The Velocity Addition Formula

Velocity Addition Formula

Updated: December 8, 2025

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, per 1905 Sections §3 and §4:

x′ = gamma • ( x – vt )

t′ = gamma • ( t – vx / c² )

Lorentz Transforms – addition, from modern textbooks:

x = gamma • ( x′ + vt′ )

t = gamma • ( t′ + vx′ / c² )

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

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

Precursor to numerator ( v + w ) in the v-a-f is the spatial Lorentz Transform.  Two problems:  1) The LT’s of 1905 are subtractive, not additive.  2) The spatial LT is addition or subtraction of distances whereas ( v + w ) is an addition of speeds.

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 document provides an LT derivation.   In particular, the plus-sign algebra of Halliday & Resnick Chapter 37 is not shown.

#### See also

Text book:
Paul A Tipler, Physics Fourth Edition, Chapter 39
Equation 39-1b: “inverse”
Equation 39-18a: v-a-f by means of quotient

Web page
https://galileo-unbound.blog/2023/10/18/relativistic-velocity-addition-einsteins-crucial-insight/

the Fizeau experiment, also in 1920-Chapter XIII

####  2 + 2 = 3

Question: How do you add two plus two and get three?

Answer A: 2 = 1.5 and 2 = 1.5 and 1.5 + 1.5 = 3

Answer B: 2 + 2 = 4 and 4 = 3

#### Verbalize

How do we verbalize?  Let’s have an abbreviation for Special Relativity: spezielle Relativitätstheorie, SRT.

For starters, speed variables v and w can be projectile names, projectile-v and projectile-w. 

Then use subscripts to distinguish Galilean speed-V from relativistic speed-V. 

Galilean sum of speeds v and w is speed-Vg subscript-g:

Vg = v + w

Relativistic sum of speeds v and w is speed-Vsrt, subscript-srt:

Vsrt = ( v + w ) / ( 1 + vw/c² )

Misleading.  If you revert to Galilean mode, and get rid of the reduction factor, you have with speed-Vg on the left-hand-side, and right-hand-side speed-w is simply speed-w.  What happens in SRT mode?  Speed-V becomes speed-Vsrt.  What about speeds v and w?  Speed-w is on a moving platform subject to length contraction and time dilation so we have in effect, speed-wsrt.  Whereas speed-v ( on the ground ) is not subject to either special effect and is speed-vg.  Can you add these speeds?

#### Transformation

The v-a-f equation is written backwards.  SRT equations are usually written that way.  A standard equation written left → to → right is an “identity”:

Identity: 2 + 2 → 4  is simply 2 + 2 = 4

An SRT equation written left ← to ← right is an “assignment” operator.

Assignment: 3 ← 2 + 2

But it’s more than that: it’s “transformation.”  Speeds v and w are both equal to 2 miles per hour.  Speed sum 2 mph + 2 mph equal to 3 mph is not merely addition, it is also transformation.

#### The distributive law

How do you transform 2 mph + 2 mph into 3 mph? You divide by the dimensionless reduction factor.

Reduction factor ( 1 + vw/c² ) = 4/3 > 1

3 mph ← ( 2 mph + 2 mph ) / ( 4/3 )

What is division by ( 4/3 ) doing?

The distributive law:

2 / ( 4/3 ) + 2 / ( 4/3 ) = 4 / ( 4/3 )

Side-A: 2 / ( 4/3 ) + 2 / ( 4/3 )

Side-B: 4 / ( 4/3 )

Algebraically equal, sides A and B are different physical situations.

Side-A:  You reduce the speeds of projectiles v and w from 2 mph to 1.5 mph and add:

3 ←  1.5 + 1.5  ← 2 / ( 4/3 ) + 2 / ( 4/3 )

Side-B: You add projectile speeds v = 2 and w = 2 with resultant sum 4 mph, and then reduce 4 mph by divisor ( 4/3 ) with resultant 3 mph.

3 ← 4 / ( 4/3 ) ← ( 2 + 2 ) / ( 4/3 )

#### Which is it?

Which side, A or B, is inherent in the v-a-f? 

Vsrt = ( 2 + 2 ) / ( 4/3 )

Halliday & Resnick is Side-B:  Spatial and temporal LT’s do a speed quotient.  The numerator adds speeds v and w unchanged, and then the denominator factors up terminal time of projectile-Vg by 4/3 > 1, thereby factoring down speed-Vg = 4 mph by 3/4, and speed-Vg = 4 becomes speed-Vsrt = 3 mph.

The reduction factor rf = ( 1 + vw/c² ) is greater than 1, which factor increases time on the terminal pavement clock doing speed-V, so that speed-V is speed-Vsrt, and speed-Vsrt ( requiring more time ) is then less than speed Vg. 

Section §5 is Side-A:  Quote, “It is worthy of remark that v and w enter into the expression for resultant velocity in a symmetrical manner.”  Unquote.  Seemingly you factor down speeds v and w ( not clocks ) by 3/4 and then add.  3/4 of 2 mph plus 3/4 of 2 mph is 1.5 mph plus 1.5 mph equal to 3 mph.

#### Another A-B dichotomy

The Galilean equation:

Vg = v + w

As before, subscript-g is Galilean.  Per 1920 Chapter VI, speed-v is the “carriage” on the “embankment,” speed-w is the walker on the carriage, aka the “man” on the carriage, and speed-Vg is walker-w on the embankment.   We use the special case where nothing is equal to speed-c.  We can easily contrive another Galilean walker directly on the embankment, walker Vge, going side-by-side, double file, with walker-Vg ( who is walker-w ):

Vge = Vg

What happens when you impose SRT on walker-Vg, while walker-Vge remains Galilean?  Speed-Vg becomes speed-Vsrt:

Vsrt = ( v + w ) / ( 1 + vw/c² )

How does walker-Vsrt relate to walker-Vge?  You have two scenarios:

Scenario-A: Walker-Vsrt, going slower than walker-Vge, falls behind and arrives at a ground destination later than walker-Vge.

Scenario-B: Walker-Vsrt and walker-Vge are side-by-side, double file. But, arriving at one ground destination at one moment, walker-Vsrt has a terminal clock reading more time than the Galilean terminal clock of walker-Vge.  Taking more time to get some place you are going slower.

Roughly, Scenario-A is Section §5 of “symmetric” while Scenario-B is college algebra where the denominator in the LT quotient is the terminal clock.

Which is it?

#### Transform a stationary clock

Section §4 does time dilation.  The equation used, taken over from Section §3, is a Lorentz Transform, in particular the subtractive temporal, not the spatial.

t′ = gamma • ( t – vx / c² )

The Michelson-Morley device moves down the road at speed-v.  A clock ( time-t′ ) is fastened to the rear of the device.  A pavement equation is shown: x = vt.  It looks like an equation of motion.  It isn’t.  It selects a stationary pavement milepost, with stationary clock-t, which is momentarily adjacent to the moving MMD-rear clock.  With x = vt in the LT, milepost clock-t at point-x looks at clock-t′ of speed-v and clock-t′ displays less time.  An equation is shown ( t′ = t / gamma, ) rendering t′ < t.  Gamma and “local time” in the LT combine to give this result.

What happens when the carriage goes down the road at speed-v, all speeds less than speed-c ?  A clock keeping time-t′ is at the front of the device ( stationary on the device ).  This clock will determine arrival time-t′ of walker-w on the floor of device, and that in turn renders speed-w.  At the moment of arrival, front-end clock-t′ looks at the adjacent pavement clock keeping time-t. How does clock-t′ figure out clock-t ?  According to college algebra gamma has divided out, so nothing remains in the additive temporal except Lorentz’s “local time” with a plus-sign.

t = gamma • ( t′ + vx′ / c² )
↓
t = t′ + vx′ / c²

Clock-t on the left-hand-side does speed-V of walker-w on the pavement, and since walker-V is clocked in by more than simple Galilean time, he is slowed down.  We have walker-Vsrt not walker-Vg.

Additive time:

1) Time transformation involves the front of the moving device, not its rear. 2) Stationary time is transformed, not moving time. 3) H&R transformation occurs without gamma. 4) From the point-of-view of the primed observer, the unprimed clock to be transformed moves with negative speed-v, not positive speed-v.  5) The transformed clock shows more time, not less time.

Is any of this stuff any good?  Hard to say.  But one thing is obvious.  None of it appears in Chapter XIII.   In Chapter XIII, walker-w is a bi-directional ray-of-light having speed w = c/n in Fizeau’s water.  The question of Chapter VI is not answered.

####  Distances not speeds

Start with the subtractive spatial LT which does length contraction in Section §4.  Moving at speed-v, the center of a “rigid sphere” crosses the starting line at time-zero.  Equation:

x′ = gamma • ( x – vt )

Only the longitudinal radius is effected. “The X dimension appears shortened.”  The result, “viewed from the stationary system” is an “ellipsoid.”

Corresponding is a one-dimensional Michelson-Morley device of fixed size-L, and speed-v.  Device size will appear shorter.  The temporal LT need not enter into it because device size-L is viewed at time-zero.

Point-vt is the rear of the device of ground speed-v and point-x is the front of the device, always distance-L in front of point-vt: x = vt + L .  Section §4 observes length-L at time t = 0 in which case: x′ = gamma • L .  But the equation holds at any time simply on the basis of algebra:

x′ = gamma • ( x – vt )
↓
x′ = gamma • ( ( vt + L ) – vt )
↓
x′ = gamma • L

It’s very confusing because left-hand-side-x′ is not a transformation of right-hand-side-x.  It’s transformation of classical ( L = x – vt ) into an SRT distance.  But neither of pavement distances x or vt are transformed.

Galileo: x′ = x – vt

As in SRT, left-hand-side-x′ is not a point on the ground.  Distances x and vt are transformed into the fixed length-L of the MMD at speed-v, so left-hand-side-x′ is a length, not a point.

x′ = L = x – vt

Re-arranged:  x = vt + x′

Point-x on the left-hand-side is the ground location ( at time-t ) of front of the MMD as the MMD moves along at speed-v, length-x′ being a constant.  Carriage walker-w from Chapter VI, doing length L = x′ of the MMD, then ends up at road location-x which is the sum of two distances.  Just what Chapter VI requires.

SRT:  x′ = gamma • ( x – vt )

x′ / gamma = x – vt

Re-arranged: x = vt + x′ / gamma

Problem: right-hand-side term x′ / gamma doesn’t really exist.  An equation of the sort L′ = L / gamma is not shown in Section §4.  R-h-s term x′ / gamma is a phenomenon but it is not a value.  The moving MMD here in SRT is a phenomenon of length L / gamma ( it appears shorter than L;  L′ < L ) but the value of right-hand-side x′ / gamma is simply Galilean-L. The moving MMD retains its original length VALUE.  It does not retain its original size.  A contradictory situation which does not appear in the subtraction equation.  Distances vt and x′ / gamma cannot be added. 

The LT doing a subtraction of distances cannot be adapted to addition of distances, “composition” of distances according to Section §5.  The v-a-f is addition of speeds, but Chapter VI says explicitly that you have to add distances.  You can’t in SRT mode.