# Muons experiment, strange logic

1. Feb 28, 2012

### Tantalos

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.phy-astr.gsu.edu/hbase/relativ/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.phy-astr.gsu.edu/hbase/relativ/muon.html#c1.
And I am confused .
Which was first? The chicken or the egg?

2. Feb 28, 2012

### ghwellsjr

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. Feb 28, 2012

### Tantalos

So how the chicken appeared on the earth if there was no egg?

4. Feb 28, 2012

### ghwellsjr

5. Feb 28, 2012

### Jonathan Scott

That experiment is all about the decay time and how many muons are left (assuming that no more are being created and the existing ones are decaying). There's no length contraction involved.

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. Feb 28, 2012

### Tantalos

So based on the presented data we can only exclude the possibility that the chicken could originate from one egg. But the origin of the chicken still remains unknown.
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. Feb 28, 2012

### ghwellsjr

MMX was not performed in a single reference frame. If it had been, they might have concluded that they were stationary in the ether. But since they repeated the experiment in different reference frames, and since they continued to believe in a fixed stationary ether, they merely concluded that their experiment was modified by moving through the ether in such a way that they couldn't detect the ether wind. That modification, they surmised, caused the length of their apparatus to shrink along the direction of motion through the ether.

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 built-in 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. Feb 28, 2012

### Staff: Mentor

Notice that the velocity of the muons is not measured in this experiment. It is inferred from the decay times and the flux.

Now, you may erroneously assume that the decay times of the relativistic muons is the same as at rest, in which case:
If instead you assume that the decay times of the relativistic muons are time dilated then you get:
Note on the first HyperPhysics page that you linked to, if you scroll down to the bottom, there is the following comment: "In an experiment at CERN by Bailey et al., muons of velocity 0.9994c were found to have a lifetime 29.3 times the laboratory lifetime.". So the first calculation is based on an erroneous assumption, and is therefore a "garbage in garbage out" calculation. There is no chicken and egg, just a mistake which has been noticed and corrected.

9. Feb 29, 2012

### Tantalos

http://es.wikipedia.org/wiki/Experimento_de_Michelson_y_Morley (I put the spanish version which contains calculations).

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, c-v in the arm parallel to the earth movement.
The light source emits a waveform A*sin(ωt-kx), 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. Mar 1, 2012

### ghwellsjr

I think you misunderstand how MMX worked. The semi-silvered mirror in the middle is both the source and the receiver for both legs of the experiment. It appears to me that you are thinking when one of the legs is aligned along the direction of motion through the ether, the time that it takes for the light to get from semi-silvered mirror to the top mirror is different than it takes for the light to make the return trip. Although this may be true, M&M had no hope of capitalizing on this because they had no means by which to measure that difference. Instead, they measured the difference in a pair of light round-trips at right angles to each other.

11. Mar 1, 2012

### Tantalos

No, I am not assuming different time or length. I am assuming that the light is behaving as the Maxwell equations describe, that is the speed of light is the same in all directions and equal c. This is in accordance with Maxwell equation because none of them is dependent on direction.

12. Mar 1, 2012

### ghwellsjr

Maxwell believed that his equations showed that the one-way speed of light was relative to an absolute fixed field and supported the idea of an absolute fixed ether. He proposed a method to measure the motion of the earth through this ether. Michelson saw his proposal and came up with his own idea to make a measurement of the two-way speed of light which would also be capable of detecting the motion of the earth through the ether.

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 one-way 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. Mar 1, 2012

### Tantalos

Michelson made a mistake in his calculations, he assumed the phase shift was only proportional to the time the wave needs to travel the distance L. But in reality phase shift is proportional to the distance from the source (how many wavelengths fit in that distance)
It is proportional to the time only in stationary case in which the distance is given by c*t in case of ligth.

14. Mar 1, 2012

### ghwellsjr

Since there was no significant phase shift at any time, how could a mistake in calculation have any effect on the outcome of the experiment?

15. Mar 4, 2012

### Tantalos

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 space-time 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/Transformaci%C3%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. Mar 4, 2012

### ghwellsjr

MMX was not trying to measure the time or speed of light traveling in different directions. It turns out, although no one knew it at the time, that that kind of measurement is impossible to make. They thought they couldn't make that measurement because they did not have the technology available to make it so they made a different measurement. What they were measuring was the difference in the time for light to make two round-trips at ninety degrees apart. They thought that if one of the round-trips was aligned with the motion through the ether, then it would take a different amount of time than the round-trip at ninety degrees to it.
There isn't a connection. It is only length contraction that they employed to explain the null result of the experiment. Time dilation came later as a necessary feature to go along with length contraction.
It can be used for that but it isn't limited to that. The Lorentz transformation more generally transforms the coordinates of any and all events in one Frame of Reference to the corresponding coordinates for the same events in another Frame of Reference moving with respect to the first one. It doesn't matter if there is an observer at rest in the first one or if there is an observer at rest in the second one.
This is describing the standard configuration for using the Lorentz transformation, which is the one most commonly used, but other forms of it can be used in more complicated situations.
Like I said, you don't have to assume that one twin is stationary in the first frame and the other twin is stationary in the second frame, but if you do, then they can never reunite and there will never be any way to compare their ages.

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 at-rest 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√(1-v2/c2)

where τ, tau, represents the time dilation of the traveling twin and t is the normal time for the at-rest 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 √(1-v2/c2).

17. Mar 4, 2012

### Tantalos

I am not assuming that one of the twins is stationary, but that one is moving with a different speed than the other. But this situation is symmetric, when A is moving away from B, then also B is moving away from A. Both can make the calculation of their ages and A comes to the conclusion that B will be younger, then B will make the calculation and to him A should be younger.

When the twin wants to return then his speed changes to -v. Then we cannot make a new calculation because their origins of their reference frames are not in the same place.

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. Mar 4, 2012

### ghwellsjr

Like I said, this is true if the twins always maintain a single velocity.

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.
Only if we don't put them in the same place. Just like we don't have to presume that an observer/object/clock is stationary in a Frame of Reference, we don't have to put any observer/object/clock at the origins and we certainly don't have to put both observers/objects/clocks at the same origin.

So if we make the origins of the two Frames of Reference be the turn-around 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 turn-around 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 non-zero 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.
It cannot according to Einstein's formulation of a Frame of Reference in his 1905 paper and he doesn't make that assumption. Where did you get that idea?

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.

19. Mar 4, 2012

### Tantalos

I mean that if there are two reference frames A and B in relative movement to each other, then A is thinking that he is stationary and B is moving, but B is thinking the opposite, that he stationary and A is moving, since there is no preferred stationary reference frame.
So each one of them can make the age calculation of the other. A will come to the result that B will be younger than him, but B will come to a result that A should be younger than him. That's a contradiction.

I agree, we don't need them to be in one place. The condition however is that the coordinate of their reference frames at time t = t' = 0 is x = x' = 0. The observers need not be at the origins and as long as v = const we can use the Lorentz transformation to calculate forward and back the space-time coordinates of some events at which they register the events. The events can happen anywhere at any time in any reference frame.
Knowing the space-time coordinates in one reference frame we can calculate the coordinates that registered an observer in another reference frame.

Yes, but the travelling twin still needs to change his direction of movement in order to return to the other twin. Imagine what will see the stationary twin when he looks at clock of the travelling twin. When the twin moves with speed v then of course the stationary twin will see a lag between his clock and the moving clock. Now use the Lorentz formula for time coordinate to check what will happen when the travelling twin suddenly stops. The lag suddenly vanishes, the clocks will show the same time. It's because of violation of the v = const condition. Changing direction of movement is also a violation of this condition.

This means that the travelling twin jumps from his reference frame to another that is moving in opposite direction. The catch is once again the condition t = t' = 0 at x = x' = 0 which is not met in this case.

This is mentioned in the last 4 lines at the end of page 4. Einstein here assumes a movement along a curved line, possibly a closed one, which of course require to change direction during that movement.

20. Mar 4, 2012

### ghwellsjr

It would be a contradiction if you were claiming that two different Frames of Reference concluded that each twin was younger than the other at the end of the scenario, that is, after they reunited, but as long as they both remain at a constant speed with respect to each other, then it's no different than two different Frames of Reference concluding that one twin is at rest and only the other twin is moving. Do you see that as contradictory?
Agreed, as long as by "observers" you mean the "twins".
The lag that you are talking about does not disappear when the traveling twin turns around. The stationary twin will continue to see the traveling twin's clock running slow for most of the scenario until that lag gets used up and then he will see the clock running fast. However, the traveling twin sees this change in the stationary twin's clock at the moment of turn around and it is this difference that confirmes the same conclusion that an analysis based on a Frame of Reference comes to.
I agree you shouldn't do an analysis in which a twin jumps frames--so don't do that. Use just one frame and there will be no frame jumping, correct?
Yes, of course. But you say this as if there is something wrong.