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QPage 458: Q11: HC Verma: Concepts of Physics - Volume 2 The Special Theory of Relativity Page 458: Q11 Solution: Suppose Swarglok heaven) is in constant motion at a speed of…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently invert the Lorentz factor by applying $\Delta t_0 = \gamma \Delta t$ instead of $\Delta t = \gamma \Delta t_0$, causing them to calculate a shorter time for the Earth frame. Always identify the proper time $\Delta t_0$ (measured in the rest frame of the moving object, Swarglok) and multiply by $\gamma$ to find the dilated time in the stationary frame (Earth).

Suppose Swarglok heaven) is in constant motion at a speed of 0.9999c with respect to the earth. According to the earth's frame, how much time passes on the earth before one day passes on Swarglok?

Step-by-Step Verified Solution

**Answer:** Time on earth = 70.7 days. **Step-by-Step Explanation:** v=0.9999cv = 0.9999c. γ=1/1−0.99992=1/1−0.9998=1/0.0002=70.7\gamma = 1/\sqrt{1 - 0.9999^2} = 1/\sqrt{1 - 0.9998} = 1/\sqrt{0.0002} = 70.7. Time dilation: Δt=γΔt0=70.7×1\Delta t = \gamma \Delta t_0 = 70.7 \times 1 day = 70.7 days.

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