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2. Solution to Thepretical Problem 2


Theoretical 2: Solution
Relativistic Correction on GPS Satelitte
Part A. Single accelerated particle 1. The equation of motion is given by F = = d (γmv ) dt ˙ mcβ
3

(1)

(1 ? β 2 ) 2 F = γ 3 ma, where γ = √
1 1?β 2

(2)

and β = v c . So the acceleration is given by a= F . γ3m (3)

2. Eq.(3) can be rewritten as dβ F = 3 dt γ m t dβ F = dt 3 mc 0 (1 ? β 2 ) 2 c β 1 ? β2 = Ft mc
Ft mc Ft 2 mc

β 0

(4) . (5)

β=

1+ 3. Using Eq.(5), we get
x t

dx =
0 0

F tdt m 1+ ? ? 1+
Ft 2 mc

x=

mc2 F

Ft mc

2

? ? 1? . (6)

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)

Relativistic Correction on GPS Satelitte

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Theoretical 2: Solution
Relativistic Correction on GPS Satelitte
If the particles velocity changes with respect to the S’ frame, then the velocity in the S frame is also change according to dv = dv = dv u+v ? uv 1 + c2 1 + uv c2 1 dv 2 γ 1 + uv c2
2. 2

udv c2 (8)

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 c2
3.

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


0

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)

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Theoretical 2: Solution
Relativistic Correction on GPS Satelitte
6. The time dilation relation is dt = γdτ. From eq.(16), we have γ= Combining this equations, we get
t 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 c
2

? ? 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 c
gt0 c gt0 c gt0 2

gt0 gτ 1 + sinh c c gt0 gτ gτ = 1 + sinh ± cosh . c c c ?

+ sinh

gτ c (22)

Using initial condition t = 0 when τ = 0, we choose the negative sign gt0 gτ gτ = 1 + sinh ? cosh c c c c ? gτ t0 = 1?e c . g
c As τ → ∞, t0 = g . So the clock reading will freeze at this value.

(23)

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Theoretical 2: Solution
Relativistic Correction on GPS Satelitte
2. 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 AC
2

sinh

gτ0 x + c c gτ0 gτ0 sinh + cosh ?1 c c e
gτ0 c

(25)

?1

(26) (27)

ln

gt +1 . c

3. The!position!of!the!particle!is!given!by!eq.!(5).!! Figure 1: Minkowski Diagram ! ! ! !! AB ! AC = = 1 ? β 2 L. ! cos θ ! 2. The position of the particle is given by eq.(6). ! ! ! ! Relativistic Correction on GPS Satelitte ! ! ! !

! ! ! ! ! ! ! ! ! ! ! !

ct#

ct’#

C! !! A! !! B! x#

x’#

(28)

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Theoretical 2: Solution
Relativistic Correction on GPS Satelitte
!! !" ! !! ! !"′! ! ′!

!! ! ! !! !

!

Figure 2: Minkowski Diagram Part D. Two Accelerated Particles 1. τB = τA . 2. From the diagram, we have tan θ = β = ct2 ? ct1 . x2 ? x1

(29)

Using eq.(6), and eq.(19) along with the initial condition, we get gτ1 c2 cosh ?1 , g c c2 gτ2 x2 = cosh ? 1 + L. g c x1 = Using eq.(16), eq.(19), eq.(30) and eq.(31), we obtain gτ1 tanh = c L+ = gL gτ1 sinh c2 c gL gτ1 sinh c2 c So C1 =
gL . c2 gL c2

(30) (31)

c
c2 g

c g

2 sinh gτ c ?

c g

1 sinh gτ c

2 cosh gτ c ?1 ?

c2 g

1 cosh gτ c ?1

gτ1 2 sinh gτ c ? sinh c

gτ1 2 + cosh gτ c ? cosh c gτ2 gτ1 gτ2 gτ1 = sinh cosh ? cosh sinh c c c c g = sinh (τ2 ? τ1 ) . c

(32)

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! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !

gτ 2 gτ ? sinh 1 c c ! !! = gL gτ 2 gτ 1 + cosh ? cosh c2 c c gL gτ gτ gτ gτ gτ sinh 1 = sinh 2 cosh 1 ? cosh 2 sinh 1 2 c c c c c c Theoretical 2: Solution Using!identity!relation,!the!last!equation!is!simply!to! gL gτ g Relativistic ! (15)! sinh 1 = Correction sinh (τ 2 ? τ 1 ) !! on GPS Satelitte 2 c c c sinh
!!

t2!

t1!

!!

!! x1! x2!

! Figure 3: Minkowski Diagram for two particles 3. From!the!length!contraction,!we!have! x ?x L' = 2 1 3. From the length contraction, we have γ1 ! !! x2?? x1 dτ ? d L dx dx 1 x ? x d γ ′ 2 2 1 2 1 1 L == ? ? 1τ dτ dτ 1 ? dτ 1 ? γ 12 dτ 1 ?γd ? γ1 2 1
dL = dτ1 dx2 dτ2 dx1 ? dτ2 dτ1 dτ1 1 x2 ? x1 dγ1 ? . 2 γ1 dτ1 γ1

(33) (34)

Take derivative of eq.(30), eq.(31) and eq.(32), we get gτ1 dx1 = c sinh , dτ1 c dx2 gτ2 = c sinh , dτ2 c gL gτ1 g cosh = cosh (τ2 ? τ1 ) 2 c c c The last equation can be rearrange to get
gL 1 cosh gτ dτ2 c c2 = + 1. dτ1 cosh g c (τ2 ? τ1 )

(35) (36) dτ2 ?1 . dτ1 (37)

(38)

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Theoretical 2: Solution
Relativistic Correction on GPS Satelitte
From eq.(29), we have x2 ? x1 = c c(t2 ? t1 ) = 1 β1 tanh gτ c c gτ2 c gτ1 sinh ? sinh g c g c . (39)

Combining all these equations, we get dL1 = dτ1
gL gτ gτ2 c2 cosh c1 gτ1 gτ2 c sinh ? c sinh + c sinh c cosh g c c ( τ ? τ ) 2 1 c

1 1 cosh gτ c

?

c2 gτ2 gτ1 sinh ? sinh g c c

1 1 gτ1 tanh c cosh2

gτ1 c

g gτ1 sinh c c (40)

2 sinh gτ dL1 gL c . = dτ1 c cosh g c (τ2 ? τ1 )

So C2 =

gL c .

Part E. Uniformly Accelerated Frame 1. Distance from a certain point xp according to the particle’s frame is L = L = x ? xp γ
c2 g1 τ cosh g1 c ? 1 ? xp τ cosh g1 c
2

(41)

c c2 g + xp L = ? 1 g1 τ . g1 cosh c c For L equal constant, we need xp = ? g . 1
2

(42)

2. First method: If the distance in the S’ frame is constant = L, then in the S frame the length is 1 + β2 Ls = L . (43) 1 ? β2 So the position of the second particle is x2 = x1 + Ls cos θ ? 2 c ? g1 t1 1+ = g1 c x2 = c2 +L g1 1+ (44)
2

? ? 1? + L 1 + g1 t1 c (45)

g1 t1 c

2

?

c2 . g1

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! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !
2

!
2 Theoretical ?c ? ? g t ?2: cSolution x2 = ? + L ? 1 + ? 1 1 ? ? !! ? Correction c ? g1 ? g1 ? Relativistic on GPS 2

Satelitte

(17)!

!

t2! L" t1! !!

! x1! x2! ! Figure 4: Minkowski Diagram for two particles the!time!of!the!second!particle!is! ct 2 = ct1 + LS sin θ
The time of the second particle is

? ? g1t1 ? ? ct2 = ct? + Ls sin cθ 1L ? 2 ? ? 2 ? 1tβ 1? 1g + ? ? β 1+ ? ? = ct + L 1 ? ? 2 ! !! + β 2 c ? 1 ? β ?1 = ct1 + ? ? g1 L ct2 = t1 ? c + . ? c 1 ? ? 2 ? ? Substitute eq.(47) to eq.(45) to get g1t1 ? ? 1+ ? ? ? ? c ? ? ? ?

(46)

(47)

!

x2 =

c2 g1 L g ? 1+ ? 1 t2 + L = ct 1 + L ? ? 2 1 2 gct c !!1 + g12 1 ? c ?
c

2

?
2

c2 g1 c2 . g1

(18)!
(48)

Substitute!eq.(18)!to!eq.(17)!to!get! 2
x2 =

c +L g1

1+

g1 1+
g1 L c2

t2 c

?

From the last equation, we can identify g2 ≡ g1 1+
g1 L c2

.

(49)

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

?

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 = c2
c L+ g 1 g1
2

1+

g1 L c2

.

(51)

3. The relation between the time in the two particles is given by eq.(47) t2 = t1 1 + c2 g2 τ2 sinh g2 c g2 τ2 sinh c g2 τ2 dτ2 dτ1 Part F. Correction for GPS 1. From Newtons Law GM m = mω 2 r r2 r= gR2 T 2 4π 2
1 3

g1 L c2 c2 g1 L g1 τ1 = 1+ 2 sinh g1 c c g1 τ1 = sinh c = g1 τ1 g1 g1 L = =1+ 2 . g2 c

(52) (53)

(54) (55)

r = 2.66 × 107 m. The velocity is given by v = ωr = 2πgR2 T
1 3

(56)

= 3.87 × 103 m/s.

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Theoretical 2: Solution
Relativistic Correction on GPS Satelitte
2. The general relativity e?ect is dτg =1+ dt dτg =1+ dt After one day, the di?erence is ?τg = gR2 R ? r ?T c2 Rr = 4.55 × 10?5 s. (59) ?U mc2 gR2 R ? r . c2 Rr (57) (58)

The special relativity e?ect is dτs = dt = 1? 1? v2 c2 2πgR2 T 2πgR2 T
2 3

(60) 1 c2 1 . c2 (61)

1 ≈1? 2 After one day, the di?erence is 1 ?τs = ? 2

2 3

2πgR2 T

2 3

1 ?T c2

(62)

= ?7.18 × 10?6 s. The satelite’s clock is faster with total ?τ = ?τg + ?τs = 3.83 × 10?5 s. 3. ?L = c?τ = 1.15 × 104 m = 11.5km.

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