Length Contraction in the Muon Experiment

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This'll be a long post, but I beg your patience because it shows an important point.

This is wrong. The 10,000 m in the Earth frame will be seen in the muon frame as 5,000 m. Length contraction, remember. Also, the muon will see no contraction of it's own frame, of course.

Again, not 20,000 m but 5,000 m making it... 0.866c. (And, in the muon frame the Earth is the one travelling.)

I'll explain, but first a bit of notation:

Let "ee" stand for "Earth meter as viewed from the Earth frame".
Let "em" stand for "Earth meter as viewed from the muon frame".
Let "mm" stand for "Muon meter as viewed from the muon frame".
Let "me" stand for "Muon meter as viewed from the Earth frame".

(The first letter indicates in which frame some length is in, and the second letter indicates which frame the length is being viewed from.)

The argument you are making is the following: At the time the Earth observer sees the muon pop into existence, there's 10,000ee to it. The Earth observer calculates that the muon will see these 10,000em as just 5,000mm in the muon frame. Now, the Earth observer sees 5,000me as just 2,500ee, and concludes that the 10,000em must be seen as 20,000mm in the muon frame!

Now, of course, there is a glaring inconsistency in what the Earth observer predicts the muon as seeing the distance as being (5,000mm? 20,000mm?), meaning that something is horribly wrong.

The error you make lies in the part in red. The problem is that you suppose that because the Earth observer sees the muon frame as contracted wrt. the Earth frame, then so will the muon. However, as we know, this is not the case! You're using a length contraction of the muon frame that the Earth observer sees in the Earth frame, directly in the muon frame, and that's not allowed (because you can't change frames without the Lorentz transforms). The muon will not see the Earth as length dilated as your analysis claims.

This is also suggested by the special notation: The dilation of 10,000em to 20,000mm is based on a relationship that relates 2,500ee to 5,000me. The units are different, meaning that such a dilation is a non sequitur; it is unjustified (though it doesn't seem so at first glance). Length contraction has the following consequence quantitatively when speaking of distances: ee to me means multiply the number by gamma, me to ee means divide by gamma. Similarly mm to em means multiply by gamma, and (the pertinent case) em to mm means divide.

When relating 10,000em to 20,000mm you multiplied by gamma where you should have divided. Simple error, paradox solved.

The distance to the Earth from the muon is frame dependent. That's really all there is to it. If two frames agree on any (non-zero) distance at any time, then they must be the same frame (modulo translation and rotation and such) due to light speed being constant in all frames. That means no relative velocity, which is certainly not the case here.

Exactly the same holds true in the Earth (as you know), yet you proceed to claim

No. Length contraction is what happens to objects at rest in the other frame when viewed from your own frame (in which they are moving). Nothing about the muon moving at relativistic speed relative to the Earth can make the meter contract in the Earth frame - how could that possibly be so? (Especially considering that there are infinitely many reference frames...)

2inq, saying that the distance from the Earth to the muon is 10,000m in both frames when they have a relative velocity is incompatible with the basic postulates of str, so if you insist on this then you are not doing an str analysis. In the notation I introduced above: Either there's 10,000mm from the muon to the surface of the Earth, or there's 10,000ee, but not both. Str forbids it.

 

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Aer essentially asked my question of Janus58. I will respond later according to how Janus58 answers.
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by funkstar:
"This is wrong. The 10,000 m in the Earth frame will be seen in the muon frame as 5,000 m. Length contraction, remember. Also, the muon will see no contraction of it's own frame, of course."
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Yes, length contraction. That's very ambigeous, the reason I introduced the '20,000 meter' figure to open discussion. Seems simple to say the 10,000 meters are contracted to 5,000 meters in the muon's frame of reference, as seen by the Earth observer. The 'distance' is cut in half, correct? Half of 10,000 meters IS 5,000 meters, 5,000 of THE SAME LENGTH METERS. If the 'meter' is contracted by 50%, it will take
20,000 of them to equal 10,000 Earth frame meters. So, contrary to common belief, the meter itself is not contracted, the contracted DISTANCE consists of FEWER meters, correct? Can't have it both ways. If the meter is contracted by 1/2 PLUS the distance is contracted by 1/2, then you end up with 10,000 'half-length' meters in a distance that is contracted by 1/2. The muon can't transverse that distance in any shorter period of time UNLESS the muon clock beats at half the rate of an Earth clock. That has been my contention all along, length contraction is due to a slower clock only, the distance SEEMS less because the muon can transverse 10,000 meters in half the time relative to the Earth clock. Time beats slow in the relativistically moving muon's frame of reference only.
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by funkstar:

"The argument you are making is the following: At the time the Earth observer sees the muon pop into existence, there's 10,000ee to it. The Earth observer calculates that the muon will see these 10,000em as just 5,000mm in the muon frame. Now, the Earth observer sees 5,000me as just 2,500ee, and concludes that the 10,000em must be seen as 20,000mm in the muon frame!"
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No, funkstar, that is not my arguement. You have '10,000 meters/Earth frame' on the brain. BEGIN the exercise in the MUON frame of reference, not as transformed from an Earth frame. Think the muon will measure the Earth's atmosphere as 5,000 of its meters? That's fine, no problem. NOW, do your Lorentz transformations TO the Earth frame of reference. The number in this FIRST transform becomes 2,500 meters. See?
When beginning in the muon frame of reference, the muon sees the EARTH frame as contracted, half his own measurement FROM HIS REST FRAME. The Earth is in motion, the universe is in motion, relative to the muon's rest frame. ALL is contracted relative to the muon's own meter. The muon's frame of reference does not consist of a tiny 'box' that he is in, it includes the whole universe. This 'whole universe' rest frame of reference is relative to the approaching 'whole universe' frame of the Earth. Or do you believe the muon's rest frame only encompasses the tiny 'point' of its existance? This misconception arises because when beginning in the Earth frame, THE MUON ONLY is the one moving relative to the Earth and the rest of the universe.

 

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Do you remember my post where I explained this?

 

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2inquisitive - I understand what you mean with your explainations as they are similar to what I have brought up. Though I don't always agree with the way you try to explain it, I believe we are standing on similar ground. I, like you, am awaiting the answers from Janus58

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Shit. I just got myself confused...

 

Lovely feeling, no?

 

If I tell you that a planet just came into existence, travelling at 0.98c, 2000m over your head, what must the planet see as the distance between us?

 

Don't know - will wait for Janus58

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398m

Betcha.

 

Thanks, Aer, I am happy at least someone got the gist of what I was trying to explain. I have never had a physics class in my life and I know I do a very poor job of trying to explain what I am thinking. I will bow out now and let the real physicists discuss physics.

 

2inquisitive,
Imagine two 10km rods.
One rod is on the ground, pointing up into the atmosphere.
The other rod is floating beside the Muon, with the Muon at one end and the other end pointing toward the approaching Earth.

Consider these events:
A - the ends of the rod pass each other
B - The end of the muon rod reaches the ground
C - the muon reaches the ground
D - The Muon passes the upper end of the Earth rod


According to special relativity:

In the Earth frame, what is the order and timing of events?
In the Muon frame, what is the order and timing of events?

 

Yes

No. The muon sees the 10,000m deep atmosphere contracted to 5,000m. The Earth observer still sees it as 10,000m.

Correct.

I don't understand what you mean, here.

As viewed from the Earth frame, yes.

However,as viewed from the muon frame 10,000 Earth frame meters are only 5,000 meters long. Not 20,000. There's no ambiguity about this, whatsoever. One is in the Earth frame, the other is in the muon frame. You simply can't use a measurement from one frame in the other frame.

Here's the problem, in a nutshell: Something that is 5,000m long (and at rest!!!) in the muons frame of reference will be seen as 2,500m long from the Earth frame. However, the atmosphere is not at rest in the muon frame! When you apply this length contraction to find that the muon also sees the distance as 20,000m, you are (I suspect unknowingly) assuming that the atmosphere is at rest in the muon frame, which is false. When you apply length contraction twice (10,000 -> 5,000 -> 2,500) the measurements will no longer describe the same object, because you are assuming it to be at rest in both reference frames! You see?

Well, a meter is a meter. But the muon will not agree that something which is measured by the Earth observer to be 1 meter long, is in fact 1 meter long. The muon will measure it to be 1/gamma meters long.

Fewer meters as seen from the muon. Not from the Earth.

Yes, I can, because you're mixing frames. Can't do that.

No. A 20,000 proper (rest) m long object in the muon frame would appear to be 10,000 m long in the Earth frame. A 10,000 proper m long object in the Earth frame appears 5,000 m long in the muon frame. You're taking length contraction that the Earth observer sees of the muon frame, and applying it in the muon frame directly. That's not correct.

But the muon is not traversing that distance as seen from the Earth frame! As seen from the Earth the muon starts at 10,000m with speed 0,866c and that's it!

Those 10,000 meters are an Earth frame object, namely the depth of the atmosphere. Earth frame objects are length contracted in the muon frame. Hence the muon sees the depth of the atmosphere as 5,000 m. Not 20,000, never 20,000.

In the muon frame, time is ticking normally.

That's because the atmosphere is a Earth frame object! Length is frame dependent! Saying that something is x m long without specifying a frame is non-sensical!

Anyhoop, on to the meat:

Ok.

Nope. You apparently didn't even see fit to try to do it, because what does the Lorentz transforms say? I'll do it for you.

Let S be the muon frame and S' be the Earth frame. Let S and S' coincide such that (x,t)=(0,0) (these are spacetime coordinates from the muon frame, with meters and seconds as units) and (x',t')=(0,0) (these are spacetime coordinates from the Earth frame, with meters and seconds as units) coincide (that is to say, describe the same spacetime event). Let x'=0 mark the surface of the Earth (at all times t') in S'. Given the above synchronization, in the muon frame S at time t, x=vt marks the position of the surface of the Earth. Agree?

Now, the depth of the Earth's atmosphere is measured (somehow) in the muon frame to be 5000m deep. That means that in the muon frame the position of the edge of the atmosphere is described at time t as x=vt+5000. Agree?

So, so your question translated to math is the following: At arbitrary t, what are the spacetime coordinates (of which we are only interested in the x' component) of the edge of the atmosphere in S' when transformed from S? Remember (x',t')=(0,t') describes the surface of the Earth in S', so if we do this we have the proper (rest) length of the atmosphere, i.e. the length that the Earth observer will see (because he is at rest wrt. the atmosphere.) as predicted by str. Your position is that the Lorents transform will yield 2500.

Well, to change from S to S' the Lorentz transforms relate x' and x by

x' = gamma(x - vt)

Using vt+5000 for x, and gamma=2 this comes out as

x' = gamma * (vt -5000 - vt) = gamma * 5000 = 10,000. (Note the time independence.)

Game, set and match.

 

yes

yes

5000m

10,000m you have to rearrange the forumula to L' = L/sqrt(1-v^2/c^2) because you used the formula above to get the answer previous to this one by using the formula in its original form, you have to reverse it to get to get the right answer.

10,000m

upper atmosphere

5000m

The answers come out the same. My prior knowledge of the measurements made in the Earth frame have no bearing on those measurements or the measurements I make. The instant I see the Earth and its atmosphere, I will measure its atmosphere as being 5000m thick, and I will then know that it is 10,000 meters thick as measured in the Earth frame.(assuming that I know how to do the transforms. If I don't then I will just measure the Atmosphere as 5000m thick and leave it at that.

 

Aer, to reiterate:

How deep is the atmosphere as measured in the Muon frame?
5,000m.

What length does this transform to as measured in the Earth frame using the above formula for v=.866c?
An object at rest(!!!) in the muon frame that is 5,000 (proper) m long will appear 2,500m long from the Earth. That's what the length contraction formula is for.

The atmosphere is not at rest in the muon frame, however, so to get the proper length we multiply by gamma instead to get....

How deep is the Earth atmosphere then according to the Earth frame?
10,000m.

 

Nice - you multiply by gamma instead to get what you think you should get! There is no reason for you to multiply by gamma just because the atmosphere is "moving" take an f'ing freeze frame picture.

 

This works mathematically - doesn't work practically. By the same argument, you must reverse the time equation as well and get that the muon sees more time on the Earth clock than it's own! Sorry, don't buy the reverse argument - try again, I am not going to accept BS.

 

I am well aware that the relativity of simultaneity offers a "way out" for special relativity regarding the length contraction issue - but even that shows to be faulty when you try to do the experiment in the Muon frame and ignore any measurements you have in the Earth frame.

 

I think you are all missing something important. In this "experiment" one of the objects changes velocity. Velocity is important. Just as the change in velocity breaks the symmetry in the "twin paradox". If you start in the earth frame, the muon appears at 10000m. You calculate that the muon sees 2000m.

An equally valid situation is that the muon is in existence and the earth is created 2000m above the muon moving at .98c. The earth will see 398m to the muon.

Do the experiment without creating objects in the upper atmosphere. The muon is accelerating (changing velocity) it's just not obvious.

Work with two objects with an obvious velocity history. This is still relative velocity, but the delta v affects the symmetry of the situation.

 

Yes, there is. That's what length contraction is all about! Only moving objects length contract. If something is moving in my frame it is length contracted by gamma. To get it's proper length I therefore have to multiply it's apparent length by gamma. Seeing as you post about the validity of the derivation of the Lorentz transformation equations I'm frankly surprised that I have to explain something like this to you...

Anyway, is the atmosphere moving in the muon frame? Yes, it is. Is the atmosphere moving in the Earth frame? No, it's not. Therefore, if the muon sees the moving atmosphere as having length l, then the length contraction formula predicts it as having length gamma*l in it's own frame, which is the same as the Earth frame.

That's how it's done.

From what frame? Time is relative to the frame, so what the muon says is the state of the universe at time t doesn't have a parrallel in the Earth frame. That is, there are no such times t and t' such that the state of the universe in frame S at time t, and the state of the universe at time t' in S' coincide!

Relativity of simultaneity of relativity, remember?