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Muons experiment, strange logic 
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#1
Feb2812, 12:00 PM

P: 46

In the muons experiment the flow is measured at top of a mountain and at the base of the mountain.
The velocity of the muons was calculated http://hyperphysics.phyastr.gsu.edu...iv/muonex.html The velocity is 5 times the speed of light, which is of course against the relativity theory. The scientists downgraded the velocity to make it compliant with the relativity theory. Now with the "new" velocity" the flow measurement is not consistent with the calculation any more, as one would expect . So time dilatation and length contracton has been assumed to make it consistent once again http://hyperphysics.phyastr.gsu.edu...v/muon.html#c1. And I am confused . Which was first? The chicken or the egg? 


#2
Feb2812, 12:13 PM

PF Gold
P: 4,792

The chicken must have come first and she must have had fertilized eggs in order to produce both male and female offspring in order to continue the species. Obviously, if the egg had come first, it could only have produced a rooster or a hen, neither of which could have continued the species.



#3
Feb2812, 12:24 PM

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#4
Feb2812, 12:50 PM

PF Gold
P: 4,792

Muons experiment, strange logic



#5
Feb2812, 01:19 PM

PF Gold
P: 1,166

What it's saying is that if moving muons decayed at the same rate as muons at rest, then by the time they reached the bottom, there wouldn't have been enough muons left to account for the observed rate unless they were travelling impossibly fast. However, if we assume that they were travelling at nearly c (which is consistent with other experiments) and that special relativity velocity time dilation applies, slowing down the decay, then we find the adjusted decay rate nicely matches the experimental observations. 


#6
Feb2812, 01:29 PM

P: 46

Similar thoughts came upon me when I read about the Michelson y Morley interference experiment. The experiment shows that there is no ether wind, but says little about the nature of light in other reference frames. Now the muons experiment should dispel the mystery but manipulating the measurement brings even more confusion. 


#7
Feb2812, 01:46 PM

PF Gold
P: 4,792

Eventually, the explanation was extended to the idea that clocks would also run slower as they moved through the ether in such a way as to cause any measurement of the speed of light to get the same answer no matter how much length contraction and time dilation occurred. So the exact same explanation is used to explain how the muons with a half life of 1.56 microseconds could survive long enough to make it to the ground because their builtin clock is running five times slower as they travel through the ether. That seems pretty straight forward to me. What is the mystery that you feel needs to be dispelled? What manipulation of the measurement do you see? Why are you confused? 


#8
Feb2812, 02:27 PM

Mentor
P: 17,545

Now, you may erroneously assume that the decay times of the relativistic muons is the same as at rest, in which case: 


#9
Feb2912, 11:58 AM

P: 46

The MMX uses interference as its principle so it measures phase difference between the two arms. They thought that the movement of earth along the ether should produce a phase shift resulting from the velocity addition c+v, cv in the arm parallel to the earth movement. The light source emits a waveform A*sin(ωtkx), kx represents phase shift at distance x from the source. So for the perpendicular arm x is equal to the calculated length R1 in the experiment. For the parallel arm let's make first some calculations of moving source and receiver. If the source is moving toward receiver with velocity v the waveform is: A*sin(ωt  k*(x  v*t)), x is the distance of the receiver and source at time t=0. k=(2*∏)/λ = ω/c, this gives: A*sin(ωt + ωt*(v/c)  k*x) = A*sin((1 + v/c)*ωt  k*x). The receiver now sees the light with Doppler shift, the phase shift is still k*x. The case when source and receiver is moving with velocity v: A*sin(ωt  k*(x + v*t  v*t)), x is the distance of the receiver and source at time t=0. The receiver sees no Doppler shift, phase shift is k*x. In the MMX when we consider the parallel arm, the first source is the mirror in the middle and the receiver is the mirror at the top of the picture. The mirror on the top reflects the light and is also a source with the receiver being the mirror in the middle. Applying our former calculation the phase shift is 2*k*x. It is consistent with the results of the MMX. Why was the MMX not explained in this way? 


#10
Mar112, 12:39 AM

PF Gold
P: 4,792




#11
Mar112, 10:01 AM

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#12
Mar112, 11:47 AM

PF Gold
P: 4,792

But since the experiment did not measure any ether wind, the scientists of the day explained that by saying that the length of the arm aligned with the direction of motion through the ether was contracted, a perfectly acceptable explanation that fit all the facts and Maxwell's equation. It wasn't until Einstein came along and postulated what you are suggesting, that the arms did not change length and that the oneway speed of light was the same in all directions, another interpretation that also fit all the facts and Maxwell's equation. But there is no experiment that supports one interpretation over the other. You are free to pick whichever one you want. 


#13
Mar112, 03:18 PM

P: 46

It is proportional to the time only in stationary case in which the distance is given by c*t in case of ligth. 


#14
Mar112, 03:45 PM

PF Gold
P: 4,792




#15
Mar412, 02:24 AM

P: 46

Till now I understand, the outcome was that light travels in both direction with the same speed, but Michelson made the calculation for the case there was a fixed ether.
But I don't understand the connection between this experiment and the time dilatation. The Lorentz transformation transforms spacetime coordinates of an event that has noticed one observer to the ones that has noticed another observer moving with a different speed. But after the Spanish Wikipedia http://es.wikipedia.org/wiki/Transfo...B3n_de_Lorentz there is a condition on when to use it: "Las transformaciones de Lorentz dicen que si el sistema O está en movimiento uniforme a velocidad V a lo largo del eje X del sistema O y en el instante inicial (t = t = 0) el origen de coordenadas de ambos sistemas coinciden, entonces las coordenadas atribuidas por los dos observadores están relacionadas por las siguientes expresiones:" "The Lorentz transformation says that if a system O is in uniform movement with speed V along the X axis of the system O and at initial time (t = t = 0) the origins of the two systems are also at the same place, then the coordinates of an event are related with:" (the Lorentz formulas). In other words, as it pertains to the twin paradox: The travelling twin must never change the speed and direction of his movement, otherwise his age calculation will not be valid. Can someone explain? 


#16
Mar412, 03:20 AM

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P: 4,792

But you can pick the frame in which just one twin is at rest and the other twin is traveling, not in his own frame, but in the frame of the atrest twin. The you can use the formula that Einstein worked out in section 4 of his 1905 paper introducing Special Relativity which looks like this: τ=t√(1v^{2}/c^{2}) where τ, tau, represents the time dilation of the traveling twin and t is the normal time for the atrest twin. So if you consider the traveling twin to always be traveling at speed v in the rest frame of the other twin, in other words, from the time he leaves until he gets back, he is always traveling at "v", although his direction can be changing, the ratio of their accumulated ages is simply √(1v^{2}/c^{2}). 


#17
Mar412, 05:24 AM

P: 46

Einstein is assuming that the reference frame can change its speed or direction and the Lorentz transformation still can be used. Which is true it can or cannot? 


#18
Mar412, 08:52 AM

PF Gold
P: 4,792

But if you are going to use the "v" in the Lorentz transform to apply to the difference in speed between the twins, even if they are both "moving" (whatever that means), then don't you mean that in the two reference frames, one of them is stationary and the other is moving? Of course you don't have to do this but if you don't, you will have to be more precise in what you mean because there are an infinite number of ways to interpret what you are talking about. So if we make the origins of the two Frames of Reference be the turnaround event for the traveling twin, then we won't introduce any spurious offsets in the calculation of his aging during his trip. But we don't even have to do that, we can continue to use the same Frame of Reference in which the "traveling" twin was at rest during the outbound portion of his trip and during which the other twin was actually traveling away from him at some constant speed "v" throughout the entire scenario. Then at the turnaround event, the "traveling" twin has to go from rest to a speed greater than "v" in order to catch up to the other twin. It doesn't matter which Frame of Reference we use to analyze the entire scenario of the Twin Paradox. The easiest one to use is the one they both start out at rest and in which only the traveling twin has a nonzero speed because the calculation is trivially simple but if you like to torture yourself, you can pick a different Frame of Reference and endure the more complicated calculations but you will end up with the same age difference in the two twins between the time they separate and the time they reunite. In fact, if you look at the end of section 4 of his paper, you will see that he introduces the Twin Paradox and its solution to the world for the first time, although he doesn't call it that and he doesn't use twins, he uses a pair of clocks. And he assigns a single Frame of Reference to both clocks which he calls K, the stationary system. 


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