Time Dilation

[Pete]

You are claiming relative measurements on the same Δt' representing the same events?

Different events, chinglu. Didn't you read my post?

And, I do not care the start time on the clock. LT calculates elapsed times.

No, a single Lorentz transform calculates a single clock reading at a single event. To get an elapsed time you need two readings.

I have performed these calculations several times and I never end up with two different answers for t' or t. Both frames agree on my numbers. Do you know why I do not end up with contradictions? Because I am doing it correctly.

chinglu, the contradictions that you produce are because you are doing it incorrectly.

Finally, I find it amazing you have two different answer for Δt' and think you are correct.

I don't find it at all amazing that I get two different answers to two different questions.

 

[Pete]

James is right, chinglu. It looks bad for you that you keep pretending to haven't answered him when you have not.

 

[Pete]

Synchronization makes no difference. Say one clock had t1 from the unprimed view and another had t2 when the origins are the same. It does not matter. LT is not about times on the clocks. it is about elapsed times.

Of course synchronization matters. That's not just SR, that's everyday practical fact.
Consider a flight from Hong Kong to Dubai.
You take off at 9:15, and land at 11:15.
How much time elapsed on the flight?

And the simple lorentz transforms we're using in this thread are most definitely about times on clocks. Your repeated insistence otherwise is why you are getting contradictions. You're doing it wrong.

 

[chinglu]

It is quite clear in post #117 that no times were "added" to any clocks. Three events are analysed in post #117. The time intervals between pairs of those events are given in both the primed and unprimed frames.

There are no errors in post #117.

Let me ask you a direct yes-no question:

Do you agree that if a frame records a proper time interval between two events, then any other frame will record a longer time interval between the same pair of events?

Yes or no?

I will wait until you answer this question. Until then, I will not be interacting with you any further.

What is this childish game after it has been proven you cannot make the frames agree on Δt' with your silly failed scheme.

The space time interval is invariant.

 

[chinglu]

Different events, chinglu. Didn't you read my post?


No, a single Lorentz transform calculates a single clock reading at a single event. To get an elapsed time you need two readings.

chinglu, the contradictions that you produce are because you are doing it incorrectly.

I don't find it at all amazing that I get two different answers to two different questions.

You do not understand relativity.

It is assumed the start event is the origins being the same for the standard configuration.

So, since the origins being the same is the start event, you get two different answers for Δt' from the start event of the origins being the same.

That is a contradiction.

 

[chinglu]

James is right, chinglu. It looks bad for you that you keep pretending to haven't answered him when you have not.

I have answered him as I have you.

You 2 get 2 different Δt' values from the time the origins were the same.

This means both of you failed.

 

[Pete]

What is this childish game after it has been proven you cannot make the frames agree on Δt' with your silly failed scheme.

I certainly can, chinglu. James did it in post #117. It is you who keeps getting it wrong.
Your problem is that you are confusing two different spacetime intervals.

For the interval beginning at x'=k, t'=0 and ending at x'=k, x=0, all frames agree that the primed clock at x'=k elapses dt' = -k/v

For the interval beginning at x'=k, t=0 and ending at x'=k, x=0, all frames agree that the primed clock at x'=k elapses dt' = -k/vγ^2

It is assumed the start event is the origins being the same for the standard configuration.

Wrong. The start event is at x'=k, not at x'=0.
You are confusing two different start events:
x'=k,t'=0
x'=k,t=0

Which of these events do you choose to be the start event?

I have answered him as I have you.

And yet again you fail to answer the simple question in his last post.

 

[chinglu]

I certainly can, chinglu. James did it in post #117. It is you who keeps getting it wrong.
Your problem is that you are confusing two different spacetime intervals.

For the interval beginning at x'=k, t'=0 and ending at x'=k, x=0, all frames agree that the primed clock at x'=k elapses dt' = -k/v

For the interval beginning at x'=k, t=0 and ending at x'=k, x=0, all frames agree that the primed clock at x'=k elapses dt' = -k/vγ^2

1) You are confused, The Minkowski metric does not tell you about timing, it tells you about the different between the timing and the timing of light.

2) Next, you are taling about events of [x'=k, t'=0], [x'=k, t=0], [x'=k, x=0] and [x'=k, x=0].

This is a garbled mess of confusing space-time and space.

Wrong. The start event is at x'=k, not at x'=0.
You are confusing two different start events:
x'=k,t'=0
x'=k,t=0

The start event is the 2 origins being the same. You do not understand this stuff.

And yet again you fail to answer the simple question in his last post.

I have answered it over and over and provided the correct math as well. All my results agree with LT and are LT invertible. You and James are weak at this and have no idea what you are doing.

This is simply about a clock at k moving toward the unprimed origin in the view of the unprimed frame.

From the view of the primed frame, it is about the origin moving toward the clock at k. The start is the origins being the same and the end is the unprimed origin begin the same as k.

It is that simple and you 2 keep introducing additional events that only get you confused and have nothing to do with the probem.

Further, James R argued that time is always time dilated Yet you said,

For the interval beginning at x'=k, t'=0 and ending at x'=k, x=0, all frames agree that the primed clock at x'=k elapses dt' = -k/v

For the interval beginning at x'=k, t=0 and ending at x'=k, x=0, all frames agree that the primed clock at x'=k elapses dt' = -k/vγ^2

Note your γ^2 differential which is not in any mainstream interpretation.

In short, your post is completely false.

If you want to understand what is correct, simply read my math. If you deviate from that, you know you are wrong.

 

[Pete]

1) You are confused, The Minkowski metric does not tell you about timing, it tells you about the different between the timing and the timing of light.

2) Next, you are taling about events of [x'=k, t'=0], [x'=k, t=0], [x'=k, x=0] and [x'=k, x=0].

This is a garbled mess of confusing space-time and space.

You are certainly confused, chinglu.
Do you not see how setting x'=k, x=0 (for example) is enough to define a single event?

The start event is the 2 origins being the same. You do not understand this stuff.

Chinglu, do you understand that an event is a particular time at a particular place?

You said that you want to find the elapsed time on the primed clock at x'=k.
Therefore the particular place of the start event is at x'=k, not at x'=0.

What's so difficult to understand about this?

You seem to want to consider this interval instead:

• Start event: at the two origins coinciding: [x=0,t=0], [x'=0,t'=0]
• End event: when x'=k meets the x origin: [x=0,t=-vk/γ], [x'=k,t'=-k/v]

Do you see that this interval corresponds to an elapsed time on the unprimed clock at x=0, not the primed clock at x'=k?
Look:
There are two different primed clocks at the start and events of this interval. One at x'=0, one at x'=k.
The only clock that is at both the start and end events is the unprimed clock at x=0.

If you want to consider the elapsed time on a clock, you need to look at that clock for more than an instant. Is that so hard to understand?

This is simply about a clock at k moving toward the unprimed origin in the view of the unprimed frame.

From the view of the primed frame, it is about the origin moving toward the clock at k. The start is the origins being the same and the end is the unprimed origin begin the same as k.

Then why are you refusing to look at the primed clock at x'=k at the start?
Why are you only looking at the clocks at x=0 and x'=0 at the start?

Further, James R argued that time is always time dilated Yet you said,

For the interval beginning at x'=k, t'=0 and ending at x'=k, x=0, all frames agree that the primed clock at x'=k elapses dt' = -k/v

For the interval beginning at x'=k, t=0 and ending at x'=k, x=0, all frames agree that the primed clock at x'=k elapses dt' = -k/vγ^2

So what's the problem?

Note your γ^2 differential which is not in any mainstream interpretation.

You said you could do maths, chinglu. Work it through and see what you find.

 

[chinglu]

You are certainly confused, chinglu.
Do you not see how setting x'=k, x=0 (for example) is enough to define a single event?

Sure that is the terminating event. But, I can't see your times.

Chinglu, do you understand that an event is a particular time at a particular place?

You said that the primed clock is at x'=k.
Therefore the particular place of the start event is at x'=k.

What's so difficult to understand about this

The start event is the origins being the same. That has spacial separation to x'=k. So, x'=k is not an agreed upon start event. That is the source of your confusion. Both frames cannot have 2 different spacially separated start events based on the relativity of simultaneity. This is confusing to the beginner.

 

[Pete]

Sure that is the terminating event. But, I can't see your times.

So use the lorentz transform to calculate them.
x'=k, x=0
...LT...
t'=-kv, t=-vk/γ
Is that so hard?

Perhaps you'd like to reconsider the post that confused you.

The start event is the origins being the same. That has spacial separation to x'=k. So, x'=k is not an agreed upon start event.
That is the source of your confusion. Both frames cannot have 2 different spacially separated start events based on the relativity of simultaneity. This is confusing to the beginner.

Exactly!

So, by starting the interval at x'=0, you can't determine the elapsed time on the primed clock at x'=k.
Your start and end events are both at x=0, so you can only determine the elapsed time on the unprimed clock at x=0.
And all frames agree that the unprimed clock at x=0 elapses dt=-vk/γ between the two specified events.

 

[James R]

Chinglu:

Since you still haven't answered the question in post #300, I won't be directly replying to anything you've written.

A simple question waits for you in post #300. You have not been able to cope with the complexity of posts #117, #118, #165 and #177, so we'll start back at the basics with you.

 

[chinglu]

Chinglu:

Since you still haven't answered the question in post #300, I won't be directly replying to anything you've written.

A simple question waits for you in post #300. You have not been able to cope with the complexity of posts #117, #118, #165 and #177, so we'll start back at the basics with you.

Sure I have handled your posts. You just type a string of symbols and hope they are true.

I produced math over and over such that all frames agreed that the moving clock elapsed more time. You have never been able to refute my math and sit around praying you can do such math. Where is your refutation of my math? You run and hide in fear. Otherwise refute my math. If you try, everyone will laugh at you when I am done.

You produced math and I showed you that frames could not agree one way or the other.

That means your math is a failure.

 

[chinglu]

So, by starting the interval at x'=0, you can't determine the elapsed time on the primed clock at x'=k.
Your start and end events are both at x=0, so you can only determine the elapsed time on the unprimed clock at x=0.
And all frames agree that the unprimed clock at x=0 elapses dt=-vk/γ between the two specified events.

Yes, so with your way, you have physical events that cannot be decided meaning your way is a failure.

My way, everything is decided and consistent.

 

[Pete]

I produced math over and over such that all frames agreed that the moving clock elapsed more time.

Rubbish. You just agreed that the interval you are analysing only tells us the time that elapses on the unprimed clock at x=0.
You are explicitly not finding the time elapsed on any primed clock.

Yes, so with your way, you have physical events that cannot be decided meaning your way is a failure.

Rubbish.
You've been shown many times how to get the time elapsed on the primed clock, you are simply choosing not to do so.

Like I said before:

For the interval beginning at x'=k, t'=0 and ending at x'=k, x=0, all frames agree that the primed clock at x'=k elapses dt' = -k/v

For the interval beginning at x'=k, t=0 and ending at x'=k, x=0, all frames agree that the primed clock at x'=k elapses dt' = -k/vγ^2

So what's your problem?

 

[James R]

chinglu:

Sure I have handled your posts.

You can't even answer a simple yes/no question - see post #300, which remains unanswered by you.

You just type a string of symbols and hope they are true.

Well, yes. That's called writing without trolling. A new concept for you, I'm sure.

I produced math over and over such that all frames agreed that the moving clock elapsed more time.

Which one is the moving clock? How can you tell?

You have never been able to refute my math and sit around praying you can do such math.

You seem to know a lot about my religious beliefs.

As for refutation, see posts #117, #118, #165 and #177, where you were soundly trounced. You have never gone through those posts line-by-line to point out errors, because there are no errors in them.

Where is your refutation of my math? You run and hide in fear. Otherwise refute my math. If you try, everyone will laugh at you when I am done.

My refutation is in posts #117, #118, #165 and #177 of the current thread, which remain unaddressed by you. You run and hide in fear.

Challenge to chinglu:

1. Respond to the question in post #300 with "yes" or "no".
2. Go through any or all of posts #117, #118, #165 or #177 line-by-line, saying whether you agree with each part and explaining clearly any errors you can find in those posts.

Assertion:

chinglu will NEVER answer either of the above challenges, especially number 2. He will run away and hide in fear, as usual.

 

[chinglu]

Rubbish. You just agreed that the interval you are analysing only tells us the time that elapses on the unprimed clock at x=0.
You are explicitly not finding the time elapsed on any primed clock.


Rubbish.
You've been shown many times how to get the time elapsed on the primed clock, you are simply choosing not to do so.

Like I said before:

For the interval beginning at x'=k, t'=0 and ending at x'=k, x=0, all frames agree that the primed clock at x'=k elapses dt' = -k/v

For the interval beginning at x'=k, t=0 and ending at x'=k, x=0, all frames agree that the primed clock at x'=k elapses dt' = -k/vγ^2

So what's your problem?

Your ending event contains no time.

This indicates you do not know what you are doing.

 

[chinglu]

chinglu:



You can't even answer a simple yes/no question - see post #300, which remains unanswered by you.



Well, yes. That's called writing without trolling. A new concept for you, I'm sure.



Which one is the moving clock? How can you tell?



You seem to know a lot about my religious beliefs.

As for refutation, see posts #117, #118, #165 and #177, where you were soundly trounced. You have never gone through those posts line-by-line to point out errors, because there are no errors in them.



My refutation is in posts #117, #118, #165 and #177 of the current thread, which remain unaddressed by you. You run and hide in fear.

Challenge to chinglu:

1. Respond to the question in post #300 with "yes" or "no".
2. Go through any or all of posts #117, #118, #165 or #177 line-by-line, saying whether you agree with each part and explaining clearly any errors you can find in those posts.

Assertion:

chinglu will NEVER answer either of the above challenges, especially number 2. He will run away and hide in fear, as usual.

It is not up to me to sort through your failed rantings. I have though but you can't understand my findings.

But, is is up to you to refute my thread.

Refute #250.

You will fail and run and hide in fear. if you chose to attack it, many will laugh at you.

 

[James R]

chinglu fails once again to respond to my challenges in post #316.

 

[Pete]

Your ending event contains no time.

That's your event, chinglu. It's when x'=k meets x=0, remember?
And we covered this before, remember?

Sure that is the terminating event. But, I can't see your times

So use the lorentz transform to calculate them.
x'=k, x=0
...LT...
t'=-kv, t=-vk/γ
Is that so hard?

So, once again:
For the interval beginning at x'=k, t'=0 and ending at x'=k, x=0, all frames agree that the primed clock at x'=k elapses dt' = -k/v

For the interval beginning at x'=k, t=0 and ending at x'=k, x=0, all frames agree that the primed clock at x'=k elapses dt' = -k/vγ^2

So what's your problem?