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

by jinksys
Tags: clocks, motion, observers, rest, tick
 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?
 Mentor 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.
 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: $v = c \sqrt{1-\frac{t_0^2}{t^2}}$ 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.
Mentor
P: 40,889

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

 Quote by jinksys 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?
This is Example 1.1 in Beiser, right? (I just looked it up on amazon.com preview.) In any case, Beiser has it backwards. (Someone should tell him!)
P: 123
 Quote by Doc Al This is Example 1.1 in Beiser, right? (I just looked it up on amazon.com preview.) In any case, Beiser has it backwards. (Someone should tell him!)
Yep, the first example :( not a good start to a book I have to use all semester!
Mentor
P: 40,889
 Quote by jinksys Yep, the first example :( not a good start to a book I have to use all semester!
P: 123
 Quote by Doc Al That's for sure. Please point this out to your instructor.
Luckily I was able to get an international version for $20, I feel for those who bought it from the bookstore for$176. I have another question...

This is the author's formula and explanation of time dilation...

$t = \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$

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?
Mentor
P: 40,889
 Quote by jinksys An Amazon review says that he has definitions of t and t_0 backwards, is this true?
I'd say so. That formula is OK if t0 means the time elapsed on the moving clock and t means the time elapsed on the "rest" clocks. Better read that book critically! (Get another book as a backup.)
P: 123
 Quote by Doc Al I'd say so. That formula is OK if t0 means the time elapsed on the moving clock and t means the time elapsed on the "rest" clocks. Better read that book critically! (Get another book as a backup.)
I am definitely going to Borders or Barnes and Noble and getting a backup. Do you have any recommendations?
 Mentor 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.)
 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).
Mentor
P: 40,889
 Quote by jinksys 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).
Yes, both "see" the other's clock as running slow. I put "see" in quotes since relativistic effects like time dilation are what's observed after taking light travel time into account. (In order to make sense of what you see, you must factor in the time it takes for the light to reach you. You don't just go by raw observations--what you literally see.)
 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....
P: 8,006
 Quote by jinksys I am definitely going to Borders or Barnes and Noble and getting a backup. Do you have any recommendations?
I learnt from an old edition of Beiser - it's very good. But typos are indeed irritating and make things hard, so it's good to also have eg. AP French's Special Relativity, and French and Taylor's An Introduction to Quantum Mechanics, Schroeder's An Introduction to Thermal Physics. Schaum's series is usually very reliable.
 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.
Mentor
P: 40,889
 Quote by jinksys 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.
Huh? Slower that what?
 She said the larger number represents the amount of local time it will take for moving clock to show T0.
Huh? The number shown on the clock itself is the local time (in the clock's frame)!

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 t0. Will earth clocks show a larger or smaller time for that rocket to go from A to B?
P: 123
 Quote by Doc Al This is Example 1.1 in Beiser, right? (I just looked it up on amazon.com preview.) In any case, Beiser has it backwards. (Someone should tell him!)
So, for the problem to be correct with Special Relativity the 'proper time' should be t0=3600, and t should be 3601, since it was stated that the ship was moving relative to earth and the observer was at rest on earth. It looks like they set up the problem correct, since they get the correct answer, but in the problem and setup they have the times reversed.

Is this right?
Mentor
P: 40,889
 Quote by jinksys So, for the problem to be correct with Special Relativity the 'proper time' should be t0=3600 and t=3601, since it was stated that the ship was moving relative to earth and the observer was at rest on earth. It looks like they set up the problem correct, since they get the correct answer, but in the problem and setup they have the times reversed. Is this right?
Yes. It's not just a typo, though, it's more of a sloppily constructed problem. What they meant to say was something like: "Earth observers find that from 1:00:00 pm to 2:00:01pm on their clocks only 3600 seconds have passed on the ship's clock."

They compound the confusion with the first sentence of the solution: "Here t0 is the proper time interval on the earth...." huh? They should have said, "Here t0 is the proper time interval as recorded by the ship's clock..."

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