So here's a scenario I made up and I want to know if it is correct or not.
I'm was standing on the earth, l$_0$ meters away from a stationary rocket in space. The rocket then starts travelling towards me at a constant speed v. From my frame of reference the distance l$_0$ will be contracted and hence I will observe the total distance between me and the rocket to be $\frac{l_0}{\gamma}$. From the frame of reference of the rocket the distance observed will also be $\frac{l_0}{\gamma}$.
Is this correct? If so then what if the rocket was travelling away from me at the same speed, how would length be contracted in such scenario?
Also, If the the rocket was travelling to the moon and the distance between the rocket and the moon initially (i.e. when me, the rocket, and the moon are stationary) is x$_0$. What would I observe the distance x$_0$ to be?
Answer
The length $l_0$ exists in your reference frame, so you may measure it without any length contraction. The rocket does not jump towards you due to length contraction when it starts moving quickly (it will move toward you, though). The length that exists in the rocket's moving frame is the length of the rocket itself. The rocket will appear to get shorter in your frame when it starts moving quickly.
From the rocket's perspective, its own length does not change since that length exists in its frame, in which it is at rest. However, space outside of the rocket would warp when the rocket suddenly started moving at a great speed, and you would appear to get closer due to length contraction in addition to the rocket's new velocity closing the gap over time.
In conclusion, $l_0$ does not contract in the Earth's reference frame, but it does in the rocket's frame.
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