
080105, 09:27 PM #221
 Posts
 2,250
V<sub>A</sub> = .373c  just use the solve function.

080105, 10:01 PM #222
funkstar, hopefully you will have patience with me, but I honestly want to get this right.
1) Set up reference frames S and S' for the muon and Earth. Let them coincide on some spacetime event.
Now, we need to do this to be able to relate spacetime coordinates between the frames.
To make it easy on ourselves let the Earth travel to the right along the xaxis in frame S. We'll ignore the y and z components, because they are uninteresting.
Let (x,t)=(0,0) in the muon frame S and (x',t')=(0,0) in the Earth frame S' describe the same event.
In the Earth frame S', let the position x'=0 describe the surface of the Earth.
So letting S and S' coincide on (0,0) means that they both point to the same point event ("On the surface of the Earth, when the Earth clock reads zero") when asked to point at the event described by (0,0) in their respective frames.
2) Find a function f(t) to describe the position of the surface of the Earth in the coordinates of frame S.
Well, in frame S the Earth is travelling along the xaxis to the right with speed v. We know it passes through (0,0) because of the way we set up the S frame. So, in the space coordinates of S (there is only the xcomponent), the surface of the Earth is described as a function of time as
f(t)=v*t.
3) Find a function g(t) to describe the position of the edge of the atmosphere in frame S.
Now, this is the interesting part. We don't know the proper length of the atmosphere so we'll have to rely on whatever measurement we get in the S frame.
Let's assume that we get the length l.
So, with f(t) in mind we easily find
g(t) = f(t) + l = v*t + l
4) Using the Lorentz transforms, find the position of g(t) in terms of the coordinates of S'
Look at the Lorentz transforms. Can you see that I set up S and S' exactly as done there? That means we can use the transforms in the form they have there. So the Lorentz transforms relates S coordinates (x,t) with S' coordinates (x',t') by the following equations:
x' = gamma (x  v*t)
t' = gamma (t  v*x/c2)
with gamma defined as usual.
First, I know that the forward transform for length contraction (as calculated from a rest frame to a moving frame) is calculated as L = L0/gamma.
So when we calculate the length that the muon must see from the rest frame of the earth we do:
L = 10000m/2 = 5000m
I know we all agree on this. Now, you provide:
x' = gamma(x  v*t)
This reads: "the distance (length) in the moving (earth) frame (x') is gamma times the length as measured in the rest (muon) frame (x)". (ignoring the elements of v*t  it matters not)
meaning that: 2 * 5000m = 10000m
Unfortunately, this is the reverse transform for length contraction. If we start from the muon frame only, with no other knowledge (all we know is that the earth is coming at us at a hell of a clip), we should naturally apply the forward transform as: x' = x/gamma which yields 5000/2 = 2500m.
The major point being, with only the information given above in blue how did you know to use the reverse transform? This is the hidden assumption I mentioned.
From the beginning I stated that the muon must see the distance to earth as 5000m and the earth must see the muon distance as 10000m. When I placed myself in the muons rest frame, I knew to use the reverse transform because I knew the the constraing FACT that the earth really saw the muon created at 10000m, because that's how we set up the scenario (in our godlike way).
If, however, you are not god, and are just a freshly created muon, how do you know whether to do the natural thing and use the forward transform based on your 5000m measurement of the oncoming earth (yielding 2500m for the distance to earth), or the reverse transform?
I claim that the determining factor is the velocity history within the bounds of the experiment (this in no way implies any MacMlike absolute references or even a third frame. It's just knowing who accelerated).
I think my animation shows that, without knowing who accelerated, you have no way of calculating the correct length contraction from either frame since you have no idea whether to use the forward or reverse transform. This also explains the asymmetry of the "twin paradox". Without that asemmetry, the "twin paradox" would not happen.
The rest of your analysis appears to use the reverse transform for length contraction.
Now, seeing as we've used nothing but measurements in the muon frame, let's see what str really predicts the length of the Earth's atmosphere to be, when starting from the muon frame.
Using (f(t),t) as our spacetime coordinates for the surface of the Earth in S, we get the following S' coordinates for the surface of the Earth:
x' = gamma (f(t)  v*t) = gamma (v*t  v*t) = 0
t' = gamma (t  v2t/c2) = t / gamma
Note the time independence of x'. That means that the surface of the Earth is at rest in S' frame. And look at t'! Time dilation...
Using (g(t),t) as our spacetime coordinates for the edge of the atmosphere in S, we get the following S' coordinates for the edge of the atmosphere:
x' = gamma (g(t)  v*t) = gamma ( v*t + l  v*t) = gamma * l.
t' = gamma (t  v * g(t)/c2)
Again note the time independence of x' for the edge of the atmosphere. That means that the edge of the atmosphere is also at rest in this frame. And what is it's length of atmosphere in the S', even when starting from a different frame S, as predicted by str?
I's gamma * l  0 = gamma * l.

080105, 10:03 PM #223

080105, 10:13 PM #224
 Posts
 2,250
Originally Posted by superluminal
ok, give me a few and I'll give you a real answer

080105, 10:24 PM #225
 Posts
 2,250
Ok, I cannot be 100% sure what funkstar will think/say first of all.
Your arguments in blue are essentially what I have been saying. However, I believe funkstar and the rest will say that we know that the muon was created at the edge of the atmosphere and the atmosphere is moving relative to the muon. This somehow justifies their approach but I am not the one to explain to you why.
The particlular reason why I am not the one to explain such a thing is because of a simple character flaw regarding science that I must admit that I have: I usually don't give a damn about anything that I haven't seen experimental confirmation of. This includes Length Contraction and the Relativity of Simulataneity unfortunately.

080105, 10:29 PM #226
Hmmm... well , treat it like a class on the foundations of relativity as currently formulated. Right?

080105, 10:32 PM #227
 Posts
 2,250
Originally Posted by superluminal

080105, 10:34 PM #228
Yes. The muon will appear squashed from the earth frame, but the distance is known to be, in fact, 10000m.

080105, 10:37 PM #229
 Posts
 2,250
Originally Posted by superluminal

080105, 10:39 PM #230
Well, we see the muon squashed and time dilated by the transforms. The distance is simply an observation of fact.
This is where the velocity history is crucial.

080105, 10:41 PM #231
Aer:
You seem to be confused about what is essentially a simple scenario.
Let's forget about exactly how the muon is created. Let's forget about the atmosphere of the Earth. Let's reduce the problem to its essentials. I will set out results for you, and you can tell me exactly if and where you disagree, ok?
An observer on the ground sees a muon coming towards him at 0.866c, with constant velocity. At some particular instant, which is designated as time zero on the ground observer's clock, the muon is seen by the same observer to be at a height of 10,000 metres. Assume, if you like, that there is a nearby mountain exactly 10,000 m high, and the ground observer can see the muon at the top of the mountain at time zero.
The ground observer says the muon's speed is 0.866c, which gives a relativistic Lorentz factor of gamma=2.
When the muon hits the ground, the ground observer's clock reads t=s/v=10,000/.866c = 0.0000385 seconds.
Now, consider things from the point of view of the muon.
The instant the muon passes the top of the mountain, an observer riding on the muon pulls out his ruler and measures the distance to the ground. It is h/gamma = 10000/2 = 5000 m. Why? Because according to the muon the mountain is moving upwards at 0.866c, and its height is therefore contracted in the direction of motion.
How long does it take the ground to reach the muon? Answer:
t' = s'/v = 5000/0.866c = 0.0000192 seconds.
So, if an observer riding on the muon sets his clock to zero as the muon passes the top of the mountain, then when the ground hits him his clock will read 0.0000192 seconds.
Comparing the travel times from the top of the mountain to the bottom we find:
t' = t/2 = t/gamma
This is time dilation.
Now, Aer, do you disagree with any of the above steps or results?
If "yes", which ones, and why?

080105, 10:42 PM #232
 Posts
 2,250
Originally Posted by superluminal
Originally Posted by superluminal

080105, 10:46 PM #233
JamesR,
If I might interject, you are absolutely correct. I believe the debate is this however. If the muon is created at "some" distance above the earth and measures that distance to be 5000m in his rest frame, what will he naturally calculate for the distance seen in the earth frame (coming at him at 0.866c)? No other info is known.

080105, 10:46 PM #234
 Posts
 2,250
Originally Posted by James R

080105, 11:07 PM #235
Hi super, allow me to add my thoughts to two of your questions to funkstar:
If, however, you are not god, and are just a freshly created muon, how do you know whether to do the natural thing and use the forward transform based on your 5000m measurement of the oncoming earth (yielding 2500m for the distance to earth), or the reverse transform?
What does this really mean?
It means that at a 5000m stick stationary with respect to the muon could be placed with one end at the muon, and one end touching the approaching Earth.
In Earth's frame, this stick will have a length of 2500m  But in Earth's frame, at the instant the muon is created the stick is not touching Earth  it's 7500m away. So, the 2500m is not a measure of the distance from Earth to the Muon at the instant of its creation.
Unfortunately, this is the reverse transform for length contraction. If we start from the muon frame only, with no other knowledge (all we know is that the earth is coming at us at a hell of a clip), we should naturally apply the forward transform as: x' = x/gamma which yields 5000/2 = 2500m.
The major point being, with only the information given above in blue how did you know to use the reverse transform? This is the hidden assumption I mentioned.
If you know the length of a moving stick, you apply the forward transform to find its stationary length.
So the real question is... Is the atmosphere a moving stick or a stationary stick?
Another question you could think about is:
What meaning would a 10000m stick poking in front of the muon have? (ie what if the muon also had an "atmosphere"?)

080105, 11:11 PM #236
Another approach would be to consider the two events of interest:
A  muon created
B  muon reaches Earth.
In the muon frame, there is no distance between these two events.
In the Earth frame, there is. So symmetry is broken.
Can you restore symmetry?
Could you add a third Event corresponding to "Earth created" which would restore symmetry? (hint  use two muons, give each an atmosphere)
Where and when would that event occur in each frame?

080105, 11:13 PM #237
 Posts
 2,250
Originally Posted by Pete

080105, 11:17 PM #238
 Posts
 2,250
Originally Posted by Pete

080105, 11:19 PM #239Originally Posted by Aer
Observations in Muon frame:
A  (0m,0s)  muon created
B  (0m,0.0000192s)  muon reaches Earth, which struck at v=0.866c
Calculation:
Distance Earth travelled between A and B = 0.866 x 3x10<sup>8</sup>m/s x 0.0000192 s = 5000m
Lorentz Transform to Earth frame:
A  (0m, 0s)  muon created
B  (10000m, 0.0000384s)

080105, 11:20 PM #240
 Posts
 2,250
Originally Posted by Pete
Bookmarks