Time Dilation
Moving clocks run slow.
Formula
Variables
How it works
This equation says that the time interval measured by an observer who sees a clock moving at speed v is longer than the interval measured by someone moving with the clock. The quantity \u0394t0 is the proper time, meaning the shortest possible time interval between two events measured in the frame where both events happen at the same place. The factor 1/\u221a(1 - v^2/c^2) comes from special relativity: space and time mix so that all inertial observers agree on the speed of light, which forces moving clocks to be observed as ticking more slowly.
When to use it
Use this when an object, particle, or spacecraft is moving at a significant fraction of the speed of light and you need the time interval seen in a different inertial frame. It is common in particle physics, astrophysics, and relativistic travel problems.
Worked example, step by step
A muon has a proper lifetime of \u0394t0 = 2.20 \u00b5s in its own rest frame and is moving at v = 0.95c relative to the lab. Step 1: Write the time dilation formula: \u0394t = \u0394t0 / \u221a(1 - v^2/c^2) Step 2: Substitute the speed: \u0394t = 2.20 \u00b5s / \u221a(1 - 0.95^2) Step 3: Square 0.95: 0.95^2 = 0.9025 Step 4: Find the denominator: \u221a(1 - 0.9025) = \u221a(0.0975) \u2248 0.312 Step 5: Divide: \u0394t = 2.20 \u00b5s / 0.312 \u2248 7.05 \u00b5s Final answer: The lab measures the muon to live for about 7.05 microseconds.
Common mistakes to avoid
- Using the moving clock's measured time as \u0394t0 instead of the proper time
- Forgetting that v must be compared with c, so the formula uses v^2/c^2, not just v^2
- Applying the formula at everyday speeds where the effect is so tiny that it is usually negligible
Did You Know?
GPS satellites must correct for time dilation, or their clocks would drift enough to ruin navigation within a day.
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