Time dilation in a 9mm bullet

TIME DILATION IN A 9mm BULLET

Consider a bullet of diameter 9mm [0.3543 in] and mass 7.5 g [0.2645 oz]. When the bullet is shot from the gun, it lasts only for 5 second before finally touching the ground. The average velocity of the 9mm bullet from the gun is 300 m/s which is approximately 0.874M [Mach]. According to the theory of Special Relativity, the bullet will undergo changes in length, mass and time. The length contracts while time and mass increase. The change in length and mass are negligible, but the change in time is considerable.

The time dilation is useful as it increases the life of bullet slightly greater than 5 second.

Let the time measured by the bullet itself be T = 5 second

The bullet’s clock will move slower than the observer’s clock. The time measured by an outside observer is given by,

T’ = T / [√1-v²/c²]

T – Proper time measured by the bullet itself

T’ – Time measured by an outside observer

c – Speed of light [3*10⁸] m/s

v – Velocity of bullet, v = 300 m/s

Substituting all values in the above equation, we get the observer’s time to be

T’ = 5.0000000000025 s

The time difference is T’ – T = 2.5picosecond

Thus when the bullet is shot it gains roughly 2.5picosecond before it finally hits the target.

DISTANCE TRAVELED:

The distance traveled by the bullet without considering time dilation is,

Distance x = speed of bullet [v] * time T

               x = 300*5

               x = 1500m

The distance traveled by the bullet considering time dilation is,

Distance x = speed of bullet [v] * time T’

               x = 300*5.0000000000025

True distance traveled x = 1500.00000000075m

Thus the bullet travels an additional 0.75nanometer due to time dilation.