Twin Paradox: Definitive Proof That It's SR?

[SeanF]

“The ‘time’ of an event is that which is given simultaneously with the event by a stationary clock located at the place of the event, this clock being synchronous, and indeed synchronous for all time determinations, with a specified stationary clock.”

The two clocks are “synchronous” in the theory, ie “running at the same rate”, because in the theory they are both “stationary” relative to one another.

While moving relatively, Clock 1 and Clock 2 are not “synchronous”.

??? Sam5, that doesn't even make any sense at all now. Just think about it for a minute. How could the "time of an event" be defined by a "stationary clock located at the place of the event," when all you know about that clock is that it is running at the same rate as "a specified stationary clock"? Doesn't it stand to reason that he must mean the clock at the event is synchronized with the specified clock? What good does it do to define the "time" of an event based on a clock when you have no idea what time that clock is showing?

And if you understood the math of SR (it just keeps coming back to that, doesn't it?) you would already know that clocks which are not moving relative to each other are always running at the same rate, but they're not always synchronized.

[Sam5]

??? Sam5, that doesn't even make any sense at all now. Just think about it for a minute. How could the "time of an event" be defined by a "stationary clock located at the place of the event," when all you know about that clock is that it is running at the same rate as "a specified stationary clock"?

Doesn't it stand to reason that he must mean the clock at the event is synchronized with the specified clock? What good does it do to define the "time" of an event based on a clock when you have no idea what time that clock is showing?

You are getting "synchronized" and "synchronous" mixed up.

He’s talking about the clock at the event (let’s say at a track race in Los Angeles), is running at the same rate as a clock in New York, if the two clocks are in the same frame and are synchronous, ie, running at the same rate. He’s just saying that all clocks in the same frame run at the same rate, no matter what their local (Los Angeles or New York) times read. Since the K frame is “stationary” for some time, the observers at A and B have time to determine, by means of light signals, that the clock at A and the clock at B are running at the same rate. This is how the K frame observers at both the clock at A and the clock at B know their clocks are running at the same rate (synchronously, ie “synchron”), although, because they are separated by some distance, the observer at A (Los Angeles) sees a different clock read-out at the clock at B (which is in New York), when he sees the light signals from clock B.

This is not to be confused with the separate light signals that will later be moving from the clock at A’ in frame K1, when those signals go from the K1 frame clock at A’ to the observers at A and B in K.

The signals between A and B in the K frame will always show the same distance and the same synchronism between the K observers’ clocks at A and B (and all the clocks at A and B) when seen by either the A or B observer, but they will show the different "local times" because they are separated in space by some distance. The synchronism (synchron) will be the same in frame K, to all frame K observers, although the local time (the local clock face readout) will be different.

These same rules apply to all the K1 observers who are looking at clocks in the K1 frame.

But they don’t apply when the K1 and K observers look into each other’s frames, by means of the “transformation” light signals that are going between the two frames.

[Tensor]

Dagnabit, I go for a one day business trip and you guys add two pages to the thread.

CM and Tensor both looked at the Ann and Bob website and said it fit with their understanding of SR. They're not getting their info from it.

Thank you "Spacetime Phyics". And thank you Taylor and Wheeler.

But, apparently, you can't even understand the K-8 websites.

#-o

[SeanF]

What good does it do to know that two clocks are running at the same rate if you don't know what the difference is in the times they are showing? When you're talking about New York and LA, it's established that there's three hours difference. What's the difference between the clocks in the SR paper? How can Einstein figure out how far behind one clock will be at the end of the experiment if he doesn't know how far behind it is at the beginning?

We imagine further that at the two ends A and B of the rod, clocks are placed which synchronize with the clocks of the stationary system, that is to say that their indications correspond at any instant to the "time of the stationary system" at the places where they happen to be.

Would you agree that in this sentence he is setting up these clocks as being synchronized?

[Sam5]

What good does it do to know that two clocks are running at the same rate if you don't know what the difference is in the times they are showing? When you're talking about New York and LA, it's established that there's three hours difference. What's the difference between the clocks in the SR paper? How can Einstein figure out how far behind one clock will be at the end of the experiment if he doesn't know how far behind it is at the beginning?

He does that in the first part of Section 1, when he first sends a light signal from A to B and reflects it back to A. If the travel times in both directions are the same, then the two clocks are “synchronous”, ie, running at the same rate. So, with that knowledge, we can find out the local times on all the different clocks inside a stationary frame by his standard formula:



So that’s no problem. What he mainly wants is all the clocks in one system running at the same rate everywhere in that system. What their local times are is not important as long as they are running at the same rate.

[Sam5]

We imagine further that at the two ends A and B of the rod, clocks are placed which synchronize with the clocks of the stationary system, that is to say that their indications correspond at any instant to the "time of the stationary system" at the places where they happen to be.

Would you agree that in this sentence he is setting up these clocks as being synchronized?

He is setting up these two clocks as being synchronized with clocks “in the stationary system”.

[SeanF]

What good does it do to know that two clocks are running at the same rate if you don't know what the difference is in the times they are showing? When you're talking about New York and LA, it's established that there's three hours difference. What's the difference between the clocks in the SR paper? How can Einstein figure out how far behind one clock will be at the end of the experiment if he doesn't know how far behind it is at the beginning?

He does that in the first part of Section 1, when he first sends a light signal from A to B and reflects it back to A. If the travel times in both directions are the same, then the two clocks are “synchronous”, ie, running at the same rate. So, with that knowledge, we can find out the local times on all the different clocks inside a stationary frame by his standard formula:

The travel times he uses are calculated based on the times on both clocks. The first trip is calculated using the time the light leaves A as displayed on Clock A and the time the light arrives at B as displayed on Clock B. The clocks need to be displaying the same time at the same time for that to work. They are synchronized.

[Sam5]

Tensor, I’m trying to find a single English word that means “after a certain moment in time, the two clocks are no longer not synchronous”. Do you think I should say, “the dis-synchronicity stops” or “the non-synchronism stops”?

[SeanF]

We imagine further that at the two ends A and B of the rod, clocks are placed which synchronize with the clocks of the stationary system, that is to say that their indications correspond at any instant to the "time of the stationary system" at the places where they happen to be.

Would you agree that in this sentence he is setting up these clocks as being synchronized?

He is setting up these two clocks as being synchronized with clocks “in the stationary system”.

Oh, heck, I left off the last sentence of that paragraph I quoted. It is:

These clocks are therefore "synchronous in the stationary system."

D**n, how about that. Looks like Einstein uses "synchronous" where you use "synchronized."

[Sam5]

The travel times he uses are calculated based on the times on both clocks. The first trip is calculated using the time the light leaves A as displayed on Clock A and the time the light arrives at B as displayed on Clock B. The clocks need to be displaying the same time at the same time for that to work. They are synchronized.

They can’t both be “displaying the same time at the same time” to both observers because they are separated in distance and the observers are relying on delayed light signals to read the time shown on the distant clock.



Ta 12:05

Tb 12:10

Ta’ 12:15

10 – 5 = 5

15 – 10 = 5

I’ve never had any problem with this. Sometimes I use stopwatch times anyway. I think he is thinking that he knows the reading on both clocks simultaneously. But that doesn’t mean observers at A and B do. They can’t and have clock b read (to the clock a observer) what the clock a observer sees on his own clock, while at the same time having the clock b observer seeing the clock a time as being the same that is on his clock b clock.

Einstein’s only interested in being sure that all the clocks in one system are “synchronous”. It doesn’t matter if they are all “synchronized” or not. You can’t “synchronize” all of them to each other at the same time, if they are all separated by great distances, because of the different travel times of the light from and to each of the different clocks. You can have all the clocks set up to read the same thing for only one observer, but they can’t all read the same thing for all observers if they are all separated by distances, but they can all be “synchronous” to all observers no matter how much they are separated.

He says, “We have not defined a common "time" for A and B, for the latter cannot be defined at all unless we establish by definition that the "time'" required by light to travel from A to B equals the "time" it requires to travel from B to A.”

So, he’s only interested in the light travel times. He wants to be sure that A and B are not moving relative to one another and are “synchronous”. He doesn’t need them to be “synchronized”, because we can’t synchronize two clocks at the same time for both observers at the same time, if the clocks are separated and if the observers are relying only on light signals to and from the clocks to tell them what the read-out is on the other clock.

[Sam5]

Oh, heck, I left off the last sentence of that paragraph I quoted. It is:

These clocks are therefore "synchronous in the stationary system."

D**n, how about that. Looks like Einstein uses "synchronous" where you use "synchronized."

In this particular case, he means that side by side “at the places where they happen to be” they do “synchronize” with the clocks in the “stationary system”.

That’s what he means by, “that is to say that their indications correspond at any instant to the “time of the stationary system”, "at the places where they happen to be.” The reason he said “at the places where they happen to be”, means they see no light travel time when they look at the nearest clock in the “stationary system”.

And then he says, “These clocks are therefore “synchronous in the stationary system.”” And because of what he describes, they are “synchronous” with the clocks in the stationary system.

It’s ok to “synchronize” clocks that are side by side. But you can’t “synchronize” both of two clocks that are separated by distances.

The standard English translation reads “synchronize”, which is ok here. Anna Beck’s translation reads “synchronous”. Einstein’s German original reads “sind synchron”, which means “are synchronous”. If they are side by side with two other clocks in the “stationary system”, ie, “at the places where they happen to be”, and if they are “synchronous” with the clocks in the “stationary system”, then they are “synchronized” with the clocks in the “stationary system”, “at the places where they happen to be”.

I know it’s a bit tricky, but please try to keep up with me.

By the way, I am trying to translate Anna Beck’s English translation of Einstein’s German original back into his German original and I'm trying to translate his German original into the correct English. Do you think I should translate “zeitbestimmungen” as “time regulations” or “time determinations”?

[SeanF]

The travel times he uses are calculated based on the times on both clocks. The first trip is calculated using the time the light leaves A as displayed on Clock A and the time the light arrives at B as displayed on Clock B. The clocks need to be displaying the same time at the same time for that to work. They are synchronized.

They can’t both be “displaying the same time at the same time” to both observers because they are separated in distance and the observers are relying on delayed light signals to read the time shown on the distant clock.



Ta 12:05

Tb 12:10

Ta’ 12:15

10 – 5 = 5

15 – 10 = 5

I’ve never had any problem with this.

Sam5, take two clocks, one set to Central Time and one set to Eastern Time, and separate them by one light-hour.

First pulse:
tA=0:00
tB=2:00
t'A=2:00

Second pulse:
tA=2:00
tB=4:00
t'A=4:00

Third pulse:
tA=4:00
tB=6:00
t'A=6:00

Both clocks are consistently ticking off two hours between pulses, and yet at no time does t'A-tB=tB-tA. They are both running at the same rate, and yet they do not satisfy Einstein's equation for defining "synchronous"! Once again, you clearly do not have the slightest idea what you are talking about.

tA is the time DISPLAYED ON CLOCK A at the moment the signal was sent from clock A. t'A is the time DISPLAYED ON CLOCK A at the moment the returned signal was received at clock A. Therefore t'A-tA is the total time ELAPSED ON CLOCK A for the round trip. Since we have defined that the time out was the same as the time back, that means that (t'A-tA)/2+tA, the midpoint of the two A times, was the time DISPLAYED ON CLOCK A at the moment the signal was bounced back.

So, (t'A-tA)/2+tA was the time DISPLAYED ON CLOCK A at bounce-back and tB was the time DISPLAYED ON CLOCK B at bounce-back. If they are equal to each other, if (t'A-tA)/2+tA=tB, then BOTH CLOCKS WERE DISPLAYING THE SAME THING AT THE SAME TIME.

(t'A-tA)/2+tA=tB
(t'A-tA)/2=tB-tA
t'A-tA=2(tB-tA)
t'A-tA=2tB-2tA
t'A+tA=2tB
t'A+tA=tB+tB
t'A=tB+tB-tA
t'A-tB=tB-tA

Therefore, if t'A-tB=tB-tA (Einstein's equation), then we know that BOTH CLOCKS WERE DISPLAYING THE SAME TIME AT THAT MOMENT. Doing these bounce-backs continuously lets us know that BOTH CLOCKS ARE CONSISTENTLY DISPLAYING THE SAME TIME AT THE SAME TIME.

That is what Einstein is defining as synchronization!

[Sam5]

Therefore, if t'A-tB=tB-tA (Einstein's equation), then we know that BOTH CLOCKS WERE DISPLAYING THE SAME TIME AT THAT MOMENT. Doing these bounce-backs continuously lets us know that BOTH CLOCKS ARE CONSISTENTLY DISPLAYING THE SAME TIME AT THE SAME TIME.

I've already said that. I showed that in my first calculation of



You are so intent on trying to trip me up, you don’t take time to study what I say.

I said:

“we can’t synchronize two clocks at the same time for both observers at the same time, if the clocks are separated and if the observers are relying only on light signals to and from the clocks to tell them what the read-out is on the other clock.

This means that two observers separated by distance can’t read the same time on each other’s clock at the same time, but that doesn’t mean the clocks can’t have the same time on their faces at the same time.

Your central and eastern clock example is ridiculous. If the clocks are synchronous but read different times the observers can calculate the time reading difference out of the equation, so you central and eastern time difference is meaningless. I’ve been telling you that all the clocks in the stationary frame are set to the same time, even though the observers, separated by distance, won’t read the same time on all their clocks at the same time. You are just making up a bunch of baloney that doesn’t mean anything. I’ve been writing a lot of informative posts that I haven't had time to post, because I’ve been busy answering your ridiculous questions and responding to your meaningless thought experiments.

You need to calm down, slow down, and relax.

[Sam5]

SeanF,

I wrote the following about 2 hours ago as part of a longer post, but I haven't had time to proofread the whole text because I've had to respond to your long series of stupid questions.

----

Part of the problem with understanding SR properly is that we all can “see” (or imagine) all the clocks in the “stationary system” all reading the same time, such as 12:00:00, no matter how long the x axis is, and no matter what the distance between the clocks is, and we all know that all the clocks read the same time at the same time, but observers in the “stationary system” can’t tell that about all the clocks in the “stationary system” all at the same time, since they have to rely on light signals to receive the information about the read-outs on each clock, and that takes time. That’s why they need to be sure their clocks are all “synchronous”.

[Sam5]

we know that BOTH CLOCKS WERE DISPLAYING THE SAME TIME AT THAT MOMENT. Doing these bounce-backs continuously lets us know that BOTH CLOCKS ARE CONSISTENTLY DISPLAYING THE SAME TIME AT THE SAME TIME.

That is what Einstein is defining as synchronization!

That’s what I said a couple of hours ago in this post:

Ta 12:05

Tb 12:10

Ta’ 12:15

10 – 5 = 5

15 – 10 = 5

When Ta is at 12:05, so is Tb.

When Tb is at 12:10, so is Ta

When Ta’ is at 12:15, so is Tb

I think you are about to go bonkers.

[Sam5]

SeanF,

While you are calming down and relaxing, you might enjoy reading this:

During the past couple of hours I’ve done a comparison of the original German, the standard English translation, and the Anna Beck translation.

Interestingly, the standard translation reads, “Thus with the help of certain imaginary physical experiments we have settled what is to be understood by synchronous stationary clocks located at different places, and have evidently obtained a definition of "simultaneous," or “synchronous,” and of “time.”"

But, the original German sentence does [/b]not[/b] use the two words “simultaneous” and “synchronous”. It uses the single word “gleichzeitig”, which Babblefish translates as, “at the same time”. This would seem to match the word “simultaneous”, rather than “synchronous”.

Although the standard translation uses the two English words “simultaneous” and “synchronous,” for his single word “gleichzeitig” (“at the same time”), Anna Beck’s translation uses the single English word “synchronous” rather than “simultaneous”. So, both the English translations are at variance with the original German phrase.

So, there seems to be a discrepancy in the two English translations, with neither one being a translation of exactly what he wrote, since he wrote “gleichzeitig”, which means “at the same time”, which I think means “simultaneous” or “simultaneously”.

For our standard English translation of another phrase, we have “for all time determinations”, whereas he uses the word “Zeitbestimmungen”, which seems to be a combination of the German word for “time” and the German word for “regulations”, so, he seems to be saying “for all time regulations”, rather than “for all time determinations”.

His phrase “ruhenden Uhr” is translated as “stationary clock”, but it actually means in German “resting clock”. I’ve also noticed that in other parts of his paper, he actually said “resting” in German, but the English translation is given as “stationary”.

The standard English translation says, “The ‘time’ of an event is that which is given simultaneously”, and in German the word he uses for “simultaneously” is “gleichzeitig”, meaning “at the same time”.

So, there are some differences between the original German and the English translations.

Anna Beck’s translation says, “and that for all time determinations is in synchrony with a specified clock at rest,” I understand “in synchrony” to mean “synchronous”, ie “running at the same rate”, which is what “synchron” means in German. His term “synchron lauft” at the very end of the sentence translates as “run synchronous” or “run synchronously”.

Down were he says, “Observers moving with the moving rod would thus find that the two clocks were not synchronous”, he uses the term “nicht synchron” for “not synchronous”, with “nicht” meaning “not”.

In this sentence, “If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous;” he uses the word “synchron”, meaning “synchronous”, not “simultaneous” and not “synchronized”. His phrase is: “synchron gehende Uhren vorhanden.”

Anna Beck translates that as, “If at the points A and B of K there are located clocks at rest which, observed in a system at rest, are synchronized.” So, she is translating “synchron” as “synchronized” rather as “synchronous”, whereas he uses the word “synchron”, which means “synchronous”, while the German word “synchronisiert” means “synchronized”. The standard English translation translates the word “synchron” properly.

Later in the sentence the standard translation says, “then on its arrival at B the two clocks no longer synchronize”. His phrase is: “Uhren nicht mehr synchron”, which translates literally as “Clocks not more synchronous,” but the German phrase “nicht mehr” directly translates as “no longer”, so we have Einstein saying the “[the] Clocks [are] no longer synchronous.” Anna Beck translates this as, “the two clocks will no longer be synchronized”.

[SeanF]

Sam5,

First of all, if you are not fluent in the German language (which you clearly are not), you are wasting your time translating his paper. You can't just use BabelFish or a German-to-English dictionary and get meaningful results.

Secondly, you said:

They can’t both be “displaying the same time at the same time” to both observers because they are separated in distance

and

What he mainly wants is all the clocks in one system running at the same rate everywhere in that system. What their local times are is not important as long as they are running at the same rate.

Then, when I showed you that what he's actually doing is showing that an observer stationary relative to the clocks, wherever they are, can agree that the clocks are all showing the same thing at the same time, you said:

That’s what I said a couple of hours ago.

I'm seriously beginning to think you are schizophrenic or something.

In the "peculiar consequence" thought experiment, when Einstein says:

If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous;

Those last two words mean "are displaying the same times at the same time." Agree or not?

[Sam5]

SeanF,

Look, I’m not going to play along with your constant nagging anymore. You are just trying to divert attention away from the paradox of the SR theory.

But you have seen the paradox in your own mind, and now you know exactly what I’ve been trying to explain to everyone for the past 4 years.

You said it right here in this post of yours:

He's wondering about considering k the "stationary system" in this sense. I think he's still a little unclear about Einstein using "stationary system" as a name for K, not a general concept.

If the clocks are synchronized in K, then they're not in k. This will be true regardless of whether K or k is considered "at rest," because it's how the experiment is set up. If the clocks were synchronized in k, then they wouldn't be in K - in that case, the 2nd Clock would lag behind the 1st when they met.

This means that if you start from the beginning of Section 1 of the Kinematical part of the SR theory and go all the way through to the end of Section 4 and say that the K frame is the “stationary” one, the clock that starts at A will “run slow” in the “peculiar consequence” thought experiment.

But if you start out at the beginning of Section 1 and go all the way through to the end of Section 4 and say that the k frame is the “stationary” one, then the clock at B will “run slow”, and THAT reveals the paradox of the 1905 SR theory, because while all the K frame observers “see” the k frame clocks “run slow” during the relative motion, all the k frame observers will “see” all the K frame clocks run slow, and that leads to the paradox of BOTH sets of frame clocks being “seen” by an equal number of observers “running slow”, because both sets of observers in both frames see themselves and their own clocks as being inside the “stationary” system, because the theory discusses only “relative motion” between the two systems.

Hundreds of physics professors and professional physicists around the world have seen this very same irreconcilable paradox. Some saw it when they first read the theory in college, while other saw it when they studied the theory on their on, later in life. They discuss it among themselves. They know it exists. I’ve discussed it with some of them at different universities around the country.

But the urban legend that there is “no paradox” in SR theory is just too great, so if any of the professors who see the paradox and know the reason for it, try to mention it in scientific papers, they get attacked by the majority of physics professors who don’t see the paradox, and they get their papers rejected.

But things are changing in physics and astronomy. Already, just within the past couple of years, more astronomers and physicists have acknowledged that that “c” is NOT a “speed limit” for moving masses on the large universal scale. This tells them that something is wrong with the Kinematical part of the SR theory. The “speed limit” seems to apply only under certain circumstances but not under others. A careful study of many of Einstein’s papers reveals that the “speed limit” seems to apply to a mass traveling through a strong gravitational field, but it does not apply to distant galaxies that seem to be moving away from the earth at velocities of faster than “c”, since they are not travelling through strong gravitational fields. They carry their own gravitational fields with them, but they don’t travel through any other strong fields.

This makes perfect sense, and it actually explains why certain parts of the Electrodynamical part of the SR theory are correct. That part is based on the 1904 Lorentz theory in which an “ether” put up a “resistance” to the motion of masses here at the earth. This “ether” in Einstein’s 1911 theory is none other than the gravitational fields of astronomical bodies.

The main part of the SR theory that is wrong is the first half, the Kinematical part, but many things he said in the Electrodynamical part are correct, because he used the c-regulators (ie, the “fields”) of the two frames through which he moved atoms and electrons, and that motion through the c-regulators (through the “fields”) causes different things to happen inside atoms. Apparently, without realizing it, his SR theory IS an “ether” theory and the “ether” is his c-regulator, which, apparently, in real life is the gravitational fields of astronomical bodies. In the SR theory, he called these separate bodies “inertial frames”, and he thought of them as “inertial frames”, not astronomical bodies, but if we apply his “inertial frame” concept to individual astronomical bodies, then we can easily transform the Electrodynamical part of the SR theory into the fundamental GR theory.

Between 1905 and 1911 he began to realize the importance of these gravitational fields in the regulation of certain electrodynamical phenomena that occur when atoms and masses move through the fields and when they are subjected to the full strengths of the fields while resting at the surfaces of astronomical bodies, and that’s how he was able to predict in 1911 that atomic clocks would slow down on the surfaces of large astronomical bodies.

So, understanding the paradox of the Kinematical part of SR theory doesn’t in any way “destroy” “Einstein relativity”, it just enhances it and makes it more understandable overall.

A lot of the things that I’ve been talking about here, such as things about the distant galaxies and their high earth-relative speeds, have been turning up on physics and astronomy websites during just the past 2-3 years. So some of what I said 2 and 3 years ago, has now become part of mainstream cosmology. This is going to happen more and more as more physics and astronomy professors quietly retire the Kinematical part of the SR paper and concentrate more on the Electrodynamical part and on General Relativity, which considers all the gravitational, electric, and magnetic fields of the universe, and how they interact.

What I think you need to be doing is explaining the paradox of the Kinematical part of GR theory to your friends, and then go on from there to study the General Relativity theory. If you continue to think that “c” is always a “speed limit” everywhere, and that light never travels at earth-relative speeds of less than or greater than “c”, you will never be able to understand the GR theory properly or some of the new discoveries being made by astronomers and physicists that reveal the Kinematical part of SR theory contains flaws and contains a serious paradox.

[SeanF]

SeanF,

Look, I’m not going to play along with your constant nagging anymore. You are just trying to divert attention away from the paradox of the SR theory.

But you have seen the paradox in your own mind, and now you know exactly what I’ve been trying to explain to everyone for the past 4 years.

You said it right here in this post of yours:

He's wondering about considering k the "stationary system" in this sense. I think he's still a little unclear about Einstein using "stationary system" as a name for K, not a general concept.

If the clocks are synchronized in K, then they're not in k. This will be true regardless of whether K or k is considered "at rest," because it's how the experiment is set up. If the clocks were synchronized in k, then they wouldn't be in K - in that case, the 2nd Clock would lag behind the 1st when they met.

This means that if you start from the beginning of Section 1 of the Kinematical part of the SR theory and go all the way through to the end of Section 4 and say that the K frame is the “stationary” one, the clock that starts at A will “run slow” in the “peculiar consequence” thought experiment.

But if you start out at the beginning of Section 1 and go all the way through to the end of Section 4 and say that the k frame is the “stationary” one, then the clock at B will “run slow”, and THAT reveals the paradox of the 1905 SR theory, because while all the K frame observers “see” the k frame clocks “run slow” during the relative motion, all the k frame observers will “see” all the K frame clocks run slow, and that leads to the paradox of BOTH sets of frame clocks being “seen” by an equal number of observers “running slow”, because both sets of observers in both frames see themselves and their own clocks as being inside the “stationary” system, because the theory discusses only “relative motion” between the two systems.

It is a coordinate transformation, not a clock change! I asked you to tell me what you believed those equations at the end of Section 3 were for, and you never answered me because you don't understand them. You don't know what they are used for, you don't know what they mean, and you don't know what you are talking about.

The theory does only discuss relative motion between two systems. Moving Clock 1 relative to the line AB is the same as moving the line AB relative to Clock 1. However, moving the line AB relative to Clock 1 is not the same as moving the line AB relative to Clock 2, and that's why there's no paradox.

[Sam5]

It is a coordinate transformation, not a clock change!

There is no such thing as a “clock change” due to "relative motion". The “clock change” in the Kinematical part of the SR theory is theoretical and fictional. Real clock changes in real life are due to mechanical, electrodynamic, or thermodynamic factors, new forces or changes in the strength of the forces placed on the macro-scale mechanical or on the micro-scale internal electrodynamical workings of the atoms inside the clocks.

The only way a clock can change rates is by “feeling” a greater or lesser mechanical or electrodynamical “force” on its vibrating or oscillating mechanism.

Put your electronic wristwatch inside the freezer of your refrigerator for several days and it will slow down its tick rate. This because its electronics feel a different strength of energy forces while cold. Take it out of the freezer and it will speed up again because it feels a stronger energy force.

[SeanF]

It is a coordinate transformation, not a clock change!

There is no such thing as a “clock change” due to "relative motion".

Seems almost like I just said that - see the "not a clock change" in my post? Doesn't mean the moving clock doesn't run slower, though.

[kilopi]

I wrote the following about 2 hours ago as part of a longer post, but I haven't had time to proofread the whole text because I've had to respond to your long series of stupid questions.

I'm seriously beginning to think you are schizophrenic or something.

Careful, careful.

[Sam5]

It is a coordinate transformation, not a clock change!

There is no such thing as a “clock change” due to "relative motion".

Seems almost like I just said that - see the "not a clock change" in my post?

Well, you are learning.

Doesn't mean the moving clock doesn't run slower, though.

Stop backsliding. The only reason a "moving clock" might "run slower" is because it is moving through fields or it is being influenced by changing field potential, effects caused by electromagnetic changes at the clock, or because of accelerative effects.

Essentially, Faraday discovered the electrodynamically-based "relative motion"/"clock slow-down and speed-up" effect in the early 19th Century when he conducted his "moving coil/moving magnet" experiments. But Faraday didn't know much about atoms and their inner workings. I'll explain the "relativity" significance of the Faraday discovery to you later. To Einstein, it was very important, and he mentioned part of it in the first paragraph of his 1905 paper.

In the meantime, you need to move on. You need to get your mind out of Annalen der Physik 17 and move on to Annalen der Physik 18.

Look at what Einstein says in issue #18, in his paper, ”Ist die Tragheit eines Korpers von seinem Energiegehalt abhangig?”:

Anna Beck translation:

”The laws governing the changes of state of physical systems do not depend on which one of two coordinate systems moving in uniform parallel translation relative to each other these changes of state are referred to.”

This sentence could use a good text editor. It should actually read like this:

”The laws governing the changes of state of physical systems do not depend on which one of two coordinate systems (moving in uniform parallel translation relative to each other) these changes of state are referred to.”

This means exactly what I’ve been telling you, that it doesn’t matter if we consider the K frame or the k frame as the “stationary” one, the laws of physics are the same in each frame.

This issue 18 paper is his famous E=mc^2 paper.

The guy got his basic ideas from the Lorentz electrodynamic theories, but he took them further and did more with them than Lorentz did. Einstein’s “Lorentz ethers” were basically the “fields” of astronomical bodies that move through space with the bodies. His “inertial frames” turned out to be astronomical bodies (or smaller atomic bodies) which act as "frames" of reference that are moving within or being influenced by local fields or by different kinds of interactions with the fields of other bodies.

[SeanF]

It is a coordinate transformation, not a clock change!

There is no such thing as a “clock change” due to "relative motion".

Seems almost like I just said that - see the "not a clock change" in my post?

Well, you are learning.

Doesn't mean the moving clock doesn't run slower, though.

Stop backsliding. The only reason a "moving clock" might "run slower" is because it is moving through fields or it is being influenced by changing field potential, effects caused by electromagnetic changes at the clock, or because of accelerative effects.

What if time itself is different?

[Sam5]

I'm seriously beginning to think you are schizophrenic or something.

Lol.

No, it’s just that this stuff is a little difficult to understand and to explain, and the words we both use aren’t always the best words to use. So sometimes, we can be agreeing about something but expressing our ideas in different ways, so we might not realize we are agreeing.

Your comments about the clocks “slowing down” and “not slowing down” at the same time might be interpreted by someone, who has no idea what you are talking about, as being somewhat “schizophrenic”. But I prefer the term “going bonkers”, which simply means you sometimes tend to get a little mixed up, because all of this “relative motion” stuff is a little complicated.

When that happens to me, I’ve got to shut down my computer and go out to the local mall and shop for a while.

When I posted:

Ta 12:05

Tb 12:10

Ta’ 12:15

10 – 5 = 5

15 – 10 = 5

I thought you would automatically realize that when Ta = 12:05, so does Tb = 12:05, but evidently you didn’t realize it when you first read that post of mine.

[SeanF]

I'm seriously beginning to think you are schizophrenic or something.

Lol.

No, it’s just that this stuff is a little difficult to understand and to explain, and the words we both use aren’t always the best words to use. So sometimes, we can be agreeing about something but expressing our ideas in different ways, so we might not realize we are agreeing.

Your comments about the clocks “slowing down” and “not slowing down” at the same time might be interpreted by someone, who has no idea what you are talking about, as being somewhat “schizophrenic”. But I prefer the term “going bonkers”, which simply means you sometimes tend to get a little mixed up, because all of this “relative motion” stuff is a little complicated.

When that happens to me, I’ve got to shut down my computer and go out to the local mall and shop for a while.

When I posted:

Ta 12:05

Tb 12:10

Ta’ 12:15

10 – 5 = 5

15 – 10 = 5

I thought you would automatically realize that when Ta = 12:05, so does Tb = 12:05, but evidently you didn’t realize it when you first read that post of mine.

No, what I didn't understand is why you claimed that it didn't mean the clocks were synchronized, merely that they were running at the same rate, and then posted this as if it backed up your claim.

[Sam5]

What if time itself is different?

Forget it! That’s just a myth that grew out of the combination of Newton’s concept of “absolute time”, and Einstein’s early 1905 concepts of kinematically-induced “time dilation” effects that he thought took place inside “relatively moving” “inertial frames”. The guy actually talked about his “wristwatch” in SR theory, as if it were an “ideal clock”! Then he went on to describe the electrodynamic effects on atoms moving around inside fields. Well, his “wristwatch” timing mechanism is not an atomic timing mechanism. What affects his wristwatch tick rates are large macro-scale thermodynamic effects and macro-scale moving-mass/acceleration effects. His overall “wristwatch” time can actually speed up at the same time the harmonic oscillation rates of the atoms inside his wristwatch are slowing down. He apparently did not think of that when he wrote "On the Electrodynamics of Moving Bodies". If he had, he would have never mentioned his “wristwatch” or compared its large-scale mechanical “time” to the kind of electrodynamical “time” ticked out, generated, or measured by means of the internal workings of moving atoms that are experiencing field potential changes.

[SeanF]

What if time itself is different?

Forget it! That’s just a myth . . .

Doesn't matter. You claimed there was a paradox in the 1905 paper, and "it's just a myth" doesn't prove there's a paradox.

[Celestial Mechanic]

Put your electronic wristwatch inside the freezer of your refrigerator for several days and it will slow down its tick rate. This because its electronics feel a different strength of energy forces while cold. Take it out of the freezer and it will speed up again because it feels a stronger energy force.

Nonsense. The watch slows down because the chemical reaction in the battery powering the watch slows down and supplies less power to the watch.

[Sam5]

Put your electronic wristwatch inside the freezer of your refrigerator for several days and it will slow down its tick rate. This because its electronics feel a different strength of energy forces while cold. Take it out of the freezer and it will speed up again because it feels a stronger energy force.

Nonsense. The watch slows down because the chemical reaction in the battery powering the watch slows down and supplies less power to the watch.

Exactly!

That’s a thermodynamical/electrodynamical time dilation effect due to the cold environment, and you don’t have to move your watch through space in order to get it to tick more slowly!

That’s why the internal closed-system “time flow” rate of a frozen embryo stops when the embryos are frozen and the embryo can be thawed and allowed to return to its normal time-flow rate many decades after it was conceived.