Clock thought experiment

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This is a thought experiment I came up with this evening. It must surely have been considered before but I couldn't find any reference and I've come up with it independently.

NASA has positioned a clock in Earth's thermosphere that can be viewed from outer space with a powerful enough telescope and activates it just as the time hits New Year's on 1 Jan 2020. Aliens live on planet XYZ one light year away from earth. They have been monitoring earth for a while and notice the clock as soon as it is activated. They send a being in a spaceship towards earth at close to the speed of light. The being constantly watches the clock as he approaches Earth (except when rotation of earth means the clock isn't visible to him). Since the light from the clock takes one year to reach planet XYZ, the aliens first notice the clock when it is 1 Jan 2021 on Earth, although they will see the time on the clock as 1 Jan 2020 on XYZ. The alien travelling towards the earth will take 1 year to reach it, so when he arrives to Earth, the clock will show 1 Jan 2022. He has been watching the clock go from 1 Jan 2020 to 1 Jan 2022 in the year it takes him to reach earth from the moment he first saw the time on the clock. My questions are, how does he observe two years' worth of clock ticks in the space of 1 year, and does he perceive the clock ticking as we do on Earth, ie one second in the space of one second?
 

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George Jones
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The alien sees the Earth's clock almost infinitely fast. Because of Lorentz contraction, the alien takes an almost infinitely short time (according to the alien) to get to the Earth.

In terms of mathematics, the alien sees the Earth's clock ticking faster (setting ##c = 1##) by the Doppler shift fact of

$$\sqrt{\frac{1+v}{1-v}}.$$

The time taken by the alien (according top the alien) for the trip is distance/speed, where the distance (according to the alien) is a Lorentz contraction of 1 light-year, i.e., ##\sqrt{1-v^2}## light-years. The time that the alien sees elapse on the Earth's clock is (rate of ticking seen) times time for trip. Putting this together gives

$$\sqrt{\frac{1+v}{1-v}} \frac{\sqrt{1-v^2}}{v} = \sqrt{\frac{1+v}{1-v}} \frac{\sqrt{\left(1-v\right) \left(1+v\right)}}{v} = \frac{1}{v} +1.$$

In the limit that ##v## approaches one (light speed), the time that the alien see elapse on the Earth's clock is 2 years.

Edit: Note that this agrees with the results according to people on Earth. According to Earth, there is no Lorentz contraction for the alien's trip, i.e., the aliens travel a distance of 1 (light-years), and the time (according Earth) for the alien's trip is distance/speed ##=1/v##. According to Earth, the alien's trip started one year after the clock turned on, so, again, ##1/v +1##.
 
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Thank you! Your answer is explained really well and makes complete sense to me. I've only read about relativity in popular science media but maybe I should pick up a textbook and study the details.
 

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