NobodySpecial said:Seems like the standard explanation and the simplest way of deriving the lorentz-fitzgerald formula, what's the problem ?
chinglu1998 said:...t' > t which means more time on primed clock which is moving.
Dr Lots-o'watts said:Why do you write t? It's delta t. You're looking for trouble.
Yes, it's correct. In the primed frame where the clock is moving, the clock is running slow so it takes more time to tick forward by a given amount. For example, if a clock is moving at 0.6c in my (primed) frame, it takes 25 seconds in my frame for the clock to tick forward by 20 seconds of its own time.chinglu1998 said:What?
Delta t then.
Do you have a comment on the OP? Is WIKI correct?
JesseM said:Yes, it's correct. In the primed frame where the clock is moving, the clock is running slow so it takes more time to tick forward by a given amount. For example, if a clock is moving at 0.6c in my (primed) frame, it takes 25 seconds in my frame for the clock to tick forward by 20 seconds of its own time.
DaleSpam said:I strongly and consistently recommend against ever using either the time dilation or length contraction formulas. It is too easy to use it incorrectly, and it doesn't save a significant amount of effort.
I recommend always using the complete Lorentz transform. This avoids any risk of using the simplified formulas when they do not apply, and it automatically simplifies when appropriate.
JesseM said:Also, I don't understand your use of the phrase "more time on primed clock which is moving". Motion is relative, there is no objective truth about which is "moving". The time dilation formula assumes you have a clock at rest relative to the unprimed frame, and which is moving relative to the primed frame, and you want to know how much time goes by in the unprimed frame when the clock ticks forward by a given amount.
I think you're confused about how relativity works, no frame is declared "moving" or "stationary" in any absolute sense, you can only talk about relative motion. The primed frame is moving relative to the unprimed frame, and the unprimed frame is moving relative to the primed frame. Clocks at rest in the unprimed frame are running slow in the primed frame (and the wiki article is specifically considering a clock at rest in the unprimed frame), just as clocks at rest in the primed frame would be running slow in the unprimed frame (though the time dilation equation in the wiki article doesn't deal with this case).chinglu1998 said:The example said the primed frame was the moving frame. And the unprimed frame is taken at the stationary system of coordinates. So, that means the clocks in the primed frame are moving relative to the unprimed frame.
What do you mean by "rest frame"? Do you understand my point that no frame is declared to be "at rest" in any absolute sense, you can only talk about a frame as the "rest frame" of some particular object? In the wiki, the unprimed frame is the rest frame of the clock, and the primed frame is the rest frame of the observer.chinglu1998 said:The following have the unprimed frame as rest frame.
No, it says the unprimed frame is the rest frame of the clock being considered in the time dilation equation. Some of the links you gave do use a different convention, but in any case it's purely a matter of convention which coordinates we use to label time in the clock's own rest frame, it's usually the primed frame but it could be the unprimed frame.chinglu1998 said:Yet, WIKI claims the opposite.
JesseM said:I think you're confused about how relativity works, no frame is declared "moving" or "stationary" in any absolute sense, you can only talk about relative motion. The primed frame is moving relative to the unprimed frame, and the unprimed frame is moving relative to the primed frame. Clocks at rest in the unprimed frame are running slow in the primed frame (and the wiki article is specifically considering a clock at rest in the unprimed frame), just as clocks at rest in the primed frame would be running slow in the unprimed frame (though the time dilation equation in the wiki article doesn't deal with this case).
JesseM said:What do you mean by "rest frame"? Do you understand my point that no frame is declared to be "at rest" in any absolute sense, you can only talk about a frame as the "rest frame" of some particular object? In the wiki, the unprimed frame is the rest frame of the clock, and the primed frame is the rest frame of the observer.
No, it also says the unprimed frame is the rest frame of the clock being considered in the time dilation equation, just like the other links you gave.
While you were posting this I was editing my post to read:chinglu1998 said:Then is it possible you compare the equations?
WIKI says t' = t γ
The others say t' = t / γ.
Can you see the difference?
No, it says the unprimed frame is the rest frame of the clock being considered in the time dilation equation. Some of the links you gave do use a different convention, but in any case it's purely a matter of convention which coordinates we use to label time in the clock's own rest frame, it's usually the primed frame but it could be the unprimed frame.
JesseM said:While you were posting this I was editing my post to read:
"Taken as stationary" relative to what? Better to say it's the rest frame of the observer, I think.chinglu1998 said:I can't see how you conclude I said any frame is at absolute rest.
I said when the frame is taken as stationary.
The phrase "taken as stationary" doesn't seem to add any meaning here. I would just say that from the perspective of the unprimed frame, the clocks in the primed frame are moving and beating slower.chinglu said:So, I think you are claiming, if the unprimed frame is taken as stationary, the moving clocks in the primed frame are beating slower.
Or you could just note that the position of the source in the unprimed frame doesn't change with time.chinglu1998 said:You can tell the unprimed frame is at rest with the light source because there is no light abberation (angle to the light beam).
Again the words "stationary" and "moving" are completely unnecessary and confusing. I would just say "clocks at rest in the unprimed frame, when viewed from the primed frame are time dilated"chinglu1998 said:So, the unprimed frame is considered stationary and the primd frame is considered the moving frame. Clocks therefore, when viewed from the stationary frame are time dilated.
"At rest" is meaningless unless you specify what frame something is at rest relative to. If I am at rest relative to the primed frame, then clocks at rest relative to the unprimed frame are moving relative to me.chinglu1998 said:For example, when do you say you are in the moving frame and the clocks in the other frame are at rest?
Here count the number of times Einstein uses "stationary". 62 times.JesseM said:"Taken as stationary" relative to what? Better to say it's the rest frame of the observer, I think.
It is the rest frame of the light clock whose time dilation we want to determine, as seen in the frame of an observer (the primed frame). Regardless of what symbol convention is used, the time dilation equation always has the following form:chinglu1998 said:I agree with you the convention is important.
But, the fact that the unprimed frame has no light abberation clearly indicates that frame is at rest with the light source which is being used to determine time dilation
No it didn't, it only said the unprimed frame is the one where the clock is at rest. Likewise, the primed frame is the frame of the observer, so it is obviously the frame where the observer is at rest. When the article refers to the "moving observer", they only mean the observer is moving relative to the clock.chinglu1998 said:and that plus the fact WIKI said the unprimed frame is at rest
As usual, it is physically meaningless to say a given object is "moving" or "at rest" unless you are really saying it is moving or at rest relative to something. It's true that some authors just talk about a "stationary frame" and a "moving frame" but it is always clear from the context that they are implicitly defining motion relative to a particular object or observer. Obviously this convention is getting you all confused because you fail to understand the unspoken "relative to", so I'd suggest that from now on you never use phrases like "stationary" or "moving" unless they are followed by "relative to X", where X is some object or observer.chinglu1998 said:I am not sure what other evidence I could produce that the unprimed frame is not moving and any clock in the primed frame is.
But he makes clear at the start that he is defining "stationary" and "moving" relative to a particular arbitrarily-chosen coordinate system:chinglu1998 said:Here count the number of times Einstein uses "stationary". 62 times.
I think it has a clear meaning.
http://www.fourmilab.ch/etexts/einstein/specrel/www/
It's just a verbal convention Einstein uses to shorten the presentation, but it's not like it's a basic part of relativity terminology that you must use such a convention, in fact most authors don't pick one frame to label "stationary" and another to label "moving", they just say "stationary relative to X" and "moving relative to X" etc.Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good. In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the "stationary system.''
If a material point is at rest relatively to this system of co-ordinates, its position can be defined relatively thereto by the employment of rigid standards of measurement and the methods of Euclidean geometry, and can be expressed in Cartesian co-ordinates.
You forgot the light abberation. The unprimed frame does not see it but the primed frame does.JesseM said:No it didn't, it only said the unprimed frame is the one where the clock is at rest. Likewise, the primed frame is the frame of the observer, so it is obviously the frame where the observer is at rest. When the article refers to the "moving observer", they only mean the observer is moving relative to the clock.
As usual, it is physically meaningless to say a given object is "moving" or "at rest" unless you are really saying it is moving or at rest relative to something. It's true that some authors just talk about a "stationary frame" and a "moving frame" but it is always clear from the context that they are implicitly defining motion relative to a particular object or observer. Obviously this convention is getting you all confused because you fail to understand the unspoken "relative to", so I'd suggest that from now on you never use phrases like "stationary" or "moving" unless they are followed by "relative to X", where X is some object or observer.
JesseM said:But he makes clear at the start that he is defining "stationary" and "moving" relative to a particular arbitrarily-chosen coordinate system:
It's just a verbal convention Einstein uses to shorten the presentation, but it's not like it's a basic part of relativity terminology that you must use such a convention, in fact most authors don't pick one frame to label "stationary" and another to label "moving", they just say "stationary relative to X" and "moving relative to X" etc.
Uh how did I forget it? I just said "the unprimed frame is the one where the clock is at rest", and the clock is a light clock, so this is exactly the same as saying the unprimed frame is the one where the light source is at rest.chinglu1998 said:You forgot the light abberation. The unprimed frame does not see it but the primed frame does.
They said the unprimed frame was at rest relative to the clock/light source ('In the frame where the clock is at rest'), which it is. Why do you have a problem with this?chinglu1998 said:So, the unprimed frame is at rest with the light source and the WIKI said the unprimed frame is at rest.
They used the primed frame to denote "the frame of reference of a moving observer traveling at the speed v" (relative to the clock). Do you deny that this "moving observer" is at rest relative to the primed frame?chinglu1998 said:I am not confused, you are. There is nothing in the WIKI article to claim the primed frame is at rest relative to anything. If you can find it, prove it.
He's not in error, because he said at the outside he was defining "stationary" relative to a particular (arbitrary) choice of frame. That's my whole point, that terms like "at rest" and "moving" are only meaningful when they are (explicitly or implicitly) defined relative to some particular frame or object, and there is no requirement that you always define them relative to the same object throughout the problem, you are free to switch between saying an object is "moving relative to frame X" and "at rest relative to frame Y" if you so choose. Do you think there is some rule that says this is forbidden?chinglu1998 said:And Einstein used the term stationary 62 times in his 1905 article. So, I will continue to use it. I suggest you read the paper before you condemn the usage and indicate why Einstein was in error.
What do you mean "assume Euclidean geometry"? Can you give a numerical example?chinglu1998 said:When you take a frame as stationary, you assume the Euclidian Geometry.
And can you give a numerical example of "apply the Minkowski geometry"? Normally I would say that Minkowski geometry is defined in terms of the metric, which says that the proper time between two events with coordinate separations \Delta t, \Delta x, \Delta y, \Delta z is given by the formula \sqrt{\Delta t^2 - (1/c^2)*(\Delta x^2) - (1/c^2)*(\Delta y^2) - (1/c^2)*(\Delta z^2)}. But this is true in all frames, even if we choose to label one of the frames as "stationary" like Einstein does. So I have no idea why you think Minkowski geometry doesn't apply in what you call the "stationary frame".chinglu1998 said:When you assume another frame is the moving frame, you apply the Minkowsky Geometry.
Again, a numerical example of what you mean by "breakdown of SR" would be helpful.chinglu1998 said:This is a vital distinction such that if the distinction is not clear, complete and total breakdown of SR occurs.
The article defines "stationary" and "moving" relative to the clock. But as I said, the time dilation equation always has the following form:chinglu1998 said:Anyway, from any point of view the WIKI article is taking the unprimed frame as stationary. If you see evidence to the contrary, please post it.
JesseM said:Uh how did I forget it? I just said "the unprimed frame is the one where the clock is at rest", and the clock is a light clock, so this is exactly the same as saying the unprimed frame is the one where the light source is at rest.
I do not. I just understand everything points to the unprimed frame as at rest. Check all my links. It is standard convention that the unprimed frame is at rest. And, WIKI confirms this.They said it was at rest relative to the clock/light source ('In the frame where the clock is at rest'), which it is. Why do you have a problem with this?
They used the primed frame to denote "the frame of reference of a moving observer traveling at the speed v" (relative to the clock). Do you deny that this "moving observer" is at rest relative to the primed frame?
Yes, and the WIKI article is clear the stationary frame is the unprimed frame. Its conclusions are in error. Do you agree with them?He's not in error, because he said at the outside he was defining "stationary" relative to a particular (arbitrary) choice of frame. That's my whole point, that terms like "at rest" and "moving" are only meaningful when they are (explicitly or implicitly) defined relative to some particular frame or object, and there is no requirement that you always define them relative to the same object throughout the problem, you are free to switch between saying an object is "moving relative to frame X" and "at rest relative to frame Y" if you so choose. Do you think there is some rule that says this is forbidden?
The way a clock ticks in different frames has nothing to do with the totally arbitrary labels "moving" and "at rest" which we may apply to different frames. As I keep telling you, the basic idea of the time dilation equation is this:chinglu1998 said:So, how are you supporting the notion that t' = t γ as posted on WIKI? Everthing says there the unprimed frame is at rest. This equations implies the moving frame (primed frame) beats faster and that is an error.
JesseM said:What do you mean "assume Euclidean geometry"? Can you give a numerical example?
And can you give a numerical example of "apply the Minkowski geometry"? Normally I would say that Minkowski geometry is defined in terms of the metric, which says that the proper time between two events with coordinate separations \Delta t, \Delta x, \Delta y, \Delta z is given by the formula \sqrt{\Delta t^2 - (1/c^2)*(\Delta x^2) - (1/c^2)*(\Delta y^2) - (1/c^2)*(\Delta z^2)}. But this is true in all frames, even if we choose to label one of the frames as "stationary" like Einstein does. So I have no idea why you think Minkowski geometry doesn't apply in what you call the "stationary frame".
This is a vital distinction such that if the distinction is not clear, complete and total breakdown of SR occurs.
The article defines "stationary" and "moving" relative to the clock. But as I said, 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
This is true regardless of whether you label the clock's frame as "stationary" or label the observer's frame as "stationary" or don't label either frame as "stationary". All that's required to have a correct expression of the time dilation equation is that it match the form above. Do you disagree? Can you think of a scenario where the time in the observer's frame and the time in the clock frame would not be related by the equation above, perhaps because we switch which is considered "stationary" and which is considered "moving"?
JesseM said:The way a clock ticks in different frames has nothing to do with the totally arbitrary labels "moving" and "at rest" which we may apply to different frames. As I keep telling you, the basic idea of the time dilation equation is this:
(time interval in frame of observer who sees clock in motion) = (time elapsed on clock)*gamma
If you make the arbitrary choice to label the clock's frame as "moving" and the observer's frame as "stationary", then it can be interpreted in this way:
(time between two readings on the clock as measured in stationary frame) = (time between two readings on the clock as measured in moving frame)*gamma
On the other hand, if you make the equally arbitrary choice to label the clock's frame as "stationary" and the observer's frame as "moving", then it can be interpreted in this way:
(time between two readings on the clock as measured in moving frame) = (time between two readings on the clock as measured in stationary frame)*gamma
Both are valid, given the aforementioned arbitrary choices about which frame to label as "stationary" and which to label as "moving". Do you disagree? If you do, then it seems you believe that stationary vs. moving does have some absolute significance, since you think we are not equally free to label the clock's frame as "moving" or "stationary" (in which case you are very confused about how relativity works, since in relativity there is no way to detect any absolute motion). On the other hand, if you agree both are valid given the right arbitrary choice of labels, then you should be able to see that the wiki article was simply adopting the second convention.
There is no coordinate system where all the dynamical predictions of Newtonian mechanics hold good (for example, Newtonian dynamics would predict that a clock ticks at the same rate regardless of its velocity). But if you pick some specific features of Newtonian mechanics which still hold in relativity, like the fact that an object moving at constant velocity remains at that velocity unless acted on by a force, then these features hold in all inertial frames in relativity, not just one. If you disagree, please name a specific Newtonian law which you think holds in one inertial frame (the 'stationary' one) but not others.chinglu1998 said:You are not going to win in this direction just because you made an error about WIKI.
Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good.2 In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the ``stationary system.''
http://www.fourmilab.ch/etexts/einstein/specrel/www/
The "Newtonian mechanics hold good" in the Euclidean geometry.
Uh, I think you are forgetting about the relativity of simultaneity. A "light sphere" is just the set of points in spacetime that light from a flash has reached at a single moment in time in one frame, but if you map this set of points to a different frame, then in that frame they will consist of events at different times. If instead you consider the light cone as is normally done in relativity, then if you look at the subset of events on that cone which all occur at some time T in the unprimed frame, their positions in the unprimed frame will form a sphere satisfying x2 + y2 + z2 = c2 T2, and likewise if you look at a different subset of events on that cone which all occur at some time T' in the primed frame, their positions in the primed frame will form a sphere satisfying x'2 + y'2 + z'2 = c2 T'2. You really are very confused if you think that light expands spherically in one inertial frame but elliptically in a different inertial frame--that would violate both the first and second postulates of special relativity! This issue of light spheres vs. light cones and why each frame sees the light as a sphere at any single instant was discussed extensively on this thread, if you're interested.chinglu1998 said:Uh, try selecting points from the light sphere in the stationary frame and mapping them with LT.
You will note the points on the light sphere satisfy x² + y² + z² = c² t².
Uh, this is a Euclidean object. Do you now understand the Euclidean nature of the stationary system of coordinates?
Now, let's look at the Minkowsky interpretation.
ξ² + η² + ς² = c² τ²
τ = ( t - vx/c² )γ
ξ = ( x - vt )γ
η = y
ς = z.
This is not Euclidean because the radius of this light sphere is not the same in all directions.
Yes, of course the wiki labels unprimed as the stationary frame, because they are defining "stationary" vs. "moving" relative to the clock and the clock is at rest in the stationary frame. And as I said:chinglu1998 said:Oh, no I do not agree with absolute positions. I know the conventions. I just know every scrap of evidence on the WIKI problem proves the unprimed is the stationary frame.
Do you agree that this equation is correct or do you think it's wrong? If you agree it's correct, you can see that with t' as time in moving frame and t as time in stationary frame, this is exactly the same as the wiki's t' = t*gammaOn the other hand, if you make the equally arbitrary choice to label the clock's frame as "stationary" and the observer's frame as "moving", then it can be interpreted in this way:
(time between two readings on the clock as measured in moving frame) = (time between two readings on the clock as measured in stationary frame)*gamma
JesseM said:There is no coordinate system where all the dynamical predictions of Newtonian mechanics hold good (for example, Newtonian dynamics would predict that a clock ticks at the same rate regardless of its velocity). But if you pick some specific features of Newtonian mechanics which still hold in relativity, like the fact that an object moving at constant velocity remains at that velocity unless acted on by a force, then these features hold in all inertial frames in relativity, not just one. If you disagree, please name a specific Newtonian law which you think holds in one inertial frame (the 'stationary' one) but not others.
Uh, I think you are forgetting about the relativity of simultaneity. A "light sphere" is just the set of points in spacetime that light from a flash has reached at a single moment in time in one frame, but if you map this set of points to a different frame, then in that frame they will consist of events at different times. If instead you consider the light cone as is normally done in relativity, then if you look at the subset of events on that cone which all occur at some time T in the unprimed frame, their positions in the unprimed frame will form a sphere satisfying x2 + y2 + z2 = c2 T2, and likewise if you look at a different subset of events on that cone which all occur at some time T' in the primed frame, their positions in the primed frame will form a sphere satisfying x'2 + y'2 + z'2 = c2 T'2. You really are very confused if you think that light expands spherically in one inertial frame but elliptically in a different inertial frame--that would violate both the first and second postulates of special relativity! This issue of light spheres vs. light cones and why each frame sees the light as a sphere at any single instant was discussed extensively on this thread, if you're interested.
chinglu1998 said:Oh, no I do not agree with absolute positions. I know the conventions. I just know every scrap of evidence on the WIKI problem proves the unprimed is the stationary frame.
I also know you will type thousands of pages rather than simply admitting you are wrong.
I agree the Wiki article is a little confusing when one is used to seeing t' = t / y. I tried to edit it to clarify a while back with the moving mirrors part, but my clarification ended up adding another mirror that didn't exist within the diagrams and probably complicated it more in the long run, so it was dropped. I might try again some other way at some point.chinglu1998 said:Then is it possible you compare the equations?
WIKI says t' = t γ
The others say t' = t / γ.
Can you see the difference?
JesseM said:Yes, of course the wiki labels unprimed as the stationary frame, because they are defining "stationary" vs. "moving" relative to the clock and the clock is at rest in the stationary frame. And as I said:
Do you agree that this equation is correct or do you think it's wrong? If you agree it's correct, you can see that with t' as time in moving frame and t as time in stationary frame, this is exactly the same as the wiki's t' = t*gamma
I'm not "refuting" him, I'm just saying you have to understand he was writing for an audience of physicists who would not jump to the conclusion that "equations of Newtonian mechanics hold good" was meant to imply that every possible dynamical equation in Newtonian mechanics holds--for example, the instantaneous action-at-a-distance of Newtonian gravity obviously doesn't hold in relativity, nor do rigid body equations which assume that a force applied to the back end of an object will instantaneously accelerate the front end. Do you disagree, and think he was talking about every conceivable Newtonian equation?chinglu1998 said:I am going to quote Einstein again. You are refuting Einstein?
Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good.2 In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the ``stationary system.''
http://www.fourmilab.ch/etexts/einstein/specrel/www/
So you agree that in every inertial frame, if we look at the position of the light at a single simultaneous instant in that frame, the position of the light at that instant forms a sphere in that frame satisfying an equation like x2 + y2 + z2 = c2 T2 in the coordinates of that frame? If you do agree, I don't see why you use this example to try to prove that there is only one frame where "Euclidean geometry" holds while in the other ones Minkowski geometry (not 'Minkowsky') holds.chinglu1998 said:I wrote the Minkowsky interpretation of the Euclidian light sphere mapped from the stationary system of coordinates did not have an equal radius in all directions.
That implies the relativity of simultaneity. Seemed obvious to me.
yuiop said:I agree with Jesse and others that it is best to avoid the concept of the velocity of a reference frame. It is better to think in terms of the velocity of a clock relative to a given reference frame without any care for the velocity of the reference frame itself. When people say "in the rest frame" it is really shorthand for "the reference frame in which in the clock (or particle) is at rest". As Dalespam said, it probably best to use the Lorentz transformation:
t' = \frac{ t - v x/c^2}{\sqrt{1-v^2/c^2}} = ( t - v x/c^2) \gamma
where t' is the time interval measured by clocks at rest in S' and t is the time interval measured by clocks at rest in the unprimed S frame. v is the velocity of S and S' relative to each other. Note that there is no implication of whether S' or S is at rest.
Now if x = 0 in the transformation then the time interval measured in the unprimed frame is the time measured by a single clock that is at rest in the unprimed frame which is the proper time denoted by t_0 and the time dilation equation can be written as:
t' = t_0 \gamma
where it is understood that t_0 is the proper time measured by a single clock. The above equation can also be written as:
t = t_0 \gamma
because proper time has an unambiguous definition, the prime symbol no longer has any significance in this formulation. The time interval t is the time interval measured by multiple clocks at rest in a reference frame moving with velocity v relative to the rest reference frame of the single clock that measures the proper time t_o between two events. Using the primed notation can cause confusion, because there is no universal convention for which frame is primed and the context of the equation has to be carefully examined.
darkhorror said:The problem is you keep on getting caught up on the name of the frames. Let's say I have two frame the stationary frame and the moving frame. The moving frame is moving at .6c compared to stationary frame. Now from the moving frame of reference how much fast is the stationary frames clock moving?
grav-universe said:I agree the Wiki article is a little confusing when one is used to seeing t' = t / y. I tried to edit it to clarify a while back with the moving mirrors part, but my clarification ended up adding another mirror that didn't exist within the diagrams and probably complicated it more in the long run, so it was dropped. I might try again some other way at some point.
Anyway, the time dilation formula can be taken either way, only depending upon which frame we consider at rest, or which is our frame of reference. Usually we take the unprimed frame to be our frame of reference which is at rest while the unprimed frame is moving, but in the Wiki article, it is the other way around. It is still correct either way, however, since the primed frame is time dilating in the same way from the perspective of the unprimed as the unprimed frame is from the perspective of the primed frame.
In other words, there is no absolute "one frame is time dilating to the other", where either t = t' / y or t' = t / y, but each time dilates relative to the other in the same way, depending upon which frame's perspective we are taking. To see this, as mentioned before in another thread, we should really use four variables, say tAA, tAB, tBA, and tBB, not just the two t and t', since the dual purpose of the same two variables becomes unclear. With the four variables, tAA will be the time that frame A observes of frame A's own clock, tAB will be the time that frame A observes of frame B's clock, and so forth. Using these, we can now see that we actually have two sets of equations, depending upon which frame is doing the observing. For frame A, it is tAB = tAA / y. For frame B, it becomes tBA = tBB / y.
Sure. Suppose the relative velocity between the primed frame and the unprimed frame is 0.6c (so gamma=1.25), and we have a clock at rest in the unprimed frame at position x=0. Then if the event of the clock reading a time of 0 happens at t=0, the event of the clock reading a time of 20 happens at t=20. So, the time between these events in the unprimed frame, \Delta t, equals 20. According to the Lorentz transformation, the times of these events in the primed frame will be given by t' = 1.25*(t - 0.6x/c). So since the event of the clock reading 0 happened at x=0, t=0 in the unprimed frame, in the primed frame it happens at time t'=1.25*(0 - 0)=0; and since the event of the clock reading 20 happens at x=0, t=20 in the unprimed frame, in the primed frame it happens at time t'=1.25*(20 - 0)=25. So the time between these events in the primed frame, \Delta t', equals 25. Therefore just as the wiki says, the time dilation equation relating the time in each frame between two events on the worldline of a clock at rest in the unprimed frame would be:chinglu1998 said:Can you show me numbers?
I expect to see the t' less than t because of time dilation. I can post Einstein'ss equation again also that verifies this.
JesseM said:I'm not "refuting" him, I'm just saying you have to understand he was writing for an audience of physicists who would not jump to the conclusion that "equations of Newtonian mechanics hold good" was meant to imply that every possible dynamical equation in Newtonian mechanics holds--for example, the instantaneous action-at-a-distance of Newtonian gravity obviously doesn't hold in relativity, nor do rigid body equations which assume that a force applied to the back end of an object will instantaneously accelerate the front end. Do you disagree, and think he was talking about every conceivable Newtonian equation?
So you agree that in every inertial frame, if we look at the position of the light at a single simultaneous instant in that frame, the position of the light at that instant forms a sphere in that frame satisfying an equation like x2 + y2 + z2 = c2 T2 in the coordinates of that frame? If you do agree, I don't see why you use this example to try to prove that there is only one frame where "Euclidean geometry" holds while in the other ones Minkowski geometry (not 'Minkowsky') holds.
JesseM said:Sure. Suppose the relative velocity between the primed frame and the unprimed frame is 0.6c (so gamma=1.25), and we have a clock at rest in the unprimed frame at position x=0. Then if the event of the clock reading a time of 0 happens at t=0, the event of the clock reading a time of 20 happens at t=20. So, the time between these events in the unprimed frame, \Delta t, equals 20. According to the Lorentz transformation, the times of these events in the primed frame will be given by t' = 1.25*(t - 0.6x/c). So since the event of the clock reading 0 happened at x=0, t=0 in the unprimed frame, in the primed frame it happens at time t'=1.25*(0 - 0)=0; and since the event of the clock reading 20 happens at x=0, t=20 in the unprimed frame, in the primed frame it happens at time t'=1.25*(20 - 0)=25. So the time between these events in the primed frame, \Delta t', equals 25. Therefore just as the wiki says, the time dilation equation relating the time in each frame between two events on the worldline of a clock at rest in the unprimed frame would be:
\Delta t' = \Delta t * 1.25
DaleSpam said:chinglu1998, the wiki article is correct, but thanks for providing a wonderful example of why the time dilation formula should always be avoided.
chinglu1998 said:This is about the OP and I am not making any claims one way or the other about rest or moving.
I simple reads articles and see what we have.
Are you claiming the WIKI article has the primed frame as stationary?
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. The second postulate of special relativity states that the speed of light is constant in all frames, which implies a lengthening of the period of this clock from the moving observer's perspective. That is to say, in a frame moving relative to the clock, the clock appears to be running more slowly.
DaleSpam said:Because you have gone on for three pages in confusion about a formula when you could have avoided the confusion entirely simply by not using the formula .
The wiki article is correct because it has the right formula given the convention used and clearly identified in the article. There is no right or wrong about if the primed frame is stationary or moving, it is only a convention and having identified the convention they provide the correct formula. Your 6 links are simply not relevant, except perhaps to show that other authors choose a different convention.chinglu1998 said:Why?
darkhorror said:That is exactly what I am saying.
from the wiki article
They are using the frame of reference of the moving observer for the t'. So yes the primed frame is "stationary" in it's own frame of reference.
The wikipedia article says that the unprimed frame is the rest frame of the light clock, this implies that no aberration will be observed in the unprimed frame (as is indicated in the upper picture accompanying the section in question).chinglu1998 said:The WIKI article said the unprimed frame is the rest frame.
Light abberation says the unprimed frame is the rest frame.
$\Delta{}t'$ is the period of one cycle of the clock in the frame in which the clock is in motion (the one corresponding to the lower picture with aberration). The equation in question relates the period of the clock in the frame in which it is observed to move with its period in the clock's rest frame. Since frequency is inversely proportional to period, the same equation implies that \Delta{}n=\frac{\Delta{}n'}{\gamma} where n denotes the number of cycles/ticks of the clock in the appropriate frame (following the convention of the wikipedia article). Thus, the equation says that a moving clock will be observed to tick less frequently in a frame in which it moves (i.e., all else being equal, moving clocks are observed to "run slow" compared to ones at rest, as expected).chinglu1998 said:But, it has the primed frame beating faster
