By Don Koks, 2022.

# How do you add velocities—relativistically and otherwise?

As any missile designer will tell you, adding velocities is generally more subtle than most people think.  The reason is that the idea of adding velocities is actually very rich, and contains details that must be understood fully.

Here's an example.  Suppose we have three point-sized objects, $A, B, C$, each of which might be moving relative to the others.  The usual velocity addition that we are all raised on says "The velocity (vector) of $C$ relative to $A$ equals the velocity of $C$ relative to $B$ plus the velocity of $B$ relative to $A$".  We can write this using bold-faced vector notation as $$\label{simple-addition} \boldsymbol{v}_{CA} = \boldsymbol{v}_{CB} + \boldsymbol{v}_{BA}\,.$$ This is sufficient for school exercises, which always use point-sized objects; but out there in the big wide world of non point-sized objects, things are not so simple.  First, whenever we specify a velocity, we must indicate who measured it.  That is the meaning of the word "frame": we must indicate the relevant frame.  Note that frames are not coordinates; rather, the word "frame" means "the observer who measures the velocity".  In simple examples or exercises, we usually measure all the velocities in a single frame.  We'll indicate this frame by a superscript, and so write equation \eqref{simple-addition} more properly as $$\label{addition-with-frame-specified} \boldsymbol{v}^F_{CA} = \boldsymbol{v}^F_{CB} + \boldsymbol{v}^F_{BA}\,,$$ where $F$ is the frame in which the velocity has been specified.  It could be the frame of any of $A$, $B$, or $C$, or it might be some other frame.  Equation \eqref{addition-with-frame-specified} is what we are really applying when we depict velocity vectors as arrows, adding them tip to tail.  It is the "per unit time" version of adding infinitesimal displacements in tip-to-tail fashion.  That idea is fundamental to how we portray space, and it can always be done in a given frame.  So \eqref{addition-with-frame-specified} is always true—even in relativity!  (We'll get to that later.)  But it is not the answer to the questions that are most often asked about adding velocities in relativity.

To see how things get more complicated even outside relativity, suppose that $A$ and $C$ are at rest relative to each other, and suppose that $B$ is no long point sized; instead, $B$ is a spinning satellite.  We know, say, the velocity of $B$ (i.e., of its centre of mass) relative to $A$, as measured by $A$—"in the frame of $A$": $\boldsymbol{v}^A_{BA}$.  Typically, we might also know the velocity of $C$ relative to $B$ as measured by $B$ (not as measured by $A$): $\boldsymbol{v}^B_{CB}$.  We want to know the velocity of $C$ relative to $A$ as measured by $A$, or $\boldsymbol{v}^A_{CA}$: $$\label{velocities-known-and-unknown} \boldsymbol{v}^A_{BA} \text{ and } \boldsymbol{v}^B_{CB} \text{ are known. What is } \boldsymbol{v}^A_{CA}\text{ ?}$$ Now we don't have enough information to write \eqref{addition-with-frame-specified}, because the superscripts of the velocities in \eqref{velocities-known-and-unknown} are not all the same: we cannot just add $\boldsymbol{v}^B_{CB}$ to $\boldsymbol{v}^A_{BA}$.

To see that more fully, suppose that $A$, the centre of mass of $B$, and $C$ are all at rest relative to each other.  In that case, $\boldsymbol{v}^A_{BA} = \boldsymbol{v}^A_{CA} = \boldsymbol{0}$.  But $B$ is spinning, and so it measures $C$ to be moving in a circle relative to $B$; hence $\boldsymbol{v}^B_{CB}$ is a non-zero function of time.  For this situation, it's clear that $$\label{spinning-satellite} \boldsymbol{v}^A_{CA} \ne \boldsymbol{v}^B_{CB} + \boldsymbol{v}^A_{BA}\,.$$ We might have expected that, since unlike the situation in \eqref{addition-with-frame-specified}, the superscripts in \eqref{spinning-satellite} are not all the same.  But we can certainly say $$\label{spinning-satellite-1} \boldsymbol{v}^A_{CA} = \boldsymbol{v}^A_{CB} + \boldsymbol{v}^A_{BA}\,.$$ Typically then, we will use the known $\boldsymbol{v}^B_{CB}$ to calculate the unknown $\boldsymbol{v}^A_{CB}$.  Outside the subject of relativity, if $B$ is not changing its orientation in $A$'s frame, then things are easy, because then for any points $P, Q$, we have $$\label{non-spinning-satellite} \boldsymbol{v}^A_{PQ} = \boldsymbol{v}^B_{PQ}\,.$$ But even outside relativity, the orientation of $B$ is often changing in the frame of $A$; and then $\boldsymbol{v}^A_{PQ} \ne \boldsymbol{v}^B_{PQ}$ and we have some work to do.  Such a frame change is an advanced exercise that requires knowledge of the time-dependent orientation of the satellite $B$ relative to $A$, and you can find it done almost exclusively in engineering books on vehicle motion.  (Strangely, few physicists and even fewer mathematicians are ever introduced to this subject.)  Be aware that such analyses can be difficult to understand in those books, for at least three reasons.  The first reason is that engineers are usually calculating with computers; but computers can only work with numbers, and so engineers must choose a coordinate system to quantify the vectors.  Those coordinates needn't be related to the frame.  Yes, expressing a velocity in coordinates needs four pieces of information to be included in the notation of the velocity vector: the identity of point "1" whose velocity is required relative to point "2", in frame "3", expressed using coordinates "4"!  So, you must be adept at changing coordinates.  (Which just goes to show: never use the widespread expression "coordinate frame", because coordinates and frames are different things, whose distinct meanings must be appreciated to understand this subject.)  The second reason that these books can be hard to read is that most use all manner of complicated notation to encapsulate those four pieces of information, and some are more successful at doing that than others.  And the third reason is that the authors of those books were also raised on the simplistic equation \eqref{simple-addition}, and so are sometimes just a little bid muddled themselves.

## The Relativistic Case

We will only treat the special-relativistic case, and only use inertial frames.  The only other well-defined frame in special relativity is the uniformly accelerated frame, which involves a level of difficulty that is outside the aim of this FAQ.  Also, in general relativity, the notion of velocity becomes somewhat ill defined.

In special relativity, we tend not to focus on equation \eqref{addition-with-frame-specified} because we seldom have all of its terms.  Instead, typically we must answer the question in \eqref{velocities-known-and-unknown}.  Forget about rotating satellites: rotation in relativity has always been a very difficult subject.  Here we'll just treat inertial observers $A$ and $B$ as points, and ask how the three velocities in \eqref{velocities-known-and-unknown} are related.  In fact, let's even forget about the full three spatial dimensions, since the point of this FAQ is to explain where the standard expression for "velocity addition" in introductory relativity texts comes from—and that expression usually uses a single space dimension and fixed velocities.  So we'll be content to do likewise, and will drop the bold font for the velocities.

In that case, picture observer $A$ who says "$B$ is moving relative to me with velocity $v^A_{BA}$ equal to some $u$".  Likewise, $B$ says "$C$ is moving relative to me (along the same single direction, of course) with velocity $v^B_{CB}$ equal to some $v$".  Observer $A$ now asks "What will I measure to be the velocity of $C$ relative to me, $v^A_{CA}$?".  This is called "adding velocities in relativity", but you can see that we're not adding velocities in a single frame [like \eqref{addition-with-frame-specified}] at all; instead we are addressing \eqref{velocities-known-and-unknown}.  We know that outside relativity, for the spinning-$B$ case, \eqref{spinning-satellite} says that we cannot simply add such velocities.  Within relativity, even though nothing is spinning here, it turns out that we still cannot turn \eqref{spinning-satellite} into an equality: we just cannot add $v^B_{CB}$ and $v^A_{BA}$ to get $v^A_{CA}$.  The reason is that in relativity, because different frames measure time and space differently, we cannot generally replace $\boldsymbol{v}^B_{CB}$ with $\boldsymbol{v}^A_{CB}$ by saying that they are equal like we did above in \eqref{non-spinning-satellite} for the non-relativistic non-spinning case.

Instead, in relativity we give observer $A$ one set of coordinates (say cartesian $t,x$) and observer $B$ another set of coordinates (say cartesian $t',x'$).  These are not just any old cartesian coordinates; they are special in that $A$ gives a single value of $t$ to all events that $A$ says are simultaneous, and likewise $B$ gives a single value of $t'$ to all events that $B$ says are simultaneous.  If $A$ and $B$ are moving relative to each other, then they have different standards of simultaneity: their frames really are distinct.  Switching frames is now a matter of switching coordinates; and for that, we rely on the Lorentz transform, which you can find derived in all introductory textbooks on relativity.  When $B$ moves relative to $A$ with a fixed velocity $u$ in $A$'s $x$ direction, the Lorentz transform says $$t' = \gamma(t - ux/c^2) + \text{constant},\quad x' = \gamma(x - ut) + \text{constant},$$ where $c$ is the speed of light in an inertial frame, and $\gamma \equiv 1/\sqrt{1-u^2/c^2}$.  Equivalently, the inverse of this is $$\label{lorentz-transform} t = \gamma(t' + ux'/c^2) + \text{constant},\quad x = \gamma(x' + ut') + \text{constant}.$$

Now consider that $A$ (who uses coordinates $t,x$) says that $C$ undergoes a displacement $\mathrm{d}x$ in a time $\mathrm{d}t$, whereas $B$ (who uses primed coordinates) says that $C$ undergoes a displacement $\mathrm{d}x'$ in a time $\mathrm{d}t'$.  It follows that $$v^A_{CA} = \mathrm{d}x/\mathrm{d}t\,,\quad v^B_{CB} = \mathrm{d}x'/\mathrm{d}t' = v\,,\quad v^A_{BA} = u\,.$$ We require $\mathrm{d}x/\mathrm{d}t$.  To get at that, consider writing the infinitesimal version of \eqref{lorentz-transform}: $$\label{lorentz-infinitesimal} \mathrm{d}t = \gamma(\mathrm{d}t' + u\,\mathrm{d}x'/c^2)\,,\quad \mathrm{d}x = \gamma(\mathrm{d}x' + u\,\mathrm{d}t')\,.$$ Dividing the second expression in \eqref{lorentz-infinitesimal} by the first gives $$\label{combining-velocities} {\mathrm{d}x\over\mathrm{d}t} = {\mathrm{d}x' + u\,\mathrm{d}t'\over \mathrm{d}t' + u\,\mathrm{d}x'/c^2} = {\mathrm{d}x'/\mathrm{d}t' + u\over 1 + u(\mathrm{d}x'/\mathrm{d}t')/c^2} = {v + u\over 1 + uv/c^2}\,.$$ Let's put \eqref{combining-velocities} into words in a sort of transitive construction that acts as a prompt to getting the solution:

• $A$ says "$B$ has velocity $u$ relative to me".
• $B$ says "$C$ has velocity $v$ relative to me".
• Then $A$ says "$C$ has velocity ${u\,+\,v\over 1\,+\,uv/c^2}$ relative to me".
Naïvely, we might've thought that the combined velocity would be $u+v$, but that isn't quite right.  It is almost correct when $u$ and $v$ are much less than $c$.  For example, when $A$ says that $B$ is moving at velocity $u=0.01c$ (relative to $A$), and $B$ says that $C$ is moving at velocity $v=0.02c$ (relative to $B$), then $A$ says that $C$ is moving with a velocity (relative to $A$) of $${u + v\over 1 + uv/c^2} = {0.01c + 0.02c\over 1.0002} \simeq 0.029994c \simeq 0.03c\,.$$ But when $u = v = 0.9c$, their combined velocity is $${u + v\over 1 + uv/c^2} = {0.9c + 0.9c\over 1.81} \simeq 0.99448c\,.$$ You can play around with the numbers to ascertain that the speed $|(u + v)/(1 + uv/c^2)|$ can never be greater than $c$.

On another note, let's return to \eqref{addition-with-frame-specified}, which, remember, is also true in relativity.  Set frame $F$ to $A$ and rearrange to give $$\boldsymbol{v}^A_{CB} = \boldsymbol{v}^A_{CA} - \boldsymbol{v}^A_{BA}\,.$$ In one dimension (thus dropping the bold font), suppose that $A$ (an inertial observer) says that $B$ is a particle moving to the right of $A$ (the positive $x$ direction) with speed 60% that of light: $v^A_{BA} = 0.6c$.  Also, $A$ says that $C$ is a particle moving to the left of $A$ with the same speed: $v^A_{CA} = -0.6c$.  Then $A$ says that $C$ moves relative to $B$ with velocity $$\label{speed-is-1.2c} v^A_{CB} = v^A_{CA} - v^A_{BA} = -0.6c - 0.6c = -1.2c\,.$$ Einstein would have had no problem with this speed being greater than $c$, because \eqref{speed-is-1.2c} does not say that either particle claims that the other is moving away with speed $1.2c$.  Einstein said that nothing can move with speed greater than $c$ relative to us in our frame when we are inertial; but that's a different scenario to the above two-particle case.  There, $A$ is not saying that anything is moving relative to $A$ at a speed of $1.2c$; remember, $A$ has already said that $B$ and $C$ have speeds of $0.6c$ relative to $A$.  Rather, $A$ is saying that one particle is moving relative to the other (not $A$) at a speed of $1.2c$.  And that's okay; Einstein would have no problem with it.  This idea trips many people up.  If you always ascertain who is doing the measuring (i.e., which frame is which, or "what are the superscripts?"), you will always get things right.

And out of interest, just how fast does each particle claim that the other is moving?  Describe the scenario as follows:

• $C$ says "$A$ has velocity $0.6c$ relative to me".
• $A$ says "$B$ has velocity $0.6c$ relative to me".
• Then $C$ says "$B$ has velocity $(0.6c+0.6c)/1.36 \simeq 0.88c$ relative to me".
Now swap the roles of $B$ and $C$:
• $B$ says "$A$ has velocity $-0.6c$ relative to me".
• $A$ says "$C$ has velocity $-0.6c$ relative to me".
• Then $B$ says "$C$ has velocity $(-0.6c - 0.6c)/1.36 \simeq -0.88c$ relative to me".
As expected, these speeds of $0.88c$ are less than $c$, as required by Einstein.