Clocks, rulers... and an issue for relativity

[Tim88]

 

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.

 

It sounds to me that you are rephrasing the following remark by Langevin on p.48, 49 of his paper (I already gave the link):

 

We will see the appearance of this absolute character of acceleration in another form.

[..]

In particular, in this reference system [that is in uniform motion] the two events considered may be taking place at the same point, in relation to which a portion of matter has traveled a closed cycle and has come back to its starting point thanks to its non-uniform motion. And we can say that for observers related to that portion of matter, the time period elapsed between the departure and return, i.e. the proper time of the portion of matter will be shorter than for observers who would have stayed connected to the reference system in uniform motion. That portion of matter [ will have aged less between its departure and its return than if it had not been accelerating, i.e. as if it had remained stationary relative to a reference system in uniform translation.

 

What is your issue with that?

[Celeritas]

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.

 

robinpike,

 

Nice to make your acquaintance. If I may try and contribute here ...

 

I’ll likely repeat much of what others have already said. First, while it may seem as such to you, it is certainly not the case. There are no preferred frames. The principle of relativity, and first postulate, states that … mechanics hold equally good in any inertial frame. Since everyone uses the same formulae from their own POV, no frame can be preferred, for otherwise they would have to start calculations from the preferred frame to attain correct solns. Not required by relativity. While inertial frames are more convenient to work with than non-inertial frames, they are not preferred. Accurate spacetime solns are attainable from any POV.

 

wrt physically real ... A moving contracted ruler is measured as such, just as the proper length of a ruler is measured as such. If real is to be taken as “exists as per measurement”, then moving contracted lengths are as real as everything else that is measured. Its only that per relativity, the moving contracted length and stationary proper length co-exist, as per POV as a function of relative velocity between the measuring apparatus and that being measured.

 

wrt like tick rate ... While I prefer to stick to interval lengths, one interpretation of relativity is this ... The rate at which time passes by me per me, is the same rate at which time passes by you per you, and no matter what our relative motion. This is commonly referred to as the rate of PROPER TIME. We do not know what that rate is, but it is assumed invariant whatever it is. For lack of any better definition, we refer to it as (for example) 1 sec per sec, which is somewhat pointless since there is no absolute reference for motion, let alone time's rate. However, for identical clocks made in a common inertial frame by the perfect vendor, they must tick at the same rate whenever they happen to arrive in any common inertial frame, since the rate of proper time is the same for all inertial frames. This is why the twins’ clocks tick at the same rate when within a common inertial frame, eg before B’s departure and upon B’s return. While they tick at differing rates relatively when in relative motion (per relativity), it remains true that the rate of proper time is the same, always. The proper rate of tick may be viewed similarly as the proper length of a ruler. A ruler’s proper length never changes, ever, even though relatively moving observers record it length-contracted. Similarly, the proper rate of tick never changes, ever, even though relatively moving observers record your tick rate as slower. A body never changes in and of itself by the gazing of a passer-by.

 

wrt differing aging ... Given the rate of proper time is the same for all (no matter what), then the only way in which B can age less than A is if he travels a shorter path-length than twin A thru 4-space over the defined spacetime interval. He who travels the shorter distance, must accrue the least seconds by his own clock. He ages the least (eg twin B). Remember, each twin holds himself the zero reference for all motion, and simply awaits the 2 events to come to him. The SPACETIME INTERVAL is defined as Event 1 → Event 2 ...

 

Event 1 … twins depart

Event 2 … twins re-unite

 

Each twin holds himself as passing only thru time, and this makes his own clock his measuring device for distance traversal thru spacetime (4-space). One’s clock, is his own odometer for such. The twins start at a same-point-in-spacetime, and end at a different same-point-in-spacetime. B ages less because he travels the shorter path between the 2 events! This description did not consider the frame-to-frame relations as per the LTs (that follows below), but is completely consistent with it ...

 

During relative motion the relativistic effects exist between the frames, with differentials in relative tick rate governed by the LTs (and by the relative geometry of their reference frames, given acceleration). Per each, their clock rates differ during their relative motion, and they age per the seconds accrued by their own clock's hands. The only difference in this description, is that we consider the relative rate of time (per the LTs) induced by the Lorentz symmetry of spacetime, versus only the invariant proper rate of time in 4-space (which we are not fully privied to by casual observation). The result is just the same.

 

So tick rate is one thing, and accrued seconds (aging) is another thing. The former being a function of v (and geometry), the latter being a function of the former and the world-line length over the spacetime interval.

 

There are no preferred frames, and no paradoxes exist, wrt relativity theory. Great minds have not toppled the theory in 116 years, and countless have tried.

 

Hope this helps.

 

Best regards,

Celeritas

Edited September 5, 2016 by Celeritas

[robinpike]

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.

[J.C.MacSwell]

 

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.

 

Yes. It is correct. Just be aware that it is not an inertial frame or Earth frame.

 

A vehicle is a very common frame of reference, generally limited to the immediate space.

 

Proper context is everything, whether implied or assumed.

[michel123456]

 

robinpike,

 

Nice to make your acquaintance. If I may try and contribute here ...

 

I’ll likely repeat much of what others have already said. First, while it may seem as such to you, it is certainly not the case. There are no preferred frames. The principle of relativity, and first postulate, states that … mechanics hold equally good in any inertial frame. Since everyone uses the same formulae from their own POV, no frame can be preferred, for otherwise they would have to start calculations from the preferred frame to attain correct solns. Not required by relativity. While inertial frames are more convenient to work with than non-inertial frames, they are not preferred. Accurate spacetime solns are attainable from any POV.

 

wrt physically real ... A moving contracted ruler is measured as such, just as the proper length of a ruler is measured as such. If real is to be taken as “exists as per measurement”, then moving contracted lengths are as real as everything else that is measured. Its only that per relativity, the moving contracted length and stationary proper length co-exist, as per POV as a function of relative velocity between the measuring apparatus and that being measured.

 

wrt like tick rate ... While I prefer to stick to interval lengths, one interpretation of relativity is this ... The rate at which time passes by me per me, is the same rate at which time passes by you per you, and no matter what our relative motion. This is commonly referred to as the rate of PROPER TIME. We do not know what that rate is, but it is assumed invariant whatever it is. For lack of any better definition, we refer to it as (for example) 1 sec per sec, which is somewhat pointless since there is no absolute reference for motion, let alone time's rate. However, for identical clocks made in a common inertial frame by the perfect vendor, they must tick at the same rate whenever they happen to arrive in any common inertial frame, since the rate of proper time is the same for all inertial frames. This is why the twins’ clocks tick at the same rate when within a common inertial frame, eg before B’s departure and upon B’s return. While they tick at differing rates relatively when in relative motion (per relativity), it remains true that the rate of proper time is the same, always. The proper rate of tick may be viewed similarly as the proper length of a ruler. A ruler’s proper length never changes, ever, even though relatively moving observers record it length-contracted. Similarly, the proper rate of tick never changes, ever, even though relatively moving observers record your tick rate as slower. A body never changes in and of itself by the gazing of a passer-by.

 

wrt differing aging ... Given the rate of proper time is the same for all (no matter what), then the only way in which B can age less than A is if he travels a shorter path-length than twin A thru 4-space over the defined spacetime interval. He who travels the shorter distance, must accrue the least seconds by his own clock. He ages the least (eg twin B). Remember, each twin holds himself the zero reference for all motion, and simply awaits the 2 events to come to him. The SPACETIME INTERVAL is defined as Event 1 → Event 2 ...

 

Event 1 … twins depart

Event 2 … twins re-unite

 

Each twin holds himself as passing only thru time, and this makes his own clock his measuring device for distance traversal thru spacetime (4-space). One’s clock, is his own odometer for such. The twins start at a same-point-in-spacetime, and end at a different same-point-in-spacetime. B ages less because he travels the shorter path between the 2 events! This description did not consider the frame-to-frame relations as per the LTs (that follows below), but is completely consistent with it ...

 

During relative motion the relativistic effects exist between the frames, with differentials in relative tick rate governed by the LTs (and by the relative geometry of their reference frames, given acceleration). Per each, their clock rates differ during their relative motion, and they age per the seconds accrued by their own clock's hands. The only difference in this description, is that we consider the relative rate of time (per the LTs) induced by the Lorentz symmetry of spacetime, versus only the invariant proper rate of time in 4-space (which we are not fully privied to by casual observation). The result is just the same.

 

So tick rate is one thing, and accrued seconds (aging) is another thing. The former being a function of v (and geometry), the latter being a function of the former and the world-line length over the spacetime interval.

 

There are no preferred frames, and no paradoxes exist, wrt relativity theory. Great minds have not toppled the theory in 116 years, and countless have tried.

 

Hope this helps.

 

Best regards,

Celeritas

Question:

We must be aware of 2 effects, time dilation and length contraction. The 2 effects are linked together. So when I observe an object time dilated, he is also length contracted. Doesn't that mean that its path is also length contracted? IOW that the object will take less time to travel a smaller distance, which looks perfectly sensible (it happens to me all the time ) IOW that the 2 effects should cancel each other?

 

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

Because, now that I am thinking about it, if they don't cancel exactly, the velocity as measured by the traveler will not be the same as the velocity measured by the staying at rest observer.

Edited September 5, 2016 by michel123456

[Tim88]

[..]

 

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.

 

The space-time interval of the un-accelerated clock is in a space-time plot depicted as a straight line: this straight line between two events is shorter than a curved line between those same events, independent from your perspective.

 

Yes. It is correct. Just be aware that it is not an inertial frame or Earth frame.

 

A vehicle is a very common frame of reference, generally limited to the immediate space.

 

Proper context is everything, whether implied or assumed.

 

See post #18 : in common SR jargon, "frame" stands for a collection of universal inertial reference systems with virtual rulers and clocks that are in relative rest - and assumed to be in true rest by convention. It is therefore perfectly correct to say that we "change frames" if we used to treat our former state of motion as being in rest, and if after acceleration we next treat our new state of motion as being in rest instead.

 

Although the word frame is a common English word, it can therefore cause confusion to use it for a car or a rocket in arbitrary motion (but as you say, proper context is everything!). One can also use frame, in an extension of SR, for accelerated reference systems; however there are issues with that and needlessly introducing such frames creates unnecessary complexity.

Question:

We must be aware of 2 effects, time dilation and length contraction. The 2 effects are linked together. So when I observe an object time dilated, he is also length contracted. Doesn't that mean that its path is also length contracted? IOW that the object will take less time to travel a smaller distance, which looks perfectly sensible (it happens to me all the time ) IOW that the 2 effects should cancel each other?

 

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

Because, now that I am thinking about it, if they don't cancel exactly, the velocity as measured by the traveler will not be the same as the velocity measured by the staying at rest observer.

 

Objects are predicted to contract if they change velocity on the way from A to B. Obviously their contraction, as measured on Earth, cannot contract for example the distance between London and Paris! We even would have different contractions at the same time of that distance if objects with different speeds were flying in-between.

 

However you are almost right: it's just a matter of picking the right reference frame. According to measurements in a moving frame, the distance between London and Paris is contracted. Thus, in the example with muons from space: according to our measurements their radioactive "clock" is time dilated, so that some can reach the Earth before falling apart. As reckoned from their rest frame, instead the height of the atmosphere is strongly reduced. And then your calculation works: v = less time / less distance.

 

See for a nice sketch http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html

 

PS. You may be interested in Bell's "spaceship paradox" which was designed to show how physical length contraction (but pure length contraction only from one perspective) can be, if it is prevented. Some of his colleagues thought that he was wrong and that space must contract, a bit like you suggested. But that's a topic by itself.

 

[edit: corrected an error in phrasing, happily without consequences]

Edited September 6, 2016 by Tim88

[robinpike]

Thank you all for your help so far. Here is a very basic example of the issue that I am struggling to understand.

 

If a series of locations in space all have a clock permanently based at their locations (like an array of dots positioned throughout space) and these locations are all stationary with respect to each other, then a travelling clock can travel between these locations, and each time it halts at one, it will find that its rate of time is the same as the clock stationed at that location.

 

If during its travels, the travelling clock synchronizes the time on its clock with a stationary clock, then on returning to that location, the travelling clock will have less time on its clock than the stationary clock.

 

This loss in time is attributed to the travelling clock taking a shorter route through space-time on its round trip away and back to the stationary clock. Let's say in this example, the stationary clocks are all moving through space-time at a 'rate of 1' and when the travelling clock journeys from one location the another, it moves at a speed of ‘8’ with respect to the stationary clocks speed of ‘0’ and its route through space-time is at the 'rate of 0.8' with respect to the stationary clocks ‘rate of 1’.

 

So far so good.

 

Since there is nothing special about this array of stationary clocks in space, another array of clocks in space can be considered, again all not moving with respect to each other, but nonetheless they are moving with respect to the first array of clocks. Let’s say they are moving at speed ‘8’ with respect to the first array of clocks. This means that their route through space-time is also at the ‘rate of 0.8’ with respect to the first array of clocks route through space-time, which is at the ‘rate of 1’. However, there is nothing special about the first array of clocks, and so it is equally valid to consider the second array of clocks as being stationary, and the first array of clocks to be moving. In which case, it is equally as valid to say that the second array of clocks are moving through space-time at the ‘rate of 1’ and the first array of clocks are moving through space-time at the ‘rate of 0.8’ with respect to the second array of clocks route through space-time, which is at the ‘rate of 1’.

 

Because travelling clocks lose time when they perform a round trip back to a stationary clock, the assumption is that the change in rate through space-time is a real change.

 

In which case, the first question… Although no preference can be made between the first or second array of clocks, and although their routes through space-time are different (because the clocks are in physical different locations in space?) is it correct to say that their rates through space-time are at the same rate as each other?

[swansont]

If a series of locations in space all have a clock permanently based at their locations (like an array of dots positioned throughout space) and these locations are all stationary with respect to each other, then a travelling clock can travel between these locations, and each time it halts at one, it will find that its rate of time is the same as the clock stationed at that location.

 

The rate will be the same after it stopped, but the rate will not have been the same while it was moving. So it will show that less time has elapsed for the moving clock. If you compare the clock without stopping or slowing, the rates will disagree.

 

If during its travels, the travelling clock synchronizes the time on its clock with a stationary clock, then on returning to that location, the travelling clock will have less time on its clock than the stationary clock.

 

A moving clock can't synchronize with a stationary clock. They can, however, set their times to the same value.

 

This loss in time is attributed to the travelling clock taking a shorter route through space-time on its round trip away and back to the stationary clock. Let's say in this example, the stationary clocks are all moving through space-time at a 'rate of 1' and when the travelling clock journeys from one location the another, it moves at a speed of ‘8’ with respect to the stationary clocks speed of ‘0’ and its route through space-time is at the 'rate of 0.8' with respect to the stationary clocks ‘rate of 1’.

 

So far so good.

 

Since there is nothing special about this array of stationary clocks in space, another array of clocks in space can be considered, again all not moving with respect to each other, but nonetheless they are moving with respect to the first array of clocks. Let’s say they are moving at speed ‘8’ with respect to the first array of clocks. This means that their route through space-time is also at the ‘rate of 0.8’ with respect to the first array of clocks route through space-time, which is at the ‘rate of 1’. However, there is nothing special about the first array of clocks, and so it is equally valid to consider the second array of clocks as being stationary, and the first array of clocks to be moving. In which case, it is equally as valid to say that the second array of clocks are moving through space-time at the ‘rate of 1’ and the first array of clocks are moving through space-time at the ‘rate of 0.8’ with respect to the second array of clocks route through space-time, which is at the ‘rate of 1’.

 

Because travelling clocks lose time when they perform a round trip back to a stationary clock, the assumption is that the change in rate through space-time is a real change.

 

In which case, the first question… Although no preference can be made between the first or second array of clocks, and although their routes through space-time are different (because the clocks are in physical different locations in space?) is it correct to say that their rates through space-time are at the same rate as each other?

No, their rates aren't the same. Each array sees the other as moving slow. They see their own rate as 1 and the other array as 0.8 (they will also see the other array as length contracted by the same factor)

 

Making the scenario more convoluted is unlikely to aid in your understanding.

[Tim88]

[..]… Although no preference can be made between the first or second array of clocks, and although their routes through space-time are different (because the clocks are in physical different locations in space?) is it correct to say that their rates through space-time are at the same rate as each other?

 

My mind boggles over "rates through space-time" ... Consequently I cannot know if a sentence containing that phrase is correct or not.

Can you rephrase your question using phrasing like "clock rates according to" ?

[VandD]

 

My mind boggles over "rates through space-time" ...

My mind too.

Briefly:

In 4D spacetime there are only events. Ticking events of a clocks. No ticking rate. A clock has only a 'ticking rate' when a reference frame is introduced. Then we can compare which event of a clock happens simultaneously with a tick event of another clock, and thus conclude that a clock has or has not a slower 'ticking rate' relative to the other clock.

[michel123456]

 

The space-time interval of the un-accelerated clock is in a space-time plot depicted as a straight line: this straight line between two events is shorter than a curved line between those same events, independent from your perspective.

 

See post #18 : in common SR jargon, "frame" stands for a collection of universal inertial reference systems with virtual rulers and clocks that are in relative rest - and assumed to be in true rest by convention. It is therefore perfectly correct to say that we "change frames" if we used to treat our former state of motion as being in rest, and if after acceleration we next treat our new state of motion as being in rest instead.

 

Although the word frame is a common English word, it can therefore cause confusion to use it for a car or a rocket in arbitrary motion (but as you say, proper context is everything!). One can also use frame, in an extension of SR, for accelerated reference systems; however there are issues with that and needlessly introducing such frames creates unnecessary complexity.

 

Objects are predicted to contract if they change velocity on the way from A to B. Obviously their contraction, as measured on Earth, cannot contract for example the distance between London and Paris! We even would have different contractions at the same time of that distance if objects with different speeds were flying in-between.

 

However you are almost right: it's just a matter of picking the right reference frame. According to measurements in a moving frame, the distance between London and Paris is contracted. Thus, in the example with muons from space: according to our measurements their radioactive "clock" is time dilated, so that some can reach the Earth before falling apart. As reckoned from their rest frame, instead the height of the atmosphere is strongly reduced. And then your calculation works: v = less time / less distance.

 

See for a nice sketch http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html

 

PS. You may be interested in Bell's "spaceship paradox" which was designed to show how physical length contraction (but pure length contraction only from one perspective) can be, if it is prevented. Some of his colleagues thought that he was wrong and that space must contract, a bit like you suggested. But that's a topic by itself.

 

[edit: corrected an error in phrasing, happily without consequences]

I went looking your link (as I always do)

 

Please explain the following:

What bothers me is the comment green underlined

One observer sees time dilation, the other sees length contraction, but neither sees both.

 

From what I understand

As seen from Earth, the muon traveled 10 km in 34μs

As seen from the muon, the muon traveled 2km in 6.8 μs

 

There is a factor γ of 5 that is applied for converting one frame to the other. The velocity is the same for both observers.10 km/34μs=2km/6.8 μs

 

BUT

From the Earth frame, the muon is calculated to have been time dilated, and thus it must have been contracted too, and its path too.

IOW from the Earth both effects must have been observed calculated. I don't understand the comment underlined in green.

Edited September 6, 2016 by michel123456

[swansont]

I went looking your link (as I always do)

 

Please explain the following:

Screen Shot 09-06-16 at 07.04 PM.JPG

What bothers me is the comment green underlined

From what I understand

As seen from Earth, the muon traveled 10 km in 34μs

As seen from the muon, the muon traveled 2km in 6.8 μs

 

There is a factor γ of 5 that is applied for converting one frame to the other. The velocity is the same for both observers.10 km/34μs=2km/6.8 μs

 

BUT

From the Earth frame, the muon is calculated to have been time dilated, and thus it must have been contracted too, and its path too.

IOW from the Earth both effects must have been observed calculated. I don't understand the comment underlined in green.

From the earth's frame, the moving object itself would be length contracted, but that is irrelevant to the observation, since we're looking at the path, not the particle. Nothing else is moving, so there is no other length contraction.

[michel123456]

From the earth's frame, the moving object itself would be length contracted, but that is irrelevant to the observation, since we're looking at the path, not the particle. Nothing else is moving, so there is no other length contraction.

I really don't understand. Because if you take it like this, neither time dilation was observed.

And i think that since the path as seen from Earth is 10km, neither the particle would have been observed length contracted. It is not compatible to observe a contracted muon traveling 10 km.

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

After more thinking, if the muon is a sphere in its own frame, the earthling would have to observe the muon elongated 5 times, not contracted.

[Mordred]

What is length of a pointlike particle?

 

Answer zero length. A contracted zero is still zero. You can't have length expansion of something that has no length.

 

The length contraction does occur but is meaningless for the length of the muon itself.

 

It is the path length itself that has meaning.

 

The pointlike particle dimension aspects is a common confusion. So don't feel bad in this case. Your questions were valid ones.

 

As Swansort stated the length of the particle is meaningless. It is length of the path that is important.

Edited September 6, 2016 by Mordred

[michel123456]

What is length of a pointlike particle?

 

Answer zero length. A contracted zero is still zero. You can't have length expansion of something that has no length.

 

The length contraction does occur but is meaningless for the length of the muon itself.

 

It is the path length itself that has meaning.

 

The pointlike particle dimension aspects is a common confusion. So don't feel bad in this case. Your questions were valid ones.

 

As Swansort stated the length of the particle is meaningless. It is length of the path that is important.

Tragic. Minus one. I am getting angry.

Edited September 6, 2016 by michel123456

[Mordred]

lol remember a particle has also a wavelength. The change in wavelength is reflected in the relativistic doppler formula. Which shows the time dilation change in wavelength.

[michel123456]

From the earth's frame, the moving object itself would be length contracted, but that is irrelevant to the observation, since we're looking at the path, not the particle. Nothing else is moving, so there is no other length contraction.

My post was also about Swansont comment.

[Mordred]

So was mine

So was mine. If you wish to give me a minus one feel free. Reputation points is meaningless to me. I post to help others understand. Not to inflate my ego...

[michel123456]

My post was also about Swansont comment.

 

 

So was mine

So was mine. If you wish to give me a minus one feel free. Reputation points is meaningless to me. I post to help others understand. Not to inflate my ego...

1.You didn't have the same comments on SwansonT post.

2. Is SwansonT post correct, or no?

3. Is my post correct, or no?

 

Your post was slippery like a fish, you are avoiding the question.

[Mordred]

Swansorts answer is correct. A muon has no length to contract or expand.

 

I honestly don't see how you read Swansort's answer as being different...

 

If it was an object with a length then yes it too would length contract.

Edited September 6, 2016 by Mordred

[michel123456]

Swansont wrote:

From the earth's frame, the moving object itself would be length contracted, but that is irrelevant to the observation, since we're looking at the path, not the particle. Nothing else is moving, so there is no other length contraction.

I showed it would be elongated. Exactly the contrary of what Swansont posted.

 

 

Please explain where I get it wrong.

[Mordred]

How do you elongated from a length contraction (only) formula?

[swansont]

I really don't understand. Because if you take it like this, neither time dilation was observed.

And i think that since the path as seen from Earth is 10km, neither the particle would have been observed length contracted. It is not compatible to observe a contracted muon traveling 10 km.

If you measure the distance from the earth' surface to where the muons were created in the atmosphere, the distance is 10 km. Since we are stationary, that length is not contracted.

 

I don't understand the reasoning that's behind your claim to the contrary.

 

After more thinking, if the muon is a sphere in its own frame, the earthling would have to observe the muon elongated 5 times, not contracted.

 

Screen Shot 09-06-16 at 07.56 PM.JPG

I have no idea how you reach this conclusion. I don't know where that picture came from or the reasoning behind it.

Swansont wrote:

I showed it would be elongated. Exactly the contrary of what Swansont posted.

 

Please explain where I get it wrong.

You "showed" it but did not explain it. Until you explain it, there's no way to show where you went wrong, only that you are wrong.

 

—————

 

To sidestep the point particle issue, let's say the particle in question was 1 cm long. The earth would see it as being 0.2 cm long. Which has no effect in the analysis of how far it traveled. 10 km is still 10 km.

[michel123456]

From the link you provided it appears that

 

1. As seen from Earth, the muon traveled 10 km in 34μs

2. As seen from the muon, the muon traveled 2km in 6.8 μs

 

That is the situation as given by the equations.

 

In the diagram the muon is represented as a black dot. It has been overscaled for the sake of understanding.

 

Here below, on the left side the situation as observed by the muon, on the right side as observed by the Earth.

In the middle is the scaling of the muon, as observed by the Earth.

If you take as granted that the muon (the object) is an ellipsoid, then on the left side it is contracted (it is a circle).

 

 

From this, if I am correct, the observer on Earth does not directly observe a contracted object. The contracted object is given by the equations and describe what the muon observes. More something like a scaling, since length & time work together.

[Mordred]

Sorry your going to need to show those calculations. The length contaction only occurs along the x axis. Which we can assume as the axis connecting events. You also need to specify which observer corresponds to which emitter.

 

I still don't see how you get elongated. If you post your math we can spot the error.

Edited September 6, 2016 by Mordred