vx) , then if this is v⋅x, and x stays the same, then there would be a difference if something were going away from the other at v or going towards each other at -v. I seem to recall that there is no difference, indicating that the scalars are what is meant, but in the Wiki article, there is a section in which they are, so I am obviously overlooking something.
The matrix for a boost in a direction ##\vec{v}=c \vec{\beta}## (i.e., the reference frame ##\Sigma'## moving in direction ##\vec{v}## against the reference Frame ##\Sigma##) is given by
$$\hat{\Lambda}=({\Lambda^{\mu}}_{\nu})=\begin{pmatrix} \gamma & -\gamma \vec{\beta}^{\text{T}} \\ -\vec{\beta} & \gamma \hat{P}_{\parallel} + \hat{P}_{\perp} \end{pmatrix}.$$
Here the ##3\times 3## projection operators are defined by
$$\hat{P}_{\perp} = \frac{1}{\beta^2} \vec{\beta} \vec{\beta}^{\text{T}}, \quad \hat{P}_{\perp}=\hat{1}-\hat{P}_{\parallel}$$
and ##\gamma=1/\sqrt{1-\vec{\beta}^2}##.

The boost of the spacetime coordinates are thus given by
$$\begin{pmatrix} c t' \\ \vec{x}' \end{pmatrix}=\begin{pmatrix} \gamma c t - \vec{\beta} \cdot \vec{x} \\ \vec{x}+(\gamma-1) \frac{\vec{\beta}(\vec{\beta} \cdot \vec{x})}{\beta^2} - \gamma \vec{\beta} c t \end{pmatrix}.$$

Thanks very much, PeroK, robphy amd vanhees 71. There are a number of points that I must reflect on, so thanks for being patient.

First, PeroK and robphy:

If I understand what you are saying, the projection from one reference frame to another (which is what the relative velocity is) is not affected by the fact that one reference frame is at a positive or negative angle to the other, which corresponds to the objects (not the reference frames) in collinear motion going towards or away from each other, so that in such cases the v in the (t-vt) is positive. When not collinear, that is another case.

(The ideas that the Lorentz boost is a change of reference frame is more-or-less clear, including in the twin paradox; it is in attempting to calculate the precise values, using the formulas I cited, that prompted this question.)

To vanhees 71. Your presentation apparently covers the cases when the movement is not collinear. It will take me somewhat longer to work through that (my mathematics is limited), so I may be back with further questions later. As a side point, I am a bit unclear about your phrasing
vanhees71 said:
reference frame Σ′\Sigma' moving in direction →v\vec{v} against the reference Frame Σ
Do you mean one reference frame rotating to another? If so, why is the direction expressed as a velocity? (Sorry if this sounds like a stupid question, but my knowledge in this area does not even permit me to know whether it is a stupid question.)

I'm not sure what you mean by "rotating". The usual terminology for Lorentz transformations is that the special sort, which is not rotating is called a boost, and that's the one I've given above for a general relative three-velocity between frames ##\Sigma'## and ##\Sigma##; ##\vec{v}## is the three-velocity of the reference frame ##\Sigma'## as measured in ##\Sigma##.

Thank you, vanhees71 and SiennaTheGr8. All this is very helpful, and I shall work on it.

Here $v$ is a vector component, i.e. ##\vec{v}=v \vec{x}##, and it can thus be negative or positive.

SiennaTheGr8 said:
I don't think that works when we write ##t^\prime = \gamma (t - vx)## and ##t = \gamma (t^\prime + vx^\prime)## and say that ##v## is the same quantity in both equations.
Why don't we work an example:

Assume frame ##S'## is moving in the x-direction relative to frame ##S## at velocity ##\vec v = -\frac 3 5 \hat {\vec x}##. This defines ##v = -\frac 3 5##. Note that in this case ##v## cannot be a magnitude or a speed as it is negative. The transformation is:
$$t' = \gamma(t - vx) = \gamma(t + \frac 3 5 x)$$
And the inverse transformation is:
$$t = \gamma(t' + vx') = \gamma(t' - \frac 3 5 x')$$

By using a standard velocity reference (like a standard angle reference), one can display various group properties of the transformation. I think this becomes difficult if speeds (magnitudes) or changes-of-reference are used instead.

PeroK
SiennaTheGr8 said:
The standard-configuration coordinate-transformation is good for learning, though, don't you think?

Whatever it becomes, I prefer that the physics and the mathematics lead to the correct equations and answers without further input [as if a computer could finish the calculation]... by this, I have in mind the following situation: a calculation is done and the sign turns out wrong... so someone's "physical intuition" is invoked to flip the last errant minus sign. I want the physics and the math from first principles to guide my intuition.

So, if the equations lead to a magnitude in that spot, great...
If I want to use magnitude in place of a signed quantity for simplicity, great... just make it clear from the onset and use it consistently.

Depending on how far one goes, one might consider the implication of a notation or choice of convention... is it too restrictive or too clumsy to discuss a further topic?
For example, if there is a third observer...

By the way, none of this really has anything to do with relativity.
or any issue involving components vs. magnitudes.

Do we want to to talk about positive and negative components of forces?
Or do we want to talk about some magnitudes "upward" and some magnitudes "downward"?

SiennaTheGr8
SiennaTheGr8 said:
I don't think that works when we write ##t^\prime = \gamma (t - vx)## and ##t = \gamma (t^\prime + vx^\prime)## and say that ##v## is the same quantity in both equations.
##v## is the same quantity in these equations. As I repeatedly said, you can formulate the general Lorentz boost in a basis-independent way using 3-vectors:
$$c t'=\gamma (c t-\vec{\beta} \cdot \vec{x})$$
$$\vec{x}'=\vec{x}+(\gamma-1) \hat{\beta} (\hat{\beta} \cdot \vec{x})-\gamma \vec{\beta}c t.$$
Here ##\vec{\beta}=\vec{v}/c##, where ##\vec{v}## is the velocity of the inertial frame ##\Sigma'## with respect to the inertial frame ##\Sigma##, ##\hat{\beta}=\vec{\beta}/|\vec{\beta}|## the corresponding direction vector, and ##\gamma=1/\sqrt{1-\beta^2}## the Lorentz factor.

It's easy to see that the inverse transformation takes the same form but the substitution ##\vec{\beta} \rightarrow -\vec{\beta}##:
$$c t=\gamma (c t'+\vec{\beta} \cdot \vec{x}')$$
$$\vec{x}=\vec{x}'+(\gamma-1) \hat{\beta} (\hat{\beta} \cdot \vec{x}')+\gamma \vec{\beta}c t'.$$

etotheipi and PeroK
SiennaTheGr8 said:
My objection is the redundancy that's introduced when standard configuration is stipulated ("primed frame moves in positive unprimed ##x##-direction") ... the ##v## in ##t^\prime = \gamma (t - vx)## and ##t = \gamma (t^\prime + vx^\prime)## is most simply understood as the relative speed of the frames, a positive value that both observers agree on.

There's no stipulation that ##v## be positive. The stipulation is that the two frames agree on the direction of the x-axis.

PS If you allow ##S'## to define its x-axis pointing in the opposite direction, then you may have to allow for appropriate transformation in the other coordinates. E.g. if ##y' = y##, then you would need ##z ' = -z## or accept that ##S'## has a left-handed coordinate system.

SiennaTheGr8 said:
Of course one can use the negative of the relative speed, but I don't know why one would!

Anyway, I'm in good company—Taylor & Wheeler in Spacetime Physics treat the boost parameter in the standard-configuration Lorentz-transformation as the relative speed of the frames: https://www.google.com/books/edition/Spacetime_Physics/PDA8YcvMc_QC?hl=en&gbpv=1&pg=PA99

One last question. If we have a frame ##S'## moving with speed ##v## in the positive x-direction (in frame S); and another frame ##S''## moving with speed ##v## in the negative x-direction, then how would you avoid having the same transformation from ##S## to both ##S'## and ##S''##?

etotheipi
SiennaTheGr8 said:
Yes, I'm fully aware that you can formulate the general Lorentz boost in this basis-independent way, and I think it's quite nice. (As I clarified in my prior post, by "same quantity" I was assuming that each observer would be using the velocity of the other frame relative to their own, in which case there's no sign-change.)

My objection isn't related to the basis-independent formulation. My objection is the redundancy that's introduced when standard configuration is stipulated ("primed frame moves in positive unprimed ##x##-direction") and then one says that the boost parameter in coordinate-transformations is not the relative speed ##v## but rather the ##v_x## component. With the directional information already encoded in the standard-configuration setup, the ##v## in ##t^\prime = \gamma (t - vx)## and ##t = \gamma (t^\prime + vx^\prime)## is most simply understood as the relative speed of the frames, a positive value that both observers agree on.
Of course, if ##\Sigma'## moves relative to ##\Sigma## with ##\vec{v}##, then ##\Sigma## moves relative to ##\Sigma'## with ##-\vec{v}##. That's the same as in Newtonian mechanics. The reason for this "reciprocity relation" is the assumed isotropy and homogeneity of space for any inertial observer, which holds for both the Newtonian and the special-relativistic spacetime descriptions.

robphy said:
I think ##(t - x v_x)## is okay...
that is, what is often written as ##v## assumes that ##\vec v=v_x \hat x##
and we write (perhaps confusingly) ##v## for ##v_x##.

Indeed. This confused me. For seasoned professionals it is easy to see the intended meaning, and I guess you can also sort of work out it needs to be signed (i.e. @PeroK's ##S'##/##S''## example), but it would be a lot clearer at least for pedagogical purposes if wikipedia were to use ##v_x## or better still just use the vector form @vanhees71 gave. Since ##v## to me implies a magnitude. Of course, once you get the hang of it then dropping the little '##x##' might save you a few precious milliseconds...

etotheipi said:
Indeed. This confused me. For seasoned professionals it is easy to see the intended meaning, and I guess you can also sort of work out it needs to be signed (i.e. @PeroK's ##S'##/##S''## example), but it would be a lot clearer at least for pedagogical purposes if wikipedia were to use ##v_x## or better still just use the vector form @vanhees71 gave. Since ##v## to me implies a magnitude. Of course, once you get the hang of it then dropping the little '##x##' might save you a few precious milliseconds...
Here's my take on it, for what it's worth.

We have two reference frames moving relative to each other. Usually, but not always, we take the x-axis to be along the direction of relative motion. If we choose a common origin (time and space), then we can instantaneously have a common set of coordinate axes (##x, y, z##). Note that if the relative motion is not along one of these axes, then we cannot have shared axes.

It is possible, but more limited, to assume that ##S'## moves in the positive x-direction in frame ##S##. Instead, we assume that ##S'## moves with velocity ##v = v_x##, where ##v## can be positive or negative. This implies that frame ##S## moves with velocity ##-v## in the x-direction frame ##S'##.

With these assumptions we have the standard transformation (##c = 1##):
$$t' = \gamma(t - vx), \ x' = \gamma(x - vt), \ y' = y, \ z' = z$$
And:
$$t = \gamma(t' + vx'), \ x = \gamma(x' + vt')$$
And you can check these work for both positive and negative ##v##.

vanhees71, SiennaTheGr8 and etotheipi
PeroK said:
With these assumptions we have the standard transformation (##c = 1##): $$t' = \gamma(t - vx), \ x' = \gamma(x - vt), \ y' = y, \ z' = z$$ And: $$t = \gamma(t' + vx'), \ x = \gamma(x' + vt')$$

And I guess the idea is that you can switch the labels of S and S' (and everything that comes along with it, like ##t/t'## etc.) and the form of the transformation from the new S to the new S' has to be the same (since otherwise we've got a real problem...!). It's then evident that the second one is a ##-(-v)##.

etotheipi said:
And I guess the idea is that you can switch the labels of S and S' (and everything that comes along with it, like ##t/t'## etc.) and the form of the transformation from the new S to the new S' has to be the same (since otherwise we've got a real problem...!). It's then evident that the second one is a ##-(-v)##.

Yes, as long as you remenber that ##v## is the velocity of ##S'##.

etotheipi
SiennaTheGr8 said:
Of course one can use the negative of the relative speed, but I don't know why one would!

Anyway, I'm in good company—Taylor & Wheeler in Spacetime Physics treat the boost parameter in the standard-configuration Lorentz-transformation as the relative speed of the frames: https://www.google.com/books/edition/Spacetime_Physics/PDA8YcvMc_QC?hl=en&gbpv=1&pg=PA99
SUMMARY of below: I think Taylor is thinking velocity, but occasionally uses speed for velocity.

If he meant speed throughout, the statement on p111 below
$v_{rel}$ is treated as a positive quantity.
as a speed seems unnecessary. But I think he's saying that to remind the reader that the rocket has positive-velocity
[earlier on p111]
$v_{rel}$, stands for relative speed of the two frames (rocket moving in the positive x-direction in the laboratory).
So, I think Taylor really means velocity.
details:

While Taylor does use "speed" on Page 99 (bolding mine),
Think of a sparkplug at rest at the origin of a rocket frame that moves with speed $v_{rel}$ relative to the laboratory. The sparkplug emits a spark at time t' as measured in the
and again on Page 100
they do depend on the relative speed $v_{rel}$
He used "velocity" earlier on page 97 (referring to equations on page 98.. screenshot below.. which are identical to those on page 102 [Eq. L-10b]) (bolding mine)
Following are the Lorentz transformation equations. Here is the $v_{rel}$ relative velocity between rocket and laboratory frames. For our convenience we lay the positive x-axis along the direction of motion of the rocket as observed in the laboratory frame and choose a common reference event for the zero of time and space for both frames.

The inverse transformation is (from page 103)

In his concluding paragraph on page 103, he says (bolding mine)
A simple but powerful argument from symmetry leads to the same result. The symmetry argument is based on the relative velocity between laboratory and rocket frames. With respect to the laboratory, the rocket by convention moves with known speed in the positive x-direction. With respect to the rocket, the laboratory moves with the same speed but in the opposite direction, the negative x-direction. This convention about positive and negative directions — not a law of physics! — is the only difference between laboratory and rocket frames that can be observed from either frame. Lorentz transformation equations must reflect this single difference. In consequence, the “inverse” (laboratory-to-rocket) transformation can be obtained from the “direct” (rocket-to-laboratory) transformation by changing the sign of relative velocity, $v_{rel}$, in the equations and interchanging laboratory and rocket labels (primed and unprimed coordinates). Carrying out this operation on the Lorentz transformation equations (L-10) yields the inverse transformation equations (L-11).

In the summary on page 111,
he concludes (bolding mine)
The equations that transform rocket coordinates (primed coordinates) to laboratory coordinates (unprimed coordinates) have the form

(eq. L-10b)

where $v_{rel}$, stands for relative speed of the two frames (rocket moving in the positive x-direction in the laboratory). The inverse Lorentz transformation equations transform laboratory coordinates to rocket coordinates:

(eq L-11b)

in which $v_{rel}$ is treated as a positive quantity. In both these sets of equations, coordinates of events are measured with respect to a reference event. It is really only the difference in coordinates between events that matter, for example $x_2-x_1=\Delta x$ for any two events I and 2, not the coordinates themselves. This is important in deriving the Law of Addition of Velocities.

I know that Morin also uses plain ##v##'s etc. for components of velocity unless there's something in more than 1 dimension happening. I think it just seems to be a convention in the field.

SiennaTheGr8 said:
And I don't have a link, but I get Jackson also: "the frame ##K^\prime## is moving in the positive ##z## direction with speed ##v##, as viewed from ##K##" (section 11.3.A in the 3rd edition).

Jackson's 11.3.A begins
Consider two inertial reference frames К and K' with a relative velocity v between them.
and does say (bolded and colored by me, as you've quoted)
the frame K' is moving in the positive z direction with speed v, as viewed from K.
So, that's not "just speed"... independent of direction.
It's "a signed speed"... "positive-z with speed".

Eq. 11.17 says
$\vec \beta =\frac{\vec v}{c} \quad \beta=|\vec \beta|$
which he uses in the boost transformations in 11.16 and 11.18.
But finally for the generalized case, eq 11.19 says
$x_0'=\gamma(x_0-\vec \beta \cdot \vec x)\\ \vec x'=\vec x + \frac{(\gamma-1)}{\beta^2}(\vec \beta \cdot \vec x)\vec\beta -\gamma \vec \beta x_0$

Still sounds like velocity to me.

etotheipi
PeroK said:
Yes, as long as you remenber that ##v## is the velocity of ##S'##.
...as measured in ##S## (just to be pedantic ;-)).

PeroK said:
A boost is a change of reference frame. There is no sense in which one reference frame is moving "away from" or "towards" another reference frame. The relationship is one of relative velocity.

Likewise, the relative velocity of two objects is not defined by whether they are moving towards each other or not. An object may move towards you, past you and away from you, but its relative velocity remains the same throughout.

The sign of the relative velocity (positive or negative) is determined by whether you take the object to be moving in the positive x-direction or negative x-direction; not whether it's moving towards you or away from you.
If I throw you another mag I hope and trust that you don't suppose that the difference between uprange and downrange doesn't matter.

Summary:: The v and x in the factor (t-vx/c^2) in the Lorentz boost: if v is the speed and x is the distance, then it doesn't matter if the movement is away or towards or anything else, the boost will be the same, but if vx is the dot product of velocity and a spatial position vector, then the direction makes a difference So which is it?

The Wikipedia article on Lorentz transformations is a bit confusing by its using speed and velocity almost interchangeably: of course γ (Gamma) stays the same, but (letting c=1) t'=γ(t-vx) , then if this is v⋅x, and x stays the same, then there would be a difference if something were going away from the other at v or going towards each other at -v. I seem to recall that there is no difference, indicating that the scalars are what is meant, but in the Wiki article, there is a section in which they are, so I am obviously overlooking something.
If velocity is important does this speed apply to the nucleus of the atom?

PeroK
bill hart said:
If velocity is important does this speed apply to the nucleus of the atom?
Why are you asking? What's this atom you are thinking of doing?

A couple of general points: the Lorentz transforms relate coordinates in two different frames of reference. Thus they apply to measurements of anything made using those two frames. However you need to be careful with atoms because you need a relativistic quantum theory to describe them, and naive application of special relativity to tiny billiard balls may well give you misleading results.

Somewhere in the opposing direction of velocity in the atom there must be a deference between polarities. If so what causes the difference between pos and neg. it appears to be difference in clockwise and anti-clockwise. One produces a specific charge. And this seems to depend on the outside charge applied. What is this? Therefore it seems that it must be an outside force. This force must have a starting point. Either outer space or influence from the sun. As applied it must drive the nucleus in one or the other direction. And that influence or velocity (spin) seems to determine or result in almost all of our envirenment. SPIN amounts to everything. A black hole in the begging is a direction movement, but when spin is encoded it changes polarity, and alters the total (NET)strength of the BH.
Here directional rotation towards or away from is reduced to a strong force or a weak force. and is altered to or involves the work that BH does.
SPIN is applied to everything! Here right hand or left hand rules must be the deciding factor. So if the (rule of charges) is applied and we use have the intelligence to use it we can override the assumption that there remains can we or can't we!

we have no control. Not so all of this implies that as energy it can be altered to do what mankind needs or desires.

weirdoguy

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