Twin paradox (Pete and MacM)

[MacM]

Hi Mac,
You can scrap that last test - I was working on the mistaken notion that something Paul T posted was something that you posted. Sorry.

Not a problem. My pleasure.

 

[Pete]

Nice try but try again.

Like I said...

Your interpretation of what you said is different to James' interpretation of what you said.

 

[MacM]

Hahaha. This matter should have been settled even the first time I pointed out the problem with your formula. You really have reading difficulty, don't you? I said in the post following your erroneous time dilation equation: "Even if you gave the correct equation..." and proceed using the corrected equation to illustrate my point.

Oh, my gosh...your equation was WRONG? You don't know what WRONG mean? I gave you what the equation should be written and said not even one word about notation usage. It was your stupidity that lead you to suspect that my comment was about notation usage.

That's your own problem. Whatever excuse you give unable to change the fact that you had given wrong equation. Do you still want to argue about it?

You bet your ass. You make a big deal out of a transcription error but fail to address the issue. How about facing the issue?

Can't do that can you? It would make you look really stupid after all these months. The fact that I rountinely write it correctly and have performed the calculation numerous times correctly get ignored and you try to make a typo mean far more than it does but ignore the fact that you do not understand Relativity. :bugeye:

http://www.sciforums.com/attachment.php?attachmentid=3337&stc=1

The day I really give you advice would be a disaster day.

Nothing new everytime you have tried to give advice it is a disaster.

There is no mistake to admit. You are the only one who make that silly claim. I have never seen in any book about relativity mentioning such a stupid idea.

Well, I can certainly agree with you on that. I suppose then according to you I am the first person to actually figure out Relativity??? Hardly.

I know I don't write English as perfect as you, the kindergarten English teacher who was sacked because of incompetency. My earlier post mostly concerning some basic math used in SR. Gosh, it math...not poetry. Why do you think its grammar really that crucial? The only problem I know is your lack of intelligence to comprehend math even the simplest one such as, couldn't get rid of that wrong time dilation equation from my mind. Hey, may be the equation is actually correct, but for MacM's relativity!

Actually it is not your lack of good manners but your inability to actually address the issue that bothers me. You continuously run off at the mouth and say nothing, just as in this thread. Now address the issue of reciprocity in Relativity.

Cut the bulshit. We already know that you are actually talking about MacM's relativity and therefore so much difference from the version that we know.

You pathetic fool. Now address the issue of reciprocity mandated by Relativity.

I remember this equation. It is a parabola open to the right, isn't it? It's remind me on my old school day. I never know it has anything to do with relativity. But, it is the math of MacM's relativity apparently. Good. [/quore]

Since you clearly are having trouble understanding Relativity, I thought I would simplify things a bit for you. Notice how you only get a curve if you look at one half of the problem?

Same thing in Relativity. A = B = A. There is no nonlinear relationship in time between clocks due to relative velocity.

It was an illegal math work, If you ask me. For those who has good math skill, please help cracking this silly math problem.

When cornered just jpretend to be dumb. But it doesn't get you to far.

You certainly do. Firstly, you invented MacM's relativity. Secondly, you invented that illegal killer math proof. Lastly, you have pasta pot.

You forgot the most important one.

I also get the last laugh. You idiot Relativity mandates no differeance in clock readings when brought back to a rest frame and directly compared.

Go ahead show us otherwise wise guy.

 

[MacM]

Like I said...

Your interpretation of what you said is different to James' interpretation of what you said.

I would like to see James R's comment on this. I see little room for confusion of the issue.

 

[Pete]

X = (X^2)^2 = X^4
Y = (Y^2)^2 = Y^4

X^4 = Y^4

X = Y.

Damn it is linear. Just like clocks in relative motion where the same function must be applied to both clocks.

Don't bother me further until you acknowledge this fact.

Hi Mac,
You seem to be having difficulty realising exactly what quantities are represented in SR equations. Essentially, the times and distances in the reciprocal situation correspond to different spacetime intervals, meaning you can't simply equate them with each other.

I'd like to repost something from an old thread, and see if it helps you see exactly how the reciprocity in SR works. The scenario is that two tickers, A and B are moving apart from each other at some speed, and we are attempting to specify the distance between them.


The distance between reference points in ticker A's frame are contracted in ticker B's frame, and not contracted in ticker A's frame.

The distance between reference points in ticker B's frame are contracted in ticker A's frame, and not contracted in ticker B's frame.

This apparent paradox is possible because distance is a spacetime interval between simultaneous events, and when a distance is specified in two different frames, there are actually two different spacetime intervals being specified.

In the case of specifying the distance between ticker A and ticker B , there are actually four distinct spacetime intervals that could be specified (only considering the two relevant frames).

Note that spacetime intervals are not relative. All observers will agree on the magnitude (in length units) of any spacetime interval.


So, when describing the distance between ticker A and ticker B, we must be careful to specify exactly what spacetime interval we mean.

Here are some events that define end points of distinct intervals that are valid inter-ticker distances:
Spacetime events:
A - ticker A clips an object that is comoving with ticker B and 1 unit from ticker B in ticker B's frame
B - ticker B clips an object that is comoving with ticker A and 1 unit from ticker A in ticker A's frame
C - ticker B's position at the time of event A in ticker B's frame
D - ticker B's position at the time of event A in ticker A's frame
E - ticker A's position at the time of event B in ticker A's frame
F - ticker A's position at the time of event B in ticker B's frame

Here are the four intervals, and the distances to which they correspond:
Distances:
Distance in ticker B's frame between inter-ticker reference events in ticker B's frame = interval A-C = 1
Distance in ticker A's frame between inter-ticker reference events in ticker B's frame = interval A-D = 1/gamma
Distance in ticker A's frame between inter-ticker reference events in ticker A's frame = interval B-E = 1
Distance in ticker B's frame between inter-ticker reference events in ticker B's frame = interval B-F = 1/gamma

(gamma = Lorentz factor)

Order of Events
In ticker B's frame, events occur in the following order: D, BF (simultaneous), AC (simultaneous), E

In ticker A's frame, events occur in the following order: F, AD (simultaneous), BE (simultaneous), C


Here are two diagrams showing the two reference frames, with the events labelled.

The green lines are the four distances specified.
<img src="/attachment.php?attachmentid=3338&stc=1">
<HR>
<img src="/attachment.php?attachmentid=3339&stc=1">


Exercise:
From these diagrams, determine which spacetime intervals correspond to dilated and undilated times for each frame.

 

[Pete]

I would like to see James R's comment on this.

[post=692979]Glad to be of service[/post]

 

[Pete]

If Y=0 then X=0.
If Y=1 then X=1.

In other words, MacM's conclusion that X=Y is false.

James, are you drinking?

 

[MacM]

MacM:

There is a lot to reply to here.

First, regarding my Special Relativistic time dilation and length contraction thread:

You have tried to drag the discussion from the current thread into that thread. As noted at the top of that thread "This thread is for those who want to learn how a couple of the well-known effects from Einstein's Special Theory of Relativity can be derived from first principles."

You are free to discuss the details of the derivation in that thread if you wish, but it is not appropriate to make assertions about the theory of relativity there without mathematical proof, since that thread specifically disproves your assertions.

It is funny indeed that you can post how others don't understand Relativity but given the fact that your post fails to fully explain Relativity and the reciprocity issue means you either don't understand Relativity or you choose to simply ignore it, so as to continue to teach BS.

It is a FIAT post which you do not want challenged. But it is your post and I will stay out of it. But I think it is a sad showing on your part.

It is not innuendo. My post was solidly mathematical. It infers nothing. Each step follows inexorably from what has gone before. There are no lies in that post. If there were, you should be able to point out where they are.

Either point out which step is wrong or shut up. Unsupported assertions are worthless.

Does you post consider and explain the issue of reciprocity? No. It therefore is nothing more than my X = Y^2 problem. A false conclusion by condsidering only half of the issue.

You misunderstand. The post you keep referring to regarding my "admission" considered a different scenario to the one presented in the original "two clock" problem in this thread, and also different to the "2 car" problem used in my derivation.

I agreed that there would be no difference in clock readings in a situation where the clocks were started and stopped symmetrically. To be explicit, let me give you that scenario:

"Two clocks, A and B, set out from the origin, with clock A travelling at +0.45c and clock B travelling at -0.45c relative to a stationary observer. All motion is at constant speed for the whole time. When each clock reads a pre-agreed time in its rest frame, that clock sends out a light pulse towards the other clock. Each clock calls the time the pulse is sent out time "zero". Each clock continues to run locally until the pulse is received from the other clock, at which time the local clock is stopped. The two clocks are then brought back together and their elapsed times compared."

In this scenario, both clocks will show the same final reading.

Precisely. That has also been the entire arguement for 1 1/2 years. However, you don't need the light signal to tell them to shut off. That is an unnecessary step. Each can simply shut down by i.e. - 3,600 seconds reading at their local time and the clocks will still agree upon being returned and compared in the same rest frame. Your light signal is nothing more than a confusion factor.

This scenario differs from the 2-clock scenario presented earlier in this thread in that 2 different signals are used to stop the clocks. In contrast, in our original scenario, and in my derivation thread, only one signal is used. The scenario presented in the paragraph above is a symmetrical one, and so no overall time difference is recorded. The orginal scenario is not symmetrical, so the effects of time dilation are not masked there.

You are confusing "Symmetrical" with "Reciprocity". Reciprocity simply means what ever happens to one clock happens to the other with regard to constant relative velocity or equal acceleration curves.

If we reverse the direction of the light signal, we get the opposite result. That is correct. But we should expect that. The situation is not symmetrical.

It is not merely a matter of reversing the test, it is recognizing that it exists even if you are not computing it such that your result is a one sided view of the situation. You have reduced your result to a perception and not the reality. The reality is that the opposite is true and the other clock has slowed by an equal amount, hence no net time dialtion affect between clocks.

The very suggestion that TD occurs becomes meaningless unless you assume both clocks are slower in accordance to some UT. But there is no measureable TD between such clocks.

That would complicate matters, and is unnecessary. It is easier to deal with the actual times, rather than just the perceived times, and, after all, it is the actual times that we are interested in. On the other hand, the perceived times can be used to derive the relativistic Doppler equation. I found that out by accident when I was working out the mathematics, but I didn't think it was important to include in my post, since that was not the point of the post.

So to take a factual view of the overall situation is an over complication? :bugeye:

No. As explained, the scenario is not symmetrical.

And as I have explained this is not merely a matter of "Symmetry". It is a matter of "Reciprocity".

A pan of evaporating water is a clock, and it does measure time. It just doesn't do it very well, because it is too suceptible to outside influences. A good clock is only influenced by time, ideally. Of course, in reality, no clock is perfect. Some are better than others. A pan clock is not a good clock, but it is a clock. An atomic clock is a better clock, though still not perfect. We have to live with the fact that we can't measure time perfectly, but that doesn't mean that what we measure isn't time.

It is far from being shown, much less proven, that clocks are measuring time. Clocks are processes nothing more.

* 1 Joule is not 1 m/s². I gave you the correct units above.
* m/s is a velocity.
* m/s² is an acceleration.

Your assertion that different units are required is useless. You're just waffling. You have no idea how time could be an "energy flow". It just sounded good to you when you made it up. It is a useless and self-contradictory concept.

You the above J = Kg*m/s^2 but went on to make a false (or unsupported conclusion) statement that a second is not the same as a Joule hence time cannot be energy.

Your statement is gibberish and has no meaning. Nobody said a second and a Joule were the same thing. But change is induced by energy and it is change that produces the concept of time.

A second is an arbitrary interval quantisizing change. A planck time however would be the minimal increment of time and represents a quantum of change or minumum quantum of energy to affect change.

In the same sentence, you say time is both energy and an energy flow. Which is it? You're contradicting yourself. You're confused. Poor MacM.

No you are confused since you lack the ability to have vision. (I don't mean that derogatorily)

Light is the only thing whose speed we know is a constant. That is why it is easiest to use light signalling in many examples. But there's nothing special about it as an information carrier. Information can be carried by all kinds of things - sound, light, messages cut into a stick, etc.

I agree with everything except your confidence that you know anything about light and its "Apparent" invariance. But that has nothing to do with the reciprocity conflict within Relativity. Lets stick to the issue.

So, can two events at different locations and different times exist in the same reference frame? Yes or no?

I am inclined to think no. Because of the time shift.

Let's do this properly, shall we?

X = Y²
Y = X²

Using the first equation in the second one, we get:

Y = X² = (Y²)² = Y⁴

Rearranging:

Y⁴ - Y = 0
Y(Y³ - 1) = 0

So, either Y = 0 or Y = 1.

If Y=0 then X=0.
If Y=1 then X=1.

In other words, MacM's conclusion that X=Y is false.

It's just bad maths.

I'll note that you haven't pointed out any errors in my algebra but have shown a different correlation. A correlation that actually gives the same result. It has been done to try and mask the issue which I have pointed out.

If X=1 and Y = 1 then X = Y

If X = 0 and Y = 0 then X = Y

Nice try but my math and point stand.

Wrong, as pointed out by Paul T.

Correct and I have pointed out the transcription error which transposed t1 and sqrt. Do not pretend that I have not written this formula many times correctly or that I have correctly computed TD many times using the formula.

Don't try and make a typo the issue. None of my work has used an incorrect formula. This is an unworthy dodge. Address the issue of "Reciprocity".

Those grade school kids are obviously a bit brighter than you, MacM. They can spot your errors.

Again you do yourself no favor by playing games over typos and not addressing the reciprocity issue.

Now that fun time is over suppose you tell us where there is any time dilation between clocks due to and after a test of relative velocity between clocks and the clocks are returned for comparison in the same rest frame.

We will go from there.

 

[James R]

I corrected the mistake, Pete.

X=Y is true, but X and Y are not general values in MacM's example. They are limited to zero or one.

Either X=Y=0 or X=Y=1.

Clearly, this has no relationship to time dilation formulae, which hold for any values of t and t'.

 

[MacM]

Hi Mac,
You seem to be having difficulty realising exactly what quantities are represented in SR equations.

I appreciate your intentions. However, actually I am not having any difficulty. You apparently are. More below.

Essentially, the times and distances in the reciprocal situation correspond to different spacetime intervals, meaning you can't simply equate them with each other.

Ho-Hum. More below.

I'd like to repost something from an old thread, and see if it helps you see exactly how the reciprocity in SR works. The scenario is that two tickers, A and B are moving apart from each other at some speed, and we are attempting to specify the distance between them.


The distance between reference points in ticker A's frame are contracted in ticker B's frame, and not contracted in ticker A's frame.

The distance between reference points in ticker B's frame are contracted in ticker A's frame, and not contracted in ticker B's frame.

This apparent paradox is possible because distance is a spacetime interval between simultaneous events, and when a distance is specified in two different frames, there are actually two different spacetime intervals being specified.

In the case of specifying the distance between ticker A and ticker B , there are actually four distinct spacetime intervals that could be specified (only considering the two relevant frames).

Note that spacetime intervals are not relative. All observers will agree on the magnitude (in length units) of any spacetime interval.


So, when describing the distance between ticker A and ticker B, we must be careful to specify exactly what spacetime interval we mean.

Here are some events that define end points of distinct intervals that are valid inter-ticker distances:
Spacetime events:
A - ticker A clips an object that is comoving with ticker B and 1 unit from ticker B in ticker B's frame
B - ticker B clips an object that is comoving with ticker A and 1 unit from ticker A in ticker A's frame
C - ticker B's position at the time of event A in ticker B's frame
D - ticker B's position at the time of event A in ticker A's frame
E - ticker A's position at the time of event B in ticker A's frame
F - ticker A's position at the time of event B in ticker B's frame

Here are the four intervals, and the distances to which they correspond:
Distances:
Distance in ticker B's frame between inter-ticker reference events in ticker B's frame = interval A-C = 1
Distance in ticker A's frame between inter-ticker reference events in ticker B's frame = interval A-D = 1/gamma
Distance in ticker A's frame between inter-ticker reference events in ticker A's frame = interval B-E = 1
Distance in ticker B's frame between inter-ticker reference events in ticker B's frame = interval B-F = 1/gamma

(gamma = Lorentz factor)

Order of Events
In ticker B's frame, events occur in the following order: D, BF (simultaneous), AC (simultaneous), E

In ticker A's frame, events occur in the following order: F, AD (simultaneous), BE (simultaneous), C


Here are two diagrams showing the two reference frames, with the events labelled.

The green lines are the four distances specified.
<img src="/attachment.php?attachmentid=3338&stc=1">
<HR>
<img src="/attachment.php?attachmentid=3339&stc=1">


Exercise:
From these diagrams, determine which spacetime intervals correspond to dilated and undilated times for each frame.[/QUOTE]

Symmetry is not the issue Reciprocity is. Now after a 1 hour test at relative velocity, what will two clocks that were initially synchronized read when brought back to a common rest frame and compared?

1 hour each of course. So where is your time dilation? It doesn't exist except as a perception to the moving observer.

 

[MacM]

James, are you drinking?

Good question.

X = 1 and Y = 1 then X = Y.

 

[Pete]

Precisely. That has also been the entire arguement for 1 1/2 years. However, you don't need the light signal to tell them to shut off. That is an unnecessary step. Each can simply shut down by i.e. - 3,600 seconds reading at their local time and the clocks will still agree upon being returned and compared in the same rest frame.

There's no argument on that fact.
The argument is over your fallacious extension of that scenario.

Have you worked through my previous long post?

Are you going to bother?

 

[James R]

No, he isn't going to bother, Pete. Just as he won't bother working his way through my derivation of special relativity.

 

[MacM]

I corrected the mistake, Pete.

X=Y is true, but X and Y are not general values in MacM's example. They are limited to zero or one.

Either X=Y=0 or X=Y=1.

Clearly, this has no relationship to time dilation formulae, which hold for any values of t and t'.

Unfortunately James R, this fails to address the reciprocity issue. Mine was an example of taking data from one half of a problem and seeing a curve when in fact the relationship under reciprocity is in reality linear.

Do you now deny that a test after one hour at a constant relative velocity that two clocks when returned to a common rest frame will both still read 1 hour?

I dare you. Of course it is true (at least according to Relativity) that each clock while in relative motion sees (percieves) a shift in each others clocks but that is an illusion of motion and not a real change in time as is evident by the subsequent direct comparison.

After one hour of relative velocity testing at 0.9c:

A clock reads 3,600 seconds but believes B's clock should read 1,569.2 seconds

B clock read 3,600 seconds but believes A's clock should read 1,569.2 seconds.

Guess what what they "Believe" is not the reality and is evidenced by the fact that upon return and comparison when at rest both clocks still read 3,600 seconds.

Damn it I am right and you are wrong.

 

[MacM]

There's no argument on that fact.
The argument is over your fallacious extension of that scenario.

Have you worked through my previous long post?

Are you going to bother?

Pete, just as I told Paul t a while back. It is useless to calculate any scenarios until you acknowledge the reciprocity issue. See my post above. Answer that question. Then if You can show I am in error I will go through your problem but I am not wrong and hence it is fool hardy to calculate different schemes which may show time differentials if reciprocity is not applied.

I have not said yo don't get the "Illusion" of TD due to relative velocity but that the clocks will not support that conclusion when directly compared in a common rest frame.

Do you deny that?

 

[MacM]

No, he isn't going to bother, Pete. Just as he won't bother working his way through my derivation of special relativity.

For the same reason I just gave Pete, James. Do you deny that after one hour test at constant relative velocity that each clock will read 3,600 seconds when directly compared back in a common rest frame?

If not why are we doing these one sided (no reciprocity as mandated by Relativity) examples. There has been no arguement about the "Illusion" only the "Realty". Clock displays do not support your claim.

 

[Pete]

Do you deny that after one hour test at constant relative velocity that each clock will read 3,600 seconds when directly compared back in a common rest frame?

If each clock is stopped after one hour in their own frame, then of course they'll read 3600 seconds. Duh!

Reciprocity simply means what ever happens to one clock happens to the other with regard to constant relative velocity or equal acceleration curves.

Mac, you're reading far too much into this thing you've made up and labelled "reciprocity". I think it is part of your misunderstanding of relativity.

Please, read through my last long post. I really think it is relevant, and may help to clear up your misunderstandings, or mine, as the case may be.

If you don't understand waht I mean at any point, please specify where.
If you think it violates Relativity at any point, please specify where.
If you think it violates "reciprocity" at any point, please specify where.

 

[James R]

MacM:

It is funny indeed that you can post how others don't understand Relativity but given the fact that your post fails to fully explain Relativity and the reciprocity issue means you either don't understand Relativity or you choose to simply ignore it, so as to continue to teach BS.

I never intended to "fully explain" relativity in that thread. That would require a book, at least.

The thread derives a couple of key results of special relativity. That is all.

Your conclusion in the above paragraph is incorrect, as usual.

It is a FIAT post which you do not want challenged. But it is your post and I will stay out of it. But I think it is a sad showing on your part.

The fact is, you can't pick holes in it, so you'll do what you always do and pretend it doesn't exist.

Dishonest.

Does you post consider and explain the issue of reciprocity? No.

That was not its aim.

You now seem to want to make a confusing distinction between "symmetry" and "reciprocity". Better point out what you think the difference is, then. So, MacM, what is it?

What is the difference between symmetry and reciprocity, according to MacM?

Please make sure you define clearly what you mean by both terms.

I agreed that there would be no difference in clock readings in a situation where the clocks were started and stopped symmetrically. To be explicit, let me give you that scenario:

"Two clocks, A and B, set out from the origin, with clock A travelling at +0.45c and clock B travelling at -0.45c relative to a stationary observer. All motion is at constant speed for the whole time. When each clock reads a pre-agreed time in its rest frame, that clock sends out a light pulse towards the other clock. Each clock calls the time the pulse is sent out time "zero". Each clock continues to run locally until the pulse is received from the other clock, at which time the local clock is stopped. The two clocks are then brought back together and their elapsed times compared."

In this scenario, both clocks will show the same final reading.

Precisely. That has also been the entire arguement for 1 1/2 years.

A blatant lie.

Refer to the earlier discussion in this thread.

However, you don't need the light signal to tell them to shut off. That is an unnecessary step. Each can simply shut down by i.e. - 3,600 seconds reading at their local time and the clocks will still agree upon being returned and compared in the same rest frame. Your light signal is nothing more than a confusion factor.

So, here is MacM's next brilliant suggestion for testing SR:

1. Get two clocks, A and B. Set them to zero time.
2. Send off one clock, B, on a journey. Any journey will do.
3. Each clock is set to stop when it reads a particular, pre-agreed value.
4. Bring the clocks back together and compare their readings.
5. Surprise surprise! The readings are the same, regardless of what trip is taken.
6. MacM concludes that relativity is thus disproved.

You are a fool. Your test shows nothing about relativity. Relativity is about comparing the rates of clocks when in motion. Your silly little scheme does no such thing.

You are confusing "Symmetrical" with "Reciprocity". Reciprocity simply means what ever happens to one clock happens to the other with regard to constant relative velocity or equal acceleration curves.

So we have a FIAT declaration from MacM.

MacM says:

1. Reciprocity exists.
2. Reciprocity means the clocks always display the same time.
3. Therefore reciprocity exists.
4. Therefore the clocks always display the same time.
5. Therefore relativity is wrong.

Brilliant argument, MacM.

Circular and a complete waste of time.

The very suggestion that TD occurs becomes meaningless unless you assume both clocks are slower in accordance to some UT. But there is no measureable TD between such clocks.

Wrong. My derivation assumes no universal time, or absolute state of rest. It does not even assume that car A is at rest.

And as I have explained this is not merely a matter of "Symmetry". It is a matter of "Reciprocity".

What is the difference?

It is far from being shown, much less proven, that clocks are measuring time. Clocks are processes nothing more.

You have no idea what time is. You have no coherent definition of that concept. Hence, any proclamations from you on what clocks do or do not measure is a waste of time.

You the above J = Kg*m/s^2 but went on to make a false (or unsupported conclusion) statement that a second is not the same as a Joule hence time cannot be energy.

Let me break it down into baby terms for you, MacM.

1. A Joule is defined in terms of a second.
2. 1 Joule does not equal 1 second.
3. Therefore, 1 Joule is not the same as 1 second.
4. Therefore energy is not the same as time.

QED.

Your statement is gibberish and has no meaning. Nobody said a second and a Joule were the same thing.

Nobody except you.

But change is induced by energy and it is change that produces the concept of time.

You don't know what energy is, either.

In the same sentence, you say time is both energy and an energy flow. Which is it? You're contradicting yourself. You're confused. Poor MacM.

No you are confused since you lack the ability to have vision.

You can't answer the question, can you? Poor confused MacM. Dodging the issue as usual.

I agree with everything except your confidence that you know anything about light and its "Apparent" invariance.

You've previously agreed that the speed of light is invariant.

Do you intend to flip-flop again?

So, can two events at different locations and different times exist in the same reference frame? Yes or no?

I am inclined to think no. Because of the time shift.

According to your own definition of what a reference frame is, given above, the answer to the question is "yes". Your answer, though, is "no".

You're contradicting yourself, again.

Poor confused MacM. He isn't even sure of what he thinks.

Now that fun time is over suppose you tell us where there is any time dilation between clocks due to and after a test of relative velocity between clocks and the clocks are returned for comparison in the same rest frame.

What kind of test?

Please specify in detail how your proposed test would be carried out.

Round we go again.

 

[James R]

MacM:

Unfortunately James R, this fails to address the reciprocity issue. Mine was an example of taking data from one half of a problem and seeing a curve when in fact the relationship under reciprocity is in reality linear.

In your X=Y example, there is no "linear relationship" between X and Y. In your example, both X and Y are constant. There is no variation of Y with X.

Therefore, it is irrelevant.

Do you now deny that a test after one hour at a constant relative velocity that two clocks when returned to a common rest frame will both still read 1 hour?

Without a reference frame, your question is unanswerable.

Who measures the 1 hour?

 

[Pete]

Regarding "reciprocity"...

I think that MacM believes that the postulates of Relativity imply that two objects always have the same velocity as each other.

MacM thinks that according to SR, if two objects have a relative velocity X, then X is the "real" velocity of each object. This leads to Mac wanting to work in two reference frames at once.

Mac think that it is against the postulates of SR to work in a reference frame in which the one object is stationary, and the second has velocity X.

This is what I think he means by "reciprocity".

He invented this notion to justify his fallacious [post=683805]clock synchronization procedure[/post], as addressed [post=684082]earlier in the thread[/post]. Followups: [post=684209]MacM[/post], [post=684241]Pete[/post], [post=684248]MacM[/post], [post=684253]Pete[/post], [post=684256]MacM[/post]. I got distracted on other issues at that point... perhaps I should return to it now.

Mac, point 6 in your synchronization procedure requires that the two clocks are moving at equal speeds in opposite directions.
Do you maintain that this is dictated by relativity, and is this what you mean by "reciprocity"?