F. Revolutions in Modern Physics: Quantum Mechanics and Special Relativity · Special RelativityAdding Velocities Near the Speed of Light
F2.3 F2.1 F3About 65 minutes
Learning goals- I can explain why the Galilean rule u = u′ + v fails near the speed of light.
- I can use the relativistic velocity addition formula, including signs for direction.
- I can show that combining speeds below c always gives a speed below c, and that light always comes out at c.
- I can solve for an unknown relative velocity using the inverse formula.
Words to know
relativistic velocity addition
The rule u = (u′ + v)/(1 + u′v/c²) for combining an object's velocity u′ in frame S′ with the velocity v of S′ relative to S.
relative velocity
The velocity of one object as measured in the rest frame of another object.
speed limit c
No object with mass, and no signal carrying information, can travel at or faster than c relative to any inertial observer.
Fizeau experiment
An 1851 measurement of the speed of light in flowing water; the result matches relativistic velocity addition.
0.6c plus 0.6c
A spacecraft passes Earth at 0.60c. It launches a probe forward at 0.60c relative to the ship. How fast is the probe moving relative to Earth?
Galileo's rule says 0.60c + 0.60c = 1.20c, faster than light. But we already know from lessons 5 and 7 that the ship's clocks and rulers differ from Earth's. Velocity is distance ÷ time, so the rule for combining velocities must change too. The correct answer is 0.882c. Here is why, and how to calculate it.
The formula
Let frame S′ move at velocity v relative to frame S (along x). An object moves at velocity u′ in S′. Its velocity u in S is
u = (u′ + v) / (1 + u′v/c²)
- The numerator is the Galilean answer. The denominator is a correction that is always bigger than 1 when u′ and v point the same way.
- Signs matter. Choose a positive direction. A velocity in the opposite direction is negative.
- Low speeds: if u′ and v are much smaller than c, u′v/c² is tiny and u ≈ u′ + v. Galileo's rule is the low-speed limit, which is why it works so well in everyday life.
- Light: if u′ = c, then u = (c + v)/(1 + v/c) = c(c + v)/(c + v) = c. Postulate 2 is built in.
- Speed limit: if both |u′| < c and |v| < c, then |u| < c. You can never build up to c by adding speeds.
To find an unknown velocity in the moving frame, rearrange (the inverse formula):
u′ = (u − v) / (1 − uv/c²)
The formula comes from the Lorentz transformation equations, which combine time dilation, length contraction and the relativity of simultaneity; you will see them in first-year university physics.
Worked exampleExample 1: The probe
Find the probe's speed relative to Earth.
- Take forward as positive: v = +0.60c (ship relative to Earth), u′ = +0.60c (probe relative to ship).
- Numerator: u′ + v = 1.20c.
- Denominator: 1 + (0.60c)(0.60c)/c² = 1 + 0.36 = 1.36.
- u = 1.20c/1.36 = 0.882c.
u = 0.882c (2.65 × 10⁸ m/s), less than c.
Worked exampleExample 2: Firing backward
A ship moves at 0.80c relative to Earth and fires a probe backward at 0.60c relative to the ship. Find the probe's velocity relative to Earth.
- Forward positive: v = +0.80c, u′ = −0.60c.
- Numerator: −0.60c + 0.80c = +0.20c.
- Denominator: 1 + (−0.60)(0.80) = 1 − 0.48 = 0.52.
- u = 0.20c/0.52 = +0.385c.
0.385c forward relative to Earth (Galileo would say 0.20c).
Worked exampleExample 3: Two ships approaching
Two spaceships approach Earth from opposite sides, each at 0.70c relative to Earth. How fast does each ship's crew measure the other ship approaching?
- We want B's velocity in A's frame, so use the inverse formula. Let S be Earth and S′ be ship A, moving at v = +0.70c. Ship B has u = −0.70c in Earth's frame.
- B's velocity relative to A: u′ = (u − v)/(1 − uv/c²) = (−0.70c − 0.70c)/(1 − (−0.70)(0.70)) = −1.40c/1.49.
- u′ = −0.940c.
Each crew sees the other ship approaching at 0.940c, not 1.40c.
For a ship moving at v = 0.60c, the probe's speed relative to Earth (solid curve) bends below c no matter how fast the probe is launched. The Galilean line (dotted) passes c when u′ = 0.40c. At u′ = c the curve reaches exactly c.Worked exampleExample 4: How much does it matter at everyday speeds?
A train moves at 25 m/s and a passenger throws a ball forward at 10 m/s. By how much does the relativistic answer differ from 35 m/s?
- u = 35/(1 + (10)(25)/c²) = 35/(1 + 250/9.00 × 10¹⁶) = 35/(1 + 2.78 × 10⁻¹⁵).
- For small x, 1/(1 + x) ≈ 1 − x, so u ≈ 35(1 − 2.78 × 10⁻¹⁵).
- Difference: 35 × 2.78 × 10⁻¹⁵ ≈ 9.7 × 10⁻¹⁴ m/s.
The relativistic answer is smaller by about 1 × 10⁻¹³ m/s, completely unmeasurable. That is why Galilean addition served physics so well for 300 years.
Evidence: light in moving water
In 1851, Hippolyte Fizeau sent light through tubes of water flowing toward and away from the light, and measured the change in its speed with an interferometer. Light travels at c/n in still water, where n = 1.33. If the water flows at speed v in the same direction, Galilean addition predicts c/n + v. Fizeau found a smaller increase, about c/n + 0.44v. In the 1800s this was explained with a mysterious 'partial ether drag'.
Relativistic addition explains it with no ether at all: u = (c/n + v)/(1 + v/(nc)). For v ≪ c this is approximately c/n + v(1 − 1/n²), and for water 1 − 1/1.33² = 0.44. An old puzzle became a confirmation of special relativity.
Velocity-addition checklist
- Draw a sketch and pick a positive direction.
- Label v (frame S′ relative to S), u′ (object relative to S′), u (object relative to S).
- Use u = (u′ + v)/(1 + u′v/c²), or u′ = (u − v)/(1 − uv/c²) for the inverse.
- Work in units of c: write 0.60c as 0.60, and c² cancels.
- Check: is |u| < c? Is your answer between the Galilean result and c?
Just for funAdding 0.99c and 0.99c gives 0.99995c. Relativity is the only place where 99 + 99 is less than 100 and nobody gets marked wrong.
Hands-on activityTesting Galilean addition at home
You need: a toy car that rolls or drives steadily (a wind-up or battery car works best), a large piece of cardboard or a long baking sheet, a string, a metre stick, masking tape and a phone that can record video.
- Tape the metre stick to the floor alongside a clear path.
- Measure the toy car's speed u′ on the cardboard while the cardboard is still: time how long it takes to cover a measured distance (use the video and count frames, or use the phone's stopwatch).
- Have a partner pull the cardboard steadily along the path with the string, and measure the cardboard's speed v the same way.
- Now run the car forward on the moving cardboard. Measure its speed u relative to the floor.
- Compare u with u′ + v. Then calculate the relativistic correction u′v/c² for your speeds.
What to look for: u should be close to u′ + v, within your measurement error (often 5 to 10%). The relativistic correction is about 10⁻¹⁸ for speeds near 0.3 m/s, more than 15 orders of magnitude smaller than your measurement error. Explain why no experiment at everyday speeds could have revealed the need for relativity.
Your turn
1A ship at 0.50c (relative to Earth) launches a probe forward at 0.50c relative to the ship. Find the probe's speed relative to Earth.
u = 1.00c/(1 + 0.25) = 0.80c.
2A ship at 0.90c fires a missile forward at 0.90c relative to the ship. Find the missile's speed relative to Earth.
u = 1.80c/(1 + 0.81) = 1.80c/1.81 = 0.9945c.
3A ship at 0.50c turns on its headlight. Use the formula to find the light's speed relative to Earth.
u = (c + 0.50c)/(1 + 0.50) = 1.50c/1.50 = c.
4A ship at 0.80c fires a probe forward at 0.50c relative to the ship. Find u.
u = 1.30c/(1 + 0.40) = 0.929c.
5A ship at 0.60c fires a probe backward at 0.90c relative to the ship. Find u (forward positive).
u = (−0.90 + 0.60)c/(1 − 0.54) = −0.30c/0.46 = −0.652c (0.652c backward).
6Earth measures a particle at 0.90c. A ship moves at 0.60c in the same direction. How fast is the particle relative to the ship?
u′ = (0.90 − 0.60)c/(1 − 0.54) = 0.30c/0.46 = 0.652c.
7Why can't you reach c by launching a probe from a probe from a probe…, each at 0.9c relative to the last?
Each relativistic addition of two speeds below c gives a speed below c. Repeating the process gets closer and closer to c but never reaches it.
More practice: warm-up
1Write the relativistic velocity addition formula.
u = (u′ + v)/(1 + u′v/c²).
2What does the formula reduce to when u′ and v are much smaller than c?
u ≈ u′ + v (Galilean addition).
3Compute (0.25c + 0.25c)/(1 + 0.0625).
4What is u if u′ = c, for any v < c?
5In the inverse formula u′ = (u − v)/(1 − uv/c²), what does u′ represent?
The object's velocity relative to the moving frame S′.
More practice: on your own
1Two electrons move toward each other, each at 0.95c in the lab. What is the speed of one relative to the other?
u′ = (−0.95 − 0.95)c/(1 + 0.9025) = −1.90c/1.9025 = 0.9987c (in magnitude).
2A ship moving at 0.75c relative to Earth sees a comet approaching head-on at 0.85c (relative to the ship). What is the comet's speed relative to Earth?
Take the ship's direction as positive: v = 0.75c, u′ = −0.85c. u = (−0.85 + 0.75)c/(1 − 0.6375) = −0.10c/0.3625 = −0.276c (0.276c toward Earth's 'back' direction, opposite to the ship's motion).
3A particle moves at 0.95c in the lab. How fast must you travel in the same direction to see it moving at only 0.50c?
0.50 = (0.95 − v)/(1 − 0.95v) → 0.50 − 0.475v = 0.95 − v → 0.525v = 0.45 → v = 0.857c.
4Use the binomial approximation to show that u ≈ c/n + v(1 − 1/n²) for light in water flowing at speed v ≪ c.
u = (c/n + v)/(1 + v/(nc)) ≈ (c/n + v)(1 − v/(nc)) = c/n + v − v/n² − v²/(nc). The last term is negligible, so u ≈ c/n + v(1 − 1/n²).
5Fizeau's water flowed at about 7 m/s. By how much does relativity predict the light speed changes (n = 1.33)?
Δu ≈ v(1 − 1/n²) = 7 × 0.435 ≈ 3 m/s, out of about 2.26 × 10⁸ m/s. Fizeau's interferometer was sensitive enough to detect this.
Challenge
1Prove that if 0 < u′ < c and 0 < v < c, then u < c. (Hint: show that c − u > 0.)
c − u = c − (u′ + v)/(1 + u′v/c²) = [c + u′v/c − u′ − v]/(1 + u′v/c²) = [(c − u′)(c − v)/c]/(1 + u′v/c²). Both brackets (c − u′) and (c − v) are positive, and the denominator is positive, so c − u > 0, i.e. u < c.
2A rocket fires stage 1 to reach 0.50c relative to Earth, then stage 2 adds 0.50c relative to stage 1, then stage 3 adds 0.50c relative to stage 2. Find the final speed relative to Earth.
After stage 2: 1.00c/1.25 = 0.80c. After stage 3: (0.80 + 0.50)c/(1 + 0.40) = 1.30c/1.40 = 0.929c.
3Ship A moves at +0.60c and ship B at −0.60c relative to Earth. Earth sends a light pulse in the +x direction. What speed does each ship measure for the pulse? What would a Galilean physicist predict?
Relativistic: ship A, u′ = (c − 0.60c)/(1 − 0.60) = 0.40c/0.40 = c; ship B, u′ = (c + 0.60c)/(1 + 0.60) = c. Galilean: A would measure 0.40c and B 1.60c. Experiments agree with c for both.
Quiz
1. A ship moves at 0.60c and launches a probe forward at 0.60c relative to the ship. What is the probe's speed relative to Earth?
2. What does the relativistic velocity addition formula give when u′ = c?
3. When do the Galilean and relativistic rules give nearly the same result?
4. Two particles approach each other, each at 0.80c in the lab. What is their relative speed?
5. A ship at 0.80c fires a probe backward at 0.60c relative to the ship. What is the probe's velocity relative to Earth?
6. Which 1851 experiment, later explained by relativistic velocity addition, measured light speed in flowing water?
Big idea (F2.3): velocities do not simply add. The relativistic rule u = (u′ + v)/(1 + u′v/c²) keeps every combined speed below c and always gives c for light. At everyday speeds it reduces to ordinary addition.
How to help: Ask your teen to compute 0.5c + 0.5c both ways (Galilean 1.0c, relativistic 0.80c) and explain the difference. Ask: 'What happens if you add c to anything?'
Common difficulties:
- Sign errors: choose a positive direction first and give every velocity a sign.
- Confusing u, u′ and v. Write a word beside each: 'object vs Earth', 'object vs ship', 'ship vs Earth'.
- Forgetting to put c² in the denominator; working in units of c (0.60 instead of 0.60c) avoids this.
- Using the addition formula where the inverse formula is needed (finding a speed relative to a moving ship).