
#1
Aug2909, 03:01 PM

P: 123

I'm reading Concepts of Modern Physics by Beiser, and the chapter example says:
A spacecraft is moving relative to earth. An observer finds that in one hour, according to her clock 3601s elapse on the spacecraft's clock. What is the craft's velocity relative to earth? (This is not a homework question, I can get the correct answer.) So is this saying that for every 3600 seconds on earth, the spacecraft's clock moves 3601 seconds? Wouldn't this mean that time runs faster on the craft and not slower? 



#2
Aug2909, 03:16 PM

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P: 40,889

Seems like an ambiguously worded example. Are you quoting it exactly, word for word?
In any case, moving clocks are observed to run slow. So if she is observing a moving clock, then when she sees 1 hour pass on the observed clock, her clock may show 3601 seconds having passed. 



#3
Aug2909, 03:25 PM

P: 123

Here is the example, verbatim:
A spacecraft is moving relative to the earth. An observer on the earth finds that, between 1 and 2 pm according to her clock, 3601s elapse on the spacecraft's clock. What is the space craft's speed relative to the earth? Solving for v, I have: [latex]v = c \sqrt{1\frac{t_0^2}{t^2}}[/latex] Where t0 is the clock at rest, and t is the clock in motion. t0 has to be more than t or else the square produces an imaginary number. So from this it seems like the clock in motion always has to read faster than the stationary one. Furthormore, here is the solution it gives: Here, t0 = 3600s is the proper time interval on the earth and t = 3601 s is the time interval in the moving frame as measured from the earth. From here they provide the math solution which is 7.1 x 10^6 m/s. 



#4
Aug2909, 03:29 PM

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P: 40,889

So clocks in motion are 'slower' to observers in rest, shouldn't the clocks tick less 



#5
Aug2909, 03:31 PM

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#7
Aug2909, 03:40 PM

P: 123

This is the author's formula and explanation of time dilation... [latex] t = \frac{t_0}{\sqrt{1\frac{v^2}{c^2}}}[/latex] t0 = rest clock t = moving clock c = speed of light v = speed of clock in motion An Amazon review says that he has definitions of t and t_0 backwards, is this true? 



#8
Aug2909, 03:46 PM

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#9
Aug2909, 03:49 PM

P: 123





#10
Aug2909, 03:55 PM

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P: 40,889

I suggest you post something in the Science Book discussion forum asking for opinions. We have quite a few active instructors who might be able to recommend a good one. (Someone who's taught from a book would be able to give a useful opinion.)




#11
Aug2909, 04:06 PM

P: 123

Ok, well now that I have that sorted out, I actually do have other questions regarding time dilation.
Suppose there are two GIANT clocks, one perched upon the earth and one perched upon a spacecraft. The clocks can be seen by both observers, one on earth and one in the craft. Are the following true? To an observer on earth and spaceship's clock ticks slower than the earth's clock. To the observer in the ship, the earth's clock moves slower than his. (assuming he is at a constant speed). 



#12
Aug2909, 04:13 PM

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#13
Aug2909, 04:25 PM

P: 5,634

Yes each observer sees the other fellows time as running slow. A good way to visualize this is via a "mirror reflecting photon clock"....a simple example which reflec the longer path taken by a photon in relative motion...maybe someone can post an online diagram source....




#14
Aug2909, 11:31 PM

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#15
Sep309, 10:42 AM

P: 123

I asked my physics adviser about the time dilation equation and she said that T should be larger than T0 because the moving clock moves slower. She said the larger number represents the amount of local time it will take for moving clock to show T0.




#16
Sep309, 10:49 AM

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Ask her this question: A rocket travels past earth (point A) to planet X (point B). During that trip, the rocket clock shows a time t_{0}. Will earth clocks show a larger or smaller time for that rocket to go from A to B? 



#17
Aug2910, 04:34 PM

P: 123

Is this right? 



#18
Aug2910, 04:53 PM

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P: 40,889

They compound the confusion with the first sentence of the solution: "Here t_{0} is the proper time interval on the earth...." huh? They should have said, "Here t_{0} is the proper time interval as recorded by the ship's clock..." 


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