Superconductivity at room temperature??

[Mister Earl]

Anyone else notice that whenever someone crops up claiming to be able to overturn relativity, or quantum mechanics, it's always without math, without a testable hypothesis, and consists entirely of speculative arguments that show gross misunderstandings?

Good god. I'm a layman in science, but even I know that one does not overturn repeatedly proven science with "racecar on a train" thought experiments.

 

[ben m]

What does "has had" mean? The fact remains that a photon has the same single frequency and value for k within all inertial reference frames:

Let me introduce you to my old friends redshift and blueshift.

 

[johanfprins]

Anyone else notice that whenever someone crops up claiming to be able to overturn relativity, or quantum mechanics, it's always without math, without a testable hypothesis, and consists entirely of speculative arguments that show gross misunderstandings?

Really? Have you read the mathematical derivations in my book? There are no speculative arguments there and your accusations before ascertaining the facts, prove why you are only suitable to form part of a lynch mob.

Good god. I'm a layman in science, but even I know that one does not overturn repeatedly proven science with "racecar on a train" thought experiments.

When did I do this? What I know is that when you derive Einstein's non-simultaneity in one reference frame for two simultaneous events in another reference frame, directly from the Lorentz transformation (which mathematically totally determines all the aspects of special relativity), it is found that Einstein's celebrated thought experiment based on a passing train and two lightning strikes, violates the Lorentz transformation. It is alo found that Einstein's derivation of length contraction cannot be derived mathematically from the Lorentz transformation; and that time dilation does not actually occur on a moving clock. These are direct and meticulous mathematics applied to the Lorentz transformation which proves this: Not airy fairy thought experiments as regularly found in text books on relativity and quantum mechanics.

 

[johanfprins]

My previous statement: "The fact remains that a photon has the same single frequency and value for k within all inertial reference frames" obviously violates red and blue shifts and I was sloppy in not stating what I intended, namely that a photon only has a single value for its frequency and a single concomitant value for its its k-vector: no matter how many identical photons you measure. Thus, it cannot be a probability wave packet consisting of a superposition of different values for the frquency and the k-vector.
Furthermore "Reality Check" has written so many things which just cannot be correct, that it will be useless to argue about all of them. I will thus be selective.

So an electon moving with a speed v has a single well-defined de Broglie wavelength.

Do you deny this?

No: a photon has the same measured single frequency and value for k.

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.

Right: an electron is not a wave packet.
The expectation values for its state variables (measurable quantities) are wave packets.

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

How dumb: The boundary conditions for a free-particle is that there are no boundary conditions because there are no boundaries !

Really! What is then keeping an electron stationary (x=0 p=0) within its own inertial reference frame. Only a harmonic potential can do this. When a photon of energy h(nu) forms a electron-positron pair, the electron and the positron end up
each having a mass-energy equal to (1/2)h(nu). Only a wave equation with a harmonic potential gives such a solution. Thus the inertia of an electron is most probably caused by a harmonic potential keeping it stationary within its inertial reference frame.

The free-particle solution represents a free particle for the simple physical reasons that
There is no potential in the Hamiltonian (it is a Hamiltonian for a free particle).

Prove that this is so. A free electron has an energy m*c^2 in its own inertial refrence frame; and any wave-equation must give this energy as the solution. Thus the Schroedinger equation cannot be applied to model a free electron since it already has the mass as input. If you use what you want to solve for as input in your equations, you are goofing!

The wave function has 1 set of position coordinates and so it is for a single particle.

This is irrelevant: An entangled wave also has 1 set of coordinates even though it formed from different subwaves which before entangling had their own coordinates.

As for experiment proof: Attosecond electron wave packet interferometry
Let us hope that you do not go the crank route and whine about the experiment not being on a single electron !

I will look at this URL as soon as I have more time. But I expect that it is most probably not relevant to the argument.

I know what the theory of relativity entails.

You could have fooled me.

I can also read. I can read that the Schrödinger equation is a non-relativistic equation. That is something you seem unable to grasp.

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. So it seems that you agree with the Vatican that Galileo was wrong?

Vacuum energy[/URL] is not infinite , e.g. in QED it is 10¹⁰⁷ Joules per cubic centimeter.

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

There is no subtracting infinity from infinity. No one would be dumb enough to do that because infinity is not a number.

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

The reduction of infinities in renormalization (regulization) is basically the repackaging of series of finite terms that each sum up to infinity into equivalent series of finite terms that do not sum up to infinity.

Fudging away infinities: Just as I have said.

I do not need to define such basic terms as wave or particle.

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

And there you go again.
I d not recall claiming that that a free particle is a wave packet.
A free particle is not a wave packet.
The expectation values for its state variables (measurable quantities) are wave packets. That is why the square of the magnitude of the wave function has been given.

Probability interpretation again! I am sorry I am not so superstitious to believe such nonsense.

And I keep on pointing out that you are wrong.

It is my opinion that you are wrong.

That is ignorant. Electrons interact as waves and so they must be waves. Electrons interact as particles and so they must be particles. Eleactons have the proerties of both particle and waves.

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. This does not make the standing wave a particle. Why do you WANT to believe in the bipolar concept of wave-particle duality when it is possible to interpret all of quantum mechanics in terms of waves and waves adapting to chages in boundary conditions?

Furthermore, a free electron is not a wave packet. A moving electron has a unique measured de Broglie wavelength.
The wave function of the free electron consists of a superposition of wavelengths. The expectation value of its momentum is a non-dispursive wave packet (in contrast to the expectation value of its position which is a dispursive wave packet).

You have a probability wave that consists of a superposition of wavelengths such when you measure the wavelength you consistently obtain exactly the same value for the wavelength, and then you call the wave a "probability distribution"? It seems to me that it must be a "certainty-distribution" in this case!

A measurement of the momentum of a free electron collapses the wave function (in the Copenhagen interpretation) and you get a unique measured de Broglie wavelength.

But since it is a probability distribution you cannot get exactly the same wavelength for every electron within an ensemble of electrons all moving with the same speed v. If you obtain the latter, then you have a "certainty-distribution": Not a probability-distribution.

As a complete aside: It is the Dirac equation that more completely describes a free electron.

I do not agree with this either; but let us not wander into Dirac's autistic universe which supposedly have magnetic monopoles.

Of course if we really wanted to discuss electrons in a metal as you seem to bring up all of the time than we should look at something I used in my thesis - the Wannier representation in WFT.

The Wannier representation is but one of many which is being used to model non-ideal materials: In most cases not with great success. Let us rather stick to the assumption which works quite well for metals like gold and copper that do not superconduct: For these materials it is assumed that the electrons screen each other so well, that, to an excellent first approximation, each electron can be solved by the same so-called "free" Schroedinger equation (having the same set of coordinates) subject to the boundary conditions which are provided by the size of the material. These waves can then superpose to form wave-packets (quasi-electrons) when an electric field is applied.

 

[Ziggurat]

My previous statement: "The fact remains that a photon has the same single frequency and value for k within all inertial reference frames" obviously violates red and blue shifts and I was sloppy in not stating what I intended, namely that a photon only has a single value for its frequency and a single concomitant value for its its k-vector: no matter how many identical photons you measure.

Oh, I see: photons don't violate special relativity, they just violate the Heisenberg uncertainty principle. Got it.

A free electron has an energy m*c^2 in its own inertial refrence frame; and any wave-equation must give this energy as the solution. Thus the Schroedinger equation cannot be applied to model a free electron since it already has the mass as input. If you use what you want to solve for as input in your equations, you are goofing!

Congratulations: you've discovered that the Schrodinger equation is not Lorentz invariant. So what? Rest mass energy is irrelevant to an explicitly non-relativistic theory. The theory should only be considered as the low-velocity limit of a correct theory, but that's not a problem. Furthermore, while it would be nice to have a theory which predicts the electron's mass correctly, we don't actually need one. Even assuming such a theory exists, waiting to be found, a theory which uses it as an input can still be accurate. And it's accuracy, not truth, which we seek from scientific theories.

In other words, your objection is a complete failure. Especially since you have nothing to offer yourself which meets your own requirements.

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. So it seems that you agree with the Vatican that Galileo was wrong?

Galileo was wrong. That's why special relativity has supplanted Galilean relativity. But the Schrodinger equation obeys Galilean relativity, NOT special relativity.

Let us rather stick to the assumption which works quite well for metals like gold and copper that do not superconduct: For these materials it is assumed that the electrons screen each other so well, that, to an excellent first approximation, each electron can be solved by the same so-called "free" Schroedinger equation (having the same set of coordinates) subject to the boundary conditions which are provided by the size of the material. These waves can then superpose to form wave-packets (quasi-electrons) when an electric field is applied.

This is not what happens. Assuming no interactions, you do indeed assume that you can solve single-particle states independently. But to get the full multi-particle wave equation, you still need a set of coordinates for each electron. That is why Fermi exclusion applies: because of the symmetry requirements of these multiple coordinate sets. Without multiple coordinates, there is no Fermi exclusion. Basically, you want to use the results of quantum mechanics while discarding the actual theory itself. And you don't even realize what you're doing, because you don't even understand what the theory actually is.

 

[ben m]

When did I do this? What I know is that when you derive Einstein's non-simultaneity in one reference frame for two simultaneous events in another reference frame, directly from the Lorentz transformation (which mathematically totally determines all the aspects of special relativity), it is found that Einstein's celebrated thought experiment based on a passing train and two lightning strikes, violates the Lorentz transformation.

Then you made a mistake; post your "derivation" online I can find it for you. But I suspect I would not be the first person to point out such a mistake .... am I right?

Why am I so sure? First, I've done many of these derivations myself and there is not a problem. Second, so has everybody else. Third, the self-consistency of the Lorentz transformation is mathematically proven---finding a "violation" of it is finding two integers whose sum is irrational.

 

[Ziggurat]

finding a "violation" of it is finding two integers whose sum is irrational.

Chuck Norris can do that.

 

[sol invictus]

I'll tell you what, Johan - why don't you write down the solution to the Schrodinger equation that you think represents a single, stationary electron in empty space. I don't want words, I want an equation, like ψ(x,t)=...

Remember, |ψ(x,t)|² is the probability density for finding the particle at x, and therefore the integral of |ψ(x,t)|² over x had better equal 1.

Johan, your inability to answer the above question is noted.

What I know is that when you derive Einstein's non-simultaneity in one reference frame for two simultaneous events in another reference frame, directly from the Lorentz transformation (which mathematically totally determines all the aspects of special relativity), it is found that Einstein's celebrated thought experiment based on a passing train and two lightning strikes, violates the Lorentz transformation.

That is literally nonsense. Einstein's train/lightning strike "experiment" (which can be found here) is nothing more or less than a picturesque description of the Lorentz transformation.

It is alo found that Einstein's derivation of length contraction cannot be derived mathematically from the Lorentz transformation; and that time dilation does not actually occur on a moving clock.

Again; nonsensical gibberish. Length contraction and time dilation are part of the Lorentz transforms.

 

[johanfprins]

I have decided not to respond to posts by you Sol and Ziggurat. But since you issued a challenge, I have to respond in the interest of another physicist here in South Africa who proved to me that Einstein, and those that followed him, made serious mistakes in the interpretation of the Special Theory of Relativity.

Then you made a mistake; post your "derivation" online I can find it for you.

I am not at liberty to do this until my colleague has had his manuscript published; which I hope will be soon. Suffice to say that when you use the Lorentz transformation correctly, you will find that the following statement by Einstein, which can be found in his little book entitled RELATIVITY: The special and General Theory, page 26, is not correct: "Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding ahead of the beam of light coming from A, Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A.". Now when using the Lorentz transformation directly you will find that within ANY inertial reference frame, the light beam from B as well as the light beam from A are both still approaching the observer with speed c. If the observer could have rushed into the light beam from B, and away from the light beam coming from A, the light speed from B relative to the observer would be larger than c, and the light speed relative to the observer from A would be less than c. This violates Einstein's own postulate that light can only move with a speed c relative to any object with mass. Try the derivation yourself if you are competent; which I, however, doubt.

But I suspect I would not be the first person to point out such a mistake .... am I right?

I also reasoned like you before it was proved to me by my colleague that Einstein's reasoning was defective: I was just as astonished, but fortunately I am not a dogmatically ossified person like you are.

Why am I so sure? First, I've done many of these derivations myself and there is not a problem.

That means nothing when I judge you by the lack of competence you have displayed on this forum.

Second, so has everybody else.

This proves that it is still possible that thousands of people can make the same mistake without noticing it: Just as during the 2000 years that Ptolemy's model of the Universe held sway.

Third, the self-consistency of the Lorentz transformation is mathematically proven---finding a "violation" of it is finding two integers whose sum is irrational.

My colleague did not question the self-consistency of the Lorentz transformation, but used it to prove by impeccable mathematics that Einstein violated the self-consistency of the Lorentz transformation when he explained non-simultenaity of two simultaneous events.

 

[CriticalThanking]

Johan,

Serious questions:

How lung until your friend publishes? How long ago did he convince you? To overturn something that big would be a guaranteed Nobel prize. Does you believe your superconductivity hypotheses are true based (ETA at least partly) upon the overturning of the current understanding of Relativity?

Thanks,

CT

 

[sol invictus]

Suffice to say that when you use the Lorentz transformation correctly, you will find that the following statement by Einstein, which can be found in his little book entitled RELATIVITY: The special and General Theory, page 26, is not correct: "Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding ahead of the beam of light coming from A, Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A.". Now when using the Lorentz transformation directly you will find that within ANY inertial reference frame, the light beam from B as well as the light beam from A are both still approaching the observer with speed c. If the observer could have rushed into the light beam from B, and away from the light beam coming from A, the light speed from B relative to the observer would be larger than c, and the light speed relative to the observer from A would be less than c. This violates Einstein's own postulate that light can only move with a speed c relative to any object with mass.

That's even stupider than I expected. You've apparently never heard of the velocity addition formula.

Einstein's (very simple) argument goes as follows (it's based on the fact that c is constant): The speed of light is c in all reference frames. Therefore, if the lightning strikes are simultaneous in the frame of the embankment and the train is moving, the train observer will see one strike before the other. But since light also moves at c in the rest frame of the train, the lightning strikes cannot be simultaneous in that frame.

My colleague did not question the self-consistency of the Lorentz transformation, but used it to prove by impeccable mathematics that Einstein violated the self-consistency of the Lorentz transformation when he explained non-simultenaity of two simultaneous events.

Then your colleague is either a complete fool or a fraud, because the relativity of simultaneity is part of the Lorentz transforms. The train example is something else - it's a way to see why relativity of simultaneity is a consequence of Einstein's two postulates (that the speed of light and the laws of physics are the same in all inertial frames).

The Lorentz transforms themselves are:

t'=γ(t-vx/c²)
x'=γ(x-vt)
y'=y
z'=z


Since t' is a function of x as well as t, simultaneity is relative. Period. As I said, it's part of the Lorentz transforms.

 

[johanfprins]

Johan,

Serious questions:

How lung until your friend publishes? How long ago did he convince you?

In my last book, which was printed in September 2010, I still used Einstein's wrong reasoning. Since then I was approached by this person out of the blue complaining that he could not find anybody who was willing to read his manuscript with an open mind. 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. I believe that he has now submitted it and it got stuck with an editor who is sitting on it: No doubt looking for a reason to reject it. But it is so well written that I doubt that this will happen: On the other hand the mentality that caused Galileo to have problems is again rife at present. He wanted me to make a recommendation but this would have been to his disadvantage since it has become clear over the past 10 years that I have been blacklisted by the physics community.

To overturn something that big would be a guaranteed Nobel prize.

I also would have thought so, but as I have found out the hard way, that our mainstream physicists are far more dogmatic at present than the Vatican was in the time of Galileo.

Does you believe your superconductivity hypotheses are true based (ETA at least partly) upon the overturning of the current understanding of Relativity?

It has given further support for my arguments on quantum mechanics; and the irrelevance of quantum field theories on which the BCS model is based.

 

[johanfprins]

The Lorentz transforms themselves are:

t'=γ(t-vx/c²)
x'=γ(x-vt)
y'=y
z'=z


Since t' is a function of x as well as t, simultaneity is relative. Period. As I said, it's part of the Lorentz transforms.

Edited by LashL: 

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But why do you not apply these equations to the following situation: You have a spaceship of length L passing by within which two lights are switched on simultaneously in the nose and tail and then both beams move with a speed c towards a detector in the centre of the spaceship. Transform these events into the inertial refrence frame relative to which the spaceship moves and calculate the speeds with which the light is moving to the detector as seen from outside: You will find that the speeds are still both c.

You

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... think that this must mean that the light beams must then also switch on simultaneously in the outside refrence frame. NO THIS NOT SO: THE LIGHT IN THE TAIL SWITCHES ON BEFORE THE LIGHT IN THE NOSE AND THIS DOES NOT HAPPEN BECAUSE THE OUTSIDE OBSERVER RUSHES INTO THE LIGHT COMING FROM THE TAIL AND MOVES AWAY FROM THE LIGHT COMING FROM THE NOSE: JEEZ!!! I will not respond to your inane comments anymore.

 

[Ziggurat]

I have decided not to respond to posts by you Sol and Ziggurat.

Of course you won't. Because you can't. This is basically an admission of failure.

Suffice to say that when you use the Lorentz transformation correctly, you will find that the following statement by Einstein, which can be found in his little book entitled RELATIVITY: The special and General Theory, page 26, is not correct: "Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding ahead of the beam of light coming from A, Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A.". Now when using the Lorentz transformation directly you will find that within ANY inertial reference frame, the light beam from B as well as the light beam from A are both still approaching the observer with speed c.

Small problem: Einstein's statement is actually consistent with that claim. In fact, they're completely consistent with the entirety of the Lorentz transformations.

Seriously, try it. Pick the appropriate coordinates, and transform them. See what happens. Show us, WITH MATH, what's going on.

If the observer could have rushed into the light beam from B, and away from the light beam coming from A, the light speed from B relative to the observer would be larger than c

But that's explicitly NOT the observer being discussed, so... you're wrong. From the viewpoint of the person on the train, the speeds are indeed the same, but the emission is not simultaneous. It's a direct result of the Lorentz transformations, one which you CANNOT avoid.

How can you screw this up so badly? And why, in the absence of any math, do you expect anyone to believe you?

 

[sol invictus]

I know that it is too much to expect from your limited intellect: But why do you not apply these equations to the following situation: You have a spaceship of length L passing by within which two lights are switched on simultaneously in the nose and tail and then both beams move with a speed c towards a detector in the centre of the spaceship. Transform these events into the inertial refrence frame relative to which the spaceship moves and calculate the speeds with which the light is moving to the detector as seen from outside: You will find that the speeds are still both c.

This is essentially identical to Einstein's train experiment. The only difference is that you have the events simultaneous in the ship's frame, rather than the frame of the "embankment" (i.e. a frame in relative motion to the ship/train).

You are of course stupid enough to think that this must mean that the light beams must then also switch on simultaneously in the outside refrence frame.

Of course not, Johan. Unlike you and your friend, I understand freshman physics.

If events at two distinct points are simultaneous in one frame, they cannot be simultaneous in any frame in relative motion with respect to it. That's what Einstein showed follows from the constancy of c with his train example, and it's part of the Lorentz transforms.

 

[Ziggurat]

Edited by LashL: 

Removed quote of moderated content.

But why do you not apply these equations to the following situation: You have a spaceship of length L passing by within which two lights are switched on simultaneously in the nose and tail and then both beams move with a speed c towards a detector in the centre of the spaceship. Transform these events into the inertial refrence frame relative to which the spaceship moves and calculate the speeds with which the light is moving to the detector as seen from outside: You will find that the speeds are still both c.

Indeed you will. You will also find that in this transformed coordinate system, the emissions are no longer simultaneous, and because the ship is moving, it takes longer for light from the back to reach the middle than light from the front. In perfect agreement with the argument Einstein made, except in the example you gave, the emission was simultaneous in the external frame and therefore not simultaneous in the train frame.

You

Edited by LashL: 

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... think that this must mean that the light beams must then also switch on simultaneously in the outside refrence frame.

I'm quite sure Sol thinks nothing of the sort. In fact, this non-simultaneity is exactly what Einstein's example is demonstrating.

I will not respond to your inane comments anymore.

You already went back on that claim. And even if you make good on this "threat" now, all it will demonstrate is that you have no good arguments to offer in response.

 

[johanfprins]

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PLEASE derive Einstein's train experiment (I just tried to make it simpler for you by suggesting the spaceship) from the Lorentz transformation, if you are able to this simple mathematics, then you will see that it is NOT POSSIBLE for any observer EVER to "hasten towards a beam light coming from B" and to "ride ahead of the beam of light coming from A" It is prohibited by Einstein's own first postulate and is also inherent in the Lorentz transformation that it is impossible.

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Nowhere have I or my friend said that the two simultaneous events will also be simultaneous within another inertial reference frame. All that is claimed, and which follows directly from the Lorentz transformation is that this is NOT CAUSED by an observer "hastening into a light beam and riding ahead of the other light beam" The latter is not possible since the speed of light is always the same relative to any object no matter how fast the object is moving!!!

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[ben m]

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PLEASE derive Einstein's train experiment (I just tried to make it simpler for you by suggesting the spaceship) from the Lorentz transformation, if you are able to this simple mathematics, then you will see that it is NOT POSSIBLE for any observer EVER to "hasten towards a beam light coming from B" and to "ride ahead of the beam of light coming from A" It is prohibited by Einstein's own first postulate and is also inherent in the Lorentz transformation that it is impossible.

Nonsense. The speed of light is finite. It's a very special speed, but within any reference frame it's just a speed and it behaves like any other speed. If there's a light source 10 meters away, and I am walking towards it at speed v, the distance between me and the source is 10 - vt. (Am I forbidden from walking towards light sources?) The light travel time is not fixed at 10/c, but decreases as (10-vt)/c. It decreases a little bit even during the ~30 nanoseconds between a photon being emitted and my detecting it.

If you reverse v and walk away from the light beam, same thing. The distance increases (it's just a distance) and the light travel time also increases.

You've clearly confused yourself.

 

[W.D.Clinger]

Suffice to say that when you use the Lorentz transformation correctly, you will find that the following statement by Einstein, which can be found in his little book entitled RELATIVITY: The special and General Theory, page 26, is not correct: "Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding ahead of the beam of light coming from A, Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A.".

No, Einstein got that right. The highlighted qualification is key.

To explain where johanfprins and/or his friend went wrong, I'll need to talk about Einstein's two observers:

• Observer M' is on the train, exactly halfway between the points A and B at which lightning strikes the train.
• Observer M is on the embankment beside the railroad track, watching the train travel from his left to his right. In observer M's frame of reference, the lightning strikes at A and B are simultaneous, and observer M deduces that observer M' must have passed observer M at the instant of those lightning strikes (because M' is exactly halfway between the points A and B at which those simultaneous lighting strikes damaged the train).

To make my presentation a little more concrete than Einstein's, I'm going to assume that observer M' is situated at the exact middle of a train car, and that points A and B are the rearward and forward ends of that car.

I'm also going to assume that the train-bound observer M' is a conductor, with instantaneous reflexes, who's about to punch a ticket that some passenger is moving at a constant rate across the punch. Furthermore, I will assume that M' is so sensitive to bright light that he punches that ticket (instantaneously) whenever light from a nearby lightning strike reaches him.

With those assumptions, the question of simultaneity reduces to this: Does the observer M' punch just one hole in the ticket, or does he punch two?

Now when using the Lorentz transformation directly you will find that within ANY inertial reference frame, the light beam from B as well as the light beam from A are both still approaching the observer with speed c.

That sentence is correct. In particular, it is correct for both M and M'.

What johanfprins and/or his friend are forgetting is that, within the inertial frame of the observer M on the embankment, observer M' is moving away from the light at A and toward the light at B. Since the speed of light is finite, observer M sees that M' has travelled some distance to the right of M before M' sees the light from either A or B. In the frame of reference of observer M, therefore, the light from A (at the rear end of the car) must travel farther before it reaches M' (at the exact middle of the car) than does the light from B (at the front of the car). Hence observer M sees the light from B reach observer M' a little before the light from A.

So observer M sees M' punching two separate holes in the ticket.

If the observer could have rushed into the light beam from B, and away from the light beam coming from A, the light speed from B relative to the observer would be larger than c, and the light speed relative to the observer from A would be less than c.

No. That is where johanfprins and/or his friend go wrong. To draw the conclusion stated in that sentence, johanfprins and/or his friend must assume that the lightning strikes at A and B are simultaneous in the frame of observer M' as well as in the frame of observer M.

That's not true. Indeed, the relativity of simultaneity is the whole point (and the title!) of Einstein's chapter 9, in which this example appears.

In effect, johanfprins is arguing that M' punches only one hole in the ticket. But we already know that, as seen by observer M, M' punches two separate holes. We can assume that observer M' puts the ticket in his pocket after punching the holes, and has dinner with observer M later. When observer M tells M' that there are two holes in the ticket, M' can verify that fact by pulling the ticket out of his pocket.

Because there are two holes in the ticket, the light from A must have reached observer M' after the light from B. That refutes johanfprins while confirming Einstein. According to observer M, the lightning flashes at A and B were simultaneous. According to observer M', they were not simultaneous; the flash at B happened before the flash at A.

Simultaneity is relative.

Try the derivation yourself if you are competent; which I, however, doubt. I also reasoned like you before it was proved to me by my colleague that Einstein's reasoning was defective: I was just as astonished, but fortunately I am not a dogmatically ossified person like you are.

johanfprins and his friend are mistaken.

johanfprins made a much more serious mistake when he decided that he and his friend were right, while Einstein and everyone else were wrong. johanfprins compounded his errors by congratulating himself on not being "a dogmatically ossified person" like the tens of thousands of physicists and millions of laypersons who actually understood Einstein's argument.

Nowhere have I or my friend said that the two simultaneous events will also be simultaneous within another inertial reference frame. All that is claimed, and which follows directly from the Lorentz transformation is that this is NOT CAUSED by an observer "hastening into a light beam and riding ahead of the other light beam" The latter is not possible since the speed of light is always the same relative to any object no matter how fast the object is moving!!!

As explained above, johanfprins and/or his friend have misinterpreted the passage to which they are objecting. That took some doing, because they had to ignore Einstein's parenthetical remark, which I highlighted at the top of this post.

As seen by observer M, observer M' really is "hastening into a light beam and riding ahead of the other light beam." johanfprins and/or his friend appear to be assuming that Einstein's remark applies also to what is seen in the inertial frame of observer M'. That's a rookie mistake.

 

[johanfprins]

It is so easy to derive the situation from the Lorentz transformation: Let us return to the case which I have proposed: namely a spaceship with length L^/ moving past at a speed v: The Lorentz transformation from the spaceship (primed coordinates) to the inertial reference frame relative to which the spaceship is moving with speed v are:
x=(gamma)(x^/+vt^/)
t=(gamma)(t^/+(v/c^2)x^/)
Synchronise a clock in K when the lights switch on simultaneously within the spaceship (K^/) and when the detector in the spaceship is moving past the clock. The coordinates for the light in the tail is x^/=-L^//2, t^/=0; and the coordinates for the light in the nose of the spacship is x^/=L^//2 and t^/=0. Now if you understand high school algebra you can surely transform these coordinates into K. (will somebody do this for Sol, Ben and Ziggurat please?) When the two light beams reach the detector in the centre of the spaceship, the coordinates in the spaceship are clearly x^/=0 and the time is t^/=(L^//2)/c; and if you can do high school algebra you can surely also transform these two coordinates into K. Now, it is not difficult for Sol Ben and Ziggurat to find a grade 6 school child to next calculate for them the speed with which the light beam moves from the tail within K (and find it is c), and the light speed moving from the nose (and find it is -c). So, according to an observer which finds him/herself "by accident" at the same position as the detector was within K when the lights switched on, the magnitude of light speeds relative to him/her is the same: Namely c from the tail and -c from the nose. He is thus NOT "hastening towards" or "riding ahead of any light beam" as Einstein has reasoned. Thus, Einstein's reasoning is wrong; and this same wrong reasoning led Einstein to derive a length contraction, which does not occur within K, and a time dilation which does not occur within K^/.