robinpike

Tim88 +1 for your very explicit and understandable posts.

And I am glad that we agree that in no way an observer can make things happen in an object just by looking at it.

Because sometimes I fell getting crazy reading some very knowledgeable posters seemingly stating otherwise (but that must be me who understands badly).

IOW, I am an observer, I look in my telescope to some distant star. This has no effect upon the star.

If the star looks length contracted to me, it is merely an effect of observation, nothing else. If I start running like the flash, the star will not change length.

Michel, we appear to share some common ground on our questions about relativity.

Looking back on my questions within my topic, I realize that I have been concerned about the offered explanation(s) finishing with the shorter route through space-time as being real, but earlier in the explanation it relying on the shorter route through space-time as being only apparent - i.e. dependent on which observer is doing the looking?

Is this the same sort of concern that you have?

Thank you all, I need time to consider your posts, please bear with me. In the meantime, explaining what I look for in your answers might help with understanding why I continue to ask questions.

As Celeritas and others have kindly summarized, the different aging of a travelling clock to its stay at home version, is a consequence of the travelling clock taking a shorter route through space-time.

I only focus on when the travelling clock returns to its original stay at home reference frame, as this shows that the different aging is real (and is agreed by all observers).

I then look at what the explanation says the travelling clock did to end up taking a shorter route through space-time - and look to see if there is a logical issue with that explanation.

So what do I mean by a logical issue?

Even if the maths agrees with what is measured, is there a contradiction in the explanation?

A simple example: A and B are constants, and the explanation relies on A being greater than B, and then later on the explanation relies on A being less than B. So this is raised as a logical inconsistency in the explanation (even though the maths might agree with what is measured).

Then someone points out that there is no actual issue, as actually B is equal to the square root of a constant C, and so B can take on a positive and a negative constant value. And so all is well after all.

Relativity is a lot more complicated, and as Celeritas mentions, it has not been toppled in 116 years.

The inconsistency that I look for, does the explanation finish with the shorter route through space-time as being real, but earlier in the explanation it relies on the shorter route through space-time as being only apparent - i.e. dependent on which observer is doing the looking.

Hope this helps with why I ask these questions.

 

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.

 

 

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.

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.)

 

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.

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!

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?

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.

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?

 

 

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?

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.

Here is the issue re-iterated using co-ordinates to notate the space and time of the stay at home clock and the travelling clock.

If we use the stay at home clock's inertial frame as the base reference in space and time, we can represent the space and time co-ordinates (x,y,z,t), say starting as these values (0,0,0,t1), the two clocks starting with the same space-time co-ordinates.

When the travelling clock leaves for its round trip journey, the stay at home clock's co-ordinates vary just with a steady movement through time, say as...

• stay at home clock: (0,0,0,t1), (0,0,0,t2), (0,0,0,t3), (0,0,0,t4), (0,0,0,t5), (0,0,0,t6), (0,0,0,t7)

Whereas when the travelling clock starts its journey, it accelerates into a different inertial frame of space-time, say its co-ordinates become...

• travelling clock: (0,0,0,t1), (X2,0,0,T2), (X3,0,0,T3)

Midway in its journey at X3 distance, the travelling clock accelerates into a different inertial frame of reference of space-time in order to return home, say as...

• travelling clock (X3,0,0,T3), (X2,0,0,T4), (X1,0,0,T5)

finally accelerating (de-accelerating) into the inertial frame of reference into the stay at home clock's original position in space, but NOT time

• travelling clock returns as (0,0,0,T6)

where the travelling clock's position in time (T6) is less than the stay at home clock's position in time (t7)

For simplicity I will ignore mutual reciprocal differences between inertial reference frames, and just concentrate on the final real difference, that is the less time on the travelling clock and look at explanations of how that occurred.

Perhaps the simplest to understand is to suggest that the rate of time slowed down while the travelling clock made its journey, but this is at odds with the premise that all inertial frames of reference are of equal standing, for this reason. Let's say as the travelling clock moved from the stay at home's inertial frame of reference, its rate of time slowed down. But then at its halfway point in its journey, when it accelerated into its return journey inertial frame of reference, its rate of time must slow down again - as there is no way to differentiate this change in inertial reference frames as any other. And finally, when it returns to the stay at home clock's inertial frame of reference, again its rate of time must slow down. But this is at odds with the final position, because although the travelling clock has lost time, its rate of time is nonetheless the same as the stay at home clock's rate of time.

So let's try another explanation.

When the travelling clock changes inertial frames of reference, its rate of time never changes, but instead its rate through space-time changes, it getting moving through space-time in fewer clock ticks than the stay at home clock. But again this has the same issue: if changing inertial frames of reference cause the travelling clock to move through space-time quicker, there is no change in inertial reference frame by which it can return to the stay at home clock's rate of movement through space-time.

With the premise that all inertial frames of reference are of equal standing, there seems to be no method by which the travelling clock can lose time AND return to the stay at home clock's rate of time / movement through space-time?

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.

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.

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?

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?

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.

 

 

The ruler can only be in its own frame. It never moves to a different frame. Observers in other frames measure it as having a different length.

That is a good explanation as it avoids leading to logical contradictions.

The OP question asked why the time on the traveller's clock became out of sync with the stay at home clock, but the length of the ruler did not. Why does length and time behave differently?

The way this has been explained to me is that the rate of time on the travelling clock does NOT change - in the same way that the length of the ruler does not change. What changes is the distance of the journey of the travelling clock through 'space-time'.

Although maybe counter-intuitive - the travelling clock takes a SHORTER distance through space-time, and hence less time is shown on its clock when it returns than the non travelling clock. At no point did the rate of time on the travelling clock slow down.

 

 

 

Nothing physical changes. That implies it goes from one state to another, and it doesn't. The frame changes, and the length depends on the frame.

So just to be clear, the physical length of a ruler does not change when it moves from one frame of reference to another, it is the ability to measure that length of the ruler that changes?

You need to think of what is meant by synchronized and clock rate.

 

The clock will no longer be synchronized. However both length and clock rate will return to match the original conditions.

 

Albiet the clock wont be synchronized.

Mordred, when you say 'both length and clock will return to match the original conditions', can you elaborate a bit more please as to what, if anything, happens to the ruler?

Is a change to the ruler's length just an apparent change - not an actual physical change?

I ask, because if the ruler does get physically shorter on its journey away from our frame of reference - by what action allows it to return to 'our synchronized original length'?

DrmDoc, if we were to have a soul, does that mean animals have a soul as well? If not, why not? What keeps the soul attached to a person's body? Could a soul move from one person to another person, or to an animal? Could two souls occupy the same body?

Has anyone here ever truly thought about how we exist, how we are self aware.When i think about this deeply, i realize something thats very strange, atoms.

we are all made up of atoms, tiny ball like atoms.hypothetically speaking, if you were to separate and isolate a single atom, would you have the reason to believe that this atom is conscience and is aware of its self? Probaly not. how ever, if enough of these atoms come together, in just the right way, they create a being that is intelligent and self aware, humans.when i think of it like this, it all seems to insane to be reality. But yet, here we are.

so essentially, we are all just trillions upon trillions of these tiny little things we call a atom, that is what i am, that is what you are, that is what everything else is as well.

i dont understand how this is possible.

i cant be the only here that thinks there has to be more to all then this, am i?

we have to be missing something, or maybe im just the one missing something l m a o.

 

 

 

xxsolarxx, that is a very astute question. Atoms interact with other things and yet atoms are not aware of themselves. But we, who are made only of atoms, are somehow able to be aware of ourselves and the universe. So how is that possible?

Even if we take something as sophisticated as say the Google self drive car, although it responds to what is going on around itself, it has no consciousness, no awareness of itself.

A couple of things that seem to be relevant to awareness, is the ability to put together questions, and the ability to distinguish between what is true and what is false. So an interesting thought, if we were to devise a computer that on its own could put together questions, and could distinguish between what is true and what is false, would it eventually become self aware too?

But we have not been designed (evolution is blind), so how have we managed to end up (as other animals have too) of having consciousness?

Suppose we were to consider the possibility that we do not have free will? In which case, can we have consciousness too? (After all individual atoms do not have free will, and we are a collection of atoms.) Would it be possible for us to distinguish between what is true and what is false if we are 'hard wired'? Without free will, what if our 'wiring' has logic errors in it? We would be doomed to be unaware of that logic error.

But evolution produces life that is able to distinguish between what is true and what is false, whether we are 'hard wired' or not. Any life, that by random chance of evolution, ends up with its 'hard wiring' such that it can distinguish between truth and non truth correctly, will tend to survive over life that does not.

It is either evidence for local hidden variables or for quantum entanglement being a non-local effect.

 

The evidence points toward it being a non-local effect and not being local hidden variables.

 

We go where the evidence points, not where we think it would have made more sense for the evidence to point.

Yes that is sensible. So the evidence of the end result supports quantum entanglement as 'what is being measured' agrees with what was predicted. But did that prediction mention anything about distance?

So my question - or concern - is around that, the results being unaffected by distance.

My understanding of quantum mechanics / entanglement is limited.

First, just out of interest... Did the theory of quantum mechanics predict that entanglement will be unaffected by distance? Or is it the case that the initial theory made no judgement / statement about distance, it was initially unknown, and the theory has been updated to agree with experimental fact by including it as a principle of the theory?

Does quantum mechanics now have a mechanism as to why distance has no relevance to the outcome of the experiment?

If not, i.e. the mechanism is still unknown, is there any surprise (as a general consensus of the physics community), that distance has no relevance to the outcome of the experiment?

Thanks

 

Not really, after all that is what is predicted by quantum theory (non locality).

 

 

A "local" [hidden] variable is one that is limited by the speed of light (that is what "local" means). So local hidden variables would be affected by distance.

Ah yep, thank you Strange and Imatfaal - so that wasn't a very good start by me then!

So starting again...

The fact that the experimental results for entanglement are not affected by distance, doesn't this suggest that there is no entangled state. That at the point of creation, the particles are already set in their states, as then distance becomes irrelevant?

Does such a result raise any concerns as to the validity of quantum entanglement as the explanation?

So despite the evidence that it is impossible for it to be the result of local hidden variables, you think it must be local hidden variables?

 

You're essentially saying that the fact that quantum entanglement isn't affected by distance proves that it can't be non-local, which doesn't make any sense.

Erm, no? I'm saying the opposite...

Despite the logical argument saying that local hidden variables cannot be the answer, the physical evidence suggests otherwise.