ELECTROMAGNETISM & RELATIVITY - Chapter 8, Exercise 1 Solution
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Lorentz Factor and Time Dilation at 0.6c

PROBLEM
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A particle travels at 0.6c. Calculate its Lorentz factor, and then
calculate how much time passes for a stationary observer if 5 s pass
in the particle's own rest frame.

SOLUTION
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Step 1: Calculate the Lorentz factor.

    gamma = 1 / sqrt(1 - v^2/c^2)
    v/c = 0.6, so (v/c)^2 = 0.36

    gamma = 1 / sqrt(1 - 0.36)
    gamma = 1 / sqrt(0.64)
    gamma = 1 / 0.8
    gamma = 1.25

Step 2: Calculate the dilated time using delta t' = gamma x delta t.

    delta t = 5 s (proper time, in the particle's own rest frame)
    delta t' = 1.25 x 5
    delta t' = 6.25 s

ANSWER: The Lorentz factor is 1.25, and 6.25 s pass for the
stationary observer while 5 s pass in the particle's own rest frame.

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WHY THIS WORKS AS AN ANSWER

At 0.6c, the Lorentz factor comes out to a clean 1.25 because
1 - 0.6^2 = 0.64 is itself a perfect square (0.8^2), a common choice
in worked physics examples for exactly this reason. The result
confirms the chapter's own central claim: time genuinely runs slower
for the moving particle from the stationary observer's own
perspective - 5 s of "particle time" corresponds to a longer 6.25 s
of "observer time," a real, measurable 25% difference at this speed,
smaller than the chapter's own 0.8c example (which gave a roughly 67%
difference) since time dilation grows more dramatically as speed
approaches c.