# A quick SR question

1. Sep 5, 2006

### Swapnil

Suppose I am an observer stationed at the origin of frame O, watching a frame O' pass by me at some velocity u. If I look at the clocks stationed on various different positions on the frame O' (at some fixed instant on my watch), will all the clocks show a different time?

2. Sep 5, 2006

### Janus

Staff Emeritus

3. Sep 5, 2006

### franznietzsche

Are you assuming that the observer in O' thinks they are synchronized?

Then yes, definitely.

4. Sep 6, 2006

### Meir Achuz

"All" is too inclusive. Some clocks will show the same time.

5. Sep 6, 2006

### shaan

acc 2 relativity ...yes coz of diff. velocities(speed in this case)

6. Sep 6, 2006

### Swapnil

Which ones? You mean the ones that are "situated" on a plane perpendicular to the direction of motion?

7. Sep 6, 2006

### bernhard.rothenstein

I think it depends on what is the meaning of "I look"
sine ira et studio

8. Sep 6, 2006

### Swapnil

Ofcourse, when I say "I look" or "I see," I am implying that I see the event at the exact time it happens (without any delays), which is the same time as my watch since all the clocks in my frame are synchronized.

9. Sep 6, 2006

### robphy

"Of course" is inappropriate in this context.

In relativity, with a finite speed of light, "I look" and "I see" (i.e. optical observations [of some events lightlike separated from the reference event]) are distinct from "I determine to be simultaneous [some events spacelike-separated from the reference event]". As usual, in this subject, a more precise and unambiguous use of language is needed. This is what bernhard.rothenstein was seeking.

10. Sep 7, 2006

### RandallB

YOU cannot do that, not at some one fixed time on your watch, you can only 'see' one clock near you, all other clocks in O' cannot be observed directly by you at that time. But you can ask other observers in your frame who are on your time to check the clock in O' at your selected fixed time in O and send that info back to you.
Then yes ALL of them will have a different time - NONE of them will have the same time.

ALSO if you watch the local clocks going by you in O' you will see they appear to be running FAST vs. your local clock.
But again if you mark that first clock going by and ask your other observers in your frame to report back to you the time on that one clock when it passes them. Including the time they observed it goiing by in your O frame. Their reports will show you that that clock is running SLOW vs. your clock.

Simple math if you want to work out all the numbers on your own.
Just alot of work but it may be helpful to you.

11. Sep 7, 2006

### Swapnil

According to Meir Achuz, some clcoks will show the same time but according to RandallB none of the clocks will have the same time. Who is right here??

12. Sep 7, 2006

### RandallB

What good is having someone tell you - See for yourself
Added note:Of course if you have a pile of clocks stacked on top of each other to look at in the O’ frame they will all read the same time – these hardly count as “some clocks" showing the same time.

Last edited: Sep 8, 2006
13. Sep 7, 2006

### bernhard.rothenstein

Have a critical look at
arxiv
Physics, abstract
physics/0511062
From: Bernhard Rothenstein [view email]
Date: Mon, 7 Nov 2005 16:35:17 GMT (265kb)

Illustrating the relativity of simultaneity
Authors: Bernhard Rothenstein, Stefan Popescu, George J. Spix
Subj-class: Physics Education

We present a relativistic space-time diagram that displays in true magnitudes the readings (daytimes) of two inertial reference frames clocks. One reference frame is the rest frame for one clock. This diagram shows that two events simultaneous in one reference frames are not compulsory simultaneous in the other frame. This approach has a bi-dimensional character.
Full-text: PDF only
sine ira et studio

14. Sep 7, 2006

### Galileo

You were right. You will observe the same time on all the clocks situated on a plane perpendicular to the motion. So if O' goes in the x-direction, all clocks on a plane x=constant have the same time.

15. Sep 8, 2006

### Meir Achuz

I take (but, not "clearly") "show" to include the time delay
(\delta t=d/c) for the light from the clock to reach your eye. With this meaning, all clocks at the same distance from you will show the same time as each other. Clocks at different distances will show different times.
There is no SR in this situation, only the time delay from EM theory.
Whether any clocks show the same time as your watch depends on the histories of you and the other frame, which would include SR and GR effects.

16. Sep 8, 2006

### RandallB

?????? Don't need SR - - excect for SR effects ?????

Niether EM or GR are useful here. Use SR

Last edited: Sep 8, 2006
17. Sep 8, 2006

### Swapnil

OK, I don't see why are you guys trying to take my words literally? Just like in most SR problems, I am assuming all ideal conditions. Let me just the rephrase the Q&A on this discussion with a little more clarity.

Question: An observer A is stationed at the origin of frame O, and is observing a frame O' passing by him at some speed u in some arbitrary direction. All the clocks in the frame O are synchronized according to an observer in the frame O and all the clocks in the frame O' are synchronized according to an observer in the frame O'. If observer A observes the clocks (which are infinitesimally tiny) "stationed" on various different positions in the frame O' at some fixed instant of time in the frame O, will all the clocks "show" a different time according to observer A?

Answer: NO, only SOME of the clocks "situated" in the frame O' will show a different time according to observer A. ALL the clocks which are "situated" on a plane PERPINDICULAR to the direction of motion will "show" the EXACT same time according to observer A.

This is the right interpretation, am I correct?

Last edited: Sep 8, 2006
18. Sep 8, 2006

### pervect

Staff Emeritus
My \$.02 on how to express this:

In frame O, one creates a cubic array of clocks all of which are synchronized according to the Einstein convention.

In frame O', moving with respect to O, clocks in a plane perpendicular to the motion will remain synchronized in frame O'. Clocks in a line parallel to the motion will no longer be sychronized in frame O'.

This follows directly from the Lorentz transform, written in geometric units for an object moving in the x direction as:

t' = gamma*(t - beta*x)
x' = gamma*(x - beta*t)
y'=y
z'=z

We can see that t' does not depend on y or z Thus the reading on the clocks in frame O' will be the same as the reading on the clocks in frame O.

But because t' depends on x, we can't saay the same for a line of clocks along the direction of motion.

19. Sep 8, 2006

### Meir Achuz

I was wrong in my original post to leave out the LT that Prevect gives.
But until Swapnil gives a clear definition of "observes" that contradicts my understanding of "observes", I would include the time delay due to the finite speed of the light you "observe".

20. Sep 8, 2006

### JesseM

In SR I was under the impression that the convention is that "observe" refers to what is measured in your inertial reference frame, as opposed to "see" which includes light signal delays. For example, see jtbell's post here.