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Is this possible or am I wrong?
So we have 2 light beams, A and B, pointed one towards the other, Now a particle of the A beam travels at the speed of light( c ), in the terrestrial refference; a particle of the B beam, also. Which means that a random particle travels a fixed distance in a time t.
But if we'd take the refference to be one of the beams, say A beam, then the Earth would travel at c and a particle of the B beam would travel twice c, right? which is not possible and, according to the physics laws(nothing can travel faster than c), which means than time would divide by 2 in order to decrease the speed by 2 (so the B particle travels at c). Now if time is divided by 2, then the time needed for a particle of B to travel the same distance d would be t/2.
Which means that a particle of B travels a fixed distance d in both t time and t/2 time.
How is this possible?
1 Answer
- oldprofLv 78 years agoFavorite Answer
Easy. Both space AND time adjust to keep C at C no matter what.
So when the speed of light is dS/dT = C in "normal" space time and ds/dt = C in adjusted space time because the speed of light is always...always...C, then when ds = 1/2 dS, dt must = 1/2 dT. QED.
In other words, time on the moving platform is running 1/2 the rate as normal time at the reference frame where the speed V of that platform is measured against. [See source.]
Source(s): dt = dT sqrt(1 - (V/C)^2) where V is the speed of the moving platform (e.g., spaceship), C is the invariant light speed, dT is the rate of time for the observer who sees the ship going V, and dt <= dT is the rate of time the observer sees pass on the platform. [STOR]