4. Consider the following systems, a frame S’ is moving with respect to another frame S, with velocity u in the x direction. If a particle is moving in the S’ frame with velocity v also in x direction, then the particle velocity in the S frame is given by v= u+v . 1 + uv c2 (7)

The time in the S’ frame is t , so the time in the S frame is given by t=γ t + ux c2 , (9)

so the time change in the S’ frame will give a time change in the S frame as follow dt = γdt The acceleration in the S frame is given by a= a 1 dv = 3 dt γ 1 + uv c23.

1+

uv c2

.

(10)

(11)

If the S’ frame is the proper frame, then by de?nition the velocity v = 0. Substitute this to the last equation, we get a (12) a = 3. γ Combining Eq.(3) and Eq.(12), we get a = 5. Eq.(3) can also be rewritten as dβ g = 3 γdτ γ dβ g = 2 1?β c c β 1? β2 = gτ c (14)τ

F ≡ g. m

(13)

β 0

dτ0

ln

1 1? β2

+

(15)

gτ 1+β =ec 1?β

β e

gτ c

+ e?

gτ c

? e? gτ β = tanh . c =e

gτ c

gτ c

(16)

Relativistic Correction on GPS Satelitte

Page 2 of 10

Theoretical 2: SolutionRelativistic Correction on GPS Satelitte6. The time dilation relation is dt = γdτ. From eq.(16), we have γ= Combining this equations, we gett τ

(17) gτ . c

1 1 ? β2

= cosh

(18)

dt =0 0

dτ cosh

gτ c (19)

c gτ t = sinh . g c

Part B. Flight Time 1. When the clock in the origin time is equal to t0 , it emits a signal that contain the information of its time. This signal will arrive at the particle at time t, while the particle position is at x(t). We have c(t ? t0 ) = x(t) ? c t ? t0 = ? 1 + g t0 2 ? t= 2 1?gt0 c gt0 c

(20) gt c2

? ? 1? (21)

.

When the information arrive at the particle, the particle’s clock has a reading according to eq.(19). So we get c gτ t0 2 ? sinh = g c 2 1? 0= 1 2 cgt0 c gt0 c gt0 2

Theoretical 2: SolutionRelativistic Correction on GPS Satelitte2. When the particles clock has a reading τ0 , its position is given by eq.(6), and the time t0 is given by eq.(19). Combining this two equation, we get x= c2 g 1 + sinh2 gτ0 ?1 . c (24)

The particle’s clock reading is then sent to the observer at the origin. The total time needed for the information to arrive is given by c g c = g c t= g c τ0 = g t= The time will not freeze. Part C. Minkowski Diagram 1. The ?gure below show the setting of the problem. The line AB represents the stick with proper length equal L in the S frame.?β The length AB is equal to 1 L in the S’ frame. 1+β 2 The stick length in the S’ frame is represented by the line AC2

Theoretical 2: SolutionRelativistic Correction on GPS SatelitteAs for con?rmation, we can subsitute this relation to the second particle position to get c2 x2 = g2 1+ g2 t2 c2

?

c2 . g1

(50)

Second method: In this method, we will choose g2 such that the special point like the one descirbe in the question 1 is exactly the same as the similar point for the proper acceleration g1 . For ?rst particle, we have xp1 g1 = c2 For second particle, we have (L + xp1 )g2 = c2 Combining this two equations, we get g2 = g2 = c2c L+ g 1 g12