Thread: Length Contraction in the Muon Experiment

1. Originally Posted by superluminal
Ok I found your error. Give me a while...
On a scale of 1-10, how bad was it? Other than the acceleration issue - I still don't see any possible counter argument. Anyway, I await your response.

2. OK People, here is the math - we are going to first work in the Earth frame and conclude what the Muon must see in it's own frame according to Special Relativity.

Then we are going to start in the Muon frame from our conclusions above and we are going to predict what the Earth must see in it's own frame - again, according to Special Relativity.

First the Earth frame: The average distance from the surface of the Earth to where a muon is created is 10,000 m, from the Earth frame. The muon travels at a velocity of .98c for it's entire existence. Thus to reach the Earth's surface, the average life of the muon as seen in the Earth frame is t = x/v = 3.4e-5 s. This is too long of a life for the average muon - luckily, Special Relativity says that the muon's time will be dilated as seen from the Earth frame. The following MATLAB code will be used for the calculations to see what Special Relativity says of the Muon Frame:
10000 earth-frame meters.

----------MATLAB CODE----------
c=3e8;
v=.98*c;
g=1/sqrt(1-(v/c)^2);

disp(' '); disp('Measurements from Earth frame:');
Lo=10e3
To=Lo/v

disp('Muon Predictions:');
L=Lo/g
T=L/v
----------------------------------
-------------RESULTS-------------
Measurements from Earth frame:
Lo = 10000
To = 3.4014e-005
Muon Predictions:
L = 1.9900e+003
T = 6.7686e-006
----------------------------------
A-ok.

Thus, in the Muon's frame, the length to the Earth's surface is 2000m upon when the muon was created. The average life of the muon is 6.8e-6 s, well within the life of the average muon based on it's half-life.
2000 muon-frame meters. (calculated)

Now for analysis from the Muon's frame: The average distance from the surface of the <strike>Earth</strike> muon to where the <strike>muon</strike> Earth is created is 2,000 m, from the Muon frame. The Earth travels at a velocity of .98c for the entire time the muon exists. Thus to reach the Muon, the time between which the muon is formed and the time the Earth's surface hits the Muon in the Muon's frame is t = x/v = 6.8e-6. Now to see what Special Relativity predicts the Muon will say the Earth frame should see:
Ok? (if we're being strict with our frame perspectives).

----------MATLAB CODE----------
c=3e8;
v=.98*c;
g=1/sqrt(1-(v/c)^2);

disp(' '); disp('Measurements from Muon frame:');
Lo=2e3
To=Lo/v

disp('Earth Predictions:');
L=Lo/g
T=L/v
----------------------------------
-------------RESULTS-------------
Measurements from Muon frame:
Lo = 1.9900e+003
To = 6.7686e-006
Earth Predictions:
L = 396.0000
T = 1.3469e-006
----------------------------------

Thus, in the Earth's frame, the length from the Earth's surface to the Muon is 396m upon when the muon was created. The time on Earth is 1.3e-6 s.
No.

In special relativity, an event that occurrs, occurrs in all frames, it's just the timing that is in question (simultaneity). So, in the earth frame the muon creation event occurred at T=0 at 10000m (given). This can never change in the earth frame.

So far so good.

The only given at the beginning is that the muon was IN FACT created at 10000m in the earth frame.

You then go on to correctly calculate what the muon would see in its frame.

You then proceed to use the CALCULATED muon-frame distance (Lo=2e3m) to back-calculate and see what the earth should see. This is incorrect, and you can see that this process will lead to ever decreasing distances as you calculate each frame!

You are violating the initial conditions of the system. All you can ever know is what is given as FACT at the start of the exercise.

So, a new problem:

An earth is created 2000m (a given FACT) above the surface of the muon, travelling at 0.98c. We calculate, correctly, that the earth will see us (muon) as 396m away in its frame.

You cannot then use the CALCULATED value to deduce the original given FACTS.

But alas, the Earth predictions from the Muon frame according to Special Relativity do not match what we actually know happens in the Earth frame. - Where is this analysis wrong?

3. The upshot is that you need to pick a frame and stay there. You cannot calculate anything for the original frame from a calculated frame.

You can calculate what the muon will see, but what we see is etched in reality and can't change. There is only one primary frame in any calculations.

4. Originally Posted by superluminal
You then proceed to use the CALCULATED muon-frame distance (Lo=2e3m) to back-calculate and see what the earth should see. This is incorrect, and you can see that this process will lead to ever decreasing distances as you calculate each frame!
Yes it would.

Originally Posted by superluminal
You are violating the initial conditions of the system. All you can ever know is what is given as FACT at the start of the exercise.
I never assumed anything other than what special relativity tells me.

Originally Posted by superluminal
So, a new problem:

An earth is created 2000m (a given FACT) above the surface of the muon, travelling at 0.98c. We calculate, correctly, that the earth will see us (muon) as 396m away in its frame.
And you don't see a problem with this? That is, we used this exact same analysis to determine that the muon was 2000m away when the Earth was created as you describe it. Get what I am saying?

Originally Posted by superluminal
You cannot then use the CALCULATED value to deduce the original given FACTS.
But it leads to contradiction - are you imposing that we cannot first assume the Muon frame? What if that was the frame in which the experiment was determined and the 2000m and 6.8e-5s were the FACTS as you put them.

5. Originally Posted by Aer
Now to see what Special Relativity predicts the Muon will say the Earth frame should see:

----------MATLAB CODE----------
c=3e8;
v=.98*c;
g=1/sqrt(1-(v/c)^2);

disp(' '); disp('Measurements from Muon frame:');
Lo=2e3
To=Lo/v

disp('Earth Predictions:');
L=Lo/g
T=L/v
----------------------------------
-------------RESULTS-------------
Measurements from Muon frame:
Lo = 1.9900e+003
To = 6.7686e-006
Earth Predictions:
L = 396.0000
T = 1.3469e-006
----------------------------------

Thus, in the Earth's frame, the length from the Earth's surface to the Muon is 396m upon when the muon was created. The time on Earth is 1.3e-6 s.

But alas, the Earth predictions from the Muon frame according to Special Relativity do not match what we actually know happens in the Earth frame. - Where is this analysis wrong?
You're not being meticulous enough in your analysis. While the Earth observer will see 1990 Muon meters as 396 Earth meters, he will not agree that this is the Muon's height. Why is this so, when you've calculated the length contraction? It's because you're treating space as independent of time.

You have to use the Lorentz transforms:

In keeping with the link's notation let S be the Muon's frame, S' be the Earth frame, moving along the x-axis with speed 0.98c wrt S, such that at time t=t'=0, x=x'=0. Assume gamma=5.025.

Now, the muon pops into existence in muon coordinates (x,t)=(1990,0). This means that at time 0 for the muon, there's 1990m to the Earth surface located at x'=0 at all times, and specifically at t=t'=0, when x=x'=0.

We now use the Lorentz equations to translate this to the Earth frame, and we get (x',t')=(10,000,3.27e-5). Note here that the Earth observer and the muon do not agree on when the muon came into existence! This is extremely important to note, because it shows the frame dependence of the initial condition! In other words, because time and space are not independent, from the earth frame, saying that the muon has height 10,000 at time 0, is not the same as what this thought experiment states.

At muon spacetime coordinates (x,t)=(1990,6.50e-6) the surface hits the muon. Again, using the Lorentz transforms this translates to the Earth frame as (x',t')=(0,1.32e-4). That is, even though the muon only saw the Earth travel 1990m, the Earth observer saw the Muon travel 10,000.

In other words, switching frames does not mean ever decreasing lengths, as is implieed in your argument.

6. No I don't see a problem and it is not a contradiction. And I did assume the muon frame in my post, but it's NOT the same physical situation anymore.

So, a new problem:

An earth is created 2000m (a given FACT) above the surface of the muon, travelling at 0.98c. We calculate, correctly, that the earth will see us (muon) as 396m away in its frame.

You cannot then use the CALCULATED value to deduce the original given FACTS.
They are two mutually exclusive and valid physical situations.

7. funkstar - yes.

8. The mutual length contraction is just as weird as mutual time dilation, but it does not lead to any inconsistencies in str*.

*In theory, inconsistencies are impossible, because str is nothing more than R^4 with the Minkowski metric.

9. Originally Posted by funkstar
Note here that the Earth observer and the muon do not agree on when the muon came into existence!
I figured simultaneity had to be a part of this somehow - length contraction and disagreement on events being simultaneous go hand in hand. Neither of the two have been proven by experiment - but they haven't been disproven either. I think the goal here should be to determine an experiment that is practicle to conduct which would confirm it one way or the other. If you want to argue on this experimental proof point - then I will ask you to first point to an experiment to back up your arguments. I'd love to read it.

10. Originally Posted by funkstar
*In theory, inconsistencies are impossible, because str is nothing more than R^4 with the Minkowski metric.
I thought there had to be some explaination - which is why I asked the question seeking out where the analysis went wrong. Some others on this forum need to learn this trait before they go around claiming a theory to be invalid (uh-HUM - not going to name names)

11. Just for illustration purposes, let's try these animations. The first shows events in the same frame as the Earth and the other in the smae frame as the Muon. Each has a measuring rod with which it measures distances. The Earth's measuring rod reaches from surface to the top of the atmosphere (10,000m). The muon measuring rod is also 10,000m as measured in its own frame. The muon is shown as a yellow dot. The measuring rods exists thorughout both animations but the Muon doesn't appear until the left end of the Muon's rod reaches the upper edge of the Earth's atmosphere.

In order to make things easy to see, the thickness of the atmosphere is not to scale, and I reduced the relative velocity to .866 c.

The first animation shows things from the Earth frame.

The muon rod approaches from the left, and due to length contraction, is half the length of the Earth Measuring Rod. At the instant the left end of the rod hits the atmosphere the muon is created. At this instant, the Moun is at the End of the Earth measuring rod, and the Earth measures the distance as 10,000m (note at this point, the right end of the Muon's rod has not yet reached the Surface of the Earth.)

It continues on until it hits the surface of the Earth.

In the second Animation:

We see things from the same frame in which the Muon is in. In this case, the Earth rushes towards the left, and undergoes length contraction.) Again, the muon appears when the atmosphere reaches the left end of the Muon measuring rod. In this frame however, the Muon measuring rod hits the Earth's surface before the muon is formed. Thus when the Muon measures its distance to the Earth at the instant of its creation, it finds that it is less than 10,000 m away from the surface.( note however that, at that same exact instant, it is at the left end of the Earth measuring rod, which we know is 10,000m long as measured from the Earth.

Again, the Earth continues on until it hits the Muon.

Perfect symmetry of length contraction, The muon sees everything in the Earth frame as contracted and the Earth sees everything in the muon frame as contracted by the same factor. and we conclude that when the muon is created it measures its distance from Earth as less than what the Earth measures the distance to the muon to be when the muon is created.

12. Nice!

13. I agree, very nice illustration, Janus58!

Now a couple of questions. In your first illustration, drawn from the Earth observer's frame of reference, the Earth observer sees the muon created at 10,000 earth-meters
above the surface. This same observer sees the muon's meterstick (rod) contracted by
50% due the the relative velocity of .866c. According to the Earth observer, he thinks the muon must transverse 20,000 of the muon's meters to reach the surface, since the
muon's meters are half as long as his. This is also illustrated by the animation. 10,000 Earth meters equal 20,000 muon meters according to the Earth observer.

In your second illustration, the muon observer's frame of reference, when the muon is created at the top of the approaching Earth's atmosphere, part of his meterstick has already passed throught the atmosphere, and the muon observer sees 5,000 of his
(the muon observer's) meters left until the Earth's surface strikes him. The muon observer, according to your illustration, thinks 5,000 of his meters are equal to 10,000 of the Earth's meters, again because the Earth's meters are 50% the length of his own meter due to the .866c velocity of the approaching Earth.

Question. Since the relative velocities are identical in both illustrations, the defining event that leads to different measurements and relativity of simultaneity is the creation point of the muon, which is in the same x,y,z and t coordinates in both reference frames. Without this event, the distances in both frames of reference would be symmetrical, correct? Is this not the same as using a 'third' frame of reference to
obtain correct results?

Edit: By the way, we can fill-in the time component to this illustration at a later time. The suspense is great, isn't it?

14. Originally Posted by 2inquisitive
I agree, very nice illustration, Janus58!

Question. Since the relative velocities are identical in both illustrations, the defining event that leads to different measurements and relativity of simultaneity is the creation point of the muon, which is in the same x,y,z and t coordinates in both reference frames. Without this event, the distances in both frames of reference would be symmetrical, correct? Is this not the same as using a 'third' frame of reference to
obtain correct results?
Without two events, The creation of the muon and the muon striking the Earth, we have no time interval to be measured in either frame, and no way to compare clocks between frames, and thus no experiment.

The creation of the muon is a spacetime event, it is not a reference frame by any usage of the term. So no, this is not the same as using a third reference frame.

I could introduce a third reference frame, say, one in which the Earth and Muon are approaching each other from the right and left at equal velocites, but it would come up with the same results as to the respective readings on the muon and Earth clock when the muon hits the surface. I could introduce an infinite number of reference frames, but they would all come up with the same final result.

15. Perhaps you didn't notice it, but your illustration does not illustrate perfect symmetry.
I pointed it out when I filled in the numbers.

The first illustration, calculated from the Earth's frame of reference, has the muon created at 10,000 Earth meters. The second illustration, calculated using the muon's
frame of reference as the beginning frame, agrees with this number, 10,000 Earth meters. This number is symmetric and agreed upon in both illustrations.

The first illustration, calculated from the Earth's frame of reference, has the 10,000 Earth meters equal to 20,000 muon meters. The Earth observer states the muon has
20,000 of its OWN meters (the muon's) to transit to reach the surface. The second illustration, calculated using the muon's frame of reference as the beginning frame, has the Earth surface located 5,000 of the muon's meters away when the muon is created due to relativity of simultaneity. This is equal to 10,000 of the Earth's meters. Using the Earth as the beginning frame, the muon must travel 20,000 muon meters to reach the surface. Using the muon frame as the beginning frame, the Earth must travel 10,000 Earth meters to reach the muon. These frames are not symmetric. The defining event, the creation of the muon at the x,y,z coordinates, must be used to explain this asymmetry. Its coordinates constitute a third frame of reference without
which both scenarios in your illustrations WOULD be symmetric, i.e. each would measure 10,000 of their own meters and state the moving frames meters were contracted to equal 20,000 of their own.

There is no problem assigning time intervals to the illustrations, we have the velocity of the muon/Earth and the distances to the muon/Earth in both illustrations. Yes, I know why you balk at assigning time intervals to the illustrations. Everyone can think about this for awhile, see what you can come up with that makes sense.

16. Animations zoomed in and with time indexes added. Note that I do not have the Muon clock starting until the muon is created. The time indexes are not in any standard time unit, just note that each clock's tick rate would be the same if all the clocks were put int he same frame. The Earth's clock's (one at each end of the Earth's measuring rod) are sychronized in the Earth frame. and are indexed such that the left clock reads zero at the instant the muon is created next to it. (so that for convenience, both clocks next to the muon at the moment of creation read zero. )

The animations pause at the begining, when the Muon is created, when the right end of the muon measuring rod reaches the surface of the Earth and when the muon reaches the center of the Earth so that you can read and compare the clock readings.
To be technically correct, the second animation should be half the length of the first, but I time expanded it so that it would be easier to keep track of the clocks.

I could have extended the Muon measuring rod out to 20,000 km in length and in that way shown that it would hit the surface of the Earth when the Surface clock read zero in both frames, but it would have meant not being able to zoom in as much to get it all in,(as it is, I let the upper atmosphere end of the Earth measuring rod exit the frame before the second animation ends.) and it would not be consistant with the first animation.

From the Earth Frame

From the Muon frame

17. Excellent, Janus58.

18. This is getting to be fun!

OK, now time to get serious.
Janus58, in your very first illustration, from the Earth frame of reference:
c = 300 meters/microsecond, in all frames according to Special Theory. I will hereafter
use the symbol 'm' to refer to a meter and the symbol 'us' to refer to a microsecond.
You stated the muon traveled at .866c in your illustrations. That equals 259.8 m/us.
10,000 m divided by 259.8 m/us = 38.49 us, the time according to the Earth observer in the Earth's frame of reference for the muon to reach the surface.
This same Earth observer in Earth's frame of reference will calculate the muon 'sees' 20,000 meters to the Earth surface due to length contraction of the muon's meter in the muon's frame of reference (again, as calculated from Earth's frame)
This same Earth observer in Earth's frame of reference will state the muon's clock is ticking at half the rate of Earth's clock due to time dilation of the muon's clock in the muon's frame of reference. Thus the Earth observer in Earth's frame of reference will calculate the muon travels 20,000 muon-meters in 19.25 microseconds of the muon's clock, equal to 1038.96 meters/microsecond. That is 3.46c. Whoops!

Now, do you realize why I kept saying 'Earth observer in the Earth's frame of reference? Because in your second illustration, you are using an Earth observer TRANSFORMED to the muon's frame of reference. The Lorentz transforms are not noted in your illustration, but they are there nontheless. That is why the 'muon' only 'sees' 5,000 meters to the Earth's surface. If the second illustration truly started in the muon's frame of reference with NO PRIOR HISTORY, the muon would see the Earth approachin at .866c and at a distance of 10,000 meters away. A meter is defined by the speed of light and cannot contract in the muon's own frame of reference. The distance to the approaching Earth is the same 10,000 meters, the meters are contracted in EARTH'S frame of reference according to the muon in its own frame of reference. No 'contracted' atmosphere if STARTING from the muon's frame of reference. And 'relativity of simultaneity' does not come into play yet with no prior history in the frame.

19. Originally Posted by 2inquisitive
This is getting to be fun!

OK, now time to get serious.
Janus58, in your very first illustration, from the Earth frame of reference:
c = 300 meters/microsecond, in all frames according to Special Theory. I will hereafter
use the symbol 'm' to refer to a meter and the symbol 'us' to refer to a microsecond.
You stated the muon traveled at .866c in your illustrations. That equals 259.8 m/us.
10,000 m divided by 259.8 m/us = 38.49 us, the time according to the Earth observer in the Earth's frame of reference for the muon to reach the surface.
This same Earth observer in Earth's frame of reference will calculate the muon 'sees' 20,000 meters to the Earth surface due to length contraction of the muon's meter in the muon's frame of reference (again, as calculated from Earth's frame)
This same Earth observer in Earth's frame of reference will state the muon's clock is ticking at half the rate of Earth's clock due to time dilation of the muon's clock in the muon's frame of reference. Thus the Earth observer in Earth's frame of reference will calculate the muon travels 20,000 muon-meters in 19.25 microseconds of the muon's clock, equal to 1038.96 meters/microsecond. That is 3.46c. Whoops!
That figure has no physical significance.[quote] In order to do a proper transform, you have to decide which event you are going to transform from (The Surface clock reading Zero and being opposite the 20,000m mark of the Muon measuring rod, or the muon appearing at the 10,000 m mark of the Earth rod and its clock rading Zero.)

In the first case, when you transform to the Muon frame at this instant the left end of the measuring rod is 20,000m from the Earth surface and the time in that frame is -57.74us before the Muon is created. the muon takes 19.24us to cross the rest of the distance for a total time of 76.98us. 20,000/76.98 = 259.8m/s = .866c

In the second case the muon time =0 and the muon finds itself 5000m from the earth surface it takes 19.24us seconds to cross that 5000m. 5000/ 19.24 =259.8m/s =.866c.

Now, do you realize why I kept saying 'Earth observer in the Earth's frame of reference? Because in your second illustration, you are using an Earth observer TRANSFORMED to the muon's frame of reference. The Lorentz transforms are not noted in your illustration, but they are there nontheless. That is why the 'muon' only 'sees' 5,000 meters to the Earth's surface. If the second illustration truly started in the muon's frame of reference with NO PRIOR HISTORY, the muon would see the Earth approachin at .866c and at a distance of 10,000 meters away.
This is complete gobbly-gook. the second animation is completely independent from the earth observer, it only based on what an observer traveling in the same frame as the muon woul determine. You have managed to completely confuse yourself.[quote]
A meter is defined by the speed of light and cannot contract in the muon's own frame of reference. The distance to the approaching Earth is the same 10,000 meters, the meters are contracted in EARTH'S frame of reference according to the muon in its own frame of reference. No 'contracted' atmosphere if STARTING from the muon's frame of reference.
That is patently ridiculous. Length contraction happens to anything that is moivng relative to your frame.
The formula for that contraction is:
L = L' * sqrt(1-v²/c²
where L' is the lenght as measured from the the relatively moving frame.
Is the Earth moving relative to the Muon?
Yes.

Does the atmosphere move with the Earth?
Yes.

How deep is the atmosphere as measured in the Earth frame?
10,000m

What length does this transform to as measured in the Muon frame using the above formula for v=.866c?
5000m.

How deep is the Earth atmosphere then according to the Muon frame?
5000m

Where does the muon form?
At the upper limit of the atmosphere

How far is the muon from the Surface of the Earth when its forms as measured in its own frame?
5000m.

This is true no matter what frame you start from!
And 'relativity of simultaneity' does not come into play yet with no prior history in the frame.
Relativity of simultaneity comes into play anytime you compare clocks in relative motion which are separated by a distance that is measured parallel to the line of relative motion, "prior history" or not.

20. Originally Posted by Janus58
A meter is defined by the speed of light and cannot contract in the muon's own frame of reference. The distance to the approaching Earth is the same 10,000 meters, the meters are contracted in EARTH'S frame of reference according to the muon in its own frame of reference. No 'contracted' atmosphere if STARTING from the muon's frame of reference.
The formula for that contraction is:
L = L' * sqrt(1-v²/c²
where L' is the lenght as measured from the the relatively moving frame.
Is the Earth moving relative to the Muon?
Yes.

Does the atmosphere move with the Earth?
Yes.

How deep is the atmosphere as measured in the Earth frame?
10,000m

What length does this transform to as measured in the Muon frame using the above formula for v=.866c?
5000m.

How deep is the Earth atmosphere then according to the Muon frame?
5000m
Quick sequential set of questions:

Is the Muon moving relative to the Earth?

Does the atmosphere move with the Earth?

How deep is the atmosphere as measured in the Muon frame?

What length does this transform to as measured in the Earth frame using the above formula for v=.866c?

How deep is the Earth atmosphere then according to the Earth frame?

Where does the muon form?

How far is the muon from the Surface of the Earth when its forms as measured in its own frame?

Originally Posted by Janus58
This is true no matter what frame you start from!
Then start with the muon frame - with the questions above, knowing no information from the Earth frame - that is, you are an observer in the Muon frame taking measurements - you have no prior knowledge of any measurements in the Earth frame.

I'm not saying you are wrong - I would just like to see how you apply special relativity.

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