Clocks, rulers... and an issue for relativity

[robinpike]

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.

[Strange]

This is the same question about the (measured) change in units versus the elapsed/distance time.

 

You are comparing a ruler with a clock. You should be comparing a ruler with a metronome and an odometer with a clock.

 

In the case of a ruler and a metronome, an observer in another frame of reference will see them foreshortened and running slower. But they will both return to the same units of measurement when returned that to that frame of reference.

 

In the case of an odometer or a clock, they will both show that the total distance travelled / time passed is different for the "stationary" and the moving observer.

[StringJunky]

In my thread Clocks and Rulers, I, ignorantly, conflated the ruler and clock as directly equivalent when I should have used an odometer instead of a ruler to compare properly with a clock..

Edited August 15, 2016 by StringJunky

[Strange]

It isn't immediately obvious that they are measuring different things, though.

[StringJunky]

It isn't immediately obvious that they are measuring different things, though.

No, it isn't, that's why I asked the question. One doesn't come across odometers everyday except if one drives a car - I call it a 'mileometer'! -. In fact, I didn't know the word until yesterday. . it sounds like a device for smelling things with.

Edited August 15, 2016 by StringJunky

[TakenItSeriously]

Think of it this way. The clock is not preserving time dilation because it was never dialated in your own frame.

 

What the clock is preserving is the length contraction you experienced in your travels. So every time you experienced x% length contraction of the Earth frame you're travel time was shorter by x% as well and that is what the clock was preserving.

[Endy0816]

@StringJunky: Happens all the time

 

Only reason I thought of odometers was that I had to work on one before. Wouldn't have had made the connection otherwise. Definitely unintuitive.

 

 

This is the same question about the (measured) change in units versus the elapsed/distance time.

 

You are comparing a ruler with a clock. You should be comparing a ruler with a metronome and an odometer with a clock.

 

In the case of a ruler and a metronome, an observer in another frame of reference will see them foreshortened and running slower. But they will both return to the same units of measurement when returned that to that frame of reference.

 

In the case of an odometer or a clock, they will both show that the total distance travelled / time passed is different for the "stationary" and the moving observer.

Metronome, good example.

Edited August 15, 2016 by Endy0816

[robinpike]

This is the same question about the (measured) change in units versus the elapsed/distance time.

 

You are comparing a ruler with a clock. You should be comparing a ruler with a metronome and an odometer with a clock.

 

In the case of a ruler and a metronome, an observer in another frame of reference will see them foreshortened and running slower. But they will both return to the same units of measurement when returned that to that frame of reference.

 

In the case of an odometer or a clock, they will both show that the total distance travelled / time passed is different for the "stationary" and the moving observer.

 

The concern is simpler than that. This is about the travelling clock losing time - a physical, real loss in time compared to the stay at home clock - and explaining how the clock returns to the same rate of time without requiring a special frame of reference.

 

First of all the assumption is that the loss in time for the travelling clock is caused by a real change in something. If this assumption is incorrect, then please explain.

 

Here are the possibilities that I can think of for what could change for the travelling clock...

 

(1) If it is because the clock click rate slows down as the travelling clock goes from one frame of reference to another, then movement into a special frame of reference is required in order to return the clock tick rate back to the stay at home clock's click rate. Thus contradicting the premise in relativity that all frames of reference are of equal standing.

 

(2) If it is because distance becomes shorter as the travelling clock goes from one frame of reference to another, then again, movement into a special frame of reference is required in order to return the clock's shorter route through space back to the stay at home clock's passage through space. Again thus contradicting the premise in relativity that all frames of reference are of equal standing.

 

(3) If it is because the route through space-time becomes shorter as the travelling clock goes from one frame of reference to another, then again, by which change of reference frame allows the clock to get back to the stay at home clock's longer route through space-time?

 

(4) Any other explanation...

 

Hope this elucidates the issue?

Edited August 15, 2016 by robinpike

[swansont]

It doesn't require a special frame. You just need to return to the starting frame. It is the change in reference frame that causes the asymmetry. If you only analyzed the clocks in inertial frames, each would see the other clock as running slow.

[robinpike]

It doesn't require a special frame. You just need to return to the starting frame. It is the change in reference frame that causes the asymmetry. If you only analyzed the clocks in inertial frames, each would see the other clock as running slow.

 

This concern applies to the mechanism by which the travelling clock ends up with less time to its stay at home clock, and by what ever mechanism is given, surreptitiously making the stay at home frame special.

 

For, if leaving the stay at home frame causes something to change, then returning to the stay at home reference should repeat that same change - not reverse it? Since both are just actions moving from one frame to another.

 

For example, if moving from one frame to another causes the travelling clock to have a shorter route through space-time, then since no frame is special, then when the travelling clock returns to the original frame, that too should be treated as the travelling clock moving from one frame to another - and thus cause it to have a shorter route through space-time? How can those two actions end up being different without making the stay at home frame special?

Edited August 15, 2016 by robinpike

[Strange]

https://en.wikipedia.org/wiki/Twin_paradox

[swansont]

This concern applies to the mechanism by which the travelling clock ends up with less time to its stay at home clock, and by what ever mechanism is given, surreptitiously making the stay at home frame special.

 

For, if leaving the stay at home frame causes something to change, then returning to the stay at home reference should repeat that same change - not reverse it? Since both are just actions moving from one frame to another.

 

For example, if moving from one frame to another causes the travelling clock to have a shorter route through space-time, then since no frame is special, then when the travelling clock returns to the original frame, that too should be treated as the travelling clock moving from one frame to another - and thus cause it to have a shorter route through space-time? How can those two actions end up being different without making the stay at home frame special?

It's "special" because it's inertial, just like an infinite number of other frames.

 

The twins paradox begins with the clocks in different inertial frames. You need to be more familiar with the science you are critiquing.

 

 

(But the reverse of speeding up is slowing down, not speeding up more)

[studiot]

Robin, try this explanation of the twins.

 

Call the stay at home twin A and the travelling twin B.

 

A never leaves Earth.

So A never arrives anywhere else.

 

This is the crux of the difference.

 

B leaves Earth

B arrives somewhere else.

B returns to Earth.

 

So B travels within the universe from one place to another, visiting places in between.

 

If we were to suggest that what B sees is equivalent to what A sees, but the other way round, then we would be just plain wrong.

If we suggest that B sees the rest of the universe recede and then return to B, it would imply that the rest of the universe must arrive somewhere else.

But B can determine when he returns, after consultation with A, that A never went anywhere.

So at least a part of the rest of the universe did not go anywhere (ie did not arrive somewhere else).

Suggesting that the universe recedes is the same as suggesting that B travels outside the universe.

 

What swansont is suggesting in post#12 is that A's journey is not through space at all.

It is only through time.

In terms of Frames, A is in frame (x,y,z,t) and B is in Frame (X,Y,Z,T)

 

A's journey is (0.0,0,0) ; (0,0,0,t₁) ; (0,0,0,t₂) ; (0,0,0,t₃) ; (0,0,0,t₄) ; (0,0,0,t₅) ; etc

 

On the other hand B sees his journey as (0,0,0,0) ; (X₁,0,0,T₁) ; (X₂,0,0,T₂) ; (X₁,0,0,T₂) ; (0,0,0,T₅)

 

Setting the X axis along B's journey to Betelgeuse, which he reckons is X₂ distant.

This makes the travel along the Y and Z axes all 0.

 

The fact the two frame origins were momentarily in coincidence at the start is the result of synchronisation, but since there is relative motion between the frames there is no reason to expect them to coincide at any other event.

 

What is happening is that they coincide in space and time the first event due to synchronisation , but the second time they coincide in space, but not in time since t₅ is not the same as T₅.

[MigL]

You haven't got a clear understanding of frames either.

The travelling clock does not go from one frame to another at any point.

It is always in its own frame. But it can be observed from a multitude of other frames. One of which, is the stay-at-home clock's.

And none of them are special or preferred.

[michel123456]

Robin, try this explanation of the twins.

 

Call the stay at home twin A and the travelling twin B.

 

A never leaves Earth.

So A never arrives anywhere else.

 

This is the crux of the difference.

 

B leaves Earth

B arrives somewhere else.

B returns to Earth.

 

So B travels within the universe from one place to another, visiting places in between.

 

If we were to suggest that what B sees is equivalent to what A sees, but the other way round, then we would be just plain wrong.

If we suggest that B sees the rest of the universe recede and then return to B, it would imply that the rest of the universe must arrive somewhere else.

But B can determine when he returns, after consultation with A, that A never went anywhere.

So at least a part of the rest of the universe did not go anywhere (ie did not arrive somewhere else).

Suggesting that the universe recedes is the same as suggesting that B travels outside the universe.

 

What swansont is suggesting in post#12 is that A's journey is not through space at all.

It is only through time.

In terms of Frames, A is in frame (x,y,z,t) and B is in Frame (X,Y,Z,T)

 

A's journey is (0.0,0,0) ; (0,0,0,t₁) ; (0,0,0,t₂) ; (0,0,0,t₃) ; (0,0,0,t₄) ; (0,0,0,t₅) ; etc

 

On the other hand B sees his journey as (0,0,0,0) ; (X₁,0,0,T₁) ; (X₂,0,0,T₂) ; (X₁,0,0,T₂) ; (0,0,0,T₅)

 

Setting the X axis along B's journey to Betelgeuse, which he reckons is X₂ distant.

This makes the travel along the Y and Z axes all 0.

 

The fact the two frame origins were momentarily in coincidence at the start is the result of synchronisation, but since there is relative motion between the frames there is no reason to expect them to coincide at any other event.

 

What is happening is that they coincide in space and time the first event due to synchronisation , but the second time they coincide in space, but not in time since t₅ is not the same as T₅.

Great explanation! One of the clearest ever.

 

Only, only the bold part should be stated otherwise because it leads to confusion: if they coincide in space and not in time that mean that they do not meet together. (their world lines cross but they do not meet*)

You surely meant something else: that they do indeed meet but that their measure of time is different. If you didn't mean that then you have a problem and that would make me very glad (no offense intended).

 

*don't ask me how that can be possible

Edited August 16, 2016 by michel123456

[studiot]

Great explanation! One of the clearest ever.

 

Only, only the bold part should be stated otherwise because it leads to confusion: if they coincide in space and not in time that mean that they do not meet together. (their world lines cross but they do not meet*)

You surely meant something else: that they do indeed meet but that their measure of time is different. If you didn't mean that then you have a problem and that would make me very glad (no offense intended).

 

*don't ask me how that can be possible

 

Thank you for the endorsement.

Indeed I meant that they meet in space, but their measure of time is different.

 

As I mentioned (x,y,z,t) perhaps some further clarification is in order, since they are not truly independent coordinates.

 

Let us consider a simpler system to start with containing just p and q ie (p,q) for generality.

 

If p and q are truly independent we can pick a point with any value of p and then choose any value of q we like without reference to the value of p.

Or if you like, for every value of p there is a point corresponding to every value of q.

Doing this of course generates the the whole pq plane.

 

But

 

Suppose I introduce a constraint, say p² + q² = k², where k is a constant.

 

Without the constraint the system has 'two degrees of freedom' ie we can independently vary two variables.

 

With the constraint, x and y edit p and q oops! are no longer independent.

In fact you may recognize the common equation of a circle and the points we are able to choose from lie on a circle of radius k.

 

Most people automatically assume that (x,y,z,t) coordinates have four degrees of freedom as with the pq plane.

 

But in a similar (though more complicated) manner there is a constraint introduced by relativity which is

 

s² = x² + y² + z² -c²t²

 

So there are less than four degrees of freedom available and the variables are not truly independent.

 

Edited August 16, 2016 by studiot

[robinpike]

You haven't got a clear understanding of frames either.

The travelling clock does not go from one frame to another at any point.

It is always in its own frame. But it can be observed from a multitude of other frames. One of which, is the stay-at-home clock's.

And none of them are special or preferred.

 

MigL, aplogies, my understanding of frames is not technical. Here is a non relativistic example of what I was trying to describe, is this a more correct use of the term frames?

 

I am driving in my car along the freeway at 50 mph. My frame is the inside of my car - let's call this Frame-A.

 

I am driving alongside another car, also travelling at 50 mph - let's call this Frame-B.

 

These two frames are different - hence they have different names - but for all intents and purposes they are almost equivalent to each other.

 

Hopefully, I am using the concept of frames correctly now?

 

A third car comes up from behind travelling at 60 mph - let's call the frame that that car is in, Frame-C.

 

As the car comes alongside my car, I accelerate up to 60 mph - but since I am still sitting in my car - FrameA - I have not changed frames, my frame is still Frame-A.

 

However, my Frame-A is no longer equivalent to Frame-B, for now it has become equivalent to Frame-C.

 

Is this still the correct use of the term frame?

 

So, I should not have said that the travelling clock changes from one frame to another, that was incorrect. But rather I should have said that the travelling clock's frame remains the same - but what changes is what other frames, its frame becomes equivalent too?

 

 

 

 

Studiot, thanks for your post - I need some time to think how to describe my concern from logic in words, to logic in maths.

Edited August 16, 2016 by robinpike

[swansont]

MigL, aplogies, my understanding of frames is not technical. Here is a non relativistic example of what I was trying to describe, is this a more correct use of the term frames?

 

I am driving in my car along the freeway at 50 mph. My frame is the inside of my car - let's call this Frame-A.

 

I am driving alongside another car, also travelling at 50 mph - let's call this Frame-B.

 

These two frames are different - hence they have different names - but for all intents and purposes they are almost equivalent to each other.

 

Hopefully, I am using the concept of frames correctly now?

 

A third car comes up from behind travelling at 60 mph - let's call the frame that that car is in, Frame-C.

 

As the car comes alongside my car, I accelerate up to 60 mph - but since I am still sitting in my car - FrameA - I have not changed frames, my frame is still Frame-A.

 

However, my Frame-A is no longer equivalent to Frame-B, for now it has become equivalent to Frame-C.

 

Is this still the correct use of the term frame?

 

So, I should not have said that the travelling clock changes from one frame to another, that was incorrect. But rather I should have said that the travelling clock's frame remains the same - but what changes is what other frames, its frame becomes equivalent too?

 

 

 

 

Studiot, thanks for your post - I need some time to think how to describe my concern from logic in words, to logic in maths.

No, that's not correct. Frames are defined by velocities, not containers.

 

Frame A and B are the same. When you accelerate to 60 mph, you have changed frames.

[StringJunky]

No, that's not correct. Frames are defined by velocities, not containers.

I mentioned in my thread of the same that a lot of people are thinking this in terms of discrete space around/within a frame being affected when it's actually the observer and the signal that should be considered.

[robinpike]

No, that's not correct. Frames are defined by velocities, not containers.

 

Frame A and B are the same. When you accelerate to 60 mph, you have changed frames.

 

Swansont, thank you - that is my understanding of frames. I am not sure why MigL suggested that moving from one frame to another was an incorrect use of the term frame?

 

The example of using the car was a simple way to describe the velocity of the frame - I didn't mean to imply the frame was the space inside the car.

Edited August 16, 2016 by robinpike

[swansont]

Swansont, thank you - that is my understanding of frames. I am not sure why MigL suggested that moving from one frame to another was an incorrect use of the term frame?

 

The example of using the car was a simple way to describe the velocity of the frame - I didn't mean to imply the frame was the space inside the car.

Regardless of the source of the misconception, when you change velocity, you change your (inertial) frame.

I mentioned in my thread of the same that a lot of people are thinking this in terms of discrete space around/within a frame being affected when it's actually the observer and the signal that should be considered.

And anything moving with the same velocity (which might be zero)

[StringJunky]

And anything moving with the same velocity (which might be zero)

Yes. Right.

[MigL]

If you are sitting in your car , whether at 0, 50, 60 mph, or even at 0.99c, everything will behave and be measured exactly the same, inside your ( very fast ) car. Because The inside of your car is always the same frame. It does not change.

( for simplicity we disregard acceleration )

The observer in the car next to you may make measurements of events in your frame, and the results could be totally different to his own frame, or they could be alike depending on whether he is in a different frame or the same one.

If he is in the car beside you matching your velocity, he is in the same frame.

If he is at rest WRT you he will measure relativistic effects in your frame ( time, length and mass changes of varying degrees ), and you in his.

As has been pointed out, a car or container doesn't define a frame

[StringJunky]

...If he is in the car beside you matching your velocity, he is in the same frame.

If he is at rest WRT you he will measure relativistic effects in your frame ( time, length and mass changes of varying degrees ), and you in his.

Shouldn't they be the same frame; If the guy in the car next you is moving at the same velocity or at rest WRT you. It looks like to me, atm, you saying the same thing two different ways.

Edited August 16, 2016 by StringJunky

[michel123456]

Thank you for the endorsement.

Indeed I meant that they meet in space, but their measure of time is different.

 

As I mentioned (x,y,z,t) perhaps some further clarification is in order, since they are not truly independent coordinates.

 

Let us consider a simpler system to start with containing just p and q ie (p,q) for generality.

 

If p and q are truly independent we can pick a point with any value of p and then choose any value of q we like without reference to the value of p.

Or if you like, for every value of p there is a point corresponding to every value of q.

Doing this of course generates the the whole pq plane.

 

But

 

Suppose I introduce a constraint, say p² + q² = k², where k is a constant.

 

Without the constraint the system has 'two degrees of freedom' ie we can independently vary two variables.

 

With the constraint, x and y are no longer independent.

In fact you may recognize the common equation of a circle and the points we are able to choose from lie on a circle of radius k.

 

Most people automatically assume that (x,y,z,t) coordinates have four degrees of freedom as with the pq plane.

 

But in a similar (though more complicated) manner there is a constraint introduced by relativity which is

 

s² = x² + y² + z² -c²t²

 

So there are less than four degrees of freedom available and the variables are not truly independent.

Wonderful! You should publish.

 

So, the 4D spacetime is not full, isn't it? Like a circle is a part of a plane, the equation that you posted (s² = x² + y² + z² -c²t²) describes what kind of geometric figure?

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

Sorry but I feel the thread is splitting in 2 very interesting parts.