The Velocity Addition Formula
Velocity Addition Formula
Updated: May 2, 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 )
The H & R derivation is shown neither in EDoMB-1905 nor in Relativity-1920. In particular, the gamma spatial is not shown.
#### See also
Paul A Tipler, Physics Fourth Edition, Chapter 39
Equation 39-1b: “inverse”
Equation 39-18a: v-a-f by means of quotient
#### EDoMB Section §4 — subtraction
Section §4 has an object of “fixed length” travelling at speed-v. Let the object be a train of length-L on track-X. The stationary train might reside between track mileposts J = mile-2 and K = mile-5. Train caboose is at mile-2 and train front is at mile-5. Either of two subtractive equations of distance express a train of length L = 3 miles:
K – J = L ( 5 – 2 = 3 )
L = K – J ( 3 = 5 – 2 )
The former is simple algebra. The latter equation is “transformation,” transforming the 3-mile gap between post J = 2 and post K = 5 into train length L = 3. Basically, equation-KJ is ordinary left → to → right algebra, whereas equation-L is right ← to ← left equation logic, right ← to ← left transformation of The Lorentz Transforms.
What happens when you bring in motion-v, relativity, and gamma ? Gamma is simply a dimensionless factor, for example value 1.6, namely kilometers per mile.
4.8 ← 1.6 • ( 5 – 2 ) ← gamma • ( K – J )
Train-L, now moving with speed-v, is train L*, and its caboose is momentarily at post J = 2 miles. Question, how long is train-L*, or by the same token, where is the front of train L* ?
Option 1: Left-hand-side 4.8 is train-length L* = 4.8 miles, such that train-L of length L = 3 miles transforms into train L* of length L* = 4.8 miles.
Option 2: Left-hand-side 4.8 is 4.8 kilometers, such that train-L of length L = 3 miles transforms into train L* of length L* = 4.8 kilometers. Stationary-L and moving-L* have the same length, they are just calibrated differently. 4.8 kilometers = 3 miles.
Option 3: Left-hand-side 4.8 is a distance of 4.8 kilometers on left-hand-side track-X*, which starred track at the particular moment has its origin at mile J = 2 of track-X. Train-L is given length-3 on starred ( moving ) axis-X*, length L = 3 miles having been transformed into L* = 3 kilometers. Then, from the point-of-view of track-X, train L* appears to have length L / gamma or 3 / 1.6 = 1.875 miles. Moving train-L* quote “appears shorter” according to Section §4.
Question: Which of these options is to be used in Section §4 ? Answer, the algebra cannot decide. Only the mad scientist can decide, and he sees that length contraction must have Option 3.
#### 1920 Chapter VI — addition
Track-X is now an “embankment.” The train, stationary or moving, is now a “carriage.” As in the train model, the stationary carriage has length L = 3 miles. The moving carriage has ( momentarily ) its rear bumper at embankment mile J = 2. Implicitly and perhaps incorrectly, the moving carriage is likewise of length L = 3 miles, and at the given moment its front bumper is at embankment milepost K = 5 miles.
In simple mechanical terms walker-w goes length L = 3 miles on the floor of the moving carriage, and during the same interval of time the carriage’s rear-bumper goes from embankment mile-0 to mile J = 2. At time-zero, walker-w at the rear carriage bumper is likewise at embankment mile-0, ending up at embankment mile K = 5 simply because the carriage’s front bumper ends up at mile K = 5. In simple mechanical terms, walker-w covers 5 miles on the embankment, though he walks only on the carriage floor.
J + L = K ( 2 + 3 = 5 )
K = J + L ( 5 = 2 + 3 )
The latter equation is right ← to ← left transformation algebra.
Question: What on the right-hand-side is transformed into the walker’s embankment excursion on the left-hand-side ? Is it his on-carriage walk of distance L = 3 or his total traverse J + L ?
Question: Does moving carriage size-L suffer L* < L of Section §4 ? Where then is walker-w at the end of time?
What happens when you bring in relativity with gamma = 1.6 ?
8 ← 1.6 • ( 2 + 3 ) ← gamma • ( J + L )
Option 1: Left-hand-side 8 has walker-w doing 8 embankment miles rather than 5 embankment miles.
Option 2: Left-hand-side 8 has walker-w doing 8 embankment kilometers, same as embankment mile-5, only calibrated as kilometers.
Option 3: Left-hand-side 8 kilometers is recalibration of embankment K = 5 miles. Walker-w must traverse distance-5 on the embankment as 5 kilometers, a distance of 3.125 miles.
As before, only the mad scientist can decide what is going on here, and he sees that the v-a-f must have Option 2. Addition-option-2 is not the inverse of subtraction-option-3.
Giancoli Physics Sixth Edition Appendix E, quote: “inverse.”
#### LT distance-addition
Chapter XIII-1920 has a version of Galilean-distance-addition:
Shown: x = ( v + w ) • t
The equation works If you watch your step.
As stated, neither Chapter XIII nor Section §5 show the H & R derivation, with its LT-distance-addition numerator:
Not shown: x = gamma • ( x′ + vt′ )
A mud hole.
#### Material posted as blog comments
Galilean addition
Distributive law
Section §3
Inverse
Addition options
Apples and oranges
#### Blog comment – Galilean addition
ReplyDeleteThree versions of Galilean spatial addition.
1920-Chapter XIII. Both right-hand-side terms unprimed.
↓
(1) x = vt + wt
Halliday & Resnick Chapter 37.
Both right-hand-side terms primed.
↓
(2) x = x′ + vt′
Google: unsw "Galilean transform equations"
Mixed units on the right-hand-side.
↓
(3) x = x′ + vt
H&R eq (37-20)
↓
x′ = x – vt
H&R says equation (37-20) can be solved for “x”, yielding eq (2). Wrong. Solving for “x” yields UNSW equation (3), i.e., mixed units right-hand-side.
#### Blog comment – distributive law – add speeds
ReplyDeleteWe gotta step away from distances for a moment and add speeds. Why? “Symmetry.”
Galilean sum V-unstarred: V = v + w
V-a-f sum with V-star: V* = ( v + w ) / ( 1 + vw/c² )
Let’s set aside the carriage-embankment model and revert to a train model of simpler vocabulary.
A train is 100 miles long. Beginning at the joint origin, track mile-0 and time-0, the caboose goes 100 track miles in 1 hour, speed v = 100 miles per hour. Likewise, beginning at the joint origin, walker-w does train floor length of 100 miles in 1 hour, speed-on-train of w = 100 mph.
The train front likewise goes 100 miles in 1 hour, from track mile 100 to track mile 200, again speed v = 100 mph. At time 1 hour, floor walker-w ends up at track terminal mile 200. Having started at track mile-0, the walker’s metaphysical track distance is 200 miles. According to classical kinematics, his track speed is 200 miles in 1 hour or V = 200 mph.
Suppose the v-a-f must reduce speed V = v + w = 200 mph to V* = 150 mph. Distributive law:
¾ • ( v + w ) = ¾ • v + ¾ • w
¾ • ( 100 + 100 ) = ¾ • 100 + ¾ • 100
¾ • 200 = ¾ • 100 + ¾ • 100
150 mph = 75 mph + 75 mph
The v-a-f is which side of the distributive law, left-hand-side or right-hand side? Obviously, left-hand-side.
Section §5 quote: “It is worthy of remark that v and w enter into the expression for resultant velocity in a symmetrical manner.”
The quoted remark calls for right-hand-side 75 + 75.
#### Blog comment – Section §3
ReplyDeleteDeriving the subtractive LT’s, Section §3 conjures up a huge equation which looks like a barrel of snaggle toothed fishheads. On its lengthy left-hand-side are two terms which are added:
x′ / ( c – v ) + x′ / ( c + v )
Each term is distance divided by speed, quotients which are time. The sum of terms is time.
Question: How can you have denominator ( c + v ) when Section §5 says it is but value-c ?
Paradoxes are piling up like the Lord Protector’s mountain of dead dogs.
#### Blog comment – inverse
ReplyDeleteIf such a thing as “inverse” existed in 1905, then clearly subtractions in Sections §3 and §4 are precursors and addition is the inverse.
Wiki: inverse function
A track is 5 miles long. A train of length 3 miles has moved 2 miles down the track, caboose at mile-2 and front at mile-5.
Galilean precursor: x′ = x – 2
3 = 5 – 2
Galilean inverse: x = x′ + 2
5 = 3 + 2
Let’s leave physical interpretation for another blog comment and do some substitution algebra which “inverse” requires.
Substitute x′ = 5 – 2 in the inverse.
Correct and required: 5 = [ 5 – 2 ] + 2
Bring in gamma = 1.6
Gamma precursor: x′ = 1.6 • ( x – 2 )
4.8 = 1.6 • ( 5 – 2 )
Gamma inverse: x = 1.6 • ( x′ + 2 )
8.0 = 1.6 • ( 3 + 2 )
Substitute x′ = 1.6 • ( 5 – 2 ) in the inverse.
Not as required: 12.8 = 1.6 • ( [ 1.6 • ( 5 – 2 ) ] + 2 )
Substitution must yield x = 8.0 on the left-hand-side. The precursor cannot be inverted if gamma is included.
#### Blog comment: addition options
ReplyDeleteLet’s use carriage miles 2, 3 and 5. For easier vocabulary we have a track and a train rather than embankment and carriage.
At time t = 1 hour, whether we have subtraction or addition, Galileo or Lorentz, the caboose of the train is at track mile-2.
Subtraction: A moving train, length 3 ordinary miles, appears momentarily between track mile-2 and track mile-5. Subtraction ( 5 – 2 ) = 3 is mostly simply 3 track miles. Galilean subtraction transforms the three 3 track miles into 3 overhead train miles, signified by right ← to ← left equation logic: 3 = 5 – 2. Lorentz transformation steps in next and, with Option-3, the gamma reduction factor transforms left-hand-side train length-3 into train length-1.875.
Addition: Again, moving train-3 appears momentarily between track mile-2 and track mile-5. Track segment-3 is added to caboose distance-2 yielding track distance-5. Galilean addition transforms the 3 track miles into 3 overhead train miles, signified by right ← to ← left equation logic: 5 = 2 + 3. Lorentz transformation steps in. Which Option is it: 1, 2 or 3? Let’s say Option-3 as in subtraction, gamma = 1.6 is in effect reduction factor ( 1 / 1.6 ). How do you add?
Option-A: 3.125 = ( 1 / 1.6 ) • ( 2 + 3 )
Option-B: 3.125 = ( 1 / 1.6 ) • 2 + ( 1 / 1.6 ) • 3
Option-C: 3.125 = 2 + ( 1 / 2.666 ) • 3
Option-C seems weird. The fact is however, relativity does not deal with stationary things, it does not transform stationary track distance-2 doing the caboose, so all of train length-3 must be reduced to length-1.125. Dubious.
#### Apples and oranges
ReplyDeleteThe v-a-f of 1905-Section §5 shows explicitly addition of speeds. Relativity-1920 Chapters VI and XIII have explicitly addition of distances as forerunner. Chapter XIII does not explicitly do the numerator transformation of a sum of distances. Let’s do fruits instead of distances.
## Subtract 2 apples from one bowl of 5 apples.
Algebra: \5 apples – 2 apples/ = \3 apples/
Galileo: \3 oranges/ ← \3 apples/ ← \5 apples – 2 apples/
Lorentz: \3 tangerines/ ← \5 apples – 2 apples/
Interpret: Left-hand-side output bowl of 3 apples is transformed into a bowl of 3 oranges and then into a bowl of 3 tangerines, smaller than 3 oranges. Lorentz contraction.
## Add one bowl of 2 apples and one bowl of 3 apples.
Algebra: \2 apples/ + \3 apples/ = \5 apples/
Galilean options shown next: 5 right-hand-side apples in two bowls are transformed: into one bowl of 2 apples and 3 oranges – or – into one bowl of 5 oranges:
\2 apples + 3 oranges/ ← \2 apples/ + \3 apples/
\5 oranges/ ← \2 oranges + 3 oranges/ ←\2 apples/ + \3 apples/
Both are good Galilean schemes, but you cannot subsequently apply Lorentz to the bowl of mixed of apples and oranges, only to a bowl of 5 oranges, with resultant 5 tangerines. Therefore the original Galilean must transform not only the input bowl of 3 apples, but also the input bowl of 2 apples. That version of the additive Galilean is not the inverse of the subtractive Galilean.
Escape clause: In textbook quotient-V, spatial-gamma and temporal-gamma ( being the same dimensionless factor ) divide out, so you can forget the gamma effect. Right? Nope, gamma is still there doing its thing in space-time.