Modern PhysicsSpecial Relativity

Relativistic Velocity Addition

Velocities combine as u′ = (u − v)/(1 − uv/c²), so no speed passes c.

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A ship moving at $0.60c$ fires a probe forward at $0.60c$ relative to the ship. Galileo would say the probe moves at $1.20c$ relative to Earth, which can't be right. The Lorentz transformation gives the correct rule.

Key idea

If frame $S'$ moves at $v$ relative to $S$, an object moving at $u'$ in $S'$ (along the same line) moves in $S$ at $$u = \frac{u' + v}{1 + \dfrac{u'v}{c^2}}.$$ The reverse is $u' = \dfrac{u - v}{1 - uv/c^2}$. If $u'$ and $v$ are below $c$, so is $u$; and if $u' = c$, then $u = c$.

Why it works

Follow the object for a short time. In $S'$ it moves $dx'$ in $dt'$, with $u' = \frac{dx'}{dt'}$. The inverse Lorentz transformation gives $dx = \gamma(dx' + v\,dt')$ and $dt = \gamma\left(dt' + \frac{v\,dx'}{c^2}\right)$. Divide, and then divide the top and bottom by $dt'$: $$u = \frac{dx}{dt} = \frac{u' + v}{1 + u'v/c^2}.$$ The denominator, the sign of disagreeing clocks, is what keeps the sum below $c$. For light, $u' = c$: $u = \frac{c + v}{1 + v/c} = c$, as postulate 2 demands.
The speed seen from the ground for an object moving at $u'$ in a frame moving at $v$. The everyday rule (dashed) crosses $c$; the relativistic curve bends and never reaches it.

Worked examples

Worked example: A probe from a fast ship

A ship moving at $0.60c$ fires a probe forward at $0.60c$ relative to the ship. How fast does the probe move relative to Earth?
  1. $u = \frac{0.60 + 0.60}{1 + (0.60)(0.60)}c = \frac{1.20}{1.36}c$.

Answer: $u = 0.882c$

Worked example: Two ships approaching

Two ships approach Earth from opposite sides, each at $0.80c$ in Earth's frame. How fast does one move as measured from the other?
  1. Use the frame of ship A, which moves at $v = +0.80c$. Ship B moves at $u = -0.80c$ in Earth's frame.
  2. $u' = \frac{u - v}{1 - uv/c^2} = \frac{-0.80 - 0.80}{1 - (-0.80)(0.80)}c = \frac{-1.60}{1.64}c$.

Answer: $0.976c$ (toward A), not $1.6c$

Worked example: A probe fired backward

The ship at $0.60c$ fires a probe backward at $0.80c$ relative to the ship. What is the probe's velocity relative to Earth?
  1. $u' = -0.80c$ and $v = +0.60c$.
  2. $u = \frac{-0.80 + 0.60}{1 + (-0.80)(0.60)}c = \frac{-0.20}{0.52}c = -0.385c$.

Answer: $-0.385c$: moving backward relative to Earth, faster than Galileo's $-0.20c$

An old experiment fits this rule. In 1851 Hippolyte Fizeau sent light through flowing water. Light moves through still water at $c/n$; with the water flowing at $v$, its speed went up by only about $v\left(1 - \frac{1}{n^2}\right)$, not the full $v$. Put $u' = c/n$ into the addition rule and expand for small $v$: you get exactly that, more than fifty years before Einstein explained it.

Common mistake

Adding speeds the everyday way and getting more than $c$. The everyday rule $u = u' + v$ is the low-speed limit of the relativistic one, good when $\frac{u'v}{c^2}$ is tiny. Check any answer: no material object's speed can come out at or above $c$, and light must come out at exactly $c$.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.

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This is one of the top rungs of Waves, Optics and Modern Physics.

Sources and further reading

  • OpenStax University Physics Volume 3, 5.6 Relativistic Velocity Transformation
  • MIT OpenCourseWare 8.20 Introduction to Special Relativity (January IAP 2005)
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