# I Special relativity: frames of reference

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1. Mar 19, 2017

### Dimani4

Hi people,

I have a question about the frame of references.
Let's have an example:
First case: Jill on rocket and Jack stationary on Earth. Jill moves relatives to Jack 0.6c (1.8*10^8m/s). The distance is 18*10^8m. At the zero time Jack and Jill synchronize their clocks. Then Jill starts to move. When Jill arrives 18*10^8 m Jack's clock shows 10sec but Jill's clock shows 8 sec. Here we can say Jill while she moves sees distance less than 18*10^8m in factor of 0.8 then for her she moves a distance of 18*10^8*0.8 then 18*10^8*0.8/1.8*10^8m/s=8ses as she sees at her clock. That's what happens from the point of Jack's view. Jack's clock shows 10sec while when he observes Jill's clock he sees 8sec.

2. Second case. Now let us see what happens from the point of view of Jill. As for her she doesn't know she's moving then for her Jack is moving (with Earth) and she's stationary. Now when Jill is stationary and Jack moves away from her with velocity of 0.6c. What time will be on Jill's and Jack's clock after the same distance as in the First case?

The clocks of them should show exactly the same time as was in the previous case. Isn't it? As for me now Jill's clock will show 10sec and Jack's 8 sec.

Thank you.

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2. Mar 19, 2017

### pervect

Staff Emeritus
Already, there is an obstacle to answering your question here. The notion of clock synchronization is frame-dependent in special relativity, see any of the threads on "Einstein's train".

A quote from Einstein on the point in question, which is also given the more formal name of "The relativity of simultaneity", the name for the general principle derived from the specific example of Einstein's Train.

So the question doesn't have an answer without more information - you need to give more details on in what frame the clocks are to be synchronized (or alternatively, an operational procedure to perform the synchronziation). Until this is cleared up the question doesn't have a well-defined answer.

3. Mar 19, 2017

### Dimani4

thank you.

Lets say we synchronize the clocks at the moment when Jill starts to move or in the second case the Earth starts to move that should be the same case as the first.

4. Mar 19, 2017

### Bartolomeo

Each of them, Jill and Jack must conduct measurements from their own rest frame.

It is not quite correct to say, that Jill’s clock dilates relatively to Jack clock. It is better to say, that Jill’s clock dilates relatively to Jack’s rest frame and vice versa.

Rest frame must have at least two clocks. Let’s say one is in origin and second one at point x or – x of x axis.

If Jill moves in Jack’s frame, Jack must have one clock at point of Jill’s departure and another one at point of Jill’s arrival, 18*10^8m away.

Then Jack synchronizes these clocks by beam of light, admitting that velocity of light is c (Einstein synchronization convention). Then, in Jack’s rest frame, these clocks show the same time.

Then Jack compares readings of Jill’s clock with his clock Jack1 when these clocks are in immediate vicinity. Let’s say both clocks show 12 PM.

When clock Jill arrives to clock Jack2 they compare readings again. In immediate vicinity. Clock Jill show 3 PM and clock Jack2 shows 6 PM.

Observer Jack (rather a family of observers Jack) concludes, that clock Jill dilates.

Jill repeats the same procedure. Puts two clocks at points of departure and arrival, synchronizes them and compares readings with Jack’s clock. Jill concludes that Jack’s clock dilates (measures shorter period of time).

https://en.wikipedia.org/wiki/Observer_(special_relativity)

Animation is here: https://en.wikipedia.org/wiki/Time_dilation#/media/File:Time_dilation02.gif

See chapter time dilation:

http://www.pstcc.edu/departments/natural_behavioral_sciences/Web Physics/Chapter039.htm

5. Mar 19, 2017

### Staff: Mentor

You need to be more careful in how you state that. A more precise statement would be "Using the frame in which Jack is at rest, Jill's clock read 8 seconds at the same time that Jack's clock reads 10 seconds"; and this is the basis for Jack's observation that Jill's clock is running slow relative to his. (Do note that if Jack is watching Jill's clock through his telescope, he won't actually see Jill's clock reading 8 seconds until his own clock reads 16 seconds - he has to allow for the 6 seconds of light travel time).

However, if we use the frame in which Jill is at rest and Jack is moving backwards at .6c, it is not true that Jill's clock read 8 seconds at the same time that Jack's clock read 10 seconds. In that frame, Jack's clock read 6.4 seconds at the same time that Jill's clock read 8 seconds, and we conclude that it is Jack's clock that is running slow relative to Jill's.

And this is why @pervect referred you to the relativity of simultaneity....

6. Mar 19, 2017

### Dimani4

Thank you.

Now it's more clear. You say in the first case the Jill is moving Jack should put 2 clocks at the beginning and in the end (rest frame). The next one when Jack is moving Jill puts two clocks at the beginning and in the end. So in the first case Jack will see 10sec in his clock and 8 sec in Jill's clock. When Jack is moving Jill will see Jack's clock dilates (running slowly) and shows 8sec and her clock will show 10sec. Right?

7. Mar 19, 2017

### Dimani4

Thank you. I should to digest it. :)

8. Mar 19, 2017

### Bartolomeo

Correct.

9. Mar 19, 2017

### Staff: Mentor

Note that Jill is not an inertial observer. There is no expectation that there should be any symmetry or equivalence between Jack and Jill's frames

10. Mar 19, 2017

### Staff: Mentor

Hmmmm... I interpreted the original problem statement as Jack and Jill colocated until Jill experiences instantaneous acceleration to .6c followed coasting at that speed, which is of course equivalent to Jill moving at a constant speed and the clocks being synchronized at the moment that and Jack move past one another.

11. Mar 19, 2017

### Staff: Mentor

I interpreted them as being initially separated and then Jill suddenly accelerating towards Jack and meeting later. However in either case Jill is not inertial and her frame is not expected to be symmetrical to Jack. It is a misapplication of a symmetry principle

12. Mar 19, 2017

### Dimani4

That's right. I'm talking only about inertial frames of reference.

13. Mar 19, 2017

### Staff: Mentor

Then you cannot have Jill start moving if you want to talk about her frame. She must be inertial, just like Jack. Meaning moving at the same velocity forever.

Last edited: Mar 19, 2017