Will O' Clocks Show Different Times?

[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?

 

[Janus]

Short answer: yes.

 

[franznietzsche]

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?

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

Then yes, definitely.

 

[Meir Achuz]

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?

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

 

[shaan]

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

 

[Swapnil]

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

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

 

[bernhard.rothenstein]

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?

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

 

[Swapnil]

I think it depends on what is the meaning of "I look"

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.

 

[robphy]

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.

"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.

 

[RandallB]

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?

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 a lot of work but it may be helpful to you.

 

[Swapnil]

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

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.

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??

 

[RandallB]

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??

What good is having someone tell you - See for yourself

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

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.

 

[bernhard.rothenstein]

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??

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

 

[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.

 

[Meir Achuz]

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?

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.

 

[RandallB]

I There is no SR in this situation, only the time delay from EM theory.
Whether any clocks show ...would include SR and GR effects.

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

Niether EM or GR are useful here. Use SR

 

[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?

 

[pervect]

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.

 

[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".

 

[JesseM]

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".

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.

 

[RandallB]

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'.

No – the laws of physics do not change in O’ just because O’ is ‘moving’.
Parallel or perpendicular mean nothing to O’ in O’ ALL clocks in O’ are synchronized and simultaneous with each other.

The point of SR is simultaneity is relative.
Observers in O will not agree with those in O’ that they have simultaneous synchronized clocks as any of the O’ clocks going by a test point in O will never and have never displayed a simultaneous synchronized (same) time to direct observations in O.
(O’ will reply to O “same to you and your out of sync clocks”)

Swapnil do the math – use .6c or .8 c and distances that give easy numbers to work with and you will see the point.

 

[JesseM]

No – the laws of physics do not change in O’ just because O’ is ‘moving’.
Parallel or perpendicular mean nothing to O’ in O’ ALL clocks in O’ are synchronized and simultaneous with each other.

pervect was talking about how clocks at rest in O (and synchronized in O) would appear in O'. Clocks in O parallel to the direction of relative motion would be out of sync in O', and clocks perpendicular to the direction of motion would be in sync.

 

[RandallB]

pervect was talking about how clocks at rest in O (and synchronized in O) would appear in O'. Clocks in O parallel to the direction of relative motion would be out of sync in O', and clocks perpendicular to the direction of motion would be in sync.

What he was thinking and what he though he was saying is one thing but what he said was:

In frame O', ... . Clocks in a line parallel to the motion will no longer be sychronized in frame O'.

It would be OK if he added "as observed from O".

 

[JesseM]

What he was thinking and what he though he was saying is one thing but what he said was:

In frame O', ... . Clocks in a line parallel to the motion will no longer be sychronized in frame O'.

It would be OK if he added "as observed from O".

That wouldn't make sense, because he was talking about clocks at rest in O as observed in O'. The full context makes it pretty clear:

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'.

I don't think there's any ambiguity that the "clocks" referred to in the second bold part were the same ones he brought up in the first bold part, and the second bold part is talking about how these clocks will appear in frame O'.

 

[aberrated]

I don't understand somethings.

like the question never said u is high enough to be close to c ... if not, then will the clocks still show different times?

 

[Janus]

I don't understand somethings.

like the question never said u is high enough to be close to c ... if not, then will the clocks still show different times?

It doesn't matter what value u has, there will still be a difference between the clocks in one frame as determined by the other frame. At low values of u (not close to c), the difference will be very very small, but it will still be there

 

[pervect]

What he was thinking and what he though he was saying is one thing but what he said was:
It would be OK if he added "as observed from O".

Huh?

Synchronization is frame dependent. That's the whole point I was trying to make. Therefore it is correct to say that the clocks are synchronized in one frame, and not synchronized in another. Synchronization has no "absolute" meaning, one must specify a frame before the term "synchronized" has any meaning.

While I have certainly written unclear, ambiguous, or wrong things by accident in a short post, I don't see the problem with what I wrote above. Rather, it appears to me that RandallB does not correctly appreciate the frame-dependence of the idea of synchronization.

 

[Swapnil]

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.

I don't know why you guys are still having discussion on this. I think what I have written above clearly describes what is going on with no/minimal ambiguity.

OK, so there might be a little ambiguity in the word "observes," (like Meir Achuz said) but like JesseM pointed out, in SR the term "observes" is part of a proper convention.

 

[RandallB]

That wouldn't make sense, because he was talking about clocks at rest in O as observed in O'.

Sorry I didn’t pick up on his changing the OP question from

clocks stationed on various different positions on the frame O'

and observing them from O ; to a problem with various clocks in O and observing them from O’ where there was previously no observer
– I don’t think pervect made the introduction of the new observer(s) clear.

 

[bernhard.rothenstein]

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

do you think that at a given time, due to the fact that the light signal used in order to synchronize the clocks, propagates with finite speed, not all the clocks being yet sybchronized.
sine ira et studio

 

[aberrated]

Suppose O' is a frame that observes an infinite grid with all the clocks synchronized to O' itself.

Now, O is a frame (me) who has a relative speed with respect to O' (thus all the other clocks of the grid as well)

How will my observation of the times be affected by my motion?
Specifically, suppose the origin (both space and time) was the event when O' and O touched each other

Assuming O is going towards the x-axis of the grid, what will the observed time of a general grid point be according to O?

 

[jtbell]

Suppose O' is a frame that observes an infinite grid with all the clocks synchronized to O' itself.

Now, O is a frame (me) who has a relative speed with respect to O' (thus all the other clocks of the grid as well)

Further let the x-axes of the two grids be parallel, likewise for the y- and z-axes, and let the relative speed v of O' with respect to O be in the +x direction.

How will my observation of the times be affected by my motion?
Specifically, suppose the origin (both space and time) was the event when O' and O touched each other

I assume you mean that the origins (x=y=z=0 points) of the two grids coincide at t=t'=0.

Then two clocks whose x'-coordinates differ by $\Delta x'$ in O' (and which are synchronized in O'), will be observed in O to be out of synchronization by an amount

$$\Delta t' = \frac{v \Delta x'}{c^2}$$

with the clock that is in the "forward" position (with respect to the motion of the clocks in O), having a reading that is behind the other one.

So if the clocks are 1 light-second apart in O', and are moving at 0.5c (0.5 light-seconds per second) in O, then in O they will be observed to have readings that are 0.5 second apart, and decreasing as you proceed in the +x direction. All O' clocks with the same x' but different y' and z' will have the same reading in O.