Clocks, rulers... and an issue for relativity

[VandD]

Here?

Edited August 27, 2016 by VandD

[robinpike]

 

I'm not sure I understand the problem. The moving clock appears to run slower because it is moving (relative to the other observers). When it stops moving then it is observed to run at the same speed as the observer's (stationary) clocks. Because they are all in the same frame of reference.

 

The equivalence of all frames of references doesn't seem relevant to that: you have one frame that is considered to be moving, and the clocks in that frame appear to be running slower.

 

The equivalence comes in when we look at the frame of reference of the moving clock. While that clock is in an inertial frame moving relative to the other observers, the clock's owner will see the clocks on the space stations running slow. Equivalent!

 

In other words, it makes no difference which inertial frame you consider to be moving and which to be stationary; each will observe the other's clock running slow.

 

But... And it is is a big but...

 

BUT, your example does not deal with inertial frames. The clock is not always in an inertial frame. It is accelerated and then slowed. That is why it accumulates less total time. (See the full "twin paradox" explanation for details.)

 

 

No, but it is dependent on the "route" the object has taken through space time.

 

If two people drive from New York to San Francisco but one drives via New Orleans and the other via Chicago, you wouldn't be surprised that their odometers read different distances.

 

Strange, the issue is that at the beginning of its journey, the travelling clock shortens its current route through space-time [that is from its current 'normal' route to the 'short' route], whereas at the end of its journey, the travelling clock lengthens its current route through space-time [that is from its current 'short' route back to the 'normal' route].

 

But in each case the travelling clock is simply moving from one inertial reference frame to another. If the acceleration at the beginning caused the travelling clock to go on a shorter route from its current route, then the acceleration at the end should do the same to whatever its current route is at that moment - that is make the travelling clock go from its 'short' route to now an even 'shorter route' through space-time?

 

But it doesn't do that - so that means that those two inertial frames under consideration are not of equal standing. Note that this has nothing to do with what direction the clock travels in. Nor am I suggesting that the equations of relativity don't work / cannot be used.

 

Here is the issue described in detail.

 

The two clocks start side-by-side, both travelling through space-time together, same space co-ordinates, same rate through space-time.

 

The travelling clock starts its journey by applying acceleration and moves away from the stay at home clock. The travelling clock is now going through space-time on a 'shorter' route than the stay at home clock. The travelling clock returns and applies acceleration to halt its movement and it stops next to the stay at home clock. The two clocks are again in the same space co-ordinates and moving through space-time at the same rate. That is the travelling clock is now going through space-time on the same 'longer' route as the stay at home clock - and that is the contentious bit if the various inertial frames of reference are of equal standing.

 

Regardless of the logical reasoning of the above, because the travelling clock has spent time on a shorter route through space-time, it has less time on its clock than the stay at home clock (exactly as the equations of relativity calculate). Einstein was correct in that it is not necessary to differentiate between inertial reference frames in order for the calculations of relativity to work. They work for the stay at home inertial reference frame, they work for the travelling clocks reference frame, and they work for any observer's reference frame. But to conclude that all inertial reference frames are of equal standing extends that logic incorrectly.

Edited August 30, 2016 by robinpike

[Endy0816]

You are forgetting the need to turn around.

[Mordred]

When you truly think about it the "at rest frame" is meaningless. All frames are equivalent to inertial frames.

 

So why would you believe one is preferred over the other.

 

If you think about it "a preferred frame" is equally meaningless.

 

You don"t base conclusions and definitions from one type of scenario, they must work in every scenario as much as possible.

Edited August 30, 2016 by Mordred

[swansont]

Strange, the issue is that at the beginning of its journey, the travelling clock shortens its current route through space-time [that is from its current 'normal' route to the 'short' route], whereas at the end of its journey, the travelling clock lengthens its current route through space-time [that is from its current 'short' route back to the 'normal' route].

 

But in each case the travelling clock is simply moving from one inertial reference frame to another. If the acceleration at the beginning caused the travelling clock to go on a shorter route from its current route, then the acceleration at the end should do the same to whatever its current route is at that moment - that is make the travelling clock go from its 'short' route to now an even 'shorter route' through space-time?

 

But it doesn't do that - so that means that those two inertial frames under consideration are not of equal standing. Note that this has nothing to do with what direction the clock travels in. Nor am I suggesting that the equations of relativity don't work / cannot be used.

 

But you are saying that, if you claim there should be additional slowing of the clock. The equations say that the clock rate will speed up as the speed is reduced to zero.

 

Your view that any acceleration "shortens the route through spacetime" is naive and incorrect. The clock rate is dependent on the speed. If the acceleration lowers the speed, the clock rate increases, according to some remote, stationary observer.

[Mordred]

Think about it this way. There is no way one can state "my observer measurements are more accurate".

 

Regardless of scenario, all observer measurements are equally accurate.

[michel123456]

You are forgetting the need to turn around.

Yes

 

In fact I often get the impression that the result from the maths indicate what is happening when the traveler reaches point C on the following

 

It cannot be that points C and D are treated mathematically as the same. There must be an entry in the formulas that make a distinction between those 2 points, because the one is here at home and the other is far far away.

[swansont]

Yes

 

In fact I often get the impression that the result from the maths indicate what is happening when the traveler reaches point C on the following

TwinParadox3.jpg

 

It cannot be that points C and D are treated mathematically as the same. There must be an entry in the formulas that make a distinction between those 2 points, because the one is here at home and the other is far far away.

 

 

It's that the event at point B is not accommodated in the formulas that is at issue. SR assumes all parties are in, and stay in, their own inertial frame

[imatfaal]

Think about it this way. There is no way one can state "my observer measurements are more accurate".

 

Regardless of scenario, all observer measurements are equally accurate.

 

Can we not go further and say that if a model can possibly give rise to different predictions then we know that the model is flawed (or incorrectly applied outside the models zone of application)?

[swansont]

Can we not go further and say that if a model can possibly give rise to different predictions then we know that the model is flawed (or incorrectly applied outside the models zone of application)?

For classical events, yes. For values, no, precisely because many observable/measurable things are relative.

 

For quantum events, no (in a way). Since QM speaks in probabilities, you can have multiple predictions. But the probability values have to be consistent.

[J.C.MacSwell]

 

Can we not go further and say that if a model can possibly give rise to different predictions then we know that the model is flawed (or incorrectly applied outside the models zone of application)?

Seems like where Newtonian Physics was prior to Einstein...

[Tim88]

[..]

The travelling clock starts its journey by applying acceleration and moves away from the stay at home clock. The travelling clock is now going through space-time on a 'shorter' route than the stay at home clock. The travelling clock returns and applies acceleration to halt its movement and it stops next to the stay at home clock. The two clocks are again in the same space co-ordinates and moving through space-time at the same rate. That is the travelling clock is now going through space-time on the same 'longer' route as the stay at home clock - and that is the contentious bit if the various inertial frames of reference are of equal standing.

[..]

What is contentious about that?

[robinpike]

You are forgetting the need to turn around.

Yes, I was simplifying, but leaving it out doesn't affect the validity of the argument. The basics of the issue is between leaving the stay at home reference frame, and returning to the stay at home reference frame. Both are caused by accelerations, but the first triggers a change from the current route to a shorter route through space-time, whereas the second triggers a change from the current route to a longer route through space-time.

 

When you truly think about it the "at rest frame" is meaningless. All frames are equivalent to inertial frames.

 

So why would you believe one is preferred over the other.

 

If you think about it "a preferred frame" is equally meaningless.

 

You don"t base conclusions and definitions from one type of scenario, they must work in every scenario as much as possible.

My examples of the issue have not come across very well - below is another go at this...

 

But you are saying that, if you claim there should be additional slowing of the clock. The equations say that the clock rate will speed up as the speed is reduced to zero.

 

Your view that any acceleration "shortens the route through spacetime" is naive and incorrect. The clock rate is dependent on the speed. If the acceleration lowers the speed, the clock rate increases, according to some remote, stationary observer.

The issue is how one acceleration is able to shorten the route through space-time, whereas another acceleration is able to increase the route through space-time. I appreciate that the equations predict (agree) with that - I am not contesting that at all. My issue is with how that difference comes about through acceleration?

 

[..]

The travelling clock starts its journey by applying acceleration and moves away from the stay at home clock. The travelling clock is now going through space-time on a 'shorter' route than the stay at home clock. The travelling clock returns and applies acceleration to halt its movement and it stops next to the stay at home clock. The two clocks are again in the same space co-ordinates and moving through space-time at the same rate. That is the travelling clock is now going through space-time on the same 'longer' route as the stay at home clock - and that is the contentious bit if the various inertial frames of reference are of equal standing.

[..]

What is contentious about that?

The issue is with how acceleration causes the two opposite changes to the route through space-time to come about.

 

 

Here is the scenario using the standard 'travelling twin' scenario. For simplicity, I have left out reference to apparent changes in clock times and only mention the physical changes.

 

At the start the two clocks are next to each other in the same reference frame and have the same time. At this point, the two clocks are travelling through space-time at the same rate.

 

The travelling clock then accelerates away from the stay at home clock and leaves the reference frame of the stay at home clock. This action causes the travelling clock to go through a shorter route through space-time than the stay at home clock's route through space-time. The travelling clock then switches off that acceleration and coasts at a constant rate through space-time, faster than the constant rate through space-time of the stay at home clock.

 

This continues until the travelling clock turns around and prepares for the return trip by applying acceleration back towards the stay at home clock's position in space. This causes the travelling clock's route through space-time to become longer. At the point when the travelling clock becomes (momentarily) stationary with respect to the stay at home clock's reference frame, the travelling clock's route through space-time is at the same rate as the stay at home clock's route through space-time, but the two clock's no longer have the same time - the travelling clock's time has physically lost time with respect to the stay at home clock's time. And of course at this point in the journey, the two clocks are no longer in the same part of space.

 

With the acceleration back to the stay at home clock's position still being applied, the travelling clock's route through space-time now goes back to being shorter again. The travelling clock then switches off that acceleration and coasts at a constant rate through space-time, faster than the constant rate through space-time of the stay at home clock.

 

Finally, on reaching the stay at home clock's position in space, the travelling clock applies acceleration in the same direction as when it first stated its journey. This time though, that action causes the travelling clock's route through space-time to become longer, until the travelling clock switches off that acceleration and finishes at rest with respect to the stay at home clock, and the two clocks are once again travelling through space-time at the same rate and back to being in the same part of space. The travelling clock's time has now lost even more physical time with respect to the stay at home clock's time, than at the half-way point in its journey.

 

Now to the issue. See that part where the travelling clock applies acceleration to turn back to the stay at home clock. The acceleration is applied steadily but the travelling clock's rate through space-time first increases and then decreases, the flipping point being when it becomes stationary with respect to the stay at home clock's reference frame.

 

And also, when the travelling clock starts its journey, and ends its journey. The travelling clock applies acceleration in the same direction, but at the start that acceleration causes the travelling clock to shorten its route through space-time, whereas at the end that acceleration causes the travelling clock to lengthen its route through space-time.

 

So how does acceleration do that? (Note that I am not contesting that acceleration does indeed do it.)

Edited September 1, 2016 by robinpike

[Endy0816]

Measurements are the same when you are parallel(4d).

 

Based on the ship's velocity vector and applied acceleration you'll impact whether you end up parallel, separating or coming together.

 

Late here, not 100% sure this is clear. I'm sure any of our experts in residence can explain it better though

[Mordred]

robinpike conceptually you have somewhat an accurate descriptive. However lets use a more accurate terminology. By spacetime route I assume you are referring to the length of the World line. Which is the affine connection between event A and event B (observer/emitter) (ds^2). Am I correct on this?

Measurements are the same when you are parallel(4d).

 

Based on the ship's velocity vector and applied acceleration you'll impact whether you end up parallel, separating or coming together.

 

Late here, not 100% sure this is clear. I'm sure any of our experts in residence can explain it better though

I for one am not sure what your expressing here lol. The first line makes sense on parallel transport to Pythagoras theory correlations. After that you lost me. Edited September 1, 2016 by Mordred

[Tim88]

[..]

 

My examples of the issue have not come across very well - below is another go at this...

 

The issue is how one acceleration is able to shorten the route through space-time, whereas another acceleration is able to increase the route through space-time. I appreciate that the equations predict (agree) with that - I am not contesting that at all. My issue is with how that difference comes about through acceleration?

 

The issue is with how acceleration causes the two opposite changes to the route through space-time to come about.

 

[..] At the start the two clocks are next to each other in the same reference frame and have the same time. At this point, the two clocks are travelling through space-time at the same rate.

 

The travelling clock then accelerates away from the stay at home clock and leaves the reference frame of the stay at home clock. This action causes the travelling clock to go through a shorter route through space-time than the stay at home clock's route through space-time. The travelling clock then switches off that acceleration and coasts at a constant rate through space-time, faster than the constant rate through space-time of the stay at home clock.

[..]

Now to the issue. See that part where the travelling clock applies acceleration to turn back to the stay at home clock. The acceleration is applied steadily but the travelling clock's rate through space-time first increases and then decreases, the flipping point being when it becomes stationary with respect to the stay at home clock's reference frame.

 

And also, when the travelling clock starts its journey, and ends its journey. The travelling clock applies acceleration in the same direction, but at the start that acceleration causes the travelling clock to shorten its route through space-time, whereas at the end that acceleration causes the travelling clock to lengthen its route through space-time.

 

So how does acceleration do that? (Note that I am not contesting that acceleration does indeed do it.)

 

I suspect that the lack of understanding all has to do with how you phrase it. As Langevin already indicated, the central issue is a change in velocity. Using the term "acceleration" instead of "change in velocity" obscures a key point, and disconnects the description of the calculation; the result is a lack of insight. I strongly suggest to try to rephrase your account accordingly.

 

We then find, depending on your choice of inertial reference system, that the acceleration implies that the traveling clock undergoes a change of speed which causes a corresponding change in clock rate. If we use for convenience the handy but somewhat misleading jargon "Observer" for "standard SR reference system", then his is qualitatively agreed on by all Observers. What also all Observers agree on, is that at the end the clock is again moving at the same speed as before (neglecting planetary motion). Note also that the clock always remains in any chosen reference system; its speed is changed with respect to the chosen system. BTW, of course we can jump reference systems midway, but that merely leads to mathematical exercises which confirm that the Lorentz transformations work.

 

If you choose for example the stay-at-home reference system (approximating it to be inertial and neglecting effects from gravitation) then the traveler accelerates to a speed which is maintained most or all of the time (in Langevin's example the traveler doesn't even stop but slings around a star); in comparison, the speed of the stay-at-home is zero. The accelerations themselves are only important insofar as they bring about these speed changes. Thus, from the perspective of the stay-at-home frame, the traveler clock had a constant higher clock rate all or most of the travel period.

 

The "the two opposite changes" in clock rate correspond perfectly with the two opposite changes in speed.

 

It is instructive to consider other reference systems, for example the one in which the travel is in rest in the outgoing leg. From that perspective the clock rate first is increased to its proper rate, and then on the way back home it is strongly decreased as the traveler tries to catch up with the Earth. Then when the traveler decelerates, the clocks are again ticking at the same slowed down rate.

 

Is there still something counter-intuitive in that description?

 

PS. it may also be enlightening to carefully read the first description and the related physical insight concerning time dilation by Einstein in 1905 - § 4 of

http://www.fourmilab.ch/etexts/einstein/specrel/www/

 

(we now know that it needs to be corrected for the effect of gravitational potential, but that's not important for the physical insight)

Edited September 1, 2016 by Tim88

[swansont]

 

The issue is how one acceleration is able to shorten the route through space-time, whereas another acceleration is able to increase the route through space-time. I appreciate that the equations predict (agree) with that - I am not contesting that at all. My issue is with how that difference comes about through acceleration?

 

The issue is with how acceleration causes the two opposite changes to the route through space-time to come about.

 

 

 

Acceleration can speed you up and it can slow you down. (or neither, since it can also change your direction) If you speed up, relative to some chosen frame, your clock runs slower. If you slow down, the clock rate increases. It's not all that complicated: your clock rate depends on v.

 

The presence of an acceleration removes your ability to treat yourself as being at rest in the analysis. Uniform motion is relative, but acceleration is not. If you have accelerated, then it is clear that you are the one who has changed from one frame to another.

Now to the issue. See that part where the travelling clock applies acceleration to turn back to the stay at home clock. The acceleration is applied steadily but the travelling clock's rate through space-time first increases and then decreases, the flipping point being when it becomes stationary with respect to the stay at home clock's reference frame.

 

And also, when the travelling clock starts its journey, and ends its journey. The travelling clock applies acceleration in the same direction, but at the start that acceleration causes the travelling clock to shorten its route through space-time, whereas at the end that acceleration causes the travelling clock to lengthen its route through space-time.

 

So how does acceleration do that? (Note that I am not contesting that acceleration does indeed do it.)

 

Don't focus on the acceleration, focus on the speed. The speed dictates what the clock does.

[swansont]

!

Moderator Note

 

Discussion hijack about reality has been split

http://www.scienceforums.net/topic/98048-relativity-and-shared-realities-split-from-clocks-rulers/

 

[studiot]

 

robinpike post#1

 

But that is not true of travelling clocks. A travelling clock that does a round trip no longer shows the same time as the stay at home clock - it shows less time.

If this were to be explained as above - that the travelling clock physically ticks at a slower rate, then it runs into the same problem as described above- how does its physical rate of ticks return to the stay at home's rate of ticks?

The loss in time of the travelling clock has been explained to me as having travelled a shorter DISTANCE through space-time, with the travelling clock not changing the rate of its ticks.

Please can someone step through this change in distance in space-time and how it does not run into the same issue as described above? That is, if the change into another frame of reference produces a shorter distance through space-time, how does changing back to the original frame of reference produce a longer distance through space-time? thanks.

 

 

Robin, this question of yours has spawned at least one more thread and become very messy, which is why I originally avoided it.

 

I did make a couple of posts containing questions designed to help you find the right view,

But you have avoided answering them.

 

When contemplating these questions it is very important to keep track of which clock is where and which system you are measuring in.

A wrong placement here leads to embarrassing paradoxes and misunderstanding.

 

In particular you should always find out which clock is at all the events - There will in general be only one of these.

 

If you are interested I will work through the Twins using largely logic, although a little easy maths will be needed.

 

 

The main thing to get the correct time differences, is to know which reading on which clocks can be used directly and which have to be transformed.

Logic is need for this, not maths.

 

But I will only do this as a discussion as I am not prepared to do all the work here.

 

So

 

In the Twins there are no clocks in the Earth system that are present at every event.

 

Can you say which clock or clocks are at all the significant events in the the history?

 

We will also be following this from swansont as it is a highly significant hint. +1

 

 

 

swansont

Don't focus on the acceleration, focus on the speed. The speed dictates what the clock does.

Edited September 2, 2016 by studiot

[studiot]

Since you have visited the forum a couple of times since last posted, do I take it that you have lost interest in this thread of yours?

[robinpike]

 

 

Acceleration can speed you up and it can slow you down. (or neither, since it can also change your direction) If you speed up, relative to some chosen frame, your clock runs slower. If you slow down, the clock rate increases. It's not all that complicated: your clock rate depends on v.

 

The presence of an acceleration removes your ability to treat yourself as being at rest in the analysis. Uniform motion is relative, but acceleration is not. If you have accelerated, then it is clear that you are the one who has changed from one frame to another.

 

Don't focus on the acceleration, focus on the speed. The speed dictates what the clock does.

 

 

 

 

Robin, this question of yours has spawned at least one more thread and become very messy, which is why I originally avoided it.

 

I did make a couple of posts containing questions designed to help you find the right view,

But you have avoided answering them.

 

When contemplating these questions it is very important to keep track of which clock is where and which system you are measuring in.

A wrong placement here leads to embarrassing paradoxes and misunderstanding.

 

In particular you should always find out which clock is at all the events - There will in general be only one of these.

 

If you are interested I will work through the Twins using largely logic, although a little easy maths will be needed.

 

 

The main thing to get the correct time differences, is to know which reading on which clocks can be used directly and which have to be transformed.

Logic is need for this, not maths.

 

But I will only do this as a discussion as I am not prepared to do all the work here.

 

So

 

In the Twins there are no clocks in the Earth system that are present at every event.

 

Can you say which clock or clocks are at all the significant events in the the history?

 

We will also be following this from swansont as it is a highly significant hint. +1

 

 

 

Since you have visited the forum a couple of times since last posted, do I take it that you have lost interest in this thread of yours?

 

Hi Studiot, I am still keen to continue the discussion! The only reason why I haven't replied until now is that I didn't want to post a new question that was irrelevant or ill thought through.

 

Yes, I do want to step through a specific scenario, but first, can I confirm a general issue that seems to follow on from what Swansont mentioned: "The speed dictates what the clock does".

 

My understanding of this resolved my concern that I had around how does the travelling clock return to its original rate through space-time.

 

So that's good. However, that resolution seems to lead to this logical issue...

 

In the universe, nearly everything has a speed with respect to us. After all, we are orbiting the sun, the sun is orbiting the galaxy, and the galaxy is itself probably moving through space with respect to other galaxies. So in that case, since all these other things in the universe have speed with respect to our inertial frame of reference, they are all moving on a shorter route through space-time than we are. Or to put it another way, none of these things are moving on a longer route through space-time than us. We are moving on the longest route through space-time of all the things in the universe.

 

That we are travelling on the longest route through space-time compared to everything else is not a logical issue in itself (although it is highly unlikely and so suggests an issue somewhere?). The issue is that any other inertial frame of reference can deduce that they are on the longest route through space-time. And since when clocks move, they really do go on a shorter route through space-time, it is not just whose point of view is being used, the longest route through space-time is a physical situation that cannot be shared by everything in the universe?

 

Note that this issue is not the same as asking the question of, who is moving? as the answer to that question can be switched from observer to observer without causing a logical issue.

Edited September 5, 2016 by robinpike

[swansont]

That we are travelling on the longest route through space-time compared to everything else is not a logical issue in itself (although it is highly unlikely and so suggests an issue somewhere?). The issue is that any other inertial frame of reference can deduce that they are on the longest route through space-time. And since when clocks move, they really do go on a shorter route through space-time, it is not just whose point of view is being used, the longest route through space-time is a physical situation that cannot be shared by everything in the universe?

 

Note that this issue is not the same as asking the question of, who is moving? as the answer to that question can be switched from observer to observer without causing a logical issue.

 

But they are the same, since it all ties back to speed. All inertial observers see everyone else's clock as being slower, while not seeing it in their own clock, because they view themselves as being at rest.

[robinpike]

 

But they are the same, since it all ties back to speed. All inertial observers see everyone else's clock as being slower, while not seeing it in their own clock, because they view themselves as being at rest.

 

But there is an issue here. A shorter route through space-time is not a mutual effect that is just a point of view, dependent on who is the observer. A shorter route through space-time is a physical thing that can be seen when the travelling clock returns to its original inertial frame of reference.

[swansont]

 

But there is an issue here. A shorter route through space-time is not a mutual effect that is just a point of view, dependent on who is the observer. A shorter route through space-time is a physical thing that can be seen when the travelling clock returns to its original inertial frame of reference.

 

 

When a clock returns to its original frame it has undergone an acceleration. It is no longer in the same inertial frame. You have changed the conditions of your scenario when you introduce an acceleration.

 

As long as the clocks remain in their inertial frame, they will see all other such clocks as running slow.

[studiot]

I was about to say that I will leave it to swansont to discuss routes through spacetime and geodesics with you, but I see he has already replied.

 

+1 you are nearly there with this question, but think about this.

 

Think of a globe.

 

The lines of longitude are the shortest paths between two points, but they are all of the same length and there are many of them.

 

They are given a special name - 'geodesics'.

 

By contrast only one parallel of latitude (the equator) is a shortest path.

Every other parallel is longer than a great circle (geodesic) between these same two points.

 

If there are multiple shortest paths, why can there not be multiple longest paths?

 

Another thing about paths.

The paths assume no 'wiggle'.

A wiggle is a deviation or detour in some sense from 'straight on'.

So a trip from London to Edinburgh via Leeds is (more or less) straight on, but

a trip via Cardiff incorporates a wiggle and via Dublin a bigger wiggle.

 

Wiggles are not possible in one dimension, but as soon as you get two or more dimensions they become longer paths than a direct one in the same way that the sum of two sides of a triangle is always greater than the third.

 

However I regard this question of yours as a diversion or wiggle so I will leave it to yourself and swansont to carry on.

When you are ready to go back to the twins, let me know.

Edited September 5, 2016 by studiot