狹義相對論問題(請用中文回答)20點

Let v1, v2 be the velocities of the jets relative to the ground, and c be the speed of light. A moving clock runs slower than a stationary one. And they are related by time dilation formula
T = t / Sqrt(1 - v^2/c^2)

1.
When observed on the ground,
let t be a time interval of the ground clock, T1 and T2 be the time intervals of the jets.
T1 = t / Sqrt( 1 - v1^2/c^2 )
T2 = t / Sqrt( 1 - v2^2/c^2 )

2.
When observed in one of the jets, say the 1st one,
velocity of the ground = v1
velocity of the 2nd jet = u = (v1 + v2) / (1 + v1v2/c^2)

Let t1 be the time interval observed in jet 1. Let T and T2 be time intervals on the clocks on the ground and jet 2.
T = t1 / Sqrt( 1 - v1^2/c^2 )
T2 = t1 / Sqrt( 1 - u^2/c^2 )

The crucial point is that the observed phenomenon is called "time dilation", whcih is a result of the constant speed of light to eah observer irrespective they are at rest or moving at a uniform speed. The "constant speed of time" is one of the postulates upon which the Theory of Special Relativity is built up.

是不是兩個原子鐘各自比較?
還是與地面的鐘作比較?
我想這和地球自轉無關, 只考慮兩個鐘的相對運動