# Relativistic length contraction

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• leonid.ge
In summary, the observer will see the rod that goes along the rocket burning faster than the rod that oscillates across the rocket.

#### leonid.ge

Hello!
I have a question.
If there is a wooden rod which burns certain time, and an astronaut inside rocket lights two such rods: one oriented along the rocket's length and the other goes across the rocket, and an observer see the rocket passing by with a relativistic speed. Will the observer see that the rod which goes along the rocket burns faster because it's length contracted?

And if we have in the rocket two clocks which use spring pendulum, and one clock is oriented so its spring oscillates along the rocket and in the other clock it oscillates across the rocket - will the first clock go slower than the second (for the observer watching the rocket passing by with relativistic speed)?

No and no.

The first case is a bit trickier though as it cannot be solved without taking additional care about relativity of simultaneity. The rod oriented along the rocket will burn in different times depending on whether it burns front to back or back to front.

FactChecker and topsquark
leonid.ge said:
Will the observer see that the rod which goes along the rocket burns faster because it's length contracted?
No. Example burning back to front:
The rest-frame of the observer shall be the unprimed frame ##S##.
The rest-frame of the rocket the primed frame ##S'##.

##\Delta x'##:= rest-length of the rod, which goes along the rocket.
##\Delta t'##:= time of the burning with reference to the rocket's rest-frame.
##u'##:= velocity of the fire with reference to the rocket's rest-frame.
##u##:= velocity of the fire with reference to the observer's rest-frame.
##v##:= velocity of the rocket with reference to the observer's rest-frame.

Time of the burning (back to front) with reference to the observer's rest-frame, considering length-contraction:
##\Delta t = \frac{\Delta x'}{\gamma (u-v)} = \frac{\Delta x'}{\gamma}\frac{1}{\frac{u'+v}{1+u'v/c^2}-v} = \frac{\Delta x'}{\gamma} \frac{1+u'v/c^2}{u'(1-v^2/c^2)} = \gamma \frac{\Delta x'}{u'} (1+\frac{u'v}{c^2}) = \gamma(\Delta t' + \Delta x' \frac{v}{c^2})##
This is the inverse Lorentz-transformation for time.

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