Actual earth measurement contradicts measurement predicted by special relativity

A particular muon lives to the ripe old age of 4.5 microseconds (three half lives). As it sits in its chair, ticking off the picoseconds, it observes the earth streaking by at 99.99999995% light speed. What will be the elapsed time of the muon's life, as measured on the earth clock?

As I understand SR, the time measured in the earth frame should be much less than 4.5 microseconds. Yet the actual measurement of muons in the atmosphere is much more than 4.5 microseconds.

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 Quote by GregAshmore As I understand SR, the time measured in the earth frame should be much less than 4.5 microseconds.
You have it backwards. According to the earth frame, the muon lives much longer than 4.5 microseconds. (Moving clocks run slow.)
 Mentor Okay, I'm getting annoyed. Stop (mis)using variants of "contradict" in thread titles.

Actual earth measurement contradicts measurement predicted by special relativity

 Quote by George Jones Okay, I'm getting annoyed. Stop (mis)using variants of "contradict" in thread titles.
If the word is appropriate, it should be used.

 Quote by Doc Al You have it backwards. According to the earth frame, the muon lives much longer than 4.5 microseconds. (Moving clocks run slow.)

In every example of special relativity, the reading on the moving clock is less than the reading on the stationary clock. Take the clock on the spaceship in the twin paradox, for example. So why isn't the clock in the lab reading lower than the clock in the muon's chair?

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 Quote by GregAshmore In every example of special relativity, the reading on the moving clock is less than the reading on the stationary clock.
And this situation is no different. In the earth frame, the muon acts as a moving clock.
 So why isn't the clock in the lab reading lower than the clock in the muon's chair?
If you want to understand what an observer riding along with the muon would say about those clock measurements made on earth, you need to understand the relativity of simultaneity as well as time dilation. (I'd suggest reading Spacetime Physics... but that's not working, is it?)

Time dilation is symmetric. The muon observer will say that earth clocks are running slowly. But he'd also say that earth clocks are out of synchronization, which explains how earth observers can measure a longer time. (Note: The time measurements made on earth take place at different locations (since the muon moves), but time measurements made in the muon frame all take place at the same location. That makes all the difference.)

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 Quote by GregAshmore If the word is appropriate, it should be used.
And "contradict" is not appropriate if the mistakes and misconceptions are yours. I am not going to continue bandying words. At the top of the Special & General Relativity forum, there two stickies. Read them. Also read the rules to which you agreed when you registered here. From the sticky with title "IMPORTANT! Read before posting"
 This forum is meant as a place to discuss the Theory of Relativity and is for the benefit of those who wish to learn about or expand their understanding of said theory. It is not meant as a soapbox for those who wish to argue Relativity's validity, or advertise their own personal theories. All future posts of this nature shall either be deleted or moved by the discretion of the Mentors.
Any further threads that have a misused variant of "contradict" in their titles will be deleted, and will remain deleted until they are appropriately titled.

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 Quote by GregAshmore If the word is appropriate, it should be used.
But it hasn't been appropriate the last three times you used it. If you're having trouble understanding something, you could just say so and ask for help.

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 Quote by GregAshmore Maybe. I gave myself a headache thinking about this on the way to work. In every example of special relativity, the reading on the moving clock is less than the reading on the stationary clock. Take the clock on the spaceship in the twin paradox, for example. So why isn't the clock in the lab reading lower than the clock in the muon's chair?
In the lab frame, the muon is moving, so to measure the amount of time between the muon clock reading 0 and the muon clock reading 4.5 microseconds, the lab frame requires two synchronized clocks at different locations to make local readings of the lab time as the muon passes each of these synchronized clocks. If you switch to the muon's frame, each of these lab clocks is running slow, but they are also out-of-sync due to the relativity of simultaneity, with the second clock the muon passes being way ahead of the first clock the muon passes at any given instant in the muon frame. So despite the fact that the two clocks are individually running slow in the muon frame, this explains (from the perspective of the muon frame) why the difference in the readings of each clock at the moment the muon passes it can be much greater than 4.5 microseconds.

 Quote by bcrowell But it hasn't been appropriate the last three times you used it. If you're having trouble understanding something, you could just say so and ask for help.
I give you my word that in future posts I will avoid the word "contradict", or any form of it.

I will take this opportunity to point out that there is an incontrovertible contradiction in Taylor-Wheeler. Therefore, the word was correctly used in that posting:
 1. The narrative in figure 3-1 contradicts the text in section 2.7 (pg 39): "Location and time of each event is recorded by the clock nearest to that event." .... as stated in 2.7: "We do not permit the observer to report on widely separated events that he himself views by eye. The reason: The travel time of light."
In the Train Paradox explanation of the relativity of simultaneity, the observer on the ground makes a judgment of simultaneity based on what he sees by his eye. That is a direct contradiction of what is permitted, according to Taylor-Wheeler.

The observer on the ground then goes on to make a prediction about what will happen in the train frame based on the combination of what he saw with his eyes in the ground frame and the motion of the train relative to him. Your opinion is that this is perfectly consistent with the principle of relativity. I contend that at the very least this procedure assumes the thing that it is trying to prove--a particular space-time relationship between the two frames. Once that assumption is removed, there is no way to determine the time of the strikes in the train frame.

Of course, you are entitled to your opinion. In fact, you are more entitled to your opinion than I am to mine. Nevertheless, my opinion is based on logic. In scientific endeavor disputes such as this are settled by experiment. The problem in relativity is that it is very difficult to set up the conditions and the instruments to collect the required data.

I have not challenged any of the experimental results. Nor have I challenged the primary conclusions drawn from them: time dilation and mass-energy equivalence. What I have challenged are the details which have not been tested, but which are held with certainty nonetheless. In my opinion, as someone who has been surprised many times at the difference between what I expected to see in a test and what actually happened, it is important to distinguish between what is known from experience and what is inferred.

 Quote by JesseM In the lab frame, the muon is moving, so to measure the amount of time between the muon clock reading 0 and the muon clock reading 4.5 microseconds, the lab frame requires two synchronized clocks at different locations to make local readings of the lab time as the muon passes each of these synchronized clocks. If you switch to the muon's frame, each of these lab clocks is running slow, but they are also out-of-sync due to the relativity of simultaneity, with the second clock the muon passes being way ahead of the first clock the muon passes at any given instant in the muon frame. So despite the fact that the two clocks are individually running slow in the muon frame, this explains (from the perspective of the muon frame) why the difference in the readings of each clock at the moment the muon passes it can be much greater than 4.5 microseconds.
This is my problem: The muon isn't passing any clocks. It is at rest in its frame, watching the world go by. That is one of the two fundamental postulates of relativity--the other being that the laws of nature have the same form for every resting observer.

If the problem were posed in the usual manner, as a resting observer measuring the elapsed time as an object moves between two positions, and the task was to calculate the elapsed time in the frame of the moving object, the answer would be smaller than the time measured in the rest frame.

What is the reason that the usual answer is not correct in this case? How is one to distinguish whether the usual answer applies, or not?

I'll look again at what Taylor and Born have to say about the relativity of simultaneity.

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Quote by GregAshmore
I will take this opportunity to point out that there is an incontrovertible contradiction in Taylor-Wheeler. Therefore, the word was correctly used in that posting:
 1. The narrative in figure 3-1 contradicts the text in section 2.7 (pg 39): "Location and time of each event is recorded by the clock nearest to that event." .... as stated in 2.7: "We do not permit the observer to report on widely separated events that he himself views by eye. The reason: The travel time of light."
In the Train Paradox explanation of the relativity of simultaneity, the observer on the ground makes a judgment of simultaneity based on what he sees by his eye. That is a direct contradiction of what is permitted, according to Taylor-Wheeler.
No, not if you understand their actual meaning in that quote. What they are saying there is just that an observer cannot judge whether two events are simultaneous simply by looking at whether the light from each event is seen simultaneously. But if the observer also takes into account "the travel time of light" for an event which happened at a known distance, then a calculation based on both when the events are "seen with the eye" and based on the distances of each event can certainly be used to judge whether the events are simultaneous. For example, if I see the light from one explosion 10 light-years away in 2010, and I see the light from another explosion 15 light-years away in 2015, with times and distances measured by my clock and a ruler at rest relative to myself, I can judge that both explosions happened simultaneously in the year 2000 in my rest frame, even though what I "saw with my eye" was that one occurred 5 years after the other. This is certainly not in contradiction with the Taylor-Wheeler quote above, because they are just making the point that it's a mistake to say seeing two events simultaneously is automatically equivalent to saying they happened simultaneously in my frame (and my example actually illustrates this since one is true but not the other!)

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 Quote by GregAshmore This is my problem: The muon isn't passing any clocks. It is at rest in its frame, watching the world go by.
Sure, and the two clocks in the lab frame pass it in succession. I didn't mean "pass it" to suggest anything about which was moving, I just meant that there was a moment the muon clock and the first lab clock passed one another, and then another moment the muon clock and the second lab clock passed one another. Even if you analyze things from the perspective of the muon frame where these two clocks are running slow, if you take into account that they are also out-of-sync in the muon frame, you still end up with the prediction that the difference in readings between the two lab clocks at the moment they pass the muon is greater than 4.5 microseconds.
 Quote by GregAshmore If the problem were posed in the usual manner, as a resting observer measuring the elapsed time as an object moves between two positions, and the task was to calculate the elapsed time in the frame of the moving object, the answer would be smaller than the time measured in the rest frame.
No, all frames always agree about all local measurements, specifically what any two clocks read at the moment they pass next to one another. If you have a pair of clocks A and B at rest and synchronized in the lab frame and a pair of clocks A' and B' at rest and synchronized in the muon frame, then the two lab clocks can be used to measure the time elapsed in the lab frame on a single muon clock, and likewise the two muon clocks can be used to measure the time elapsed in the muon frame on a single lab clock. All frames will agree on local facts like what times A and A' showed at the moment they passed one another, and the result will be that the lab clock is measured to run slow in the muon frame and the muon clock is measured to run slow in the lab frame.

Let's pick an example with easier-to-deal with numbers. Say we have a pair of clocks A' and B' on a rocket which is moving at 0.8c relative to the lab frame, and the clocks are 20 light-seconds apart in the rocket rest frame, meaning they are 12 light-seconds apart in the lab frame due to length contraction. Also these two clocks are synchronized in the rocket's own frame, which means the rear clock is running ahead of the front clock by 0.8*20 = 16 seconds in the lab frame (relativity of simultaneity). Meanwhile we also have two clocks A and B at rest in the lab frame, and 20 light-seconds apart in the lab frame, and synchronized in the lab frame. Suppose before any of them pass one another, they are arranged like the "diagram" below, with B' to the left of A' and A' to the left of A and A to the left of B, and with B' and A' traveling to the right:

B'...A'---> A...B

So, first A' will pass A, and let's suppose A reads t=0 and A' reads t'=0 at that moment. So at t=0 in the lab frame, B' is 12 light-seconds away from A and it reads t'=16 seconds. Moving at 0.8c, it takes 12/0.8 = 15 seconds in the lab frame to reach A, so A reads t=15 when B' passes it, but B' has only elapsed 0.6*15 = 9 seconds in that time, so since it started reading t'=16 it will read t'=16+9=25 at the moment it passes A.

So, so far we have these local facts:

*When A and A' pass, A reads t=0 and A' reads t'=0
*When A and B' pass, A reads t=15 and B' reads t'=25

Now since A and B are 20 light-seconds apart in the lab frame, and A' passed A at t=0 and is moving at 0.8c, A' will pass B at t=20/0.8=25 seconds in the lab frame. At this moment B reads t=25 and since A' read t'=0 when it passed A and is running slow by a factor of 0.6, when A' passes B, A' reads t'=25*0.6=15 seconds.

Likewise since B' passed A at t=15 in the lab frame, B' will pass B 25 seconds later in the lab frame, when B reads t=15+25=40 seconds. B' will tick forward by 25*0.6=15 seconds in this time, but since B' already read t'=25 seconds at the moment it passed A, that means B' will read t'=25+15=40 seconds when it passes B.

So, we have the following local facts:
*When B and A' pass, B reads t=25 and A' reads t'=15
*When B and B' pass, B reads t=40 and B' reads t'=40

So, the full listing of local facts about all 4 passing events is:

1. When A and A' pass, A reads t=0 and A' reads t'=0
2. When A and B' pass, A reads t=15 and B' reads t'=25
3. When B and A' pass, B reads t=25 and A' reads t'=15
4. When B and B' pass, B reads t=40 and B' reads t'=40

Both frames agree about these local facts (that's a basic principle of relativity, there is always complete agreement about local facts like this). But notice that if either frame uses a pair of their own clocks to measure the time elapsed on one of the other frame's clocks, they conclude the other frame's clock is running slow! For example, in the lab frame, if we look at local events 1 and 3, we find that the rocket clock A' elapsed 15 seconds in the time between passing lab clock A and passing lab clock B, while A read t=0 as A' passed it and B read t=25 as A' passed it, meaning that in the lab frame it took 25 seconds for A' to tick forward by 15 seconds. But in the rocket frame, if we look at local events 1 and 2, we find that the lab clock A elapsed 15 seconds in the time between passing rocket clock A' and passing rocket clock B', while A' read t'=0 as A passed it and B' read t'=25 as A passed it, meaning that in the rocket frame it took 25 seconds for A to tick forward by 15 seconds. The situation is entirely symmetrical, as you can see by the numbers in the four local passing events!

 You might find it helpful to take a look at the illustrations I did for this thread showing two rulers moving at relativistic speeds relative to one another, each with clocks placed at each ruler-marking that are synchronized in the ruler's rest frame. You can see from the diagrams how length contraction, time dilation and the relativity of simultaneity all work together to make it possible for the situation to be completely symmetrical, with each frame saying that the other ruler is contracted and that the clocks on it are slowed-down and out-of-sync, without there being any contradictions in their predictions about local events like what times a given pair of clocks will read at the moment they pass next to one another.
 I'm working on it.

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 Quote by GregAshmore A particular muon lives to the ripe old age of 4.5 microseconds (three half lives). As it sits in its chair, ticking off the picoseconds, it observes the earth streaking by at 99.99999995% light speed. What will be the elapsed time of the muon's life, as measured on the earth clock? As I understand SR, the time measured in the earth frame should be much less than 4.5 microseconds. Yet the actual measurement of muons in the atmosphere is much more than 4.5 microseconds. What have I missed?
Let us imagine we have a clock hovering just above the Earth atmosphere that is sychronised with another clock at ground level in the Earth frame. (Ignore any difference in clock rates due to gravitational time dilation as this will be negligable in this example). When both these clocks are showing zero a muon is created near the top clock and speeds towards the ground. The velocity of the muon means it takes (say) 9 microseconds of Earth frame time to reach the ground. The Earth frame sees the muon's clock as ticking slowly and only 4.5 microseconds elapse on the muon clock by the time it reaches the ground so it is still "alive" and is detected at ground level. The muon sees itself as statioary and sees the the Earth clocks whizzing past it. It observes that the Earth clocks are ticking slowly relative to its own clock and as far as it is concerned only (say) 2.25 microseconds elapse on the Earth clocks between the top clock and ground clock passing it. So how does the muon see 2.25 seconds elapse on the Earth clocks when the Earth observer says 9 microseconds elapses on the Earth clocks, during the muon's atmospheric transition? The answer is in the differences of simultaneity. As far as the muon is concerned, when the top clock read zero, the ground clock already had 6.75 seconds elapsed on it. By the time the ground clock arrived at the muon, the ground clock's display advanced from 6.75 to 9 microseconds meaning that only 2.25 seconds elapsed on the Earth clock as far as the muon is concerned.

 Quote by yuiop Let us imagine we have a clock hovering just above the Earth atmosphere that is sychronised with another clock at ground level in the Earth frame. (Ignore any difference in clock rates due to gravitational time dilation as this will be negligable in this example). When both these clocks are showing zero a muon is created near the top clock and speeds towards the ground. The velocity of the muon means it takes (say) 9 microseconds of Earth frame time to reach the ground. The Earth frame sees the muon's clock as ticking slowly and only 4.5 microseconds elapse on the muon clock by the time it reaches the ground so it is still "alive" and is detected at ground level. The muon sees itself as statioary and sees the the Earth clocks whizzing past it. It observes that the Earth clocks are ticking slowly relative to its own clock and as far as it is concerned only (say) 2.25 microseconds elapse on the Earth clocks between the top clock and ground clock passing it. So how does the muon see 2.25 seconds elapse on the Earth clocks when the Earth observer says 9 microseconds elapses on the Earth clocks, during the muon's atmospheric transition? The answer is in the differences of simultaneity. As far as the muon is concerned, when the top clock read zero, the ground clock already had 6.75 seconds elapsed on it. By the time the ground clock arrived at the muon, the ground clock's display advanced from 6.75 to 9 microseconds meaning that only 2.25 seconds elapsed on the Earth clock as far as the muon is concerned.
If one accepts the "pre-charge" of 6.75 seconds on the ground clock, everything adds up very nicely. I need to do my homework on this to make sure I have it, but the math isn't difficult.

For me, there is still an unresolved conceptual problem.

According to Einstein, Born, and Taylor-Wheeler, the aging process (as T-W put it) in the moving object is slower than the aging process in the resting object. (This is inferred from the fact that the clock in the moving object ticks at a slower rate.) Each object has the right to claim that it is moving. Therefore the aging process for each object is slower than it is for the other.

This, of course, is the Twin Paradox, but without the distraction of the turn-around.

I do not know how to resolve this difficulty. The obvious approach is to declare, as Born does regarding the absolute ether and absolute simultaneity, that a unique state which can be claimed with equal right by multiple entities can have no physical meaning: There is no such thing as slowed aging.

But that declaration causes other problems. And it is too late this evening to think about them.

 Quote by GregAshmore For me, there is still an unresolved conceptual problem. According to Einstein, Born, and Taylor-Wheeler, the aging process (as T-W put it) in the moving object is slower than the aging process in the resting object. (This is inferred from the fact that the clock in the moving object ticks at a slower rate.) Each object has the right to claim that it is moving. Therefore the aging process for each object is slower than it is for the other. This, of course, is the Twin Paradox, but without the distraction of the turn-around. I do not know how to resolve this difficulty. But that declaration causes other problems. And it is too late this evening to think about them.
them.[/QUOTE]

The muon experiments are a perfect example of the one way experiment Einstein described in the first few lines of Part IV of his 1905 paper - This involves two clocks initially synchronized while at rest in the same frame - the accelerated clock will have logged less time when it reaches the distant non-accelerated clock - the reciprocal experiment is never performed - it is assumed that Relativety is correct and that neither clock has a right to a preferred frame - but what happens if a clock is already in motion and it passes a point on the earth - and we read it as it passes then use some signaling to start a distant clock allowing for the light travel time - the question becomes whether the two clocks will measure different times when they pass - we never perform this experiment - there is no acceleration - and there is no way to distinguish which clock is moving - so can you have a meaningful actual time difference when they meet. SR is independent of accelerations - but in every experiment where actual objective age difference is measured, there is acceleration somewhere (either at start up or turn around). In other words, in a totally symmetrical situation involving a pion already in motion at the time we take its measure by a lab clock, will the result correspond to the case where the clocks were initally synced while at rest in the same frame. If yes - then it seems that you would have a way of detecting absolute motion - but if the clocks read the same when the meet - the initial changing of frames due to acceleration becomes the real factor that contributes to the age difference.