The formula for time dilation in special relativity is given by:
Δt' = Δt / √(1 - (v^2 / c^2))
Where: Δt' is the time interval measured in the moving frame of reference (where time appears dilated), Δt is the time interval measured in the stationary frame of reference, v is the relative velocity between the two frames of reference, and c is the speed of light in a vacuum, which is approximately 299,792,458 meters per second.
In your scenario, if one person sees twice as much time pass compared to another person, it means that the time dilation factor (Δt' / Δt) is 2. Let's assume that the two observers are in relative motion, and the faster observer measures a speed of v1, while the slower observer measures a speed of v2.
We can set up the equation as follows:
2 = Δt / Δt' = √(1 - (v1^2 / c^2)) / √(1 - (v2^2 / c^2))
To solve this equation, we need to know the values of v1 and v2, as well as the speed of light c. Please provide the specific values for the velocities, and I can assist you in calculating the corresponding time dilation.