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Around the earth with one clock, and another clock stays put

  1. Nov 4, 2014 #1
    Hi all

    I just started doing some research on special relativity and I figured something.

    If two people both reset their watches. One person goes around the earth once, doesn't matter at what speed, and the other person stays there. The watch of the person that went around the earth will run 0.134 seconds earlier.

    Is this correct?
     
  2. jcsd
  3. Nov 4, 2014 #2

    jedishrfu

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  4. Nov 4, 2014 #3

    A.T.

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    It does matter how fast they go, and even which direction they go around.
     
  5. Nov 4, 2014 #4

    Nugatory

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    How did you calculate this? It's not right, but without seeing your calculations we won't be able to help you spot the error in them.
     
  6. Nov 4, 2014 #5
    Sorry, I'm realy new to this.
    Also, excuse me for my bad English...

    Let's assume going around the earth is just 25 000 miles.
    So for light to go around the earth it would take 25 000 miles / ( 186 000 miles / second) = 0.134 seconds.


    What I don't understand is that you say the speed that you travel at is important in this example.

    Let's say if he goes around the earth at 1/2 of the speed of light. His watch will go half as slow relativily to the person standing still for 0.134 * 2 seconds. So his watch should run 0.134 seconds earlier than the watch of the person standing still.

    If he goes around the earth at any other speed, the total time difference will stay 0.134 seconds.

    Let's say he goes at 1/4 of the speed of light. His watch will go 1/4 slower relativily to the watch of the person standing still for 0.134 * 4 seconds. Which again is 0.134 seconds.

    So, what is wrong with this that the speed at which you travel does have an impact on the time difference.

    It's probably a stupid question...
     
  7. Nov 4, 2014 #6

    Nugatory

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    You have the wrong relationship between speed and time dilation. If the distance travelled is ##L## at a speed ##V## according to the earthbound observer, the time taken in the trip will be ##T=L/V## and the time elapsed on the traveller's watch will be ##\sqrt{T^2-(L/c)^2}##.

    (Actually the above contains a few simplifying assumptions, but it's good enough to show the that the relationship is more complicated than just "half the speed, half the time dilation" as you are assuming).
     
    Last edited: Nov 4, 2014
  8. Nov 4, 2014 #7
    Thanks a lot Nugatory. They didn't seem to mention that in the documentary video's I watched.

    So basically, the 0.134 seconds difference is only correct if the speed you travel = c.

    I know it's technically impossible to travel faster than c because of the mass, but I heard, IF you can, you'd be able to go back in time.
    According to that formula that's not possible.
    http://www.mathsisfun.com/data/func...800&ymin=-0.2122&ymax=0.4413&aval=2.000&uni=0

    Anyway , my problem is solved :p
    Thanks guys!
     
  9. Nov 4, 2014 #8

    pervect

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    This is correct, but a bit terse.

    Unfortunately the problem is not correctly solved yet :(, for the original poster has misinterpreted the answers.

    It takes about .133 seconds for light to travel around the Earth's equator. But that number isn't the answer to the OPs question.

    Ignoring the time dilation and other effects due to GR, we can consider the time dilation depends on the bodies total velocity in the special-relativity--like earth centered inertial frame. We conclude the time dilation of the clock moving in the same direction as the Earth's rotation caries it is greater than the time dilation of the clock moving in the opposing direction.

    Now we can ask - what happens in the limit of low and high velocities. The high velocity limit is that the proper time on the carried clock approaches zero, in the limit as the velocity approaches infinity it takes no proper time to travel around the Earth. So lets look at the low velocity limit which should be more interesting.

    If we take a series expansion of the well-known SR time dilation formula ##1/ \sqrt{1-( \frac{u+v}{c} )^2} ## as a series in v , with u being some additional velocity due to the Earth's rotation, we get

    ##\approx \frac{1}{\sqrt{1-u^2/c^2}} - \frac{uv}{c^2}## when u << c


    In this low velocity limit , if we let ##r_e## be the radius of the earth and ##T_e## be the period of rotation of the Earth (a sidereal day,if one wishes to be precise), and multiply the difference in time dilation by the trip time, we get for the total discrepancy

    [tex]\frac{ \left( 2 \pi r_e \right)^2} {c^2 T_e}[/tex] = (.133 sec)^2 / (1 sidereal day)

    which turns out to be about 200 nanoseconds if I've calculated it right. I'd check the number (also the formulae) against other sources before considering the problem "solved" though.

    Wiki might be helpful, http://en.wikipedia.org/wiki/Sagnac_effect discusses some of the issues, but they don't ask the same question the OriginalPoster does, so the answer doesn't apply directly .
     
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