SME
University Physics III: Waves, Optics & Modern · Problem type
Relativistic velocity addition
Use this for “A ship moving at 0.7c fires a probe forward at 0.5c relative to the ship. How fast does the probe move as seen from the planet?”
Diagram
u
=
v + u′
1 + vu′
=
0.6 + 0.6
1 + 0.6 × 0.6
=
1.2
1.36
u = 0.8824c
Adding the speeds, v + u′ = 1.2c, would beat light (dashed); the true world line stays inside the light line.
This is an example. Type your own values to start a new problem.
u
c
=
c
+
c
1
+
×
uG
Sum without relativity, in c
worked out (not typed)
Formulas
u
=
v
+
u
′
1 +
v
×
u
′
0.882353
=
0.6 + 0.6
1 + 0.6 × 0.6
uG
=
v
+
u
′
1.2 = 0.6 + 0.6
Assumptions
•
Frame S′ moves at v along +x past frame S; the object moves along the same line at u′ in S′ and at u in S.
•
Every speed is in units of c, so u = (v + u′) ÷ (1 + vu′ ÷ c²) reads u = (v + u′) ÷ (1 + vu′); a minus sign means toward −x.
•
When v and u′ are small the bottom is near 1 and u ≈ v + u′, the Galilean sum; below c plus below c always stays below c.
Step-by-step
WE KNOW
v = 0.6, u′ = 0.6
FIND
object’s speed in S, in c (u), sum without relativity, in c (uG)
Step 1 · Find object’s speed in S, in c (u)
Use u = (v + u′) ÷ (1 + v × u′)
Add
the
speeds,
then
divide
by
1
+
vu′:
the
bottom
grows
with
both
speeds
and
keeps
u
below
c.
u
=
0.6 + 0.6
1 + 0.6 × 0.6
u
=
1.2
1 + 0.36
u
=
1.2
1.36
u
=
0.882353
Step 2 · Find sum without relativity, in c (uG)
Use uG = v + u′
What everyday speeds would give: the plain sum, which can pass c.
uG = 0.6 + 0.6
uG = 1.2
Check
Put the numbers back in. Both sides should match.
0.882353
=
0.6 + 0.6
1 + 0.6 × 0.6
✓
1.2 = 0.6 + 0.6 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Special relativity
Main lesson
Relativistic energy and momentum
Use this for “A proton (rest energy 938.3 MeV) moves at 0.8c. Find its total energy, its kinetic energy and its momentum in MeV/c.”
The Lorentz transformation
Use this for “A rocket passes Earth at 0.8c. An event happens 1,200 m away along its path, 2,000 m of ct after they meet. Where and when is it in the rocket’s frame?”