Time Dilation and the Lorentz Factor

Calculate the Lorentz factor ($\gamma$) for a spaceship traveling at $60\%$ the speed of light ($v = 0.6c$).

1. Identify the ratio: $\frac{v}{c} = 0.6$. 2. Square the ratio: $(0.6)^2 = 0.36$. 3. Subtract from 1: $1 - 0.36 = 0.64$. 4. Take the square root: $\sqrt{0.64} = 0.8$. 5. Find the reciprocal: $\gamma = \frac{1}{0.8} = 1.25$.

Result: Time for the traveler moves $1.25$ times slower than for a stationary observer.

A muon is a particle that lives for $2.2 \mu s$ ($t_0$) when at rest. If a muon travels at $0.99c$ (where $\gamma \approx 7.09$), how long does it live from the perspective of a laboratory scientist?

1. Identify proper time: $t_0 = 2.2 \mu s$. 2. Identify $\gamma$: $7.09$. 3. Apply the formula: $t = \gamma t_0 = 7.09 \times 2.2 \mu s$. 4. Calculate: $t \approx 15.6 \mu s$.

Result: The scientist sees the muon live over 7 times longer than it 'should' because of its high velocity.

An astronaut travels to a star 10 light-years away at $v = 0.95c$. How much time passes for (a) Earth and (b) the Astronaut?

1. Earth's perspective ($t$): $t = \frac{distance}{velocity} = \frac{10 \text{ ly}}{0.95c} \approx 10.53 \text{ years}$. 2. Calculate $\gamma$: $\gamma = \frac{1}{\sqrt{1 - 0.95^2}} \approx 3.20$. 3. Astronaut's perspective ($t_0$): Since $t = \gamma t_0$, then $t_0 = \frac{t}{\gamma}$. 4. Calculate: $t_0 = \frac{10.53}{3.20} \approx 3.29 \text{ years}$.

Result: While a decade passes on Earth, the astronaut only ages about 3.3 years.