<center>That Crazy Velocity History Thingy
or
I Don't Mind Being Wrong But I Hate Being Misunderstood</center>

Hi Aer, funkstar, JamesR, and Pete.

Something has been bothering me about our muon/earth distance discussions which ended (for me) with the admission that I was mistaken in some fashion. I felt initially that funkstar had a "hidden assumption" in his use of the Lorentz transforms that caused my (and Aer's) assertions to be mistaken. He apparently didn't and all seemed correct.

However, there was a hidden assumption, which I believe lead to an incorrect and misleading understanding of the usage of the Lorentz transforms.

I reproduce funkstars analysis here for reference:
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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 x-axis 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 x-axis 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 x-component), 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.

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.
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Now my analysis with Y = 2:

I will ignore the time element since we are not progressing past T = 0. All measurements are, after all, being made in the instant of muon creation. Simultaneity is not a consideration since there is only one event at x=x'=0 and t=t'=0 where the spacetime coordinates coincide.

In the earth frame: (earth stationary (x,t) and muon moving (x',t'))

x' = Y (x - v*t)

and since t = 0

x' = Y(x)

Meaning 10000 earth meters (x), transformed to the muon frame would appear as 20000m (x') to the muon since the muon frame is contracted wrt the earth frame. This makes sense.

These are coordinate transforms. I.e., what would 10000 earth meters actually transform to in the muon frame? 20000 muon meters.

This however is not the question we are asking.

We wish to know, what would 10000 earth meters look like from the moving muon frame?

We then apply length contraction which is calculated directly from the Lorentz Transforms:

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/imgrel/lencon.gif

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

L' = L/Y

L' = 10000m/2 = 5000m

Which we know is correct. The muon sees the distance to the earth as contracted.

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Now, In the muon frame: (muon stationary (x,t) and earth moving (x',t'))

All we know is our measured length x = 5000m

x' = Y(x)

Meaning 5000 muon meters (x), transformed to the earth frame would appear as 10000m (x') to the earth since the earth frame is contracted wrt the muon frame.

These are coordinate transforms. I.e., what would 5000 muon meters actually transform to in the earth frame? 10000 earth meters.

This however is not the question we are asking.

We wish to know, what would 5000 muon meters look like from the moving earth frame?

We again apply length contraction as derived from the Lorentz Transforms:

L' = L/Y

L' = 5000m/2 = 2500m

This is in fact what the muon must calculate. It's wrong, but the muon knows nothing of this. The situation is symmetrical from an SRT standpoint. There are no preferred frames.
Or are there? Sort of. Non-inertial phases break the symmetry of the spacetime trajectory of a particle. The earth did not accelerate. The muon did. Therefore, knowing this, it can correctly deduce that the earth must see 10000m not 2500m.

This is where I coined the unfortunate phrase "velocity history". I mean to say "Acceleration History" even though the actual acceleration has nothing to do with it. We know that assymetry due to acceleration is vital since it explains the "twin paradox".

If you're still not sure, just do funkstar's exact analysis from the initial earth frame.

I for one understood what you meant from the begining - makes perfect sense IMO.

Good. Then I'm not necessecarily going crazy.

Well, I wouldn't go that far...

Yes, I suppose that's a bit of a stretch...

We wish to know, what would 5000 muon meters look like from the moving earth frame?

We again apply length contraction as derived from the Lorentz Transforms:

L' = L/Y

L' = 5000m/2 = 2500m

This is in fact what the muon must calculate. It's wrong, but the muon knows nothing of this.

It's not wrong. It's right. 5000 muon metres, as viewed from Earth, look like 2500 Earth metres.

There's nothing wrong with that calculation.

This is where I coined the unfortunate phrase "velocity history". I mean to say "Acceleration History".


Damn and here I thought I was first here to point out the reciprocity falicy of SRT and refered to velocity (acceleration) history as being required to do calculations. :D

You are on the right track SL and you would do well to ignore James R's rhetoric saying all is fine because it agrees with SRT.

Your problem here is that you are accepting an unsupported assumption of SRT and forgetting that it is derived by ignoring the tick rate of the clock in motion which is claiming the spatial contraction.

MacM:

Please stay out of this thread. superluminal needs answers from people who know something about relativity.

superluminal, there's no inconsistency, but you're confused about what applying the Lorentz length contraction formula means.

Let the Earth frame be S' and the muon frame be S. S' moves to the right in S, and S moves to the left in S'. Let them coincide on the origos.

Now, we need just one set of coordinates designating the spacetime point where the muon is created. Let's start from Earth frame S', then. [Edit: This choice of frame is insignificant. Going back and forth between the frames is easy, and yields consistent result. I choose the Earth frame because I've worked in the muon frame in the quoted analysis, and I want to show how it doesn't matter.]

At time t' = 0, muon created at x=10,000.

The Lorentz tranforms for going from S' to S are

x = &gamma; (x' + vt')
t = &gamma; (t' + vx'/c²)

which yields (v=0.866c, &gamma;=2) S coordinates

x = 20,000
t = 5.773e-5

These are the coordinates specifying where and when the muon was created in S. Now, does this mean that the muon will see 20,000 muon meters to the Earth at the moment of it's creation? No, because in the moun frame the Earth is moving. And where is the Earth('s surface) at muon time t = 5.773e-5, i.e. at the time of the muons creation in the muon frame? Well, in the muon frame it's described in S by

f(t) = v*t

so it's at

x = 5.773e-5 * 0,866c = 14998 (so let's say 15000)
t = 5.773e-5

So at the time of the muons creation in the Earth frame there's 10,000m to it, and at the time of the muons creation in the moun frame there 5,000m to the Earth.

So instead of appearing as 20,000m the 10,000m appears as 5,000m. Why is this? It is because time matters.

Now, you're saying that this is incompatible with the Earth seeing length contraction of the moun frame, because the Earth should see 5,000 muon meters as 2,500m. But there's no contradiction in that. Let's try going back to the Earth frame S' with the coordinates for the Earth at the muons creation as seen from the muon S. The transforms read

x' = &gamma;(x - vt)
t' = &gamma;(t - vx/c²)

so we get

x' = 2 * (15000 - 0.866c*5.773e-5) = 3.49 (so let's say 0)
t' = 2 * (5.773e-5 - 0.866c*15000/c²)= 2.886e-5

and where is the muon at Earth time t'=2.886e-5? Well, it's described in the Earth frame by

g(t')=10000-vt'

x' = 10000 - 0.866c*2.886e-5 = 2329

or approximately 2500m from the Earth (there are accumulating rounding errors by using &gamma;=2 and so on.) So 10,000m in the Earth frame appear as 5,000m in the muon frame, and 5,000m in the muon frame appear as 2,500m in the Earth frame. Length contraction is perfectly symmetric. No preferred frame necesssary.

The reason you think there's a contradiction here is because you're treating a dynamic distance which is time-dependent as a proper length which isn't. Of course this gets you in trouble, because it's just wrong. You can't ignore time in str except in very few cases, such as the length of proper objects. So, yes, the relativity of simultaneity does come into it, and, no, the acceleration history is not necessary. You also didn't point out what my hidden assumption was. You'll notice that the creation of the muon isn't even mentioned in the first analysis, so there's no acceleration of anything. Yet I was still able to conclude that the Earth (specifically, Earth's atmosphere) is length contracted when viewed from a frame with relative motion. Your argument above amounts to saying that because the muon frame is also length contracted in the Earth frame, then there's a contradiction. This simply isn't so, because you (again) ignored time when you shouldn't, as my above analysis shows. Conclusion: Trust the Lorentz transforms. Don't trust your intuition.


Note: If you start from the muon frame and say that at time t=0 the muon is created at x=5000, as one might be tempted to, the specific coordinates used in the experiment will change (because we now use a different set of S coordinates to designate the creation of the muon). So spacetime coordinates from these two experiments will not be comparable. However, the conclusions about length contraction will not change.

funkstar:

The reason you think there's a contradiction here is because you're treating a dynamic distance which is time-dependent as a proper length which isn't. Of course this gets you in trouble, because it's just wrong. You can't ignore time in str except in very few cases, such as the length of proper objects.

This is what has been kicking my ass from the very beginning. I have definitely been unconciously applying the same rules (intuition?) to the dynamic space between objects, as well as objects themselves.

You ever have the feeling that something isn't quite right but you ignore it because it dosen't seem to make any difference? We've spent so little time (over the last year) discussing length contraction and I thought I had it figured.

I'm going to sleep on it but I think (99.99287%) that I agree with you. If I understand correctly now:

- The Lorentz transforms are indeed coordinate transforms (which I knew) from one inertial system to another.

- The length contraction formula (derived from the LT) yields the percieved contraction of an object with proper length in another frame (and now that I look at it I can even see it in the derivation. The result is time independent...)

- I have been applying the length contraction formula to the dynamic space between objects which is completely wrong.

I'm having an excellent feeling of comfort with all this now. Once again, that sentence in red above nailed it funkstar. Thanks again!

Damn, I feel like a kid with a new toy!

superluminal, you're welcome. That last post sums it up nicely, and I'm glad to see it!

MacM:

Please stay out of this thread. superluminal needs answers from people who know something about relativity.

Pardon me. Screw you. When you give a physics response to the issues I have raised then and only then can you claim to understand physics.

It's wrong, but the muon knows nothing of this. The situation is symmetrical from an SRT standpoint. There are no preferred frames.

Right. There is no asymmetry. The Earth will see all of the muon's scales as being length contracted, and the muon will see the Earth's scales as length contracted. Perfect symmetry. And this isn't wrong. What <I>is</I> wrong is to assume that the muon will see an Earth observer measure the distance between itself and the Earth's surface, as less than the distance that the muon will measure. The source of the mistake is this:

Simultaneity is not a consideration since...

Wrong. Simultaneity <I>has</I> to be considered. I'll show this through another example. A paradox I read about a few years ago. In this scenario, a man standing on a platform sees a train going by at near light speed. And it is length contracted, so that now it's only a foot long. He takes out a ruler, which is also a foot long, and drops it flat on the platform, just as the train goes by. The ends of the train coincide with the ends of the ruler, and he concludes that the train is now 1 foot long. The paradox is that a man in the train will see the <I>platform</I>, and the scale, as length contracted, but not the train. So how can the short ruler measure the long train? Simultaneity gives the answer. In the platform frame, both ends of the ruler hot the ground simultaneously. But in the train's frame, one end hits the ground first, and this coincides with the front of the train, and the second end hits the ground only after a while, when the back of the train gets near it. Now, the first time I read this, it seemed pretty stupid to me. Why the heck does the guy have to <I>drop</I> the ruler? Let him place it on the ground long before the train even comes near. I don't know if there is someone here who thinks the same way... maybe some junior member (I too was a junior back then.). The reason it isn't stupid is this. Even if you don't drop the ruler, you have to see <I>when</I> the ends of the train conincide with those of the ruler. This process is not simultaneous in all frames.
The same case applies here. The muon does see the Earth's scales to be length contracted. It <I>can</I> say that a 5000m scale as seen by the muon will be seen as a 2500m one by Earth. But it <I>cannot</I> say that the Earth will measure the current distance between them as 2500m.

Off topic link: http://www.sciforums.com/showthread.php?t=47770

Thanks Rosnet.

Then we are all in agreement?

This all goes back to what I said: Neither length contraction or the simultaneity of relativity have been proven! Yet the justification for both, length contraction and the relativity of simultaneity are: the relativity of simultaneity and length contraction respectively :bugeye:

These are not abstract concepts here. In 100 years - one would expect direct evidence of one or the other! Yet there is none. Perhaps all attempts to measure them have been pushed under the rug as failures (probably thought their methods were wrong) and then they continue to go on to confirm another time dilation effect which has already been directly observed 8239 times.

You probably still don't understand why I want to go bang my head against the wall everytime length contraction is explained to be a result of the relativity of simultaneity and vice versa :rolleyes: :eek:

Aer:

These are not abstract concepts here. In 100 years - one would expect direct evidence of one or the other! Yet there is none. Perhaps all attempts to measure them have been pushed under the rug as failures (probably thought their methods were wrong) and then they continue to go on to confirm another time dilation effect which has already been directly observed 8239 times.

And hidden crashed UFO's at Area 51? Sorry.

And hidden crashed UFO's at Area 51? Sorry.
What do UFO nuts (i.e. MUFON - I think that is the right abbrev.) have to do with what I said? Not Sorry. :m: :D

When you give a physics response to the issues I have raised then and only then can you claim to understand physics.

The university which granted my degrees thinks otherwise.

This all goes back to what I said: Neither length contraction or the simultaneity of relativity have been proven! Yet the justification for both, length contraction and the relativity of simultaneity are: the relativity of simultaneity and length contraction respectively :bugeye:

Two separate proofs of length contraction: http://www.sciforums.com/showthread.php?p=833683

Two separate proofs of length contraction: http://www.sciforums.com/showthread.php?p=833683
You wouldn't know proof if it hit you in the head. LINK TO EXPERMENTAL OBSERVATION. That BS is just conforming to your beliefs in SR.

And yes, everything you wrote in that post is correct according to SR (well, I am assuming you didn't screw up). That just proves that SR predicts length contraction - this is not the same as proof in science!

This all goes back to what I said: Neither length contraction or the simultaneity of relativity have been proven!

"The first clear example of time dilation was provided over fifty years ago by an experiment detecting muons. These particles are produced at the outer edge of our atmosphere by incoming cosmic rays hitting the first traces of air. They are unstable particles, with a "half-life" of 1.5 microseconds (1.5 millionths of a second), which means that if at a given time you have 100 of them, 1.5 microseconds later you will have about 50, 1.5 microseconds after that 25, and so on. Anyway, they are constantly being produced many miles up, and there is a constant rain of them towards the surface of the earth, moving at very close to the speed of light. In 1941, a detector placed near the top of Mount Washington (at 6000 feet above sea level) measured about 570 muons per hour coming in. Now these muons are raining down from above, but dying as they fall, so if we move the detector to a lower altitude we expect it to detect fewer muons because a fraction of those that came down past the 6000 foot level will die before they get to a lower altitude detector. Approximating their speed by that of light, they are raining down at 186,300 miles per second, which turns out to be, conveniently, about 1,000 feet per microsecond. Thus they should reach the 4500 foot level 1.5 microseconds after passing the 6000 foot level, so, if half of them die off in 1.5 microseconds, as claimed above, we should only expect to register about 570/2 = 285 per hour with the same detector at this level. Dropping another 1500 feet, to the 3000 foot level, we expect about 280/2 = 140 per hour, at 1500 feet about 70 per hour, and at ground level about 35 per hour. (We have rounded off some figures a bit, but this is reasonably close to the expected value.)

To summarize: given the known rate at which these raining-down unstable muons decay, and given that 570 per hour hit a detector near the top of Mount Washington, we only expect about 35 per hour to survive down to sea level. In fact, when the detector was brought down to sea level, it detected about 400 per hour! How did they survive? The reason they didn't decay is that in their frame of reference, much less time had passed. Their actual speed is about 0.994c, corresponding to a time dilation factor of about 9, so in the 6 microsecond trip from the top of Mount Washington to sea level, their clocks register only 6/9 = 0.67 microseconds. In this period of time, only about one-quarter of them decay.

What does this look like from the muon's point of view? How do they manage to get so far in so little time? To them, Mount Washington and the earth's surface are approaching at 0.994c, or about 1,000 feet per microsecond. But in the 0.67 micro-seconds it takes them to get to sea level, it would seem that to them sea level could only get 670 feet closer, so how could they travel the whole 6000 feet from the top of Mount Washington? The answer is the Fitzgerald contraction---to them Mount Washington is squashed in a vertical direction (the direction of motion) by a factor of 1/sqrt(1-v²/c²), the same as the time dilation factor, which for the muons is 9. So, to the muons, Mount Washington is only 670 feet high---this is why they can get down it so fast! "

- http://landau1.phys.virginia.edu/classes/109/lectures/srelwhat.html

How else you can explain muon travelling 6000 feet just in .67 microsec (dilated time) if not length contracted for muon frame?!

This all goes back to what I said: Neither length contraction or the simultaneity of relativity have been proven! Yet the justification for both, length contraction and the relativity of simultaneity are: the relativity of simultaneity and length contraction respectively :bugeye:


Simultaneity is not justified by length contraction. The two postulates alone suffice to prove this. The invariance of the speed of light leads dirctly to relativity of simultaneity. We again take our train and platform example. This time, there is someone standing right in the middle of the train. And there are two bombs, each fixed at one end of the train. They'll explode if a laser beam hits the detector attached to them. The train is in uniform motion (to the right, say) with respect to the platform. Now, look at things from the train's frame of reference. At some arbitrary moment in this frame, the person in the middle of the train shines a laser towards each bomb. Both beams reach the bombs at the same time, and the explode simultaneously. Let's look at the same thing from the platform frame of reference. Don't think about time dilation or length contraction. Just remember that the speed of light in this frame is still 'c'. So, according to the platform observer, at some moment in time, the person in the middle of the train shines a laser beam towards each of the bombs. The beam which goes to the right has a greater distance to travel, because the train would have moved to the right in the interval. On ther other hand, the beam which goes left, will meet the bomb at the left end faster, for the same reason. So in the platform frame, the bomb at the left end of the train explodes first. So these events are not simultaneous. Now, this isn't changed even if we bring in time dilation and length contraction.

Since the two postulates have been tested out quite well, there is no problem. And
I don't think any new postulate is going to mess things up.

...Assume someone standing right in the middle of a moving train moving has placed two bombs, one at each end of the train. They'll explode when light from his flash bulb reaches them. He observes simultaneous explosions shortly after setting off the flash bulb.

Two observers, standing on the ground right next to a bomb when it explodes do not die at the same time. In their reference frame, the flash moving towards the rear of the train explodes its bomb before the one at the front of the train explodes. (During the light transit time, the bomb at the rear of the train has been moving towards the on coming light, while the one at the front of the train was tring to race away from the on coming light of the flash.)... I have taken the liberty to shorten your post and modify it, but give you the credit for the idea. I think it is a good way to convence any reasonable person that what is simultaneous in one frame is not simultaneous in another frame whenever one frame is moving wrt the other.

Further more, the effect is not due to the time for light to travel to the observer - I had the two (now dead) stationary observers stand right next to the bombs when they exploded. (Perhaps they were the unlucky ones of hundreds standing shoulder-to-shoulder along the tracks.) If one wants to add a third stationary observer, standing mid way between the two that get killed, that third observer would observer them die at different times. (The two delays for him to see them die, due to finite speed of light, are equal.)