Clocks, rulers... and an issue for relativity

[michel123456]

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

Thank you for the clear answer.

However I have seen diagrams of the twins paradox that do not show that.

A little digging and back.

[Mordred]

you will see different rates but tbe observer at rest or in his own reference frame will always see his clock as the fastest clock.

Edited August 25, 2016 by Mordred

[michel123456]

from here http://www.einsteins-theory-of-relativity-4engineers.com/twin-paradox-2.html

 

 

The diagram on the right.

On the return trip.

_Traveling is at middle trip 2011, Earth time is 2015

 

_from there and after, for the Earth 2 years pass, from 2015 to 2017

_during those 2 years, the traveling is observed aging 4 years.

 

That is twice the rate.

 

Or am I wrong again?

 

---------------

(edit) there are many such diagrams, this one was took at random.

(2nd edit) the same double rate appears on the left diagram but not so evidently.

Edited August 24, 2016 by michel123456

[swansont]

 

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?

 

 

Accelerations do not involve inertial frames. When an acceleration is involved, you can no longer apply a caveat true only for inertial frames.

[Mordred]

yes different rates... however the sender of the greetings still sees his own clock as the fastest.

from here http://www.einsteins-theory-of-relativity-4engineers.com/twin-paradox-2.html

 

TwinParadox2.jpg

 

The diagram on the right.

On the return trip.

_Traveling is at middle trip 2011, Earth time is 2015

 

_from there and after, for the Earth 2 years pass, from 2015 to 2017

_during those 2 years, the traveling is observed aging 4 years.

 

That is twice the rate.

 

Or am I wrong again?

 

---------------

(edit) there are many such diagrams, this one was took at random.

(2nd edit) the same double rate appears on the left diagram but not so evidently.

[swansont]

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.)

 

I think the issue also is with the turnaround of the twins paradox, when the signals start arriving on earth close together. That seems to be an added confusion that's irrelevant to this issue. Your example of breaking the trip up into distinct parts is likely a better approach.

[Tim88]

A key premise of relativity is that there is not a fundamental frame of reference that is above all other frames of reference.

 

But this seems to lead to a logical conflict when explaining how clocks lose time when they travel in a round trip (the travelling twin scenario). Please can the steps of how the travelling clock loses time with respect to the stay at home clock be walked through, with particular focus on why the following is not a problem...

 

A recent post discussed travelling clocks and rulers.

 

The conclusion with regards to rulers, was that a travelling ruler does not physically change its length compared to a stay at home ruler. Although each frame of reference sees the other ruler as being shorter in length, this is mutual - and the other frame of reference sees the other ruler as being shorter. The travelling ruler does not actually change its physical length - and this is why when it returns to the stay at home ruler, their lengths remain unchanged.

 

IF the travelling ruler were to get PHYSICALLY SHORTER as it moved to its different frame of reference as it starts its journey, then this would cause a conflict with the initial premise (that there is not a fundamental frame of reference). For, in order for the ruler to return to its original length on return to the stay at home ruler, it would need to INCREASE its length as it returned to the stay at home frame of reference. But there is nothing special about that change in reference frame to the original change in reference frame when it started its journey - so it can't in one get physically shorter, and then in the other get physically longer, i.e. return to its original; length.

 

However, rulers do not physically change their lengths as they move from one frame of reference to another, so not an issue!

 

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.

"rulers do not physically change their lengths as they move from one frame of reference to another":

probably you mean, the lengths of rulers that change in velocity does not physically change.

If with "physical" you mean the same as I do, then that can't be correct, as moving clocks that are co-moving with such a ruler tick slower; the two cannot be logically disconnected! For the inverse argument, see the Kennedy-Thorndike experiment: assuming (physical) length contraction, the null result implied a change in the emission frequency. Since that time we got positive confirmation of the change in emission frequencies, from which we nowadays logically conclude that length contraction also has to occur (with a few plausible assumptions).

- https://en.wikipedia.org/wiki/Kennedy%E2%80%93Thorndike_experiment#Importance_for_relativity

 

It is also not an original premise that there is no "fundamental" frame of reference. As a matter of fact, SR was already worked out, on the assumption of such a frame, by Lorentz and Poincaré a few months before Einstein; and he merely declared that such an assumption is "superfluous" for the derivation.

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

 

Apart of the error indicated above, you are in good company with your reasoning, see Langevin's interpretation here, from p. 47:

https://en.wikisource.org/wiki/Translation:The_Evolution_of_Space_and_Time

 

Note one very important point however, which also is stressed in the preceding part of that same paper: all frames in uniform rectilinear motion are equivalent for doing observations, the same laws of physics hold in all of them (that's the PoR). As long as one keeps the phrasing non-philosophical (Einstein was good in that!), everyone can agree.

 

Concerning your last point, I have difficulty grasping what the issue is; but possibly the discussion by Langevin is useful for you.

PS. did you actually try to calculate an example? In my experience that is an almost necessary step for understanding.

You will then find that if you transpose your reference frame to one that is co-moving with the outbound going astronaut:

1. his wristwatch now appears to tick at the normal rate; you may think that this will prevent his watch to loose as much time as from the point of view when you use the Earth's rest frame as reference (or even that the earth clocks will seem to end up loosing more time).

2. however after turn-around, on the inbound trip his watch ticks much slower - even more so than if you would use the earth's rest frame, because now the speed that you assign to the astronaut on the inbound trip is much higher.

3. As a result (that's a design feature of the Lorentz transformations, in order to fulfill the PoR), the total amount that the astronaut's watch will be behind at the event of arriving home, is the same no matter what inertial reference frame you use.

Edited August 24, 2016 by Tim88

[swansont]

probably you mean, the lengths of rulers that change in velocity does not physically change.

If with "physical" you mean the same as I do, then that can't be correct, as moving clocks that are co-moving with such a ruler tick slower;

 

Not to someone in that frame. The clock and ruler are the same.

[michel123456]

yes different rates... however the sender of the greetings still sees his own clock as the fastest.

That is not true.

On the right diagram, outbound trip.

The traveler goes from 2007 to 2008, 1 year has passed for him, while 2 years have passed on Earth. So he observes the clock on Earth ticking twice faster than his one.

Edited August 25, 2016 by michel123456

[Mordred]

no, the reference frame of rest is the vertical column. If Jim is the observer on the diagram on the left. His clock is along the y axis. Not the diagonal. That diagonal line is Pams clock. Which if you draw a vertical line at 2012 to 2011 you can see Pams clock running slower than Jims clock.

 

For Pams clock the Pam is the sender image on right. Same thing Pam sees Jims clock running slower.

 

AT no point does an observer see another clock running faster than his own...

 

DO NOT CONFUSE time elapsed with time rate. They are two very different units of measure. The first is accumulated time. The second is time per a unit.

 

The sender of the greeting Jim on the left sees his clock as the fastest. The sender on the right Pam sees her clock running the fastest.

yes different rates... however the sender of the greetings still sees his own clock as the fastest.

Jim sender on the left diagram sees his own clock as fastest. Pam sender on right diagram sees her clock as fastest. The quoted section is accurate.

Edited August 25, 2016 by Mordred

[michel123456]

no, the reference frame of rest is the vertical column. If Jim is the observer on the diagram on the left. His clock is along the y axis. Not the diagonal. That diagonal line is Pams clock. Which if you draw a vertical line at 2012 to 2011 you can see Pams clock running slower than Jims clock.

 

For Pams clock the Pam is the sender image on right. Same thing Pam sees Jims clock running slower.

 

AT no point does an observer see another clock running faster than his own...

 

DO NOT CONFUSE time elapsed with time rate. They are two very different units of measure. The first is accumulated time. The second is time per a unit.

 

The sender of the greeting Jim on the left sees his clock as the fastest. The sender on the right Pam sees her clock running the fastest.

Here?

For the time elapsed observed to go from 2011 to 2015 (4 years) while the observer lives 2 years (from 2015 to 2017), doesn't that mean that the traveling clock is observed ticking twice as fast? IOW that the time rate is observed faster?

 

?????????????????????

Edited August 25, 2016 by michel123456

[J.C.MacSwell]

Gravitational effects aside, an observer in an inertial frame will never observe/calculate someones clock to be running faster. However, they can observe/see a blueshifted transmission from a clock that will make it appear as such due to the diminishing distance of transmission. When they allow for that they will measure the approaching clock to be running slower.

[Janus]

Here?

TwinParadox2b.jpg

For the time elapsed observed to go from 2011 to 2015 (4 years) while the observer lives 2 years (from 2015 to 2017), doesn't that mean that the traveling clock is observed ticking twice as fast? IOW that the time rate is observed faster?

 

?????????????????????

I believe that Mordred and you are using using "see" in different ways.

 

see

sē/

verb

1.

perceive with the eyes; discern visually.

"in the distance she could see the blue sea"

 

2.

discern or deduce mentally after reflection or from information; understand.

"I can't see any other way to treat it"

You are using definition 1 and Mordred is using definition 2. Definition 2 is used most often in Relativity. Mostly because we are more interested with what is happening with another clock than what we visually see happening.

For example in the above, the reason Jim visually sees Pam's clock ticking faster is that the distance between Pam and Jim is decreasing, and each successive signal takes less time to cross the distance between them. This effect is in addition to the Relativistic time dilation that we are more interested in and in most cases simply ignored.

[Mordred]

Here?

TwinParadox2b.jpg

For the time elapsed observed to go from 2011 to 2015 (4 years) while the observer lives 2 years (from 2015 to 2017), doesn't that mean that the traveling clock is observed ticking twice as fast? IOW that the time rate is observed faster?

 

?????????????????????

Ok even without accounting for seperation distance mentioned above. Lets look at those two numbers.

 

4 years and 2 years. This represents the total time elapsed. This isn't rate of time. Rate of time is the time for each second to tick. Total time is the accumulation of those ticks.

 

It is the ticks of the clock that synchronizes when the clock is in the same reference frame. Not the total time elapsed.

 

As mentioned above there is also a difference between time it takes to recieve each tick, when you involve direction. As mentioned you need to account for this.

 

This is where understanding the formulas used to generate those graphs becomes important.

 

its also important to keep track of who the observer is...

I believe that Mordred and you are using using "see" in different ways.

 

 

You are using definition 1 and Mordred is using definition 2. Definition 2 is used most often in Relativity.

I suspect this is common, a person with familiarity with the subject can more readily deduce the correct relations. Speaking of relations. Just a side note any time I see a linear graph I instantly think [latex]y=mx+c [/latex].

 

I tend to do this also with common waveforms (gotta love Laplace transformation).

 

There is a lesson I learned years ago.

 

"The most complex problems can be simplified if you reduce it to its individual parts"

 

when I look at a graph I start reducing it.

Edited August 25, 2016 by Mordred

[michel123456]

I believe that Mordred and you are using using "see" in different ways.

 

 

You are using definition 1 and Mordred is using definition 2. Definition 2 is used most often in Relativity. Mostly because we are more interested with what is happening with another clock than what we visually see happening.

For example in the above, the reason Jim visually sees Pam's clock ticking faster is that the distance between Pam and Jim is decreasing, and each successive signal takes less time to cross the distance between them. This effect is in addition to the Relativistic time dilation that we are more interested in and in most cases simply ignored.

So you agree that the observer "visually sees" the moving clock ticking faster. Because of the direction of motion.

 

If this effect is not so interesting ("ignored"), why is the other part, the outbound, important?

And what is this whole diagram about if it is not about relativistic effects???? It is supposed to explain the twins paradox, which is a relativistic effect.

[Mordred]

the effect isn't ignored. You need to account for doppler shift to get the time dilation portion. You seem to have difficulty seperating the individual cause and effects.

 

change in clock rates due to time dilation

 

change in percieved clock rates due to distance change.

Edited August 25, 2016 by Mordred

[Xerxes]

Though Mordred's logic is compelling, I am left with an uncomfortable feeling.

 

Namely, granted that 2 clocks with shared coordinates with respect to which they are equally at rest (or co-moving) they function identically. And yet the conclusion seems to be that their relative accumulated time (i.e. their relative "ages") depends upon their respective kinematic histories.

 

And given that these histories may not always be accessible, are we to conclude that elapsed time i.e. age, is a fiction?

 

Sorry if this is a foolish question- my excuse is I am not a physicist

[Tim88]

Not to someone in that frame. The clock and ruler are the same.

It's similar with the changing magnetic field of an accelerating electron. Instead of claiming that such concepts as time dilation and certain magnetic fields are "unphysical", we more commonly say that they are "relative". And the effects of a change in velocity are in certain ways "absolute", as Langevin so nicely explained.

[Strange]

And given that these histories may not always be accessible, are we to conclude that elapsed time i.e. age, is a fiction?

 

 

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

 

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

[michel123456]

the effect isn't ignored. You need to account for doppler shift to get the time dilation portion. You seem to have difficulty seperating the individual cause and effects.

 

change in clock rates due to time dilation

 

change in percieved clock rates due to distance change.

Dear Mordred

With all my respect

If this diagram is accurate, the observer at rest observes the traveling clock ticking faster while on its route back.

OTOH it has been said repeatedly that it is not possible, that the observer at rest will always observe the traveling clock ticking slower independently of the direction of travel. Which must be totaly wrong if I understand clearly.

 

 

You cannot state that the diagram does not count for relativistic effects since the number of time stamps upon the traveling path comes out from a relativistic formula.

You cannot state that the equations do not take count of the direction, it would be a terrible mistake.

And you cannot state that the effects of time dilation must be separated from effects due to distance change, that would be a mistake too, because the distance do change.

 

IOW how do you expect me to understand anything from all this mess.

[swansont]

One concept you have to consider comes from Einstein clock synchronization. In a single frame you synchronize remote clocks by accounting for how long the signal took to get to you. i.e. if a clock sends out a signal at noon and you are one light-second away, you account for that second when setting the time on your clock.

 

That concept doesn't go away when one clock is moving. The frequency of the arriving signals is not indicative of the frequency of the clock when there is relative motion. In the spacetime diagram you will notice that the dilation (4/5) is not reflected in the signal rate of either the outbound or inbound clock, which change by factors of 2 relative to the earth clock.

[michel123456]

One concept you have to consider comes from Einstein clock synchronization. In a single frame you synchronize remote clocks by accounting for how long the signal took to get to you. i.e. if a clock sends out a signal at noon and you are one light-second away, you account for that second when setting the time on your clock.

 

That concept doesn't go away when one clock is moving. The frequency of the arriving signals is not indicative of the frequency of the clock when there is relative motion. In the spacetime diagram you will notice that the dilation (4/5) is not reflected in the signal rate of either the outbound or inbound clock, which change by factors of 2 relative to the earth clock.

I don't understand the bold part.

The time dilation is introduced de facto. There are 8 time stamps intervals for the moving clock, and 10 for the standing clock. That represents time dilation.

This time dilation is observed by the standing at rest observer: the observation is represented by the diagonal simultaneity lines.

[Mordred]

Those time stamps contain two details. Not one. The time dilation portion of each time stamp is the gamma factor. (Lorentz boost). If the ship is moving at constant velocity the Lorentz boost itself is constant.

 

The other detail in those time stamps is seperation distance. This is the additional delay. This seperation distance portion isn't time dilated. It simply increases the time of a signal to reach a to b. Not due to dilation but due to change in distance.

Edited August 26, 2016 by Mordred

[swansont]

I don't understand the bold part.

The time dilation is introduced de facto. There are 8 time stamps intervals for the moving clock, and 10 for the standing clock. That represents time dilation.

This time dilation is observed by the standing at rest observer: the observation is represented by the diagonal simultaneity lines.

8/10 = 4/5

 

But if you look at the rate of the clock signals being received on earth, what do you see? Every 2 years, or every 1/2 year. The rate at which the signals are received do not tell you the actual time dilation. The signal rate indicates the Doppler shift. This is explained in the link you provided.

[Mordred]

The time dilation portion is

[latex]\Delta\acute {T}=\frac{\Delta T_o}{\sqrt {1-\frac{v^2}{c^2}}}[/latex]

or alternately [latex]\acute{T}=\gamma T_0[/latex]

so if you set the velocity as constant the time dilation portion will stay constant. The additional delay is due to seperation distance. Not time dilation.

The spacetime interval which is defined as the seperation distance between events includes both distance and time deviation.
The time dilation length contraction is a path deviation. Without time dilation this portion is simply the distance. With time dilation the distance follows a curved path. The intervals in your link has two influences upon distance. Spatial seperation and curvature due to dilation. It is the curvature influence were intetested in.

look at each specific formula for the distance of seperation.



https://en.m.wikipedia.org/wiki/Time_dilation

honestly this has been mentioned several times. ((by myself and others)) SO I have no idea how one can miss the seperation distance itself. Unfortunately the link that includes those graphs did not mention the pertinant detail of seperation distance. (at least it didn't go into the mathematical detail of how to distinquish between the Doppler portion (toward/away from) redshift to the time dilation influence on the Doppler). Those formulas are on the link above.

Nor does that link resolve the twin paradox itself. (or even describe the actual paradox) which is resolved by Pams rotating frame/acceleration/deceleration.

 

lets try this angle....

 

[latex]\acute{x}=\gamma(x-vt),\acute{y}=y,\acute{z}=z,\acute{t}=\gamma(\frac{t-vx}{c^2})[/latex]

[latex]\gamma=(1-\frac{v^2}{c^2})^{-\frac{1}{2}}[/latex]

 

[latex]\acute{x}=\gamma(x-vt)[/latex] if [latex]\gamma=0[/latex] then the seperation along the x axis is simply the spatial components, as there is no length contraction or time dilation.

[latex]\acute{x}=x,\acute{y}=y,\acute{z}=z,\acute{t}=t[/latex]

 

so separation between events follows Pythagoras theory, in the former case as the x axis and time axis changes due to gamma, Pythagoras theory no longer holds true. (this deviation is your length contraction and time dilation)

Edited August 26, 2016 by Mordred