TIME
DILATION IN MOUNT EVEREST
INTRODUCTION
The Earth performs
circular motion once every 24 hours as it rotates about its own axis. Hence
every object bound to the planet performs circular motion with the same angular
velocity ω as that of Earth. Consider the tallest mountain in the world ‘Mount
Everest’ which is 8848 m above sea level. This mountain also performs circular
motion. Consider three observers one on ground level, one at the center of
earth and another on the top of the mountain. The observer on the top performs
a circular motion of greater radii with respect to the ground level in the same
time which means the top observer is moving faster than ground level observer.
The ground level observer also performs a circular motion relative to the
observer in center of Earth. So according to the Special Theory of Relativity,
a clock at the top of mountain would run slower than that on the ground level.
We’ll find the time gained by top observer and ground level observer relative
to the observer in the center of Earth.
ASSUMPTIONS
- Earth is a perfect homogenous sphere.
- The effect of Gravitational time dilation is negligible.
CALCULATION
The Angular velocity is
given by,
ω = 2π/T [rad/s]
T – Rotational time
period [T ≈ 24 hours = 86400s]
ω = 2*3.14/86400
ω = 7.2722*10-5
rad/s
The angular velocity is
same at all points on earth since it doesn’t exhibit differential rotation. But
the tangential velocity on surface varies with the distance from the center.
The greater the distance, higher is the velocity. It is important to note that
since ground level observer is also performing a circular motion, he will also
experience time dilation relative to observer at the center of Earth.
The tangential velocity
of observer at center of Earth is given by,
v = [R]*ω [m/s]
R – Average radius of
earth [R = 0 m]
v = 0 m/s
The tangential velocity
of ground observer is given by,
v = [R]*ω [m/s]
R – Average radius of
earth [R = 6371000m]
v = 6371000*7.2722*10-5
v = 463.3118 m/s
The tangential velocity
of mountain summit observer is given by,
v = [R + h]*ω [m/s]
R – Average radius of
earth [R = 6371000m]
h – Height of the
mountain [h = 8848 m]
v = 6379848*7.2722*10-5
v = 463.9553 m/s
According to the
Special Theory of Relativity, the time dilation equation is,
t' = t/γ [s]
t’– Mountain Top observer’s time. [s]
t1 – Ground
level observer’s time. [s]
t2 -
γ – Relativistic gamma
factor, γ = 1/√ [1-(v/c) 2]
c - Velocity of light
[c = 3*108 m/s]
t' = t*√ [1-(v/c)
2]
Center of Earth
observer time
t2 = t since v = 0
Ground level observer
time
t1 = t*√
[1-2.3850*10-12]
t1 = t*√ [0.999999999997615]
t1 = t* 0.999999999998808
Mountain top observer
time
t' = t*√
[1-2.3916*10-12]
t' = t*√ [0.999999999997608]
t' = t* 0.999999999998804
CONCLUSION
We can observe that ground level and top
observer’s time aren’t the same with respect to center of Earth observer’s time
which proves that time dilates at a height. We’ll consider 5 different t2 values and calculate t1 and t' values. The larger the t2 the more is the difference between t1, t2 and t’. Thus
mountain top observer will gain some time over ground level and maximum time
over center of earth observer. In fact mountain top observer will gain 37.7
microsecond in one year with respect to center of Earth observer.
Time
|
t2 [Center of Earth observer] (s)
|
t1
[Ground level observer] (s)
|
t' [Mountain Top observer] (s)
|
||
1
minute
|
60
|
|
|
||
1
hour
|
3600
|
|
|
||
1
day
|
86400
|
|
|
||
1
month
|
2592000
|
|
|
||
1
year
|
31536000
|
|
|
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