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.



Comments

  1. #### Blog comment – Galilean addition

    Three 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.

    ReplyDelete
  2. #### Blog comment – distributive law

    Galilean 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.

    ReplyDelete

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