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WIKI and Time Dilation

by chinglu1998
Tags: dilation, time, wiki
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chinglu1998
#73
Jan2-11, 11:39 PM
P: 182
Quote Quote by IsometricPion View Post
As with the calculation in my previous post, [tex]$\Delta{}t'=\gamma{}\Delta{}t\Rightarrow{}\Delta{}n'=\frac{\Delta{}n}{\ gamma}$[/tex], so in each frame clocks in motion tick slower than clocks at rest (no matter which frame is taken to be at rest).
OK, how do you explain the WIKI article.

Would you please use the experiment in the article to prove reciprocal time dilation?
JesseM
#74
Jan2-11, 11:40 PM
Sci Advisor
P: 8,470
Quote Quote by chinglu1998 View Post
Tes, I start to think WIKI is correct. The absolute light abberation causes a longer path for the light travel for the primed framer vs the unprimed frame in an absolute sense.

Let me take the unprimed frame as stationary. This promed frame light beam still longer.

cēt'ē = vēt'ē + cētē

cēt'ē - vēt'ē = cētē

t'ē( cē - vē) = cētē

t' = tγ

Wow, when taking the unprimed frame as stationary with absolute light abberation, the moving clock beats faster just like WIKI said.
No, any clock in the moving (primed) frame must be running slow as measured in the stationary (unprimed) frame. I showed why this is not incompatible with the wiki's time dilation equation in post #62.

You have to define the precise meanings of t and t' in your equations. t' doesn't deal with any single "moving clock", rather it is the time in the primed frame between the event of the light being emitted from the bottom of the light clock (which is at rest in the unprimed frame and therefore moving relative to the primed frame) and the event of it hitting the top of the light clock (this event will occur at a different position in the primed frame). To measure the time of these two events in the primed frame using clocks, you would need a pair of clocks at different positions in the primed frame. As I explained in #62, each of these primed clocks is individually running slow as measured in the unprimed frame, but the unprimed frame nevertheless agrees that t' is greater than t because of the way the two primed clocks are out-of-sync in the unprimed frame.
darkhorror
#75
Jan2-11, 11:41 PM
P: 140
Quote Quote by chinglu1998 View Post
I did. Are they starting at (0,0,0,0) for both?
It was a very simple question, you don't have to do any math. Just read the question. If you have a space ship moving away from earth at .6c. From the ship's frame of reference which clock is moving slower the one of the ship or the one on earth?
grav-universe
#76
Jan2-11, 11:41 PM
P: 429
Quote Quote by chinglu1998 View Post
If you do the math on WIKI, you will find if you take the unprimed frame as rest with respect to the primed frame, then t' = t γ.

If you take the primed frame as rest with respect to the unprimed frame, then t' = t γ.

So either way, it is absolute that t' = t γ. Why is this? This light beam path for the primed frame is absolutly longer than that of the unprimed frame because of light abberation. This is not a relative issue.
From the perspective of the primed frame, we have t = t' y. From the perspective of the unprimed frame, we have t' = t y. The unprimed frame is comparing the period for the light pulse to travel between the mirrors in the primed frame to the period that the primed frame measures for the same pulse to travel to the same mirrors in the primed frame also. The primed frame, however, is comparing the period for a different pulse to travel between the mirrors in the unprimed frame to the period the unprimed frame measure for that same pulse in the unprimed frame. These are two completely different sets of circumstances.
chinglu1998
#77
Jan2-11, 11:45 PM
P: 182
Quote Quote by JesseM View Post
There is no single "moving clock". 25s is the time measured in the primed ('moving') frame between two events on the worldline of the clock at rest in the unprimed ('stationary') frame. But these events occur at two different positions in the primed frame, so you'd need a pair of clocks at rest and synchronized in the primed frame to assign time-coordinates to both events in a local way, after which you could figure out their difference in time-coordinates. That's how position and time coordinates are supposed to be assigned in SR, using a lattice of rulers and clocks at rest relative to one another and synchronized using the Einstein synchronization convention, as illustrated here:



Einstein also discusses the idea that coordinate times and coordinate positions should be defined using local measurements on a set of rulers and synchronized clocks in sections 1 and 2 of the 1905 paper you linked to.


As always, "moving" and "stationary" are arbitrary labels. Changing the labels does not magically change any facts about the amount of coordinate time between two specific events in different frames. Instead of playing these silly word-games, why don't you go through my analysis using the Lorentz transformation and figure out specifically where you think I made an error? Do you disagree that if a clock is at rest at x=0 in the unprimed frame, then the coordinates of it showing a time of 0 would be x=0,t=0 while the coordinates of it showing a time of 20 would be x=0,t=20? Do you disagree that when we transform x=0,t=0 into the primed frame we get t'=0, and when we transform x=0,t=20 into the primed frame we get t'=25? Please address these quantitative specifics instead of retreating into vague word-games.
Your conclusions are correct. Do you understand them?

Since LT matrix invertible, then t' = 25 and t = 20 regardless of which frame is at rest.

So if unprimed at rest, t' > t.

If primed frame at rest, then t' > t.

Have you figured this out yet?
chinglu1998
#78
Jan2-11, 11:46 PM
P: 182
Quote Quote by darkhorror View Post
It was a very simple question, you don't have to do any math. Just read the question. If you have a space ship moving away from earth at .6c. From the ship's frame of reference which clock is moving slower the one of the ship or the one on earth?

I am discussing this same subject with Jesse right now.
JesseM
#79
Jan2-11, 11:49 PM
Sci Advisor
P: 8,470
Quote Quote by chinglu1998 View Post
I meant when calculating the moving frame it is Minkowsky. In the moving frame, the relativity postulate holds. So, the equation for the light sphere holds in all frames, but when calculating those framea, it does not, the calculation is Minkowsky.
I don't understand what "calculation" you refer to. If you mean we must use the Lorentz transformation to go from stationary coordinates to moving coordinates I agree, but it is equally true that we must use the Lorentz transformation to go from moving coordinates to stationary coordinates. So again I don't see why you think "Minkowski" applies to the moving frame but not to the stationary frame.
Quote Quote by chinglu1998
If you apply Minkowsky to your own coordinates for you, that is failure. I made this clear. That is called frame mixing.
I don't know what you mean by "apply Minkowski to your own coordinates for you", but whatever it means, presumably it would be equally a mistake for the stationary observer to "apply Minkowski to his own coordinates for himself" and a mistake for the moving observer to "apply Minkowski to his own coordinates for himself"? As always the situation is completely symmetrical, there's no sense in which it is correct to "apply Minkowski" in the moving frame but not to do something equivalent in the stationary frame.
Quote Quote by chinglu1998
How exactly are you going to make this equation true exclusively within a stationary fame?
Huh? I never said anything about it being exclusive to the stationary frame, I said 'this equation still holds if we are dealing with the t,x,y,z coordinates of the "stationary" frame?'--that "still" was to indicate that although it's true this Minkowski equation holds in the moving frame, it also holds in the stationary frame. It holds in all inertial frames, so again I don't see how it makes sense to say that Minkowski geometry should only be used in the moving frame and Euclidean geometry should only be used in the stationary frame. For any equation (Euclidean or Minkowski) which applies to a situation in one frame, you can find an equivalent situation in any other frame where the same equation applies.
chinglu1998
#80
Jan2-11, 11:49 PM
P: 182
Quote Quote by grav-universe View Post
From the perspective of the primed frame, we have t = t' y. From the perspective of the unprimed frame, we have t' = t y. The unprimed frame is comparing the period for the light pulse to travel between the mirrors in the primed frame to the period that the primed frame measures for the same pulse to travel to the same mirrors in the primed frame also. The primed frame, however, is comparing the period for a different pulse to travel between the mirrors in the unprimed frame to the period the unprimed frame measure for that same pulse in the unprimed frame. These are two completely different sets of circumstances.
How do you make this hsppen with the WIKI article give the fact light abberation is absolute?

That means, the clock at rest with the light source will measure t=d/c and the frame that views that clock as moving will also conclude it calculates t=d/c and it will view a light beam at an angle with a longer distance to travel. Light abberation is not reciprocal. It is only true for observers moving relative to the light source.
chinglu1998
#81
Jan2-11, 11:57 PM
P: 182
Quote Quote by JesseM View Post
I don't understand what "calculation" you refer to. If you mean we must use the Lorentz transformation to go from stationary coordinates to moving coordinates I agree, but it is equally true that we must use the Lorentz transformation to go from moving coordinates to stationary coordinates. So again I don't see why you think "Minkowski" applies to the moving frame but not to the stationary frame.
Oh, you are not familiar with topological spaces. You have a space and then a mapping to another space. That mapped space if not isomorphic is not the same as the original space.

So, in my mind, I see the two as two different objects.

There is the stationary space which has Euclidian properties.
There is the Minkowsky space which has odd properties. The light sphere in Minkowsky is off center and not spherical.

It is my view these are different. If you have your view fine. This is simply my view.


I don't know what you mean by "apply Minkowski to your own coordinates for you", but whatever it means, presumably it would be equally a mistake for the stationary observer to "apply Minkowski to his own coordinates for himself" and a mistake for the moving observer to "apply Minkowski to his own coordinates for himself"? As always the situation is completely symmetrical, there's no sense in which it is correct to "apply Minkowski" in the moving frame but not to do something equivalent in the stationary frame.
We agree.

Huh? I never said anything about it being exclusive to the stationary frame, I said 'this equation still holds if we are dealing with the t,x,y,z coordinates of the "stationary" frame?'--that "still" was to indicate that although it's true this Minkowski equation holds in the moving frame, it also holds in the stationary frame. It holds in all inertial frames, so again I don't see how it makes sense to say that Minkowski geometry should only be used in the moving frame and Euclidean geometry should only be used in the stationary frame. For any equation (Euclidean or Minkowski) which applies to a situation in one frame, you can find an equivalent situation in any other frame where the same equation applies.
Good, can you apply the Minkowski metric only to the stationary frame and none other?
JesseM
#82
Jan2-11, 11:59 PM
Sci Advisor
P: 8,470
Quote Quote by chinglu1998 View Post
Your conclusions are correct. Do you understand them?

Since LT matrix invertible, then t' = 25 and t = 20 regardless of which frame is at rest.
I still don't understand the significance of calling one frame "at rest" other than it just being a totally arbitrary label like "frame A" or "the friendly frame". Switching such arbitrary labels has no mathematical significance whatsoever, so if you switch the labels but change nothing else about the problem the mathematical analysis will be exactly the same, you don't have to "invert" the LT matrix or any other such mathematical change in your calculations.
Quote Quote by chinglu1998
So if unprimed at rest, t' > t.

If primed frame at rest, then t' > t.

Have you figured this out yet?
Since I think "at rest" is a totally arbitrary verbal label of course I agree that changing the label won't change any mathematical fact like t' > t. But this is assuming that aside from changing the labels we are leaving the physical facts of the problem unchanged--specifically, we are still dealing with a problem where we have two events on the worldline of a clock in the unprimed frame and want to know the time between these events in both frames. If on the other hand we had two events on the worldline of a clock in the primed frame, then in this case t > t' (and again this would be true regardless of which frame we called 'stationary'). As I said back in post #25, all that matters is which frame the clock we're analyzing is in (i.e. the clock whose worldline the two events are on), once you know that the time dilation equation always has the following form:

(time interval in frame of observer who sees clock in motion) = (time elapsed on clock)*gamma

Do you agree that this always works, regardless of what frame the clock is in or which frame we choose to label "stationary"?
grav-universe
#83
Jan3-11, 12:01 AM
P: 429
Quote Quote by chinglu1998 View Post
How do you make this hsppen with the WIKI article give the fact light abberation is absolute?

That means, the clock at rest with the light source will measure t=d/c and the frame that views that clock as moving will also conclude it calculates t=d/c and it will view a light beam at an angle with a longer distance to travel. Light abberation is not reciprocal. It is only true for observers moving relative to the light source.
They are two different light pulses. Light pulse 1 bounces back and forth between the mirrors in the unprimed frame, so t1 = 2 d / c. A clock in the primed frame measures t1' = (2 d / c) / sqrt(1 - (v / c)^2) for light pulse 1 to travel between the mirrors of the unprimed frame, so t1' = t1 y. Vice versely, a light pulse 2 that bounces between the mirrors in the primed frame will give t2 = t2' y. Each sees the other's clock time dilating.
JesseM
#84
Jan3-11, 12:07 AM
Sci Advisor
P: 8,470
Quote Quote by chinglu1998 View Post
Oh, you are not familiar with topological spaces. You have a space and then a mapping to another space. That mapped space if not isomorphic is not the same as the original space.
A change in coordinates cannot take you into a different "topological space" if the geometry (given by the metric) is unchanged.
Quote Quote by chinglu1998
There is the stationary space which has Euclidian properties.
There is the Minkowsky space which has odd properties. The light sphere in Minkowsky is off center and not spherical.
I have only seen physicists refer to Minkowski spacetime, not "Minkowski space" in the sense of a 3D space containing only a light sphere rather than a full 4D light cone. Is this a term you have made up? Can you give any precise mathematical definition of what you mean by "Minkowski space"?
Quote Quote by JesseM
Huh? I never said anything about it being exclusive to the stationary frame, I said 'this equation still holds if we are dealing with the t,x,y,z coordinates of the "stationary" frame?'--that "still" was to indicate that although it's true this Minkowski equation holds in the moving frame, it also holds in the stationary frame. It holds in all inertial frames, so again I don't see how it makes sense to say that Minkowski geometry should only be used in the moving frame and Euclidean geometry should only be used in the stationary frame. For any equation (Euclidean or Minkowski) which applies to a situation in one frame, you can find an equivalent situation in any other frame where the same equation applies.
Quote Quote by chinglu1998
Good, can you apply the Minkowski metric only to the stationary frame and none other?
Uh, did you read anything I wrote above? I was specifically pointing out that I never claimed "you apply the Minkowski metric only to the stationary frame and none other", this is a strawman that is the exact opposite of what I have been saying. What part of "it's true this Minkowski equation holds in the moving frame" and "it holds in all inertial frames" didn't you understand? It was you who seemed to claim that the moving frame is Minkowski but that the stationary frame is not Minkowski, while I am saying that all frames are on equal footing as far as all equations are concerned, and "stationary" vs. "moving" has no significance except as an arbitrary verbal label.
yuiop
#85
Jan3-11, 12:30 AM
P: 3,967
Quote Quote by chinglu1998 View Post
If you do the math on WIKI, you will find if you take the unprimed frame as rest with respect to the primed frame, then t' = t γ.

If you take the primed frame as rest with respect to the unprimed frame, then t' = t γ.

So either way, it is absolute that t' = t γ. Why is this? This light beam path for the primed frame is absolutly longer than that of the unprimed frame because of light abberation. This is not a relative issue.
If you take the unprimed frame as at rest with respect to the primed frame, then t' = t.

If you take the primed frame as at rest with respect to the unprimed frame, then t' = t.

The v in the gamma factor is the velocity of the two frames relative to each other. If the two frames are at rest with respect to each other then v=0 and gamma=1 and t'=t.

Quote Quote by chinglu1998 View Post
Under SR, a moving clock is supposed to beat slower. What about the WIKI article?
From the equation:

[tex]t' = t \gamma [/tex]

the elapsed time t' is greater than t for any none zero v. Agree?

t is the time measured by a single clock (A) at rest in frame S. Let us say t= 10 seconds. In frame S' clock A appears to be moving at 0.8c which is the velocity of frame S' relative to S. Let us say we have two clocks (B' and C') that are at rest in frame S' and that clock A is initially adjacent to clock B' and ends up adjacent to clock C'. The time measured by the synchronised clocks (B' and C') at rest in S' is 16.666 seconds. So the time measured in frame S' by the moving clock (A) is less than the time measured by clocks B' and C' that at rest in frame S'. This is in agreement with what you say should happen. Note that the time that elapses on clock A is the same (10 seconds) according to observers in frame S AND frame S'. The time interval of 16.666 seconds measured in S' is measured by two spatially separated clocks and is NOT the time measured by clock A according to any observer. Clock A is only running slow in a relative sense compared to the coordinate time difference measured by separate clocks.
IsometricPion
#86
Jan3-11, 01:55 AM
P: 177
Quote Quote by chinglu1998 View Post
Would you please use the experiment in the article to prove reciprocal time dilation?
The upper picture is of the light clock in its rest frame and is used to define the time interval between ticks of the clock in its rest frame. The second picture is of the same light clock in a frame in which it is moving and is used to derive the time interval between ticks in that frame. Since the velocity of the light clock is perpendicular to that of the path the light takes in the light clock, this path length is invariant (under the transformation from the unprimed frame to the primed frame). Then follow a few lines of algebra culminating in a relation between the interval between ticks in the primed and unprimed frames (this algebra only assumes that space is euclidean, the velocity is perpendicular to the light path, and that the primed position is the time-integral of the primed velocity, all of which are valid in SR). If one measures the number of ticks (N) of the unprimed clock in the primed frame during a time period [tex]$\alpha{}$[/tex] as measured by N' ticks of an equivalent clock in the primed frame, [tex]$\Delta{}t'=\gamma{}\Delta{}t\Rightarrow{}\frac{\alpha}{\Delta{}t'}=\fr ac{\alpha{}}{\gamma{}\Delta{}t}\Rightarrow{}N=\frac{\alpha}{\gamma{}\De lta{}t}\,\,\,\,N'=\frac{\alpha}{\Delta{}t}\Rightarrow{}N=\frac{N'}{\gam ma}$[/tex]
Thus, the unprimed clock ticks fewer times than an equivalent clock at rest in the primed frame for an arbitrary time period measured in the primed frame. So, a clock that ticks once per second in its rest frame would be measured to tick 4 times when traveling at 0.6c relative to a frame in which an equivalent clock at rest would tick 5 times.
DaleSpam
#87
Jan3-11, 08:09 AM
Mentor
P: 17,535
Quote Quote by chinglu1998 View Post
when calculating the moving frame ... In the moving frame ... what is moving coordinates and what is stationary ... within a stationary fame
Quote Quote by chinglu1998 View Post
which frame is at rest ... if unprimed at rest ... If primed frame at rest
You still seem to not understand that velocities are relative. None of these statements have any meaning because velocity is a relative quantity and you have not specified what the velocity is relative to in any of these! Each of these needs to be re-written to specify "at rest wrt ___", "stationary wrt ___", "moving wrt ___", etc.

VELOCITIES ARE RELATIVE
DaleSpam
#88
Jan3-11, 08:33 AM
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P: 17,535
Quote Quote by chinglu1998 View Post
There is the stationary space which has Euclidian properties.
I think you mean Galilean rather than Euclidean, but this is incorrect. All frames use the Minkowski metric to determine spacetime intervals, regardless of whether or not they are moving wrt some given frame or object.

Quote Quote by chinglu1998 View Post
There is the Minkowsky space which has odd properties. The light sphere in Minkowsky is off center and not spherical.
I think you mean the light cone. The light cone is a right cone in all frames.

Quote Quote by chinglu1998 View Post
Good, can you apply the Minkowski metric only to the stationary frame and none other?
It applies to all frames.
darkhorror
#89
Jan3-11, 09:39 AM
P: 140
To chinglu1998
Trying to get an example that might make sense to see what they are talking about.

If we have two frames of reference the stationary frame and the moving frame. moving frame is moving away from stationary frame at .6c.

In the "moving frame" frame of reference the "stationary frame" is moving at .6c, and the "moving frame" is at rest.

Then in the "stationary frame" frame of reference the "moving frame" is moving at .6c, and the "stationary frame" is at rest.

This also means that which ever frame you chose the other frame's clock is going to be ticking slower.

So if you ask which clock is moving slower you have to ask in which frame of reference. Otherwise the question is meaningless.

So lets say you have two clocks t and t'. Clock t is in the "stationary frame", t' is in the "moving frame"

Now if I put the observer in the "moving frame", to this observer clock t is ticking slower than t'.

This is basicly what they are talking about on wiki
From the frame of reference of a moving observer traveling at the speed v (diagram at lower right), the light pulse traces out a longer, angled path.
yuiop
#90
Jan3-11, 10:44 AM
P: 3,967
Let me try again with a simple example and maybe you can let me know which parts you disagree with.

Anne is at rest in frame S. Anne considers herself stationary.
Bob is at rest in frame S'. Bob considers himself stationary.
According to Bob, Anne is moving at 0.6c in the x direction.
According to Anne, Bob is moving at 0.6c in the -x direction.

The fact that Bob thinks Anne's velocity is +0.6c and Anne thinks Bob's velocity is -0.6c is not important because the velocity is squared in the gamma factor and the sign of the velocity is "lost".
The gamma factor in this case is [tex]\gamma = 1/\sqrt{1-(\pm 0.6)^2} = 1.25[/tex] from either point of view.

From Bob's point of view:

Anne passes Bob at time zero on both their clocks. (Event 1)
Bob has a brother Bob2 who is at rest with Bob in frame S'.
Bob and Bob2's clocks are synchronised.
Anne passes Bob2 when 25 seconds are showing on Bob2's clock. (Event 2)
When Anne passes Bob2, 20 seconds have elapsed on her clock.
Since Bob and Bob2's clcoks are synchronised, Bob concludes that 25 seconds elapses on his clock between events 1 and 2, when 20 seconds elapses on Anne's clock between those same two events.
Therefore according to Bob:

[tex]t_{Bob} = \gamma ( t_{Anne}) [/tex]

From Anne's point of view:

Bob is moving and passes Anne at time zero on both their clocks. (Event 1)
Anne has a sister Anne2 who is at rest with Anne in frame S.
Anne and Anne2's clocks are synchronised.
Bob passes Anne2 when 25 seconds are showing on Anne2's clock. (Event 3)
When Bob passes Anne2, 20 seconds have elapsed on his clock.
Since Anne and Anne2's clocks are synchronised, Anne concludes that 25 seconds elapses on her clock between events 1 and 3, when 20 seconds elapses on Bob's clock between those same two events.
Therefore according to Anne:

[tex]t_{Anne} = \gamma ( t_{Bob}) [/tex]

==============================

This appears to be a contradiction to the earlier conclusion that according to Bob:

[tex]t_{Bob} = \gamma ( t_{Anne}) [/tex]

but there is a slight "decepton" going on here.

In Anne's rest frame, the time [itex]t_{Anne}[/itex] in the equation [itex]t_{Anne} = \gamma ( t_{Bob} )[/itex] is really the difference between [itex]t_{Anne}[/itex] and [itex]t_{Anne2}[/itex]'s clocks while [itex]t_{Bob}[/itex] is the time measured by a single clock.
In Bob's rest frame, the time [itex]t_{Bob}[/tex] in the equation [itex]t_{Bob} = \gamma ( t_{Anne}) [/itex] is really the difference between [itex]t_{Bob}[/itex] and [itex]t_{Bob2}[/itex]'s clocks while [itex]t_{Anne}[/itex] is the time measured by a single clock.

In other words, the [itex]t_{Bob}[/itex] in [itex]t_{Bob} = \gamma ( t_{Anne}) [/itex] is not the same as the [itex]t_{Bob}[/itex] in [itex]t_{Anne} = \gamma ( t_{Bob}) [/itex]. [itex]t_{Bob}[/itex] in the first equation is measured by two spatially separated clocks (coordinate time) and [itex]t_{Bob}[/itex] in the second equation is measured by a single clock (proper time). [itex]t_{Bob}[/itex] in the first equation is the difference between events 1 and 2 while [itex]t_{Bob}[/itex] in the second equation is the difference between events 1 and 3. This is the "deception" that is hidden when using the primed notation that gives the appearance of a reciprocal time dilation relationship.

Note that the time on the right hand side of both equations is measured by a single clock that is present at both the start and finish events. This is the proper time and I have mentioned it several times before, but I repeat it again, because it is an important concept to understand. All observers agree on the proper time of a clock between two events.


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