V=dx/dt (relative velocity between two inertial observers)

[Kairos]

In special relativity, when writing the relative velocity between two inertial observers v=dx/dt, I suppose that "dt" is a coordinate time interval? But as it is not measurable, is the velocity correct and the same for both observers if each one uses his proper time intervals?

 

[PeroK]

First, you need to understand the difference between the proper time of a clock at rest in an IRF and the coordinate time in that frame:

The proper time of a single (at rest) clock is a local measurement of the coordinate time at that fixed point in space. In other words, the time measured by that clock serves as a measurement of coordiante time at that one fixed point in space.

The coordinate time is a more abstract concept of the time assigned to every event in spacetime. In terms of measurements, it's the time measured by a hypothetical grid of at-rest clocks at every point in space.

When we talk about an object having a uniform velocity (in 1D) in an IRF, we mean that it has a worldline of the form $(t, vt)$. That statement makes sense mathematically, in terms of the concept of an IRF and associated coordinates. And, in practical terms, in means that the object is at the point with x-coordinate $vt$ when the local clock at that point reads $t$. Which can be verified by a sequence of local measurements.

That is, in fact, the definition of uniform velocity. You can treat it both as a mathematical statement about coordinates and as a statement about a sequence of local measurements.

 

[Ibix]

An observer can't "use his proper time intervals" without defining a simultaneity convention. He needs two events on the other guy's worldline in order to measure his velocity, and he can only be present at (at most) one of those.

It is true that, in an inertial reference frame, the lapse of proper time of a stationary clock is the same as the lapse of coordinate time, but the clock's proper time is not defined off its worldline. You can't define the time at which something not on the clock's worldline happens solely by reference to the clock's proper time. You need some way to map the proper time on to events off the worldline. It's like using a ruler to measure where a thing is - if the thing is not touching the ruler you need some way of projecting the ruler markings to where the thing actually is. That's what a simultaneity convention does.

It's fairly easy to show that if you move with speed $v$ in my inertial rest frame then I move with speed $-v$ in yours. Write down the coordinates of two events on my worldline in my frame and Lorentz transform them.

 

[Kairos]

The coordinate time is a more abstract concept of the time assigned to every event in spacetime. In terms of measurements, it's the time measured by a hypothetical grid of at-rest clocks at every point in space

in the same rest frame?

 

[Kairos]

in an inertial reference frame, the lapse of proper time of a stationary clock is the same as the lapse of coordinate time,

This is expected to hold for each inertial observers in relative motion, but can I conclude that they have the same coordinate time?

Thanks to both of you for your help, I will nevertheless need some time to digest these notions.

 

[PeroK]

in the same rest frame?

If two clocks are moving inertially and are at rest relative to each other, then they share the same rest frame. So, yes, if that's what you mean.

 

[Ibix]

This is expected to hold for each inertial observers in relative motion, but can I conclude that they have the same coordinate time?

The existence of the Lorentz transforms that change one frame's coordinate time into another's would suggest that they do not have the same coordinate time.

 

[Kairos]

The existence of the Lorentz transforms that change one frame's coordinate time into another's would suggest that they do not have the same coordinate time.

Okay, I understand that the coordinate times of the two observers are different when they are represented in a single spacetime diagram, but if we draw two spacetime diagrams with identical coordinate units, one for each frame considered at rest, I suppose that the coordinate time intervals of the two diagrams are identical?

 

[etotheipi]

Okay, I understand that the coordinate times of the two observers are different when they are represented in a single spacetime diagram, but if we draw two spacetime diagrams with identical coordinate units, one for each frame considered at rest, I suppose that the coordinate time intervals of the two diagrams are identical?

Co-ordinate time intervals... between which two events? That you must specify in order to speak meaningfully about a time interval in the first place. Geometrically speaking, if two spacetime events have a separation vector $X^a$ then the time interval between those two events measured by an inertial observer of 4-velocity $u^a$ is nothing but $u_a X^a$. It's clear that inertial observers of different 4-velocities will measure different time intervals between those two events.

I'm not sure of the point you're trying to make with the space-time diagrams. You need to be somewhat careful in interpreting them. To construct such a diagram it's necessary to arbitrarily choose two Minkowski-orthogonal vectors - usually the time and one spatial basis vector of some chosen frame - and represent them as perpendicular arrows in the plane. Other pairs of vectors which are also Minkowski-orthogonal are then not necessary perpendicular in the plane, although they are constrained to lie symmetrically about one of the null-lines (at $\pi/4$) through the origin.

The length and direction of the separation vector between any two events, as represented in the plane, thus depends on the aforementioned choice of space-time diagram. That the projections of the separation vector onto the axes of two frames differ should not be a surprise if you've come across the Lorentz transformations, because this just reflects the fact that different inertial observers foliate spacetime into different families of spatial slices.

There are some subtleties to the subject of the reciprocity of the relative velocity. It does not hold true in special relativity! But we ought to put discussion of that on hold until you have a firmer grasp of the underlying machinery.

 

[Kairos]

There are some subtleties to the subject of the reciprocity of the relative velocity. It does not generally hold true in special relativity! But we ought to put discussion of that on hold until you have a firmer grasp of the underlying machinery.

indeed, if a relative velocity is not reciprocal, I don't understand anything because it is assumed constant in the Lorentz factor..
thanks anyway

 

[etotheipi]

That's actually not a problem. If the relative velocity of $O_1$ with respect to $O_2$ is $\mathbf{v}_{12}$, and the relative velocity of $O_2$ with respect to $O_1$ is $\mathbf{v}_{21}$, then in fact these still satisfy $\mathbf{v}_{12} \cdot \mathbf{v}_{12} = \mathbf{v}_{21} \cdot \mathbf{v}_{21}$, or otherwise expressed $||\mathbf{v}_{12}|| = ||\mathbf{v}_{21}||$. Thus the Lorentz factor still satisfies$$\gamma = - (u_1)_a (u_2)^a = \frac{1}{\sqrt{1- \mathbf{v}_{12} \cdot \mathbf{v}_{12}}} = \frac{1}{\sqrt{1- \mathbf{v}_{21} \cdot \mathbf{v}_{21}}}$$Except, in special relativity, we no longer generally have $\mathbf{v}_{12} = - \mathbf{v}_{21}$, rather we have$$\mathbf{v}_{12} = - \gamma \left( \mathbf{v}_{21} + [\mathbf{v}_{21} \cdot \mathbf{v}_{21}] \mathbf{u}_1 \right)$$although you can check that you recover the Galilean expression in the non-relativistic limit! It reflects that $\mathbf{v}_{12}$ is proportional to the orthogonal projection of $\mathbf{v}_{21}$ onto $O_2$'s local rest space.

 

[PeroK]

indeed, if a relative velocity is not reciprocal, I don't understand anything because it is assumed constant in the Lorentz factor..
thanks anyway

This is something that isn't often explained. If object $A$ is moving with velocity $\vec v$ relative to $B$ (i.e. as measured in $B$'s rest frame), then object $B$ is moving with velocity $-\vec v$ relative to $A$ (i.e. as measured in $A$'s rest frame).

The basis of the argument is the isotropy of spactime. The magnitude of one velocity cannot be larger than the other, because there's nothing to distinguish them and determine which one should be larger. And taking the direction of relative motion as a common x-axis, say, takes care of the opposite sign of the velocity.

 

[Kairos]

The magnitude of one velocity cannot be larger than the other, because there's nothing to distinguish them and determine which one should be larger.

that's what I meant. I am happy to read you!

 

[robphy]

Of course, there is the analogous problem in Galilean relativity and in Euclidean geometry.

In Euclidean geometry, the issue is handled by referring to an angle between rays from a common point, which corresponds to an oriented arc of the unit circle that is cut by those rays. Then for slopes with respect to one of the rays, we take the tangent of the angle.

So, in Minkowski spacetime geometry, one has the rapidity that is a oriented arc of the unit hyperbola.
Then for velocities with respect to one of the inertial worldlines, we take the hyperbolic-tangent of the rapidity.

As @etotheipi says, the velocity vectors are in different spaces.
That is to say, even though we colloquially say that they their relative velocities are in the "same direction" (e.g., relative motion in the x-direction), these spatial-vectors in spacetime [each orthogonal to their corresponding worldline] are not parallel.
Rather, these spacelike vectors are coplanar with the two worldlines.

 

[Kairos]

I am not sure I understand. For any two objects in uniform motion in any direction relative to each other, I assumed that in SR there is always only one possible magnitude of relative velocity and opposite vector signs arbitrarily chosen. Should I understand that it is not true?

 

[PeroK]

I am not sure I understand. For any two objects in uniform motion in any direction relative to each other, I assumed that there is always only one possible magnitude of relative velocity and opposite vector signs arbitrarily chosen. Should I understand that it is not true?

In order to get started with the study of SR, then yes that's true. The motion of one object relative to the other can be taken to define a common x-axis.

Technically, IMO, it is better to think in terms of reference frames than isolated observers.

 

[etotheipi]

As said before, it's not true in special relativity that $\mathbf{v}_{12} = - \mathbf{v}_{21}$, however it is true that $||\mathbf{v}_{12}|| = ||\mathbf{v}_{21}||$. The more intuitive way of re-stating the true relationship is $\mathbf{v}_{12} = - \gamma^{-1} \bot_{\mathbf{u}_2} \mathbf{v}_{21}$, where $\bot_{\mathbf{u}_2}$ is the orthogonal projector onto $O_2$'s local rest space.

In fact, even this equation only holds true where $O_1$ and $O_2$'s worldlines coincide. But hopefully you can see the motivation; the 4-vector $\mathbf{v}_{21}$ (canonically identified with the 3-vector $\vec{v}_{21}$ in $O_1$'s local reference space) is orthogonal to $O_1$'s worldline, whilst $\mathbf{v}_{12}$ is orthogonal to $O_2$'s worldline. Thus they are not parallel!

 

[PeroK]

As said before, it's not true in special relativity that $\mathbf{v}_{12} = - \mathbf{v}_{21}$, however it is true that $||\mathbf{v}_{12}|| = ||\mathbf{v}_{21}||$.

I don't think that's necessarily relevant to the OP's question. There's nothing inherently unique about the convention needed to support a specific algebraic form for 3D Lorentz Transformations.

To get started for uniform motion of two objects or reference frames, we can choose our axes so that $\vec v_{12} = - \vec v_{21}$

If the wordlines do not intersect, I would simply move one observer away from the origin of their reference frame.

Also, it's not necessary to consider the tangent spaces as separate vector spaces. That mathematical sophistication allows you to generalise to curved spacetime, but it's not necessary to go beyond the notion of a single vector space in flat spacetime.

 

[etotheipi]

To get started for uniform motion of two objects or reference frames, we can choose our axes so that $\vec v_{12} = - \vec v_{21}$

But this you cannot do. Vectors are geometric objects independent of any co-ordinate system, and cannot be made parallel by a mere choice of axes.

Given the three spatial basis vectors $(\vec{e}_i)$ of an observer's local reference space, the velocity of a particle relative to this observer is defined $\vec{v} = d\vec{x}/dt$ where $\vec{x} = \overrightarrow{O(t)M(t)}$ is the 3-vector joining the observer to the particle in his local reference space, which is put naturally into correspondence with a 4-vector $\mathbf{v}$ by the rule $v^i \vec{e}_i \mapsto v^i \mathbf{e}_i$.

 

[PeroK]

But this you cannot do. Vectors are geometric objects independent of any co-ordinate system, and cannot be made parallel by a mere choice of axes.

I'm defining a three-vector as a directed line segment.

 

[Vanadium 50]

This is a B-level thread. Is the OP's difficulty that things just aren't complicated enough?

 

[etotheipi]

I don't know how that clarifies it. With the observers as defined in post #11, let's say we consider $\mathbf{v}_{21} = \lambda \mathbf{e}$ where $\mathbf{e}$ is a unit vector parallel to $\mathbf{v}_{21}$. Then, $\mathbf{v}_{12} = -\lambda \mathbf{e}'$, where$$\mathbf{e}' = \gamma^{-1} \bot_{\mathbf{u}_2} \mathbf{e}$$Of course you can define two co-ordinate charts with basis vectors $(\mathbf{e}_{\mu})$ and $(\tilde{\mathbf{e}}_{\mu})$ such that $\mathbf{e}_x = \mathbf{e}$ and $\tilde{\mathbf{e}}_x = \mathbf{e}'$, and this situation would then correspond to "relative motion along the $x$-axis", but it doesn't mean that $\mathbf{v}_{12}$ is parallel to $\mathbf{v}_{21}$!

 

[Ibix]

Are your $\mathbf{v}$ supposed to be four vectors or three vectors?

 

[etotheipi]

Are your $\mathbf{v}$ supposed to be four vectors or three vectors?

Anything bold is a 4-vector

 

[Ibix]

Anything bold is a 4-vector

I think everyone else is talking about the three velocity, which pre-supposes certain things, like an inertial frame (edit: or a defined notion of "space" anyway).

 

[Kairos]

As said before, it's not true in special relativity that $\mathbf{v}_{12} = - \mathbf{v}_{21}$, however it is true that $||\mathbf{v}_{12}|| = ||\mathbf{v}_{21}||$. The more intuitive way of re-stating the true relationship is $\mathbf{v}_{12} = - \gamma^{-1} \bot_{\mathbf{u}_2} \mathbf{v}_{21}$, where $\bot_{\mathbf{u}_2}$ is the orthogonal projector onto $O_2$'s local rest space.

In fact, even this equation only holds true where $O_1$ and $O_2$'s worldlines coincide. But hopefully you can see the motivation; the 4-vector $\mathbf{v}_{21}$ (canonically identified with the 3-vector $\vec{v}_{21}$ in $O_1$'s local reference space) is orthogonal to $O_1$'s worldline, whilst $\mathbf{v}_{12}$ is orthogonal to $O_2$'s worldline. Thus they are not parallel!

Since we can always define a unique axis of reciprocal uniform motion in SR, I don't see how it cannot be parallel to itself. But I am not skilled enough to judge your argument, so I trust you.

 

[etotheipi]

I think everyone else is talking about the three velocity, which pre-supposes certain things, like an inertial frame.

Sure, but they are in correspondence as per the end of #19, so there's no loss in generality. It's just easier to write formulae with their 4-vector equivalents.

But it doesn't matter whether they are inertial observers, it still holds that their relative velocities defined in the standard manner are not parallel. Of course given suitable charts for each observer, e.g. as in #22, you can map these vectors to their respective co-ordinate spaces in such a way that the images are oppositely oriented co-ordinate vectors.

 

[etotheipi]

Okay, here's the proof, which is relatively short. Consider two observers, of 4-velocities $u_1^a$ and $u_2^a$. At the intersection of their world-lines, or if they are inertial observers, then the relative velocity of $O_2$ with respect to $O_1$, $(v_{21})^a = \frac{\mathrm{d}x^i}{\mathrm{d}t} (e_i)^a$, may be obtained via the orthogonal decomposition$$u_2^a = \gamma(u_1^a + (v_{21})^a)$$Thus, we may write two equations,\begin{align*}

u_2^a &= \gamma(u_1^a + (v_{21})^a) \quad \quad \quad(1) \\
u_1^a &= \gamma(u_2^a + (v_{12})^a) \quad \quad \quad (2)

\end{align*}Let's substitute $(2)$ into $(1)$, i.e.\begin{align*}
u_2^a &= \gamma[\gamma(u_2^a + (v_{12})^a) + (v_{21})^a] \\

\gamma^{-1} u_2^a - \gamma(u_2^a + (v_{12})^a) &= (v_{21})^a \\

-\gamma (-\gamma^{-2} u_2^a + u_2^a + (v_{12})^a) &= (v_{21})^a \\

-\gamma ((v_{12})^a + u_2^a(1 -\gamma^{-2})) &= (v_{21})^a \\

-\gamma ((v_{12})^a + (v_{12})_b (v_{12})^b u_2^a ) &= (v_{21})^a

\end{align*}This proves the result!

 

[Ibix]

But it doesn't matter whether they are inertial observers, it still holds that their relative velocities defined in the standard manner are not parallel. Of course given suitable charts for each observer, e.g. as in #22, you can map these vectors to their respective co-ordinate spaces in such a way that they appear to be oppositely oriented co-ordinate vectors.

So the point you are making is that the two inertial frames have different notions of "space", which are non-parallel 3d planes in 4d spacetime. Since my measurement of your three-velocity lies in my spatial plane, and your measurement of my three-velocity lies in your spatial plane, those vectors cannot be parallel.

For simplicity's sake, perhaps we could just observe that if I define my +x direction to be the direction you are going, and you define your -x direction to be the direction that I am going then: our y and z directions coincide (or can be made to coincide); your x and t directions lie in the plane defined by my x' and t' directions and vice versa; we measure equal three velocity magnitudes for each other; we regard those velocities to have opposite signs.

 

[etotheipi]

Yeah, absolutely. I tried to draw it, of course suppressing one dimension (beware, it's a slightly awful drawing...)

The "flat" plane is the rest space of $O_1$, and the "tilted" plane is the rest space of $O_2$. The "dotted, vertical plane" is the plane of the Lorentz boost, in which the $\mathbf{e}_0$ and $\mathbf{e}_1$ vectors have been rotated into the $\tilde{\mathbf{e}}_0$ and $\tilde{\mathbf{e}}_1$ vectors.

 

[robphy]

Geometrically speaking, reduce the complexities of a (3+1)d-Spacetime by working in the plane made by the two worldlines and define their x-axes to lie on that plane. (This is the Spacetime interpretation of “let’s assume that the observers have a relative velocity along their x-axes”.)

So, we can draw the usual (1+1)d-spacetime diagrams so that each observer can find the spatial and time components of a displacement along the other worldline to define that worldline’s spatial velocity in the observer’s frame.

Yes, this can be formalized by writing projections operations (which is what we mean when we say “find the components” in some frame).

 

[Ibix]

So the point you are making is that the two inertial frames have different notions of "space", which are non-parallel 3d planes in 4d spacetime. Since my measurement of your three-velocity lies in my spatial plane, and your measurement of my three-velocity lies in your spatial plane, those vectors cannot be parallel.

For simplicity's sake, perhaps we could just observe that if I define my +x direction to be the direction you are going, and you define your -x direction to be the direction that I am going then: our y and z directions coincide (or can be made to coincide); your x and t directions lie in the plane defined by my x' and t' directions and vice versa; we measure equal three velocity magnitudes for each other; we regard those velocities to have opposite signs.

Let's expand on this a bit.

If I aim a laser pointer in the direction that I say you are travelling, and you aim one in the direction you say that I am travelling, every frame will agree that the three-velocities of the two beams are anti-parallel (actually this is restricted to frames moving in the $\pm x$ direction, although similar statements can be constructed in any other frame). This is the sense that people mean when they say that the two velocities are opposite. We are thinking of something like a pair of trains traveling away from each other along a straight track.

However, as @etotheipi and @robphy will agree, the three-velocities of those laser beams are not the three-velocities of the two observers. If we state the (correct) usual claim formally, it says that if I define your three-velocity to be $(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt})=(v,0,0)$ then you will define my three velocity to be $(\frac{dx'}{dt'},\frac{dy'}{dt'},\frac{dz'}{dt'})=(-v,0,0)$. This is correct. However, despite the components being the same with opposite signs, it does not mean that the velocities are opposite because we are not using the same coordinate system. We need to find a way to express your measure of my three-velocity in my coordinate system if we want to compare them.

The way to do that is to note that a spacelike three-vector is three components of a four-vector with the fourth component being zero (this is the same as a two-vector $(x,y)$ in a plane being expressible as a three-vector $(x,y,0)$). My measure of your three-velocity can be written in my un-primed coordinates as a four-vector $(0,v,0,0)$ (this is a purely spacelike vector lying in my "space" plane, not your four-velocity). Your measure of my three-velocity can be written in your primed coordinates as the four-vector $(0,-v,0,0)$. Now I should apply the inverse Lorentz transforms to your measure, which maps $(0,-v,0,0)$ to $(-\gamma v^2/c,-\gamma v,0,0)$. This latter form is in my coordinate system, and is clearly not parallel to $(0,-v,0,0)$.

So, to summarise, there are good senses in which you and I can be said to be traveling in opposite directions. It is also true that if I say that you have a speed of $v$ in the $+x$ direction then you will say I have a speed of $v$ in the $-x$ direction. However, since we don't share a common definition of "space", we cannot correctly say that our three-velocities (even extending them to four vectors) are anti-parallel. We can only say that they lie in the plane defined by our two four-velocities.

 

[etotheipi]

That was nicely explained; it's not such an easy thing to get used to. We can imagine again a train traveling along some straight tracks with constant speed, and consider the usual platform-fixed $O_p$ and train-fixed $O_t$ observers, who have established co-ordinate charts relative to which the other observer is moving parallel to their respective $x$-axes.

The restricted Lorentz transformation that maps $O_p$'s basis $(\mathbf{e}_{\mu})$ to $O_t$'s basis $(\tilde{\mathbf{e}}_{\mu})$ leaves the plane $\pi = \mathrm{span} (\mathbf{e}_2, \mathbf{e}_3)$ invariant, whilst its orthogonal complement $\pi^{\bot} = \mathrm{span}(\mathbf{e}_0, \mathbf{e}_1)$ is the time-like plane of the Lorentz boost within which $\mathbf{e}_0$ and $\mathbf{e}_1$ have been rotated into $\tilde{\mathbf{e}}_0 = \Lambda(\mathbf{e}_0)$ and $\tilde{\mathbf{e}}_1 = \Lambda(\mathbf{e}_1)$ respectively.

Although our Galilean thinking might initially lead us to think that $\mathbf{e}_1 = \tilde{\mathbf{e}}_1$, that's actually not the case precisely because in space-time these vectors are rotated with respect to each-other. Thus $v \mathbf{e}_1$ is not anti-parallel to $-v\tilde{\mathbf{e}}_1$ in space-time.

Luckily I don't think there's too much potential for confusion, because it's pretty obvious what someone means when they say the relative velocities are opposite, i.e. that $\mathrm{d}x/\mathrm{d}t = - \mathrm{d}x' / \mathrm{d}t'$. But I think it's rather nice to think about Lorentz boosts in terms of space-time rotations and planes.

 

[robphy]

It's important to realize that this issue about working in the plane of these worldlines
and these non-parallel spatial axes
is NOT just about Minkowski spacetime.
This is what we do in Euclidean space!
(Look at the x and x' axes in Euclidean space.)

From this unified [Cayley-Klein geometry] point of view (admittedly viewed in historical hindsight),
one should realize that
it is the Galilean case that is the special-case, the exception, the peculiarity!
But our everyday low-relative-speed "common sense" has
misled us to believe in the Galilean structure
...and makes it hard for us
to accept the Einstein-relativity/Minkowski-spacetime viewpoint [which agrees with experiment].

 

[vanhees71]

In SRT, the "relative velocity" usually denotes a "three-velocity" like $\vec{v}=\mathrm{d} \vec{x}/\mathrm{d} t$, which is a pretty complicated object, because it's not covariant. Particularly it is NOT the spatial part of a four-vector wrt. some Minkowski-orthonormal basis.

Suppose you have two particles. Then the relative velocity is the three-velocity of particle 2 wrt. the rest frame of particle 1. The problem is to get this relative velocity when the three-velocities $\vec{v}_j=\vec{\beta}_j c$ ($j \in \{1,2\}$) of the two particles are given wrt. some arbitrary inertial reference frame.

It's of course much more convenient to answer this question, using covariant objects, i.e., scalars, vectors, and tensors. Here we just use the (normalized) four-velocities, which are defined as
$$u_1=\gamma_1 \begin{pmatrix} 1 \\ \vec{\beta}_j \end{pmatrix}, \quad j \in \{1,2 \}.$$
To find the relative velocity we just need to boost into the reference frame, where particle 1 is at rest. The corresponding Lorentz transformation is of course with the boost-velocity $\vec{v}_1$, i.e.,
$$\hat{\Lambda}=\begin{pmatrix} \gamma_1 & -\gamma_1\vec{\beta}_1^{T} \\ -\gamma_1 \vec{\beta}_1 & I+(\gamma_1-1) \frac{\vec{\beta}_1 \otimes \vec{\beta}_1}{\beta_1^2} \end{pmatrix}.$$
It's easy to check that then indeed $u_1'=\hat{\Lambda} u_1=(1,0,0,0)^T$. So we really boost into the rest frame of particle 1. Then the relative four-velocity is simply given by $u_2'=\hat{\Lambda} u_2$. After some algebra (nothing really complicated but a bit of somewhat tedious algebraic work ;-)) you find for the relative three-velocity
$$\vec{\beta}_{\text{rel}2}=\frac{\vec{u}_2'}{u_2^{\prime 0}}=\frac{1}{1-\vec{\beta}_1 \cdot \vec{\beta}_2} \left [ \vec{\beta}_2-\vec{\beta}_1 + \frac{\gamma_1}{\gamma_1+1} \vec{\beta_1} \times (\vec{\beta_1} \times \vec{\beta}_2) \right].$$
Note that the corresponding relative velocity of particle 1 is NOT the negative of this, except when $\vec{\beta}_1 \parallel \vec{\beta}_2$.