Superconductivity at room temperature??

[johanfprins]

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.

Yes if he knew relativity theory or else he would not.

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

Of course not: Just as the inquisition had no idea what Galileo was talking about when he said that even though objects on the earth can be stationary, the earth is not stationary. As I have pointed out before: You should have been a Cardinal at the Vatican in the time of Galileo. You still are unable to understand what Galileo postulated.

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

Clock rates are not frame dependent and muon lifetimes are also not frame dependent. The apparent increase in lifetime of the muon does not occur within the reference frame moving with the muon.

The fact that you categorize "longer muon lifetimes" and "slower clocks" differently reveals yet another childish misunderstanding.

I am afraid that it is you who are having a childish misundersatnding.

There is no difference between the two. All processes run slow in moving frames.

Define a "moving" reference frame to me please! All inertial refrence frames are moving since there is not one of them that is uniquely stationary: Thus, if your argument is correct, all clocks within all inertial reference frames must go slower and do this simultaneously at different rates within the same inertial reference frame; since such a clock moves with different speeds relative to clocks within all the other inertial reference frames in the universe: Can you not see that you are spouting nonsense? Your explanation can only be valid if there exists a uniquely stationary clock: And Einstein's postulates is based on the assumption that such a clock does not exist.

You could make a clock out of muons, for example.

You can make a clock out of any radioactive "particle" and this clock will keep the same time rate within any possible inertial reference frame, no matter at what speed such a reference frame moves relative to any other inertial reference frame in the universe.

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).

Childishly wrong!

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

Yes I am. Why do you people keep on referring to publications that I know in detail and which you are unable to interpret correctly. The problem does not lie with my knowledge but with the lack of logic you, Ben and Ziggurat has. In this experiment these clocks were flown around the world in aeroplanes which took off and landed all the time: So there was no constant relative motion between them. Furthermore, the data collected have HUGE experimental uncertainties in them; and are thus far from convincing: The result could have been found because the experimenters wanted to find the result. This publication is discussed in my friend's manuscript.

You have still not defined to me what you mean by stating that a clock is stationary. Stationary relative to what?

 

[sol invictus]

Clock rates are not frame dependent and muon lifetimes are also not frame dependent. The apparent increase in lifetime of the muon does not occur within the reference frame moving with the muon. I am afraid that it is you who are having a childish misundersatnding. Define a "moving" reference frame to me please!

All muons decay at a certain fixed rate in their rest frame. All muons decay more slowly in a frame in which they are moving.

Put another way, if a muon at rest with respect to a collection of mutually stationary clocks decays after 10 ticks of those clocks, a muon moving with respect to those clocks will decay after (say) 20 ticks.

If you accept that - and you said you did in a previous post - then you must accept that if a clock at relative rest to a collection of mutually stationary clocks ticks in perfect time with them, the same clock put in motion with respect to that collection of clocks should tick more slowly than them. That follows from the facts that

a) there is nothing special about muons, and

b) the moving clock must count 10 ticks before a muon moving along with it decays, which means 10 ticks of the moving clock is 20 ticks of the stationary clocks.

All inertial refrence frames are moving since there is not one of them that is uniquely stationary: Thus, if your argument is correct, all clocks within all inertial reference frames must go slower and do this simultaneously at different rates within the same inertial reference frame; since such a clock moves with different speeds relative to clocks within all the other inertial reference frames in the universe: Can you not see that you are spouting nonsense?

No, I cannot. It is true that it is not obvious how this can work out consistently, which is why many children have trouble with these concepts the first time they encounter them. And yet, all apparent paradoxes turn out to not be paradoxical after all.

The reason for this is that all these effects are nothing other than a slight generalization of perspective and simple geometry. Just as a sheet of paper looks like it has a different width when viewed from different angles (and yet this seemingly contradictory statement never leads to any problems), moving clocks run slower than stationary clocks, moving meter sticks are length contracted, and yet no true paradoxes ever do (or can) emerge.

Furthermore, the data collected have HUGE experimental uncertainties in them

As always, you are wrong. According to the page I linked to, the effect (say the east-west time difference) is something like 21 sigma. Do you know what that means?

 

[johanfprins]

Well we will have to agree to disagree and let time determine whom of us are really infantile when it comes to relativity. The fact is that all objects with mass are stationary within their own respectice reference frames. his law of Galileo cannot be ignored ever as it has been done in both Special Relativity and Quantum Mechanics. Thus a clock which is stationary within the reference frame within which the object with mass is stationary can therefore not record time at any other rate than all stationary clocks do. It only seems as if such a clock is ticking slower when the reference frame within which it is stationary moves relative to another reference frame; but in actual fact it is not really happening on that clock at all, since it remains stationary within its own ineertial reference frame rate. Thus it is not true that when you go an a long space trek that you will be younger when you return to earth. The only time that you will have to take time dilation into account is when you communicate with earth; as is done for the GPS satellites.

So if there are any intelligent people on this forum who can argue logically and want to ask questions, I will respond: But not to people like Sol, Ziggurat and Ben anymore. God help physics.

 

[sol invictus]

It only seems as if such a clock is ticking slower when the reference frame within which it is stationary moves relative to another reference frame; but in actual fact it is not really happening on that clock at all, since it remains stationary within its own ineertial reference frame rate. Thus it is not true that when you go an a long space trek that you will be younger when you return to earth.

Your position is self-contradictory. You have already agreed that muons decay more slowly when they are moving (with respect to whatever set of reference clocks we choose, such as a stationary set on earth):

"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)"

Unless you think muons are magical and time dilation only applies to them, you have therefore already agreed that the traveling twin will be younger. His lifetime - as measured by an earth clock or the age of his twin - is longer, just like the muons's.

So if there are any intelligent people on this forum who can argue logically and want to ask questions, I will respond: But not to people like Sol, Ziggurat and Ben anymore. God help physics.

Don't you find it just a little odd that you cannot find anyone on this forum that agrees with you, Johan? When you are the only sane person on earth, you are insane.

 

[The Don]

Well we will have to agree to disagree and let time determine whom of us are really infantile when it comes to relativity.

Let us all agree that you are wrong.

....snip....It only seems as if such a clock is ticking slower when the reference frame within which it is stationary moves relative to another reference frame; but in actual fact it is not really happening on that clock at all, since it remains stationary within its own ineertial reference frame rate. Thus it is not true that when you go an a long space trek that you will be younger when you return to earth. The only time that you will have to take time dilation into account is when you communicate with earth; as is done for the GPS satellites.

You do realise that they have actually done the experiment, sent one clock on a journey, kept a second, synchronised, clock stationary and observed the difference

http://en.wikipedia.org/wiki/Hafele–Keating_experiment

So if there are any intelligent people on this forum who can argue logically and want to ask questions, I will respond: But not to people like Sol, Ziggurat and Ben anymore. God help physics.

Before argument, maybe some maths.


eta.....

The Don
BSc (Hons.) Physics
Bristol

 

[ben m]

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.

Yes, if the bomb was previously at rest in the ground frame, you have to accelerate it to get it to move in the ground frame.

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:

Sure there was; some time in the past, the airplane took off (and climbed, and accelerated, and turned, etc.), all of which involved acceleration forces applied to the bomb. In both cases, the bomb's velocity with respect to the ground, is v0 + integral(a dt).

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.

Sure. Notice that the helicopter's falling bomb is also stationary in the airplane's reference frame.

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.

In relativity, positions and velocities are not invariant. They depend on the observer. An observer on the ground says the bomb's v_x is positive, one on the airplane says v_x = 0. You can find other observers who will say v_x is negative. So what? Why does the airplane's frame get labeled "real" and everyone else sees "illusions"?

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.

Nope. Wrong. In this case, you simply forgot about the forces on the bomb on an airplane taking off. More generally, you're writing down two different problems and complaining that there's a difference between them. Of course there's a difference between them. One problem (the helicopter bomb) tracks an object that changes x-velocity mid-problem. The other problem (the airplane bomb) tracks an object that doesn't. Fine.

Note that all observers agree. The airplane observer says "the airplane bomb had v_x = 0 before release and v_x = 0 after release, so F=0." The ground observer says "the airplane bomb had v_x = 300mph before release and v_x = 300mph after release, so F = 0." No disagreement there.

The airplane observer says "the helicopter bomb had v_x = -300 mph before release, and 0 mph after release, so there was a force." The ground observer says "the helicopter bomb had v_x = 0 before release and v_x = +300mph after release". No disagreement there.

Basically every object with mass which is not being accelerated by a force, is STATIONARY:

There's always an inertial frame where that is true, yes. But there's nothing special about it.

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.

Nope. You seem to think that there's a symmetry enforcing this. There's a symmetry, true---each of these frames is equally valid---but that symmetry does not enforce clock agreement.

 

[johanfprins]

Let us all agree that you are wrong.

You are entitled to believe that. It is anybody's inalieniable God-given right to make a fool of himself. I have done that sometimes in the past, but at present people like you Sol, Ben, Ziggurat etc. are showing me that I have all along been an amateur when it came to making a fool of myself.

You do realise that they have actually done the experiment, sent one clock on a journey, kept a second, synchronised, clock stationary and observed the difference

http://en.wikipedia.org/wiki/Hafele–Keating_experiment

So they have sent a clock on a linear path with a constant speed v through a region of space with no gravity, and brought the clock back to compare with a clock which was not sent along? Amazing feat I must agree.

Before argument, maybe some maths.

I agree. Just a pity that even when I do the math, as I have done above with length contraction and time dilation, I am run down by argument without any mathematics. Do the following mathematics for me please: Synchronise the clocks at the origins of K and K^/ and calculate the distance D between the origins D=vt from the perspective of K and D^/=vt^/ from the perspective of K^/. Are these distances the same or different? If the clocks are not keeping the exact same time they must be different. Use some mathematics and calculate the difference between D and D^/ for me. And if you cannot and do not post the results here; we can all agree that you know nothing about physics or mathematics. You will then do everybody a favour if you just shup-up and go away.

The Don
BSc (Hons.) Physics
Bristol

You don't say! From which primary school in Bristol did you obtain this degree?

 

[Ziggurat]

So they have sent a clock on a linear path with a constant speed v through a region of space with no gravity, and brought the clock back to compare with a clock which was not sent along? Amazing feat I must agree.

Time dilation doesn't require a linear path. It doesn't require a constant speed either. Changing the direction of motion makes no difference, and changing the magnitude of velocity can be handled rather easily with a bit of calculus. It's exactly the same as calculating the length of a curve. And gravitational affects can be (and were) accounted for, so they pose no problem. And even if you think gravity wasn't accounted for properly, it is not possible for gravitational effects to account for the sign difference in the outcome for east vs. west travel. Only time dilation can.

 

[sol invictus]

Do the following mathematics for me please: Synchronise the clocks at the origins of K and K^/ and calculate the distance D between the origins D=vt from the perspective of K and D^/=vt^/ from the perspective of K^/. Are these distances the same or different?

You cannot even ask a coherent question. Are the distances the same or different when?

If the clocks are not keeping the exact same time they must be different.

Again, no, that obviously does not follow. For your passing spaceships example (which was at least clearly formulated), I showed you precisely how it works mathematically.

In frame K, the moving ship K' is Lorentz contracted and its clocks are running slow. Those two effects are equal in magnitude, so that the times each ship's clock reads at the end of the experiment are equal - as must be the case by the symmetry of the problem.

 

[ben m]

Synchronise the clocks at the origins of K and K^/ and calculate the distance D between the origins D=vt from the perspective of K and D^/=vt^/ from the perspective of K^/. Are these distances the same or different? If the clocks are not keeping the exact same time they must be different. Use some mathematics and calculate the difference between D and D^/ for me.

OK, let's take observer K to have measured the position of his origin at t0 (event A) and the position B of the other origin when his clock reads t0 (event B). Let's add a third event C representing the clocks and origins syncing up. The four-vectors of A and B in {t,x,y,z} format are:

A = {t0,0,0,0}
B = {t0, v*t0,0,0}.
C = {0,0,0,0}

These same two events are seen by K' to have four-vector positions A' and B' and C', now as vectors showing {t',x',y',z'}

A' = {t0*gamma,-t0*v*gamma,0,0}
B' = {1/gamma*(t0 - t0*v^2/c^2),0,0,0}
C' = {0,0,0,0}.

What do you want to measure? Let's see, K' would measure the K-origin's speed by comparing A' and C' (two points along the K-origin's trajectory). change in x' divided by change in t' gives between these events gives ... v. So the velocity symmetry works fine.

Do you want to measure the distance between the origins? You have to say when do you want to do that---the distance keeps changing! At t=0 the distance is zero, later on it gets larger. You might say: "I want to measure the distance, using the K' ruler, at the moment when t = t0. Just subtract Ax' from Bx', and the difference in x' is -t0*v*gamma." Does that break the symmetry? Does that show K' and K "disagreeing" inappropriately on something they should agree on?

But wait a second---first of all, that's actually not how K' would measure a spatial distance. A' and B' are not at the same t'; there's a delay between them according to the t' clock that K' is using.

(Imagine you tried to measure the length of a bus this way. "I measured the coordinates of the bus's rear bumper; it was at Broadway and 41st. An hour later I measured the coordinates of the bus's front bumper, it was at 4th Ave and 110th St. I subtract the two coordinates, and delta-x is 69 blocks, so the bus is 69 blocks long." It's just nonsense.)

More importantly in this case, t' is not the same as t. There is not necessarily a conflict between two people, one of whom said "We were 2 meters apart when my clock read 2 seconds" and the other of whom said "we were 4 meters apart when my clock read 4 seconds". Positions are not invariant even in Galilean relativity.

K' would, in fact, measure the distance by taking two x' measurements at the same t'. For example, if K' waits until his own clock reads t0 (the same thing the K clock read to begin with), the two origins can be located at spacetime points D' and E' (again in {t',x',y',z'} format)

D' = {t0,0,0,0}
E' = {t0,-v*t0,0,0}
C' = {0,0,0,0} as before

and that is the sense in which K' would make the statement "the other origin is a distance v*t away from me at time t". The only subtle thing about it is that, yes, the value K' plugs in for "v" is the same as the value K plugged in, as we learned from the first set of events we considered (we learn v from B-C, we learn v' from A'-C', and v' = v.)

 

[RussDill]

Well we will have to agree to disagree and let time determine whom of us are really infantile when it comes to relativity. The fact is that all objects with mass are stationary within their own respectice reference frames. his law of Galileo cannot be ignored ever as it has been done in both Special Relativity and Quantum Mechanics. Thus a clock which is stationary within the reference frame within which the object with mass is stationary can therefore not record time at any other rate than all stationary clocks do. It only seems as if such a clock is ticking slower when the reference frame within which it is stationary moves relative to another reference frame; but in actual fact it is not really happening on that clock at all, since it remains stationary within its own ineertial reference frame rate. Thus it is not true that when you go an a long space trek that you will be younger when you return to earth. The only time that you will have to take time dilation into account is when you communicate with earth; as is done for the GPS satellites.

<ad hominem snipped

Wow, you have quite a strong misunderstanding there. You are either ignoring experimental evidence, or misunderstanding it. Enjoy your (hopefully) upcoming Eureka moment.

 

[RussDill]

So they have sent a clock on a linear path with a constant speed v through a region of space with no gravity, and brought the clock back to compare with a clock which was not sent along? Amazing feat I must agree.

So....you agree that one clock ticked off fewer seconds than the other clock. But are still somehow saying that "It only seems as if such a clock is ticking slower when the reference frame within which it is stationary moves relative to another reference frame; but in actual fact it is not really happening on that clock at all, since it remains stationary within its own ineertial reference frame rate."

Interesting...hopefully you can see the contradiction in your statement and the gears are turning

ETA: nevermind....statements like 'because they have been "converted and washed in the blood of the holy triumviarate"' are always a bad bad sign.

 

[The Don]

You don't say! From which primary school in Bristol did you obtain this degree?

I obtained my degree from the University of Bristol

 

[The Don]

I've gone back to the start of the thread to remind me how all of this started.

Below are my initial musings and johanfprins' responses

Why is he (self) publishing books rather than papers ?

Because all my manuscripts have been blocked by the mainstream physicists for the last 7 years. And when I mean blocked I mean blocked since not a single editor or "peer reviewer" gave any reason based on sane physics-arguments.

Looking back at this thread, the responses to johnfprins and his reaction indicate to me that he would reject any criticism as being not based on sane physics

Why isn't he prominently associated with a major university or other research institute ?

I have been associated with the University of the Witwatersrand and the University of Pretoria: But after discovering superconduction at room temperature I was excommunicated because the theoretical physicists working on superconduction claimed that it cannot be superconduction since they cannot model it in terms of the wrong models which were proposed by Ginzberg and Landau and by Bardeen, Cooper and Schrieffer. They were just not willing to allow experimental facts to prove that they have been barking up the wrong tree for more than 50 years and even awarded Nobel Prizes for models which are physically nonsense.

Given his ability to do the maths, his unwillingness to learn and his rudeness to those who don't agree with his theories, I can now appreciate why former colleagues may now be giving him a wide berth

Why is he promoting via a website ?

What other option do I have?
Not when the so-called "experts" on superconduction advise the world that I must be wrong in order to protect their own wrong dogma. As one of them wrote to me: "You will one day get recognition for having discovered superconduction at room temperature, but this not a suitable time". I makes me think of an old white man who said to me in 1972: " I know Apartheid will have to end in South Africa, but not in my time. My children will have to handle that". It is disgraceful that the physicists in charge of superconduction research have also become just as bigoted.

But like Malan, he'll go to his deathbed believing he was right, despite the evidence to the contrary

 

[johanfprins]

OK let us settle this

There has been a lot of nonsense posted on this thread; and even I made a mistake. The greatest nonsense which has been posted, is to argue that Drude's equation can be applied to a superconductor by setting the resistivity term equal to zero: This is what the London brothers have done. Instead of trying to argue against ignorance, I decided to rephrase the section in my book which deals with a superconducting element within an electric circuit so that it will hopefully also be clear to those who only have half a mind.
This can be read at:

www.cathodixx.com/pdfs/ChargeInjection.pdf

I made the mistake to argue that the localised fermion-orbitals that are responsible for superconduction can polarise while a current flows through the superconductor. I cannot believe that I could have argued in this manner except to state that I was still thinking within the wrong paradigm that Brian Josephson has posulated when he modelled tunnelling. It is clear that only when a current is not flowing can the orbitals polarise: As I have postulated from the start a superconductor is a Mott-type insulator which can superconduct when the density of the Mott-type orbitals is high enough. In a metal the required insulator forms when the Fermi-level moves through a mobility-edge, while in a ceramic this insulator builds up more gradually: hence the appearance of a pseudogap. The onset of normal conduction through a Josephson junction, is of course an insulator-to-metal phase transition.
To see how Josephson tunnelling really works:

www.cathodixx.com/pdfs/SuperTunnelling.pdf

To see that my mechanism can model ALL superconductors EVER discovered; and does not require Cooper pairs to do so::

www.cathodixx.com/pdfs/SingleMechanism.pdf

For recent experimental evidence that Cooper pairs do not exist, see:

www.physorg.com/news/2011-06-superconductivity-side-unmasked.html

Now I propose that Ben m, Sol Invictus, and Ziggurat first do a bit of reading before any one of them again jumps in with both feet firmly planted in his mouth. AND PLEASE spare me the infantile jabberring of W D Clinger and the so-called "Don".

Then I have asked many times on this forum for explanations which were just ignored: I will repeat two:
1. What is a particle? Define it please!
2. When is a clock stationary and when is it moving?

Thanks.

 

[ben m]

For recent experimental evidence that Cooper pairs do not exist, see:

www.physorg.com/news/2011-06-superconductivity-side-unmasked.html

That's experimental evidence that Cooper pairs are not always phonon-mediated. In conventional superconductors, the cooper pairs are bound by a phonon interaction. In (at least some) unconventional superconductors like the pnictides, the cooper pairs are probably bound by an electronic interaction.

They're still cooper pairs, they're still bosons.

It is impossible to believe that you have not heard of this; Doug Scalapino has been talking about it for ten years. What's up?

The technique you are using---linking to a random abstract, pretending it overturns conventional models, and ignoring the paper's own contents stating insist that it doesn't---is one crackpot behavior you hadn't fallen into before. It's an especially tiresome crackpot behavior, please don't go there.

 

[Mashuna]

Yay! Johan's back! I've been stuck reading religious crazy people posting, it's about time for some science.

 

[Ziggurat]

There has been a lot of nonsense posted on this thread; and even I made a mistake.

*A* mistake? No, johan, you have made many, many, many mistakes.

I made the mistake to argue that the localised fermion-orbitals that are responsible for superconduction can polarise while a current flows through the superconductor. I cannot believe that I could have argued in this manner except to state that I was still thinking within the wrong paradigm that Brian Josephson has posulated when he modelled tunnelling. It is clear that only when a current is not flowing can the orbitals polarise: As I have postulated from the start a superconductor is a Mott-type insulator which can superconduct when the density of the Mott-type orbitals is high enough.

You are continuing your pattern of jumping from wrong to wrong. You come up with one ridiculous notion, claim that anyone who criticizes it is just ignorant, finally concede that you were wrong when it becomes clear even to you that the explanation doesn't make sense, and then jump on to the next nonsensical theory. Your new version will be no better than your old version. It will just be wrong in a different way. The only uncertainty is how long it takes you to abandon this wrong idea in favor of the next flavor of wrong.

Now I propose that Ben m, Sol Invictus, and Ziggurat first do a bit of reading before any one of them again jumps in with both feet firmly planted in his mouth. AND PLEASE spare me the infantile jabberring of W D Clinger and the so-called "Don".

The problem is not that we haven't done enough reading. The problem is that you haven't understood what you've read. Ben's takedown of your use of that Shimojima paper is exactly correct.

 

[johanfprins]

That's experimental evidence that Cooper pairs are not always phonon-mediated. In conventional superconductors, the cooper pairs are bound by a phonon interaction. In (at least some) unconventional superconductors like the pnictides, the cooper pairs are probably bound by an electronic interaction.

BS!! There are no Cooper pairs

They're still cooper pairs, they're still bosons.

They are not: They are fermions. I asked you not to jump in with both feet in your mouth. But it seems thst you like to chew your toes!

It is impossible to believe that you have not heard of this; Doug Scalapino has been talking about it for ten years. What's up?

Oh I have heard that ad neaseum: But it is wrong. The fact that Doug Scalapino wants to believe in fairytales does not make it a fact. Poor Doug. So wrong but so certain of himself: tsk! tsk!

The technique you are using---linking to a random abstract, pretending it overturns conventional models, and ignoring the paper's own contents

I did not ignore the contents.

stating insist that it doesn't---is one crackpot behavior you hadn't fallen into before. It's an especially tiresome crackpot behavior, please don't go there.

Time will prove who the treal crackpots are! You and Doug Scalapino are in the front of the line!

 

[johanfprins]

*A* The problem is not that we haven't done enough reading. The problem is that you haven't understood what you've read. Ben's takedown of your use of that Shimojima paper is exactly correct.

Well why do you attack before reading what I have written? Talking about crackpots.