Remember the answer that broke everything — 0.7c + 0.5c = 1.2c, faster than light? This is where we fix it. When several things move at speeds near c, you can’t just add velocities any more. Instead you use the relativistic velocity addition equation, which cleverly bends every answer so it always stays below the cosmic speed limit.
📘 What you need to know
Galilean velocity addition (u = u′ + v) only works when speeds are much less than c
For speeds near c, use the relativistic velocity addition equations instead
Stationary frame: u = (u′ + v) ÷ (1 + u′v/c2)
Moving frame: u′ = (u − v) ÷ (1 − uv/c2)
The result is always less than c — nothing can be pushed past the speed of light
If one speed is c (light), the answer comes out as exactly c
The signs on top and bottom always match: both minus, or both plus
Why simple addition fails
On the Galilean page, adding velocities was just arithmetic: a ball thrown forward on a moving train adds the train’s speed to the ball’s. That works perfectly for trains and swimmers. But push the speeds up near c and it produces nonsense.
Take the rocket from the postulates page: flying at 0.7c, it releases a probe forward at 0.5c relative to itself. Simple addition claims a stationary observer sees 1.2c — impossible. The relativistic equation gives a very different, perfectly legal answer: 0.89c. Close to light speed, but never over it.
Simple addition overshoots the speed limit (1.2c). The relativistic equation pulls the answer back to 0.89c — fast, but safely below c.
Here’s the intuition for the weird-looking formula. The top of the fraction is just the old Galilean sum, u′ + v. The bottom is a “brake” that only kicks in when speeds get large: at low speed u′v/c2 is almost zero, so the brake does nothing and you’re back to simple addition. Near c the brake grows and holds the answer under the limit. Same equation, self-correcting.
The velocity addition equations
Let’s fix the labels, exactly as before: u is the object’s velocity in the stationary frame S, u′ is its velocity in the moving frame S′, and v is the speed of the moving frame itself. Picture Person E on Earth and Person F flying past in a rocket, who fires a missile forward — each measures the missile at a different speed.
Person F fires a missile forward at u′ (in the rocket frame). Person E on Earth measures it at u. Relativistic velocity addition connects the two.
To find the speed in the stationary frame (what E measures) from the moving-frame speed:
Velocity in the stationary frameu = (u′ + v) ÷ (1 + u′v/c2)
And to go the other way — finding the speed in the moving frame (what F measures) from a stationary-frame speed — both signs flip from plus to minus:
Velocity in the moving frameu′ = (u − v) ÷ (1 − uv/c2)
Notice the sign discipline: the sign on the top and the sign on the bottom are always the same. If it’s a plus on top, it’s a plus underneath. Get one to disagree and the answer is wrong.
u′ moving frame
+ v ÷ brake
u stationary frame
< c always
✓ under limit
WE 1
A spaceship travels at 0.60c relative to the ground and launches a shuttle forward at 0.50c relative to the ship. How fast does the ground see the shuttle moving?
Step 1 — identify: u′ = 0.50c, v = 0.60c, want u
u = (u′ + v) ÷ (1 + u′v/c²)
Step 2 — substitute (c² cancels with c-units)u = (0.50 + 0.60) ÷ (1 + 0.50×0.60)u = 1.10 ÷ (1 + 0.30) = 1.10 ÷ 1.30u = 0.85cGalilean would say 1.10c — impossible. The brake pulls it back to a legal 0.85c.
WE 2
A rocket moves at 0.60c relative to the ground. A probe is released from the back of the rocket at 0.82c relative to the rocket (i.e. backwards). What is the probe’s velocity relative to the ground?
Step 1 — backwards means negative: u′ = −0.82c, v = 0.60c
u = (u′ + v) ÷ (1 + u′v/c²)
Step 2 — substitute (signs match top and bottom)u = (−0.82 + 0.60) ÷ (1 + [−0.82]×0.60)u = −0.22 ÷ (1 − 0.492) = −0.22 ÷ 0.508u = −0.43cThe probe still moves backward relative to the ground — the rocket’s forward speed isn’t enough to reverse it.
WE 3
Ship A moves at 0.80c relative to the ground. A second ship, B, moves at 0.30c in the same direction. How fast does A move relative to B?
Step 1 — B is the moving frame: v = 0.30c, u = 0.80c, want u′
u′ = (u − v) ÷ (1 − uv/c²)
Step 2 — substituteu′ = (0.80 − 0.30) ÷ (1 − 0.80×0.30)u′ = 0.50 ÷ (1 − 0.24) = 0.50 ÷ 0.76u′ = 0.66c“Relative to B” means B is the observer, so we use the u′ equation (the moving-frame one).
WE 4
Two ships approach head-on. The ground measures ship P moving east at 0.70c and ship Q moving west at 0.70c. What is the velocity of Q as measured by P?
Step 1 — set east +ve. Q’s ground speed u = −0.70c; frame P has v = +0.70c
u′ = (u − v) ÷ (1 − uv/c²)
Step 2 — substitute (mind the double negatives)u′ = (−0.70 − 0.70) ÷ (1 − [−0.70]×0.70)u′ = −1.40 ÷ (1 + 0.49) = −1.40 ÷ 1.49u′ = −0.94cGalilean would give −1.40c — impossible. Even two near-light ships closing head-on stay under c relative to each other.
Light always comes out as c
Here’s the most beautiful check of all. Put light itself into the equation — set u′ = c for a beam fired from the moving frame — and no matter what v is, the answer is exactly c. For example with v = 0.7c: (c + 0.7c) ÷ (1 + 0.7) = 1.7c ÷ 1.7 = c. The equation was practically built to protect the second postulate.
Quick recap: Relativistic velocity addition replaces simple addition near c: u = (u′ + v) ÷ (1 + u′v/c2). The dividing “brake” keeps every result below c, and light always transforms to exactly c.
🛠️ Solving a velocity addition problem
Label everything. Which speed is u (stationary frame), u′ (moving frame), and v (frame speed)?
Which equation? Want the stationary-frame speed → use u. Want a moving-frame speed → use u′.
Set a positive direction and give opposite motions a minus sign before substituting.
Match the signs. Top and bottom always carry the same sign (both + or both −).
Sanity-check. The magnitude must come out below c. If it doesn’t, hunt for the slip.
💡 Top tips
Keep speeds in units of c. The c2 in the denominator cancels, so answers land neatly in c.
Signs match. The operator on top mirrors the one on the bottom — always.
“Relative to X” = X’s frame. That tells you whether you want u or u′.
Answer must be under c. A result of c or more is a guaranteed error — unless the object is light.
Low speeds? The brake vanishes and you’re back to plain u = u′ + v.
⚠ Common mistakes
Using simple addition (u′ + v) at relativistic speeds — that only holds when v « c
Letting the top and bottom signs disagree — they must match
Mixing up u and u′, or the wrong equation for the frame asked about
Dropping a minus sign for objects moving in the opposite direction
Accepting an answer of c or greater for an ordinary object — impossible
That’s the last piece of “how frames convert into each other” — positions, times, and now velocities all handled relativistically. From here we stop converting and start exploring the consequences. Next comes the space-time interval: a quantity that, remarkably, every observer agrees on even while they argue about distances and times. It’s the bedrock the famous effects are built on.
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