The Velocity Addition Formula

Velocity Addition Formula

Updated: April 11, 2026

On the Electrodynamics of Moving Bodies

Einstein-EDoMB-1905-Section §5

https://www.fourmilab.ch/etexts/einstein/specrel/www/

Einstein: Relativity, The Special and the General Theory, 1920, Chapter VI page 19, Chapter XIII page 45

https://www.google.com/books/edition/Relativity/n8QKAAAAIAAJ?hl=en&gbpv=1

Both sources show the velocity-addition-formula as it is today.

Shown: V = ( v + w ) / ( 1 + vw/c² )

#### Lorentz Transforms

Lorentz Transforms – subtraction, shown in EdoMB-1905 Section §3 and used in Section §4 ( length contraction and time dilation. ) Shown in Halliday & Resnick as equations 37-2I:

x′ = gamma • ( x – vt )

t′ = gamma • ( t – vx / c² )

Lorentz Transforms – addition, shown in H&R as equations 37-22.  Shown neither in Section §5 ( v-a-f ) nor in Chapter XIII ( v-a-f ):

x = gamma • ( x′ + vt′ )

t = gamma • ( t′ + vx′ / c² )

#### Transition from subtraction to addition

Halliday & Resnick, 10th Edition, Chapter 37-2. 

H&R has the two LT sets shown here: subtraction with unprimed on the right-hand-side and addition with primed on the right-hand-side. Given the subtractive set, how do you get the additive set?  Quote: “Simply solve the subtractives for x and t.”  The implied algebra is not shown.  Galilean spatials without gamma:
 
Subtract:  x′ ← x – vt

Two pavement mileposts are on the right-hand-side.  Though not a milepost, left-hand-side output is a pavment segment, a pavement reality.  Re-arrange and add:

Add: x ← x′ + vt

On the right-hand-side, milepost-vt and segment-x′ are pavement realities.  Left-hand-side output-x is the right-hand-side input milepost-x of subtractive mode.  In Galilean mode, unprimed milepost-vt and segment-x′ can be added to place composite milepost-x on the pavement. The equation is correct.

When you bring in gamma, you cannot add primed-x′ and unprimed-vt.  “Simply by solving for x and t,” you can find an equation which does not have mixed units on the right-hand-side, and you can add in gamma mode.  We won’t show the gamma mode derivation.  It is long and complicated. See comment.  Result:

Add: x ← x′ + vt′

The right-hand-side is addition of two primed mileposts on axis-X′. Doable. It is not addition of mixed units, and yields the needed algebraic value of milepost-x.  But even in simple Galilean mode the addition cannot be the reality of a third milepost.  You cannot add two mileposts and come up with a third milepost.  You can only add an extension to one milepost and come up with a second milepost.  The equation is a falsehood.

#### Textbook derivation of the v-a-f

Halliday & Resnick, 10th Edition, Chapter 37-4.

Let x′ be of speed-w:  x′ = wt′

Spatial:  x = gamma • ( wt′ + vt′ )

Temporal:  t = gamma • ( t′ + vwt′ / c² )

Speed is distance divided by time:

V = x / t = gamma • ( w + v ) • t′ / gamma • ( 1 + vw/c² ) • t′

gamma divides out:

V = x / t = ( w + v ) • t′ / ( 1 + vw/c² ) • t′

Time-t′ divides out:

V = ( v + w ) / ( 1 + vw/c² )

EDoMB-1905 Section §5 and Relativity-1920 Chapter XIII invoke the Lorentz Transforms but neither provide an LT derivation of the v-a-f.   In particular, Section §5 begins with motion and transitions to distance. Reversal.  Everything in SRT is reversal if not dyslexia.

Motion  ←  ( w + v )
↓
Distance  ←  ( w + v ) • t′

#### Abbreviations and notation

SRT = spezielle Relativitätstheorie, Special Relativity

CLS = classical mechanics, a term from Chapter XIII.

rf = reduction factor, denominator ( 1 + vw/c² ) > 1

Let’s change a notation without changing anything on the ground.  Bring in an asterisk for the v-a-f output:

CLS sum:  V  ← ( v + w )

SRT sum:  V*  ← ( v + w ) / rf

#### Chapter VI

Note: As per modern notation and Section §5 we use symbol-V for output velocity. Chapters VI and XIII use symbol-W.

A carriage “travels” an embankment at speed-v.   A man “traverses” the floor of the carriage at speed-w.  That same man has velocity-V on the embankment.

New vocabulary:  The carriage is still the “carriage.”  The embankment is the “pavement.”  The man of speed-w on the carriage is “walker-w” of speed-w.  

Chapter VI stipulation:  Time is 1.  Let’s have miles, hours and miles per hour, so time is 1 hour.  The carriage-rear goes a distance on the pavement at speed-v, elapsed time being 1 hour.  Speed-v is then also a number which represents distance on the pavement.  For example: 5 mph for 1 hour is 5 miles. 

Another stipulation:  Walker-w does the length of the carriage ( from back to front ) during that same 1 hour.

Blog number: The carriage has length 10 miles.  Walker-w traverses the length of the carriage ( 10 miles ) in 1 hour. Walker-w then has speed 10 mph on the carriage floor.

Blog number:  The carriage has speed v = 1 mph.  The carriage-rear necessarily travels a distance of 1 mile, from mile-0 to mile-1, and the carriage-front travels from mile-10 to mile-11.

Classical:  Walker-w must end up at the carriage-front-10 which location at time-1 is pavement-11.  Walker-w, who has speed w = 10 on the carriage, has velocity V = 11 on the pavement:

CLS sum:  V  ← ( v + w )

11 mph ← ( 1 mph + 10 mph )

It’s hard to see why output velocity-V on the pavement should be anything other than V = 11 mph, unless it so happens that the speed-of-light, ie speed-c, is 10 mph. Walker-w has speed-c on the carriage: w = c.  It’s obvious that velocity V = v + c on the pavement cannot be 11 mph, greater than speed-c.  We must have velocity V* = 10 mph.

SRT sum: V* ← ( v + w ) / rf

rf = 1.1 ← 1 + 1/10 ← 1 + v/c

10 mph ← ( 1 mph + 10 mph ) / 1.1

With speed w = c = 10 mph, walker-w cannot start at pavement-0 and arrive at pavement-11 at time-1, rendering velocity V = 11 mph > c.  One of Nixon’s egregious minions arriving “at that point in time.”

You can change the photon’s velocity by changing its time of journey, duration of journey.  You can change that duration by “adjusting” ( c.f. Section §4 ) terminal clock time.  Change terminal time from 1 hour to 1.1 hours.  Quotient 11 miles over 1.1 hours is 10 mph.

Question: Which terminal clock has face-time increased by the factor 1.1 ?  Which terminal clock does velocity-V* ?    Heaven only knows.

Is it the carriage front clock, always at carriage-10, or the stationary clock at pavement mile-11 ?

Changing the carriage front clock (which has pavement speed-v ) is like time dilation on the rear clock of Section §4, except that face-time on the front clock is “adjusted” upward not downward.  Might be something that can be done, but adjustment of the front clock does not adjust velocity-V on the pavement.

So you adjust the pavement clock.  Double jeopardy. As before, the destination clock ( stationary on the pavement ) must be adjusted upward not downward, a reversal of Section §4.  Then, also contrary to Section §4, the moving clock must adjust the stationary clock.

Notice in our discussion of Chapter VI we made no use of unprimed~primed symbols.  That notation is not in Chapter VI, only later in Chapter XIII where the v-a-f is supposedly derived, and it's the visceral impact of elemental chaos.  Roughly, primed and unprimed are Section §5, although primed is Greek.

H&R Chapter 37-4 has the primed axis on the right-hand-side.  The r-h-s axis is necessarily the input axis and reasonably the stationary axis.  Possibly Chapter 37 is a reversal: the primed r-h-s axis is moving, and that axis demands a transformed value of velocity-V* on the stationary ( unprimed ) axis. Legerdemain.

####  Length contraction

The carriage goes down the pike at speed-v.  Length contraction, based on speed-v and gamma, is in effect.  At time of 1 hour, when the rear of the carriage is at mile-1, the front of the carriage is shy of mile-11.

The H&R derivation of the v-a-f transitions from a quotient with gamma in numerator and denominator to a quotient with no gammas.  When two gammas are not shown, is that to say they are present but to null effect in the quotient, or more simply, the two gammas have simply disappeared?  Length contraction and time dilation have disappeared. 

Bertrand Russell:  “Mathematics is the subject where we never know what we are talking about, nor whether what we are saying is true.”

#### Primed-unprimed

Let’s paraphrase primed/Greek-unprimed of Section §5, first paragraph, transition from speed to distance already cited.

Opposite H&R, the primed-system has motion-v and unprimed-system is stationary.  An equation of motion is not shown.  Meanwhile, on the primed system, a “point” has equation of motion:

Shown:  x′ = wt′

“Required: the motion of point-x′ relative to the unprimed-system,” unquote.

Shown: x = [ ( w + v ) • t ] / rf

Revised: x = ( wt + vt ) / rf

These are distance equations, not speed equations.

Item: Point x′ = wt′ cannot be rewritten as x′ = wt.

Item: Distances wt′ and vt are on different axes.  In CLS mode these distances can be added.  In SRT mode, distances of mixed units cannot be added.

Item: The sum ( wt + vt ) is the ambiguous addition previously outlined: Is term-wt a location on an axis or a value added to location-vt?  Without primed notation, term-wt is best understood as a location just like location-vt.   The “composition” of distances requires value added.

Big AL, what’s happenin, Man?

####  Double file

Set up a special case wherein walker-w on the carriage has classical velocity-V on the pavement and all speeds are less than speed-c, ie, V < c:

V = v + w

Contrive walker-u, a solitary no-carriage guy who has pike velocity U = V.

U and V are velocities of a couple of guys going down the pike double file, side-by-side. 

What happens when walker-w on the pike is v-a-f output-V* ?

V* = ( v + w ) / rf

Are walkers U and V* going down the pike double file or is walker-V* falling behind?  Them sombitches !

####  Obviously correct

The v-a-f is obviously correct in some sense. When walker-w on the carriage becomes photon-c on the carriage, pavement velocity is obviously V* not V.

Wrong:  V = ( v + c ) > c

Right: V* = ( v + c ) / ( 1 + v/c ) = c

What happens when V = ( v + w ) < c

Why do the Perp Walk when we ain’t got nobody doin no harm?

#### The speed of sound

Section §3 says more or less explicitly that classical one-way outward bound photon speed on the floor of the Michelson-Morley device is speed c – v, not speed-c.  An ether drag effect.  In other words, the photon left to its own devices behaves like the speed of sound in air.  Section §3 does not deal with one-way speed on the carriage.  Best guess: Lorentz’s “local time” transforms classical one-way on-deck photon speed from speed c – v to speed-c.

On the moving MMD, SRT steps up a one-way on-deck photon from speed c – v to speed-c.  You then have, according to classical mechanics, a photon of speed c + v on the pavement.  To solve that problem, you put the v-a-f to work on the pavement, stepping down pavement speed from c + v to speed-c.

A formula transforms up on the device, and then a different formula transforms down on the pavement.  Wouldn’t it be better if just one Whac-A-Mole did both jobs at once?  What equation might that be?  We haven’t the foggiest notion.