Time Dilation

[James R]

chinglu:

I assume the frames agree on the start event when the frames are the same.

An event at the origin is one event. An event at x=-k is a different event. Those two spatially-separated events can only be simultaneous in one frame. They cannot be simultaneous in two frames.

Note that your calculations try to cross frames and predict some time on the clock in the primed frame when calculating in the unprimed frame. We do not need this. We simply look at each frame's calculations individually.

If you're only doing calculations in one frame, then you never need to worry about Lorentz transformations etc. But from the start of this thread you have been trying to compare time intervals in TWO frames. That requires translating events from one frame to another. And that means you need to make sure you're actually using the same event in both frames.

Your constant mistake has been to assume that two different events are actually the same event.

This method you use above is called frame mixing.

The only method I have used is to apply the Lorentz transformation correctly to the events you have specified. You, on the other hand, have continually tried to bypass the hard work, which has meant that you have made the same error over and over again.

Complete explanations of where you went wrong can be found in posts #117, #118, #165 and #177.

 

[rpenner]

You are less than honest.

Suitable for this debate mutatis mutandis.

http://atheismresource.com/wp-content/uploads/Debate-Flow-Chart.jpg

 

[chinglu]

chinglu:



An event at the origin is one event. An event at x=-k is a different event. Those two spatially-separated events can only be simultaneous in one frame. They cannot be simultaneous in two frames.



If you're only doing calculations in one frame, then you never need to worry about Lorentz transformations etc. But from the start of this thread you have been trying to compare time intervals in TWO frames. That requires translating events from one frame to another. And that means you need to make sure you're actually using the same event in both frames.

Your constant mistake has been to assume that two different events are actually the same event.



The only method I have used is to apply the Lorentz transformation correctly to the events you have specified. You, on the other hand, have continually tried to bypass the hard work, which has meant that you have made the same error over and over again.

Complete explanations of where you went wrong can be found in posts #117, #118, #165 and #177.

What we have thus far is we both agree in the unrprimed frame,

Δt = (-k/γ)/v

What we have thus far is we both agree in the unrprimed frame,

Δt' = (-k)/v

Now, we both agree the origns same are same event.

Next, we both agree when k and unprimed origin same, same event.

So, we are both in agreement both frames can agree on the start event and the end event.

Your calculations in #117, #118 do not equal to your conclusions of
Δt = (-k/γ)/v
Δt' = (-k)/v

So, your own calculations are contradicting your own statements.

As we can see, we compare the time intervals of the start and end events.

We have no choice but to conclude

Δt'/γ = Δt

This mean, when primed frame is stationary, moving origin is time dilated or Δt' = Δtγ

When unprimed frame is stationary, moving k is time expanded or Δt'/γ = Δt.

This is also consistent with the invertibility of the LT matrix.

Last,

If coordinate k, then apply Lorentz Transforms

t = ( t' + vx/c² )γ
t = ( (-k)/v + v(-k)/c²)γ
t = -k/v( 1 + v²/c²)γ = -(k/γ)/v

Now, go other way.
t' = ( t - vx/c² )γ
t' = ( (-k/γ)/v - v(0)/c² )γ = -k/v

We can see my calculations are the same as Lorentz transform and the results are correct with invertible LT matrix.

Your calculations do not calculate
Δt = (-k/γ)/v
Δt' = (-k)/v

You calculations do not consistent with invertibility of the LT matrix.

 

[chinglu]

Suitable for this debate mutatis mutandis.

http://atheismresource.com/wp-content/uploads/Debate-Flow-Chart.jpg

Not sure what this has to with math.

You calculations are not Lorentz Matrix invertible. So your math wrong.

 

[rpenner]

Not sure what this has to with math.

You calculations are not Lorentz Matrix invertible. So your math wrong.

It's remarkable that you would make this claim in light of Post #18 where the Lorentz transform takes S to S' and the inverse Lorentz transform takes S' to S for both events and intervals.

Indeed, in Post 147, Appendix [2], it was demonstrated that the inverse Lorentz transform composed with the Lorentz transform is always the identity transform, and you do not begin to explain how this could not be the case.

 

[chinglu]

It's remarkable that you would make this claim in light of Post #18 where the Lorentz transform takes S to S' and the inverse Lorentz transform takes S' to S for both events and intervals.

Indeed, in Post 147, Appendix [2], it was demonstrated that the inverse Lorentz transform composed with the Lorentz transform is always the identity transform, and you do not begin to explain how this could not be the case.

Here is normal Lorentz Transform inversion.

Let x' and t' be given and time-like interval.
x = ( x' + vt' ) γ
t = ( t' + vx'/c² ) γ

Now, given t and x have been calculated as above,
x' = ( x - vt ) γ

Plug in above x and t as calculated.
x' = ( [ ( x' + vt' ) γ] - v[( t' + vx'/c² ) γ] ) γ
x' = γ²( ( x' + vt' ) - v( t' + vx'/c² ) )
x' = γ²( x' - v²x'/c² ) )
x' = γ²( x' ( 1 - v²/c² ) )
x' = x' .
This is invertability.

But, your claim, if unprimed rest frame, always t = t' γ since moving frame interval is always time dilated.
Let's try your theory.

x' = ( x - vt ) γ
Plug in x = ( x' + vt' ) γ but, according to your theory, must plug in t = t' γ always.
x' = ( ( x' + vt' ) γ - v(t' γ)) γ
x' = γ²( ( x' + vt' ) - vt')
x' = γ² x'
1 = γ² Only if v = 0. So, your theory fails Lorentz Transform inversion test just like I said.

 

[James R]

chinglu:

There's little point in my repeating myself. You need to go back and read posts #117 and #118. If you like, you can also look at #165 and #177.

Get back to me if you have any questions about those.

What we have thus far is we both agree in the unrprimed frame,

Δt = (-k/γ)/v

What we have thus far is we both agree in the unrprimed frame,

Δt' = (-k)/v

You haven't specified which events you used to calculate these two time intervals. I did, in post #117. We can't "agree" about this unless you are specific about which events you used.

Next, we both agree when k and unprimed origin same, same event.

No. An event at x'=-k MUST be different to an event at the origin x'=0. Two different spatial locations = two different events.

So, we are both in agreement both frames can agree on the start event and the end event.

Yes. They can agree if you specify which events you're talking about.

Your calculations in #117, #118 do not equal to your conclusions of
Δt = (-k/γ)/v
Δt' = (-k)/v

So, your own calculations are contradicting your own statements.

No. There's no contradiction. Nor have you shown any.

You calculations do not consistent with invertibility of the LT matrix.

Then go through post #117 and show exactly where I made a mistake. Go through it line by line.

We've had almost 100 posts since then, and you've produced nothing new.

 

[rpenner]

But, your claim, if unprimed rest frame, always t = t' γ since moving frame interval is always time dilated.
Let's try your theory.

x' = ( x - vt ) γ
Plug in x = ( x' + vt' ) γ but, according to your theory, must plug in t = t' γ always.
x' = ( ( x' + vt' ) γ - v(t' γ)) γ
x' = γ²( ( x' + vt' ) - vt')
x' = γ² x'
1 = γ² Only if v = 0. So, your theory fails Lorentz Transform inversion test just like I said.

My claim (which the second table of post 18 makes explicit) is that if clock is at rest in the unprimed frame, then: Δx = 0, and each tick of the clock has a certain amount of time Δt > 0, while in the primed coordinates the same object is seen by the moving observer as moving at speed -v, or Δx' = −vΔt', and the moving observer sees the time between ticks as Δt' = γΔt > Δt.

This is entirely consistent with the Lorentz transforms, which simplify with Δx = 0 as:
Δx' = γ(Δx − vΔt) = −vγΔt = −vΔt'
Δt' = γ(Δt − vΔx/c²) = γΔt
Likewise entirely consistent with the inverse Lorentz transforms of Δx' = −vΔt'
Δx = γ(Δx' + vΔt') = γ(−vΔt' + vΔt') = 0
Δt = γ(Δt' + vΔx'/c²) = γ(Δt' −v²Δt'/c²) = Δt'/γ

 

[chinglu]

My claim (which the second table of post 18 makes explicit) is that if clock is at rest in the unprimed frame, then: Δx = 0, and each tick of the clock has a certain amount of time Δt > 0, while in the primed coordinates the same object is seen by the moving observer as moving at speed -v, or Δx' = −vΔt', and the moving observer sees the time between ticks as Δt' = γΔt > Δt.

This is entirely consistent with the Lorentz transforms, which simplify with Δx = 0 as:
Δx' = γ(Δx − vΔt) = −vγΔt = −vΔt'
Δt' = γ(Δt − vΔx/c²) = γΔt
Likewise entirely consistent with the inverse Lorentz transforms of Δx' = −vΔt'
Δx = γ(Δx' + vΔt') = γ(−vΔt' + vΔt') = 0
Δt = γ(Δt' + vΔx'/c²) = γ(Δt' −v²Δt'/c²) = Δt'/γ

You are now consistent with Lorentz Matrix invertibility.

Let's take unprimed as stationary and use your calculations. that mean primed frame moving and use information in the rest frame to calculate the moving frame.

You have
Δt' = γ(Δt − vΔx/c²) = γΔt, thus Δt'/γ = Δt.

By your own post, the moving frame is time expanded or beats faster than the at rest unprimed frame.

quod erat demonstrandum

 

[chinglu]

chinglu:

There's little point in my repeating myself. You need to go back and read posts #117 and #118. If you like, you can also look at #165 and #177.

Get back to me if you have any questions about those.



You haven't specified which events you used to calculate these two time intervals. I did, in post #117. We can't "agree" about this unless you are specific about which events you used.



No. An event at x'=-k MUST be different to an event at the origin x'=0. Two different spatial locations = two different events.



Yes. They can agree if you specify which events you're talking about.



No. There's no contradiction. Nor have you shown any.



Then go through post #117 and show exactly where I made a mistake. Go through it line by line.

We've had almost 100 posts since then, and you've produced nothing new.

There is little point in repeating myself. I told you step2 is an error for the problem.

You assigned a negative elapsed time to the k coordinate from the view of the unprimed frame. The unprimed frame does not care the time on that clock. It has nothing to do with anything.

the unprimed frame is only timing the arrival of k to the unprimed origin in which you already confessed that time is
Δt = (-k/γ)/v

You also confessed

Δt' = -k/v

The ratio Δt'/Δt = γ which means the moving clock k beats time expanded.

Do not forget, the ratio Δt'/Δt = γ is based on confessions you made.

 

[James R]

There is little point in repeating myself. I told you step2 is an error for the problem.

You assigned a negative elapsed time to the k coordinate from the view of the unprimed frame. The unprimed frame does not care the time on that clock. It has nothing to do with anything.

If you wish to compare time intervals for two events in two different frames, you need to make sure you're dealing with the same events in both frames. You did not do that.

I have explained your errors quite clearly in posts #117, #118, #165 and #177. You have not directly addressed any of those posts.

There's no point in talking to you further. You are unteachable.

Goodbye.

 

[chinglu]

If you wish to compare time intervals for two events in two different frames, you need to make sure you're dealing with the same events in both frames. You did not do that.

I have explained your errors quite clearly in posts #117, #118, #165 and #177. You have not directly addressed any of those posts.

There's no point in talking to you further. You are unteachable.

Goodbye.

Wrong. I did that over and over. The start is origin same. The end is k same with unprimed origin.

I asked you over and over if this end event could not be agreed upon. You never answered. The fact is it can.

I clearly set up intervals that you also agreed upon.

That proved dt'/dt = γ.

I explained to you over and over your statement 2 in your posts has not issue on the problem.

 

[rpenner]

But, your claim, if unprimed rest frame, always t = t' γ since moving frame interval is always time dilated.
Let's try your theory.

x' = ( x - vt ) γ
Plug in x = ( x' + vt' ) γ but, according to your theory, must plug in t = t' γ always.
x' = ( ( x' + vt' ) γ - v(t' γ)) γ
x' = γ²( ( x' + vt' ) - vt')
x' = γ² x'
1 = γ² Only if v = 0. So, your theory fails Lorentz Transform inversion test just like I said.

My claim (which the second table of post 18 makes explicit) is that if clock is at rest in the unprimed frame, then: Δx = 0, and each tick of the clock has a certain amount of time Δt > 0, while in the primed coordinates the same object is seen by the moving observer as moving at speed -v, or Δx' = −vΔt', and the moving observer sees the time between ticks as Δt' = γΔt > Δt.

This is entirely consistent with the Lorentz transforms, which simplify with Δx = 0 as:
Δx' = γ(Δx − vΔt) = −vγΔt = −vΔt'
Δt' = γ(Δt − vΔx/c²) = γΔt
Likewise entirely consistent with the inverse Lorentz transforms of Δx' = −vΔt'
Δx = γ(Δx' + vΔt') = γ(−vΔt' + vΔt') = 0
Δt = γ(Δt' + vΔx'/c²) = γ(Δt' −v²Δt'/c²) = Δt'/γ

You are now consistent with Lorentz Matrix invertibility.

I think you mean that I have always been consistent with the use of Lorentz transforms and inverse Lorentz transforms, since I quote you results from post 18. But thank you for realizing this a mere 170 posts later.

Let's take unprimed as stationary and use your calculations. that mean primed frame moving and use information in the rest frame to calculate the moving frame.

It is a mistake to call the frames or clocks or objects as moving or stationary except in relationship to a labelled frame. Every distinct event has a unique set of space-time coordinates in every inertial frame -- these sets of coordinates are like the names of objects and every frame is a different language. Information doesn't flow between the frames on an event-by-event basis. The Lorentz transform is a dictionary which for a given pair of frames lets us look up the name (coordinates) of and event in the other frame.

You have
Δt' = γ(Δt − vΔx/c²) = γΔt, thus Δt'/γ = Δt.

I have Δx between two events equal to zero in the unprimed frame and therefore in a different frame where Δx' is not equal to zero, then Δt' > Δt. In the most important case for this, the two events are a clock measuring 11:59:59 and the same clock measuring 12:00:00, so that Δx = 0 and Δt = 1 second means the clock is not moving for the unprimed observer and it measures one second in one second.

Since γ ≠ 0, Δt' = γΔt and Δt'/γ = Δt represent the same algebraic statement. Importantly since v ≠ 0 means γ > 1, both Δt' = γΔt and Δt'/γ = Δt mean Δt' > Δt.

By your own post, the moving frame is time expanded or beats faster than the at rest unprimed frame.

The frame is not time-expanded, nor does it beat. The frame is a system of space-time coordinates assigned to events. If the unprimed frame is the coordinate system where we agree that a certain clock is stationary, then the primed frame is the frame where we have agreed that the clock is moving.

The moving clock is time dilated, which means each tick of the moving clock needs a longer time than the unmoving clock. Δt' > Δt. But this means the beats slower since faster and slower refer to how many ticks pass per amount of time. So if the start and end events are ticks of the the clock, then Δt' > Δt means that the primed frame measures the lower tick frequency for the moving clock than then unprimed frame measures for the stationary clock, 1/Δt' < 1/Δt.

For the same two events, the primed coordinates see the clock move, since Δx' ≠ 0, and since Δt' > 1 second, the "moving clock" is seen in the primed coordinates to need more than 1 second to move from 11:59:59 to 12:00:00. Therefore, you have gotten it backwards. If Δx = 0 and v ≠ 0, then γ > 1 and Δt' = γΔt means Δt' > Δt, so that if Δt represents the time between two ticks of a stationary clock, then Δt' represents the time between two ticks of a moving clock and Δt' > Δt means that the rate of the ticks is slower for the moving clock 1/Δt' < 1/Δt.

quod erat demonstrandum

Arrogant and wrong when remedial concepts like what "beats faster" or "frame" or "event" or "interval" have yet to have been mastered.

 

[James R]

chinglu:

If you can find a mistake in posts #117, #118, #165 and #177, then get back to me. Don't write one or two lines saying I have said something I have not said. You need to go through post #117 line by line and show me the point where I made an error, if any such error exists. Otherwise, don't bother writing to me again.

I have wasted enough time on you. After four separate explanations of the same point, plus innumerable posts in between breaking it down into bite-sized chunks for you, you seem incapable of understanding a simple point. Either you're stupid, or your a troll. Either way, you're a waste of my time.

 

[chinglu]

I think you mean that I have always been consistent with the use of Lorentz transforms and inverse Lorentz transforms, since I quote you results from post 18. But thank you for realizing this a mere 170 posts later.

Let's look at your post #18 all the way to the right with time. You said, for this same exercise, t' = t/γ for unprimed frame stationary and t' = tγ for primed frame take stationary. Afrer all these pages, you still do not understand this is not LT invertible. That is the proof I wrote for you. This mean you cannot even tell when you step into a contradiction.

It is a mistake to call the frames or clocks or objects as moving or stationary except in relationship to a labelled frame. Every distinct event has a unique set of space-time coordinates in every inertial frame -- these sets of coordinates are like the names of objects and every frame is a different language. Information doesn't flow between the frames on an event-by-event basis. The Lorentz transform is a dictionary which for a given pair of frames lets us look up the name (coordinates) of and event in the other frame.

So, what is the semantics of this dictionary? When you have my understanding, you will not step into contradiction. Anyway, show readers you know the semantics.

I have Δx between two events equal to zero in the unprimed frame and therefore in a different frame where Δx' is not equal to zero, then Δt' > Δt. In the most important case for this, the two events are a clock measuring 11:59:59 and the same clock measuring 12:00:00, so that Δx = 0 and Δt = 1 second means the clock is not moving for the unprimed observer and it measures one second in one second.

Please explain if Δx = 0, how are you judging the start and stop of events? In other words, if hold clock with Δx = 0, how do you know when to stop time? Do you get a message from God? You need to specify this.

Since γ ≠ 0, Δt' = γΔt and Δt'/γ = Δt represent the same algebraic statement. Importantly since v ≠ 0 means γ > 1, both Δt' = γΔt and Δt'/γ = Δt mean Δt' > Δt.

Where exactly do you have reciprocal time dialation, since that is not SR? You seem to be non-mainsteam here which may make you a crackpot. Now, if you try to prove reciprocal time dilation, I will run you into a contradiction using the invertibility of the LT matrix.

The frame is not time-expanded, nor does it beat. The frame is a system of space-time coordinates assigned to events. If the unprimed frame is the coordinate system where we agree that a certain clock is stationary, then the primed frame is the frame where we have agreed that the clock is moving.

I can say what I want. I say the frame beats a time beat. Einstein said the frame has but one time. Are you refuiting this also?

It is essential to have time defined by means of stationary clocks in the stationary system, and the time now defined being appropriate to the stationary system we call it ``the time of the stationary system.''
http://www.fourmilab.ch/etexts/einstein/specrel/www/

The moving clock is time dilated, which means each tick of the moving clock needs a longer time than the unmoving clock. Δt' > Δt. But this means the beats slower since faster and slower refer to how many ticks pass per amount of time. So if the start and end events are ticks of the the clock, then Δt' > Δt means that the primed frame measures the lower tick frequency for the moving clock than then unprimed frame measures for the stationary clock, 1/Δt' < 1/Δt.

Can you please show this is consistent with the invertibility of the LT matrix given the same events between the frames? No you can not.

Arrogant and wrong when remedial concepts like what "beats faster" or "frame" or "event" or "interval" have yet to have been mastered.

Too bad. You continue to contradict yourself after all these pages where I showed you reciprocal time dilation is inconsistent with the invertibility of the LT matrix.

I just keep hammering you over and over on this and you never learn.

 

[chinglu]

chinglu:

If you can find a mistake in posts #117, #118, #165 and #177, then get back to me. Don't write one or two lines saying I have said something I have not said. You need to go through post #117 line by line and show me the point where I made an error, if any such error exists. Otherwise, don't bother writing to me again.

I have wasted enough time on you. After four separate explanations of the same point, plus innumerable posts in between breaking it down into bite-sized chunks for you, you seem incapable of understanding a simple point. Either you're stupid, or your a troll. Either way, you're a waste of my time.

You have agreed that

t' = k/v.

and

t = (k/γ)/v

for the same events for the 2 frames.

Is there any way possible way you can at least admit what you agreed to?

 

[James R]

chinglu:

You give me the spacetime coordinates for the two events I have supposedly agreed about. Make sure you give the coordinates of both events in both the unprimed and primed frame. Then we can talk further. Please express the coordinates in the form $$(x,t)=$$ whatever and $$(x',t')=$$ whatever.

 

[chinglu]

chinglu:

You give me the spacetime coordinates for the two events I have supposedly agreed about. Make sure you give the coordinates of both events in both the unprimed and primed frame. Then we can talk further. Please express the coordinates in the form $$(x,t)=$$ whatever and $$(x',t')=$$ whatever.

Yes, for better use x' = -k where k > 0

Event 1
(x,t)=( 0,0 )
(x',t')=( 0,0)

Event 2
(x,t)=( 0,(-k/γ)/v )
(x',t')=( k,-k/v )

Event 1 is always true for standard configuration.

Event 2

Calculate from x frame. (x,t)=( 0,(-k/γ)/v )
x' = ( x - vt )γ

x' = ( 0 - v((-k/γ)/v )γ
x' = k

t' = ( t - vx/c² )γ

t' = ( (-k/γ)/v - v(0)/c² )γ

t' = -k/v


Calculate from x' frame. (x',t')=( k,-k/v )

x = ( x' + vt' )γ

x = ( k + v(-k/v) )γ

x = 0

t = ( t' + vx'/c² )γ

t = ( -k/v + vk/c² )γ

t = k/v( -1 + v²/c² )γ

t = (-k/v)/γ

 

[rpenner]

If 0 < v < c, event 2 happens now before event 1. Also, event 2 happens at x' = k, not x' = -k.

 

[chinglu]

If 0 < v < c, event 2 happens now before event 1. Also, event 2 happens at x' = k, not x' = -k.

Also, event 2 happens at x' = k, not x' = -k

I may have made a mistake, I have before. I have above,

Event 2
(x,t)=( 0,(-k/γ)/v )
(x',t')=( k,-k/v )

x' = ( 0 - v((-k/γ)/v )γ
x' = k

No event 2 does not happend before event 1.

Event 1 is the originss being same.

Event 2 is the unprimed origin and k being same.