Length Contraction in the Muon Experiment

When beginning from the Earth observer's frame of reference, the Earth observer sees the muon's clock running slow. Does this Earth observer also see the distance to the muon contracted?

When beginning from the muon observer's frame of reference, the muon observer sees the Earth's clock running slow. Does the muon observer also
see the distance to the Earth contracted?

The muon is at rest in its frame of reference anf the Earth is approaching it.
The Earth is at rest in its frame of reference and the muon is approaching it.

 

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2inq:

No.

Yes.

Yes.

I could be wrong about everything. I haven't slept for 26hrs.

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The first two statements were exactly symmetrical. Why a 'no' to the first and a 'yes'
to the second statement?

 

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2inq,

Can you please go and look at the "Visual SRT 1" OP? and read it because this is exactly why I posted it.

 

OK, I read and responded to your other thread. That thread does not answer the question I asked of you in this thread. Why a yes and a no to summetrical situations? There is no acceleration phase (non-inertial phase) in either situation. Why can these two inertial frames not be treated identically?

 

2inq:

I've told you twice, already: The muon start 10,000 Earth meters up in the atmosphere - there's no disagreement on this!. Both the Earth observer and the muon agree that there's a big "10 clicks" sign right where the muon pops into existence. However, due to length contraction of the Earth and it's atmosphere, because the Earth is moving in the muon frame, the muon sees those 10,000 Earth meters as (say) just 600.

Now, you're asking why the Earth observer doesn't see the 10,000 meters as contracted. Why would he? The 10,000 meters is a distance measurement in the Earth frame. Nothing about a relativistic muon can make that contract for a Earth observer.

The reason the situation is not symmetrical, is that you cannot say that the distance from the Earth's surface to the muon is invariant between frames.

Think of it this way: You start out with an implicit measurement in the Earth frame, namely that the muon originates 10,000 Earth meters up in the atmosphere. This cannot be reversed to say that the Earth is 10,000 muon meters away in the muon frame! To do so would be using a measurement from one frame directly in a different frame, and you can't do that in str: You have to use the Lorentz transforms to "translate" between frames.

D'you see?

 

2inq,

The muon is a massive particle and is created and accelerated to 0.987c with enormous energy. The earth does nothing. So in the animation, the muon is A and the earth is B.

Yes?

 

Right.

This is from the Visual SRT 1 thread. I was trying to address that same issue with this statement.

 

No, funkstar, you and superluminal still don't get it. I am starting the exercise FROM THE MUON'S FRAME OF REFERENCE, not the Earth observers.

More detail. A proton sees the Earth approaching it at approximately .987c.
The proton interacts with the Earth's atmosphere and decays into high-energy pions, they also see the Earth approaching at appox. .987c. The pions quickly decay into a high-energy muon and two neutrinos, which also
see the Earth approaching at .987c. The muons are not accellerated TO .987c, they are created at that velocity relative to the approaching Earth. The muon will see the distance to the Earth as 10,000 meters, its meters. The muon will see the Earth clock as ticking much slower than its own clock. By its own clock, the muon will live 2.2 microseconds before decaying, the muon will see only a small fraction of that time elapse on the Earth clock before it dies due to decay. The muon believes the Earth will see a contracted distance between them due to the Earth's velocity, still measuring 10,000 meters but those meters are contracted, equal to only 600
of the muon's meters. Understand? I started and completed the exercise from the muon's point of view. According to the muon's point of view, the Earth's clock will only tick off about .13 microseconds while its own proper time will elapse 2.2 microseconds and it will decay before the Earth reaches it. That is what I mean by beginning the exercise in the 'wrong' frame.

 

2inq,

When the muon is created it will see the distance to the earth as 600m. It will tick off its 2.2us life and decay as it hits the surface. The end.

Ok. It decays at 2.2us when it hits the earth, 10000m away. That's an effective speed of:

10000/2.2e-6 = 454 x 10⁷m/s = 15 times the speed of light. Still think you're right?

 

by superluminal:

2inq,

When the muon is created it will see the distance to the earth as 600m. It will tick off its 2.2us life and decay as it hits the surface. The end.
===============================================================

As previously stated by funkstar, both frames of reference agree there are 10,000 meters to the Earth's surface when the muon is created. The Earth is the frame moving in the muon's rest frame. Why should the muon see a contracted distance when it is not the one moving?
----------------------------------------------------------------------------------

by superluminal:


Ok. It decays at 2.2us when it hits the earth, 10000m away. That's an effective speed of:

10000/2.2e-6 = 454 x 107m/s = 15 times the speed of light. Still think you're right?
==============================================================

I am following the Special Theory method of the assuming the observer is the one at rest and the other frame is the one moving. I just started from the muon's frame instead of the Earth's frame as is explained in STR examples. When beginning the example, the observer in the first frame of reference always 'sees' the clock ticking slowly in the other frame, the frame which is moving. Then, STR switches to the second frame of reference and states this observer 'sees' the distance as contracted, again the 'other' frame is the one in motion. You ask me if I think I'm right? Can you not keep track of the posts, superluminal? I said early on that Special Theory blows up when the exercise is begun in the 'wrong' frame of reference. This is what I am illustrating.
Now to your assertion that the muon will see the distance to the Earth as 600 meters
when it is created and it will hit the surface of the Earth in 2.2 microseconds. Remember, light travels one meter in 1/299,792,458 seconds in ANY frame according to Special Theory. That means a muon's meter is 16.7 times as long as an Earth meter. No 'shorter' meters there! There are FEWER meters. Think about it, it also means the speed of light is not constant, but is frame dependent.

The only way that makes sense is if the muon clock is the one beating slower from EITHER frame of reference, due to the fact that the muon IS the one with greater velocity. Just as in General Relativity, the slower clock keeps its status as the clock beating slower and the faster clock (Earth's) is always seen as the one beating relatively faster. But again, the speed of light would have to be frame dependent here also. The speed of light IS frame dependent in General Relativity, a light clock in low Earth orbit will beat relatively slower than a light clock in high Earth orbit.

 

SL,

I posted a paper with experimental data included. A paper that you referenced in your previous post. You haven't a clue what you are doing do you SL? You didn't even know what the paper was you submitted with your gross condescending bullshit? Who gave you the references you post, your little propaganda research team? Most of the links you provided were just advertisements, did you know that?

You respond with obscenities. But then what else is to be expected? What's the matter, a little unsure of yourself that you have to wave off a scientific discussion using an all too familiar belligerent mouth [hand]?

Muons are not created at "10,000" meters, muons are created throughout the total volume of the atmoswphere.

How do you and your robot friend you defend, who isn't able to answer for itself, handle actual experimental results, data that is, in your religious discussions? I get it, simply ignore it right? Like Aer. If it suggests a threat to SRT crunch the messenger. That's the drill your've picked here isn't it SL, you and your friend that is too personally wounded with its big bad emotional boo boo to respond with anything even approaching sensibility?


Geistkiesel

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​

 

I never said that! In fact, I repeatedly emphasized to you that the 10,000 meters is in the Earth frame, and that due to the relative velocity of the Earth frame wrt. the muon frame, those 10,000 Earth meters become 600 muon meters, because of length contraction!

Don't you get it yet? You can't use a distance measurement from the Earth frame directly in the muon frame!

 

by funkstar:

"I've told you twice, already: The muon start 10,000 Earth meters up in the atmosphere - there's no disagreement on this!. Both the Earth observer and the muon agree that there's a big "10 clicks" sign right where the muon pops into existence. However, due to length contraction of the Earth and it's atmosphere, because the Earth is moving in the muon frame, the muon sees those 10,000 Earth meters as (say) just 600."
==========================================================

I repeat: I am beginning the exercise FROM THE MUON'S FRAME OF REFERENCE. Don't you get it? The Earth observer or his measurement of distance has not even entered the exercise yet. No 'Earth meters'. The muon is at rest in its frame and the Earth is approaching THE MUON. Consider an example of two particles in open space approaching each other at .987c. When the particles are 10,000 meters apart, does one proclaim that the distance is only 600 meters in its frame? WHICH ONE? No, both will measure the distance as 10,000 meters according to Special Theory. You are choosing a preferred frame by insisting the distance is only contracted in the muon's frame of reference. That is a no-no.

 

As measured from which Frame?

The one which isn't the one that measures the distance as 10,000 meters.

No, you are the one trying to infer a prefered frame of reference when you said above that the two muons are 10,000 meters apart at some point without specifying which frame that measurement will be made. Try thinking of it this way: Each muon has a measuring tape that it carries with it and extends 10,000 meters ahead of it (as measured in its own frame). At the instant a muon is opposite the end of the other muon's tape, where is the other muon with respect to the first moun's tape according to the first muon? Do this for both Muons.

 

Aer, I don't know where I didn't make myself clear in the other posts, but why do you still say that light from the edge of the universe reaches the muon? It doesn't. In the muon's own frame, it's life is very short (the normal length). So, just when it reaches the earth, it dies. Any light, which isn't seen as reaching it in the earth frame, doe not reach it in it's own frame either. How can you not understand this. I'l give my earlier diagram here also.
Length contraction accounts for both the ability of the muon to reach the earth (in the muon frame), <I>and</I> the <I>inability</I> of light from beyond to reach the muon. It is clear enough from the diagram.

<Img src= "http://www.geocities.com/virusmakermad/Muon.GIF"></Img>

Yellow: Source; Red: Photon; Brown: Muon; Black platform: Earth;

In fig. B. instead of showing the earth and the source move towards the Muon, as really happens in that frame, I've superimposed two images from t=0 and t=t on top of each other in such a way that the Muon appears to move.

 

So did I.

Yes, there are. It is undisputably so, because there's an imaginary altitude balloon at Earth height 10 km with "Height 10km" on it right where the muon comes into existence (with relativistic speed wrt the Earth immediately; no acceleration necessary.)

I never disputed this. In fact, I wrote it explictly several times.

Yes, because that measurement of 10,000m is frame dependent! Don't you know that distances are relative?

Do the math as per Janus58's example. It should clear it up for you, as well as enlightening you to the relativity of simultaneity (it's the pole-in-the-barn "paradox" in light disguise.)

No, they won't. There's no way in hell they can both measure the distance as being the same if they are in relative movement, without violating the basic postulates of str.

Absolutly not! All I'm saying is that the Earth meters are contracted in the muon frame which is perfectly in line with str. You, on the other hand, are assuming a frame invariant distance between the muon and the Earth which is in direct contradiction with str. Not once have I claimed that the Earth observer won't see the muon meter as length contracted! However, that is a vastly different claim than what you are saying must happen! Do you understand why it is nonsensical to claim that because the muon is moving in the Earth frame, the Earth meter must contract for the Earth observer?

Again, you can't switch frames in str except via the Lorentz transforms. Your use of a measurement from the Earth frame in the muon frame without using the transforms is not allowed. And that measurement most definitely comes from the Earth frame, because we know that muons originate in the atmosphere, right next to that damned balloon!

 

OK, after careful examination I have come to the conclusion that you are ALL wrong (in one regard or another).

This is the Earth frame, you know this right?

No. This is wrong.

A meter is a meter is a meter. A meter is not redefined from one frame to another.

This is ambiguous, are you talking about what the Earth observer must conclude the muon sees?

Again, I think you are confusing frames.

How can you say No to one and Yes to the other? You are assuming your own interpretation to each question differently than the other.

This is precisely the problem they are having with my question as well.

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Where are you pulling these numbers? This is what initially confused me in your very first response. It should be 2000m, not 600m.

You must mean Earth Meter = distance as measured in meters in the Earth rest frame and Muon Meter = distance as measured in meters in the Muon rest frame? Again, why 600m? Is my source wrong?

The other must 'see' the other's spatial coordinates as contracted.

As I said, your explaination is inadequate as I see it. I merely started a new thread to get other explainations from other people - sorry if this is offensive to you.

 

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:

----------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
----------------------------------

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.



Now for analysis from the Muon's frame: The average distance from the surfaceof the Earth to where a muon 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:


----------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?

 

Aer,

"Thus, in the Earth's frame, the length from the Earth's surface to the Muon is 396m"

If muons are observed by balloon instruments in the earth frame, to detect muon creation at 10000m, how can you say this. I haven't analyzed your post yet, but you are making an invalid assumption somewhere.

As funkstar has pointed out several times, the speed of a muon cannot affect distances in the earth's frame.