WIKI and Time Dilation: The Possible Error in Relative Velocity

[chinglu1998]

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

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]

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/

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?

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.

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]

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.

Sure I know how to follow this convention.

Anyway, this OP is about whether WIKI has made an error on said link. What do you say?

 

[chinglu1998]

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?

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?

 

[chinglu1998]

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.

Yes, no one said any absolutes.

The WIKI article said the unprimed frame is the rest frame.

Light abberation says the unprimed frame is the rest frame.

But, it has the primed frame beating faster. This is an error.

 

[JesseM]

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.

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:

$$\Delta t' = \Delta t * 1.25$$

 

[chinglu1998]

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?

Let us not bring in gravity and acceleration. We are talking SR.

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.

I did not say there was only one frame where the Euclidean geometry holds true for the light sphere. I said in the context of the stationary system, that holds true.

You can take any frame as stationary and the light postulate holds true, ie x² + y² + z² = c² t²

 

[chinglu1998]

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:

$$\Delta t' = \Delta t * 1.25$$

What?

The stationary clock elapses 20s and the moving clock elapses 25s?

So under your new theory, moving clocks elapse more time than stationary clocks? How is that time dilation?

Why not simply admit you were wrong about the WIKI article? It would save you lots of additional errors.

 

[Dale]

chinglu1998, the wiki article is correct, but thanks for providing a wonderful example of why the time dilation formula should always be avoided.

 

[chinglu1998]

chinglu1998, the wiki article is correct, but thanks for providing a wonderful example of why the time dilation formula should always be avoided.

Why?

 

[darkhorror]

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?

That is exactly what I am saying.

from the wiki article

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.

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.

 

[chinglu1998]

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 .

Dale, I know how to do it either way.

But, since I am confused and the article said the unprimed frame is stationary and I provided 6 links that reversed the WIKI article, can you explain your position?

 

[Dale]

Why?

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]

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.

That's fine.

So I must always assume when an article says the unprimed frame is at rest, it is not actually at rest, the other frame is. And, when the article says the primed frame is moving, it is actually not moving but at rest. Do I understand you correctly?

 

[IsometricPion]

The WIKI article said the unprimed frame is the rest frame.

Light abberation says the unprimed frame is the rest frame.

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).

But, it has the primed frame beating faster

$$\Delta{}t'$$ 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).

 

[darkhorror]

That's fine.

So I must always assume when an article says the unprimed frame is at rest, it is not actually at rest, the other frame is. And, when the article says the primed frame is moving, it is actually not moving but at rest. Do I understand you correctly?

You don't assume anything you read it to see what it's actually saying. There are two frames of reference it doesn't matter they they are called. But when you look at them in there own frame of reference and aren't accelerating they are at rest in there own frame of reference.

So if I say I have two frames a moving and a stationary frame. Then say from the frame of reference of the moving frame, that frame isn't actually moving in it's own frame.

 

[JesseM]

What?

The stationary clock elapses 20s and the moving clock elapses 25s?

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.

So under your new theory, moving clocks elapse more time than stationary clocks? How is that time dilation?

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.

 

[Dale]

So I must always assume when an article says the unprimed frame is at rest, it is not actually at rest, the other frame is. And, when the article says the primed frame is moving, it is actually not moving but at rest.

This is a meaningless pair of sentences. Every time you use the term "at rest" you must specify "at rest with respect to ___", and every time you use the word "moving" you must specify "moving with respect to ___".

 

[yuiop]

Sure I know how to follow this convention.

Anyway, this OP is about whether WIKI has made an error on said link. What do you say?

Well it is hard to say because it appears someone is editing (and making a bit of a mess of - no "t" on the RHS of the equations) the first part of the article http://en.wikipedia.org/wiki/Time_d...nce_of_time_dilation_due_to_relative_velocity as we speak.

Further down http://en.wikipedia.org/wiki/Time_dilation#Time_dilation_due_to_relative_velocity the Wiki article states that:

$$\Delta t' = \gamma \Delta t$$

where Δt is the time interval between two co-local events (i.e. happening at the same place) for an observer in some inertial frame (e.g. ticks on his clock) – this is known as the proper time..

which is in exact agreement with what I was saying. Do you agree that part is correct?

<EDIT> O.K. it appears the WIki article has been edited yet again back to its original form. It now states that:

$$\Delta t' = \frac{ \Delta t}{\sqrt{1-v^2/c^2}}$$

and makes it absolutely clear from the definition $\Delta t = 2L/c$ and the accompanying diagrams that $\Delta t$ is the proper time measured in the rest frame of the light clock. This means the Wiki article is unambiguous and correct.

 

[JesseM]

Let us not bring in gravity and acceleration. We are talking SR.

SR deals with acceleration--it can deal with the behavior of accelerated objects even if you only analyze them from the perspective of inertial frames, but in any case modern physicists would say that even if you use accelerated frames you're still in the domain of "SR" as long as there is no spacetime curvature.

Do you agree that Einstein's statement should not be interpreted to mean that all Newtonian laws "hold good", but only some of them like the law of inertia? Yes or no?

I did not say there was only one frame where the Euclidean geometry holds true for the light sphere. I said in the context of the stationary system, that holds true.

Even if you choose to label one frame as "stationary" and another as "moving", it is still true in the moving frame that the light postulate holds true in the coordinates of that frame (light has a coordinate velocity of c in all directions in the moving frame), and that in the coordinates of that frame the position of the light at any specific time T' satisfies x'² + y'² + z'² = c² T'². So you still haven't given any non-confused explanation of what you meant in post #23 when you said "When you take a frame as stationary, you assume the Euclidian Geometry.
When you assume another frame is the moving frame, you apply the Minkowsky Geometry.
This is a vital distinction such that if the distinction is not clear, complete and total breakdown of SR occurs."

 

[chinglu1998]

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).

Agreed.

$$\Delta{}t'$$ 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).

Agree with you conclusion about SR but not WIKI.

You have already agreed the article claims the clock at rest with the light source has no light abberation. Let's take that clock as stationary. That is legal right?

Now, the clock that sees the light abberation is moving compared to the unprimed frame clock. So let's agree WIKI is correct.

Let's say as everyone here is claiming that WIKI has this correct.

So, t' = tγ.

This means when the unprimed frame is at rest and observering the to clock, it is moving and beating faster than it rest clock.

 

[darkhorror]

To chinglu1998

Lets say there is a spaceship that's moving away from Earth at .6c.

From the ships frame of reference is the Earth's clock moving faster or slower than the one on the ship?

 

[chinglu1998]

This is a meaningless pair of sentences. Every time you use the term "at rest" you must specify "at rest with respect to ___", and every time you use the word "moving" you must specify "moving with respect to ___".

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.

 

[chinglu1998]

To chinglu1998

Lets say there is a spaceship that's moving away from Earth at .6c.

From the ships frame of reference is the Earth's clock moving faster or slower than the one on the ship?

Under SR, a moving clock is supposed to beat slower. What about the WIKI article?

 

[darkhorror]

Under SR, a moving clock is supposed to beat slower. What about the WIKI article?

So from my question which one is moving slower? the Earth or ship?

 

[chinglu1998]

SR deals with acceleration--it can deal with the behavior of accelerated objects even if you only analyze them from the perspective of inertial frames, but in any case modern physicists would say that even if you use accelerated frames you're still in the domain of "SR" as long as there is no spacetime curvature.

Yes, I know the SR acceleration equations.

Do you agree that Einstein's statement should not be interpreted to mean that all Newtonian laws "hold good", but only some of them like the law of inertia? Yes or no?

I never took him to mean all. So, I agree with you. But, I specifically focused on the Euclidian measurements of the stationary frame and that part is not refutable.

Even if you choose to label one frame as "stationary" and another as "moving", it is still true in the moving frame that the light postulate holds true in the coordinates of that frame (light has a coordinate velocity of c in all directions in the moving frame), and that in the coordinates of that frame the position of the light at any specific time T' satisfies x'² + y'² + z'² = c² T'². So you still haven't given any non-confused explanation of what you meant in post #23 when you said "When you take a frame as stationary, you assume the Euclidian Geometry.
When you assume another frame is the moving frame, you apply the Minkowsky Geometry.
This is a vital distinction such that if the distinction is not clear, complete and total breakdown of SR occurs."[/QUOTE

In my post and statement "I assume the Euclidian Geometry" in the stationary system. Do you prove otherwise?

This is not Minkowsky. If you attempy to apply Minkowsky to stationary system, you get failure.

 

[JesseM]

Now, the clock that sees the light abberation is moving compared to the unprimed frame clock.

There is no single "clock that sees the light abberation", there is an observer in whose frame the light aberration occurs (the primed frame), but if you wanted to define the coordinates of the primed frame using clocks you'd need a system of multiple clocks at rest in the primed frame, as I mentioned above in post #52. It really is much clearer if we only talk about a single physical clock (as the wiki article does), and all other times are understood to be coordinate times in some frame.

Let's say as everyone here is claiming that WIKI has this correct.

So, t' = tγ.

This means when the unprimed frame is at rest and observering the to clock, it is moving and beating faster than it rest clock.

"observing the to clock"? Is that a typo? If you meant to say "moving clock", then again there is no single moving clock. In my numerical example, if you had a single moving clock A' in the primed frame at position x'=0 and another moving clock B' at position x'=-12, then when clock A' reads t'=0 it will be passing the stationary clock which also reads t=0, and when clock B' reads t'=25 it will be passing the stationary clock which reads t=20, so this is why we say the time between the two readings on the stationary clock was 25 seconds in the moving frame. However, in the stationary frame it's not that these two clocks are "beating faster", rather it's that they're out-of-sync due to the relativity of simultaneity. In the stationary frame, at the time clock A' reads t'=0 (and A' is passing next to the stationary clock at this moment), clock B' already reads t'=9. Then after 20 seconds in the stationary frame, both of these clocks have ticked forward by 20/1.25=16 seconds, so clock A' reads t'=0+16=16 while clock B' reads t'=9+16=25, and this is the moment clock B' is passing next to the stationary clock. So despite the fact that A' and B' were both ticking slow in the stationary frame, the stationary frame still agrees that A' read t'=0 when it passed the stationary clock, and B' read t'=25 when it passed the stationary clock 20 seconds later in the stationary frame.

 

[chinglu1998]

So from my question which one is moving slower? the Earth or ship?

Can you give me coordinates?

 

[darkhorror]

Can you give me coordinates?

Look what I asked earlier

Let's say there is a spaceship that's moving away from Earth at .6c.

From the ships frame of reference is the Earth's clock moving faster or slower than the one on the ship?

 

[Dale]

the unprimed frame as rest with respect to the primed frame
...
the primed frame as rest with respect to the unprimed frame

What are you talking about? In the Wiki article the unprimed frame is never at rest wrt the primed frame and conversely the primed frame is never at rest wrt the unprimed frame.

 

[chinglu1998]

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.

c²t'² = v²t'² + c²t²

c²t'² - v²t'² = c²t²

t'²( c² - v²) = c²t²

t' = tγ

Wow, when taking the unprimed frame as stationary with absolute light abberation, the moving clock beats faster just like WIKI said.

 

[chinglu1998]

What are you talking about? In the Wiki article the unprimed frame is never at rest wrt the primed frame and conversely the primed frame is never at rest wrt the unprimed frame.

So where is the third frame to apply the velocity add equations?

 

[chinglu1998]

Look what I asked earlier

I did. Are they starting at (0,0,0,0) for both?

 

[Dale]

So where is the third frame to apply the velocity add equations?

Why would you need a third frame?

 

[JesseM]

I never took him to mean all. So, I agree with you. But, I specifically focused on the Euclidian measurements of the stationary frame and that part is not refutable.

But you never explained why you think we can't also assume the same Euclidean laws hold in the moving frame, or why doing so would mean that "complete and total breakdown of SR occurs."

Even if you choose to label one frame as "stationary" and another as "moving", it is still true in the moving frame that the light postulate holds true in the coordinates of that frame (light has a coordinate velocity of c in all directions in the moving frame), and that in the coordinates of that frame the position of the light at any specific time T' satisfies x'² + y'² + z'² = c² T'². So you still haven't given any non-confused explanation of what you meant in post #23 when you said "When you take a frame as stationary, you assume the Euclidian Geometry.
When you assume another frame is the moving frame, you apply the Minkowsky Geometry.
This is a vital distinction such that if the distinction is not clear, complete and total breakdown of SR occurs."[/QUOTE

In my post and statement "I assume the Euclidian Geometry" in the stationary system. Do you prove otherwise?

No, because I was talking about the moving frame, as I think was pretty clear from the bolded statements above. Do you disagree that in the moving frame, "the position of the light at any specific time T' satisfies x'² + y'² + z'² = c² T'²"? Isn't this "Euclidean geometry"?

This is not Minkowsky. If you attempy to apply Minkowsky to stationary system, you get failure.

You have never given any mathematical illustration of what you mean by "apply Minkowsky to stationary system" (and as I told you, the name is "Minkowski" with an i) or why this leads to "failure". As I said in post #25, this is what most physicists mean by "Minkowski 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".

Do you disagree that the metric $$d\tau^2 = dt^2 - (1/c^2)dx^2 - (1/c^2)dy^2 - (1/c^2)dz^2$$ is what physicists normally mean by "Minkowski geometry", or that this equation still holds if we are dealing with the t,x,y,z coordinates of the "stationary" frame?

By the way, I notice you skipped my post #52 to reply to post #55. Hopefully you will go back to #52 and address it, particularly my request for an explanation of the precise step in my mathematical derivation that you think contains an error.