Superconductivity at room temperature??

[sol invictus]

Try again and actually use the Lorentz transformation, Johan. Including the off-diagonal terms. Write down the four-vector coordinates, in the ground rest frame, of the two flashes and of the "light arrives at detector" event. Transform these three coordinates into the train rest frame. Look at the result.

That is impossible - Johan is mathematically incompetent. But his mistake here is more basic, it's a mistake in logic. He thinks that because Δt=Δt', there cannot be time dilation or length contraction.

But this is obviously wrong. Δt is a length divided by a speed. If one clock measures a contracted length, and the other is running slow, they can (and do) agree.

And indeed, in either frame K or K', that's exactly what happens - one captain measures a contracted length with a normal clock, while the other measures a non-contracted length with a slow clock. Of course which is which depends on whether we describe things from K or K', but in either case all physical results are identical.

 

[johanfprins]

But this is obviously wrong. Δt is a length divided by a speed. If one clock measures a contracted length, and the other is running slow, they can (and do) agree.

Just answer the following questions with a simple yes or no:
1. When a rod of length L passes you with a speed v will it contract?
2. When a rod with the same length L passes an astronaut within his spaceship with the same speed v relative to the spaceship, will it contract?
3. Will the contraction of these rods of equal length be different on earth than in the spaceship?

And indeed, in either frame K or K', that's exactly what happens - one captain measures a contracted length with a normal clock, while the other measures a non-contracted length with a slow clock. Of course which is which depends on whether we describe things from K or K', but in either case all physical results are identical.

Both captains are measuring within their own reference frames not even knowing what happens on the other spaceship. Both measure an object (the other spaceship) with length L₀ passing by. Both reference frames must give the same result for the same experiment. Thus if the one captain measures a contraction in the spaceship passing him, the other captain will also measure the same contraction for the other spaceship passing him. Thus if the clock of the second captain is for some voodoo reason running slower, he/she will not measure the same length contraction and this violates Einstein's postulate that the same experiment must give the same result, no matter within which inertial refrence frame the same experiment is done.

 

[sol invictus]

Just answer the following questions with a simple yes or no:
1. When a rod of length L passes you with a speed v will it contract?

Yes.

2. When a rod with the same length L passes an astronaut within his spaceship with the same speed v relative to the spaceship, will it contract?

Yes.

3. Will the contraction of these rods of equal length be different on earth than in the spaceship?

Neglecting some tiny effects, no.

Both captains are measuring within their own reference frames not even knowing what happens on the other spaceship. Both measure an object (the other spaceship) with length L₀ passing by. Both reference frames must give the same result for the same experiment. Thus if the one captain measures a contraction in the spaceship passing him, the other captain will also measure the same contraction for the other spaceship passing him.

Yes.

Thus if the clock of the second captain is for some voodoo reason running slower, he/she will not measure the same length contraction

No. As I've shown you now at least three times, both clocks read exactly the same:

"If one clock measures a contracted length, and the other is running slow, they can (and do) agree.

And indeed, in either frame K or K', that's exactly what happens - one captain measures a contracted length with a normal clock, while the other measures a non-contracted length with a slow clock. Of course which is which depends on whether we describe things from K or K', but in either case all physical results are identical."

and this violates Einstein's postulate that the same experiment must give the same result, no matter within which inertial refrence frame the same experiment is done.

Wrong, as above.

 

[johanfprins]

No. As I've shown you now at least three times, both clocks read exactly the same. If one clock measures a contracted length, and the other is running slow, they can (and do) agree.

When two identical spaceships pass each other, on which spaceship is the clock running at a slower rate and on which at the normal rate?

And indeed, in either frame K or K', that's exactly what happens - one captain measures a contracted length with a normal clock, while the other measures a non-contracted length with a slow clock.

So there are inertial reference frames relative to which a rod does not shrink when it moves: At least you got that right, but it is so relative to all inertial reference frames.

Of course which is which depends on whether we describe things from K or K', but in either case all physical results are identical."

So when the captain in K^/ is switching on his clock it is not really he who is measuring from his perspective but the captain in K who is doing the measurement from his perspective? Wow!!! Now that is really spooky! And obviously nonsense.

 

[ben m]

When two identical spaceships pass each other, on which spaceship is the clock running at a slower rate and on which at the normal rate?

Each thinks the other is running slower.

 

[Ziggurat]

When two identical spaceships pass each other, on which spaceship is the clock running at a slower rate and on which at the normal rate?

In BOTH ships, they view their own clocks as running normal and the other clock as running slow. It is indeed symmetric in that regard.

So there are inertial reference frames relative to which a rod does not shrink when it moves: At least you got that right, but it is so relative to all inertial reference frames.

Nope.

Here's how it works. The length of each ship (in its own reference frame) is L₀, and the velocity of the ships with respect to each other is v This gives us a gamma of 1/sqrt(1-v²/c), which we'll use to simplify the expression for both length contraction and time dilation.

Captain 1 and captain 2 sync their clocks when their noses pass. What does captain 1 observe? He observes that the length of ship 2 is contracted L'=L₀/gamma. Since it's going at velocity v, he measures the time that ship 2's tail takes to reach him is t = L'/v = L₀/(v*gamma).

Now, since his OWN length is not contracted, captain 1 observes that it takes longer for the nose of rocket 2 to reach his own tail. The time this takes is simply t = L₀/v. But what does captain 2's clock read when the nose of 2 reaches the tail of 1? Well, according to captain 1, captain 2's clock is running slow, so captain 2's clock should read t' = t/gamma = L₀/(v*gamma).

So both clocks will read the same thing ( L₀/(v*gamma) ) when those clocks reach the tail of the other rocket. We obtain one number by using length contraction and the other by using time dilation, but they produce the same result, which is what we need to maintain the symmetry of the scenario. This is possible (and completely consistent) precisely because of the relativity of simultaneity. The two pilots do not agree on which of the two spatially separated events (nose 1 reaching tail 2 or nose 2 reaching tail 1) happened first. Captain 1 thinks nose 1 reached tail 2 before nose 2 reached tail 1, and captain 2 performs the same observations the other way around (ship 1 is contracted and time dilated) to find that nose 2 reaches tail 1 before nose 1 reaches tail 2. Simultaneity is relative, so that disagreement in the sequence of spatially separated events is not a problem.

In fact, if you discard both length contraction and time dilation, those events must happen simultaneously in both reference frames. But that's not special relativity, that's Galilean relativity. And you've already agreed to the relativity of simultaneity, which means you've already abandoned Galilean relativity (where simultaneity is absolute). It is not possible to have relative simultaneity without BOTH length contraction and time dilation.

This is really basic special relativity. Yet it has eluded you completely. Whatever capacity you may once have had for understanding this stuff, it's quite clear that you have lost it. It's tragic, really, and I'm sorry for your loss.

 

[johanfprins]

In BOTH ships, they view their own clocks as running normal and the other clock as running slow. It is indeed symmetric in that regard.

Good we are finally getting somewhere. So you now agree that on both ships the clocks are actually running at the same normal rate. It only seems to the one captain AS IF the other captain's clock is running slower and vice versa. Thanks you finally grabbed it!:

It is not possible to have relative simultaneity without BOTH length contraction and time dilation.

Wrong! There is no length contraction of a passing rod at all, and the time dilation is an apparent dilation which does not actually occur within K^/ even though the captain within K thinks that it is occuring within K^/; and vice versa. What is actually happening is that that the rod (say stationary within K^/) becomes longer and that the time along this longer rod becomes dependent on position. This can easily be checked by transforming the simultaneous positions of the nose and tail of the stationary rod within K^/ into their corresponding non-simultaneous positions that they have within K. It is then found that at any instant in time the perception within K is created that the time at the tail of the rod is ahead of the time at the nose of the rod, even though this is not the actual case when you travel with the rod.

This is really basic special relativity. Yet it has eluded you completely. Whatever capacity you may once have had for understanding this stuff, it's quite clear that you have lost it. It's tragic, really, and I'm sorry for your loss.

It is more tragic that I was taught to believe the same nonsense that you, Sol and Ben are still spouting, and that I was so stupid that I lapped it up. The tragedy is that you still believe that the time in a passing spaceship K^/ actually passes slower within the spaceship, while in reality it is only a relativistic perception within K that this is so: It does not actually happen on the spaceship.

It is the same stupidity to conclude that a ball which is launched straight up within a passing spaceship K^/, actually follows a parabolic path within the spaceship, while it is only the perception within K that gives a parabolic path. This is what relativity does: It gives a distorted view of what is acttually happening within a moving inertial reference frame. GET IT?

 

[Ziggurat]

Good we are finally getting somewhere. So you now agree that on both ships the clocks are actually running at the same normal rate. It only seems to the one captain AS IF the other captain's clock is running slower and vice versa.

No. The moving clocks actually run slower in the reference frame where they are moving. This is not some optical illusion. This is hard requirement of special relativity. And you cannot reconcile the relativity of simultaneity (which you have already agreed to) with any other view.

But since you feel differently, show us the math. Give us the various coordinates in the different reference frames, and show how those coordinates transform from one frame to another. Do the math.

But I know that this request is made in vain. You've become just another "Einstein was wrong" internet crank. And internet cranks never do the math. You can't do it with relativity, you can't do it with quantum mechanics, hell, you can't even do it with simple circuit energy conservation.

 

[Mashuna]

This is really basic special relativity. Yet it has eluded you completely. Whatever capacity you may once have had for understanding this stuff, it's quite clear that you have lost it. It's tragic, really, and I'm sorry for your loss.

It's kind of sad to read Johan's posts now.

 

[ben m]

It is the same stupidity to conclude that a ball which is launched straight up within a passing spaceship K^/, actually follows a parabolic path within the spaceship, while it is only the perception within K that gives a parabolic path.

Wait, no---you have it backwards. The ball you're talking about is actually following a parabolic path, and the OTHER GUY who says it's straight up and down is the one who has a perception problem. Because he's moving, presumably, and carrying all of these rulers and nonsense. I'm standing still with my rulers, so I'm right about the real trajectory.

</sarcasm>

 

[sol invictus]

When two identical spaceships pass each other, on which spaceship is the clock running at a slower rate and on which at the normal rate?

The rate of slowing is determined by the velocity. If you work in frame K, the clock on ship K' is running slow. If you work in frame K', the clock on ship K is running slow. If you work in a frame where both ships are moving, both clocks are running slow (although not necessarily at the same rate).

So there are inertial reference frames relative to which a rod does not shrink when it moves: At least you got that right, but it is so relative to all inertial reference frames.

What? I didn't say anything like that. If the rod is moving, it's length is contracted.

So when the captain in K^/ is switching on his clock it is not really he who is measuring from his perspective but the captain in K who is doing the measurement from his perspective? Wow!!! Now that is really spooky! And obviously nonsense.

I have no idea what you are talking about.

 

[Reality Check]

That is what I said, or intended to say, and this means that, when using the probability interpretation for quantum mechanics, a photon cannot be a summation of different frequencies and their concomitant k's.

You are wrong. As I stated before

• So an electon moving with a speed v has a single well-defined de Broglie wavelength.
• No: a photon has the same measured single frequency and value for k.

This is the bit of quantum mechanics that states when you measure an observable then you get a single value.

The Copenhagen interpretation is basically that the meaning of the wave function is that its modulus squared is a probability of measuring that single value.

This is according to the Copenhagen probability interpretation which is wrong.

Where is your evidence that the Copenhagen interpretation is wrong?
You may be correct, e.g. the many worlds interpretation may be right.
That does not affect the fact than in QM, a measurement measures a single value.

Really! What is then keeping an electron stationary (x=0 p=0) within its own inertial reference frame.

That is ignorant. The electron is not stationary. It has momentum. p is not zero. x changes.

A free electron has an energy m*c^2 in its own inertial refrence frame
...
Thus the Schroedinger equation cannot be applied to model a free electron since it already has the mass as input.

Really determined to display your ignorance, johanfprins .
One more time: The Schroedinger equation is non-relativistic.
In it there is no energy mass equivalence. The classical (non-relativistic) mass is just plugged into the Hamiltonian.

This is irrelevant:

It is relevant: A wave function with one set of position coordinates by defintion describes one particle.
It would take an ignorant or deluded person to think that you can describe multiple particles by a wave functions that ignores all by one of the particles.
So the question becomes: johanfprins

• Can you understand that a single particle wave function describes a single particle?
• Or do you want to remain ignorant that a single particle wave function describes a single particle?
• Or do you just want to remain deluded that a a single particle wave function describes multiple particles?

When a body with mass moves, the movement is a relativistic effect, since the body is in reality stationary within the inertial refrence frame moving along with it.

One more time: The Schroedinger equation is non-relativistic.

Only after it has been renormalised: Before renormalisation it is infinite.

Whoops - you are right at last!

But this is what is in effect being done when you renormalise to get QED.

You are amost right but still oh so wrong !
No one tries to subtract infinity from infinity in renormalization. That operation is ill-defined (if you try to treat infintity as a number and subtract it from itself then you come to the conclusion that 0 = 1)

So your knowledge of physics is based on concepts that you cannot define? This sounds like VOODOO to me!

That is idiotic: my knowledge of physics includes my understanding of standard definitions of basic terms which Wikipedia does a good job of explaining (not so good on more complex stuff!).
Your inability to understand standard stuff like:

• The Schroedinger equation is non-relativistic.
• A single particle wave function describes a single particle.

indicates your lack of understanding of the basics of physics.

It is my opinion that you are wrong.

The actual science that you are ignorant of, are ignoring or are deluded about states that you are wrong. See above.

Again nonsense: A localised standing wave within an inertial reference frame will be observed to move like a particle within another inertial refrence frame that passes the standing wave.
...

Again gibberish:

The Wannier representation is but one of many which is being used to model non-ideal materials: In most cases not with great success.
...gibberish snpiped....

In actual fact the Wannier representation is used to model "ideal" materials.
It is real life (non-ideal) materials that show the limits of models based on the Wannier representation. The problem I came across in my thesis was the measured giant peaks in the thermopower of nearly magnetic materials (e.g RhFe wheere Fe is in the dilute limit). The theory could produce peaks but much smaller.

 

[Reality Check]

You see present day physics is under the control of an inquisition: Typical examples are people like Ben, Sol and Ziggurat: I decided to read the manuscript and was gobsmacked.

Here we go: The good old crank colleague (or johanfprins) who has a paranoid delusion that there is a conspiracy to suppress their work!

All you have to do publish a paper in the scientific literature (rather than in the massive swamp of crank books that there is out there) is convince a couple of reviewers who are experts in the field that there are no obvous flaws in your theory.
So the question is: Why can you or your colleague not produce a theory that can convince a couple of scientists that it is correct?

Could it be that your colleague 's theory is totally bogus since he cannot grasp that the Lorentz transformations means that simultaneity is relative?

P.S. Einstein was the second person to think up the train + lightning thought experiment, the first was Daniel Frost Comstock, 7 years earlier in 1910.

 

[Reality Check]

Good we are finally getting somewhere. So you now agree that on both ships the clocks are actually running at the same normal rate. It only seems to the one captain AS IF the other captain's clock is running slower and vice versa.

Just to emphasize the point that Ziggurat makes in his reply to this bit of your post: Time dilation is a real physical effect. Both captains measure that the other clock is running slower.

The standard experimental example of this is the increase in the half-life of short-lived particles when they are moving fast. You can also just measure it using clocks.
Experimental basis of Special Relativity

The Ives and Stilwell Experiment
H.E. Ives and G.R. Stilwell, “An Experimental Study of the Rate of a Moving Atomic Clock”, J. Opt. Soc. Am. 28 pg 215–226 (1938); JOSA 31 pg 369–374 (1941).
This classic experiment measured the transverse Doppler effect for moving atoms.
Otting, Physik. Zeitschr. 40, 681 (1939).
-
Hasselkamp et al., Z. Physik A289 (1989), pg 151.
A measurement that is truly at 90° in the lab. Agreement with SR to an accuracy of a few percent.

See also Mandelberg and Witten.
Measurements of Particle Lifetimes
Rossi and Hoag, Physical Review 57, pg 461 (1940). Rossi and Hall, Physical Review 59, pg 223 (1941). Rasetti, Physical Review 60, pg 198 (1941). Redei, Phys. Rev. 162 no. 5 (1967), pg 1299.
Various measurements of the lifetimes of muons.
See also: Bailey et al.
Durbin, Loar and Havens, Physical Review 88, pg 179 (1952).
-
D. Frisch and J. Smith, “Measurement of the Relativistic Time Dilation Using Mesons”, Am. J. Phys. 31 (1963) 342.
Measurements of the lifetimes of pions. An interpretation was given by: Terell, Nuovo Cimento 16 (1960) pg 457.
Greenberg et al., Phys. Rev. Lett. 23 no. 21 (1969), pg 1267.
More accurate measurement of pion lifetimes.
Ayres et al., Phys. Rev. D3 no. 5 (1971), pg 1051.
Measurements of pion lifetimes, comparison of positive and negative pions,
etc.
Burrowes et al., Phys. Rev. Lett. 2 (1959), pg 117.
Measurements of Kaon lifetimes.

 

[johanfprins]

No. The moving clocks actually run slower in the reference frame where they are moving.

You have just claimed correctly above (probably for the first time in your life that you made a correct physics statement) that owing to symmetry the clock in reference frame K must keep exactly the same time in reference frame K^/; and now you deny your own statement. Amazing!

This is not some optical illusion.

In fact it is a type of illusion; which my new friend classify as a type II relativistic illusion: It has an effect, but it does not really occur within either of the two reference frames. You also have a type I relativistic illusion, like the parabolic path within one inertial refrence frame when projecting a ball vertically within the other refrence frame.

This is hard requirement of special relativity.

No it is not!

And you cannot reconcile the relativity of simultaneity (which you have already agreed to) with any other view.

Oh yes you can: As you will see when (or rather if) the inquisition council in charge of modern physics, allows his manuscript to be published.

But since you feel differently, show us the math. Give us the various coordinates in the different reference frames, and show how those coordinates transform from one frame to another. Do the math.

I am not a plagiarist and therefore I am asking you to wait patiently for the publication of my new friend's manuscript. The math is done meticulously and in detail. In fact it is so simple that I am amazed that somebody as arrogant as you could not have thought it out by himself.

But I know that this request is made in vain. You've become just another "Einstein was wrong" internet crank. And internet cranks never do the math. You can't do it with relativity, you can't do it with quantum mechanics, hell, you can't even do it with simple circuit energy conservation.

I hope that you will have the decency to apologise once the impeccable proof is there that you are wrong: But this will obviously not happen since you are a "flat earther" who will argue to his death that the earth cannot be a sphere.

We had another ESKOM power failure today, and the lights only came on a few minutes ago. Off to bed till later: Nighty night!

 

[sol invictus]

You have just claimed correctly above (probably for the first time in your life that you made a correct physics statement) that owing to symmetry the clock in reference frame K must keep exactly the same time in reference frame K^/; and now you deny your own statement. Amazing!

In the experiment you described, both clocks will end up reading the same. Nothing in Zig's (correct) post is in conflict with that, nor has he denied it anywhere that I've seen.

Your problem, Johan, is that you've failed to understand anything about this - even the most simple aspects. As a result, you see non-existent contradictions all over the place.

 

[ben m]

Let me just quote from a college freshman textbook. (Bold added, italics original)

The time dilation effecti is completely symmetric. If a clock C at rest in S is observed by S', then S' concludes that clock C is running slow. Each observer believes that the other's clock is running slower than the ones at rest in the reference frame of the observer. Time dilation is often summarized bt the phrase, "moving clocks run slow". It is useful to remember this phrase, but do so with caution. The phrase indicates that a clock moving relative to an frame containing an array of synchronized clocks will be found to run slow when timed by those clocks. That is, only in the sense of comparing a single moving clock with two separate synchronized stationary clocks can we declare that "moving clocks run slow".

In other words, the standard relativity lesson for 18 year olds includes the admonition that relativity sounds confusing and paradoxical if you treat it casually, descriptively, or sloppily, but that everything works according to the precise meanings of things.

 

[Ziggurat]

You have just claimed correctly above (probably for the first time in your life that you made a correct physics statement) that owing to symmetry the clock in reference frame K must keep exactly the same time in reference frame K^/; and now you deny your own statement. Amazing!

I made no such claim. I said that both clocks will have the same reading when they reach the tail of the other rocket. But in the reference frame of EITHER rocket, the two events (nose 1 and tail 2 meeting, or nose 2 and tail 1 meeting) are not simultaneous. It is precisely because of time dilation (plus length contraction) that we can reconcile the symmetry of the scenario with the relativity of simultaneity. Throw out time dilation and length contraction, and you're back to Galilean relativity with absolute simultaneity.

No it is not!

Yes, it is. Any alternative is no longer special relativity. If you want to claim that special relativity is wrong, go ahead. But you don't get to deny what it is that the theory claims.

Oh yes you can: As you will see when (or rather if) the inquisition council in charge of modern physics, allows his manuscript to be published.

Oh please. Stop with the paranoia. If you think it can't get past peer review (and I suggest you consider the rather more pedestrian reason for why a paper might get rejected), you can always publish on the web.

I am not a plagiarist and therefore I am asking you to wait patiently for the publication of my new friend's manuscript.

Such results are always just around the corner.

The math is done meticulously and in detail.

Given your complete inability to do any math, your word is of no value in this regard. And you can't even use your scrumples about plagiarism as a defense in this regard: you can't even do standard textbook math from already-published sources.

I hope that you will have the decency to apologise once the impeccable proof is there that you are wrong

If such proof is ever forthcoming, I certainly will. But it would need to be actual proof, and nothing you have offered on any topic even slightly resembles that.

Furthermore, you are in absolutely no position to lecture me about decency. When I first started talking to you in this thread, I was civil and polite in my disagreements with you. Yet you repeatedly insulted me without provocation. I asked you multiple times to refrain from insults, and you never could. You have lost all right to complain to me about decency. Perhaps with your failing memory these events have passed from your mind, but I remember them.

 

[johanfprins]

Let me try again: Consider a hovering helicopter situated above a cloud and launching a bomb with a horizontal speed v₀. To obtain this launch-speed the bomb must be subjected to a force causing it to accelerate before leaving the helicopter. If this force is not there, the bomb will fall straight down from the helicopter to the ground. But since the force is there an observer on the gound will see the bomb appearing from the cloud and following a parabolic path down to earth.

Now consider an airplane flying with a speed v₀ and releasing a bomb (without any horizontal acceleration) at precisely the same position above the cloud at which the helicopter had hovered. This bomb will follow the same path that the bomb (launched by the helicopter) had followed; but now there is no launch-force involved: However, an observer looking at the path of the bomb relative to the reference frame attached to the ground, must conclude that there must have been a horizontal launch force: Even though there is not such a launch-force since the bomb is falling straight down within the reference frame attached to the airplane: i.e. the reference frame within which the aeroplane is stationary.

Thus, two points that must be noted: (i) The parabolic path is generated by the movent of the aeroplane relative to the ground: It is thus a relativistic effect which is ONLY observed from the ground and not from the aeroplane relative to which the bomb is falling straight down. My friend classified this relativistic behaviour as a type I "relativistic illusion". "Illusion" in this case does not imply that the path is not real relative to the ground, but that it would not have been there if it were not for the relative motion. The path is very real, as many soldiers have found out during the war. (ii) But the launch force is not real: It does not really occur within any of the two reference frames. This has been classified by my friend as a type II "relativistic illusion": In this case it is a real illusion since the force does not exist at all within either reference frame: But it is required to calculate the path as observed relative to earth.

Now consider a muon created in a laboratory: It moves with a very slow speed relative to the laboratory, and is thus near-stationary relative to the laboratory. We then measure a lifetime (delta)t(rest) for the muon. Now consider a muon created within cosmic rays which approaches the earth at a very high speed: When we now measure its lifetime with a clock on the earth one obtains that (delta)t(speed) and that (delta)t(speed)>>(delta)t(rest). If it were possible to travel down to earth with the cosmic muon, it will be at rest relative to such an observer : His/her clock will thus measure the lifetime (delta)t(rest). Thus, the longer measurement recorded by an observer on earth is, like the parabolic path above, a type1 I relativistic illusion; since it is only observed relative to earth and NOT on a clock travelling with the cosmic ray muon. But the observer on earth has to rationalise the increase in lifetime that he/she measures for the cosmic ray muon, and thus argues that a clock travelling with the muon must be slower than a clock on earth: However, the latter rationalisation is exactly the same as conluding that a bomb dropped from an aeroplane has been launched by a horizontal force. Just as in the latter case, the assumption that a clock moving with the muon is slower is a type II relativistic illusion, since this slowing down of a clock cannot occur within the reference frame travelling with the muon and also not on earth. It does not really occur at all!

Now let us return to our two spaceships moving past each other: Which one of them is stationary, and which one of them is moving? One can only give an answer to this question if there were a unique stationary inertial refrence frame within our universe. Eisntein's postulates states that such a reference frame does not exist. Thus motion itself is a type I relativistic illusion (remember Zeno?), since the spaceship moving, with a speed v relative to the observer within the other spaceship, is not moving within its own inertial reference frame, and vice versa. Basically every object with mass which is not being accelerated by a force, is STATIONARY: Once you realise this incontestable conclusion, you will also realise that two clocks moving with a speed v relative to each other must keep exactly the same time since both are stationary within their respective inertial refrence frames.

 

[sol invictus]

L
Now consider an airplane flying with a speed v₀ and releasing a bomb (without any horizontal acceleration) at precisely the same position above the cloud at which the helicopter had hovered. This bomb will follow the same path that the bomb (launched by the helicopter) had followed; but now there is no launch-force involved: However, an observer looking at the path of the bomb relative to the reference frame attached to the ground, must conclude that there must have been a horizontal launch force:

No, obviously not. The observer could simply conclude that the bomb wasn't at rest at the time is was "launched", but instead had some initial horizontal motion.

Indeed, force and acceleration are (at least in Galilean relativity) independent of reference frame.

(i) The parabolic path is generated by the movent of the aeroplane relative to the ground: It is thus a relativistic effect which is ONLY observed from the ground and not from the aeroplane relative to which the bomb is falling straight down. My friend classified this relativistic behaviour as a type I "relativistic illusion". "Illusion" in this case does not imply that the path is not real relative to the ground, but that it would not have been there if it were not for the relative motion.

A better term is "frame dependent".

(ii) But the launch force is not real: It does not really occur within any of the two reference frames. This has been classified by my friend as a type II "relativistic illusion": In this case it is a real illusion since the force does not exist at all within either reference frame: But it is required to calculate the path as observed relative to earth.

It is not required, so I have no idea what you are talking about here.

Now consider a muon created in a laboratory: It moves with a very slow speed relative to the laboratory, and is thus near-stationary relative to the laboratory. We then measure a lifetime (delta)t(rest) for the muon. Now consider a muon created within cosmic rays which approaches the earth at a very high speed: When we now measure its lifetime with a clock on the earth one obtains that (delta)t(speed) and that (delta)t(speed)>>(delta)t(rest). If it were possible to travel down to earth with the cosmic muon, it will be at rest relative to such an observer : His/her clock will thus measure the lifetime (delta)t(rest). Thus, the longer measurement recorded by an observer on earth is, like the parabolic path above, a type1 I relativistic illusion; since it is only observed relative to earth and NOT on a clock travelling with the cosmic ray muon.

That's correct, yes. Clock rates, lifetimes, etc. are frame-dependent quantities.

But the observer on earth has to rationalise the increase in lifetime that he/she measures for the cosmic ray muon, and thus argues that a clock travelling with the muon must be slower than a clock on earth: However, the latter rationalisation is exactly the same as conluding that a bomb dropped from an aeroplane has been launched by a horizontal force. Just as in the latter case, the assumption that a clock moving with the muon is slower is a type II relativistic illusion, since this slowing down of a clock cannot occur within the reference frame travelling with the muon and also not on earth. It does not really occur at all!

The fact that you categorize "longer muon lifetimes" and "slower clocks" differently reveals yet another childish misunderstanding. There is no difference between the two. All processes run slow in moving frames. You could make a clock out of muons, for example.

Once you realise this incontestable conclusion, you will also realise that two clocks moving with a speed v relative to each other must keep exactly the same time since both are stationary within their respective inertial refrence frames.

And yet, they don't. Moving muons decay slower than stationary muons, so moving muon-clocks run slow (as to all other types of clocks).

Are you aware of this experiment, Johan? http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/airtim.html