Repositioning clocks: continuing from the switch paradox [Archive]

calebhoilday

Jul31-10, 08:06 PM

starthaus

Jul31-10, 08:48 PM

JesseM

Jul31-10, 09:45 PM

In this hypothetical, an experiment is going to be conducted concerning the repositioning of clocks.

two clocks are placed at the leading end of a vessel, that is 1 light second long, to an observer at rest relative to it.

This vessel is shot into space from an observation station at (0.75)^0.5. As these clocks occupy essentially the same position, it can be safely assumed that these clocks in both the vessel frame and the observing station frame, that they are synchronised to each other enough that the difference is negligible between the frames.

One of the clocks is going to be shot from the leading end to the trailing end at a velocity of -(0.75)^0.5 according to the vessel or decelerate to a velocity of 0 to the observing station, where upon hitting the trailing end will return to the velocity of the vessel.
OK. In the ship's frame, the ship's length is 1 light-second, so the clock will take a time of 1 light-second/0.866c = 1.1547 seconds to reach the back in the ship's frame. During this period of 1.1547, the clock was traveling at 0.866c so it was running slow by sqrt(1 - 0.866^2) = sqrt(1 - 0.75) = 0.5, meaning it only elapsed a time of 0.5*1.1547 = 0.57735 seconds. Meanwhile the front clock was at rest in this frame, so it elapsed the full time of 1.1547 seconds. So, by the time the back clock reaches the back and comes to rest again in the ship's frame, the clock at the back is behind the clock at the front by a time of 1.1547-0.57735=0.57735 seconds.

In the observation tower frame, the ship is moving at 0.866c so its length is shrunk by a factor of 0.5, meaning its length is 0.5 light-seconds. So when one clock comes to rest in this frame and the back of the ship approaches it at 0.866c, the back of the ship will catch up to this clock in a time period of 0.5 light-seconds/0.866c=0.57735 seconds. During this time the clock was at rest so it ticked a full 0.57735 seconds. But during this time the clock at the front was moving at 0.866c so it was slowed down by a factor of 0.5, meaning it only ticked a time of 0.5*0.57735=0.288675 seconds. So, in the observation tower frame, by the time the clock reaches the back, it is ahead of the clock at the front by a time of 0.57735-0.288675=0.288675 seconds.
Before this occurs predictions based on special relativity are conducted as to what the difference between the clocks will be in each frame.

(Trailing - leading) in observing frame = duration of event in frame x (dilation of trailing - dilation of leading)
In observing frame:

duration = 0.57735 seconds
dilation of trailing = 1
dilation of leading = 0.5

So, 0.57735*(1 - 0.5) = 0.288675, which is indeed what I got for (trailing-leading)
(Trailing - leading) in vessel frame = duration of event in frame x (dilation of trailing - dilation of leading)
In vessel frame:

duration = 1.1547 seconds
dilation of trailing = 0.5
dilation of leading = 1

So, 1.1547*(0.5 - 1) = -0.57735 seconds, which is again what I got for (trailing-leading)
special relativity suggests that any frame of reference can be treated as though it is an absolute frame, considering itself to have a velocity of 0. For relativity to hold when one translates the goings on in a particular frame to another, it must be consistent with what the frame would predict.

for this situation it means, that the difference in clocks in the observing station frame converted into the vessel frame through the use of Lorentz transformations,
How can you convert a "difference" using the Lorentz transformation? The Lorentz transformation only transforms the coordinates of localized events from one frame to another.

For example, let's assume both clocks read 0 at the moment they depart from one another. That means that the clock that goes to the back will read a time of 1.1547 seconds when it reaches the back wall--call that event A. Then since the clock is now 0.288675 ahead of the clock at the front in the observation tower frame, that must mean that if we let event B be the event of the front clock reading 1.1547-0.288675=0.866 seconds, then this event B is simultaneous with event A in the observation tower frame. So if we let A be at the origin of the observation tower frame, meaning A has coordinates x=0 and t=0, then B must also have coordinate t=0, and since the ship is 0.5 light seconds long in this frame B must have coordinate x=0.5.

So in the observation tower frame, we have A (clock reaching back wall and reading 1.1547 seconds) at x=0, t=0 and B (clock at front wall reading 0.866 seconds) at x=0.5, t=0. If we use the Lorentz transformation with gamma=2 to find the coordinates of these events in the vessel frame, event A has coordinates x'=0, t'=0 while event B has coordinates:

x' = gamma*(x-vt) = 2*(0.5 - 0.866*0) = 1
t' = gamma*(t-vx/c^2) = 2*(0 - 0.866*0.5) = -0.866

So in the vessel frame, at t'=-0.866 seconds, event B occurs, which is the event of the clock at the front reading 0.866 seconds. And the clock is at rest in this frame, so 0.866 seconds later at t'=0, the clock at the front reads 0.866 + 0.866 = 1.732 seconds. Call this event C. We know that event A also occurred at t'=0 in this frame, so event A is simultaneous with event C in the vessel frame. And remember that A was the event of the clock at the back reading 1.1547 seconds, so in this frame (trailing - leading) = 1.1547 - 1.732 = -0.5773 seconds. And that's just what I got for (trailing - leading) in the vessel frame by the earlier method of figuring out the velocity and time for the clock to go from front to back in the vessel frame! So you see, everything works out consistently.
(((1-VS)/|S-V|)*(((1-((S-V)/(1-VS))^2)^0.5)-1))=(((1-V^2)/(V-S))*(((1-S^2)^0.5)-((1-V^2)^0.5))-V

V: velocity of vessel / speed of light
S: velocity of trailing clock / speed of light
You need to explain your work, I have no idea what the basis for this complicated equation is.

JesseM

Jul31-10, 09:47 PM

This vessel is shot into space from an observation station at (0.75)^0.5. As these clocks occupy essentially the same position, it can be safely assumed that these clocks in both the vessel frame and the observing station frame, that they are synchronised to each other enough that the difference is negligible between the frames..
This is impossible. You keep making the same mistake over and over.
No mistake there, he was just assuming the two clocks remained synchronized with one another as long as they were right next to each other at the front of the ship (and so following exactly the same worldline, and elapsing the same proper time to any point on that worldline), they do get out-of-sync as soon as one heads towards the back of the ship while the other remains at the front.

starthaus

Jul31-10, 10:06 PM

No mistake there, he was just assuming the two clocks remained synchronized with one another as long as they were right next to each other at the front of the ship (and so following exactly the same worldline, and elapsing the same proper time to any point on that worldline), they do get out-of-sync as soon as one heads towards the back of the ship while the other remains at the front.

One of the clocks is going to be shot from the leading end to the trailing end at a velocity of -(0.75)^0.5 according to the vessel or decelerate to a velocity of 0 to the observing station, where upon hitting the trailing end will return to the velocity of the vessel.

In order for this to happen they need to be moved away from each other in the frame of the rocket. This means that they are no longer sychronized in the frame of the launcher.

Fredrik

Jul31-10, 10:08 PM

If you disagree, show your calculations...

I showed you a calculation in the other thread.

http://www.physicsforums.com/showthread.php?p=2818150

JesseM

Jul31-10, 10:55 PM

In order for this to happen they need to be moved away from each other in the frame of the rocket. This means that they are no longer sychronized in the frame of the launcher.
Yeah, but I don't think he was saying they would stay synchronized once they were moved away from each other, in fact his whole analysis says they wouldn't. When he said "As these clocks occupy essentially the same position, it can be safely assumed that these clocks in both the vessel frame and the observing station frame, that they are synchronised to each other enough that the difference is negligible between the frames", I think he was just talking about the period of time when both clocks were together in the front of the rocket as it was shot away from the observation station, not about what happened after the two clocks began to move away from each other (the quoted paragraph was from before he mentioned moving them apart).

calebhoilday

Aug1-10, 12:15 AM

JesseM has the understanding down of the situation

"So, 0.57735*(1 - 0.5) = 0.288675, which is indeed what I got for (trailing-leading)" JesseM

If 0.288675 seconds passed in the observer frame then half as much passed in the vessel

then you need to adjust that based on the relativity of simultaneity equation which is simply V in as your dealing with a vessel that 1 light second long. Once this is done you will have the same answer as mine. You can also find if S was -(0.75)^0.5 instead of 0 congruency would emerge.

calebhoilday

Aug1-10, 12:44 AM

i dilated time in a place i shouldn't have and the result is congruent. Least I get it now.

JesseM

Aug1-10, 01:21 AM

JesseM has the understanding down of the situation

"So, 0.57735*(1 - 0.5) = 0.288675, which is indeed what I got for (trailing-leading)" JesseM

If 0.288675 seconds passed in the observer frame then half as much passed in the vessel
Not true, the time dilation equation only says that if two events happened at the same position but different times in the vessel frame (like different ticks of a clock at rest in the vessel frame), then the time between them in the vessel frame is half as much the time between these same events in the observer frame (i.e. more time passes between those two ticks of the vessel clock as seen in the observer frame, so the clock is running slow as seen by the observer). Likewise if two events happen at the same position and different times in the observer frame, the time in between them in the observer frame is half the time between them in the vessel frame. But 0.288675 isn't the time between two events at the same position in either frame. In fact it isn't really a time between a pair of events at all, instead 0.288675 is the amount the trailing clock is ahead of the leading one in the observing frame, so if at some moment the leading clock reads T seconds, then at the same moment the trailing clock reads T+0.288675 seconds.
then you need to adjust that based on the relativity of simultaneity equation which is simply V in as your dealing with a vessel that 1 light second long.
Relativity of simultaneity says that if two clocks are a distance L apart and synchronized in their mutual rest frame, then they are out-of-sync by vL/c^2 in the frame that sees them moving at v. But the clocks at the front and back of the ship aren't actually synchronized in the ship frame, so this wouldn't be the correct equation to tell you how out-of-sync they are in the observer frame.
Once this is done you will have the same answer as mine.
My math is correct, if it disagrees with your it's because yours is incorrect. If you disagree, you need to either explain your own reasoning in a more clear step-by-step manner (as I said, you just wrote down a giant complicated equation without explaining how you derived it), or point out where the flaw is in my own analysis.

calebhoilday

Aug1-10, 03:41 AM

i dilated time in a place i shouldn't have and the result is congruent. Least I get it now. what i meant is that your answer is congruent as you didn't dilate time. This was bad math on my part.

i'll be getting rid of the time dilation from the formula soon and then we will be on the same page.

The parameters I used are big and my assumption that as you slow the velocity in which you reposition clocks, time dilation will become negligible, it is possible that the parmeters are just a sweet spot of congruency. My assumption could turn out to be similar to martingale strategy, as the assumption may only work if the velocity of clock repositioning is 0 in the vessel frame.

We shall soon see, ill be editing the formula and using the parameters of s=0.8 v=0.9 in the first post.

calebhoilday

Aug1-10, 04:03 AM

turns out the assumption was bad. But now at least im informed

starthaus

Aug1-10, 08:41 AM

Austin0

Aug1-10, 08:54 AM

Relativity of simultaneity says that if two clocks are a distance L apart and synchronized in their mutual rest frame, then they are out-of-sync by vL/c^2 in the frame that sees them moving at v. .

They are out of sync by \gamma(v)\frac{vL}{c^2}

L as defined above, is a distance in the clocks rest frame. Your equation would be correct applied to a length as measured in a relative frame , yes??

starthaus

Aug1-10, 08:57 AM

L as defined above, is a distance in the clocks rest frame. Your equation would be correct applied to a length as measured in a relative frame , yes??

no, L=proper length in the equation I posted.

JesseM

Aug1-10, 10:38 AM

They are out of sync by \gamma(v)\frac{vL}{c^2}
No, if you have two clocks a distance L apart in their rest frame, and two events on their worldlines which occur simultaneously in their rest frame, then these same two events are separated by a time of \gamma(v)\frac{vL}{c^2} in the observer's frame where the clocks are moving at speed v. But that's not telling us the amount they're out-of-sync in the observer's frame! To find that, we need to find two events on each worldline that are simultaneous in the observer's frame, and then find the time interval between them in the rest frame of the clocks (since the observer judges how out-of-sync the clocks are by looking at what each clock reads at a single moment in his own frame).

In the observer's frame, the clocks are only separated by a distance of L/gamma(v), so suppose in the observer's frame the two simultaneous readings occur at these coordinates:

x=0, t=0
x=L/gamma, t=0

In that case the time of the first event in the clock frame is t'=0, while the time of the second event is:

t' = gamma*(0 - vL/gamma*c^2) = -vL/c^2

So if the clocks are synchronized in their own rest frame, the difference in readings for these two events will be vL/c^2.

starthaus

Aug1-10, 10:45 AM

No, if you have two clocks a distance L apart in their rest frame, and two events on their worldlines which occur simultaneously in their rest frame, then these same two events are separated by a time of \gamma(v)\frac{vL}{c^2} in the observer's frame where the clocks are moving at speed v.

It is very simple, really:

t'=\gamma(v)(t-vx/c^2)

so:

dt'=\gamma(v)(dt-vdx/c^2)

If the events are simultaneous in F, then dt=0 so:

dt'=-\gamma(v)vdx/c^2

If the events are simultaneous in F' then dt'=0 and

dt=vdx/c^2

In the observer's frame, the clocks are only separated by a distance of L/gamma(v), so suppose in the observer's frame the two simultaneous readings occur at these coordinates:

I was quite clear in my answer (http://www.physicsforums.com/showpost.php?p=2822474&postcount=15) that L=proper length, i.e. L=dx

JesseM

Aug1-10, 11:59 AM

It is very simple, really:

t'=\gamma(v)(t-vx/c^2)

so:

dt'=\gamma(v)(dt-vdx/c^2)

If the events are simultaneous in F, then dt=0 so:

dt'=-\gamma(v)vdx/c^2

If the events are simultaneous in F' then dt'=0 and

dt=vdx/c^2
Yes, I agree with all of that. The point is that if you want to know how out-of-sync two clocks are in F' if the clocks are at rest and synchronized in F, and have a proper distance of dx in their rest frame, then you have to consider simultaneous readings in F' and figure out the time between them in F. So, the last equation tells you how out-of-sync the clocks are in F'.

Likewise, if you had two clocks at rest and synchronized in F', and a proper distance dx' apart in F', then if you wanted to know how out-of-sync they are in F, you'd have to set dt=0, then you'd get dt'=vdx'/c^2. So regardless of what frame the clocks are at rest and synchronized in, if L is the proper distance between them in their own rest frame, it is always true that they are out-of-sync by vL/c^2 in the frame that sees them moving at speed v (in a direction parallel to the axis between them).

starthaus

Aug1-10, 01:21 PM

Yes, I agree with all of that. The point is that if you want to know how out-of-sync two clocks are in F' if the clocks are at rest and synchronized in F, and have a proper distance of dx in their rest frame, then you have to consider simultaneous readings in F' and figure out the time between them in F.

Why would I do such a thing ? In F dt=0.

So, the last equation tells you how out-of-sync the clocks are in F'.

I don't know why you persist in this error. I gave you the correct equation of desynchronization in F'.

Likewise, if you had two clocks at rest and synchronized in F', and a proper distance dx' apart in F', then if you wanted to know how out-of-sync they are in F, you'd have to set dt=0, then you'd get dt'=vdx'/c^2.

You can definitely do that but you are doing things backwards. Are you doing this because you are intent on justifying your initial formula? If this is what is driving you, it is easier to do it correctly:

dx'=\gamma(dx-vdt)
dt'=\gamma(dt-vdx/c^2)

If dt=0 (events simultaneous in F) then:

dx'=\gamma*dx
dt'=-\gamma*vdx/c^2=-vdx'/c^2

i.e. events are not simultaneous in F' and they are "desynchronized" (out of sync) by the amount dt'

JesseM

Aug1-10, 03:56 PM

Why would I do such a thing ? In F dt=0.
dt=0 for what two events? F' is the observing frame, if you want to figure out how out-of-sync the clocks are in F', you have to pick simultaneous readings in F'. For example, suppose the clocks are 10 light-seconds apart in their rest frame F, and synchronized in their rest frame. And suppose the clocks are moving at 0.8c in the frame of the observer F'. Would you then agree that at the same moment F' measures the leading clock reading T=0, he also measures the trailing clock to be reading T=8? And that this is what it means for F' to say the two clocks are out-of-sync by 8 seconds in F'? So if event 1 is the leading clock reading T=0, and event 2 is the trailing clock reading T=8, then for these events dt=8 but dt'=0 (while dx=10 and dx'=6).
I don't know why you persist in this error. I gave you the correct equation of desynchronization in F'.
I think you're just confused about what "desynchronization" means. Do you disagree that in my example above, we would say that in frame F' the two clocks are desynchronized by 8 seconds?

starthaus

Aug1-10, 08:15 PM

dt=0 for what two events?

Obviously, in F. The formulas (http://www.physicsforums.com/showpost.php?p=2822729&postcount=19) are self-evident.

For example, suppose the clocks are 10 light-seconds apart in their rest frame F, and synchronized in their rest frame. And suppose the clocks are moving at 0.8c in the frame of the observer F'. Would you then agree that at the same moment F' measures the leading clock reading T=0, he also measures the trailing clock to be reading T=8?

dx=10 , v=0.8c, \gamma=1/0.6

dt'=\gamma vdx/c^2

Do you disagree that in my example above, we would say that in frame F' the two clocks are desynchronized by 8 seconds?

Yes, I disagree. See above for the correct calculations. The correct anser is not 8s, as you claim but 8/0.6=13.33s

JesseM

Aug1-10, 11:08 PM

dt=0 for what two events?
Obviously, in F.
I didn't ask "what frame", I asked "what two events" (events are frame-independent). For example, I picked the two events "leading clock reads a time of T=0" and "trailing clock reads a time of T=8".
For example, suppose the clocks are 10 light-seconds apart in their rest frame F, and synchronized in their rest frame. And suppose the clocks are moving at 0.8c in the frame of the observer F'. Would you then agree that at the same moment F' measures the leading clock reading T=0, he also measures the trailing clock to be reading T=8?
dx=10 , v=0.8c, \gamma=1/0.6

dt'=\gamma vdx/c^2
That's not an answer to my question. I asked "Would you then agree that at the same moment F' measures the leading clock reading T=0, he also measures the trailing clock to be reading T=8?"

Event 1: leading clock reading T=0
Event 2: trailing clock reading T=8

In frame F (where the clocks are at rest, synchronized, and have a separation of 10 light seconds) these events could have coordinates:

Event 1: x=0 light-seconds, t=0 seconds
Event 2: x=10 light-seconds, t=8 seconds

Using the Lorentz transformation equation t'=gamma*(t - vx/c^2), you can see that in the observer's frame F' (where the clocks are moving at 0.8c) both of these events occur at the same time-coordinate t'=0. So, that's why I say "at the same moment F' measures the leading clock reading T=0, he also measures the trailing clock to be reading T=8." Please tell me if you agree or disagree with this statement, and if you disagree please tell me where you think my analysis above goes wrong.

starthaus

Aug2-10, 12:16 AM

I didn't ask "what frame", I asked "what two events" (events are frame-independent). For example, I picked the two events "leading clock reads a time of T=0" and "trailing clock reads a time of T=8".

That's not an answer to my question. I asked "Would you then agree that at the same moment F' measures the leading clock reading T=0, he also measures the trailing clock to be reading T=8?"

Event 1: leading clock reading T=0
Event 2: trailing clock reading T=8

In frame F (where the clocks are at rest, synchronized, and have a separation of 10 light seconds) these events could have coordinates:

Event 1: x=0 light-seconds, t=0 seconds
Event 2: x=10 light-seconds, t=8 seconds

Using the Lorentz transformation equation t'=gamma*(t - vx/c^2), you can see that in the observer's frame F' (where the clocks are moving at 0.8c) both of these events occur at the same time-coordinate t'=0.

No, they don't. You are asked to find out the time separation between the two events, that are simultaneous in F (dt=0) as measured from frame F'. There is nothing entitling you to set t'=0.

So, that's why I say "at the same moment F' measures the leading clock reading T=0, he also measures the trailing clock to be reading T=8." Please tell me if you agree or disagree with this statement, and if you disagree please tell me where you think my analysis above goes wrong.

I don't understand why you have such a great difficulty in understanding the basic equations. Your analysis goes wrong in a simple attempt to calculate the time separation of two events that are simultaneous in frame F (dt=0) as measured in frame F' (where they are separated by dt'=\gamma v dx/c^2). This is especially disturbing in the context of having showed you not only the formulas but also their derivation.

JesseM

Aug2-10, 12:26 AM

I don't understand why you have such a great difficulty in understanding the basic equations. Your analysis goes wrong in a simple attempt to calculate the time separation of events.
What part of my analysis is wrong? Do you disagree that if my two events have coordinates (x=0, t=0) and (x=10, t=8) in the F frame, then these events both have t'=0 in the F' frame?
This is especially disturbing in the context of having showed you not only the formulas but also their derivation.
Your equation dt' = gamma*v*dx/c^2 is incorrect for those two events, because dt was nonzero for the events I picked. The full Lorentz transformation equation is

dt' = gamma*(dt - v*dx/c^2)

If dt is not zero, you can't reduce this to dt' = gamma*v*dx/c^2. For the two events I chose (the event of the leading clock reading T=0, and the event of the trailing clock reading T=8), dt was not zero, here dt=8.

So, I ask again: "Would you then agree that at the same moment F' measures the leading clock reading T=0, he also measures the trailing clock to be reading T=8?" It's a simple yes-or-no question.

starthaus

Aug2-10, 12:36 AM

What part of my analysis is wrong? Do you disagree that if my two events have coordinates (x=0, t=0) and (x=10, t=8) in the F frame, then these events both have t'=0 in the F' frame?

Your equation dt' = gamma*v*dx/c^2 is incorrect for those two events, because dt was nonzero for the events I picked. The full Lorentz transformation equation is

dt' = gamma*(dt - v*dx/c^2)

...which is precisely what I showed you earlier (http://www.physicsforums.com/showpost.php?p=2822598&postcount=17).

If dt is not zero,

...but the discussion has ALWAYS (http://www.physicsforums.com/showpost.php?p=2822598&postcount=17) been about simultaneous events in F. That is dt=0.

you can't reduce this to dt' = gamma*v*dx/c^2.
But we are talking about dt=0. I pointed out that your repeated dropping (http://www.physicsforums.com/showpost.php?p=2822591&postcount=16) of \gamma in the formula for dt' , for this PARTICULAR situation (i.e dt=0), is a mistake. Why are you now changing the goalposts to talk about dt>0?
BTW, even if you want to change the talk to dt>0., your dropping of \gamma is STILL a mistake.

Austin0

Aug2-10, 01:48 AM

...which is precisely what I showed you earlier (http://www.physicsforums.com/showpost.php?p=2822598&postcount=17).

...but the discussion has ALWAYS (http://www.physicsforums.com/showpost.php?p=2822598&postcount=17) been about simultaneous events in F. That is dt=0.

But we are talking about dt=0. I pointed out that your repeated dropping (http://www.physicsforums.com/showpost.php?p=2822591&postcount=16) of \gamma in the formula for dt' , for this PARTICULAR situation (i.e dt=0), is a mistake. Why are you now changing the goalposts to talk about dt>0?
BTW, even if you want to change the talk to dt>0., your dropping of \gamma is STILL a mistake.

Desynchronization is not dependant on specific events but can be calculated purely as a function of velocity and spatial separation between clocks in their own rest frame. Yes??

The separation between clocks was clearly defined as distance in their rest frame in JesseM's original statement,( that you took exception to) , so his equation is totally correct within that context.
The equation you posted is only correct as applied to the spatial separation between the clocks as measured in a relative frame. The "L" in your equation is not the same L as defined by JM
Yes???
If you think it would apply to L as per JM i.e. proper length, then either you are simply wrong or I have to relearn relative simultaneity.

starthaus

Aug2-10, 08:26 AM

If you think it would apply to L as per JM i.e. proper length, then either you are simply wrong or I have to relearn relative simultaneity.

it is the latter

JesseM

Aug2-10, 08:38 AM

...but the discussion has ALWAYS (http://www.physicsforums.com/showpost.php?p=2822598&postcount=17) been about simultaneous events in F. That is dt=0.
No, you have been talking about simultaneous events in F, but I have not. I have been talking about how "out-of-sync" or "desynchronized" the clocks are in the frame of the observer F'. If the observer F' measures the leading clock reading T=0 at the same moment (in frame F') that the trailing clock reads T=8, it makes sense to say that in frame F' the two clocks are "out-of-sync by 8 seconds" (with vL/c^2 = 8). This is how other authors talk, for example this page (http://galileo.phys.virginia.edu/classes/252/synchronizing.html) says at the bottom:
Equally important, the clocks—which are synchronized by an observer on the train—appear unsynchronized when viewed from the ground, the one at the back of the train reading vL/c^2 seconds ahead of the clock at the front of the train, where L is the rest length of the train (the length as measured by an observer on the train).
I trust anyone reading this discussion can understand this point about what it means to say two clocks are "out-of-sync" by a given amount, even if you can't, so if you refuse to answer my simple question "Would you then agree that at the same moment F' measures the leading clock reading T=0, he also measures the trailing clock to be reading T=8?", I will end this discussion here.

Austin0

Aug2-10, 08:46 AM

it is the latter

Given an inertial frame F1 with two clocks separated by 10 units of 1 ls ,,as measured in that frame and an observation frame F2 ......v= 0.8 c
WOuld you question that as observed and measured in F2 that the degree of desynchronization between the clocks in F1 would be 0.8* 10 = (+) or( - ) 8.0 sec. ????

You are maintaining that the desynch would be 1.666667 * 0.8*10 =13.3333 ????

starthaus

Aug2-10, 09:15 AM

Given an inertial frame F1 with two clocks separated by 10 units of 1 ls ,,as measured in that frame and an observation frame F2 ......v= 0.8 c
WOuld you question that as observed and measured in F2 that the degree of desynchronization between the clocks in F1 would be 0.8* 10 = (+) or( - ) 8.0 sec. ????

You are maintaining that the desynch would be 1.666667 * 0.8*10 =13.3333 ????

I am not "maintaining". 13.33ns is the correct result.

yossell

Aug2-10, 09:19 AM

I am not "maintaining". 13.33ns is the correct result.

How can you say that this is the correct result without maintaining it?

starthaus

Aug2-10, 09:25 AM

How can you say that this is the correct result without maintaining it?

Because I have proved (http://www.physicsforums.com/showpost.php?p=2823106&postcount=21) it. "Maintaining" has a connotation of claiming without proof.

starthaus

Aug2-10, 09:48 AM

No, you have been talking about simultaneous events in F, but I have not. I have been talking about how "out-of-sync" or "desynchronized" the clocks are in the frame of the observer F'. If the observer F' measures the leading clock reading T=0 at the same moment (in frame F') that the trailing clock reads T=8, it makes sense to say that in frame F' the two clocks are "out-of-sync by 8 seconds" (with vL/c^2 = 8). This is how other authors talk, for example this page (http://galileo.phys.virginia.edu/classes/252/synchronizing.html) says at the bottom:
.

Excellent, so the author, on one hand calculates t_B-t_F=\gamma vL/c^2, as it should be. Then, he waives his hands and declares :

so that although the ground observer measures the time interval between the starting of the clock at the back of the train and the clock at the front as \gamma vL/c^2 seconds, he also sees the slow running clock at the back actually reading vl/c^2 seconds at the instant he sees the front clock to start.

So, the author is confused of what the moving observer actually measures <shrug>. He calculates correctly and claims one thing and then he contradicts himself and claims another thing.

JesseM

Aug2-10, 10:25 AM

So, the author is confused of what the moving observer actually measures <shrug>. He calculates correctly and claims one thing and then he contradicts himself and claims another thing.
No, he's not contradicting himself, he's just talking about different measurements involving different pairs of events. If he "measures the time interval between the starting of the clock at the back of the train and the clock at the front", then he's measuring the time interval between the events of each clock starting, i.e. the time between the event "clock at back reads T=0" and the event "clock at the front reads T=0". This time is indeed gamma*vL/c^2. On the other hand, if he measures the clock at the back "at the instant he sees the front clock to start", then here "at the instant" refers to simultaneity in his own frame, so he must be looking at the event on the back clock's worldline which is simultaneous with the event of the front clock starting according to his own frame's definition of simultaneity. As I already showed, this would be satisfied if he picked the events "clock at back reads T=8" and "clock at front reads T=0".

Another example of an author using the phrase "unsynchronized by" to refer to the difference between a pair of clock readings which are simultaneous in the observer's frame (like front clock reading T=0 and back clock reading T=8 in my example, two events which are simultaneous in the observer's frame F') can be seen here (http://books.google.com/books?id=jBgu9psKkREC&q=%22unsynchronized+by%22+simultaneity&dq=%22unsynchronized+by%22+simultaneity&hl=en&ei=q99WTLuuKcT38Aa2qdmgBA&sa=X&oi=book_result&ct=result&resnum=2&ved=0CCkQ6AEwAQ):
Suppose there is a clock at the planet P synchronized in S with Homer's clock at Earth. In S', these clocks are unsynchronized by the amount L0V/c^2.

So, when both these authors talk about how "unsynchronized" clocks are, what they mean is the difference in the clock's readings for a pair of events on the clocks' worldlines which are simultaneous in the observer's frame F', not simultaneous in the clocks' frame F. This is a perfectly well-defined physical question, so unless you actually disagree that two events on the clocks' worldlines which are simultaneous in F' will show a difference in readings of vL/c^2, your objection amounts to nothing more than a semantic quibble about the meaning most physicists would attach to the English phrase "unsynchronized by".

starthaus

Aug2-10, 10:43 AM

No, he's not contradicting himself, he's just talking about different measurements involving different pairs of events. If he "measures the time interval between the starting of the clock at the back of the train and the clock at the front", then he's measuring the time interval between the events of each clock starting, i.e. the time between the event "clock at back reads T=0" and the event "clock at the front reads T=0". This time is indeed gamma*vL/c^2.

...which is exactly what I have been talking about all along. This is why mathematics is much better than natural languages. I have said all along dt=0. I did not say dt'=0.

JesseM

Aug2-10, 11:19 AM

...which is exactly what I have been talking about all along.
But it's not what I have been talking about all along, and I was the one who made the original statement in post #10 that "Relativity of simultaneity says that if two clocks are a distance L apart and synchronized in their mutual rest frame, then they are out-of-sync by vL/c^2 in the frame that sees them moving at v", which you objected to in post #13 by saying "They are out of sync by \gamma(v)\frac{vL}{c^2}". Given what I meant by the phrase "out-of-sync by" (which matches what physicists would normally mean by that type of phrase in this context), my statement was perfectly correct.
This is why mathematics is much better than natural languages. I have said all along dt=0. I did not say dt'=0.
And I never said your math was wrong, I just said that your objection to the statement "they are out-of-sync by vL/c^2" was misguided, since the statement doesn't concern events that have a time difference of dt=0 but rather events that have a time difference of dt'=0. If you acknowledge natural language is ambiguous, maybe you shouldn't be so quick to raise objections to natural language statements that you don't understand, and should instead ask questions to try to clarify what the accepted terminology is.

starthaus

Aug2-10, 01:18 PM

two clocks are a distance L apart and synchronized in their mutual rest frame,

...meaning dt=0

starthaus

Aug2-10, 01:30 PM

And I never said your math was wrong, I just said that your objection to the statement "they are out-of-sync by vL/c^2" was misguided, since the statement doesn't concern events that have a time difference of dt=0 but rather events that have a time difference of dt'=0. If you acknowledge natural language is ambiguous, maybe you shouldn't be so quick to raise objections to natural language statements that you don't understand, and should instead ask questions to try to clarify what the accepted terminology is.

If this is what you wanted to say, then you needed to phrase it precisely.
Two events separated by dt,dx in frame F are perceived simultaneously in frame F'. What is their time separation in frame F?

0=dt'=\gamma(dt-vdx/c^2)

so:

dt=vdx/c^2

The observer in frame F' does not "measure" a de-synchrnization of vdx/c^2. The observer in F' only infers a de-synchronization from the fact that the events appeared simultaneous in F'.

JesseM

Aug2-10, 02:27 PM

two clocks are a distance L apart and synchronized in their mutual rest frame,
...meaning dt=0
No, it doesn't mean that, not if we aren't considering a pair of events that are simultaneous in the clocks' rest frame, like the pair "front clock reads T=0" and "back clock reads T=8". If the clocks are synchronized in frame F, then for these two events dt=8. It's only meaningful to give a value of dt when you have first specified what two events you want to calculate the time-interval between.
If this is what you wanted to say, then you needed to phrase it precisely.
I phrased it in the standard way that other physicists phrase similar claims. Again, you might have asked what I meant before jumping in to accuse me of saying something incorrect. Anyway, I explained almost immediately what I meant in more precise language, like this from post #16:
But that's not telling us the amount they're out-of-sync in the observer's frame! To find that, we need to find two events on each worldline that are simultaneous in the observer's frame, and then find the time interval between them in the rest frame of the clocks (since the observer judges how out-of-sync the clocks are by looking at what each clock reads at a single moment in his own frame).

In the observer's frame, the clocks are only separated by a distance of L/gamma(v), so suppose in the observer's frame the two simultaneous readings occur at these coordinates:

x=0, t=0
x=L/gamma, t=0

In that case the time of the first event in the clock frame is t'=0, while the time of the second event is:

t' = gamma*(0 - vL/gamma*c^2) = -vL/c^2
I don't think this is at all ambiguous, if you had been paying attention rather than rushing to object you might have seen that I was just using "out-of-sync in the observer's frame" in a way that didn't match how you interpreted this phrase, so that your objection was a purely semantic one.
Two events separated by dt,dx in frame F are perceived simultaneously in frame F'. What is their time separation in frame F?

0=dt'=\gamma(dt-vdx/c^2)

so:

dt=vdx/c^2

The observer in frame F' does not "measure" a de-synchrnization of vdx/c^2. The observer in F' only infers a de-synchronization from the fact that the events appeared simultaneous in F'.
Again you seem to be assuming a certain use of language which doesn't match how physicists normally talk. It is standard in relativity to talk about anything that is true in one frame to be what is "measured" by an observer at rest in that frame (for example, the bottom of p. 1 of this pdf (http://web.mit.edu/sahughes/www/8.022/lec11.pdf) says "inertial observers in different frames of reference measure different intervals of time between events"), and it is also standard to say that in a frame where two clocks are moving at v they are "unsynchronized" by vL/c^2 in that frame (for example, I quoted this textbook (http://books.google.com/books?id=jBgu9psKkREC&q=%22unsynchronized+by%22+simultaneity&dq=%22unsynchronized+by%22+simultaneity&hl=en&ei=q99WTLuuKcT38Aa2qdmgBA&sa=X&oi=book_result&ct=result&resnum=2&ved=0CCkQ6AEwAQ) earlier saying "Suppose there is a clock at the planet P synchronized in S with Homer's clock at Earth. In S', these clocks are unsynchronized by the amount L0V/c^2.") In any case I didn't actually use the word "measure" to describe how out-of-sync the clocks are in frame F', for example my original post #10 which you objected to just said "Relativity of simultaneity says that if two clocks are a distance L apart and synchronized in their mutual rest frame, then they are out-of-sync by vL/c^2 in the frame that sees them moving at v."

starthaus

Aug2-10, 03:27 PM

No, it doesn't mean that, not if we aren't considering a pair of events that are simultaneous in the clocks' rest frame, like the pair "front clock reads T=0" and "back clock reads T=8". If the clocks are synchronized in frame F, then for these two events dt=8.

Synchronized means showing the same exact time, i.e. dt=0. The above does not qualify. What you mean is "running at the same rate" (which is trivial, since the clocks are at rest wrt each other). This is why natural language is not an appropriate language for physics.

In any case I didn't actually use the word "measure" to describe how out-of-sync the clocks are in frame F', for example my original post #10 which you objected to just said "Relativity of simultaneity says that if two clocks are a distance L apart and synchronized in their mutual rest frame,

...which, once again, means dt=0 (see above)

then they are out-of-sync by vL/c^2 in the frame that sees them moving at v."

I think we've beaten this one to death.

JesseM

Aug2-10, 03:44 PM

Synchronized means showing the same exact time, i.e. dt=0.
They do always show the same time at any given time in F. At t=0, the trailing clock reads T=0, and the leading clock reads T=0. At t=8, the trailing clock reads T=8, and the leading clock reads T=8. That's what any physicist would understand the phrase "synchronized in frame F" to mean--that if you compare their readings at a single time in frame F, both clocks show the same reading at this time. Do you interpret "synchronized in frame F" to mean something different?

If not, just note that in the equation dt' = gamma*(dt - v*dx/c^2), dx and dt can stand for the distance and time interval between any two events I choose, so if I choose event #1 to be "the leading clock reads T=0" and I choose event #2 to be "the trailing clock reads T=8", then since event #1 occurs at t=0 in frame F and event #2 occurs at t=8 in frame F, the dt for this specific pair of events is dt=8, despite the fact that the events both occur on the worldlines of two clocks which are "synchronized in frame F" according to the definition above.

starthaus

Aug2-10, 04:32 PM

They do always show the same time at any given time in F. At t=0, the trailing clock reads T=0, and the leading clock reads T=0. At t=8, the trailing clock reads T=8, and the leading clock reads T=8. That's what any physicist would understand the phrase "synchronized in frame F" to mean--that if you compare their readings at a single time in frame F, both clocks show the same reading at this time. Do you interpret "synchronized in frame F" to mean something different?

No, what gives you the idea?

If not, just note that in the equation dt' = gamma*(dt - v*dx/c^2), dx and dt can stand for the distance and time interval between any two events I choose, so if I choose event #1 to be "the leading clock reads T=0" and I choose event #2 to be "the trailing clock reads T=8", t

I chose "trailing clock AND leading clock at the same time T", hence dt=0

JesseM

Aug2-10, 04:37 PM

No, what gives you the idea?
OK, so do you agree that for the synchronized clocks I described, the event "leading clock reads T=0" occurs at time t=0 in frame F, and the event "trailing clock reads T=8" occurs at time t=8 in frame F, so between this specific pair of events we have dt=8, and this doesn't contradict the fact that the two clocks are "synchronized" in frame F?
I chose "trailing clock AND leading clock at the same time T", hence dt=0
I'm not talking about the events you chose, I'm talking about the events "leading clock reads T=0" and "trailing clock reads T=8". Do you disagree that in frame F dt=8 for these events, and in frame F' dt'=0 for these events? If you don't disagree with that, do you disagree that according to the definition of "unsynchronized by" I gave above, it is correct to say that in frame F' the two clocks are "unsynchronized by 8 seconds", since in frame F' the event of the leading clock reading T=0 occurs simultaneously with the event of the trailing clock reading T=8?

You may not like this definition of "unsynchronized by", but that's just a semantic quibble, and I already showed that physicists tend to use "unsynchronized" in the same way. So please tell me if you disagree that according to this definition, it is correct to say that the clocks are unsynchronized by 8 seconds in frame F'.

starthaus

Aug2-10, 04:41 PM

OK, so do you agree that for the synchronized clocks I described, the event "leading clock reads T=0" occurs at time t=0 in frame F, and the event "trailing clock reads T=8" occurs at time t=8 in frame F, so between this specific pair of events we have dt=8, and this doesn't contradict the fact that the two clocks are "synchronized" in frame F?

I'm not talking about the events you chose, I'm talking about the events "leading clock reads T=0" and "trailing clock reads T=8". Do you disagree that in frame F dt=8 for these events, and in frame F' dt'=0 for these events? If you don't disagree with that, do you disagree that according to the definition of "unsynchronized by" I gave above, it is correct to say that in frame F' the two clocks are "unsynchronized by 8 seconds", since in frame F' the event of the leading clock reading T=0 occurs simultaneously with the event of the trailing clock reading T=8?

If you insist on this choice then:

dt'=\gamma(dt-vdx/c^2)

JesseM

Aug2-10, 04:49 PM

If you insist on this choice then:

dt'=\gamma(dt-vdx/c^2)
Yes, that's the equation I already mentioned I was using, in post #41 for example:
If not, just note that in the equation dt' = gamma*(dt - v*dx/c^2), dx and dt can stand for the distance and time interval between any two events I choose, so if I choose event #1 to be "the leading clock reads T=0" and I choose event #2 to be "the trailing clock reads T=8", then since event #1 occurs at t=0 in frame F and event #2 occurs at t=8 in frame F, the dt for this specific pair of events is dt=8, despite the fact that the events both occur on the worldlines of two clocks which are "synchronized in frame F" according to the definition above.
And also earlier, in post #24:
Your equation dt' = gamma*v*dx/c^2 is incorrect for those two events, because dt was nonzero for the events I picked. The full Lorentz transformation equation is

dt' = gamma*(dt - v*dx/c^2)

If dt is not zero, you can't reduce this to dt' = gamma*v*dx/c^2. For the two events I chose (the event of the leading clock reading T=0, and the event of the trailing clock reading T=8), dt was not zero, here dt=8.

starthaus

Aug2-10, 04:53 PM

Yes, that's the equation I already mentioned I was using, in post #41 for example:

Good, because this is exactly the equation I showed early (http://www.physicsforums.com/showpost.php?p=2822598&postcount=17) in post 17 in this thread.
Now, you can get a physically measurable result from it by accepting that "synchronized" means dt=0 or you can get a non-measurable entity by continuing to insist that the "measurement" is done by imposing dt'=0. Of course, in the latter case, there is no physical measurement that can be done, just a mathematical inference since you have already imposed dt'=0.

JesseM

Aug2-10, 05:08 PM

Now, you can get a physically measurable result from it by accepting that "synchronized" means dt=0
The way physicists talk, "synchronized" is a property of the clocks themselves, not a property of the pair of events on each clocks' worldline we choose to analyze. If at every time coordinate t, both clocks read T=t in frame F (like my example where both clocks read T=0 at time coordinate t=0, and both read T=8 at time coordinate t=8), the clocks themselves are said to be "synchronized" in frame F, and that remains true of the clocks even if we choose to look at a pair of events on each clock's worldlines which occur at different times in F. If you define "synchronized" relative to the pair of events we choose as opposed to the clocks themselves, you are using language in a bizarre way that you shouldn't expect any physicist to understand without a detailed explanation.

Besides, you seem to contradict yourself, since I earlier asked:
They do always show the same time at any given time in F. At t=0, the trailing clock reads T=0, and the leading clock reads T=0. At t=8, the trailing clock reads T=8, and the leading clock reads T=8. That's what any physicist would understand the phrase "synchronized in frame F" to mean--that if you compare their readings at a single time in frame F, both clocks show the same reading at this time. Do you interpret "synchronized in frame F" to mean something different?
And you replied:
No, what gives you the idea?
But you can see that according to the definition of "synchronized in F" I give above, all that matters is that the clocks always show the same reading at any single time in F, that remains true even if we choose to analyze two events on their worldlines which don't occur at the same time in F.
or you can get a non-measurable entity by continuing to insist that the "measurement" is done by imposing dt'=0. Of course, in the latter case, there is no measurement that can be done.
You don't think it's possible for an observer in F' to measure what both clocks read at a single time t'? If not, you must be using some weird definition of "measure" which again doesn't match the terminology used by physicists. As I said earlier in post #39:
It is standard in relativity to talk about anything that is true in one frame to be what is "measured" by an observer at rest in that frame (for example, the bottom of p. 1 of this pdf (http://web.mit.edu/sahughes/www/8.022/lec11.pdf) says "inertial observers in different frames of reference measure different intervals of time between events")

starthaus

Aug2-10, 05:14 PM

The way physicists talk, "synchronized" is a property of the clocks themselves, not a property of the pair of events on each clocks' worldline we choose to analyze. If at every time coordinate t, both clocks read T=t in frame F (like my example where both clocks read T=0 at time coordinate t=0, and both read T=8 at time coordinate t=8), the clocks themselves are said to be "synchronized" in frame F, and that remains true of the clocks even if we choose to look at a pair of events on each clock's worldlines which occur at different times in F.

The point is that you don't even have to "look at them" from frame F' "(if you ever accept that "synchronized" means dt=0).

For chuckles and grins, how do you look at two clocks separated by say a proper distance of 100m whizzing past you at 0.8c? Please explain.I am an experimental physicist and I must confess that I have no clue how one can achieve what you are suggesting.

JesseM

Aug2-10, 05:30 PM

The point is that you don't even have to "look at them" from frame F' "(if you ever accept that "synchronized" means dt=0).
It doesn't mean that according to how physicists use the term "synchronized", no, it only means that at the clocks always show the same reading simultaneously in their rest frame. I can provide quotes showing them using it my way rather than your way if you like.
For chuckles and grins, how do you look at two clocks separated by say 100m whizzing past you at 0.8c? Please explain.
The standard textbook procedure for physically defining the meaning of an inertial frame's coordinates involves a grid of measuring-rods and clocks at rest in that frame, with the clocks synchronized according to the Einstein synchronization convention (http://en.wikipedia.org/wiki/Einstein_synchronisation), and with the coordinates of any event being defined in terms of local readings on this system of measuring-rods and clocks. So, suppose F' has a ruler at rest relative to him which defines his x'-axis, and at each marking on the ruler there's a clock which is also at rest in F'. Then if he's observing a pair of clocks moving relative to himself in the -x' direction, he might observe through his telescope that at the moment the leading clock passed the x'=0 light-second marker on his ruler, the leading clock read T=0, while his own clock at rest at that marker read t'=0 seconds. Then he might also observe through his telescope that at the moment the trailing clock passed the x'=6 light-second marker on his ruler, the trailing clock read T=8, while his own clock at rest at that marker read t'=0 seconds. Regardless of how long it takes the light from these events to reach his telescope (wherever he is sitting), because of these local readings on his measuring-rod/clock system, he will assign the event of the leading clock reading T=0 coordinates of x'=0, t'=0 and he will assign the event of the trailing clock reading T=8 coordinates of x'=6, t'=0. That's the standard meaning of how inertial observers "measure" the coordinates of different events in SR, using local readings on this sort of system of measuring rods and clocks.

starthaus

Aug2-10, 05:36 PM

he might observe through his telescope that at the moment the leading clock passed the x'=0 light-second marker on his ruler, the leading clock read T=0, while his own clock at rest at that marker read t'=0 seconds. Then he might also observe through his telescope that at the moment the trailing clock passed the x'=6 light-second marker on his ruler, the trailing clock read T=8,

You mean the observer manages to swivel the telescope instantaneously such that he can observe both clocks simultaneously (as mandated by your condition dt'=0) while they whizz by him at 0.8c? And with no motion blur such that he can read the clock faces?

Aaron_Shaw

Aug2-10, 05:41 PM

starthaus

Aug2-10, 05:45 PM

I'm sure one could manufacture a device like a telescope which is able to observe in two different directions at once.

Indeed. What about the motion blur? You think that they can do something about the 0.8c?
You must also solve the problem that the two light paths from the two clocks aren't equal, so you will need to adjust the clock readings (that is, after you manage to de-blur the images you are getting at 0.8c).

JesseM

Aug2-10, 05:48 PM

You mean the observer manages to swivel the telescope instantaneously such that he can observe both clocks simultaneously (as mandated by your condition dt'=0) while they whizz by him at 0.8c?
Well, for one thing the observer might be at different distances from the x'=0 mark and the x'=6 mark, in which case the light from the two events wouldn't reach him at the same time. But you can also just imagine there is a camera next to each mark that takes continuous footage of events in the vicinity of that mark, and that each camera is continuously sending data to the observer's computer via cables, or that the observer periodically flies out to the different cameras to see what they've recorded. This is just meant to be a theoretical procedure for defining each frame's coordinates anyway, as long as it'd be possible in principle you don't have to worry about how practical it would be (since in practice you can use other methods like parallex to gauge the distance of different events, and divide distance by speed of light to figure out the time interval between when the event occurred in your frame and when the light from the event actually reached you).

starthaus

Aug2-10, 05:52 PM

Well, for one thing the observer might be at different distances from the x'=0 mark and the x'=6 mark, in which case the light from the two events wouldn't reach him at the same time. But you can also just imagine there is a camera next to each mark that takes continuous footage of events in the vicinity of that mark, and that each camera is continuously sending data to the observer's computer via cables, or that the observer periodically flies out to the different cameras to see what they've recorded.

...and how does the observer ensure that the snapshots are simultaneous? How does he physically realize the condition dt'=0?

This is just meant to be a theoretical procedure for defining each frame's coordinates anyway, as long as it'd be possible in principle you don't have to worry about how practical it would be

In my profession, I do.

JesseM

Aug2-10, 05:59 PM

Indeed. What about the motion blur? You think that they can do something about the 0.8c?
You have to distinguish between theoretical statements about what would be true in idealized experiments which don't violate any fundamental laws of physics, and practical statements about what we might actually measure. In practice we've never been able to accelerate a pair of clocks to relativistic speed to observe how if they are synchronized in their own frame according to Einstein's procedure they will be out-of-sync in our frame, but we know this is a theoretical prediction of SR.
You must also solve the problem that the two light paths from the two clocks aren't equal, so you will need to adjust the clock readings (that is, after you manage to de-blur the images you are getting at 0.8c).
You don't have to worry about light delays if you are using local readings on your own set of synchronized clocks. If I am sitting at position x'=100, and at t'=94 I see the image of an event happening right next to the x'=6 marker on my ruler, and my clock at that marker reads t'=0 in the image, then I will assign that event a time coordinate of t'=0 even though I didn't actually learn about it until t'=94.

JesseM

Aug2-10, 06:02 PM

...and how does the observer ensure that the snapshots are simultaneous? How does he physically realize the condition dt'=0?
Again, by looking at local clocks at rest in his frame which are in the immediate vicinity of where the snapshot was taken, clocks which have been previously "synchronized" in the observer's frame according to the Einstein synchronization convention (http://en.wikipedia.org/wiki/Einstein_synchronisation).
In my profession, I do.
Well, things work different in theoretical physics, where physicists often explore the predictions of the laws of physics for situations that would be too difficult to realize as practical experiments for now.

starthaus

Aug2-10, 06:26 PM

Well, things work different in theoretical physics, where physicists often explore the predictions of the laws of physics for situations that would be too difficult to realize as practical experiments for now.

In other words, the experiment is not physically realizable with the convention dt'=0 for labeling the clocks as synchronized.

Austin0

Aug3-10, 01:17 AM

Desynchronization is not dependant on specific events but can be calculated purely as a function of velocity and spatial separation between clocks in their own rest frame. Yes??

If you think it would apply to L as per JM i.e. proper length, then either you are simply wrong or I have to relearn relative simultaneity.

it is the latter

I think I may hold off on any adjustments to my understanding of relative simultaniety.

Given an inertial frame F1 with two clocks separated by 10 units of 1 ls ,,as measured in that frame and an observation frame F2 ......v= 0.8 c
WOuld you question that as observed and measured in F2 that the degree of desynchronization between the clocks in F1 would be 0.8* 10 = (+) or( - ) 8.0 sec. ????

You are maintaining that the desynch would be 1.666667 * 0.8*10 =13.3333 ????

I am not "maintaining". 13.33ns is the correct result.

With your understaning of the Lorentz math it should be blindingly obvious that simply given a relative velocity of 0.8 c it can be stated with certainty that the relative degree of desynchronization between the two frames must be (+) or (-) 0.8 sec per light second of proper spatial separation of clocks . Without further information or recourse.
This applies reciprocally from either frame.

Your 13.33333 sec is only correct for specific locations x',0 and x', 16.6667=dx'= 16.6667 . I t has no general truth regarding the desynchronization between the frames and is only correct at all because you have idiosyncratically turned the conditions as stated upside down and insisted on working from the other frame.
I can accept correction when warranted but I fail to discern either point or purpose in your exercise as it educates none , apparently not even yourself IMHO

JesseM

Aug3-10, 11:14 AM

In other words, the experiment is not physically realizable with the convention dt'=0 for labeling the clocks as synchronized.
It's "physically realizable" in the sense that it could be realized without violating any laws of physics, but it's not practically realizable with present technology, no. But the point is that no experiment comparing two events on clock worldlines from the perspectives of two frames moving at relativistic velocities relative to one another is practically realizable, because we can't really achieve relativistic velocities relative to the Earth. So, it's also not practically realizable to look at two events with dt=0 in the clock rest frame and then show experimentally that the time between these events in a frame moving at relativistic velocity relative to the clocks is dt'=gamma*L*v/c^2.

starthaus

Aug3-10, 11:32 AM

It's "physically realizable" in the sense that it could be realized without violating any laws of physics, but it's not practically realizable with present technology, no.

Which means that such a definition of synchronization has no practical use.

But the point is that no experiment comparing two events on clock worldlines from the perspectives of two frames moving at relativistic velocities relative to one another is practically realizable, because we can't really achieve relativistic velocities relative to the Earth. So, it's also not practically realizable to look at two events with dt=0 in the clock rest frame and then show experimentally that the time between these events in a frame moving at relativistic velocity relative to the clocks is dt'=gamma*L*v/c^2.

This is incorrect, the synchronization definition employing dt=0 does not imply any observation of the clocks. All you need to know is that they had been accelerated slowly enough such that they did not get out of synch. For that extent, the clocks may be totally enclosed in arocket, unobservable, you will still be able to determine their desynchroonization in frame F' based only on two things : proper length L and coordinate speed v.

JesseM

Aug3-10, 04:52 PM

Which means that such a definition of synchronization has no practical use.
So do you also think the Lorentz transformation equation dt' = gamma*(dt - v*dx/c^2) has no practical use? Is there any practical way to verify that for two events that have a dx of 10 light-seconds in our frame and a dt of 8 seconds in our frame, the events would have a dt' of 0 in a frame moving at 0.8c relative to us?
This is incorrect, the synchronization definition employing dt=0 does not imply any observation of the clocks.
I don't know what "synchronization definition" you are using that's different from mine. My definition had nothing to do with a specific pair of events so I am not considering any particular value of dt, my definition is just that if two clocks show the same reading at every time-coordinate in a given frame, they are "synchronized" relative to that frame. Are you using a different definition?
All you need to know is that they had been accelerated slowly enough such that they did not get out of synch.
"Did not get out of sync" in your frame, or in the rest frame of the clocks? If you accelerate both clocks at the same coordinate acceleration in your rest frame, then both clocks will remain synchronized in your frame, which means according to my definition of synchronization above, they will become out-of-sync in their new rest frame.
For that extent, the clocks may be totally enclosed in arocket, unobservable, you will still be able to determine their desynchroonization in frame F' based only on two things : proper length L and coordinate speed v.
OK, suppose that while the rocket is accelerating up to relativistic speed, both clocks are right next to each other in the middle of the rocket. This means we can know theoretically that they remain "synchronized" in all frames (both our frame and other frames) according to my definition above, as long as they are together. Then once the rocket is coasting, there's a device which automatically pushes one clock slowly along a track towards the front of the rocket, and which automatically pushes the other slowly along a track towards the back (the device is designed to push both clocks at the same slow speed relative to the rocket). In that case, we can determine theoretically that the after reaching the front and back the two clocks should still be "synchronized" in the current rest frame of the rocket (according to my definition above), and that this means that two readings on the clocks which have a separation of vL/c^2 should occur simultaneously in our frame.

starthaus

Aug3-10, 05:44 PM

So do you also think the Lorentz transformation equation dt' = gamma*(dt - v*dx/c^2) has no practical use?

I never said such a thing, please stop putting words in my mouth and try to stick to the subject being discussed.

Is there any practical way to verify that for two events that have a dx of 10 light-seconds in our frame and a dt of 8 seconds in our frame, the events would have a dt' of 0 in a frame moving at 0.8c relative to us?

No, there isn't. You, yourself have admitted earlier (http://www.physicsforums.com/showpost.php?p=2824279&postcount=55) on that there isn't any practical way.

I don't know what "synchronization definition" you are using that's different from mine.

You are using dt'=0 which , according to your own admission is impossible to implement practically.
I am using dt=0 which is very easy to implement (and is implemented in prcatice).

OK, suppose that while the rocket is accelerating up to relativistic speed, both clocks are right next to each other in the middle of the rocket.

This is not the original definition of the problem and has nothing to do with the original experiment. In the original experiment the clocks start separated by the distance L. Please stick with the experiment under discussion.

JesseM

Aug3-10, 08:14 PM

I never said such a thing, please stop putting words in my mouth and try to stick to the subject being discussed.
I didn't put words into your mouth, I asked about what your position on this is.
No, there isn't. You, yourself have admitted earlier (http://www.physicsforums.com/showpost.php?p=2824279&postcount=55) on that there isn't any practical way.
So, do you think we can ever verify that the equation dt' = gamma*(dt - v*dx/c^2) is correct practically in a case where we pick two events with dt not equal to zero? If not, does this equation have any "practical use" if we already know dt' = gamma*v*dx/c^2 applies in cases where dt=0?
You are using dt'=0 which , according to your own admission is impossible to implement practically.
I am using dt=0 which is very easy to implement (and is implemented in prcatice).
I don't know what you mean by "implement". It is not possible in practice to ever actually measure any value for dt' if the primed frame is moving at a relativistic speed relative to us, since in practice we can't actually get a system of rulers and clocks moving at relativistic speed relative to us. But if you allow theoretical conclusions based on knowledge of the time between events dt in our frame, like the conclusion that two events with dt=0 in our frame must occur with a separation of dt'=gamma*v*dx/c^2, then there's no justification for not allowing it for events with nonzero dt. For example, you don't even need to accelerate clocks at all, you can just consider two clocks at rest in F which are 10 light-seconds apart and synchronized in F. In this case if you define your events to be "left clock reading T=0" and "right clock reading T=8", then here dx=10 and dt=8--since these clocks are at rest relative to us it isn't hard to measure their positions and times. And then as a theoretical conclusion, we know it must be true that dt' = gamma*(dt - v*dx/c^2), so if the primed frame has v=0.8 this implies dt'=0 for this particular pair of events.
This is not the original definition of the problem and has nothing to do with the original experiment.
In what post was an "original experiment" specified?
In the original experiment the clocks start separated by the distance L.
I don't remember any post that said they "start" separated by L, it was just said that the clocks are at a distance of L in their rest frame F and are moving at v relative to the observer's frame F'. In my example this will be true as soon as the mechanism moves the clocks to the front and back of the rocket. We can define the "start" as some time when the rocket is already coasting and the mechanism has finished moving the clocks.

If you somehow think we must define the "start" as before the rocket leaves Earth, in this case your example doesn't meet the specified conditions at the "start" either, since at that point the clocks weren't yet at rest in F' which was one of the conditions discussed.

starthaus

Aug3-10, 09:46 PM

But if you allow theoretical conclusions based on knowledge of the time between events dt in our frame, like the conclusion that two events with dt=0 in our frame must occur with a separation of dt'=gamma*v*dx/c^2, then there's no justification for not allowing it for events with nonzero dt.

Yes, there is. Not a theoretical disproof but a practical one, the measurement in the latter case requires the realization of the condition dt'=0 in the observer frame F'. By your own admission (http://www.physicsforums.com/showpost.php?p=2824279&postcount=55), this is not realizable.

JesseM

Aug4-10, 10:35 AM

Yes, there is. Not a theoretical disproof but a practical one, the measurement in the latter case requires the realization of the condition dt'=0 in the observer frame F'. By your own admission (http://www.physicsforums.com/showpost.php?p=2824279&postcount=55), this is not realizable.
My own claim was about the practicality of measuring dt' in frame F', which is equally impossible in practice for dt'=0 and dt'=gamma*v*L/c^2, because we can't practically get rulers and clocks accelerated to the relativistic speed of F'. Your own argument about dt'=gamma*v*L/c^2 was based on the wholly impractical notion of a rocket being accelerated to a relativistic speed, and then on a purely theoretical calculation of what would be true in F' based on what we know is true in F. If this practically impossible scenario counts as a "realization" of dt'=gamma*v*L/c^2 in your increasingly bizarre terminology, then simply having two synchronized clocks at rest in F and picking two events on their worldlines with coordinates x=0,t=0 and x=10,t=8 in F (like the event of the first clock reading T=0 and the event of the second clock reading T=8) should also count as a "realization" of dt'=0, since a purely theoretical calculation of what would be true in F' also shows that dt'=0 for this pair of events.

Likewise, if you want a "realization" of dt'=0 not just for any pair of events but for a pair of events on the worldlines of clocks actually at rest in F' (as opposed to clocks at rest in F as above), then if "realization" can include the wholly impractical notion of a rocket that can accelerate to relativistic speeds, then my thought-experiment with the rocket that has a mechanism that slowly moves the clocks to either side should also count as just as much of a "realization" as your "realization" of dt'=gamma*v*L/c^2.

starthaus

Aug4-10, 10:49 AM

My own claim was about the practicality of measuring dt' in frame F', which is equally impossible in practice for dt'=0 and dt'=gamma*v*L/c^2,

This is not true if you assume dt=0 instead of dt'=0. The impractical aspect goes away once you replace your condition dt'=0 with my condition dt=0.

JesseM

Aug4-10, 10:57 AM

This is not true if you assume dt=0 instead of dt'=0.
Oh really? If F' is moving at 0.8c relative to our frame F, do you have a practical experimental method to verify that dt'=gamma*v*L/c^2 for two events which have dt=0 in F, using clocks that are actually at rest in F' to experimentally measure dt'? Or are you resorting to a purely theoretical calculation of the value of dt' in F'?

starthaus

Aug4-10, 11:00 AM

Oh really? If F' is moving at 0.8c relative to our frame F, do you have a practical experimental method to verify that dt'=gamma*v*L/c^2 for two events which have dt=0 in F, using clocks that are actually at rest in F' to experimentally measure dt'? Or are you resorting to a purely theoretical calculation of the value of dt' in F'?

Sure I do, I am an experimentalist, remember? All I need to do is to determine v. The beauty of the method is that it works even with variable v. As long as you accept that SR is correct, the determination holds.

JesseM

Aug4-10, 12:28 PM

Sure I do, I am an experimentalist, remember? All I need to do is to determine v. The beauty of the method is that it works even with variable v. As long as you accept that SR is correct, the determination holds.
So you are just doing a theoretical calculation in F', not actually experimentally measuring dt' with clocks at rest in F'? If so, am I also allowed to just measure dx and dt for two events in F, then do a theoretical calculation to find dt' in F?

starthaus

Aug4-10, 01:55 PM

So you are just doing a theoretical calculation in F', not actually experimentally measuring dt' with clocks at rest in F'? If so, am I also allowed to just measure dx and dt for two events in F, then do a theoretical calculation to find dt' in F?

No, you are not since your experiment depends directly on dt'=0.

JesseM

Aug4-10, 02:02 PM

No, you are not since your experiment depends directly on dt'=0.
I don't know what you mean by "depends directly on". The result of the theoretical calculation is dt'=0, but I start with the measured values of dx=10 and dt=8. Similarly you start with the measured values of dx=L and dt=0, and the result of your theoretical calculation is dt'=gamma*v*L/c^2. What possible difference can you point to that allows you to say your experiment is a "realization" of dt'=gamma*v*L/c^2 but mine is not a "realization" of dt'=0? In both cases we measure a dx and dt, then calculate a dt' theoretically using the equation dt' = gamma*(dt - v*dx/c^2).

starthaus

Aug4-10, 04:17 PM

I don't know what you mean by "depends directly on".

It means that, according to your own description (http://www.physicsforums.com/showpost.php?p=2822869&postcount=20), you need to observe both clocks simultaneously in frame F'.

JesseM

Aug4-10, 04:38 PM

It means that, according to your own description (http://www.physicsforums.com/showpost.php?p=2822869&postcount=20), you need to observe both clocks simultaneously in frame F'.
No, I already explained that this was only if you wanted to measure dt' using rulers and clocks at rest in frame F'. But your definition of what it means to "realize" dt'=gamma*v*L/c^2 does not require us to measure dt', it allows us to just calculate it theoretically from a dx and dt that we have measured in F. So you can stop linking back to that post of mine, it isn't relevant to the question of why you think it's possible to "realize" dt'=gamma*v*L/c^2 but not to "realize" dt'=0. If you allow a theoretical determination of dt' from measured dx and dt to count as a "realization" in the first case, why don't you allow it in the second case? Appears to be an irrational double-standard on your part.

starthaus

Aug4-10, 04:53 PM

No, I already explained that this was only if you wanted to measure dt' using rulers and clocks at rest in frame F'. But your definition of what it means to "realize" dt'=gamma*v*L/c^2 does not require us to measure dt', it allows us to just calculate it theoretically from a dx and dt that we have measured in F. So you can stop linking back to that post of mine, it isn't relevant to the question of why you think it's possible to "realize" dt'=gamma*v*L/c^2 but not to "realize" dt'=0. If you allow a theoretical determination of dt' from measured dx and dt to count as a "realization" in the first case, why don't you allow it in the second case?

Because your experiment relies on simultaneous observation of the two clocks in frame F'. Mine doesn't.

Appears to be an irrational double-standard on your part.

There is no point in starting to sling ad-homs, let's keep the discussion civil.

JesseM

Aug4-10, 05:08 PM

Because your experiment relies on simultaneous observation of the two clocks in frame F'.
No it doesn't. In both of the experiments I mentioned in post #65, I measured two events in frame F, getting dx and dt for these events in F, then I did a purely theoretical calculation to get dt' in F' (I have already stated this several times, are you reading my posts carefully?) That's exactly what you did too, so I don't see why your experiment is a "realization" of dt'=gamma*v*L/c^2 but mine is not a "realization" of dt'=0.
There is no point in starting to sling ad-homs, let's keep the discussion civil.
My point is that there's no rational argument for treating your type of experiment as a "realization" of dt'=gamma*v*L/c^2 but not treating mine as a "realization" of dt'=0.

starthaus

Aug4-10, 06:02 PM

No it doesn't. In both of the experiments I mentioned in post #65, I measured two events in frame F, getting dx and dt for these events in F,

How do you do these "measurements"? You are located in frame F', in motion (at very high speed) wrt F.

JesseM

Aug4-10, 06:30 PM

How do you do these "measurements"? You are located in frame F', in motion (at very high speed) wrt F.
OK, I see we were originally using F' to be the observer's frame, but in more recent posts I got mixed up and started using F as the observer's frame, sorry about the confusion. But is that the definition you have been assuming in all your posts? For example, in post #62 you said:
You are using dt'=0 which , according to your own admission is impossible to implement practically.
I am using dt=0 which is very easy to implement (and is implemented in prcatice).
Were you really saying here that it's impossible to "implement" a measurement of two events in our own primed frame such that we find dt'=0 for these events? If we synchronize two clocks at rest in our primed frame using the Einstein synchronization convention, then if we consider an event that was measured to happen next to the left clock when it read T'=5 and another event which was measured to happen next to the right clock when it read T'=5, doesn't this qualify as a measurement of dt'=0 for these two events?

If you got the notation confused too, can you clarify what you meant in this comment from post #60?
All you need to know is that they had been accelerated slowly enough such that they did not get out of synch. For that extent, the clocks may be totally enclosed in arocket, unobservable, you will still be able to determine their desynchroonization in frame F' based only on two things : proper length L and coordinate speed v.
Is F' still supposed to be the observer's frame? And were the clocks accelerated from an initial state of rest in F'? If so, then if they were both accelerated at the same rate, wouldn't they just remain synchronized in F'? Or were you assuming in this comment that F was the observer's frame and F' was the rest frame of the rocket?