Clocks, rulers... and an issue for relativity

[Strange]

You can make the scenario even funnier:

take 2 twins traveling, one to the East, the other to the West (for the ease of understanding), and a 3rd observer on Earth. Will the 2 twins have the same age when coming home?

Obviously the answer is Yes.

 

 

Or maybe, no: https://en.wikipedia.org/wiki/Hafele–Keating_experiment

[michel123456]

 

 

Or maybe, no: https://en.wikipedia.org/wiki/Hafele–Keating_experiment

Right, the ease of understanding has been blown out.

Take 2 twins traveling in opposite directions.

[J.C.MacSwell]

Right, the ease of understanding has been blown out.

Take 2 twins traveling in opposite directions.

It is easy to see that one travelling East and one West is not a symmetrical comparison. Is it not?

[robinpike]

 

 

When the clock accelerates at the turnaround, there will be an accumulation of a difference in phase. The earthbound twin does not see the spaceship's clock running slow on the return trip. The difference in phase continues to accumulate..

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html

 

Hi Swansont, the issue does not lie with what happens / what is seen, but with the premise of relativity that all inertial frames are of equal standing, and that if accelerating from one frame into another frame affects time, say that slows down time - then the premise means that affect can be the only affect that can happen. There is no method by which the clock can return to its original rate of time. (Note it is the RATE of time that is the issue for the travelling clock, when it returns back to the stay at home clock at the end of its journey - how does its rate return to that of the stay at home clock's rate?) That issue applies to whatever mechanism is put forward as to how the travelling clock loses time compared to the stay at home clock. If the explanation is that the travelling clock never changes its rate of time, but rather it takes a shorter route through space-time, then again there is no method by which the clock can return to the stay at home clock's 'standard' route through space-time. The travelling clock would be forever locked into its faster route through space-time.

 

I think, Robin, that this is the beginning of your difficulty.

 

 

Yes the travelling clock accelerates into a different frame, whose origin (for time at least) is not the same as the origin for the stay at home clock

 

So the t0 and t1 etc for the travelling clock are not the same t0 t1 etc as for the stay at home clock.

 

That is why I used different letters for these frames.

 

Why is this?

 

Well suppose instead of being twins, B was already travelling in the travelling frame and just happened to be passing A at t0 in the stay at home frame.

 

B then makes the two way journey.

 

Can you tell what would be the difference between their clocks on B's return?

 

Hi Studiot, I meant to convey that the time for the travelling clock was different - indicated by using a capital T for its time co-ordinate. I didn't mean to suggest that t1 was equal to T1, nor t2 equal to T2 etc.

 

I am not sure what you mean by the travelling clock moving into a frame whose origin is not the same as the stay at home clock? But my comment to Swansont is the key point of this issue. Whatever it is that changes for the travelling clock, if the premise that all reference frames are of equal standing applies, then that change cannot result in two different affects. The travelling clock cannot move from a 'standard' route through space-time into a shorter route through space-time, and later move from a 'shorter' route through space-time to a standard route through space-time?

Edited August 23, 2016 by robinpike

[swansont]

Hi Swansont, the issue does not lie with what happens / what is seen, but with the premise of relativity that all inertial frames are of equal standing, and that if accelerating from one frame into another frame affects time, say that slows down time - then the premise means that affect can be the only affect that can happen. There is no method by which the clock can return to its original rate of time. (Note it is the RATE of time that is the issue for the travelling clock, when it returns back to the stay at home clock at the end of its journey - how does its rate return to that of the stay at home clock's rate?) That issue applies to whatever mechanism is put forward as to how the travelling clock loses time compared to the stay at home clock. If the explanation is that the travelling clock never changes its rate of time, but rather it takes a shorter route through space-time, then again there is no method by which the clock can return to the stay at home clock's 'standard' route through space-time. The travelling clock would be forever locked into its faster route through space-time.

 

 

Why? Because you say so? What analysis shows this? Because the equations of relativity disagree with you.

[J.C.MacSwell]

 

Hi Swansont, the issue does not lie with what happens / what is seen, but with the premise of relativity that all inertial frames are of equal standing, and that if accelerating from one frame into another frame affects time, say that slows down time - then the premise means that affect can be the only affect that can happen. There is no method by which the clock can return to its original rate of time. (Note it is the RATE of time that is the issue for the travelling clock, when it returns back to the stay at home clock at the end of its journey - how does its rate return to that of the stay at home clock's rate?) That issue applies to whatever mechanism is put forward as to how the travelling clock loses time compared to the stay at home clock. If the explanation is that the travelling clock never changes its rate of time, but rather it takes a shorter route through space-time, then again there is no method by which the clock can return to the stay at home clock's 'standard' route through space-time. The travelling clock would be forever locked into its faster route through space-time.

 

 

Hi Studiot, I meant to convey that the time for the travelling clock was different - indicated by using a capital T for its time co-ordinate. I didn't mean to suggest that t1 was equal to T1, nor t2 equal to T2 etc.

 

I am not sure what you mean by the travelling clock moving into a frame whose origin is not the same as the stay at home clock? But my comment to Swansont is the key point of this issue. Whatever it is that changes for the travelling clock, if the premise that all reference frames are of equal standing applies, then that change cannot result in two different affects. The travelling clock cannot move from a 'standard' route through space-time into a shorter route through space-time, and later move from a 'shorter' route through space-time to a standard route through space-time?

That's equivalent to saying that if something accelerates in one direction, it can never accelerate in the opposite and return to it's original velocity. (or original rest frame)

[Endy0816]

Hi Swansont, the issue does not lie with what happens / what is seen, but with the premise of relativity that all inertial frames are of equal standing, and that if accelerating from one frame into another frame affects time, say that slows down time - then the premise means that affect can be the only affect that can happen. There is no method by which the clock can return to its original rate of time. (Note it is the RATE of time that is the issue for the travelling clock, when it returns back to the stay at home clock at the end of its journey - how does its rate return to that of the stay at home clock's rate?) That issue applies to whatever mechanism is put forward as to how the travelling clock loses time compared to the stay at home clock. If the explanation is that the travelling clock never changes its rate of time, but rather it takes a shorter route through space-time, then again there is no method by which the clock can return to the stay at home clock's 'standard' route through space-time. The travelling clock would be forever locked into its faster route through space-time.

 

You just need to match velocities again.

 

What I do is visualize a ruler for time as changing relative. Really it isn't a clock thing, so much as it is a time thing.

 

Could be a couple of twins blowing a bubble every minute just as well. At the conclusion of the trip the counts will be different.

Edited August 24, 2016 by Endy0816

[michel123456]

That's equivalent to saying that if something accelerates in one direction, it can never accelerate in the opposite and return to it's original velocity. (or original rest frame)

He says that on the return trip, the clock must be observed with an increasing rate (the clock ticks faster) so that at the end of the trip the clicking match the clock that stayed at rest.

IOW there must exist a way to observe clocks ticking faster and not always ticking slower, as suggested by the word "dilation".

[robinpike]

Why? Because you say so? What analysis shows this? Because the equations of relativity disagree with you.

 

Hi Swansont, I have never said that the equations of relativity don't work? The question is: IF you have the premise that inertial reference frames are of equal standing, THEN there is a logical contradiction with the fact that a travelling clock loses time and yet returns to the rate of time of the stay at home clock.

 

That's equivalent to saying that if something accelerates in one direction, it can never accelerate in the opposite and return to it's original velocity. (or original rest frame)

 

You are drawing a conclusion that I have never stated - and then attributing that (your) conclusion is false.

 

The issue is that IF all reference frames are of equal standing, THEN it is not possible to slow down the rate of time by one action and then return to the original rate of time by another action.

 

If you accelerate to the right, you change your position in space (say by an amount X). If you then accelerate to the left, you change your position in space (say by an amount X again, but crucially in the opposite direction). Since you have control of the accelerations being in opposite directions, you can engineer to arrive back to your original position in space.

 

But that doesn't work for time, if you accelerate to the right, you reduce your 'passage through time' (say by an amount T). If you then accelerate to the left, you reduce your 'passage through time' (again by an amount T). So both actions REDUCE your 'passage through time'.

 

What action enables your, now reduced passage through time, to increase back to the normal 'passage through time' rate?

Edited August 24, 2016 by robinpike

[swansont]

 

Hi Swansont, I have never said that the equations of relativity don't work? The question is: IF you have the premise that inertial reference frames are of equal standing, THEN there is a logical contradiction with the fact that a travelling clock loses time and yet returns to the rate of time of the stay at home clock.

 

 

No, there is no contradiction. Rate (frequency) and accumulated time (phase) are not the same thing. Phase is the integral of frequency.

 

J.C. was correct above: mathematically this is similar to two cars that start at the same speed and position, moving in a straight line. One speeds up, and then later on slows down. It ends up at the same speed, but not at the same position, as the first car. (IOW, I've described one car passing another on a motorway, which surely can be accepted to be free of logical contradictions)

[studiot]

 

Hi Studiot, I meant to convey that the time for the travelling clock was different - indicated by using a capital T for its time co-ordinate. I didn't mean to suggest that t1 was equal to T1, nor t2 equal to T2 etc.

 

I am not sure what you mean by the travelling clock moving into a frame whose origin is not the same as the stay at home clock? But my comment to Swansont is the key point of this issue. Whatever it is that changes for the travelling clock, if the premise that all reference frames are of equal standing applies, then that change cannot result in two different affects. The travelling clock cannot move from a 'standard' route through space-time into a shorter route through space-time, and later move from a 'shorter' route through space-time to a standard route through space-time?

 

Sorry if what I said was not clear.

This is in part due to my attempt to use popular language about frames, which was perhaps a mistake.

 

So first a word about frames.

 

Frames are nothing more than a set of reference axes, indeed they come from the phrase 'frame of reference'.

Some are more some are less convenient to work in than others.

Some are Euclidian or rectangular xyz frames, some have other configurations.

 

 

Since you can describe the desired properties of any object in any frame, you can either say that the object is in all frames or that all frames are equivalent.

 

Maths is required to effect correspondence between particular frames and effect the transformation to these properties as a result of changing to new reference axes.

 

So really I should not have said (as per popular parlance) an object enters one frame, and by inference leaves another.

The object can be described (referenced) in any frame; it is we who are changing the reference, but both the source and target frame are still valid before and after the change.

 

 

 

I am not sure what you mean by the travelling clock moving into a frame whose origin is not the same as the stay at home clock?

 

So this was a very good question and picks up my overloose phraseology.

 

 

Yes the travelling clock accelerates into a different frame, whose origin (for time at least) is not the same as the origin for the stay at home clock

 

I should have said something like

 

Observations of the travelling clock are made in a different frame from those of the stay at home clock.

In particular I should have used the word in not into.

Edited August 24, 2016 by studiot

[robinpike]

Perhaps the question that I am asking is a very subtle one, and so is tricky for me to separate from the normal questions around travelling clocks. (Also I didn't mean to sound as if I was annoyed with you, J. C. MacSwell, for posing that conclusion. Many people were probably drawing that inference of what I was saying, so it helps this discussion to mention it.)

 

Here is a simple example of the issue. Two space stations are stationary with respect to each other but some distance apart. The travelling clock is on a spaceship that takes off from one space station and goes to the other space station. Once landed, it then takes off and returns back to the first space station. It continues to do this shuttle back and forth while an observer on another space station, some distance from both, watches the travelling clock. All three space stations are stationary with respect to each other.

 

First point, there is nothing special about the reference frame that the three space stations are stationary with respect to each other. I am not saying the issue lies with the concept of which reference frame is chosen.

 

Now to the crucial bit. The observer on the third space station watches the time on the travelling clock. She can see the rate of time slowing down as the travelling clock accelerates and lifts off from one space station, stays at that reduced rate, and then as it lands on the receiving space station, return back to the original rate of time, now with the time on the travelling clock behind the time on the stay at home clock on that particular space station.

 

The issue is this: When accelerating from a space station, the rate of time of the travelling clock slows down. When de-accelerating to land on a space station, the clock's rate of time speeds up (it is no longer slowed down). This is not some apparent view of the travelling clock's time, it is real as can be seen by comparing the clocks when they are together, stationary and side-by-side. So how does it obtain those two different actions to its rate of time IF all reference frames are of equal standing?

 

If you don't like the use of words 'the rate of time' slowing down / not slowing down, you can substitute 'passage through space-time' being quicker / slower, shorter / longer etc. It doesn't matter how the loss in time on the travelling clock is explained, all have the same issue IF the premise that all frames of reference are of equal standing is applied.

Edited August 24, 2016 by robinpike

[swansont]

I don't see that you've identified an issue that is problematic in any way. The clock on the spaceship will end up showing less accrued time; all observers will agree. You imply there's a problem for multiple frames of reference, but you have identified only one inertial frame in your example: the one with the space stations in it.

Perhaps the question that I am asking is a very subtle one, and so is tricky for me to separate from the normal questions around travelling clocks. (Also I didn't mean to sound as if I was annoyed with you, J. C. MacSwell, for posing that conclusion. Many people were probably drawing that inference of what I was saying, so it helps this discussion to mention it.)

 

 

The issue is this: When accelerating from a space station, the rate of time of the travelling clock slows down. When de-accelerating to land on a space station, the clock's rate of time speeds up (it is no longer slowed down). This is not some apparent view of the travelling clock's time, it is real as can be seen by comparing the clocks when they are together, stationary and side-by-side. So how does it obtain those two different actions to its rate of time IF all reference frames are of equal standing?

 

 

 

When the clock's rate increases, it only goes back to the original rate it had in the rest frame. It does not go any faster than that, so there is no chance for it to "make up" any of the "lost" time.

 

Going back to the car example. If they both go 100 kph, and one slows down to 50 kph for an hour, it will be 50 km behind the other car. If it then speeds back up to 100 kph, it doesn't make up any ground, it just stops losing ground. It remains 50 km behind. (the acceleration phases are instantaneous in this example, for simplicity). The only way for it to make up the gap is to go faster than 100 kph, but nothing analogous to that happens with the clocks.

[robinpike]

To help explain this issue, here are some fictitious examples of a simpler nature.

 

Suppose that when moved to the right, clocks runs slower. And when moved to the left, clocks run quicker. Those two effects being confirmed by experiment.

 

If in the explanation of the above, a premise is included: that all directions in space are of equal standing. Then you can see there is an issue with that premise, for how can a clock run slower when moved to the right, and run quicker when moved to the left, when according to the premise all directions are of equal standing?

 

So now let's make it a bit more complicated.

 

Suppose that when moved away from a location in space, clocks run slower. And when moved towards that location in space, clocks run quicker. Those two effects confirmed by experiment.

 

If in the explanation of the above, a premise is included: that all locations in space are of equal standing. Then you can see that there is an issue with that premise, for how can a clock run slower when moved away from THAT location in space, and run quicker when moved towards THAT location in space, when according to the premise all locations are of equal standing?

 

Note that those two examples above are fictitious, mentioned just to help understand the more complex issue under discussion.

 

But the problem with relativity that I have been discussing is a little bit more complex than the above examples, because the premise at issue does not state that all directions in space are of equal standing, nor does it state that all locations in space are of equal standing, but rather the premise states that all inertial frames of reference are of equal standing.

 

And the experimental evidence is a bit more complex as well, because when clocks move, there are apparent changes in the rate of time of the clock. The only absolute experimental evidence that we can be absolutely sure about is that when the travelling clock returns to the stay at home clock, it has lost time but it has the same rate of time as the stay at home clock.

When the clock's rate increases, it only goes back to the original rate it had in the rest frame. It does not go any faster than that, so there is no chance for it to "make up" any of the "lost" time.

 

 

That is the issue! How does it go from a slower rate of time back to the original rate of time - IF the premise that all frames of inertial reference are of equal standing is true?

Edited August 24, 2016 by robinpike

[studiot]

 

RobinPike

 

Suppose that when moved to the right, clocks runs slower. And when moved to the left, clocks run quicker.

 

 

 

Winnie the Pooh

 

Suppose they don't

[Mordred]

 

 

That is the issue! How does it go from a slower rate of time back to the original rate of time?

there is a difference between rate of time and accumulated number of ticks.

 

The duration between clock ticks restore to being the same. The number of ticks will be different.

 

You keep mixing these two up. It is the tick rate that is restored. Not the amount of time each clock records.

[robinpike]

there is a difference between rate of time and accumulated number of ticks.

 

The duration between clock ticks restore to being the same. The number of ticks will be different.

 

You keep mixing these two up. It is the tick rate that is restored. Not the amount of time each clock records.

 

Mordred, so how is the tick rate restored - IF the premise that all inertial frames of reference are of equal standing is true?

 

Winnie the Pooh

Suppose they don't

 

I like it!

[Strange]

That is the issue! How does it go from a slower rate of time back to the original rate of time - IF the premise that all frames of inertial reference are of equal standing is true?

 

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.)

Edited August 24, 2016 by Strange

[Mordred]

look at the Lorentz boost formula for time dilation. Lets start with the same reference frame (equivalent to at rest).

 

In this case between observers if both observers are at rest there is no time dilation. Both clocks tick at the same rate. When one clock goes to a different frame time dilation occurs. The tick rate is different from the clock at rest.

 

Over a period of time you count the number of ticks for each clock.

 

Now bring the second clock back to the same reference frame. You no longer have any lorentz boost. No longer have any time dilation between clocks.

 

The clocks both tick at the same rate. However the recorded number of ticks (time elapsed) will be different.

 

As I stated the accumulated time elapsed will be different. However the rate of ticks will restore.

(reading your posts this aspect is where your getting confused.)

Edited August 24, 2016 by Mordred

[michel123456]

 

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.

 

Ding (that's the sound of understanding in my hollow head)

And during the recovery, the moving clock at all instants is running slower than the clock at rest. At no moment it had to go faster.

 

Even not in the return trip?? (still not fully understanding) I mean no matter the direction of travel there is always time dilation? Or is there time dilation when going out and time "contraction" when coming back?

Edited August 24, 2016 by michel123456

[Strange]

Exactly.

 

And to undertsand why the two frames are not symmetrical, the important point is that they will differ in when the moving clock started and stopped its acceleration (i.e. relativity of simultaneity) and hence how many ticks took place in the "moving" frame of reference versus the "non moving".

[michel123456]

Ding (that's the sound of understanding in my hollow head)

And during the recovery, the moving clock at all instants is running slower than the clock at rest. At no moment it had to go faster.

 

Even not in the return trip?? (still not fully understanding) I mean no matter the direction of travel there is always time dilation? Or is there time dilation when going out and time "contraction" when coming back?

That was a question (the bold part).

[Mordred]

The two terms "dilated" and "contracted" are not symmetrical lol...

 

Dilated can mean contracted or expanded. however the Lorentz boost for both outgoing and ingoing will be contracted in both directions.

 

Ie the observer at rest watching the moving observer will always see the other clock as slower. Regardless of direction

Edited August 24, 2016 by Mordred

[J.C.MacSwell]

To help explain this issue, here are some fictitious examples of a simpler nature.

 

Suppose that when moved to the right, clocks runs slower. And when moved to the left, clocks run quicker. Those two effects being confirmed by experiment.

 

If in the explanation of the above, a premise is included: that all directions in space are of equal standing. Then you can see there is an issue with that premise, for how can a clock run slower when moved to the right, and run quicker when moved to the left, when according to the premise all directions are of equal standing?

 

So now let's make it a bit more complicated.

 

 

In your Bizarro World described it still works...right and left are relative to the observer, so no direction is preferred for the Bizzarro Physics...it is positions and directions of the clocks themselves relative to an observer that determine how they will be viewed, measured and interpreted. As long as Bizzarro Physics is self consistent all observers can reconcile their different interpretations to agree on causes and effects.

[Mordred]

Positions being properly termed events.

 

Each event observer and emitter has positions

 

[latex](\acute{ct},\acute{x},\acute{y},\acute{z})[/latex] and [latex](ct,x,y,z)[/latex]

Edited August 24, 2016 by Mordred