Need Clarification for time dilation effects

[NATURE.M]

So basically, my physics class has begun focusing on SR. And i just wanted to clarify something. I have been referencing:

http://physics.mq.edu.au/~jcresser/Phys378/LectureNotes/SpecialRelativityNotes.pdf

On page 23, it states "A clock will be observed to run at its fastest when it is stationary in a frame of reference."
I always thought that the clock that is stationary in a reference frame will run slower. Like if the clock C' is stationary relative to reference frame S'. Now S' moves at speed v relative to another reference frame S. Apparently, the clocks in S are running slow from the viewpoint of an observer in S. However, they measure the clocks in S' to be running more slowly. Is this a true statement?
If so, since he clocks in S' are stationary, shouldn't they be running faster according to the above statement? This appears contradictory to me with what was just mentioned (considering its true).

Thanks in advance.

 

[ghwellsjr]

So basically, my physics class has begun focusing on SR. And i just wanted to clarify something. I have been referencing:

http://physics.mq.edu.au/~jcresser/Phys378/LectureNotes/SpecialRelativityNotes.pdf

On page 23, it states "A clock will be observed to run at its fastest when it is stationary in a frame of reference."
I always thought that the clock that is stationary in a reference frame will run slower. Like if the clock C' is stationary relative to reference frame S'. Now S' moves at speed v relative to another reference frame S. Apparently, the clocks in S are running slow from the viewpoint of an observer in S. However, they measure the clocks in S' to be running more slowly. Is this a true statement?
If so, since he clocks in S' are stationary, shouldn't they be running faster according to the above statement? This appears contradictory to me with what was just mentioned (considering its true).

Thanks in advance.

Clocks that are stationary in a reference frame tick at the same rate as the Coordinate Time of the reference frame. Clocks that are moving according to a reference frame tick slower and so take longer to tick the same amount of time than the Coordinate Time of that reference frame. Time Dilation is the ratio of the elapsed Coordinate Time of a reference frame to the elapsed time on a clock. The faster it moves, the greater this ratio.

If a clock is inertial (moving at a constant velocity) and moving in a particular Inertial Reference Frame (IRF), you can transform to the IRF in which the clock is stationary and the Time Dilation will be 1.

So if you have two clocks in relative inertial motion, each one will have a Time Dilation factor of 1 in its own rest frame and the other clock will have a Time Dilation factor that is greater than 1.

It's a symmetrical relationship.

Does that help?

 

[NATURE.M]

I'm still a bit confused though. The following reference (pg. 7-8)
http://cosmo.nyu.edu/hogg/sr/sr.pdf initially made a lot of sense to me.
Namely, I find the following statement to make a lot of sense for the most part:

"Consider two observers, Deepto (D) and Erika (E), moving relative to one another in spaceships...D measures E’s speed to be u with respect to D’s rest frame. By the principle of relativity, E and D must observe the same speed of light, so we are forced to conclude that E will measure longer time intervals ∆t′ between the flashes in D’s clock than D will...The time intervals between flashes of D’s clock are longer as measured by E than as measured by D. This effect is called time dilation. Moving clocks go slow."
(rest of description on pg. 7 of link)

So I guess what I'm trying to ask is when they say "moving clocks go slow" are they referring to the motion of D's clock relative to E, in this particular case.
I understand the symmetry of the situation enables you to say that D measure's E's clock to go slow as well.

 

[ghwellsjr]

So I guess what I'm trying to ask is when they say "moving clocks go slow" are they referring to the motion of D's clock relative to E, in this particular case.

No, they are referring to the tick rate of the clock. That is what I said in my first response to you. The faster a clock moves in a particular reference frame, the slower it ticks which is the same as saying it takes more Coordinate Time for it to tick a given amount of its own Proper Time.

 

[Manraj singh]

https://www.physicsforums.com/showthread.php?t=737197. You can get some info in this thread. There is a link to another in it where you might get help.

 

[NATURE.M]

Oh that's makes a lot of sense. In the case of E measuring D's clock, basically the time interval between 2 tics is larger than the time interval that D observes between the two tics. I guess the way I was interpreting "moving clocks go slow" was pushing me into a hole.
Thanks alot!

 

[ghwellsjr]

Oh that's makes a lot of sense. In the case of E measuring D's clock, basically the time interval between 2 tics is larger than the time interval that D observes between the two tics. I guess the way I was interpreting "moving clocks go slow" was pushing me into a hole.
Thanks alot!

You're welcome.