Proper (and coordinate) times re the Twin paradox

[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?

 

[Mister T]

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.)

 

[Mister T]

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?

 

[Mister T]

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.

Your descriptions don't make clear the fact that events have a duration of zero time.

(t,-vt,0,0) event 1, A is located at the position -vt ,
(t,vt,0,0) event 2. B is located at the position vt,
(t,0,0,0) event 3. C is located at the position zero.

Is it you're assertion that these three events are simultaneous in C's frame?

[Edit: Note they are simultaneous in C's rest frame given my descriptions of the events.]

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 [...]

Well, now you've lost me. I thought the events were labeled 1, 2, and 3. Now you seem to be labeling them A and B.

Anyway, Events 1 and 2 are not simultaneous in A's frame if they were simultaneous in C's frame.

 

[PeterDonis]

I will address Peter's suggestion

Yes, you did; and stevendaryl gave the correct response, that yes, you have described a perfectly consistent scenario in accordance with relativity, and there is no issue with it at all. So what's the problem?

If the problem is that you simply can't believe that the scenario as you've described it is consistent, I'm sorry, we can't help you with that. Relativity is an experimental fact.

If you have some other question, then, as stevendaryl asked you, what is it? What is the specific issue you are having with the scenario as you described it? So far nobody has been able to figure out what that is.

the proper time in A has the same duration as the proper time in B - as measured within each frame.

Yes, that's correct. Proper time along a specific path in spacetime, which is what you are referring to here (proper time along A's path, and proper time along B's path) is an invariant; it's the same for all observers and in all frames. It's just geometry; it's no different than saying that the distance from New York to Washington, DC is the same no matter what coordinate grid we put on the Earth's surface.

Everything is relative: A measures propertime on her clock, B measures proper time on his clock and C measures proper time on their clock

Yes, that's correct; each clock measures proper time along its own path in spacetime.

they each measure coordinate times on the other's clocks

This is not quite correct. Nobody can directly measure times of events that are not on their paths through spacetime. What they can do is calculate what those times are, based on measurements they can make directly. So A and B and C can each calculate what the times registered on the others' clocks will be, and can relate that to coordinate times in their own frames.

 

[Dale]

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.

The proper time between the origin and events 1 and 2 are the same in every frame.

 

[Ebeb]

Yes, but any clock in its own reference frame is at rest and only measures time passing. And, in Spacial Relativity which as I understand it is where the twin paradox may be considered, identical clocks can be placed in any frame and will keep identical time - identical clocks; identical laws of science; identical conditions...
so how do they measure different times?

Because of relativity of simultaneity.
When both clocks move relative to each other, the 'identical time' you refer to, f.ex event 5 seconds proper time clock A and event 5 seconds proper time on clock B are not occurring simultaneously anymore. When they move relative to each other you are not allowed to say that "When I tick 5 seconds per my inertial ref frame, the other clock also ticks 5 times per his inertial ref frame." It's a statement that doesn't make sense anymore.
Per clock A ref frame clock B ticks slower. It means that f.ex. when clock A shows 5 seconds proper time, clock B shows only 4 seconds of its proper time (for gamma factor 1.25 for 0.6c). For clock A, clock B ticks slower, not because clock B ticks slower in its own inertial ref frame, but because of relativity of simultaneity clock A ref frame and clock B ref frame take different directions in 4D spacetime. This slows the popping up of clock B proper time events per clock A ref frame (and slows the popping up of clock A proper time events per clock B frame).

The 5 seconds clock A proper time is also a clock A coordinate time for 4 seconds clock B proper time.

You know how to read and draw spacetime diagrams?

 

[Grimble]

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

That is exactly what I think is happening; I am not saying there is anything wrong, I am first checking that what I think is happening is correct.

Your descriptions don't make clear the fact that events have a duration of zero time.

I am sorry I did not think that was necessary, as it is a fundamental fact that events are instants in time...