The Velocity Addition Formula

Velocity Addition Formula

Updated: October 18, 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. EDoMB-1905-Sections §3 and §4:

x′ = gamma • ( x – vt )

t′ = gamma • ( t – vx / c² )

Lorentz Transforms – addition from modern textbooks, not shown in Section 5:

x = gamma • ( x′ + vt′ )

t = gamma • ( t′ + vx′ / c² )

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

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

Numerator in the v-a-f is the plus-sign LT spatial, an addition of distances.

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 and Resnick Chapter 37 is not shown.

####  Special case: w = c

A laser-gun moves along the ground at speed-v.  It emits a photon of muzzle velocity w = c.

V = ( v + c ) / ( 1 + vc/c² )

V = ( v + c ) / ( 1 + v/c )

V = ( v + c ) / ( ( v + c ) / c )

V = c

Ground speed-V of the photon is speed-c just as muzzle velocity is speed-c.  Can’t be anything else.

####  Some example numbers

Speed-c is 10.

Speed-v is 2.

Use the special case of w = c = 10.

Denominator reduction factor ( 1 + vw/c² )  →  ( 1 + vc/c² )  →  ( 1 + v/c )  →  ( 1 + 1/5 ) →  6/5.

The v-a-f:  10 = ( 2 + 10 ) / (6/5)

Invert denominator reduction factor ( 6/5 ) > 1 so that it becomes numerator factor 5/6 < 1.

The v-a-f:  V = 10 = ( 5/6 ) × ( 2 + 10 ) 

Interpret simple equation:

10 = ( 5/6 ) × ( 2 + 10 )

Option-A: Add speeds-2 and -10 yielding sum-12 and then reduce by ( 5/6 ). 

10 = ( 5/6 ) × 12  ← 12 = ( 2 + 10 )

Problem: speed sum 2 + 10 = 12 cannot exist as a phenomenon.  Faster than the speed-of-light.

Option-B: Use the distributive law and distribute the reduction factor: 

10 = 1.6666  + 8.3333  ←  ( 5/6 ) × 2 + ( 5/6 ) × 10

All speeds shown are valid.  Problem: Input speeds v and w = c ( prior to transformation ) are reduced from 2 and 10 to 1.6 and 8.3.  The v-a-f is not adding what it is supposed to add.

Which is it, Option-A or Option-B ?  Either way, it is the mollycoddling of criminals.

####  Carriage and embankment

Let’s swap out the moving laser gun in favor of a carriage-on-embankment as per 1920-Chapter VI.

Rolling on the embankment, the carriage has speed v = 2 feet per second, about walking speed ( approx 2 miles per hour ).

On board the carriage, walker-w has speed w = 3 fps as judged by the carriage floor. 

According to classical mechanics, walker-w of speed w = 3 fps on the carriage has speed V = 5 fps on the embankment.

Classical: V = 5  ←  2 + 3  ←  v + w

Let the speed-of-light be c = 10 fps, less than Roger Bannister’s 15 fps, or 15 miles per hour.

The dimensionless reduction factor in the denominator of the v-a-f is:

rf = ( 1 + vw/c² )

rf = ( 1 + 2 × 3 / 10² )

rf = 1.06

Plug it into the v-a-f:

←  ( v + w ) / 1.06  ←  ( v + w ) / ( 1 + vw/c² )

V =  4.72 fps  ←  5 / 1.06  ←  ( 2 + 3 )  / 1.06

Bring in someone new, a conventional walker on the embankment, walker-Z who has speed Z = 5 fps.  His undertaking begins at the same place-time as metaphysical walker V = 4.72 fps.

Destination:  Walker-V and walker-Z go to a common fixed destination 50 feet down the track.

Conventional walker-Z with ground speed Z = 5 fps arrives at destination d = 50 feet at ground time t = 10 seconds.

How do walkers Z and V compare, when walker-V is carriage walker-w reconsidered as speed on the ground.

Option-J:  Walker-V, going slower than walker-Z ( 4.72 fps versus 5 fps ) trails behind, arriving at destination-50 at time 10.6 seconds.

Option-K:  Walkers V and Z go hand-in-hand, double file, arriving at destination-50 at one moment.  But these walkers have their own arrival clocks, clock-VT and clock-ZT.  Clock-VT reads 10.6 seconds and clock-ZT reads 10 seconds. 

Which is it, Option-J or Option-K?  College algebra is obviously Option-K, the clock model.  In particular, the temporal Lorentz Transform ( denominator ) revises the unprimed ground doing VT upward, to more time, while an adjacent unprimed ground doing ZT is not revised.  It stretches credulity.  Can carriage walker w = 3 ( reconsidered as ground walker-V ) transform just one of two stationary destination clocks? 

####  Four scenarios: P, Q, R, S

Carry over: speed-v is walker-v of 2 feet per second, speed-w is walker-w of 3 feet per second.

## Scenario-P:  Speeds v and w are positive pathway speeds.  Walker-w is 3 fps on the pathway, not 3 fps out in front of walker-v.  Speeds v and w are subtracted, not added, as per EDoMB-Section §4 which does Lorentz Contraction and time dilation. 

V = 1 fps  ←  3 – 2  ←  w – v

Just as speeds v and w are phenomena on the ground, so also is speed V = 1 an observable phenomenon, the extent to which ground speed-w is faster than ground speed-v.

## Scenario-Q:  Speed-v is now negative, moving negatively at 2 fps with respect to the starting line while walker-w moves off the starting line at positive 3 fps.  Use right-hand-side subtraction as in Scenario-P:

V = 5 fps  ←  3 + 2  ←  w – (  – v )

Speed-V = 5 fps is observable speed of divergence of walkers v and w.  It’s just possible that these speeds can be plugged into the Lorentz Transform set-up of Section §4 ( subtraction ) yielding relativistic speed of divergence V = 4.72 fps.

Problem:  This divergence is not the carriage and embankment model of 1920-Chapter 6.

## Scenario-R:  As in scenario-P, speeds v and w are positive pathway speeds, walker-w being 3 fps on the pathway, nit the negative speed in scenario-Q nor the extension of walker-v in scenario-S.   Speeds v and w on the right-hand-side are added:

V = 5 fps  ←  2 + 3  ←  v + w

Left-hand-side speed V = 5 is good algebra, but no projectile has speed 5 fps.

## Scenario-S:  Like scenario-R, speeds v and w are positive, walker-v at 2 fps and walker-w at 3 fps.  Like scenario-R, speeds are added.  Now speed w = 3 fps is not a pathway speed.  Walker-w is an extension of walker-v at speed-w, a property not shared by any of scenarios-P, Q or R.

V = 5 fps  ←  2 + 3  ←  v + w

Left-hand-side speed V = 5 is both algebra and observable motion on the ground, a version of walker w = 3 fps at 5 fps.  Walker-w is now walker-V of the carriage and embankement, as required, and the v-a-f cranks out speed V = 4.72 fps.

Queston:   The LT’s of Section §4 don’t do speed-V, rather distance x′ and time t′.  Scenario-P does not show a relativistic version of speed-V. 

Can the subtractive LT’s of Section §4 be adapted to scenario-Q, yielding relativistic divergence speed V = 4.72 ?  Maybe, but as said before, scenario-Q is not the carriage and embankment model.

Can the subtractive LT’s of Section §4 be adapted in additive scenario-S to yield relativistic compound speed V = 4.72 ?  It’s really a stretch.