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Not much to tell. During discussions in s.p.r ~ 1998 or so I realized that I did not really know the experimental basis of SR. So I did what a physicist does and I studied it. In the process I amassed a foot-high stack of papers reporting experimental tests of SR [#]. In 2001 I was offered early retirement from Bell Labs, and decided I could take it only if I had another job, and felt it should be in physics. I spent a month walking the halls at Fermilab but did not find a job [@], so I stayed on at Bell Labs. I had spent 1975-82 at Fermilab, and in 2002 one of my friends from back then offered me a position as a postdoc. My family situation could not afford such a drastic pay cut, and we negotiated that as a trial I would work 50% in physics (evenings, weekends, and vacations) while working full time at Bell Labs. The trial was spectacularly successful because I did what I do best, which is develop software tools for the design and analysis of physics experiments; my Bell Labs experiences significantly augmented this skill. One of those tools attracted the notice of my current employer, and in 2006 he made me an offer I could not refuse, including equity in his company. So I retired from Bell Labs and came back to particle physics.

[#] The FAQ page I wrote on the experimental basis of SR is essentially an index to this collection. There is no doubt that SR is correct to the accuracy it can be measured today

[@] No surprise, as once you leave the field it is impossible to return. A few years later I did so anyway (I'm not a Marine so it took five years).

Tom Roberts

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That class includes Lorentz's theory of electrons and the Tangherlini transforms. These are the most notable members of that infinite class of theories, and both have considerable literature (most of which was written before it was pointed out that they are indistinguishable from SR).

That class is the set of all theories of transforms between inertial frames such that the round-trip vacuum speed of light is c in every inertial frame. Except for SR and LET, each one has an anisotropic one-way speed of light. The ONLY difference among these theories is how coordinate clocks are synchronized, which is clearly an arbitrary human convention that cannot possibly have any physical consequences.

You just display your personal ignorance. Over and over and over....

>>	Indeed for every physical theory, one can ALWAYS construct any number of other theories that are experimentally indistinguishable from it.

>	So SR is untestable? Like the god hypothesis?

Not at all. You need to LEARN HOW TO READ. And how to make correct logical inferences from what you have read. Your claim here is completely unrelated to what I wrote.

It is only in your PERSONAL FANTASY WORLD that SR is "untestable". In the real world, SR is eminently testable, has been tested hundreds of times, and has passed every test.

>>	Fortunately we live in the real world, not your fantasy world, and scientists are not as foolish as you. The key to a theory's validity is absence of refutation, not "confirmation".

>	So the reasaon you don't contemplate measuring the speed of light from red-shifted sources and from blue-shifted sources is your fear that SR will be refuted?

Again, YOU NEED TO LEARN HOW TO READ. And how to make correct logical inferences from what you have read. Your claim here is completely unrelated to what I wrote.

As I have said many times, physicists are not at all "afraid" that SR will be refuted. Indeed we would be OVERJOYED to see an experiment that does so, because it will necessarily teach us a lot about the world we live in. Of course your FANTASIES are unrelated to that.

You concentrate only on red-shifted sources, to the exclusion of the wide experimental record confirming and supporting SR. That's downright STUPID.

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>

So I retired from Bell Labs and came back to particle physics. [#] The FAQ page I wrote on the experimental basis of SR is essentially an index to this collection. There is no doubt that SR is correct to the accuracy it can be measured today FYI: Fermilab is the premier laboratory for particle physics in the U.S.A. It is less than a mile from my home. My office is located on site, and I commute by bicycle whenever weather and schedule permit. Tom Roberts

Thanks for this personal account.

Based on your comment [#] the most important question is: What is SR? Which type of processes does SR try to explain? Which type of processes requires GR? Or require most processes both?
For me, all forms of science start with observations and experiments. This is specific clear for chemical science, medical science and physics, including particle physics.
For astronomy and celestial mechanics, the starting point is also observations. The next step is to explain these observations and experiments. Explain means to unravel and describe more details. These explanations we call a theory or a law.
It is my impression, that the starting point in SR and GR are often thought experiments because real experiments are complex or the accuracy is a problem.
If GR is defined as the area where gravitation is involved and SR where this is not the case, then it is clear, that performing, SR experiments without gravitation (GR) is only possible in theory. Any experiment which involves a rocket to test the behaviour of a clock requires, at the beginning of a experiment, acceleration. This is the most logical explanation of why the behaviour of a clock changes, compared with a clock which does not move.
The question is what does that experiment has to do with SR?
In summary, the most important issue is, to describe the experiments on which the theory of SR and GR are based, as complete and detailed as possible. The same with thought experiments. As such, it is not so much accuracy which is an issue, but the details of the experiments and observations.

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The purpose of physics isn’t really as you describe: providing explanations for processes. If you expect that, you’ll be disappointed. The purpose of physics is to discern the rules by which things in nature play. It isn’t even necessary to account for why the rules are what they are, only that we understand what those rules are.

I’ll give you a simple example: the law of conservation of momentum, one of the pillars of physics. Notice that this law does not provide any explanation for how bodies interact. Instead, it just sets a rule that the interaction must respect. Nor is there any explanation provided for WHY momentum is conserved, or for why it is this quantity that appears to be conserved and not some other.

SR and GR are very much like that. They describe a rule — in this case a kind of symmetry — that interactions must obey and which measurements will respect. This provided little insight into the details of the interaction, nor does it account for WHY nature has this symmetry — only that it does. And that is useful enough.

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There is a "one level deep" answer from Noether's theorem: when the Lagrangian of a system is invariant under translations in space, then momentum is conserved. Moreover, the theorem gives an explicit formula that tells precisely what quantity is conserved.

Note that any sensible Lagrangian for an isolated system is so invariant. This is true of Newtonian mechanics, classical electrodynamics, SR, GR, and the standard model.

Noether's theorem is extremely powerful, and is part of the foundation of modern physics: when the Lagrangian is invariant under translations in time, then energy is conserved. When the Lagrangian is invariant under rotations then angular momentum is conserved. When the Lagrangian is invariant under certain gauge transformations then charge is conserved. Etc....

Tom Roberts

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The point is taken though.

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But to most people the notion that "the LAWS of physics are the same over there as they are right here" is simpler and easier to accept than is conservation of momentum. Ditto for "the laws of physics will be the same tomorrow as they are today" vs conservation of energy, and "the laws of physics are the same when you north as they are when you face east" vs conservation of angular momentum.

Tom Roberts

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>	I'll give you a simple example: the law of conservation of momentum, one of the pillars of physics. Notice that this law does not provide any explanation for how bodies interact. Instead, it just sets a rule that the interaction must respect.

Crap. It *is* a rule, and it describes how the interaction proceeds.

Maybe if you started thinking instead of parroting anthropomorphic twaddle, you'd actually understand some of the stuff you pontificate about.

----snip----

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Wrong. You said "it *sets* a rule. And you made that monumental gaffe of anthropomorphising it.

----attempted distraction snipped----

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-- Jan

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-- Jan

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Such sentence requires a certain amount of inner reflection. The problem is most documents about SR and GR are based on oversimplifications. To solve that, I wrote, we need more detail.

>

The purpose of physics isn’t really as you describe: providing explanations for processes. If you expect that, you’ll be disappointed. The purpose of physics is to discern the rules by which things in nature play. It isn’t even necessary to account for why the rules are what they are, only that we understand what those rules are.

Part of the problem is definitions of words. To explain and to understand are both related to the same subject: science. Understanding is related to humans, what I think, what we think. Explaining is related to communication, between humans. Both require detail, lots of detail

>

I’ll give you a simple example: the law of conservation of momentum, one of the pillars of physics. Notice that this law does not provide any explanation for how bodies interact. Instead, it just sets a rule that the interaction must respect. Nor is there any explanation provided for WHY momentum is conserved, or for why it is this quantity that appears to be conserved and not some other.

In physics, there are many conservations rules. General speaking something can not appear out of empty space, and something can not disappear completely. The problem is to prove that, this is the case, requires very detailed and very accurate experiments. Sometimes as detailed as the quark level and as global as the whole universe.

>

SR and GR are very much like that. They describe a rule — in this case, a kind of symmetry — that interactions must obey and which measurements will respect. This provided little insight into the details of the interaction, nor does it account for WHY nature has this symmetry — only that it does. And that is useful enough.

I doubt if SR and GR are something like a rule. Anyway nothing in physics will obey or respect any rule (or law) Science, physics is what it is and humans will try to understand how the physical processes operate and influence each other. As a by-product, they will see that many processes are (more or less) the same and call this behaviour a law or a rule.

Nicolaas Vroom

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1)"If one clock remains in an inertial frame" means never 2)We have GPS now, so anyone can check: two clocks synchronized, separated, and rejoined remain synchronized if maintained properly.

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Before you can answer this question you first need a clear definition of what means symmetry. (and what is not)

> >	Sure. That's what "one level deep" means. But to most people the notion that "the LAWS of physics are the same over there as they are right here" is simpler and easier to accept than is conservation of momentum.

The fact that the behaviour of physical processes (chemical reactions) are everywhere (basically) the same is reasonable to accept. The specific law 'conservation of momentum' is of a whole different ball game and requires a very detailed explanation. In fact, it has to be demonstrated starting with simple experiments. (Like all laws)

> >	Ditto for "the laws of physics will be the same tomorrow as they are today" vs conservation of energy,

You cannot compare both.

> >	and "the laws of physics are the same when you north as they are when you face east" vs conservation of angular momentum.

You cannot compare both.

>	True, but this explanation presumes Lagrangians describe the dynamics and it's a mystery why they do.

Again also here you should start with a simple experiment indicating what is measured, how and what the results are. The results of many processes can be predicted using Newton's Law. If my impression is not wrong these same processes can also be predicted using Lagrangians. If that is the case a lot of the mystery is solved.

Nicolaas Vroom.

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>	On Friday, 13 December 2019 08:55:51 UTC+1, JanPB wrote:

>>	On Thursday, 12 December 2019 at 9:03:51 AM UTC-8, tjrob137 wrote:

>> >	On 12/12/19 10:18 AM, Odd Bodkin wrote:

>> > >

This just pushes the WHY question to WHY nature has such a symmetry.

>	Before you can answer this question you first need a clear definition of what means symmetry. (and what is not)

>> >	Sure. That's what "one level deep" means. But to most people the notion that "the LAWS of physics are the same over there as they are right here" is simpler and easier to accept than is conservation of momentum.

>

The fact that the behaviour of physical processes (chemical reactions) are everywhere (basically) the same is reasonable to accept. The specific law 'conservation of momentum' is of a whole different ball game and requires a very detailed explanation. In fact, it has to be demonstrated starting with simple experiments. (Like all laws)

both wrong. That conservation is nothing but same amplitude normal distribution taken from the quantum level. Summing those up gives your conservation of momentum, vectors and everything. There is nothing hidden in physics anymore.

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How do you know all these facts?

Nicolaas Vroom

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Because the clocks traveled over different paths through spacetime between the two events: departure and arrival.

This difference in their elapsed proper times is fundamentally no more surprising than two sides of a triangle totaling to a longer path length than the third side. Indeed the twin paradox with (idealized) instantaneous accelerations and inertial travel _IS_ a triangle.

But it is a timelike triangle in hyperbolic spacetime, so the two legs of the traveling twin total up to LESS path length than the one leg of the inertial twin. For timelike paths, path length is elapsed proper time.

>	The explanation is completely dependent about the innerworkings of the clock.

No, it is NOT. This is GEOMETRY, and not any effect on the clocks. The result is dependent on the paths through spacetime taken by the two clocks, and NOTHING ELSE. (Given that no clock is damaged by any accelerations.)

You keep talking around this, but after many years still fail to understand it. You need to either STUDY some good books on the subject, or find another hobby more suited to your abilities. (If you study a textbook, be sure to do the problems -- that's where the learning occurs.)

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My conclusion is: we both agree that the moving clock runs behind. The difference is in wordings. I expect that in many cases the results of your predictions of how much your clock runs behind versus my clock runs behind are the same.

> >	The explanation is completely dependent on the inner workings of the clock.

>	No, it is NOT. This is GEOMETRY, and not any effect on the clocks. The result is dependent on the paths through spacetime taken by the two clocks, and NOTHING ELSE. (Given that no clock is damaged by any accelerations.)

GEOMETRY of what? The inner workings of the clock, based on light signals, is affected because the clock travels from 'A to B' compared with a clock with travelled from 'A to A'

In fact, I have two predictions: If the moving clock runs horizontally from A to B and back to A, the results will be different if the light path between the two mirrors is horizontal or vertical. In case the light path is horizontal the mirrors are vertical. In case the light path is vertical the mirrors are horizontal.

Nicolaas Vroom.

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> >	The problem with this text is that it is very complicated. In order to understand someone must explain what means: synchronize, inertial frame, proper time and inertial clock.

>	And that’s why reading books, not a newsgroup, is essential for understanding this stuff.

More specifically, you need to be reading *textbooks* designed so that they can be used for self-study, i.e. with problems and solutions so that you can actually test your understanding.

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I agree with you in principle. The problem is why to write this text in an article about FAQ's, which should be easy to understand and be more self-supporting. Specific the line which starts with: This is a "paradox" is horrible. Experiments, ansich, on there own, can never introduce a paradox. Only when you compare an experiment with something else this comparison can be paradoxical, implying something must be wrong.

Nicolaas Vroom

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-- Jan

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You don't know what "paradox" means.

In "twin paradox" it does NOT mean "contradiction" or "inconsistency". Rather, it means:

A seemingly absurd or self-contradictory statement or proposition that when investigated or explained may prove to be well founded or true.

Tom Roberts

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Consider the book Space-Time physics edition 2 page 76 Box 3-4. Here you can read: Box 3.4 Does a moving clock really "Run slow" You keep saying "The time between clock-ticks is shorter as MEASURED in the rest frame of the clock than MEASURED in a frame in which the clock is moving" I am interested in reality, not someone's measurement. Tell me what really happens!

Here we pose two related scientific questions: 1) Are differences in clock rates really verified by experiment? 2) Does something about a clock really change when it moves, resulting in the observed change in tick rate? etc We conclude that free-float motion does not affect the structure or operation of clocks (or rods). If this is what you mean by reality, then there are really no such changes due to uniform motion etc End of text.

The most important thing that a moving clock experiment teaches you is that the number of ticks of the moving clock is less than the number of ticks of a clock at rest. Secondly, you can calculate based on these numbers and the structure of the clock what the average speed (relative to the speed of light) is of the moving clock. IMO that is all By definition, the clock at rest stays at A. The moving clock moves from A to B to A (and undergoes acceleration and deacceleration)

Nicolaas Vroom.

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In the second case, using humans is more tricky. Suppose at the start of the trip that the twins where 20 years of age and that counts reflect years. This implies at the end of the experiment that the physical age of the moving twin is 70 years, while the stay at home twin is most probably dead. I have my doubts related to the moving twin

Nicolaas Vroom.

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I think you mean Wigner, "The Unreasonable Effectiveness of Mathematics in the Natural Sciences" (1960).

Tom Roberts

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Of spacetime. That is, of the world we inhabit, modeled as a manifold with metric; the metric determines the geometry (or IS the geometry). In SR, the geometry is Minkowski, not Euclidean.

>	In fact, I have two predictions:[...]

You have not accepted and understood SR. Until you learn what it says you will remain mystified. Your "predictions" are INCORRECT in the context of SR (which is the only available context here).

In SR, light in vacuum propagates isotropically with speed c relative to every inertial frame. So the orientation of a light clock at rest in such a frame simply does not matter.

Tom Roberts

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----snip----

> >	GEOMETRY of what?

>	Of spacetime. That is, of the world we inhabit,

Bullshit. That's the "world" you assume we inhabit.

>	modeled as a manifold with metric; the metric determines the geometry (or IS the geometry). In SR, the geometry is Minkowski, not Euclidean.

> >	In fact, I have two predictions:[...]

>	You have not accepted and understood SR. Until you learn what it says you will remain mystified. Your "predictions" are INCORRECT in the

You snip them and call them incorrect without argument. And you call yourself a scientist?

>	context of SR (which is the only available context here).

Wrong. The Particle Theory of Matter is also available.

>	In SR, light in vacuum propagates isotropically with speed c relative to every inertial frame.

That statement is absolutely and utterly insane.

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Some of the experimental tests of SR refute Galilean relativity by hundreds of sigma, and confirm SR to a few parts per billion. That is not "equivocal".

In physics, experimental results consistent with the predictions of a theory does indeed mean the theory is confirmed by the experiment. That's what these words mean, in this context. Here "confirmation" is not a binary property, but is related to the accuracy/resolution of the experiment.

>>	http://math.ucr.edu/home/baez/physics/Relativity/SR/ experiments.html#Twin_paradox The best example is Bailey et al.

>	Two clocks flown around the world, one east, one west, both compared with a reference clock on the ground.

You REALLY need to learn how to read. That is not Bailey et al -- their experiment could not be more different from your description here -- they measured muons in a storage ring.

>	[... ignorant and irrelevant yammering.]

Your argument from disbelief is USELESS. Don't you have any understanding of basic logic?

And you get your "facts" wrong -- for the airplane tests, NONE of the clocks are at rest "in the same FoR", including the reference clock. Hint: the earth rotates.

To understand how clocks in airplanes can display relativistic effects is no different from any experimental analysis -- it requires one to analyze the theoretical prediction and compare to the experimental result and the accuracy of the clocks.

Today, commercial atomic clocks are accurate enough to show relativistic effects for clocks driven at highway speeds for a few hours. And NIST's ion clocks are accurate enough to display relativistic effects for relative speeds of a few meters/sec for a few seconds.

Tom Roberts

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Clearly you have no rational argument, or you would use it instead of just insulting people.

Tom Roberts

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For atomic clocks in airplanes, by the manufacturer. For muon decays, by Bailey et al (who demonstrated no effect > ~100 parts per million, from the enormous acceleration of ~ 10^18 g).

>>	To understand how clocks in airplanes can display relativistic effects is no different from any experimental analysis -- it requires one to analyze the theoretical prediction and compare to the experimental result and the accuracy of the clocks.

>	And the stability of the clocks under acceleration. When has that ever been done?

By the manufacturer. IIRC the commercial atomic clocks Bailey et al used are specified to maintain their accuracy for accelerations up to 2 g. Commercial airliners, of course, don't come near that value.

Tom Roberts

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The problem with this is that "non-inertial frame" is inherently self-contradictory -- no non-inertial coordinates can possibly be a frame, because "frame" implies mutually orthogonal coordinates, and that happens ONLY for (locally) inertial frames.

Tom Roberts

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There's no need to attempt to "ensure that the currents in the clock circuitry never vary", because that is irrelevant -- what matters is their tick rate, not internal currents. Presumably they MEASURE the variation in tick rate for various accelerations, and write the specifications accordingly.

Note that the atoms used to regulate the clock's ticking are far more robust than merely 2g. But the mechanical components and the method of maintaining the atomic beam within them are limited.

>>	For muon decays, by Bailey et al (who demonstrated no effect > ~100 parts per million, from the enormous acceleration of ~ 10^18 g).

>	Muons. Postulated particles. Give us a break.

OBSERVED particles.

Yes, you need a break -- find some hobby more suited to your abilities, as you are clearly too ignorant to understand basic physics, and too stupid to do the obvious thing to fix your deficiencies: STUDY.

"Ignorance can be cured, but stupid is forever." [attribution lost]

>	And how do the testers manage to avoid circularity as they attempt to ascertain that conformance in an accelerated environment

There is no "circularity" in MEASURING the variance in clock tick rates under various conditions.

Tom Roberts

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IMO the most important issue to understand science is not mathematics but first physics based on observations and experiments.

Have a look at page 68 of the book Space-Time physics second edition. This page Figure 3.4 shows a rocket which travels horizontal. Inside the rocket there is a clock. There is also a clock at rest. The flash, inside the clock, is reflected by means of mirrors. From the view point of the clock or rocket the flash moves perpendicular to the direction of motion. The mirrors move horizontal. Only one cycle is shown. Compared with the clock at rest the moving clock shows less counts, because the light path between a complete cycle is much longer. (2 versus 1)

Now we are going to make this experiment slightly more complicated. In stead of one clock at rest we use three clocks at rest. One in which the lightsignal moves in the x direction, one in the y direction and one in the z direction. Each depicted in figure 1-3 at page 12. All these clocks, when started simultaneous, will show the same number of counts.

We do the same with the clock inside the rocket. We use 3 clocks, all with a different lightsignal direction. In this case the moving clocks do not show the same result. Suppose the rocket moves in the x direction. In that case the clock with the lightsignal in the x direction, shows a different number of counts compared with the clock in the y and z direction. The physical difference is that the light signal of the clock in the x direction only moves in the x direction. The light signal of a clock in the y direction (figure 3.4) both moves in the y and x direction. The light signal of a clock in the z direction (figure 3.4) both moves in the z and x direction.

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Fair enough. But math is the language of physics, and cannot be avoided. When you discuss gedankens, all you have is the math of SR.

>	Have a look at page 68 of the book Space-Time physics second edition.[...]

I don't have the book, and that's 'way too complicated to bother with. Your misconceptions are very basic and don't need complicated scenarios.

>>	In SR, light in vacuum propagates isotropically with speed c relative to every inertial frame. So the orientation of a light clock at rest in such a frame simply does not matter.

>	The issue is orientation of the movement of the clock versus the initial orientation the lightsignal of clock, inside the moving clock.

But when light propagates ISOTROPICALLY with speed c, the orientation of a light clock simply does not matter. This is a very basic aspect of SR that you fail to grasp.

Tom Roberts

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Using humans is hopeless, as one cannot measure their aging with anywhere near the accuracy required. Indeed one must use highly accurate clocks for all practical realizations of the twin paradox.

>	In the first case, the result of the experiment shows that the moving clock runs slower than the clock at rest.

No, it does NOT. It shows that the traveling twin's clock ACCUMULATED fewer ticks than the inertial twin's clock ACCUMULATED. This experiment says nothing at all about how the clocks "run" (i.e. their tick rates), for a very simple reason: THE CLOCKS' TICK RATES ARE NEITHER MEASURED NOR COMPARED; only their total accumulated ticks are compared.

The difference between "running slower" and "accumulating fewer ticks" is important. As I keep saying: physics is both complicated and subtle; precision in thought and word is essential. Your thinking here is too fuzzy, which is why you are so confused.

One key principle of SR is that the laws of physics are the same in all inertial frames (the Principle of Relativity). This directly implies that the tick rate of the traveling twin's clock is THE SAME as the tick rate of the inertial twin's clock, because the physical laws that govern the clocks' ticking are the same.

Of course you have to understand how the English language works: when one discusses "that clock's tick rate", it mentions ONLY the clock, so NOTHING else can be involved -- it does NOT include the case when one observes the clock from some frame relative to which the clock is moving.

For many people, much of the difficulty and fuzzy thinking is due to an inability to get away from implicitly imagining that there is some sort of "absolute or universal frame" relative to which the ticking of clocks can always be discussed. That fictitious frame is often confused with the inertial twin's frame. The problem with this is that THIS IS RELATIVITY, and there is no such "absolute" or "universal" frame.

One CAN discuss the ticking of the traveling twin's clock as measured/observed in the inertial twin's frame. But that is NOT the clock's tick rate, it is quite different.

>	I don't see any form of a paradox in that experiment.

The usual expectation is that traveling does not affect the aging of the clock. So the "paradox" is in the APPARENT inconsistency of the traveling twin returning with less elapsed proper time than the inertial twin; with analysis, however, this APPARENT inconsistency is resolved and found to be valid. Just as in the quoted definition above.

Tom Roberts

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What you need is the mathematics that describes the inner behaviour of a clock.

> >	Have a look at page 68 of the book Space-Time physics second edition.[...]

>	I don't have the book, and that's 'way too complicated to bother with. Your misconceptions are very basic and don't need complicated scenarios.

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The tick rate of the moving "clock" (in terms of the "stationary" system x,t of inertial coordinates) is the same for any orientation of the clock. For example, when a clock of length L is oriented perpendicular to v (moving in the x direction), we obviously have for each leg of the trip dx^2/dt^2 = v^2 and (L^2 + dx^2)/dt^2 = c^2, and hence the time for a light pulse to travel from one end of the rod to the other and back is (2L/c)g where g=1/sqrt(1 - v^2/c^2).
On the other hand, if the clock is oriented parallel to v, the time from back to front of the rod satisfies dt_f = [L/g + v*dt_f]/c and the time from front to back satisfies dt_r = [L/g - v*dt_r]/c, so we have (again) the round trip time dt_f + dt_r = (2L/c)g.

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Not really. NONE of the phenomena discussed in this thread depend on the inner workings of a clock. They are all just geometrical relationships that are the same for all clocks.

>	What you also can have is that the clock moves in horizontal direction and the light signal inside the clock also moves in horizontal direction. The mathematics involved is different.

The math of the calculation does indeed depend on the orientation of the light clock, when looked at from a frame other than its rest frame, but the RESULT is the same and independent of orientation. Moreover, when looked at in its rest frame it is QUITE CLEAR that its orientation does not matter.

As I said above NONE of these phenomena depend on the inner workings of a clock -- for a light clock that includes its orientation (relative to anything).

>	What is complicated is the concept of worldline.

Not really -- you're making a mountain out of a molehill. Any persistent object, when plotted in spacetime, MUST have a worldline. That's what "persistent" and "worldline" mean.

Tom Roberts

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I fully agree with you

> >	In the first case, the result of the experiment shows that the moving clock runs slower than the clock at rest.

>

No, it does NOT. It shows that the traveling twin's clock ACCUMULATED fewer ticks than the inertial twin's clock ACCUMULATED. This experiment says nothing at all about how the clocks "run" (i.e. their tick rates), for a very simple reason: THE CLOCKS' TICK RATES ARE NEITHER MEASURED NOR COMPARED; only their total accumulated ticks are compared.

I fully agree with you. In fact, the only thing that is measured in a twin type experiment is the accumulated number of ticks of the stay at home twin versus the moving twin when they return. These numbers are different. I don't see any problem, because the start event and the end event of both these measurements are the same, to define these numbers as rates. In fact, you only need one reference frame. This is depicted in Fig 13A on page 23 of the book SpaceTime Physics I edition in the case, the light signal moves perpendicular. Fig 13A has to be modified in case of the light signal moves horizontally.

>

One key principle of SR is that the laws of physics are the same in all inertial frames (the Principle of Relativity). This directly implies that the tick rate of the travelling twin's clock is THE SAME as the tick rate of the inertial twin's clock, because the physical laws that govern the clocks' ticking are the same.

The problem with this is that IMO a twin type experiment does not involve inertial frames but only by approximation. To perform a twin type experiment accelerations and decelerations are always involved because they meet at the same starting point. A second problem is that in SR "the physical laws that govern the clock's ticking are the same" does not answer the question what these laws are. (As a side comment: the behaviour of any process is not governed by any law or mathematical equation)

>	Of course, you have to understand how the English language works: when one discusses "that clock's tick rate", it mentions ONLY the clock, so NOTHING else can be involved -- it does NOT include the case when one observes the clock from some frame relative to which the clock is moving.

IMO there is nothing wrong in claiming that the tick rate of two clocks is different, assuming the measuring period is the same. A different type of experiment is to start with a set of clocks, all synchronized in a frame at rest. When that is the case, each time when the moving clock meets a clock at rest it can be observed that the accumulated ticking count of the moving clock runs behind the clock at rest.

> >	I don't see any form of a paradox in that experiment.

>	The usual expectation is that traveling does not affect the aging of the clock.

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The inner working (the mechanics) how the clock is built or operated is very

important. See also Fig 13A on page 22.

> >	What you also can have is that the clock moves in horizontal direction and the light signal inside the clock also moves in horizontal direction. The mathematics involved is different.

>	The math of the calculation does indeed depend on the orientation of the light clock, when looked at from a frame other than its rest frame, but the RESULT is the same and independent of orientation.

The first part: Yes. The second part: No. See below.

>	Moreover, when looked at in its rest frame it is QUITE CLEAR that its orientation does not matter.

When at rest the accumulated counts will be the same. When not at rest the accumulated counts of both types of moving clocks will be different. Types in the sense of different orientations.

>	As I said above NONE of these phenomena depend on the inner workings of a clock -- for a light clock that includes its orientation (relative to anything).

The important point is that one type is in accordance with the Lorentz

transformation and the other type not.

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Okay

>	On the other hand, if the clock is oriented parallel to v, the time from back to front of the rod satisfies dt_f = [L/g + v*dt_f]/c and the time from front to back satisfies dt_r = [L/g - v*dt_r]/c, so we have (again) the round trip time dt_f + dt_r = (2L/c)g.

When the clock moves towards the right the time t1 of the light to travel from the back to the front follows the equation: c*t1 = L + V*t1 or t1= L/(c-v). In this case, both signals travel towards the right. The time t2 from the front to the back is C*t1 + v*t2 = L or t2 = L(c+v) In this case, the signals travel towards each other. The clock towards the right and the reflected lightsignal towards the left. t1+t2 = L/(c-v) + L/(c+v) = 2Lc/(c^2-v^2) = (2Lc/c^2)/(1-v^2/c^2) = (2L/c)g^2

Nicolaas Vroom

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You neglected the length contraction of the clock in the direction parallel to v. Remember, in terms of the "stationary" coordinates, the spatial length of the clock is L/g (not L). Replace L with L/g in your analysis, and you get the correct tick rate for the parallel orientation case: (2L/c)g.

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This is true only if a tick of a clock is a universal interval of time......it is NOT. The LT says that:
Delta(t’)=gamma(t) This equation says that an interval of Delta(t’) on the primed clock is worth gamma(t) on the unprimed clock. IOW, one-second on the primed clock is worth gamma-seconds on the unprimed clock.

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Consider the LHC, which is a large storage ring. In this ring there are 360 trains (on a track) all with the same length. The total length of the ring L = 2*pi*R. The length of each train is L/360. At the same time all the trains starts to move, faster, faster and faster.

Question: will there be a distance between the trains? If that is the case there is physical "length contraction" of each train and in principle it is possible to "inject" more trains in the LHC.

Nicolaas Vroom

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If the train cars are disconnected, yes, they will each undergo length contraction, and this will open gaps between the cars. If the cars are solidly connected, they will stretch until something breaks. This is just Bell's Spaceship scenario. Likewise the equipotential fields surrounding the charged particles in the LHC contract into ellipsoids in the direction of motion. This "flattening" effect has been verified experimentally.

>	If that is the case there is physical "length contraction" of each train and in principle it is possible to "inject" more trains in the LHC.

Right.

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Ok. But your analogy to "trains" on a circular track is overly simplistic and ignores essential differences between charged particle beams and train cars.

>	In this ring there are 360 trains (on a track) all with the same length.

Actually there are 2808 bunches around its circumference (each beam). They are not equally spaced, because there is a "gap" containing no bunches, which is created when they are injected from the SPS; IIRC the gap is about 10% of the circumference.

Each bunch is NOT of fixed length in its rest frame, because these are charged particles (protons), not solid structures like trains. They must be focused longitudinally (as well as transversely). That difference is important.

>	The total length of the ring L = 2*pi*R. The length of each train is L/360. At the same time all the trains starts to move, faster, faster and faster.

The protons are injected at 450 GeV, so they are never at rest in the LHC -- they enter the ring at 0.9999976 c and are then accelerated to 0.99999999999999 c, which takes ~ 25 minutes.

>	Question: will there be a distance between the trains?

There is always an empty interval between the proton bunches. At any given point around the ring the interval between them is 25 ns (except during the gap), independent of energy; their length [#] at 450 GeV is 1.7 ns, and at 6.5 TeV is 0.25 ns. That ratio in lengths is ~ 7, while the ratio of gammas is ~ 14 -- their length is determined mostly by the longitudinal focusing and dynamics, not relativistic "length contraction" (which is, of course, also present).

[#] Bunch length and interval are measured in ns, as that is how the instrumentation works. Multiply by c to get values in meters.

>	If that is the case there is physical "length contraction" of each train and in principle it is possible to "inject" more trains in the LHC.

Conservation laws of accelerator physics, plus the technology of injecting into a storage ring, mitigate against this. They also prevent closing or filling the gap that was created during injection.

If instead of the LHC one considers a gedanken with train cars moving around a circular track, then coeal5136 got it right -- as they are accelerated either gaps form between them or they stretch to the breaking point.

Tom Roberts

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>	If instead of the LHC one considers a gedanken with train cars moving around a circular track, then coeal5136 got it right -- as they are accelerated either gaps form between them or they stretch to the breaking point.

Completely nonsense. This is what I call for being nonsense. It reveals you don't undrestand your own theory. That's a mater of observation, having nothing to do with stresses etc.

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>	Tom Roberts wrote:

>>	If instead of the LHC one considers a gedanken with train cars moving around a circular track, then coeal5136 got it right -- as they are accelerated either gaps form between them or they stretch to the breaking point.

>	Completely nonsense. This is what I call for being nonsense. It reveals you don't undrestand your own theory. That's a mater of observation, having nothing to do with stresses etc.

To be more specific, as is been said around here once, as soon you deteriorate time rates, say close to speed of light etc, things gets immaterialized wrt to each other. It's actually also written in the holly ancient scriptures. Then also the above _observation_ is in itself hypothetical.

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Right.

>	The problem with this is that IMO a twin type experiment does not involve inertial frames but only by approximation.

Yes. But that is no actual problem, because it is easy to calculate the elapsed proper time of a clock moving with speed v(t) relative to an inertial frame with coordinates {x,y,z,t}:

T_elapsed = \integral sqrt(1 - v(t)^2) dt with the integral taken over the path in question. (This explicitly shows that the only quantity that matters is the clock's speed relative to the frame used for the calculation, as a function of the frame's time coordinate.)

So you can EASILY compute the elapsed proper time of the traveling twin for ANY v(t) you like -- use whatever accelerations and decelerations you think are appropriate. There's no need for any approximation here. Using the inertial frame of the stay-at-home twin, it is quite clear that T_elapsed for the traveling twin is necessarily smaller than T_elapsed for the stay-at-home twin (who has v(t)=0).

>	A second problem is that in SR "the physical laws that govern the clock's ticking are the same" does not answer the question what these laws are.

Right. SR makes no pretense to that. SR is just a theory about the geometry of the world we inhabit, not about how objects in that world behave. But for the twin paradox It DOES NOT MATTER, as that is a purely geometrical situation.

>	(As a side comment: the behaviour of any process is not governed by any law or mathematical equation)

The behavior of an object is MODELED by the laws of physics we have discovered. Insofar as they are accurate, you make a distinction without a difference.

>	IMO there is nothing wrong in claiming that the tick rate of two clocks is different, assuming the measuring period is the same.

Your opinion is WRONG. This has to do with the way the English language works: when you discuss "the tick rate of that clock", you mention ONLY the clock, so nothing else can be involved. In particular, you cannot use that phrase to refer to a measurement of the clock's tick rate from a frame relative to which it is moving, because that CLEARLY includes the relationship between clock and frame.

>	A different type of experiment is to start with a set of clocks, all synchronized in a frame at rest. When that is the case, each time when the moving clock meets a clock at rest it can be observed that the accumulated ticking count of the moving clock runs behind the clock at rest.

Sure. But this does NOT compare tick rates, it successively compares the accumulated ticks of the moving clock to the accumulated ticks of successive clocks in the frame.

Tom Roberts

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I doubt if it is that simple because in principle both sides can undergo accelerations. That means both the stay at home twin and the moving twin. To be more specific: because what is the speed v(t) In order to calculate that speed, you need a coordinate system, in fact a grid in 3D with, at as many positions as possible, clocks at fixed positions, at rest (synchronized within that stationary frame) This allows you to calculate the speed v as a function of a clock count at rest. And using the above formula, to calculate the number of counts of the moving clock at each instant when the position of the moving clock coincides with the position of a clock at rest. And, this is important, the calculated and the observed count should be the same. (what that means time-wise is in some sense unimportant)

> >	A second problem is that in SR "the physical laws that govern the clock's ticking are the same" does not answer the question of what these laws are.

>	Right. SR makes no pretence to that. SR is just a theory about the geometry of the world we inhabit, not about how objects in that world behave. But for the twin paradox, It DOES NOT MATTER, as that is a purely geometrical situation.

> >	(As a side comment: the behaviour of any process is not governed by any law or mathematical equation)

>	The behaviour of an object is MODELED by the laws of physics we have discovered. Insofar as they are accurate, you make a distinction without a difference.

Or is it the other way around? The above law is a mathematical description of the behaviour of a moving clock between two parallel mirrors. IMO this law and all other laws (or mathematical equations) are modelled based on the results of experiments. As such also different mechanical clocks require different laws.

> >	IMO there is nothing wrong in claiming that the tick rate of two clocks is different, assuming the measuring period is the same.

>	Your opinion is WRONG. This has to do with the way the English language works: when you discuss "the tick rate of that clock", you mention ONLY the clock, so nothing else can be involved.

Ok

>	In particular, you cannot use that phrase to refer to a measurement of the clock's tick rate from a frame relative to which it is moving, because that CLEARLY includes the relationship between clock and frame.

I fully agree. That's why above I only use one reference frame.

A whole different issue is if there is length contraction involved. To test that, you need a reference frame with at least three clocks (a distance l apart) and a moving rod (with the same length l at rest) and two clocks. The clocks at rest are indicated as A1, A2 and A3. The moving clocks are indicated as B1 and B2. Initially at rest B1 coincides with A1 and B2 with A2. Movent towards the right. The question is: when B1 coincides with A2 and B2 with A3 are the clock readings at both events identical? For example: when there is no length contraction the clock readings could be 100 (clock at rest) and 90 (moving clock) and the same for both events.

When there is length contraction the clock reading of the event B1, A2 could be 100 (clock at rest) and 90 (moving clock) and for the events B2, A3 something like 105 (clock at rest) and 94 (moving clock).

But also other possibilities are possible including length expansion.

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Let me explain what I understand in small steps.

1. Suppose I travel in a train at a constant speed of 100 km/hour from A to D. The total distance is 300 km. There is a point B at 100 km and a point C at 200 km. Suppose I start my trip at 12 o'clock. Because I travel with a constant speed of 100 km/hour I know that I will be at point B at 1 o'clock at point C at 2 o'clock and at point D (target) at 3 o'clock. When I arrive I check the local time. It's at 3 o'clock. Mission fulfilled.

2. This raises one important question: How do I know that I travel with an average speed of 100 km/hour? To do that I can use the following equation: v = (x(D)-x(A)) / (t(D)-t(A)). Assuming x(A) and t(A) both equal to zero you get: v = 300 km/3 hour = 100km/hour. But that is after I have my experiment.

3. Now suppose I want to know during my trip that I travel with a speed of 100 km/hour. In order to know you need a speed indicator with a needle. When the train is at rest you mark the position of the needle. This is your zero point. Now you start the train with the supposed speed of 100 km/hour and you mark the position of the needle when the average speed is reached. You continue your travel and you check that you arrive after 3 hours and that the position of the needle is more or less stable. In fact what you have done is calibrated your speed indicator during an actual experiment. Using this speed indicator every one can travel from A to D in 3 hours.

4. What is the purpose of equation T_elapsed = \integral sqrt(1 - v(t)^2) dt This equation is supposed to show that if I travel from A to D in 3 hours that, if I also have a clock on this train, then the elapsed time on this moving clock will be less. That seems easy. The problem with this equation is that the speed v(t) is a number between 0 and 1. The real speed is this number times the speed of light. This raises an important question: What is my speed (relative to the speed of light) when I'm at rest at point A. This is definitely not zero because the Earth surface is rotating, The Earth rotates. The sun rotates. The same for point D. What if point D is on the Moon and I travel on a space ship?

5. Suppose I'm at rest at point A in a space ship. (See above) What does that mean? When I use the above formula it means that when I give my engine a fixed boost towards the right, my onboard clock will run slower compared with clocks at rest. It also tells me that when I give my engine the same fixed boost towards the left my onboard clock will run slower compared with clocks at rest. In fact this behaviour is symmetrical. The more boosts I give, the faster my space ship moves, the slower my onboard clock runs.

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It calculates the elapsed proper time of a clock that moves with speed v(t) RELATIVE TO AN INERTIAL FRAME with time coordinate t. This is using units with c=1; for other units, divide v(t) by c.

>	This equation is supposed to show that if I travel from A to D in 3 hours that, if I also have a clock on this train, then the elapsed time on this moving clock will be less. That seems easy.

Yes. Or at least straightforward.

>	The problem with this equation is that the speed v(t) is a number between 0 and 1.

That is not a "problem", that is part and parcel of the equation, because it uses units with c=1.

>	The real speed is this number times the speed of light.

Hmmm. That equation uses units with c=1, so multiplying by c does not change anything. For other units, divide v(t) by c.

>	This raises an important question: What is my speed (relative to the speed of light) when I'm at rest at point A. This is definitely not zero because the Earth surface is rotating, The Earth rotates. The sun rotates. The same for point D. What if point D is on the Moon and I travel on a space ship?

You need to remember the context of that equation: IT APPLIES ONLY IN AN INERTIAL FRAME. All your mentions of rotations indicate you do not understand the context of the equation.

In practice, if you are discussing a gedanken with speeds 0.1 c or greater, the orbits of earth and sun are irrelevant and you can consider either of them to be at rest in an inertial frame -- the error from doing so is negligible. If you are considering travel within the solar system, at much smaller speeds, then you must use a locally-inertial frame, such as the ICRS.

>	5. Suppose [...] The more boosts I give, the faster my space ship moves, the slower my onboard clock runs.

No. Your clock ALWAYS runs at its usual rate, regardless of how you choose to accelerate it. But when measured by the inertial frame it runs slower. These are QUITE DIFFERENT, and you keep confusing them.

As I keep saying: relativity is subtle and complicated, and precision in thought and word is essential. You keep using insufficiently precise wordings, and confuse yourself.

But yes, the above equation is independent of acceleration (including direction), and depends only on SPEED RELATIVE TO THE INERTIAL FRAME BEING USED. Of course acceleration is needed to change that speed, but it is only the speed that appears in the equation, so only the speed matters.

>	6. Now suppose I'm not at rest at point A in a space ship.

[...]

You are still confused, because you forgot the requirement to apply that equation ONLY IN AN INERTIAL FRAME.

You keep forgetting that the CONTEXT of the equation is as important as the equation itself. The context specifies when it is valid, what units are used, and what the symbols mean.

(You are not alone in ignoring context around here.)

Tom Roberts

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Fortunately we have GPS now, so we can be absolutely sure that your moronic mumble has nothing in common with real clocks, real observers or real anything.

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> >	This raises an important question: What is my speed (relative to the speed of light) when I'm at rest at point A. This is definitely not zero because the Earth surface is rotating, The Earth rotates. The sun rotates. The same for point D. What if point D is on the Moon and I travel on a space ship?

>	You need to remember the context of that equation: IT APPLIES ONLY IN AN INERTIAL FRAME.

That immediate put a huge limitation. But the problem is the same. It is easy to claim that you are at rest in that frame, but how do you that if your speed is 0.000001 * c?

>	All your mentions of rotations indicate you do not understand the context of the equation.

See next.

>	In practice, if you are discussing a Gedanken with speeds 0.1 c or greater, the orbits of earth and sun are irrelevant and you can consider either of them to be at rest in an inertial frame -- the error from doing so is negligible.

Of course, you can see as a Gedanken assume that my speed is 0.1 c (and constant) then it is easy to use this formula. But then this problem becomes more or less mathematics. The issue is how can you show in a real experiment that your speed is 0.1 c or 0.000001 c. Only in such a case then you can check the result based on an actual experiment. When you have done that, you should also test speeds of -0.1 and -0.000001 c and test if the number of counts predicted is the same as observed. In fact, you should be able to demonstrate that the behaviour is symmetric.

>	If you are considering travel within the solar system, at much smaller speeds, then you must use a locally-inertial frame, such as the ICRS.

> >	5. Suppose [...] The more boosts I give, the faster my space ship moves, the slower my onboard clock runs.

>	No. Your clock ALWAYS runs at its usual rate, regardless of how you choose to accelerate it. But when measured by the inertial frame it runs slower. These are QUITE DIFFERENT, and you keep confusing them.

Of course, that is what I mean. In fact, as I mentioned above: You should also do that in the opposite direction and demonstrate that the behaviour symmetric is.

>	As I keep saying: relativity is subtle and complicated, and precision in thought and word is essential. You keep using insufficiently precise wordings, and confuse yourself.

I agree with you.

>	But yes, the above equation is independent of acceleration (including direction), and depends only on SPEED RELATIVE TO THE INERTIAL FRAME BEING USED. Of course, acceleration is needed to change that speed, but it is only the speed that appears in the equation, so only the speed matters.

> >	6. Now suppose I'm not at rest at point A in a space ship.

>	[...] You are still confused, because you forgot the requirement to apply that equation ONLY IN AN INERTIAL FRAME. You keep forgetting that the CONTEXT of the equation is as important as the equation itself. The context specifies when it is valid, what units are used, and what the symbols mean.

I agree that the equation applies to an inertial system. At the same time, it is very important to discuss more complex situations And not only as thought experiments.

Again, Thanks

Nicolaas Vroom

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It is not the equations that impose such limitations, they are imposed by the world we inhabit. It is quite clear that inertial frames make modeling the world simpler than using non-inertial coordinates. That's just the way it is. Live with it -- after all, you have no choice.

In practice, if you want to use some specific non-inertial coordinates, what you will do is either:
1. calculate in an inertial frame and transform to the coordinates.
2. determine the transform from some inertial frame to the coordinates, and transform the equations.
This _IS_ how it is done, because in principle and in practice we only know the equations in inertial frames.

>	But the problem is the same. It is easy to claim that you are at rest in that frame, but how do you that if your speed is 0.000001 * c?

Whenever you say "speed" you must also mention the coordinates relative to which it applies. When you don't do that, as in your last sentence here, you just make nonsense.

If you are at rest in a given inertial frame, then your speed RELATIVE TO THAT FRAME is zero. DUH!

>	The issue is how can you show in a real experiment that your speed is 0.1 c or 0.000001 c. Only in such a case then you can check the result based on an actual experiment.

Right. Well, outside of particle physics, where speeds > 0.9 c are common. Careful measurements and careful calculations are required. But that is part and parcel of doing experimental physics.

Tom Roberts