Looking to understand time dilation

[matheinste]

Yes, I'm not disagreeing with anything you said, I'm merely pointing out that we shouldn't think that it's necessary to say that the two "are at rest in the same inertial frame" because when they are at rest with each other in one inertial frame, they will be at rest with each other in all inertial frames.

That's fine.

Matheinste.

 

[Mike_Fontenot]

[...]
If no inertial reference frame can be considered "preferred" according to SR, [...]
[...]

If the traveler is perpetually inertial, then he certainly DOES have a "preference" for one particular inertial frame. There is only one inertial frame that agrees with the elementary measurements and elementary calculations that he makes, using clocks and rulers with which he is stationary. That preferred inertial frame is the one in which he is stationary.

Of course, other inertial observers (who are moving at some constant velocity with respect to the above inertial observer) will prefer a different inertial frame (the one in which THEY are stationary).

What special relativity says is that there is no single inertial frame that ALL inertial observers prefer. There is no single inertial reference frame that is UNIVERSALLY special.

Mike Fontenot

 

[Dale]

If the traveler is perpetually inertial, then he certainly DOES have a "preference" for one particular inertial frame.

Only as a matter of personal and computational convenience, not as a matter of physical necessity. And there may be some situations where a different frame is more convenient, e.g. the center of momentum frame for a collision.

There is only one inertial frame that agrees with the elementary measurements and elementary calculations that he makes, using clocks and rulers with which he is stationary.

You have never provided any substantiation for this, and it is only even possibly true if you define "elementary calculations" such that it is a tautology. Otherwise the principle of relativity ensures that all inertial reference frames will give the same predictions for the result of any given measurement and your statement is wrong.

 

[ghwellsjr]

If the traveler is perpetually inertial, then he certainly DOES have a "preference" for one particular inertial frame. There is only one inertial frame that agrees with the elementary measurements and elementary calculations that he makes, using clocks and rulers with which he is stationary. That preferred inertial frame is the one in which he is stationary.

Of course, other inertial observers (who are moving at some constant velocity with respect to the above inertial observer) will prefer a different inertial frame (the one in which THEY are stationary).

What special relativity says is that there is no single inertial frame that ALL inertial observers prefer. There is no single inertial reference frame that is UNIVERSALLY special.

Mike Fontenot

The term "preferrred frame" has a specific meaning. You are making the same mistake that I am dealing with someone else on in this thread:

https://www.physicsforums.com/showthread.php?t=442132&page=2

It doesn't mean a personal preference like "I prefer blondes". Even if everyone in the world preferred blondes, it wouldn't mean they are "UNIVERSALLY special". I hope you don't think the word "special" in Special Relativity has anything to do with the concept of "preferred". This has nothing to do with anyone's preferences. It has to do with whether we base our physics on the concept of an æther.

 

[Mike_Fontenot]

[...]
If the traveler is perpetually inertial, then he certainly DOES have a "preference" for one particular inertial frame.
[...]

Only as a matter of personal and computational convenience, not as a matter of physical necessity.
[...]

It's not a matter of physical necessity that you refrain from constantly hitting yourself in the head with a hammer, either.

The well-known results of time-dilation and length-contraction are directly available for your use (when you are an inertial observer), provided that you choose to use the usual Lorentz coordinates, in an inertial frame in which you are stationary. And the time coordinate, with that choice, corresponds to the time shown on your OWN watch. The spatial coordinate, with that choice, correponds to the lengths as reported by your OWN rulers and measuring tapes. There is a REASON why Einstein chose those coordinates, when developing his special theory.

In the spirit of general relativity, you are of course free to choose some other set of coordinates, by transforming those Lorentz coordinates in an almost unlimited number of ways, provided that the eigenvalues of the resulting metric are either {1, -1, -1, -1} or {-1, 1, 1, 1}, assuming that spacetime is everywhere flat.

In those alternative coordinate systems, your "clocks" would behave in very odd ways, compared to ordinary clocks. And likewise for your "rulers". Real work-a-day experimental physicists would seldom, if ever, choose those types of "measuring devices" to do their experiments in their laboratories. There is a REASON that the phrase "in the laboratory frame" is used so often.

[...]
And there may be some situations where a different frame is more convenient, e.g. the center of momentum frame for a collision.
[...]

Not if you're one of the two people involved in the impending collision. I've spent a lot of hours flying airplanes and gliders, constantly needing to avoid mid-air collisions. In all that time, I've NEVER used the center of mass coordinate system to help me avoid mid-air collisions.

Mike Fontenot

 

[Mike_Fontenot]

[...]
I hope you don't think the word "special" in Special Relativity has anything to do with the concept of "preferred". This has nothing to do with anyone's preferences. It has to do with whether we base our physics on the concept of an æther.
[...]

No, the term "special" in "special relativity" refers to the special case of limitless flat spacetime (no gravitational fields).

My other use of the term "special", as in "There is no single inertial reference frame that is UNIVERSALLY special", referred to the fact that in flat spacetime, there is no single inertial reference frame that is preferred by ALL inertial observers.

Mike Fontenot

 

[ghwellsjr]

You got the word "special" right, I just don't know why you can't get the word "preferred" right.

 

[ghwellsjr]

Real work-a-day experimental physicists would seldom, if ever, choose those types of "measuring devices" to do their experiments in their laboratories. There is a REASON that the phrase "in the laboratory frame" is used so often.

I thought your CADO formula was for a traveler going at substantial speeds, traveling incredible distances for almost an entire liftetime to "know" at any instant of time how old his daughter was back at home.

 

[JesseM]

And the time coordinate, with that choice, corresponds to the time shown on your OWN watch.

But your own watch can only be used to assign coordinates to events on your own worldline, an accelerating observer could come up with multiple non-inertial coordinate systems which all agree that the coordinate time between events on his worldline is equal to his own proper time between those events, but which disagree about the time between events far from him or about simultaneity of distant events. There is no physical reason to think any of these non-inertial coordinate systems more accurately represents his "own measurements" of time than any other.

The spatial coordinate, with that choice, correponds to the lengths as reported by your OWN rulers and measuring tapes.

For an inertial observer "their own" ruler presumably means an inertial ruler at rest relative to themselves. But what does it mean for an accelerating observer? Does it mean that they are using an accelerating ruler, and if so how to decide how parts far from them are accelerating (is the ruler accelerating in a Born rigid way for example?) Or are you supposing that at each moment they define "their own" ruler to be a different inertial ruler which is instantaneously at rest relative to themselves at that moment?

In those alternative coordinate systems, your "clocks" would behave in very odd ways, compared to ordinary clocks. And likewise for your "rulers".

Again this is meaningless if you haven't defined precisely what "your clocks" and "your rulers" means for an accelerating observer, and why you think the accelerating observer doesn't have a choice of how clocks far from his own worldline should move or be synchronized with one another, and likewise why he doesn't have a choice about how points on his own (accelerating?) ruler far from his own worldline should themselves be moving (remember that in SR there is no simple notion of how the back end of a 'rigid ruler' should accelerate if we know how the front end is accelerating, since there is no well-defined notion of 'rigidity' for objects undergoing arbitrary acceleration, physical rulers will behave like silly putty or slinkys when you accelerate one end)

 

[Dale]

It's not a matter of physical necessity that you refrain from constantly hitting yourself in the head with a hammer, either.
...
I've spent a lot of hours flying airplanes and gliders, constantly needing to avoid mid-air collisions. In all that time, I've NEVER used the center of mass coordinate system to help me avoid mid-air collisions.

Very cute. Completely irrelevant to the topic, but cute.

So Mike, are you going to continue to dodge the challenge and hide from the issue? After a dozen or so requests you have had plenty of opportunity but you still can't even define your terms let alone demonstrate your claim. I suspect that you know that your claim is wrong.

 

[ghwellsjr]

I think Mike has defined his terms "elementary measurements" and "elementary calculations". Look at this post:

If the traveler is perpetually inertial, then he certainly DOES have a "preference" for one particular inertial frame. There is only one inertial frame that agrees with the elementary measurements and elementary calculations that he makes, using clocks and rulers with which he is stationary. That preferred inertial frame is the one in which he is stationary.

And now this one:

The well-known results of time-dilation and length-contraction are directly available for your use (when you are an inertial observer), provided that you choose to use the usual Lorentz coordinates, in an inertial frame in which you are stationary. And the time coordinate, with that choice, corresponds to the time shown on your OWN watch. The spatial coordinate, with that choice, correponds to the lengths as reported by your OWN rulers and measuring tapes. There is a REASON why Einstein chose those coordinates, when developing his special theory.

It looks like Mike thinks that Einstein was claiming that you have to use the inertial frame in which you are stationary to make SR work.

Then, somehow, he slips from a perpetually inertial frame to a non-inertial frame when the traveler starts his voyage:

No, I use a SINGLE reference frame for the traveler during his entire voyage. I call that single reference frame the "CADO" or "{MSIRF(t)}" reference frame ... it is NOT an inertial frame.

And I'm sure he really believes this is what Einstein was promoting.

 

[Grimble]

This maybe where you are going wrong. When the traveller reaches the turnaround point he is momentarily, or for any desired length of time, at rest with respect to the stay at home and their clocks will be ticking at the same rate as each others as they are at rest with respect to each other.

The same is true when the traveller returns and halts next to the stay at home, they are at rest in the same inertial frame and so their clocks tick at the same rate as each other. And of course at that stage it is meaningful to compare clocks, and although they are now once again ticking at the same rate,

But yes I agree with what you are saying here! So where does this show that I am going wrong?

the traveller's clock shows less accumulated time.

Now this is where you lose me, What has SR and transforming measurements with LT possibly got to do with ACCUMULATED time?

LT is used to transform measurements: what a clock currently reads, not a calculation of the difference between two readings on a clock.

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.

 

[Dale]

It looks like Mike thinks that Einstein was claiming that you have to use the inertial frame in which you are stationary to make SR work.

Which is a complete misunderstanding of SR, particularly the first postulate. I think it is obvious to everyone besides Mike.

 

[matheinste]

i.e. When the traveller turns round he first comes to rest with respect to the home twin. At this moment their ages, as shown by LT, will once again be equal. Then the traveller returns and during that return journey their ages and apparent ages will differ once again but when the traveller has returned and come to rest once again they will have identical ages once again.

It is only if the traveller keeps on traveling past the home twin on his return that the age difference is apparent and it will be apparent to each of them.

Grimble

"when the traveller has returned and come to rest once again they will have identical ages once again."

"It is only if the traveller keeps on traveling past the home twin on his return that the age difference is apparent and it will be apparent to each of them."

These two statements are incorrect.

Perhaps I have missed something from your previous posts that puts these statements in context, if so forgive me and point them out.

Don't forget you cannot use one LT for the whole of the journey for the travellers non inertial frames.

Matheinste

 

[ghwellsjr]

Now this is where you lose me, What has SR and transforming measurements with LT possibly got to do with ACCUMULATED time?

LT is used to transform measurements: what a clock currently reads, not a calculation of the difference between two readings on a clock.

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 that there has only ever been one velocity relative to the frame of reference. In this case, you can use LT to calculate both the difference in clock readings between the stationary one and the moving one, and/or you can calculate the relative tick rate between the two clocks. If you then accelerate the moving clock, you have to be very careful about how you describe what is happening to get a precise unique result. Usually, people just take a short cut and assume that the accelerating clock is gradually changing to its new tick rate as calculated by LT. Then to get an idea of the difference in the clock readings, you have to integrate, or multiply the time interval that a clock was ticking at a particular rate by its tick rate to find the accumulated time at the point of acceleration. And remember, this is all being done from the one frame of reference of the stationary observer (or any other single frame of reference).

In other words, if there were only one velocity involved, you could use LT directly to calculate the difference in clock readings OR you could use the tick rate multiplied by the time interval and you will get the same answer but when you introduce a new velocity, you cannot use the former method to calculate the difference in clock readings, you have to use the later twice, once for each time interval that the clock was moving at each speed and then add them together.

 

[Dale]

The Lorentz Transform assumes that there has only ever been one velocity relative to the frame of reference.

Careful here. The velocity referred to here is the velocity between two inertial reference frames, not the velocity of some object in a given inertial frame. I am sure that you understand, but some people may misread that and think that you meant that the Lorentz transform cannot be used if an object is accelerating.

 

[hprog]

hprog, are you comfortable with the concept now or are you still confused?

Yes I am still confused.
Let me explain what I see here.
First let me say two principles I am taking out of the Special Relativity.
1) Any object in the same frame of reference - even if they are very far apart from each other - must agree on the fact that only one of them is younger, even if we have no clue who of them.
2) Any object next two each other must basically agree who is younger even if they are in different frames of reference.

Now let A and B move away in linear motion, A and B are far away and you claim that both can claim to be younger.
Now let's have C - which is using the same of frame of reference as A - next to B, and B and C will agree that they are the same age.
Yet since A and C must agree that only one of them is younger - even if they are far away -it follows that B and A must also agree that only one of them is younger, even if they are far away.

How is this fitting together with SR?

 

[Dale]

How is this fitting together with SR?

The presence of object C doesn't change the comparison of A and B. The relativity of simultaneity ensures that everything works out correctly.

The easiest way to visualize this is to draw a spacetime diagram and use the Lorentz transform to label the A- and B-frame coordinates.

 

[Mike_Fontenot]

Some members of this forum have expressed the opinion that the coordinates used, in an inertial reference frame, can be chosen in many different ways, and that all such choices are equally good.

The first part of that opinion is true. The second part isn't.

To anyone who's opinion includes that second part, I've got these questions for you.

We have all seen how the Lorentz equations (which relate the coordinates in one inertial frame to the coordinates in another inertial frame) are derived ... any textbook on special relativity will give a derivation.

1) Have you ever seen a derivation, of the equations relating two inertial frames, that use any coordinates OTHER THAN the "standard Lorentz coordinates"? (By "standard Lorentz coordinates", I mean the case where one of the four coordinates is a TIME coordinate consisting of the readings on ordinary clocks that are stationary in the given inertial frame, and the other three coordinates are SPATIAL coordinates consisting of the readings on ordinary measuring tapes that are stationary in the given inertial frame).

2) Have you ever seen equations, relating two inertial frames, WRITTEN out, and/or USED in actual calculations, where the coordinates in the equation are OTHER THAN the standard Lorentz coordinates?

3) Have you ever seen the well-known time-dilation result (that everyone has heard about, and probably often used), expressed in terms of coordinates OTHER THAN the standard Lorentz coordinates?

4) Have you ever seen the similarly well-known length-contraction result expressed in terms of coordinates OTHER THAN the standard Lorentz coordinates?

If you took a poll of all work-a-day physicists, asking them the above questions, I think you'd get very few, if any, "Yes" answers. Wonder why not?

To any forum members who answer "yes" to any of those questions: Pick one of those equally-good sets of coordinates for the two inertial frames (different in non-trivial ways from the standard Lorentz coordinates), and, using those coordinates, write out the equations relating the two inertial frames, and then state the time-dilation and length-contraction results, in terms of those coordinates. State how the postulates of special relativity would be specified in those coordinates. And then, since those coordinates are equally good, maybe you should consider writing an entire textbook that exclusively uses those coordinates.

Mike Fontenot

 

[matheinste]

Some members of this forum have expressed the opinion that the coordinates used, in an inertial reference frame, can be chosen in many different ways, and that all such choices are equally good.

Mike Fontenot

For good read valid.

Matheinste.

 

[JesseM]

Some members of this forum have expressed the opinion that the coordinates used, in an inertial reference frame, can be chosen in many different ways, and that all such choices are equally good.

That doesn't make any sense, how can the coordinates "in an inertial reference frame" be chosen in many different ways? Given an inertial observer, if you want to construct an inertial frame where that observer is at rest there is a standard procedure for doing so, your only choices are where to put the spacetime origin and what directions to orient the spatial axes. What people are arguing is that if you want to define a non-inertial frame in which the accelerating twin has a constant position coordinate, then you have an infinite variety of choices about how to do so, and they are all equally good. No one is disputing that inertial frames have a privileged role in SR, and that standard equations like the time dilation equation and the length contraction equation only are guaranteed to work in inertial frames. But your CADO equations define a non-inertial frame for any given accelerating observer, one where standard SR equations like the time dilation equation won't work. the thing everyone is arguing with you is that there is nothing that makes this particular non-inertial frame more physically "correct" than other non-inertial frames one could define for the accelerating observer.

 

[Dale]

Mike, it is not our turn to answer yet another ill-informed challenge from you, we have done that enough times already. It is your turn to answer my challenge first, and stop dodging the issue. Define your terms and demonstrate that there is any measurement where two different coordinate systems disagree on value of the predicted result.

Btw, the fact that you have not seen examples of 1-4 above clearly indicates that you should spend more time reading and learning and less time advertising your CADO.

 

[Grimble]

"when the traveller has returned and come to rest once again they will have identical ages once again."

"It is only if the traveller keeps on traveling past the home twin on his return that the age difference is apparent and it will be apparent to each of them."

These two statements are incorrect.

Perhaps I have missed something from your previous posts that puts these statements in context, if so forgive me and point them out.

Yes I agree that they are not what you were taught but in what way are they incorrect?
LT shows that when one is traveling the measurements vary as a function of their relative velocity. When they complete that part of the journey they are at rest and their relative velocity is zero and their measurements MUST then be identical once again.

Don't forget you cannot use one LT for the whole of the journey for the travellers non inertial frames.

Matheinste

But I am not doing, am I? If you read my statements I am considering two separate elements of travel. When each element completes when the two parties are at rest.

 

[Grimble]

The Lorentz Transform assumes that there has only ever been one velocity relative to the frame of reference.

No, I am sorry but I don't see that. The Lorentz transform assumes nothing and has no need to assume anything about the constancy of the relative velocity.

It is transforming the current measurements from one frame of reference to another, It is, surely, calculating what the effect of the relative velocity has upon the measurements from a moving FoR.

After all, according to the first postulate, time will be passing at identical rates, within any Inertial Frames of reference. Any clock at rest in an Inertial FoR will be measuring Proper Time according to an adjacent observer also at rest in the FoR

In this case, you can use LT to calculate both the difference in clock readings between the stationary one and the moving one, and/or you can calculate the relative tick rate between the two clocks.

Yes.

If you then accelerate the moving clock, you have to be very careful about how you describe what is happening to get a precise unique result. Usually, people just take a short cut and assume that the accelerating clock is gradually changing to its new tick rate as calculated by LT. Then to get an idea of the difference in the clock readings, you have to integrate, or multiply the time interval that a clock was ticking at a particular rate by its tick rate to find the accumulated time at the point of acceleration. And remember, this is all being done from the one frame of reference of the stationary observer (or any other single frame of reference).

But WHY are you involving acceleration? LT has not and cannot have anything to do with acceleration, or did I miss that in the equations?
LT is describing how to covert(transform) measurements from one FoR to another. It is solely concerned with a unique moment in time. It is instantaneous.
If we were to use LT to transform the measurements whilst the traveling body was accelerating we could do. It is only using the CURRENT velocity!
Integration etc. is an interesting exercise but is irrelevant in this case.

In other words, if there were only one velocity involved, you could use LT directly to calculate the difference in clock readings OR you could use the tick rate multiplied by the time interval and you will get the same answer but when you introduce a new velocity, you cannot use the former method to calculate the difference in clock readings, you have to use the later twice, once for each time interval that the clock was moving at each speed and then add them together.

No, 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.

Why does modern thinking add such a load of baggage onto a simple clear principle?

It is quite simple, after all, to show that while the stationary twin will read a different time from the traveller's clock (LT), he would read the SAME time if he measured the traveller's time with his own clock (or one at rest in his own FoR and synchronised with his clock).

 

[Dale]

"when the traveller has returned and come to rest once again they will have identical ages once again."

"It is only if the traveller keeps on traveling past the home twin on his return that the age difference is apparent and it will be apparent to each of them."

These two statements are incorrect.

Yes I agree that they are not what you were taught but in what way are they incorrect?

matheinste is correct. The Lorentz transform shows the time on a system of clocks synchronized using the Einstein synchronization convention. When the traveling twin returns to rest his clock is no longer synchronized with the stay-at-home clocks.

 

[matheinste]

Yes I agree that they are not what you were taught but in what way are they incorrect?
LT shows that when one is traveling the measurements vary as a function of their relative velocity. When they complete that part of the journey they are at rest and their relative velocity is zero and their measurements MUST then be identical once again.

.

At the end of the journey, when the two observers are at rest with respect to each other, each will observe the other's clock ticking rate and ruler lengths to be the same as each other's. Their clocks' accumulated times will not be the same, assuming the usual twin scenario. To say that the accumulated time is the same for each, when re-united, is analogous to saying, loosely, that their spatial distances traveled is the same when they are reunited, which it clearly is not. You seem to be mixing up the units of measurement, i.e. tick rates, or time intervals between ticks, and meter lengths, with accumulated time and accumualted spatial distance.

I find it strange that after so many replies by myself and others in this, and other threads, that you do not understand the phenomenon of differential ageing. The fact that you have stuck at it so long indicates a certain determination to learn. Why not begin again without any preconceptions or appeals to intuition. It really is not that difficult.

Matheinste.

 

[Mike_Fontenot]

For good read valid.

Your concise response is very insightful, profound, and thought-provoking.

Is DOES seem clear that ANY one-to-one (not necessarily linear) mapping of the standard Lorentz coordinates, for an inertial observer, CAN be used for his coordinates, if desired. They may not be at all desirable or practically useful, but are they "invalid"?

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".

Mike Fontenot

 

[Mike_Fontenot]

Some members of this forum have expressed the opinion that the coordinates used, in an inertial reference frame, can be chosen in many different ways, and that all such choices are equally good.

That doesn't make any sense, how can the coordinates "in an inertial reference frame" be chosen in many different ways?

I could have worded that better, but your next comment indicates that you nevertheless DID correctly understand the MEANING of what I said (even though you disagreed with WHAT I said, of course). A better way to have worded my statement would have been,

"Some members of this forum have expressed the opinion that the coordinates used, for an inertial reference frame in which some given perpetually inertial observer is stationary, can be chosen in many different ways, and that all such choices are equally good."

Now, regarding your response to that statement:

Given an inertial observer, if you want to construct an inertial frame where that observer is at rest there is a standard procedure for doing so, your only choices are where to put the spacetime origin and what directions to orient the spatial axes.
[...]

DaleSpam disagrees with you, in his post #57 that followed your post. Why don't you two guys get on the same page (in regard to the degree of choice available in the coordinates used in perpetually inertial reference frames), so that I won't have to argue with both of you.

Mike Fontenot

 

[Mike_Fontenot]

[...]
Btw, the fact that you have not seen examples of 1-4 above clearly indicates that you should spend more time reading and learning and less time advertising your CADO.

JesseM disagrees with you, in his post #56 that preceded your post. Why don't you two guys get on the same page (in regard to the degree of choice available in the coordinates used in perpetually inertial reference frames), so that I won't have to argue with both of you.

Mike Fontenot

 

[JesseM]

I could have worded that better, but your next comment indicates that you nevertheless DID correctly understand the MEANING of what I said (even though you disagreed with WHAT I said, of course). A better way to have worded my statement would have been,

"Some members of this forum have expressed the opinion that the coordinates used, for an inertial reference frame in which some given perpetually inertial observer is stationary, can be chosen in many different ways, and that all such choices are equally good."

Unless you're just talking about the choice of where to place the spacetime origin or how to orient the spatial axes, no one has said you have any "choice" in how to construct an inertial frame, the subject under discussion has always been non-inertial frames which is what your CADO equations deal with.

Given an inertial observer, if you want to construct an inertial frame where that observer is at rest there is a standard procedure for doing so, your only choices are where to put the spacetime origin and what directions to orient the spatial axes.

DaleSpam disagrees with you, in his post #57 that followed your post.

I think you are having reading comprehension problems, there was nothing in DaleSpam's post #57 that contradicted my comment above. Perhaps you can explain what particular comment of DaleSpam's you think conflicted with my comment about the lack of choice in constructing inertial frames? DaleSpam didn't even talk about inertial frames in post #57!

 

[Dale]

Why don't you two guys get on the same page

Both of us are on the same page in that you need to define your terms and justify your unsubstantiated claim that different coordinate systems predict different results for measurements. This is just your latest attempt to dodge that key issue.

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".

I would encourage you to go ahead and do as you suggest here, which is what I have been requesting for dozens of posts on multiple threads now.

 

[JesseM]

Dale, just so Mike is clear on this point, do you agree that this is true in flat spacetime?

Given an inertial observer, if you want to construct an inertial frame where that observer is at rest there is a standard procedure for doing so, your only choices are where to put the spacetime origin and what directions to orient the spatial axes.

 

[Grimble]

matheinste is correct. The Lorentz transform shows the time on a system of clocks synchronized using the Einstein synchronization convention.

Yes...

When the traveling twin returns to rest his clock is no longer synchronized with the stay-at-home clocks.

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?

 

[Grimble]

You seem to be mixing up the units of measurement, i.e. tick rates, or time intervals between ticks, and meter lengths, with accumulated time and accumualted spatial distance.

Can you please show me where in the Lorentz transformation eqations there is any use of terms representing accumulated measurements?
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.

I find it strange that after so many replies by myself and others in this, and other threads, that you do not understand the phenomenon of differential ageing. The fact that you have stuck at it so long indicates a certain determination to learn. Why not begin again without any preconceptions or appeals to intuition. It really is not that difficult.

Matheinste.

I am so sorry that you still believe that I don't understand exactly what you are saying. I do and I have from the moment I first came across these concepts. It is simple and straightforward.

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.

Simple logic leads to this conclusion however one thinks about it. Simple logic. Every explanation that says different has holes in it big enough to drive a bus through.

But those holes are obscured by assumptions that are seemingly accepted without any thought by everyone who understands differential ageing.

Saying go away and try again.

I have NO PRECONCEPTIONS. That is precisely the point. The accepted explanations are the ones with preconceptions that are unfounded.

Grimble

 

[JesseM]

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?

It is the acceleration, but it's not that anything special happens at the moment of acceleration, rather it's a question of the overall geometry of each clock's path through spacetime. Consider the fact that in 2D Euclidean geometry, if you have two dots A and B and two paths between them, one of which is a straight line while the other consists of two straight segments with a bend in the middle (like a V-shape), the straight path will always have a shorter length than the bent path, because a straight line is always the shortest distance between points in Euclidean geometry. So in that sense you could say the reason the bent path has a greater length is because of the bend (a change in slope as seen in any Cartesian coordinate system, analogous to a change in velocity in spacetime). But if you measure the length of each path by having a car with the odometer running driving along each path, nothing special happens to the odometer reading at the moment the car passes the bend, it doesn't suddenly jump forward by a huge amount or anything like that, the fact that this car will have significantly greater odometer reading when it reunites with the other car at point B is a function of the overall shape of the path.

In fact the geometric analogy between lengths of paths in 2D space and elapsed time on clocks in spacetime can be made fairly precise if you treat Cartesian coordinate systems on the plane as analogous to inertial spacetime coordinate systems in SR. If you draw a 2D cartesian coordinate system with xy axes on the plane where those two paths between A and B are drawn, and you again imagine cars with odometers running driving along both paths, then at any point on the path you can talk about the rate that the odometer reading of the car is increasing, not relative to time but just relative to an increase in the x-coordinate. For example, if you have a path defined by the equation y = (3/4)*x, then if you pick two points where the difference in x-coordinate, $\Delta x$, is 4 (such as x=0 and x=4), then the difference in the corresponding y-coordinate, $\Delta y$, must be 3 (in this case going from y=0 to y=3), so by the Pythagorean theorem the car's odometer must have increased by $\sqrt{\Delta x^2 + \Delta y^2 } = \sqrt{4^2 + 3^2} = 5$. If we define the slope of a path as $S = \Delta y / \Delta x$, then the amount of odometer increase as a function of increase in x can be written as $\sqrt{\Delta x^2 + \Delta y^2 } = \sqrt{\Delta x^2}*\sqrt{1 + \Delta y^2 / \Delta x^2 } = \Delta x * \sqrt{1 + S^2 }$.

This is directly analogous to the time dilation equation in relativity if we replace "amount of odometer increase relative to increase in the x-coordinate" with "amount of proper time increase as a function of the t-coordinate", and "slope = $\Delta y / \Delta x$ with "speed = $\Delta x / \Delta t$". In the case of relativity, if an object is moving at constant speed v = dx/dt, then the amount of proper time increase $\Delta \tau$ as a function of increase in t-coordinate $\Delta t$ is given by $\Delta \tau = \Delta t * \sqrt{1 - v^2 }$ (actually the full formula would be $\Delta t * \sqrt{1 - v^2 / c^2}$ but it is common to use units where c=1, like distance in light-years and time in years). You can see that this is very similar to the formula for odometer increase as a function of increase in x-coordinate, except that this formula subtracts v whereas that formula added the slope S, which has to do with the fact that the formula for distance in a plane is given by the Pythagorean theorem $\sqrt{\Delta x^2 + \Delta y^2 }$ whereas the formula for proper time along a path in spacetime is given by $\sqrt{\Delta t^2 - (1/c^2)*\Delta x^2 }$ (this difference is key to why a straight line in space minimizes the distance between points, whereas a straight path through spacetime maximizes the proper time between events).

And just as we can pick different inertial frames where the velocities of the two twins differ, and therefore their rate of time dilation differ too, so if we have two paths in a 2D plane we can draw different possible Cartesian coordinate axes on the plane, with their x and y axes oriented differently, and they will disagree on the "rate of odometer increase relative to x-coordinate" for a given car on a given section of the path it's driving along. In different Cartesian systems, the slope of a path at any given point will be different, and thus the amount of odometer increase along that path will increase by different amounts for a given increment of the x-coordinate according to the equation $\Delta x * \sqrt{1 + S^2 }$. In fact for any given point on a path, you can pick a coordinate system where the x-axis is perfectly parallel to the path at that point, so the slope S will be 0 and the odometer will increase by the same amount as the increase in x-coordinate; this is analogous to the fact that in relativity, for any given point on a path through spacetime, you can pick an inertial frame where a clock moving along that path has a velocity of 0 at that point, and in this frame the clock will tick forward by the same amount as the increase in t-coordinate as opposed to running slower as it would in a frame where its velocity was nonzero.

Another point is that since the distance along any straight segment with slope S and difference in x-coordinate between endpoints $\Delta x$ is given by the formula $\Delta x * \sqrt{1 + S^2 }$, this means that if you have a curvy path whose slope S(x) = dy/dx is varying continuously as you vary x, then you can treat the path as a series of infinitesimal straight segments with difference in x-coordinate dx, and thus the total length of the path between two points with x-coordinates x₀ and x₁ can be computed with this integral:

$\int_{x_0}^{x_1} \sqrt{1 + S(x)^2} \, dx$

The length of a path doesn't depend on which particular coordinate system you use, you can pick different Cartesian coordinate systems where the x and y axes are oriented differently and S(x) (the slope as a function of x-coordinate) is different (along with the coordinates x₀ and x₁ of the endpoints of the curve), but they will all give the same result for the integral above, and they will all agree that a straight line path between two points (one where S(x) is a constant in any Cartesian coordinate system) is shorter than a non-straight path (one where S(x) varies in any Cartesian coordinate system) between the same two points.

Similarly, in any inertial frame, if you know the velocity as a function of time v(t) for any path through spacetime, and you know the coordinate times t₀ and t₁ of the endpoints in that frame, then the elapsed proper time on a clock which follows that path will be given by:

$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2 / c^2 } \, dt$

In different inertial frames v(t) will look different, but just like with the geometrical formula, this formula will give the same result for total elapsed time no matter what inertial frame you use. And it will always be true that if you have two paths which meet at the same endpoints, then a path with constant v(t) will have a greater total elapsed proper time than a path with changing v(t) (a path involving accelerations). Nevertheless, you can see that the formula itself involves only velocity, it doesn't involve the rate of change of velocity! Similarly the geometrical formula involved only slope at each point and didn't involve the rate of change of slope, but nevertheless it's very intuitive geometrically that a path with unchanging slope will have a shorter length than a path with changing slope, since we known unchanging slope means a straight line and in Euclidean geometry a straight line is the shortest distance between points.