Looking to understand time dilation

[matheinste]

Differential ageing is something that occurs when an observer observes a moving clock.
The observer at rest observes the traveling clock to slow.
The traveling observer observes the resting clock to slow.
The only way this can be true is if neither clock actually slows and the slowing is merely an effect of measuring a moving object.

Grimble

What you describe above is not diffefential ageing but the effects of time dilation.

Matheinste.

 

[Dale]

I am so sorry if I seem to be difficult but why is the traveling clock no longer synchronised?
Is it the acceleration? How does this affect anything?
And yet for a clock at rest in another Inertial FoR that is not accelerating and has not accelerated since their clocks were synchronised the LT still works using the current velocity?

It is not the acceleration per se, but rather the fact that a clock's reading is continuous. If what you suggested were true then any slight jostling of a clock at a sufficiently large x position would cause it to jump by years. Here is an example.

Consider two systems A (unprimed), and B (primed), of inertial clocks and rods in the standard configuration with the B clocks and rods moving at 0.6 c in the +x direction relative to the A clocks. The A clocks and B clocks are related by the Lorentz transform. Now, consider a clock C which is initially synchronized and at rest next to the A clock at x=1 which impulsively accelerates at t=0 to be at rest in B. Just before the impulse t is approaching 0, and just after the impulse (by the Lorentz transform) t' is approaching -0.75. So, either the time on the clock must be discontinuous, jumping suddenly from 0 to -0.75 or the time on the clock must not be synchronized with the B clocks, remaining at 0 although the nearby B clock (which C is now at rest next to) reads t'=-0.75.

 

[matheinste]

Yes...

I am so sorry if I seem to be difficult but why is the traveling clock no longer synchronised?
?

Sychronism cannot apply to a single clock. To an observer moving with respect to a pair of clocks synched in their own common rest frame, these clocks, appear to be running slow, and out of sych, that is, showing different times. That is due to the relativity of simultaneity.

Sorry if I am stating what may be already obvious to you, but it is necessary to know that we are using the same terms to mean the same thing. Clocks ticking at the same rate, such as clocks at rest with respect to each other, and as those of the reunited twins, are not necessarily synchronised. Synchronised clocks show the same time. Before departure the twins' clocks tick at the same rate as each other, when observed by an observer at rest with them, and we will assume they have been synchronized with the usual procedure. On reuniting, their clocks are again ticking at the same rate as each other, but they are not showing the same (accumulated) time. However, if not resynched the time difference will remain constant.

I suppose it may be correct to say that clocks ticking at the same rate are ticking in synchronism, but if they are not showing the same time they are not sychronized, but it may cause confusion.

Matheinste.

 

[ghwellsjr]

Grimble, I'm going to try to answer your questions about the Lorentz Transform (LT) that have gotten buried throughout this thread. (To see each original post, click on the right arrow above the quoted post, ">".)

Maybe I see it too simply but the Lorentz Transformation Formulae are showing how the age of one twin is measured differently by the other twin as a factor of their current velocity.
...
So a change in the twins relative velocity could only affect the current measurements.

LT is a function of the CURRENT velocity. It has absolutely nothing to do with the history of how their relative speed might have varied over the course of their travel relative to one another.

The Lorentz transform assumes nothing and has no need to assume anything about the constancy of the relative velocity.
...
when the velocity changes it is the transformation that changes and the fact that there was a different velocity at some point is irrelevant. It is only the Current velocity that matters.

LT transforms the measurements that exist at a particular point in time, as a function of the relative velocity at that point in time.

If that relative velocity were different the transformed measurements would be different at that moment in time.

The above quotes show a consistent misunderstanding for what the Lorentz Transform does and how it works. The purpose of the LT is to relate times and positions between two different reference frames that have a fixed relative velocity. You cannot change the value of "v" in midstream because that will define a different reference frame and change the value of the Lorentz factor, gamma.

But you don't need to use the Lorentz Transform to understand what's going on. All you need to do is use the Lorentz factor and realize that the faster an object goes relative to a reference frame, the more time slows down for that object.

So let's consider the reference frame in which the two twins start out at rest and assume they are the same age and they will, of course, be aging at the same rate, because they are at rest with each other. Then the traveling twin takes off at some speed and ages at a slower rate, the faster the speed, the slower the rate of aging. But he won't instantly become younger, he has to spend time at this lower aging rate to accumulate a significant age difference. The longer he travels, the greater the difference between his age and the home twin.

Then at some point he slows down until he is a rest once again in the reference frame that we are analyzing this scenario in. That means he is also at rest with the home twin. You have mistakenly concluded that because they are both at rest with each other that their ages are now the same but this is not true. No one changes their age just because they accelerate or decelerate, just their aging rate. Although, in this reference frame, the traveling twin is younger than the home twin, we cannot say that this is fundamentally true because it is frame dependent, and will, in general, be different if we use a different reference frame.

To illustrate this, let's analyze the scenario from another reference frame, that of the traveling twin, while he is traveling. This reference frame will be moving constantly with respect to the first reference frame and so the twins at the beginning will also be moving in the reference frame at whatever speed the traveling twin went up to in the first reference frame. You will note that the twins are at rest with respect to each other, just as before, but now they are both aging a slower rate and staying the same age as each other. But then the traveling twin "takes off", which means he comes to rest in the reference frame and starts aging at a faster rate than the home twin. After some time, he "takes off" in the opposite direction so that he now is traveling at the same rate as the home twin and starts aging at a slower rate, the same as the home twin. So during the time that he was "traveling", which is when he was at rest in the reference frame, he aged at a faster rate than his brother and so now he will be "older" than his brother.

So I hope this has helped you to see why we cannot draw any conclusion about the actual age difference between the twins while they are separated because there is no reference frame that we can consider to be the "correct" one from which to analyze the situation. In fact, if we use the reference frame that is half way between the two I already described, we will see that the twins travel in opposite directions at the same speed and so they are always aging the same and end up at the same age (at the half-way point).

So, depending on the reference frame we pick to analyze the first half of the twin paradox, we will conclude that the home twin is younger, the same age, or older than the traveling twin, clearly not a satisfying situation to be in.

But if we continue through the second half of the twin paradox, where the traveling twin comes back to the home twin, all three reference frames will yield the same conclusion, that the traveling twin is younger.

This is pretty easy to see in the first reference frame because when the traveling twin starts his journey back home, he will once again begin aging at a slower rate, so since he was aging at a slower rate for both halves of the journey, he is younger than the home twin.

In the second reference frame, we have left both twins traveling at the same rate and aging at the same lower rate but in order for the traveling twin to get back home, he will have to go even faster than he did during the first half of the journey (with respect to the home twin) and so he will age at an even slower rate. It is not obvious that he will at this slower rate will overcome the faster rate during the first half, but at least we can see that it would be possible. You will have to work out the mathematics to prove that what I am saying is correct.

Now let's take the third reference frame in which the twins have been aging identically (accept during accleration and deceleration). For the traveling twin to get back home, he has to go at a faster speed and therefor a slower aging rate until he catces up to his brother. At least in this one, it is easy to see that the traveling twin will end up younger than the home twin.

 

[Mike_Fontenot]

Previously, I wrote (in response to Matheinste's comment, "For good, read valid.", in his response to my post at https://www.physicsforums.com/showpost.php?p=2970849&postcount=54 ):

"I believe that some of my objections to alternative coordinate SYSTEMS do go beyond the issues of what is "equally good", prudent, or practical, and do instead concern the issue of validity. But that distinction needs to be "fleshed out". "

Here are some of my thoughts on that distinction:

Start with two perpetually inertial observers, Jerry and Sue, who are moving at a constant relative velocity, v, with respect to one another. Furthermore, suppose that Jerry and Sue were momentarily co-located at the instant when they were born. Here, I'm choosing the same names for the observers that I used in this post,

https://www.physicsforums.com/showpost.php?p=2965424&postcount=70 ,

from another thread, for (hopefully) obvious reasons.

We need to set up a coordinate system that is appropriate for describing Jerry's conclusions about what's going on in the (assumed flat) universe, and likewise for Sue.

Jerry and Sue both have an intuitive sense of "time". They are each surrounded by natural objects, stationary with respect to themselves, that seem to vibrate in a very regular way. The possibility of counting those vibrations, immediately suggests itself as a way to measure this mysterious thing we perceive as "time".

Jerry can build a "clock" that counts those natural vibrations. He builds a huge pile of identically constructed clocks, all synchronized to give the same reading at any instant. For simplicity, Jerry sets each clock so that it directly displays his age. Sue also makes a huge pile of clocks, that each read her age.

Jerry also has an intuitive notion of "length". To quantify that notion, he can define a unit of length by basing it on some stationary natural object that he is co-located with. For example, he might choose a particular stationary gold rod (at some arbitrary but fixed temperature and pressure) to serve as his standard of length. And he can make a whole pile of those "rulers", each essentially alike. Sue can do the same thing.

Jerry enlists the aid of many other observers, who are initially stationary with respect to himself, to VERY slowly distribute his pile of synchronized clocks around the universe, using his gold rods to determine the spacing between the clocks. For simplicity, I'll use a one-dimensional space ... the generalization to three spatial dimensions should be obvious.

Once that is done, Jerry has his coordinate system, and he can use it to label any spatial point in the universe, at any instant of time, by specifying a pair of values (t1, x1). Sue can do the same thing, with the pair of values (t2, x2).

The above coordinates are the standard Lorentz coordinates. (The synchronization of the non-co-located clocks can also be accomplished using light pulses, instead of the infinitesimally slow distribution of the clocks, and the results are the same either way).

Now, it's clear that Jerry, IF he chooses, can use ANY coordinate pairs that have a one-to-one relationship with his standard Lorentz coordinates, for his coordinate system. But (as far as I can see), there are no advantages to doing that, and lots of obvious disadvantages. The single biggest disadvantage is that other choices don't have the intuitive meaningfulness for Jerry, that the standard Lorentz coordinates provide.

Now, suppose Jerry wants to know how old Sue is, at some instant of his life. In order for that question to have any meaning for him, Jerry would have to insist that Sue has used EXACTLY the same standards to define time and length, as he himself did. And he must insist that she has set up her coordinate system in EXACTLY the same way that he himself set up his own coordinate system.

There are two different ways that Jerry can determine Sue's current age (and each method gives exactly the same result).

The easy, quick way is just to use the standard Lorentz equations. Einstein used only two axioms to derive the standard Lorentz equations: the constancy of the magnitude of the velocity of light (invariant among different inertial frames), and the principle of (special) relativity. The principle of (special) relativity says (for example) that whatever conclusions Jerry comes to about Sue's current age (at any given time in his life), Sue must come to EXACTLY the same conclusions about Jerry's current age (at any given time in her life).

The second way, is for Jerry to receive images from Sue, which show her age at the time of image transmission. When he receives an image, he knows that the reported age is NOT her current age, because she will have aged during the transit of the image. Jerry can compute how much Sue has aged during the transit of the message, and then add that extra ageing to the age reported in the image, in order to determine her current age when the image is received.

Anyone who has never done the above calculations would greatly benefit by doing them. The calculations are elementary, but they are easy to get wrong. You'll KNOW when you've done them correctly, because your result must be consistent with the standard Lorentz equations (and with the standard time-dilation result).

Once Jerry has determined Sue's current age, using the above method, he would legitimately consider any other alternative value for Sue's age to be INVALID, not just inferior.

Suppose you ate eight grapes for breakfast, and eleven grapes for lunch. If someone told you that the total number of grapes you ate at those two meals was other than nineteen, you would consider the value they gave you to be invalid. Jerry's reaction, to being told to use some current age for Sue other than the value he calculated from first principles, is essentially no different than your reaction to being given some other total for the number of grapes that you ate.

Mike Fontenot

 

[Dale]

Now, it's clear that Jerry, IF he chooses, can use ANY coordinate pairs that have a one-to-one relationship with his standard Lorentz coordinates, for his coordinate system.

Exactly.

Start with two perpetually inertial observers, ...

The remainder of your post is not relevant to the question of simultaneity conventions in non-inertial reference frames.

 

[John232]

Time dilation happens because an observer in constant motion will observe light to travel at a constant speed in his clock and an outside observer will see it move a farther distance because the beam will travel along with the observer in motion with his velocity. Then the only way the observer in motion can measure the photon to travel at the same speed traveling a shorter distance is for his time to slow down. If the position of the photon for each observer was simulanous then one observer would measure it traveling a longer distance than the other. Then if they both watch the photon move from position simultanously one would have to have a scewed measurement of time as they watch it travel different amounts of distance.

If the d = c t and the distance is different then the only way to have a constant c is to alter time.

 

[ghwellsjr]

There are two different ways that Jerry can determine Sue's current age (and each method gives exactly the same result).

The easy, quick way is just to use the standard Lorentz equations. Einstein used only two axioms to derive the standard Lorentz equations: the constancy of the magnitude of the velocity of light (invariant among different inertial frames), and the principle of (special) relativity. The principle of (special) relativity says (for example) that whatever conclusions Jerry comes to about Sue's current age (at any given time in his life), Sue must come to EXACTLY the same conclusions about Jerry's current age (at any given time in her life).

Now let me see if I've got this right. Jerry and Sue were born at the same location at the same time but immediately separated with a constant relative speed. Jerry uses his rest frame to determine how old Sue is at any point in time. She will alway be younger than him, correct? And Sue uses her rest frame to determine how old Jerry is at any point in time and he will always be younger than her, correct? And this is because each is using their own rest frame which you call the "standard Lorentz coordinates", correct?

But wouldn't a more logical frame be the one in which they are both traveling at the same speed in opposite directions? Then, not only would their ages be reciprocal, they would be always identical, no calculation necessary, thank you. Lorentz and Einstein both would agree with me and your way is not just inferior, it is INVALID, because you are using two different reference frames and I am using just one.

 

[Mike_Fontenot]

Start with two perpetually inertial observers, ...

The remainder of your post is not relevant to the question of simultaneity conventions in non-inertial reference frames.

Oh, but it IS.

You cannot possibly understand how to determine the current age of some distant, perpetually inertial person, for the case where you choose to undergo some accelerations during your life, unless and until you fully understand how to determine the current age of that distant, perpetually inertial person, for the case where you are yourself perpetually inertial. Specifically, you need to understand how to determine the amount of ageing the distant perpetually inertial person undergoes during the transit of images she transmits, giving her age at the instant of image transmission.

Once you understand the latter case, you can proceed to an understanding of the former case, by asking, and correctly answering, the following two questions:

1) Suppose you are unaccelerated for the entire INITIAL part of your life, but then you choose to accelerate at some instant in the LATTER part of your life. The question is, "At what time in your life, BEFORE your acceleration, do you cease being an inertial observer?". The correct answer is that you remain an inertial observer up until the instant that you accelerate.

2) Suppose you are unaccelerated for the entire LATTER part of your life, but you DID accelerate at some instant very EARLY in your life. The question is, "At what time in your later life, AFTER your acceleration, do you become an inertial observer?". The correct answer is that you become an inertial observer as soon as you stop accelerating.

Once you have the correct answers to the above two questions, you can then show, using a limiting argument, that an accelerating observer's conclusions about the current age of that distant, perpetually inertial person, will be exactly the same, at each instant of the accelerating observer's life, as the conclusions of the inertial frame with which he is momentarily stationary at that instant.

That's why the CADO equation, which is derived directly from the Lorentz equations (which relate two inertial frames), is applicable to both a perpetually inertial observer, AND to an observer who accelerates in any manner whatsoever.

Mike Fontenot

 

[ghwellsjr]

You cannot possibly understand how to determine the current age of some distant, perpetually inertial person, for the case where you choose to undergo some accelerations during your life, unless and until you fully understand how to determine the current age of that distant, perpetually inertial person, for the case where you are yourself perpetually inertial.

Once you understand the latter case, you can proceed to an understanding of the former case, by asking, and correctly answering, the following two questions:

1) Suppose you are unaccelerated for the entire INITIAL part of your life, but then you choose to accelerate at some instant in the LATTER part of your life. The question is, "At what time in your life, BEFORE your acceleration, do you cease being an inertial observer?". The correct answer is that you remain an inertial observer up until the instant that you accelerate.

2) Suppose you are unaccelerated for the entire LATTER part of your life, but you DID accelerate at some instant very EARLY in your life. The question is, "At what time in your later life, AFTER your acceleration, do you become an inertial observer?". The correct answer is that you become an inertial observer as soon as you stop accelerating.

Once you have the correct answers to the above two questions, you can then show, using a limiting argument, that an accelerating observer's conclusions about the current age of that distant, perpetually inertial person, will be exactly the same, at each instant of the accelerating observer's life, as the conclusions of the inertial frame with which he is momentarily stationary at that instant.

That's why the CADO equation, which is derived directly from the Lorentz equations (which relate two inertial frames), is applicable to both a perpetually inertial observer, AND to an observer who accelerates in any manner whatsoever.

Mike Fontenot

There is a difference between an inertial person/observer/clock and an inertial reference frame. An inertial refererence frame is defined to be perpetually inertial and extend to infinity in all directions and have a particular defined fixed origin for both time and space. Persons/observers/objects/clocks do not have to remain inertial; their positions, motions, accelerations, ages, aging rates are all defined within that one and only one inertial reference frame.

Then if you want, you can define a different inertial reference frame that has its own velocity and origins with respect to the first one and you can use the Lorentz Transform to transform all the persons/observers/objects/clocks from the first one into the second one. And you can do this as many times as you want. If you want to consider an accelerating reference frame, you have to be very careful because that means in essence that you have to be doing a whole lot of Lorentz Transforms all along the way.

Whatever you define in your first inertial reference frame will dictate how everything will appear in all the other reference frames that you transform it into after that.

And I want to make it clear that you have not defined any reference frame in your example about Jerry and Sue. You merely said that they were born at the same instant in time and in the same location and that they are perpetually inertial with a fixed velocity between them.

Now you need to define a single reference frame in which to analyze whatever happens next. You have not done this. You have done what virtually all people do when they think they have come up with a paradox, they describe part of a problem in one reference frame and another part of the problem in a different reference and don't bother to correctly use the Lorentz Transform (taking into account the time and space origins) to get them into a single reference frame and, as a result, they come up with all kinds of incorrect conclusions. That is the same thing you are doing. And then you need to understand that any conclusion you come to in one reference frame is only true for that one reference frame unless you can correctly transform the entire scenario into any other reference frame and the same conclusion holds true. Then you will have an invariant characteristic that is not frame dependent and which can be considered an absolute truth as opposed to a relative truth.

If you would limit the use of your CADO process to calculate the age difference between Jerry and Sue when and if they ever reunite, then you will have a legitimate process. But since you are claiming that your scheme of having Jerry estimate Sue's age before they reunite is the only valid one (as if there is only one valid one) then you are clearly not understanding reality.

 

[Dale]

1) Suppose you are unaccelerated for the entire INITIAL part of your life, but then you choose to accelerate at some instant in the LATTER part of your life. The question is, "At what time in your life, BEFORE your acceleration, do you cease being an inertial observer?". The correct answer is that you remain an inertial observer up until the instant that you accelerate.

Following up on ghwellsjr's post, you misunderstand the difference between an observer and a reference frame. That is a very common misunderstanding since relativity is so poorly taught in most cases. Reference frames are either inertial or non-inertial, they do not "become inertial" or "cease being inertial" in any sense.

When we are talking about simultaneity the question is inherently about a reference frame. An observer has events on his worldline but no sense in which those events are simultaneous with remote events unless a specific reference frame is arbitrarily selected.

Because the choice of reference frame is so completely arbitrary there is a standard convention which is used for inertial observers: the convention is that "an observer's reference frame" refers specifically to a frame where the observer is at rest at the origin and whose coordinates form an orthonormal basis (Minkowski metric). There is no such standard convention for associating a reference frame with a non-inertial observer.

However, by the first postulate even a perpetually inertial observer's reference frame is not special in any physical way, and is only chosen as a matter of convention and convenience. The inertial observer is free to use coordinates where he is not at rest or coordinates that form a non-orthonormal basis (e.g. some other synchronization convention besides Einstein's). Sometimes such coordinates can be useful for one reason or another. There is nothing physical which requires even a perpetually inertial observer to use their reference frame for determining simultaneity.

The fact that the CADO reduces to the usual convention for an inertial observer does not make either the usual convention or the CADO anything other than a convention. In order to show that it is a physical requirement rather than merely a convention you will need to demonstrate that other conventions predict the wrong result for some physical experiment. That is what is required to demonstrate that the other conventions are not valid, but the first postulate guarantees that no such prediction can be obtained.

 

[ghwellsjr]

Following up on ghwellsjr's post, you misunderstand the difference between an observer and a reference frame. That is a very common misunderstanding since relativity is so poorly taught in most cases. Reference frames are either inertial or non-inertial, they do not "become inertial" or "cease being inertial" in any sense.

Until I got on this forum, I always thought a reference frame in SR was inertial by definition but now I have learned that you can have a non-inertial frame which I assumed means that it accelerates in any arbitrary way. Do I have that right? If so, what's it good for? How do you transform between an inertial reference frame and a non-inertial frame? Or are we simply limiting a non-inertial frame to one in which we take gravity into account but otherwise would be inertial?

 

[Grimble]

The easy, quick way is just to use the standard Lorentz equations. Einstein used only two axioms to derive the standard Lorentz equations: the constancy of the magnitude of the velocity of light (invariant among different inertial frames), and the principle of (special) relativity. The principle of (special) relativity says (for example) that whatever conclusions Jerry comes to about Sue's current age (at any given time in his life), Sue must come to EXACTLY the same conclusions about Jerry's current age (at any given time in her life).

Yes this is exactly as I see it. Neither frame is privileged and we cannot say which of them is moving.
So at any time each can come to the conclusion that time is passing more slowly for the other.
But does the passage of time actually slow for either? How can it?
Does the passage of time slow for both? Wouldn't that be meaningless?
Or does time just appear to slow as seen by the other observer? Yes, surprise, that could work.

Now let me see if I've got this right. Jerry and Sue were born at the same location at the same time but immediately separated with a constant relative speed. Jerry uses his rest frame to determine how old Sue is at any point in time. She will alway be younger than him, correct? And Sue uses her rest frame to determine how old Jerry is at any point in time and he will always be younger than her, correct? And this is because each is using their own rest frame which you call the "standard Lorentz coordinates", correct?

But wouldn't a more logical frame be the one in which they are both traveling at the same speed in opposite directions? Then, not only would their ages be reciprocal, they would be always identical, no calculation necessary, thank you. Lorentz and Einstein both would agree with me and your way is not just inferior, it is INVALID, because you are using two different reference frames and I am using just one.

No, of course they wouldn't say it was invalid. From the intermediate frame the age of each would be seen to be the same because the time dilation for each would be equal. Yet each would still see the other as younger by the same amount.

Special Relativity is all about the relativity between two inertial FoR. How can it be shown just using one?

1) Suppose you are unaccelerated for the entire INITIAL part of your life, but then you choose to accelerate at some instant in the LATTER part of your life. The question is, "At what time in your life, BEFORE your acceleration, do you cease being an inertial observer?". The correct answer is that you remain an inertial observer up until the instant that you accelerate.

2) Suppose you are unaccelerated for the entire LATTER part of your life, but you DID accelerate at some instant very EARLY in your life. The question is, "At what time in your later life, AFTER your acceleration, do you become an inertial observer?". The correct answer is that you become an inertial observer as soon as you stop accelerating.

Very good questions and very well put, Mike. It demonstrates one fundamental point very well.
(I must apologise at this point for I have lost the reference where Einstein stated that he believed but couldn't (or hadn't yet) proved that SR would work equally well in an accelerating frame.)
The point is that an accelerating frame of reference could, for any infinitessimal interval, be considered an inertial FoR.

I ask you to consider, How can the passage of time in one FoR be made to slow according to the speed, or relative speed of another FoR.

Remember they are not connected in any way.

What if instead of two there were 10 or 100 or an infinite number of FoRs? How couod the clock in one frame slow according to all of them.? By different amounts.

Any careful examination of LT shows that it is only an observed effect. Like the projection of an image depending on the distance it is projected and the angle of the screen upon which it is projected.

That a clock at rest in ANY inertial FoR measures Proper Time.
That the distances between resting clocks in ANY inertial FoR are Proper distances.
Therefore the time kept by Jerry's clock is Proper Time and the time kept by Sue's clock is Proper Time.
But the time that each reads on the other's clock is Co-ordinate time.

All that Special Relativity comes down to is that for any two bodies sufficiently far from any other bodies that gravity has no effect will each experience the reciprocal of the others measurements of their respective motions and that MUST of course include accelerations! How could it not?

There is no privileged FoR.

I can draw a set of simple diagrams that demonstrate EXACTLY how all this works. It is all VERY SIMPLE.

Grimble.

 

[Dale]

I have learned that you can have a non-inertial frame which I assumed means that it accelerates in any arbitrary way. Do I have that right?

Yes.

If so, what's it good for? How do you transform between an inertial reference frame and a non-inertial frame?

That depends on the details of the "arbitrary way" that you choose. Some might be useful for analyzing things on an accelerating rocket, others for analyzing a merry-go-round, or in a variety of curved spacetimes, etc. It just depends on the details and the purpose of the coordinate system.

 

[Mike_Fontenot]

[...]

Very thoughtful comments and questions.

[...]
Neither frame is privileged and we cannot say which of them is moving.
So at any time each can come to the conclusion that time is passing more slowly for the other.
But does the passage of time actually slow for either?

Try to make your statement, about "actual" slowing, perfectly precise. The question, of whether time "actually" slows for some particular person, doesn't have any meaning in special relativity.

[...]
Or does time just appear to slow as seen by the other observer?
[...]

It's critical to be very clear about what you mean by "to appear". It is a term which causes serious misconceptions and confusion. Sue's slow ageing, ACCORDING to Tom, is certainly NOT some kind of phony apparition, like an image in the distorted mirrors of a circus fun-house.

The slower rate of Sue's ageing, according to Tom, is as real as real can be: Sue's slower relative ageing is what Tom CORRECTLY deduces from his own elementary measurements and calculations.

[...]
I ask you to consider, How can the passage of time in one FoR be made to slow according to the speed, or relative speed of another FoR.
[...]

Sue's relative ageing is slow, ACCORDING to Tom's correct deductions. But his conclusions about her relative ageing in no way changes Sue's perception of the progression of her own life.

[...]
What if instead of two there were 10 or 100 or an infinite number of FoRs? How could the clock in one frame slow according to all of them.? By different amounts.
[...]

Sue's "rate of ageing" has meaning only as a RELATIVE quantity ... it has no absolute meaning.

Observers stationary in EACH of all those other inertial frames each CORRECTLY conclude that Sue is ageing more slowly THAN THEY THEMSELVES ARE. Those observers will generally deduce different rates of ageing for Sue, relative to their OWN ageing. They are ALL correct. That's just the way special relativity IS. But it is NOT correct to conclude that Sue's current age is a meaningless concept to them ... nothing is more meaningful to them than what they deduce from their own elementary measurements and calculations.

Mike Fontenot

 

[Grimble]

Very thoughtful comments and questions.



Try to make your statement, about "actual" slowing, perfectly precise. The question, of whether time "actually" slows for some particular person, doesn't have any meaning in special relativity.

Then taking into account that the time shown on a clock at rest in an inertial FoR. I am saying that both Sue's clock and Jetty's clock are, in fact, keeping identical time.

It's critical to be very clear about what you mean by "to appear". It is a term which causes serious misconceptions and confusion. Sue's slow ageing, ACCORDING to Tom, is certainly NOT some kind of phony apparition, like an image in the distorted mirrors of a circus fun-house.

The slower rate of Sue's ageing, according to Tom, is as real as real can be: Sue's slower relative ageing is what Tom CORRECTLY deduces from his own elementary measurements and calculations.



Sue's relative ageing is slow, ACCORDING to Tom's correct deductions. But his conclusions about her relative ageing in no way changes Sue's perception of the progression of her own life.


Sue's "rate of ageing" has meaning only as a RELATIVE quantity ... it has no absolute meaning.

Observers stationary in EACH of all those other inertial frames each CORRECTLY conclude that Sue is ageing more slowly THAN THEY THEMSELVES ARE. Those observers will generally deduce different rates of ageing for Sue, relative to their OWN ageing. They are ALL correct. That's just the way special relativity IS. But it is NOT correct to conclude that Sue's current age is a meaningless concept to them ... nothing is more meaningful to them than what they deduce from their own elementary measurements and calculations.

Mike Fontenot

But this is what I have all my difficulty with.
Yes Tom measure's Sue to be ageing slower than he is; BUT that is because he is measuring a moving clock, using the measures in the moving clock's FoR; and it is adding the change due to the movement that makes the difference.
If Tom were to measure Sue's movement using his OWN rulers and clocks he would calculate the same times and distances that Sue does.

 

[Dale]

The point is that an accelerating frame of reference could, for any infinitessimal interval, be considered an inertial FoR.

I think you are thinking of the equivalence principle which says that over an infinitesimal region of spacetime you can construct a free-falling inertial frame even in a gravitational field. What you have stated above is incorrect since the metric over an infinitesimal interval in an accelerating frame is not necessarily the Minkowski metric of inertial frames, nor necessarily even diagonal.

 

[Mike_Fontenot]

[...]
But this is what I have all my difficulty with.
[...]

You are asking the right questions. I think you're going to get it figured out, if you keep putting in the effort.

[...]
If Tom were to measure Sue's movement using his OWN rulers and clocks he would calculate the same times and distances that Sue does.

No, he wouldn't.

To understand this, you need to bite the bullet and figure out how Tom determines the amount of ageing that Sue undergoes during the transit of the images reporting her current age. Take a careful look at this posting:

https://www.physicsforums.com/showpost.php?p=2978931&postcount=75 ,

and then spend some quality time figuring out how to do those calculations. It's a non-trivial effort, but it'll be time well spent.

Mike Fontenot

 

[ghwellsjr]

No, of course they wouldn't say it was invalid. From the intermediate frame the age of each would be seen to be the same because the time dilation for each would be equal. Yet each would still see the other as younger by the same amount.

Grimble, you're not going to understand my post:

Now let me see if I've got this right. Jerry and Sue were born at the same location at the same time but immediately separated with a constant relative speed. Jerry uses his rest frame to determine how old Sue is at any point in time. She will alway be younger than him, correct? And Sue uses her rest frame to determine how old Jerry is at any point in time and he will always be younger than her, correct? And this is because each is using their own rest frame which you call the "standard Lorentz coordinates", correct?

But wouldn't a more logical frame be the one in which they are both traveling at the same speed in opposite directions? Then, not only would their ages be reciprocal, they would be always identical, no calculation necessary, thank you. Lorentz and Einstein both would agree with me and your way is not just inferior, it is INVALID, because you are using two different reference frames and I am using just one.

until you realize that it is a reaction to Mike's post:

Once Jerry has determined Sue's current age, using the above method, he would legitimately consider any other alternative value for Sue's age to be INVALID, not just inferior.

Now as to your question:

Special Relativity is all about the relativity between two inertial FoR. How can it be shown just using one?

SR is all about the relativity of an infinite number of FoR, not just two, but just one at a time. Any situation or scenario you want to consider must be analyzed in just one FoR, your choice. Then if you want, you can use the Lorentz Transform to see what it would look like from any other number of FoR, again your choice or anybody else's choice. The only things that can be considered absolute are the things that look the same in all FoR.

That's the problem with Mike's CADO scheme, he's claiming that only one FoR is VALID for each observer even though it produces intermediate results that are different in all other FoR. He is considering a single FoR for each observer as preferred and calls it the Standard Lorentz Coordinates. I did a Google search on this term and found five hits, two were Mike's on this forum and the other three were not using it like Mike is using it. He has invented the term and I'm saying that neither Lorentz nor Einstein would agree with it. Sorry you missed the sarcasm.

 

[Grimble]

I think you are thinking of the equivalence principle which says that over an infinitesimal region of spacetime you can construct a free-falling inertial frame even in a gravitational field. What you have stated above is incorrect since the metric over an infinitesimal interval in an accelerating frame is not necessarily the Minkowski metric of inertial frames, nor necessarily even diagonal.

No I am sorry but that is not at all what I am thinking!

I am thinking about Relativity. The relativity of two FoR alone in space. Gravity is not relevant here. Neither is acceleration! Now, now PLEASE don't go getting upset that I say that. Let me explain.

Relativity, and Special Relativity, are all about how the measurements in one FoR can be transformed to another FoR.

Two FoR alone in space and the only movement that can be determined is the movement of one relative to the other.

Yes, yes, yes, I know that within each frame, using some sort of accelerometer it is possible to determine what acceleration each one is experiencing but that has NOTHING to do with relativity, which is all about the fact that what ever movement one experiences relative to the other is the reciprocal of what the other experiences relative to the one.

That is why there is not term for acceleration in the LT equations.

Grimble.

 

[Grimble]

Now for seeing it all in one FoR:

Let us envisage three observers, A,B & C at rest in space.
They are situated in a straight line in the order ACB.
AC = CB = 4 light years, so AB = 8 ly
Also, because they are at rest in an inertial FoR the synchronised clocks that each possesses will be showing Proper time. (clock at rest in an inertial FoR is following its world line and is therefore measuring Proper time)

Imagine another two observers a & b each traveling at 0.8c. a is traveling from A towards C and b is traveling from B towards C.

And let us say that in our rest FoR they simultaneously pass A and B at time t0.

The observer at C will observe them both passing him at time t1 and t1 - t0 = 5 years.

Note that this has nothing to do with LT as all our observations and measurements are
made in the resting FoR of observers A,B & C.
The traveling observers, a and b, are detected at points A and C and points B and C respectively. From these observations A, B & C can determine that their speeds are equal and are each 0.8c.

Now what do a and b observe in their FoRs?
Well they too may consider them selves at rest in inertial FoRs. Therefore their clocks will also be showing proper time. And if they had imaginary synchronised clocks at rest with them and passing observer C at t0 we know that the proper distance between a and C at time t0 measured by a in a's FoR would be 4 proper light years.

Therefore a and b would each observe that they would travel 4 lt in 5 years at a speed of 0.8c.

It is only when the observers A, B and C use the measurements from a and b's FoRs that LT comes into play, for if C were to read a & b's clocks as they passed him at time t1 he would read time dilated times and distances - and calculate that a and b each traveled a distance of 2.4 (contracted) light years in a time of 3 (dilated) years.
Travelling at a speed of 2.4/3 = 0.8c.

 

[Dale]

using some sort of accelerometer it is possible to determine what acceleration each one is experiencing but that has NOTHING to do with relativity

This statement is also incorrect. The accelerometer has everything to do with relativity, it is fundamental to both of the two postulates. The Lorentz transform is a transform between two inertial frames. I.e. given one inertial frame you can find any other inertial frame via the Poincare group (which includes the Lorentz transform). But how do you know that your first frame is inertial? It cannot be determined by virtue of the Lorentz transform alone, it can only be determined by some physical experiment, such as using an accelerometer.

Only once you have performed such an experiment can you label a given coordinate system as inertial. Once you have done so you automatically know the metric in that coordinate system, and you can then determine the metric in any arbitrary non-inertial coordinate system also. Evaluated at some point the non-inertial metric will not necessarily reduce to the inertial metric.

 

[ghwellsjr]

Now what do a and b observe in their FoRs?
Well they too may consider them selves at rest in inertial FoRs. Therefore their clocks will also be showing proper time. And if they had imaginary synchronised clocks at rest with them and passing observer C at t0 we know that the proper distance between a and C at time t0 measured by a in a's FoR would be 4 proper light years.

Everything was fine up to this point and now it gets confusing. It's OK for a and b to have their own clocks at rest with them but what are you saying they are synchronized to? They cannot be synchronized to each other's clocks or the A, B, and C's clocks. If you want to use a second FoR (the one in which a is at rest) you need to transform the entire scenario into that FoR and you have to define when t0 is relative to the original t0. (It looks like you have done a big shift--which is OK--because in the orginal FoR, t0 was when a was at A and b was a B but now it appears that you have put t0 for this second FoR when a and b are at C). Is that what you really meant to do?

Finally, if you want to consider a third FoR, the one in which b is at rest, you have to do another transform.

I'm not going to comment on the rest of your post until you clarify this issue.

 

[Grimble]

Everything was fine up to this point and now it gets confusing. It's OK for a and b to have their own clocks at rest with them but what are you saying they are synchronized to? They cannot be synchronized to each other's clocks or the A, B, and C's clocks. If you want to use a second FoR (the one in which a is at rest) you need to transform the entire scenario into that FoR and you have to define when t0 is relative to the original t0. (It looks like you have done a big shift--which is OK--because in the orginal FoR, t0 was when a was at A and b was a B but now it appears that you have put t0 for this second FoR when a and b are at C). Is that what you really meant to do?

Finally, if you want to consider a third FoR, the one in which b is at rest, you have to do another transform.

I'm not going to comment on the rest of your post until you clarify this issue.

Then let me repeat and clarify for you:

Now what do a and b observe in their FoRs?
Well they too may consider them selves at rest in inertial FoRs. Therefore their clocks will also be showing proper time. And if they had imaginary synchronised clocks at rest with them and passing observer C at t0 we know that the proper distance between a and C at time t0 measured by a in a's FoR would be 4 proper light years.

Let us take the inertial frame of observer a.
If it had an imaginary clock synchronised with the clock adjacent to observer a (why would it - or even how could it - be synchronised with any thing else?) and this imaginary clock were to be passing C at time t0, then it would be exactly 4 ly ahead of a in a's FoR.

And, I thought it quite apparent that we would be considering a similar arrangement in observer b's FoR.

This part is really redundant but just emphasises that in the inertial frames of a and b, the measurements by a and b would be exactly equal to those in our rest frame or AB & C. i.e. that a and b, by their own measurements within their own FoRs, would calculate they had each traveled 4 proper lys in 5 proper years.

 

[ghwellsjr]

OK, I think I see what you are proposing, tell me if this is what you're thinking:

In a's FoR, a has a clock adjacent to himself and another synchronized clock 4 ly ahead of himself. Then when a passes next to A at t0, the other clock is passing by C, again at t0 in a's FoR.

 

[Grimble]

OK, I think I see what you are proposing, tell me if this is what you're thinking:

In a's FoR, a has a clock adjacent to himself and another synchronized clock 4 ly ahead of himself. Then when a passes next to A at t0, the other clock is passing by C, again at t0 in a's FoR.

Precisely, as both a's measurement of a distance of 4 light years and A,B and C's measurement of 4 light years are each made at rest in their individual FoRs, between two clocks that are at rest in their individual FoRs; Thus each measurement is a measurement of Proper distance. And each measurement is made wholly within a single FoR with no reference to the other FoR.

Now as each measurement is 4 Proper light years, at time t0 when a is passing A, a's second clock MUST be passing C.

So before you say anything, Length Contraction does not come into play for Either of these measurements.

 

[Grimble]

This statement is also incorrect. The accelerometer has everything to do with relativity, it is fundamental to both of the two postulates. The Lorentz transform is a transform between two inertial frames. I.e. given one inertial frame you can find any other inertial frame via the Poincare group (which includes the Lorentz transform). But how do you know that your first frame is inertial? It cannot be determined by virtue of the Lorentz transform alone, it can only be determined by some physical experiment, such as using an accelerometer.

Absolutely! You are right in what you say here, but is it as relevant as it appears to be?
SR, LT and all the rest refer to Inertial FoRs but why are they restricted to Inertial FoRs?
Is it not because it was simpler to deal with them first?
Having said that, what is it about non-inertial frames that precludes their inclusion?

Ah! But of course that is the effects of the extra forces at work that result in the FoR being non-inertial.

But, I say, all we are transforming between frames are coordinates; coordinates at one point in time. It matters not what forces are resulting in those movements and those coordinates, we are not measuring forces only positions, coordinates in 4 dimensional Space-time.

So we are transforming measurements from one FoR to another. Redefining them according to a different set of coordinates. The forces acting upon the objects being measured are not relevant to those measurements.

Or we could say that for any measurement taken the FoR concerned can be considered inertial (from the point of view of those measurements) for the instant that those measurements exist.

And as we are dealing with the movement of one FoR relevant to another we do NOT need to know anything about which if either is inertial, for either one could be, or perhaps neither depending on how one defines them.

Only once you have performed such an experiment can you label a given coordinate system as inertial. Once you have done so you automatically know the metric in that coordinate system, and you can then determine the metric in any arbitrary non-inertial coordinate system also. Evaluated at some point the non-inertial metric will not necessarily reduce to the inertial metric.

I must admit you lose me here, What exactly are you referring to when you talk of the metric? I suspect it is something more than the 4 space-time coordinates and relative velocity that we are dealing with in LT?

 

[Dale]

we are not measuring forces only positions, coordinates in 4 dimensional Space-time.

Do you really believe this? If so, perhaps you can find the x-coordinate-meter in some catalog. I know of many devices that measure distances, but I don't know of any that measure coordinates. Do you see the distinction?

The forces acting upon the objects being measured are not relevant to those measurements.

They are relevant if they affect the outcome of the measurements. When you make a measurement you are not simply mathematically determining a coordinate, you are performing a physical experiment of some sort. The outcome of that physical experiment is generally affected by the forces you refer to. A clock at rest in an rotating reference frame behaves differently from a clock at rest in an inertial frame.

I must admit you lose me here, What exactly are you referring to when you talk of the metric? I suspect it is something more than the 4 space-time coordinates and relative velocity that we are dealing with in LT?

The metric is the geometric object which connects a coordinate system to the physics. It encodes all of the information about distances, angles, durations, relative velocities, and the causal structure of spacetime. For instance, if you measure a small distance the result of the measurement is
$$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$$
Where ds is the distance, g is the metric, dx is the change in the coordinates, and the Einstein summation convention is used. So even a simple measurement of distance depends on the metric.

http://en.wikipedia.org/wiki/Metric_tensor_(general_relativity)
http://www.phy.duke.edu/~rgb/Class/phy319/phy319/node131.html
http://www.mathpages.com/rr/s5-02/5-02.htm

 

[JesseM]

Absolutely! You are right in what you say here, but is it as relevant as it appears to be?
SR, LT and all the rest refer to Inertial FoRs but why are they restricted to Inertial FoRs?
Is it not because it was simpler to deal with them first?
Having said that, what is it about non-inertial frames that precludes their inclusion?

Nothing precludes their inclusion, but the "twin paradox" is based on the idea that a moving clock runs slower than a clock at rest, with the "paradox" arising because people falsely think this should be true in the non-inertial frame of the traveling twin; in fact, the idea that a moving clock runs slower than a clock at rest is only guaranteed to hold in an inertial frame.

Or we could say that for any measurement taken the FoR concerned can be considered inertial (from the point of view of those measurements) for the instant that those measurements exist.

What do you mean by "the FoR concerned"? A measurement doesn't come attached to a frame of reference, the reading on any physical measuring instrument is a frame-independent fact which all frames should make the same prediction about.

And as we are dealing with the movement of one FoR relevant to another we do NOT need to know anything about which if either is inertial, for either one could be, or perhaps neither depending on how one defines them.

You do if you want to use them to make physical predictions about things like the elapsed time on a clock with known velocity as a function of time in some coordinate system.

I must admit you lose me here, What exactly are you referring to when you talk of the metric? I suspect it is something more than the 4 space-time coordinates and relative velocity that we are dealing with in LT?

In any coordinate system, you want to be able to calculate the frame-independent proper time along any timelike worldline (along with a frame-independent 'proper distance' along any spacelike worldline). For any given coordinate system, there is an associated metric which gives the line element which you can integrate along a worldline with known coordinates in your coordinate system and get the correct proper time/proper distance for that worldline. For example, in an inertial coordinate system in flat spacetime the line element could be written as $$ds^2 = dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2)$$ (this is just the infinitesimal form of the spacetime interval which gives proper time/proper distance along a constant velocity-path with known endpoints (t1, x1, y1, z1) and (t2, x2, y2, z2), in which case the proper time could be calculated by $$\sqrt{(t_2 - t_1)^2 - (1/c^2)[(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2]}$$)

 

[ghwellsjr]

Now for seeing it all in one FoR:

Let us envisage three observers, A,B & C at rest in space.
They are situated in a straight line in the order ACB.
AC = CB = 4 light years, so AB = 8 ly
Also, because they are at rest in an inertial FoR the synchronised clocks that each possesses will be showing Proper time. (clock at rest in an inertial FoR is following its world line and is therefore measuring Proper time)

Imagine another two observers a & b each traveling at 0.8c. a is traveling from A towards C and b is traveling from B towards C.

And let us say that in our rest FoR they simultaneously pass A and B at time t0.

The observer at C will observe them both passing him at time t1 and t1 - t0 = 5 years.

Note that this has nothing to do with LT as all our observations and measurements are
made in the resting FoR of observers A,B & C.
The traveling observers, a and b, are detected at points A and C and points B and C respectively. From these observations A, B & C can determine that their speeds are equal and are each 0.8c.

Now what do a and b observe in their FoRs?
Well they too may consider them selves at rest in inertial FoRs. Therefore their clocks will also be showing proper time. And if they had imaginary synchronised clocks at rest with them and passing observer C at t0 we know that the proper distance between a and C at time t0 measured by a in a's FoR would be 4 proper light years.

Therefore a and b would each observe that they would travel 4 lt in 5 years at a speed of 0.8c.

It is only when the observers A, B and C use the measurements from a and b's FoRs that LT comes into play, for if C were to read a & b's clocks as they passed him at time t1 he would read time dilated times and distances - and calculate that a and b each traveled a distance of 2.4 (contracted) light years in a time of 3 (dilated) years.
Travelling at a speed of 2.4/3 = 0.8c.

OK, I think I see what you are proposing, tell me if this is what you're thinking:

In a's FoR, a has a clock adjacent to himself and another synchronized clock 4 ly ahead of himself. Then when a passes next to A at t0, the other clock is passing by C, again at t0 in a's FoR.

Precisely, as both a's measurement of a distance of 4 light years and A,B and C's measurement of 4 light years are each made at rest in their individual FoRs, between two clocks that are at rest in their individual FoRs; Thus each measurement is a measurement of Proper distance. And each measurement is made wholly within a single FoR with no reference to the other FoR.

Now as each measurement is 4 Proper light years, at time t0 when a is passing A, a's second clock MUST be passing C.

So before you say anything, Length Contraction does not come into play for Either of these measurements.

I'm afraid that you've done a Galilean Transformation which is only applicable at low speeds, certainly not at the 0.8c you have proposed in your scenario. You need to perform a valid Lorentz Transformation from your orginal scenario where A, B, and C are at rest to this one where a is at rest. If you had done so, you would understand that all dimensions along the direction of a's motion are contracted, so that the distances from A to C and from C to B are not 4 ly. Also, the clocks that were synchronized in the original FoR or no longer synchronized in this new FoR and they are time dilated.

Your concept of LT as you indicated at the end of the first quote is all mixed up.

But I'm wondering, why are you interested in doing this transformation? What added information do you think it is going to provide?

 

[Grimble]

I'm afraid that you've done a Galilean Transformation which is only applicable at low speeds, certainly not at the 0.8c you have proposed in your scenario. You need to perform a valid Lorentz Transformation from your orginal scenario where A, B, and C are at rest to this one where a is at rest. If you had done so, you would understand that all dimensions along the direction of a's motion are contracted, so that the distances from A to C and from C to B are not 4 ly. Also, the clocks that were synchronized in the original FoR or no longer synchronized in this new FoR and they are time dilated.

Your concept of LT as you indicated at the end of the first quote is all mixed up.

But I'm wondering, why are you interested in doing this transformation? What added information do you think it is going to provide?

I am sorry but you are still not grasping what I am saying.
I am not doing any form of transformation, not even a Galilean one.
I am dealing with two separate sets of coordinates.
In one case A,B & C are making measurements entirely within their own set of coordinates ...
And, completely independently a is making measurements according to his own set of coordinates ...
And according to the observations made in the first FoR an event occurs where a passes A at time t0.
Then a second event occurs where a passes C at time t1.

What I am showing is that the measurements taken within each FoR are equal as absolute proper measurements.

They obviously will not be equal in either FoR where LT means that the 'remote/moving' measurement observed from each frame WILL be length contract/time dilated.
But, nevertheless the measurements of time and distance made by A of the distance AC and the time t1-t0, will by necessity have to be equal to the measurements made by a of the distance between his clocks and the time of the two events t1-t0 as these measurements are all entirely local within each FoR and NO TRANSFORMATION is involved.

 

[Grimble]

Do you really believe this? If so, perhaps you can find the x-coordinate-meter in some catalog. I know of many devices that measure distances, but I don't know of any that measure coordinates. Do you see the distinction?

No. I'm sorry but you have me there.
My understanding is that the coordinates place an event at a place and time according to the Frame of Reference and that the x coordinate is the measure of the displacement parallel to the x axis. But apparently that is not correct?

They are relevant if they affect the outcome of the measurements. When you make a measurement you are not simply mathematically determining a coordinate, you are performing a physical experiment of some sort.

Are, you mean like taking measurements of an object that is moving with a high velocity relative to you. Like the way those measurements are different from those taken with rulers and clocks adjacent to the object. The way that those measurements of a moving object are distorted as a function of the relative velocity?

 

[Grimble]

Nothing precludes their inclusion, but the "twin paradox" is based on the idea that a moving clock runs slower than a clock at rest,

But it doesn't does it? It only appears to because of the distortion caused by taking measurements at speed.

with the "paradox" arising because people falsely think this should be true in the non-inertial frame of the traveling twin;

How can you say that this isn't true? Special relativity - the movement of one relevant to the other - reciprocal movement.

in fact, the idea that a moving clock runs slower than a clock at rest is only guaranteed to hold in an inertial frame.

Yet it doesn't does it? There is nothing in Einstein's SR that says that it would, only that the observer at rest would see it run slow.

 

[JesseM]

Nothing precludes their inclusion, but the "twin paradox" is based on the idea that a moving clock runs slower than a clock at rest,

But it doesn't does it? It only appears to because of the distortion caused by taking measurements at speed.

It does run slower relative to a given inertial frame, and that isn't a "distortion" any more than any other frame-dependent observation like the observation that one object has a greater velocity than another. My point was that the twin paradox is based on falsely thinking that the frame-dependent truth "a moving clock runs slower than a clock at rest" would still work in a non-inertial frame.

with the "paradox" arising because people falsely think this should be true in the non-inertial frame of the traveling twin;

How can you say that this isn't true? Special relativity - the movement of one relevant to the other - reciprocal movement.

Nope, that's an entirely false statement if you are talking about non-inertial movement. Special relativity does not claim there is the least bit of reciprocity if one object is moving inertially and the other is moving non-inertially. Acceleration is absolute in relativity.

in fact, the idea that a moving clock runs slower than a clock at rest is only guaranteed to hold in an inertial frame.

Yet it doesn't does it? There is nothing in Einstein's SR that says that it would, only that the observer at rest would see it run slow.

The rate that a clock runs relative to coordinate time (i.e. $$d\tau / dt = \sqrt{1 - v^2/c^2}$$) is frame-dependent, but Einstein's SR says that if you have two specific events A and B on its worldline such that it moved inertially between those events, then in any inertial frame, if the coordinate time between A and B is $$\Delta t$$ and the speed of the clock as it moved between them in that frame is v, the time $$\Delta \tau$$ elapsed on the clock is $$\Delta \tau = \Delta t \sqrt{1 - v^2/c^2}$$. The time elapsed on a given worldline between two specific events on that worldline is frame-independent, so all inertial frames will agree on the value even though they disagree on the value of $$\Delta t$$ and v.

Did you read my post on the [post=2972720]geometric analogy[/post]? Do you agree that if we have two dots A and B on a 2D plane, with a straight line segment joining them, then in any Cartesian coordinate system where the difference in x-coordinate of A and B is $$\Delta x$$ and the slope $$\Delta y / \Delta x$$ is S, then the length of the line segment is $$\sqrt{\Delta x^2 + \Delta y^2} = \sqrt{\Delta x^2}\sqrt{1 + (\Delta y^2 / \Delta x^2)} = \Delta x \sqrt{1 + S^2}$$? And that this length is coordinate-independent, so different Cartesian coordinate systems with their x-y axes will all agree on the value of $$\Delta x \sqrt{1 + S^2}$$ even though they disagree on the value of $$\Delta x$$ and S? If you agree with that, perhaps you can see that this formula can also be applied to a polygonal path made up of multiple straight segments, like a V-shaped path, so for example if the path consists of a straight segment going from A to B and another straight segment with a different slope going from B to C, the total length of this path would be $$\Delta x_{AB} \sqrt{1 + S^2_{AB}} + \Delta x_{BC} \sqrt{1 + S^2_{BC}}$$. You can do the same sort of thing in SR if you want to figure out a clock's elapsed time (a frame-independent quantity) along a polygonal worldline consisting of multiple inertial segments joined by instantaneous acceleration, like the V-shaped worldline of the traveling twin in the simplest version of the twin paradox.

 

[ghwellsjr]

I'm afraid that you've done a Galilean Transformation which is only applicable at low speeds, certainly not at the 0.8c you have proposed in your scenario. You need to perform a valid Lorentz Transformation from your orginal scenario where A, B, and C are at rest to this one where a is at rest. If you had done so, you would understand that all dimensions along the direction of a's motion are contracted, so that the distances from A to C and from C to B are not 4 ly. Also, the clocks that were synchronized in the original FoR or no longer synchronized in this new FoR and they are time dilated.

Your concept of LT as you indicated at the end of the first quote is all mixed up.

But I'm wondering, why are you interested in doing this transformation? What added information do you think it is going to provide?

I am sorry but you are still not grasping what I am saying.
I am not doing any form of transformation, not even a Galilean one.
I am dealing with two separate sets of coordinates.
In one case A,B & C are making measurements entirely within their own set of coordinates ...
And, completely independently a is making measurements according to his own set of coordinates ...
And according to the observations made in the first FoR an event occurs where a passes A at time t0.
Then a second event occurs where a passes C at time t1.

What I am showing is that the measurements taken within each FoR are equal as absolute proper measurements.

They obviously will not be equal in either FoR where LT means that the 'remote/moving' measurement observed from each frame WILL be length contract/time dilated.
But, nevertheless the measurements of time and distance made by A of the distance AC and the time t1-t0, will by necessity have to be equal to the measurements made by a of the distance between his clocks and the time of the two events t1-t0 as these measurements are all entirely local within each FoR and NO TRANSFORMATION is involved.

A Frame of Reference is the same thing as a Set of Coordinates, so when you talk about different observers making measurements with their own set of coordinates it's the same thing as saying they are making measurements within the Frame of Reference in which they are at rest. I'm sure you know this.

Now here's the scenario as you have defined it:

Now for seeing it all in one FoR:

Let us envisage three observers, A,B & C at rest in space.
They are situated in a straight line in the order ACB.
AC = CB = 4 light years, so AB = 8 ly
Also, because they are at rest in an inertial FoR the synchronised clocks that each possesses will be showing Proper time. (clock at rest in an inertial FoR is following its world line and is therefore measuring Proper time)

Imagine another two observers a & b each traveling at 0.8c. a is traveling from A towards C and b is traveling from B towards C.

And let us say that in our rest FoR they simultaneously pass A and B at time t0.

The observer at C will observe them both passing him at time t1 and t1 - t0 = 5 years.

Note that this has nothing to do with LT as all our observations and measurements are
made in the resting FoR of observers A,B & C.
The traveling observers, a and b, are detected at points A and C and points B and C respectively. From these observations A, B & C can determine that their speeds are equal and are each 0.8c.

But then you make a claim about what a and b will measure or see from their Frames of Reference:

Now what do a and b observe in their FoRs?
Well they too may consider them selves at rest in inertial FoRs. Therefore their clocks will also be showing proper time. And if they had imaginary synchronised clocks at rest with them and passing observer C at t0 we know that the proper distance between a and C at time t0 measured by a in a's FoR would be 4 proper light years.

Therefore a and b would each observe that they would travel 4 lt in 5 years at a speed of 0.8c.

But, you don't realize that you cannot just make this claim without doing a Lorentz Transformation to get from your first FoR to your second FoR. You need to understand that in the original FoR, the distances involved for a and b will be contracted to 0.6 and the clocks for a and b will be running at 0.6 of normal, so a and b will measure the 4 ly distances as 2.4 ly and the 5 y times as 3 years, and yes, they will measure their own speeds as 2.4 ly / 3 y or 0.8c. They can have synchronized clocks within their FoR's but they won't be synchronized with the original FoR or with each other.