I Proper (and coordinate) times re the Twin paradox

[Battlemage!]

On the other hand, why should "I still respect the dogma" necessarily mean "I believe the dogma?"

 

[Grimble]

No, the "same physical location" implies a frame which identifies locations and determines if they are the same or not. However, what you can say is that the two events are timelike separated, which implies that there exists a reference frame where they have the same location.

OK, and a worldline is a succession of events that have a unique set of coordinates in each frame?

So, Proper time is the time measured between two points on the world line of a clock and the coordinates of those points can be different in each frame. But the time displayed by the clock is the same wherever it is viewed from, it is the measure of the time that has passed on its worldline between those two po

The interval between our two points is therefore different for each frame it is viewed from, being a function of the time displayed on the clock, and the spatial displacement in that frame.

The Spacetime interval is different because the spatial coordinate difference is excluded from the function - s² = (ct)[/SUP]2[/SUP] - x²

Therefore the Spacetime Interval is the time displayed on the clock in the frame where the clock is at rest; the clock's own frame.

 

[jbriggs444]

OK, and a worldline is a succession of events that have a unique set of coordinates in each frame?

A worldline is a continuous succession of events. Full stop. These events will generally have different coordinates in different frames.

There is no guarantee of uniqueness. The coordinate [0,0,0,0] in two different frames might happen to describe the same event on the same worldline.

So, Proper time is the time measured between two points on the world line of a clock and the coordinates of those points can be different in each frame. But the time displayed by the clock is the same wherever it is viewed from, it is the measure of the time that has passed on its worldline between those two po

The interval between our two points is therefore different for each frame it is viewed from, being a function of the time displayed on the clock, and the spatial displacement in that frame.

Here you are using "interval" to mean "difference between time coordinates for the starting and ending events on the worldline". Do not do that. Do not impose your own idiosyncratic meaning on words that already mean something else.

 

[Dale]

and a worldline is a succession of events that have a unique set of coordinates in each frame?

Yes, although this is not the defining feature, it is true.

Proper time is the time measured between two points on the world line of a clock and the coordinates of those points can be different in each frame. But the time displayed by the clock is the same wherever it is viewed from, it is the measure of the time that has passed on its worldline between those two po

yes

The interval between our two points is therefore different for each frame it is viewed from, being a function of the time displayed on the clock, and the spatial displacement in that frame.

Usually the unqualified word "interval" refers to the spacetime interval which is invariant. I think that you mean the coordinate time difference, which is usually not described using the word "interval" in order to avoid confusion.

The Spacetime interval is different because the spatial coordinate difference is excluded from the function - s2 = (ct)[/SUP]2[/SUP] - x2

If by excluded you mean subtracted then that is essentially true. If you mean something else then please clarify.

 

[Grimble]

A worldline is a continuous succession of events. Full stop. These events will generally have different coordinates in different frames.

There is no guarantee of uniqueness. The coordinate [0,0,0,0] in two different frames might happen to describe the same event on the same worldline.

unique set of coordinates

note: I specified a 'unique set of coordinates', not a 'set of unique coordinates'...

Here you are using "interval" to mean "difference between time coordinates for the starting and ending events on the worldline". Do not do that. Do not impose your own idiosyncratic meaning on words that already mean something else.

The interval between our two points is therefore different for each frame it is viewed from, being a function of the time displayed on the clock, and the spatial displacement in that frame.

Yes, I was using the word interval (unqualified) to mean an interval ( normal English usage). I certainly was not using it to mean

difference between time coordinates for the starting and ending events on the worldline

as I explicitly specified

time displayed on the clock, and the spatial displacement in that frame.

 

[jbriggs444]

note: I specified a 'unique set of coordinates', not a 'set of unique coordinates'...

Neither of which makes it clear what you consider to be unique. Possibly you simply meant that given an event and a coordinate system, there is a one to one mapping between coordinate tuples and events. One event per tuple and one tuple per event. If so then the word "unique" conveyed no useful meaning. That sort of uniqueness is taken for granted.

Yes, I was using the word interval (unqualified) to mean an interval ( normal English usage).

If you are using "interval" in a normal English sense then you owe it to us to define for us what that means in a scientific sense. We cannot know what specific meaning you intend by using the term.

Edit to add:

You responded to me stating that you used the word "interval"...

I certainly was not using it to mean
difference between time coordinates for the starting and ending events on the worldline
as I explicitly specified

and to @Dale stating that

I meant the difference between the coordinates of the two points as measured from another frame

That seems contradictory. Can you clarify?

 

[Grimble]

Usually the unqualified word "interval" refers to the spacetime interval which is invariant. I think that you mean the coordinate time difference, which is usually not described using the word "interval" in order to avoid confusion.

I meant the difference between the coordinates of the two points as measured from another frame; would it be correct to refer to that as the coordinate time difference?

If by excluded you mean subtracted then that is essentially true. If you mean something else then please clarify.

Yes, sorry, I did mean subtracted...

I must apologise for the way I refer to things with unscientific word usage, but I stopped studying physics in university in 1978, which was a while ago... and the correct usage can be a bit tricky.

 

[Herman Trivilino]

So Minkowski was wrong?

I don't see how the information you posted implies that as a conclusion. Respecting a dogma means you respect something that others believe to be true.

Time and space can still be regarded as separate entities even if the time and space coordinates are different in different frames.

 

[Herman Trivilino]

The Spacetime interval is different because the spatial coordinate difference is excluded from the function - $s^2 = (ct)^2 - x^2$.

No, the spacetime interval is the same because the spatial coordinate difference is excluded from the function - $s^2 = (ct)^2 - x^2$ whenever $x=0$. In those cases we would call both $s$ and $ct$ the proper time.

Therefore the Spacetime Interval is the time displayed on the clock in the frame where the clock is at rest; the clock's own frame.

Correct.

I must apologise for the way I refer to things with unscientific word usage, but I stopped studying physics in university in 1978, which was a while ago... and the correct usage can be a bit tricky.

If you are here to learn then there's no need to apologize for making mistakes. But realize that those errors will often get corrected by other posters, especially if they are being used to draw false conclusions, and even more especially if those false conclusions are being presented as "corrections" to valid information posted by others.

 

[Dale]

I meant the difference between the coordinates of the two points as measured from another frame; would it be correct to refer to that as the coordinate time difference?

Yes, but don't forget that when you specify a coordinate time difference in needs to be clear which coordinate system is being used.

I must apologise for the way I refer to things with unscientific word usage, but I stopped studying physics in university in 1978, which was a while ago... and the correct usage can be a bit tricky.

I understand, the concepts are so specific that even subtle terminology changes can drastically alter the intended meaning. That is why a lot of questions get seemingly contradictory answers, so it is good to check on the intended meaning

 

[Grimble]

May I thank everyone for being so patient and forbearing with me, it is very much appreciated.

 

[Grimble]

Neither of which makes it clear what you consider to be unique.

The way I understand English the difference between - "a unique set of coordinates" and "a set of unique coordinates" it is the placement of the word unique that is important:
When I say . "...unique set..." it is the set of coordinates that is unique; while putting 'unique' next to coordinates implies that I am referring to a set of coordinates where each coordinate is unique, whether that be unique in that set (which has to be true) or is unique amongst coordinates from any set.

 

[Grimble]

If you are using "interval" in a normal English sense then you owe it to us to define for us what that means in a scientific sense. We cannot know what specific meaning you intend by using the term.

Yes, I see this, but it isn't easy to try and do that for every word that may be used as so many everyday English terms have specific Scientific meanings.

It can be difficult to try and find the words to express what one means without misusing some term or other...

 

[jbriggs444]

The way I understand English the difference between - "a unique set of coordinates" and "a set of unique coordinates" it is the placement of the word unique that is important:
When I say . "...unique set..." it is the set of coordinates that is unique; while putting 'unique' next to coordinates implies that I am referring to a set of coordinates where each coordinate is unique, whether that be unique in that set (which has to be true) or is unique among coordinates from any set.

In context, we had a set of events, a set of coordinate systems, and a set of coordinate tuples, each of which has four "coordinates". There are a lot of ways to have meant "unique". More than the two that one word placement can distinguish between.

 

[Grimble]

Now, to go back to an earlier thought experiment

Think of proper time as something that we observe: Say we design our clock so that every time it ticks it punches a hole in a piece of paper somewhere inside; we start with a fresh piece of paper at event A and remove it at event B. How many holes are there in the piece of paper? That's a simple direct observation; all observers everywhere will agree about the answer without any rigamarole about reference frames or relative velocity or time dilation. We call the number of holes in the piece of paper "the proper time along the path from A to B", and it is a fact that has nothing to do with any other observers and their notions of time, distance and speed.

please shew me what is wrong with my logic here...

Twin B is at rest in his inertial frame of reference. After 10 seconds 10 holes will be punched in his paper.
Twin B is at rest in an inertial frame of reference and measures 10 seconds proper time.
After 10 seconds, Twin A, also at rest in her inertial frame of reference, has 10 holes punched in her paper and measures 10 seconds proper time.

Now the Twins are separating at 0.6c.

Each twin will measure the other twin who is moving at 0.6c to be time dilated and measure γt = 12.5 seconds (γ = 1.25, t = 10) to have passed (coordinate time?), for the other traveling twin's clock, yet still count only 10 holes in their paper...

 

[PeroK]

Now, to go back to an earlier thought experimentplease shew me what is wrong with my logic here...

Twin B is at rest in his inertial frame of reference. After 10 seconds 10 holes will be punched in his paper.
Twin B is at rest in an inertial frame of reference and measures 10 seconds proper time.
After 10 seconds, Twin A, also at rest in her inertial frame of reference, has 10 holes punched in her paper and measures 10 seconds proper time.

Now the Twins are separating at 0.6c.

Each twin will measure the other twin who is moving at 0.6c to be time dilated and measure γt = 12.5 seconds (γ = 1.25, t = 10) to have passed (coordinate time?), for the other traveling twin's clock, yet still count only 10 holes in their paper...

... back to square one, the basics of SR. No universal time, no universal simultaneity. Measuring the time of an event in one frame requires the time and position of the event in another frame.

 

[Dale]

How are you going to instruct them to punch holes in the paper? I.e. What is the stopping criteria. If your criteria is simply that they should each punch 10 holes then clearly they will each have 10 holes

Take a piece of paper and a ruler. Draw one line 10 cm long. From one end of that line, draw another line at an angle less than 45 deg, also 10 cm long. How long is each line?

 

[Grimble]

OK. With two twins let us specify the movement is measured along the mutual x axes, as is the convention in all such diagrams.

As for punching the holes, that was not my invention I borrowed that from Nugatory's post (#13). Let us say that each twin has a light clock with the mirror set at 0.5 light seconds from the light and that they punch a hole each time the light pulse returns.

 

[stevendaryl]

OK. With two twins let us specify the movement is measured along the mutual x axes, as is the convention in all such diagrams.

As for punching the holes, that was not my invention I borrowed that from Nugatory's post (#13). Let us say that each twin has a light clock with the mirror set at 0.5 light seconds from the light and that they punch a hole each time the light pulse returns.

Okay, I understand the scenario. You have two twins, each is punching one hole per second (according to their own lightclock). What's the question?

Let's assume that the twins start off together for their first hole punch. Then let's identify a few events:

• e_1: They each punch their first hole.
• e_{A2}: Alice (the first twin) punches her second hole.
• e_{B2}: Bob (the second twin) punches his second hole.

In a reference frame in which Alice is momentarily at rest, e_{A2} occurs before e_{B2}. In a reference frame in which Bob is momentarily at rest, e_{B2} occurs before e_{A2}. So what question are you asking?

 

[Dale]

As for punching the holes, that was not my invention I borrowed that from Nugatory's post (#13). Let us say that each twin has a light clock with the mirror set at 0.5 light seconds from the light and that they punch a hole each time the light pulse returns.

That isn't the question I asked. The question is how do they know when to stop punching holes? If you just instruct them to punch 10 holes and stop then of course both will have 10 holes.

 

[Vitro]

This is where it goes wrong:

...and measure γt = 12.5 seconds (γ = 1.25, t = 10) to have passed (coordinate time?), for the other traveling twin's clock...

That's not correct, each twin measures 12.5 seconds (coordinate time) to have passed on their own clock in order for the other twin's paper to show 10 holes (10 seconds of proper time).

The rule of thumb is: if you can measure it with a single clock then it's a proper time, if you need two (or more) clocks then it's a coordinate time. Alternatively, if you measure it at the same location it's a proper time, if you measure it at different locations it's a coordinate time.

 

[Nugatory]

The rule of thumb is: if you can measure it with a single clock then it's a proper time, if you need two (or more) clocks then it's a coordinate time. Alternatively, if you measure it at the same location it's a proper time, if you measure it at different locations it's a coordinate time.

And if either or both events don't happen at the location of the clock, it is always a coordinate time.

 

[Grimble]

That's not correct, each twin measures 12.5 seconds (coordinate time) to have passed on their own clock in order for the other twin's paper to show 10 holes (10 seconds of proper time).

I am sorry but I do not understand what you are saying here...

Surely each twin is measuring proper time on the clock they are holding in their reference frame; the coordinate time is that time, transformed by the Lorentz Transformation Equations, from proper time to coordinate time (multiplying it by gamma...)

While each twin, at rest in their inertial frame of reference will measure the 10 holes punched by their clocks in proper time?

 

[jbriggs444]

I am sorry but I do not understand what you are saying here...

Surely each twin is measuring proper time on the clock they are holding in their reference frame; the coordinate time is that time, transformed by the Lorentz Transformation Equations, from proper time to coordinate time (multiplying it by gamma...)

[emphasis mine]

The Lorentz transforms contain more than a multiplication by gamma. There is also a term for relativity of simultaneity.

 

[Grimble]

That isn't the question I asked. The question is how do they know when to stop punching holes? If you just instruct them to punch 10 holes and stop then of course both will have 10 holes.

I presume they will continue to punch holes until they are switched off.
The important point is that both clocks will punch 10 holes.
They are each at rest in an inertial frame of reference and so are keeping proper time for that clock. The clocks are identical, the laws of science are identical, the times measured will presumably be identical - what reason is there for them to be different?

It seems to me that if another clock C, were permanently mid way between A and B, then their relative velocities would be ^v/₂ and -^v/₂ with respect to the clock C. And C would measure the same time dilation for each A and B and the same length contraction for their frames(?)

 

[PeroK]

I presume they will continue to punch holes until they are switched off.
The important point is that both clocks will punch 10 holes.
They are each at rest in an inertial frame of reference and so are keeping proper time for that clock. The clocks are identical, the laws of science are identical, the times measured will presumably be identical - what reason is there for them to be different?

It seems to me that if another clock C, were permanently mid way between A and B, then their relative velocities would be ^v/₂ and -^v/₂ with respect to the clock C. And C would measure the same time dilation for each A and B and the same length contraction for their frames(?)

The basis of your argument is (assuming a third clock at C:)

In A's frame, A's clock reaching $10s$ coincides with C's clock reaching $8s$ (say).

In B's frame, B's clock reaching $10s$ coincides with C's clock reaching $8s$.

Therefore, in A's frame: A's clock reaching $10s$, B's clock reaching $10s$ and C's clock reaching $8s$ are all simultaneous. Hence, simultaneity is not relative and SR is wrong?

Although, given this, C's clock must also read $10s$ as well (just put another clock that stays half-way between A and C) and there's no time dilation either.

 

[Dale]

I presume they will continue to punch holes until they are switched off.
The important point is that both clocks will punch 10 holes.

As you have stated it they will both punch an infinite number of holes, not just 10.

It is important that you actually answer this question, not avoid it. They start punching holes when they are together, they each punch a hole when a local clock that they carry ticks 1 s, but how do they know when to stop?

 

[Ibix]

The clocks are identical, the laws of science are identical, the times measured will presumably be identical - what reason is there for them to be different?

Two rulers have their zero markings aligned but do not point in the same direction. The rulers are identical, the distances measured will presumably be identical - what reason is there for the 10cm marks to be in different places?

 

[Herman Trivilino]

OK. With two twins let us specify the movement is measured along the mutual x axes, as is the convention in all such diagrams.

Yes, but when you draw a spacetime diagram of the situation A's x-axis is not parallel to B's x-axis. The reason is because they are in relative motion. They can't each be present at the punching of the other's tenth hole if they were each present at starting of the other's clock.

 

[Ibix]

Two rulers have their zero markings aligned but do not point in the same direction. The rulers are identical, the distances measured will presumably be identical - what reason is there for the 10cm marks to be in different places?

To expand on this a bit - both rulers and clocks are devices for measuring intervals along lines in spacetime. Rulers can only measure spacelike intervals and clocks can only measure timelike intervals. But notice those likes. There is no unique direction in spacetime that is Time. There are a whole family of directions which are timelike. So, generally, a clock does not necessarily measure what I choose to call time anymore than rulers are restricted to measuring what I choose to call forwards or sideways.

Clocks always measure an interval, but this may be something I call a mix of time and distance. Just as a ruler always measures distance, but this may be something I call a mix of forwards and sideways. The only special thing about Minkowski space is that you can't map the direction you call the future onto any of the spatial directions (or vice versa) by rotation because of the way the geometry is defined.

 

[Grimble]

The basis of your argument is (assuming a third clock at C:)

In A's frame, A's clock reaching $10s$ coincides with C's clock reaching $8s$ (say).

In B's frame, B's clock reaching $10s$ coincides with C's clock reaching $8s$.

Therefore, in A's frame: A's clock reaching $10s$, B's clock reaching $10s$ and C's clock reaching $8s$ are all simultaneous. Hence, simultaneity is not relative and SR is wrong?

Although, given this, C's clock must also read $10s$ as well (just put another clock that stays half-way between A and C) and there's no time dilation either.

If Observer C measures the coordinate time for clock A to equal the coordinate time for clock B, when A and B are traveling at the same speed relative to C, then is this not measuring equal times for A's clock and B's clock. Are their Lorentz transformations not the same?

 

[Grimble]

As you have stated it they will both punch an infinite number of holes, not just 10.

It is important that you actually answer this question, not avoid it. They start punching holes when they are together, they each punch a hole when a local clock that they carry ticks 1 s, but how do they know when to stop?

I'm sorry but I do not understand why you are asking that. Clocks continue to work ad infinitum...
A clock does not have to stop to take a reading from it?

 

[stevendaryl]

I'm sorry but I do not understand why you are asking that. Clocks continue to work ad infinitum...
A clock does not have to stop to take a reading from it?

Could you state more precisely what your question is? You have two different observers, Alice and Bob, each punching a hole in a paper at the rate of once per second (according to their own clocks). What's your question about it? In Alice's rest frame, Bob is punching slower than Alice is. In Bob's rest frame, it's the other way around.

 

[Dale]

I'm sorry but I do not understand why you are asking that. Clocks continue to work ad infinitum...
A clock does not have to stop to take a reading from it?

No, but you do have to have some rule about when you are going to take the reading. That is what you need to consider. The clock ticks 1, 2, 3, ... 946737, ... What is the criteria used to determine which of those infinite numbers is the reading?

Please think this through, don't dismiss it.

 

[Grimble]

In context, we had a set of events, a set of coordinate systems, and a set of coordinate tuples, each of which has four "coordinates". There are a lot of ways to have meant "unique". More than the two that one word placement can distinguish between.

Something is unique or it is not unique. There is no half-way unique or partly unique...
A unique set of coordinates is unlike any other set of coordinates.
A set of unique coordinates is made up of a multitude of coordinates each of which is unique...
You are trying to determine what makes them unique.

 

[jbriggs444]

Something is unique or it is not unique. There is no half-way unique or partly unique...

Uniqueness is a relative property. It is a property of an item within a collection. If you do not specify the collection, you have not specified the property.

Backing up and making that relevant in context...

Suppose that you have a worldline composed of a continuous sequence of events all of which are timelike separated from one another. Suppose further that you have singled out a coordinate system covering that set of events. Then each position on the worldline corresponds to a unique event -- no other event is at that position on the worldline and no other position on the worldline is at that event. Further, each event has a unique coordinate. No other coordinate tuple denotes that event and no other event has that coordinate tuple.

But now suppose that no coordinate system has been singled out. There is still a one to correspondence between positions along the worldline and events. But there is no longer a unique correspondence between events and coordinate tuples. An event can be associated with many coordinate tuples since there are many possible coordinate systems. A coordinate tuple can be associated with many possible events since there are many possible coordinate systems. The property of uniqueness has been lost.

 

[PeroK]

If Observer C measures the coordinate time for clock A to equal the coordinate time for clock B, when A and B are traveling at the same speed relative to C, then is this not measuring equal times for A's clock and B's clock. Are their Lorentz transformations not the same?

The Lorentz Transformations are not the same, as A and B are traveling in opposite directions relative to C.

 

[Grimble]

The Lorentz Transformations are not the same, as A and B are traveling in opposite directions relative to C.

And how does that make a difference to the results?
Both A and B are moving away from C, either can be a positive or negative displacement depending on how the observer in C observes them...

 

[Grimble]

Uniqueness is a relative property. It is a property of an item within a collection. If you do not specify the collection, you have not specified the property.

Oh for goodness sake!
What I said was

...a worldline is a succession of events that have a unique set of coordinates in each frame?

.
and that means a set of coordinates that is different from any other set of coordinates in the same frame.
It does not matter how each coordinate is different from any other coordinate as I specifically referred to a unique set of coordinates.

 

[stevendaryl]

Could somebody please repeat what the question is?

 

[PeroK]

And how does that make a difference to the results?
Both A and B are moving away from C, either can be a positive or negative displacement depending on how the observer in C observes them...

It makes a difference to A and B. You have done a simple thought experiment:

At a certain time $t$ on C's clock, A's clock reads $t'$ and B's clock also reads $t'$. In any case, in C's frame A and B's clocks are synchronised.

Then you have made the intuitive assumption that A and B's clocks must be synchronised in each others frame.

This is a good thought experiment because:

If you understand the relativity of simultaneity, you will see the problem with this assumption.

If you don't understand the relativity of simultaneity, then you won't see the problem.

 

[Dale]

And how does that make a difference to the results?
Both A and B are moving away from C, either can be a positive or negative displacement depending on how the observer in C observes them...

If A is positive then B is negative and vice versa.

 

[Dale]

Could somebody please repeat what the question is?

Grimble does not have a well formed question that I can see and refuses to answer clarifying questions.

 

[Vitro]

Could somebody please repeat what the question is?

I'm not sure either, but I think he's making an argument that all clocks accumulate proper time at the same rate regardless of how they move relative to each other, and they only show different readings while in relative motion (because of the gamma factor) but if brought at rest in the same FoR they should always show the same (proper) time.

 

[PeterDonis]

Both A and B are moving away from C, either can be a positive or negative displacement depending on how the observer in C observes them...

Lorentz transformations don't work on displacements, they work on coordinates.

Here's my suggestion: pick an inertial frame, such as C's rest frame. Write down the coordinates of all of the events of interest in this frame, explicitly. Then write down the Lorentz transformation that goes from C's rest frame to A's rest frame. Then write down the (different!) Lorentz transformation that goes from C's rest frame to B's rest frame. Then transform the coordinates of all of the events of interest using each of these transformations, and write down the results.

Doing this will, first, help you clarify for yourself what the implications of your scenario are (I don't think you fully understand them), and second, help the rest of the posters in this thread understand what you are describing and what question you are asking.

If you are unable to complete the above exercise, then I strongly suggest closing this thread until you have taught yourself how to do so. Being able to do an exercise like the above is a basic skill in relativity, and if you don't have it, you shouldn't be posting an "I" level thread.

 

[sdkfz]

I'm not sure either, but I think he's making an argument that all clocks accumulate proper time at the same rate regardless of how they move relative to each other, and they only show different readings while in relative motion (because of the gamma factor) but if brought at rest in the same FoR they should always show the same (proper) time.

I believe you're correct. I think he wants "proper time" to be synonymous with "absolute time". (If it helps, I've been part of discussions (over many years, elsewhere) with Grimble where he strongly stuck to the claim that simultaneity is absolute, and the differing views of observers in relative motion is effectively just illusion. Any discussion with Grimble needs to account for his wanting to reject the "relative" part of "relativity", and to find something absolute underneath it all.)

 

[Herman Trivilino]

Ahhh... Now I think I understand the issue. 10 seconds of proper time for Twin A is the same as 10 seconds of proper time for Twin B. But the twins experience different amounts of proper time between departure and reunion. Theory predicts that they are different, and experiments have confirmed it. But if you don't understand the theory and refuse to accept the experimental results then you are left in a state of denial that can't be resolved.

 

[Grimble]

OK. Yes I have a problem with relativity. There seems to be something fundamental that is constantly glossed over, that we are expected to accept and believe in...

Introducing "Space and Time" Minkowski wrote

According to Lorentz every body in motion, shall suffer a contraction in the direction of its motion, namely at velocity v in the ratio [gamma]
This hypothesis sounds rather fantastical. For the contraction is not to be thought of as a consequence of resistances in the ether, but purely as a gift from above, as a condition accompanying the state of motion.

Now all that is just in my mind - I accept that. Everything should just fit seamlessly into a working model, yet each time I try something just doesn't line up. Every part works with every other but never all at once however one looks at it some part is left adrift.

I will try once more to show this by drawing an example; but first I will address Peter's suggestion

Here's my suggestion: pick an inertial frame, such as C's rest frame. Write down the coordinates of all of the events of interest in this frame, explicitly. Then write down the Lorentz transformation that goes from C's rest frame to A's rest frame. Then write down the (different!) Lorentz transformation that goes from C's rest frame to B's rest frame. Then transform the coordinates of all of the events of interest using each of these transformations, and write down the results.

C's frame. A is traveling with velocity -v, B is traveling with velocity v.
(t,-vt,0,0) event 1, A has traveled a distance -vt ,
(t,vt,0,0) event 2. B has traveled vt,
(t,0,0,0) event 3. C has remained at rest.

So for A,
t' = γ(t - (-v)(-vt)/c²
t' = γ(t - tv²/c²
t' = γt(1 - v²/c²)
t' = t/γ

And for B,
t' = γ(t - (v)(vt)/c²
t' = γ(t - tv²/c²
t' = γt(1 - v²/c²)
t' = t/γ

So as I see it ( and please explain where I am going wrong!) the proper time in A has the same duration as the proper time in B - as measured within each frame.

Everything is relative: A measures propertime on her clock, B measures proper time on his clock and C measures proper time on their clock and they each measure coordinate times on the other's clocks.

At its simplest we can take C out of Peter's thought experiment above leaving us with two events A and B separated with coordinates (t, 0 , 0 , 0) and (t, vt, 0, 0) in A's frame and (t, 0, 0, 0) and (t, -vt, 0, 0) in B's frame and calculating as above we have time t' = t/γ, for each coordinate time and of course t = γt' for the proper times as measured from either frame.

 

[stevendaryl]

C's frame. A is traveling with velocity -v, B is traveling with velocity v.
(t,-vt,0,0) event 1, A has traveled a distance -vt ,
(t,vt,0,0) event 2. B has traveled vt,
(t,0,0,0) event 3. C has remained at rest.

So for A
...
t' = t/γ

And for B,
...
t' = t/γ

So as I see it ( and please explain where I am going wrong!) the proper time in A has the same duration as the proper time in B - as measured within each frame.

Yes. Letting event 0 be the event where x=y=z=t=0, then what we conclude is:

1. The proper time between event 0 and event 1 is t/\gamma
2. The difference in coordinate times between events 0 and 1, according to A's coordinate system, is t/\gamma
3. The difference in coordinate times between events 0 and 1, according to C's coordinate system, is t
4. The proper time between event 0 and event 2 is t/\gamma
5. The difference in coordinate times between events 0 and 2, according to B's coordinate system, is t/\gamma
6. The difference in coordinate times between events 0 and 2, according to C's coordinate system, is t

Those are the facts. What is the question? Why do you think there is anything paradoxical or unclear about all this?

What we don't calculate (although we could) is:

• What is the coordinate time between events 0 and 1, according to B's coordinate system
• What is the coordinate time between events 0 and 2, according to A's coordinate system

The answer to both those questions is: \frac{\gamma' t}{\gamma} where \gamma' is the gamma factor computing using the relative speed between A and B (which will not be v but will be, using the velocity addition formula, \frac{2v}{1-\frac{v^2}{c^2}}).

So

• In A's coordinate system, event 1 takes place before event 2
• In B's coordinate system, event 2 takes place before event 1
• In C's coordinate system, the two events are simultaneous.

Please, please. Ask a question. If you just state facts, that's not a question. People will spend dozens of posts trying to guess what your point is. Why not actually say what your point is?

 

[Battlemage!]

Lorentz transformations don't work on displacements, they work on coordinates.

Here's my suggestion: pick an inertial frame, such as C's rest frame. Write down the coordinates of all of the events of interest in this frame, explicitly. Then write down the Lorentz transformation that goes from C's rest frame to A's rest frame. Then write down the (different!) Lorentz transformation that goes from C's rest frame to B's rest frame. Then transform the coordinates of all of the events of interest using each of these transformations, and write down the results.

Doing this will, first, help you clarify for yourself what the implications of your scenario are (I don't think you fully understand them), and second, help the rest of the posters in this thread understand what you are describing and what question you are asking.

If you are unable to complete the above exercise, then I strongly suggest closing this thread until you have taught yourself how to do so. Being able to do an exercise like the above is a basic skill in relativity, and if you don't have it, you shouldn't be posting an "I" level thread.

I know you're getting a lot thrown at you, but if you have time what do you think of this attempt to find velocities as measured by the observers and the consequential time measurement differences?

The coordinates of clock A are (x, t). The coordinates of clock C are (x', t'). The coordinates of clock B are (x'', t'')

Let v_A' = -v_A be the velocity of clock A relative to C (moving to the left in the negative x direction) and let v_B' be the velocity of clock B relative to C (moving to the right, in the positive x direction), with v_A' = -v_B' according to C.
If A is considered at rest, then C is moving at v_A according to A (to the right, in the positive x direction). Clock A knows that according to C, B moves at v_B' = -v_A' = v_A. So, we should be able to do a velocity transformation.

$$ v_B = \frac {v_A + v_B'}{1 + \frac{v_Av_B'}{c^2}} = \frac {v_A + v_A}{1 - \frac{v_A(v_A)}{c^2}} = \frac {2v_A}{1 - \frac{v_A^2}{c^2}}$$

So now we can calculate the time coordinate transformation from A to B, correct?

Then, the Lorentz factor from A to B would be: $γ_{AB} = \frac{1}{\sqrt{1 -\left( \frac {2v_A}{1 - \frac{v_A^2}{c^2}}\right)^2}c^{-2}}$
So, $t'' = γ_{AB} \left(t - \frac{v_B'x}{c^2}\right) = \frac{1}{\sqrt{1 -\left( \frac {2v_A}{1 - \frac{v_A^2}{c^2}}\right)^2}c^{-2}} \left(t - \frac{\left(\frac {2v_A}{1 - \frac{v_A^2}{c^2}}\right)x}{c^2}\right)$

And going the other way, $t = γ_{AB} \left(t'' + \frac{v_B'x''}{c^2}\right) = \frac{1}{\sqrt{1 -\left( \frac {2v_A}{1 - \frac{v_A^2}{c^2}}\right)^2}c^{-2}} \left(t '' + \frac{\left(\frac {2v_A}{1 - \frac{v_A^2}{c^2}}\right)x''}{c^2}\right)$So it looks to me that the only possible way for t'' to equal t according to observer A or B is if $\frac{\left(\frac {2v_A}{1 - \frac{v_A^2}{c^2}}\right)}{c^2} = 0$
How far off is that?