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

1. Aug 29, 2009

### jinksys

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. Aug 29, 2009

### Staff: Mentor

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

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. Aug 29, 2009

### jinksys

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

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.

Last edited: Aug 29, 2009
4. Aug 29, 2009

### Staff: Mentor

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

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!)

5. Aug 29, 2009

### jinksys

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

Yep, the first example :( not a good start to a book I have to use all semester!

6. Aug 29, 2009

### Staff: Mentor

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

7. Aug 29, 2009

### jinksys

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

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?

8. Aug 29, 2009

### Staff: Mentor

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

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.)

9. Aug 29, 2009

### jinksys

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

I am definitely going to Borders or Barnes and Noble and getting a backup. Do you have any recommendations?

10. Aug 29, 2009

### Staff: Mentor

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

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. Aug 29, 2009

### jinksys

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

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. Aug 29, 2009

### Staff: Mentor

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

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.)

13. Aug 29, 2009

### Naty1

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

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. Aug 29, 2009

### atyy

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

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.

15. Sep 3, 2009

### jinksys

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

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. Sep 3, 2009

### Staff: Mentor

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

Huh? Slower that what?
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?

17. Aug 29, 2010

### jinksys

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

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?

Last edited: Aug 29, 2010
18. Aug 29, 2010

### Staff: Mentor

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

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..."

19. Aug 29, 2010

### jinksys

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

Again, thanks for the all the help.

Here is an email I sent out to my classmates regarding this example. Was I correct in my statements?

If you were one of the lucky ones who've been able to acquire a textbook, I would like you to take a look at example 1.1 in the text.

Example 1.1 is set-up incorrectly. It should read:

A spacecraft is moving relative to the earth. An observer on the earth finds that between 1:00:00 PM and 2:00:01 PM, according to her clock, 3600 seconds elapse on the spacecraft's clock. What is the spacecraft's speed relative to earth?

Here t0 = 3600s is the proper time interval on the spacecraft and t=3601s as measured by a clock at rest with the observer.

Since it was stated that the spaceship was moving relative to earth, a clock in the ship's frame of reference measures the 'proper time'.

The plugging-in of numbers and solving for v part of the example is correct, however.

20. Aug 29, 2010

### Staff: Mentor

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

Sounds good to me (and much improved over the original).