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- Thread starter Wizardsblade
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from your point of view the distance between you and the clock has shrunk by half but the clock is also ticking half as fast. also there is a shift in simultaneity that occurs whenever one accelerates that complicates things.

it would be easier to calculate it from the point of view of a stationary observer. from their point of view the length contraction only affects you and is irrelevant. once you determine how much time the stationary observer measures then its trivial to convert to your time. v=0.75^0.5

btw, you didnt state whether you are moving toward or away from the clock.

it would be easier to calculate it from the point of view of a stationary observer. from their point of view the length contraction only affects you and is irrelevant. once you determine how much time the stationary observer measures then its trivial to convert to your time. v=0.75^0.5

btw, you didnt state whether you are moving toward or away from the clock.

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Because the speed of light is constant in all frames I do not think it matters what direction you are moving in. Once you have accelerated there is no difference between you moving toward/away from the clock or the clock moving toward/away from you (besides doppler effect but that is not what this question is about.)btw, you didnt state whether you are moving toward or away from the clock.

What I am trying to find out is what happens when one accelerates as far as length contraction/time dilation, and what experimental evidence there is as support. Specifically in situations such as the one above.

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do you mean 'see' the clock reach 0 or do you mean 'calculate' that the clock reaches 0 (taking time of flight for the photons into account)?

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Actually “see”. Then calculate =)

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Hi Wizardsblade,

Assuming you are moving away from the big clock:

At gamma =2 your velocity is v=sqrt(0.75) = 0.866c as stated by granpa.

When you see the big clock read 10 seconds you know the clock has reached zero locally so you accelerate at that instant. To an observer at rest with the big clock the distance the light travels to catch up with you is d=c*t seconds. This is the same as the distance you have traveled in the same time, plus your head start of 10 light seconds, so the unaccelerated observer can also say d=(v*t+10). Equating the two expressions for distance we get c*t = v*t+10. From that it is easy to work out that the time t = 10/(c-v) = 10/(1-0.866) = 74.666 seconds as measured by the non accelerated observer. The time according to you is t/2=37.333 seconds due to time dilation.

Assuming instantaneous acceleration as soon as you switch frames from stationary to 0.866c you see an offset of -17.333 seconds on the big clock due the relativity of simultaneity so you calculate the big clock does not reach zero until 17.333 seconds after you accelerate and visual signal arrives at 37.333 seconds after you accelerated, giving a total elapsed time of 20 seconds by your calculations.

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Thanks guys. Does length contraction not play a part?

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