The Velocity Addition Formula
Velocity Addition Formula
Updated: April 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 / 1.4 = 2.14 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 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.
#### 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
ReplyDeleteGalilean sum with 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 is the right-hand-side.