Clocks, rulers... and an issue for relativity

[Tim88]

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.

 

The green underlined sentences are an overly simplified version of what I stated in my post: 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, which has the same explanatory effect.

 

You noticed correctly that the muon is contracted according to the Earth's rest frame, and that clocks on earth are slowed down according to the muon's rest frame. However neither plays a role in that discussion; they are irrelevant for explaining why so many muons can reach the earth.

 

However, for explaining how the same speed can be measured "both ways", it can be useful to also look at it from "the muon's perspective"!

The Earth's rulers are then just as much length contracted as the atmosphere [edit, elaboration: this results, from the muon's perspective, in an exaggerated distance measurement!], while the Earth's clocks appear to be slowed down by a factor 5. You may think that now v = distance / time doesn't work out. However a clock high up in the atmosphere is, from that perspective, incorrectly synchronized with a clock on the surface. According to the muons, it's thanks to this incorrect clock synchronization that Earth observers still measure the same, correct speed v when clocking a muon between both clocks.

 

As a matter of fact, we have now started a variant of the calculation exercise that I advised to do in the other thread.

[Edit:] It's worth the effort to plug in numbers and verify this for yourself.

 

And possibly this makes you reconsider the sketch that you made in a following post.

Edited September 6, 2016 by Tim88

[swansont]

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

I think it's from assuming it's a sphere as measured on earth, so it would be elongated in its own frame. Which is nonsense, but as it doesn't matter to the analysis, it's not worth being bothered.

 

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).

 

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

 

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.

 

Even if it's not a point particle and you assume it's a sphere in the earth's frame (two wild assumptions, but that's OK), we would still be talking about 10^-20 m or smaller. Why does this matter to the distance traveled?

[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.

 

michel123456,

 

The extent of a clock’s own aging is marked by the time accrued by its own hands. The less time accrued by a clock over a defined spacetime interval, the shorter its traversal thru the 4-space respectively.

 

You asked … Does the moving contracted observer always travel the shorter path thru spacetime?

 

During segments of all-inertial relative motion, yes, but only per the stationary POV. However here, a comparison of the 4-space traversal lengths do not produce a valid relative aging comparison in any absolute sense, because the 2 observers can disagree with each other as to who aged more, and both are correct (because they each hold themself stationary and the other moving). That relative aging comparison is not apple-to-apple.

 

The twins scenario allows for an absolute relative aging comparison, because all observers agree on the outcome. No disagreement exists, because the same 2 events define the spacetime interval upon both worldlines. Both spacetime intervals are defined by a common pair of events (only 2 events). Both clocks are co-moving and co-located at both those events, and so any disagreement vanishes (x=0, and v=0 so no relativistic effects). So when the interval is completed, they can have no disagreement as to who aged less (over the entire interval in collective). All in the cosmos agree. Everyone, and no matter what the frame, must agree on the readout of 2 co-moving clocks in-the-same-place-at-the-same-time. The relative aging comparison is hence, absolute, given total consensus. One note here ... in all-inertial scenarios, it requires 3 or 4 events to define the 2 spacetime intervals for comparison ... which introduces the relative disagreement for the measure of space and time (and hence for relative aging as well), caused by the relative motion.

 

One more point wrt your question ... during periods of non-inertial relative motion, the twin A clock ticks faster than B's own clock within the B spacetime system. For that particular segment, A travels a longer distance thru 4-space than B (per B), even though A is always moving and length-contracted per B. This is a case where it can no longer hold true that the moving length-contracted POV travels the shorter distance thru spacetime, per the stationary POV. But again, the comparison of 4-space traversal lengths does not constitute a valid relative aging comparison between the clocks in an absolute sense, unless both worldlines possess the very same 2 events that define the interval of consideration.

 

Best regards,

Celeritas

[robinpike]

 

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.

 

 

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

 

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.

 

The purpose of using a framework of clocks, static within that framework, is to help me understand that there is no issue when a clock travels from one point to any other point in that framework, or from any point in that framework back to its starting point.

 

Yes, the synchronization of the travelling clock to a stationary clock is when it halts next to that stationary clock.

 

To avoid complications as to how separated clocks synchronize, I only consider loss in time for the travelling clock when it starts and ends its journey with the same stationary clock. It is then certain that the two clocks are at the same point in time and at the same location in space as each other, at the start of the journey and at the end of the journey. This means that the loss in time of the travelling clock must be a real physical loss in time (rather than an apparent loss in time).

 

Having a second framework that is moving compared to the first framework, helps me to understand that there is no issue when a travelling clock moves from any point to any other point within any single framework. And so there is nothing special about which single framework (frame of reference) is chosen.

 

The part that I do not understand is when a clock travels from one framework to a different framework (i.e. the frameworks are moving through space at different speeds). Ideally I would like this to be explained one step at a time.

 

So the first step (I think) is for me understand how clocks move through space-time, when moving at different speeds to each other through space.

 

Taking the given answer: clocks moving through space at different speeds, see the other as moving slow through space-time, this being mutual. Perhaps my choice of phrasing is not very good though. So perhaps it is clearer to say that each clock sees the other clock as taking a shorter route through space-time than itself?

 

 

 

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" ?

 

Sure, that wasn't a very clear phrase to use. I was trying to describe the passage though space-time and the 'rate' at which the travelling clock was making progress through its route through space-time. Does this simply equate to saying that the route through space-time is shorter or longer?

Edited September 7, 2016 by robinpike

[Celeritas]

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.

 

robinpike,

 

All relativistic effects are measurable. If measurable, then real. For otherwise, Newton mechanics would not model reality either.

 

Your concern seems to be this ...

Q) does the disagreement between observers of relative motion produce a conflict in the description of reality?

 

While POVs can differ on the relative measure of time, they never disagree on the time readout of 2 clocks when momentarily co-located in 4-space. Any observer, using his own clock and ruler (and/or light signals), can accurately predict what those 2 clocks will read at their flyby event. That, is all that is important. It matters not, that I say their flyby event occurred at x,t = 2,7 while you say it occurred at x',t' = 45,111 ... and so long as others can predict our clocks' time readouts for any event we are located at. The LTs accurately predict intersections in space and time, and the readouts of clocks at intersections, and no matter how we define our units of measure. Add, the LTs apply to the case of 2 hypothetical clocks executing a flyby anywhere in space and time, not just real clocks having done so, doing so, or soon to do so. The LTs map each point (event) of one inertial spacetime system, to a unique point of another inertial spacetime system, moving relatively or not.

 

***************************************************************************

 

3-space systems (with time implied) are defined by cartesian coordinate systems, using the euclidean (space) metric. It defines a distance between any 2 points per the Pythagorean theorem ...

 

d² = x² + y² + z²

 

d = √(x² + y² + z²)

 

Herman Minkowski (Einstein's math teacher who thought him lazy), defined a new metric that relates any 2 classical cartesian systems in relativistic motion, given the 2 relativity postulates true. It's called the Minkowski (spacetime) metric, necessary to make the 2 postulates of relativity compatible (they were not under Newton's model). It defines the distance between 2 points in 4-space. For time-like intervals (where causality is observed), the Minkowski metric is defined as ...

 

s² = (ct)² - x² - y² - z²

 

s = √( (ct)² - (x² + y² + z²) )

 

so if considering only the location of a moving body of velocity v traveling along only x (x=vt), it may be modeled as ...

 

s = √( (ct)² - (vt)² )

 

s = t√(c²-v²)

 

s = ct√(1-v²/c²)

 

s, is the length of a SPACETIME INTERVAL. It is an invariant (constant) in relativity, because no one can say a clock did not tick the precise amount its own hands indicated as it traveled thru spacetime from one event at its own location to another event at its own location. So observers of differing frames all agree on the value of s, and hence the value of tau (t') for said clock between those events ...

 

s = ct'

The reason is this ... Minkowski converted the time tau of the moving observer (t') into a spatial length, by multiplying it by c (where c=1), so it then has units of distance instead of time. This simply converts time to a 4th spatial axis, hence the term 4-space. And why not? You and I are in relative inertial motion. You (as a stationary inertial observer) hold me moving thru both space and time. Yet, I hold myself as stationary (as you do) passing only thru time. Who is right? We are both right. It may then be said that a passage thru time, is a passage thru space. If you and I are passing thru space in parallel, at the same rate, we both declare ourselves stationary, co-moving, and passing only thru time. Yet, others in the cosmos moving relatively, will disagree. Those moving at a virtual c wrt us, will claim we travel virtually only thru space, and not thru time (our clocks having slowed to a virtual zero tick rate, per they). The key words here ... RELATIVE and POV ... folks can disagree yet all be correct. No preferred POV, no paradox ...

s = ct√(1-v²/c²)

 

ct' = ct√(1-v²/c²)

 

t' = t√(1-v²/c²)

 

t' = t/γ

 

Wrt the spacetime system origins ... Time t' represents the aging of the moving clock for the defined spacetime interval. Time t represents the aging of the stationary POV's own clock wrt the defined spacetime interval, per its own POV. Those in relative motion hold themselves the stationary, and the clock traversing the spacetime interval as ticking slower. Hence they record a longer time-interval for the clock (that exists at both events) to complete its journey. That's called a COORDINATE TIME, and differing POVs will disagree on the length of said time interval (its variable per POV, not an invariant). Bottom line, the clock that exists at BOTH events marks the PROPER TIME between them in 4-space, and is the SHORTEST recordable duration between those events. All in the cosmos agree on the duration said clock records, ie it's an invariant.

So disagreements wrt simultaneity and the measure of space and time, do not prevent accurate spacetime predictions. Truely, confusion only stems from a lack of total understanding of the overall mechanism of relativity in collective. One other point ... For me personally, it was not until I realized that ... all moments in time co-exist as the inches on a ruler do (a POV we are not fully privied to), that the LT solns made complete sense to me. The only difficulty arises in explaining WHY we only perceive an ever changing NOW, that progresses only toward the future, given all moments in time co-exist. Regardless, the model works, has been long validated, and still stands.

Best regards,

Celeritas

Edited September 7, 2016 by Celeritas

[michel123456]

 

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

 

Is that correct?

[studiot]

Celeritas, nice and nice again +1

 

Just to reinforce a couple of points.

 

 

celeritas post#153

 

But again, the comparison of 4-space traversal lengths does not constitute a valid relative aging comparison between the clocks in an absolute sense, unless both worldlines possess the very same 2 events that define the interval of consideration.

In other words if the clocks are present at both events.

 

 

celeritas post#155

 

While POVs can differ on the relative measure of time, they never disagree on the time readout of 2 clocks when momentarily co-located in 4-space. Any observer, using his own clock and ruler (and/or light signals), can accurately predict what those 2 clocks will read at their flyby event.

 

Both clocks may be read by both parties at the common event, and even set the same but they are not synchronised

 

(swansont already mentioned this in one of these threads.)

Edited September 7, 2016 by studiot

[imatfaal]

Is that correct?

 

Yes.

 

From Muon's Frame the atmosphere is contracted but its own clock time is unchanged - a clock with the muons would show that time. From the Earth's frame the atmosphere (ie same FoR) is unchanged and the time measured on an earth clock is 34 nanoseconds. 34 nanoseconds is 22 halflives of a muon and should result in 1/(2^22) in fewer than 3 out of ever 10 million muons reaching surface. BUT if we apply relativistic thinking we know that the muons clock is running slow when measured from the perspective of the Earth - slow enough for the correct percentage of muons to penetrate 10km.

 

What is crucial is the number of muons which reach the earth's surface - this is measured and is an event which cannot vary. For the muons time passes as normal but the journey through the earth's atmosphere is shortened, for the earth-based scientist the journey remains the same but the time passes slower on the muon than for the earth-based clock.

 

Each phenomenom happens in the other frame.

[michel123456]

 

Yes.

 

From Muon's Frame the atmosphere is contracted but its own clock time is unchanged - a clock with the muons would show that time. From the Earth's frame the atmosphere (ie same FoR) is unchanged and the time measured on an earth clock is 34 nanoseconds. 34 nanoseconds is 22 halflives of a muon and should result in 1/(2^22) in fewer than 3 out of ever 10 million muons reaching surface. BUT if we apply relativistic thinking we know that the muons clock is running slow when measured from the perspective of the Earth - slow enough for the correct percentage of muons to penetrate 10km.

 

What is crucial is the number of muons which reach the earth's surface - this is measured and is an event which cannot vary. For the muons time passes as normal but the journey through the earth's atmosphere is shortened, for the earth-based scientist the journey remains the same but the time passes slower on the muon than for the earth-based clock.

 

Each phenomenom happens in the other frame.

Thank you.

 

So we have to do with 2 different frames, one for the muon, one for the Earth.

In both frames, the velocity is the same.

 

10 km/34μs = 2km/6.8 μs

 

Is that correct?

[swansont]

The part that I do not understand is when a clock travels from one framework to a different framework (i.e. the frameworks are moving through space at different speeds). Ideally I would like this to be explained one step at a time.

 

So the first step (I think) is for me understand how clocks move through space-time, when moving at different speeds to each other through space.

They don't move at different speeds to each other. They move at the same speed relative to each other.

 

In your scenario, when the clock starts moving, it will run slow. Since it accelerated, we know it is the one moving relative to the other clock. While there is relative motion, they will each see the other's clock rate as being slower then theirs. When the moving clock stops (reduces its relative speed to zero)the rate will increase, since clock rate depends on v. However, it was the one that underwent a change of inertial frame, so it will be the on that has accrued less time passage.

 

You can then reverse this and have the clock return to the original place, and the time difference will be even larger.

Thank you.

 

So we have to do with 2 different frames, one for the muon, one for the Earth.

In both frames, the velocity is the same.

 

10 km/34μs = 2km/6.8 μs

 

Is that correct?

Yes. You can't have one speed be different than the other. That's not even some feature for just special relativity. It's true in Newtonian physics as well.

[Tim88]

 

[..] during periods of non-inertial relative motion, the twin A clock ticks faster than B's own clock within the B spacetime system. For that particular segment, [..] A is always moving and length-contracted per B. [..]

But again, the comparison of 4-space traversal lengths does not constitute a valid relative aging comparison between the clocks in an absolute sense, unless both worldlines possess the very same 2 events that define the interval of consideration.

 

Best regards,

Celeritas

 

It's not clear to me which is A and which is B, but anyway: "does not constitute a valid relative aging comparison [..] in an absolute sense" is an understatement. No matter what can be read into a diagram, length contraction and time dilation go hand in hand in SR and are a function of speed only.

As was elaborated in the spun-off thread starting from http://www.scienceforums.net/topic/98048-relativity-and-shared-realities-split-from-clocks-rulers/?p=941488 , pretending that a clock remains length contracted but temporarily ticks faster due to what is done to another far away clock is not a valid physical description in SR. If you think otherwise, please comment there as it relates to claims about physical reality.

 

[..] If measurable, then real.

[..] Who is right? We are both right. [..]Yet, others in the cosmos moving relatively, will disagree. [..].. folks can disagree yet all be correct. [..]

 

That's a very nice presentation of the Block universe interpretation of SR.

 

I notice that we use different (and incompatible!) definitions of "real"; onlookers should keep our definitions in mind. And about the idea that folks can disagree yet all be correct, that is for me a self contradiction of terms. If I correctly understand what you try to say, I would say that folks can disagree and still all are somewhat right (but technically, all or most are wrong). That could be another thing to elaborate in the thread on "shared realities".

[..]

 

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.

 

[..]

 

I may have overlooked that part, thanks to Cerelitas putting it in bold; I now put a single word in bold. In the parallel thread I hope to have clarified why the word "only" is erroneous. Consequently it leads to the inconsistency that you arrived at. For example, if I see a turn to the left, from another perspective it may have turned to the right. The turning is not only apparent, although it depends on which observer is doing the looking. The phrase "being only apparent - i.e. dependent on which observer is doing the looking" is a logical error.

Edited September 7, 2016 by Tim88

[michel123456]

(...)

I notice that we use different (and incompatible!) definitions of "real"; onlookers should keep our definitions in mind. And about the idea that folks can disagree yet all be correct, that is for me a self contradiction of terms. If I correctly understand what you try to say, I would say that folks can disagree and still all are somewhat right (but technically, all or most are wrong).( ...)

Yes.

[VandD]

 

I notice that we use different (and incompatible!) definitions of "real"; onlookers should keep our definitions in mind. And about the idea that folks can disagree yet all be correct, that is for me a self contradiction of terms. If I correctly understand what you try to say, I would say that folks can disagree and still all are somewhat right (but technically, all or most are wrong). That could be another thing to elaborate in the thread on "shared realities".

What is this for rubbish?

 

[michel123456]

(...)

 

So we have to do with 2 different frames, one for the muon, one for the Earth.

In both frames, the velocity is the same.

 

10 km/34μs = 2km/6.8 μs

 

Is that correct?

Yes. You can't have one speed be different than the other. That's not even some feature for just special relativity. It's true in Newtonian physics as well.

 

So we have on the left side a muon traveling at speed S through 2km.

As seen from Earth, the SAME distance is not 2km but 10km. As a result the earthling concludes that time must be different for the frame of the muon, because it is obliged that S is the same for both observers. The time of the muon is dilated as seen from Earth.

 

But also what one must do is compare 10km and 2 km, by scaling the 2km distance onto the 10 km. Because it is the SAME distance (as viewed by the 2 different frames) The value of the scaling is 5 = γ

One must stretch the 2 km by a factor of 5. Because the distance as seen from Earth is 10km, not 2 km.

 

Are you following till here?

Edited September 7, 2016 by michel123456

[robinpike]

So to the next step that I want to understand. (Apologies if I my phrasing is not always as it should be - I am trying my best to use the correct terms.)

 

When the travelling clock is at rest and next to a stationary clock, the two clocks can synchronize with each other. At this point, we / all observers can note that their routes through space-time must be the same, that is they have the same spatial co-ordinates and their progress along their timelines stay abreast with each other.

 

The travelling clock then goes on its round trip, eventually returning back to the stationary clock and halting alongside that clock. At this point, we / all observers can note that their routes through space-time must again be the same, that is they have the same spatial co-ordinates and their progress along their timelines stay abreast with each other.

 

From experiment, we also know that the travelling clock will have lost time compared to the stay at home clock.

 

What I would like confirmed is how the travelling clock lost time. My naive deduction is that during its journey, the travelling clock was on a shorter route through space-time than the stay at home clock?

 

So first point, is that the correct explanation?

 

And thinking ahead, assuming that is the correct explanation, then I note this point...

 

For an observer, who by misfortune did not see the travelling clock start its journey, and only sees the travelling clock coasting through space away from the stationary clock (or towards it if it is on the return leg of its round trip), that observer is therefore unable to determine if it is the 'travelling' clock on a shorter route through space-time, or whether it is the 'stationary' clock that is on the shorter route through space-time. Despite this unfortunate observer not being able to deduce (or measure) which is which, it does not alter that it is the travelling clock that is on the shorter route through space-time.

Edited September 7, 2016 by robinpike

[Tim88]

What is this for rubbish?

 

A "rubbish" comment looks like trash to me - I'm still new here, should I press the Report button?

[VandD]

A "rubbish" comment looks like trash to me - I'm still new here, should I press the Report button?

I apologize. I was thinking 'that's rubbish', but I should simply have written 'No'.

Edited September 7, 2016 by VandD

[imatfaal]

So we have on the left side a muon traveling at speed S through 2km.

As seen from Earth, the SAME distance is not 2km but 10km. As a result the earthling concludes that time must be different for the frame of the muon, because it is obliged that S is the same for both observers. The time of the muon is dilated as seen from Earth.

 

But also what one must do is compare 10km and 2 km, by scaling the 2km distance onto the 10 km. Because it is the SAME distance (as viewed by the 2 different frames) The value of the scaling is 5 = γ

One must stretch the 2 km by a factor of 5. Because the distance as seen from Earth is 10km, not 2 km.

 

Are you following till here?

 

"Because it is the SAME distance (as viewed by the 2 different frames)" No No NO!

 

Distance is dependent on the relative speed of the frames of reference of the object and the observer. It is the same atmosphere - but not the same distance

[swansont]

So we have on the left side a muon traveling at speed S through 2km.

As seen from Earth, the SAME distance is not 2km but 10km. As a result the earthling concludes that time must be different for the frame of the muon, because it is obliged that S is the same for both observers. The time of the muon is dilated as seen from Earth.

 

But also what one must do is compare 10km and 2 km, by scaling the 2km distance onto the 10 km. Because it is the SAME distance (as viewed by the 2 different frames) The value of the scaling is 5 = γ

One must stretch the 2 km by a factor of 5. Because the distance as seen from Earth is 10km, not 2 km.

 

Are you following till here?

 

 

No, I'm not. The 2 km is the 10 km contracted by the factor of 5. You seem to be implying you must apply yet another transformation, but you don't. It's already included in the numbers.

 

The speed that the muon measures is the distance it measured (2 km) divided by the time elapsed on its clock (6.8 μs). The speed that the earth measures is the distance it measured (10 km) divided by the time elapsed on its clock (34 μs). All of these measurements take place in their own frame.

[Tim88]

I apologize. I was thinking 'that's rubbish', but I should simply have written 'No'.

 

OK With that improved phrasing (and you possibly can still edit your earlier post), michel's posts #162 and your #163 are rather amusing together as each put the exact same phrase in bold for exactly opposite reasons. It may be useful to elaborate in michel's thread.

So we have on the left side a muon traveling at speed S through 2km.

As seen from Earth, the SAME distance is not 2km but 10km. As a result the earthling concludes that time must be different for the frame of the muon, because it is obliged that S is the same for both observers. The time of the muon is dilated as seen from Earth.

 

But also what one must do is compare 10km and 2 km, by scaling the 2km distance onto the 10 km. Because it is the SAME distance (as viewed by the 2 different frames) The value of the scaling is 5 = γ

One must stretch the 2 km by a factor of 5. Because the distance as seen from Earth is 10km, not 2 km.

 

Are you following till here?

 

[retake:] I don't follow your "also", except if you try to redo what I already did in an earlier post on which you did not comment (#151). It may help to define frames by name, e.g. S0 and S1, and to clarify which perspective you use for each comment.

 

Moreover, scaling (in the sense of what you seem to mean with "stretch") does not work as in simple problems due to difference in assumed simultaneity (clock synchronisation). Mutual time dilation and length contraction cannot be understood without understanding the effect of relativity of simultaneity; but I suspect that you don't fully understand that main issue. Please clarify as it's necessary in order to understand my post #151 as well as most of this kind of discussions.

[..] So first point, is that the correct explanation? [..]

 

What kind of explanation are you looking for? If it is about how to calculate, that's what SR can tell you. If it is about what "really" happens, then it's just outside the official scope of SR (and very close to the split-off thread). Nobody can prove what hidden reality is, but we know of two interpretations that match the Lorentz transformations, as already mentioned (here or in the other thread). What can be done is to clarify those explanations. Already Celeritas (somewhat) clarified one model; the other one was (somewhat) clarified by Langevin. Notably they do not agree on all points, so that in some instances one has to specify which model one uses.

You basically have the choice between block universe and presentism.

[..]

 

For an observer, who by misfortune did not see the travelling clock start its journey, and only sees the travelling clock coasting through space away from the stationary clock (or towards it if it is on the return leg of its round trip), that observer is therefore unable to determine if it is the 'travelling' clock on a shorter route through space-time, or whether it is the 'stationary' clock that is on the shorter route through space-time. Despite this unfortunate observer not being able to deduce (or measure) which is which, it does not alter that it is the travelling clock that is on the shorter route through space-time.

 

That's wrong, if I correctly understand you. One has to shut off one's senses (close one's eyes and feeling) in order to not know who is accelerating. Thus the two parts of your sentence are not equivalent: that observer is able to determine that the traveller is not inertial from observing the turnaround.

 

[edit: additional comments]

Edited September 7, 2016 by Tim88

[michel123456]

 

"Because it is the SAME distance (as viewed by the 2 different frames)" No No NO!

 

Distance is dependent on the relative speed of the frames of reference of the object and the observer. It is the same atmosphere - but not the same distance

You are correct, it is the same atmosphere. Wrong wording from me. If the atmosphere was a solid wood, the one is measuring it 10km, the other 2km.

 

The green underlined sentences are an overly simplified version of what I stated in my post: 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, which has the same explanatory effect.

 

You noticed correctly that the muon is contracted according to the Earth's rest frame, and that clocks on earth are slowed down according to the muon's rest frame. However neither plays a role in that discussion; they are irrelevant for explaining why so many muons can reach the earth.

 

However, for explaining how the same speed can be measured "both ways", it can be useful to also look at it from "the muon's perspective"!

The Earth's rulers are then just as much length contracted as the atmosphere [edit, elaboration: this results, from the muon's perspective, in an exaggerated distance measurement!], while the Earth's clocks appear to be slowed down by a factor 5. You may think that now v = distance / time doesn't work out. However a clock high up in the atmosphere is, from that perspective, incorrectly synchronized with a clock on the surface. According to the muons, it's thanks to this incorrect clock synchronization that Earth observers still measure the same, correct speed v when clocking a muon between both clocks.

 

As a matter of fact, we have now started a variant of the calculation exercise that I advised to do in the other thread.

[Edit:] It's worth the effort to plug in numbers and verify this for yourself.

 

And possibly this makes you reconsider the sketch that you made in a following post.

OK that is your post #151.

 

The bold part is important. The entire post is important.

I don't understand especially the bold part.

 

To me:

1. the muon sees the atmosphere as 2km thick. It goes through it in 6.8 μs, at velocity S.

2. the earthling sees the atmosphere as 10km thick. He observes the muon going through it in 34μs, at velocity S.

 

Both observe the same velocity.

 

They are no crossing observations, for example

 

3. the earthling does not observe the atmosphere 2km thick and the muon at 34μs. It is wrong.

4. And as another example, the earthling does not observe the atmosphere 10 km and the muon at 6.8 μs. It is also wrong.

 

The observations are as stated in points 1 & 2 above. They are no crossing observations.

[studiot]

So to the next step that I want to understand. (Apologies if I my phrasing is not always as it should be - I am trying my best to use the correct terms.)

 

When the travelling clock is at rest and next to a stationary clock, the two clocks can synchronize with each other. At this point, we / all observers can note that their routes through space-time must be the same, that is they have the same spatial co-ordinates and their progress along their timelines stay abreast with each other.

 

Note here that the travelling clock is always 'at rest' in its own frame, even when it is moving so the underlined statement is incomplete. It is better phrased

When the travelling clock is at rest in the frame of the the stay-at-home clock, and next to it.

 

The travelling clock then goes on its round trip, eventually returning back to the stationary clock and halting alongside that clock. At this point, we / all observers can note that their routes through space-time must again be the same, that is they have the same spatial co-ordinates and their progress along their timelines stay abreast with each other.

 

From experiment, we also know that the travelling clock will have lost time compared to the stay at home clock.

 

What I would like confirmed is how the travelling clock lost time. My naive deduction is that during its journey, the travelling clock was on a shorter route through space-time than the stay at home clock?

 

So first point, is that the correct explanation?

 

Yes the two underlined statements go hand-in-hand as they amount to the same thing so I would not say that one is the cause of the other.

The proof of this is quite easy, given the Lorenz transformation.

 

And thinking ahead, assuming that is the correct explanation, then I note this point...

 

For an observer, who by misfortune did not see the travelling clock start its journey, and only sees the travelling clock coasting through space away from the stationary clock (or towards it if it is on the return leg of its round trip), that observer is therefore unable to determine if it is the 'travelling' clock on a shorter route through space-time, or whether it is the 'stationary' clock that is on the shorter route through space-time. Despite this unfortunate observer not being able to deduce (or measure) which is which, it does not alter that it is the travelling clock that is on the shorter route through space-time.

 

I can't see any good reason why a third observer could not deduce which clock is stay-at-home and which is going on a journey by sufficient observation.

Clearly this third observer must be in a class frame of his own.

This is because the traveler visits events not visited by the stay-at-home observer, as I mentioned before.

Edited September 7, 2016 by studiot

[michel123456]

Here I have made a slight change to the diagram, so that the contraction on the left side (the muon's frame) is more evident.

 

 

 

[Tim88]

[..]

"it can be useful to also look at it from "the muon's perspective"!

The Earth's rulers are then just as much length contracted as the atmosphere [edit, elaboration: this results, from the muon's perspective, in an exaggerated distance measurement!], while the Earth's clocks appear to be slowed down by a factor 5. You may think that now v = distance / time doesn't work out. However a clock high up in the atmosphere is, from that perspective, incorrectly synchronized with a clock on the surface. According to the muons, it's thanks to this incorrect clock synchronization that Earth observers still measure the same, correct speed v when clocking a muon between both clocks."

OK that is your post #151.

 

The bold part is important. The entire post is important.

I don't understand especially the bold part.

 

To me:

1. the muon sees the atmosphere as 2km thick. It goes through it in 6.8 μs, at velocity S.

2. the earthling sees the atmosphere as 10km thick. He observes the muon going through it in 34μs, at velocity S.

 

Both observe the same velocity.

 

They are no crossing observations

[..]

 

 

Right. However, "velocity" is not a directly observable property. The fundamental, standard measuring method for measuring velocity is to determine the change of position per time interval: v = dx/dt.

 

It is straightforward to analyse that measurement from the Earth's perspective: one can simply determine the times that are recorded when a muon passes two distant clocks, and together with the known distance one obtains the speed. Alternatively one can send the detection signals at near light speed to a single clock, and make corrections for the assumed transit times. By chance, here's a paper about it (just see the "background"): http://www.iontrap.wabash.edu/adlab/papers/S2010_castilow_vest_muon_speed.pdf

 

Obviously, a measurement with two clocks strongly depends on the synchronization of those two clocks; and that is typically done with so-called Einstein synchronization, using light signals. The issue is one of the first things discussed in Einstein's famous paper, and it was already discussed in the literature before SR.

- §1 of http://www.fourmilab.ch/etexts/einstein/specrel/www/

Note also the remarks near the end of §2.

 

With that, both methods (two clocks or one clock) depend on the same assumption of the one-way transit speed of light relative to the system S0 on Earth (this is also called "closing speed" due to some disagreements about meaning of words). It is natural for observers on Earth to assume that the speed of light on Earth is isotropic, and this assumption is the standard convention for setting up a reference system. The speed measurement is therefore not as objective as one may intuitively think. For what follows I will pretend that the speed is measured with two clocks, as that is easier to formulate clearly.

 

How is this muon speed measurement by Earth interpreted from a standard reference system S1 that is co-moving with the muon?

 

Here is where "relativity of simultaneity" comes in. According to S1, the earth is moving at great speed upwards so that the speed of light is not isotropic relative to the Earth - instead, it is assumed to be isotropic relative to S1. Consequently, the clocks and time-of-transit calculations by Earth are faulty from the perspective of S1 - and the obtained one-way light speed reading is faulty too. I don't know if this is totally new for you, or if you recall the effect.

 

Now back to what I tried to make clear in post #151. According to S1:

S0 (the Earth) measures its heights 5 times larger, and its clocks run 5x slower; when using the simultaneity of S1, people on Earth will therefore measure a too high muon speed. And that is what you would obtain from thoughtlessly doing v=distance/time for time measurements between two clocks. But people on Earth use a simultaneity convention that on its own leads to a too low speed. This is because the bottom clock is advanced on the top clock, so as to measure a light ray passing from the top clock to the bottom clock as passing at speed c (and similarly for a light ray going upward).

For the speed measurement, the synchronisation convention compensates for the effects from time dilation and length contraction.

 

S1 can perfectly explain how due to the three effects together - time dilation, length contraction and man-made clock synchronization - people on Earth find exactly the same relative speed between the muon and Earth as S1 finds.

In addition, S1 explains from these effects why S0 measures objects in S1 to be length contracted, and clocks in S1 to run slow - all wrong from the perspective of S1, but perfectly straightforward and not due to some mysterious perception effect that defies understanding.

 

Does that help?

 

[edit: slight corrections in phrasing]

Edited September 7, 2016 by Tim88

[swansont]

Here I have made a slight change to the diagram, so that the contraction on the left side (the muon's frame) is more evident.

 

Screen Shot 09-06-16 at 07.56 PM2.jpg

 

 

 

 

To what end? This has absolutely no bearing on the experiment.