A reply to “The Extinction of Pi: The short version”

We’re very proud to have a guest author today. If you’ve been following the comments on Ex Falso… you will have noticed Dan, who kindly offered to write replies to Mathis’ more math-based papers. He took a look at Mathis’ ‘proof’ that \pi=4 (in kinematic situations) and promptly sent us this article. We’re happy to have Dan on board and are looking forward to more articles!


 

In The Extinction of Pi: the short version Mathis claims to have a simple geometric proof that Pi=4.

Before I begin I feel I should comment on Mathis’ method of proof on this one. This is one of the rare occasions where his ideas are easy to understand as there is no excess verbiage, no ill defined terms, only one point where the verbal explanation does not quite match the diagram (and an honest mistake at that), and no appeals to unrelated or ambiguous notions like ‘velocity of the radius’, although you may find this happens more frequently in long version of The Extinction of Pi. Many mathematicians have different requirements before they will call a proof elegant. If all known proofs of a statement require long and intricate arguments or “heavy machinery” and a new proof which relies solely on elementary geometry is found, then they will for the most part agree that it is elegant. All things considered, had his conclusions been accurate or his method sound, then this is the only time I have seen Mathis present what would be called an elegant proof.

Disproof #1

Left: Mathis diagram --- Right: my diagram

To quote Mathis himself:

Given the diagram above, which is part of a circle, we let

AO = AB

This makes the angle at B equal to 45°. Which means that

DB = DC.

This must mean that, by substitution,

AD + DC = AB

Now, the distance AD + DC can be redrawn as any number of sub-distances, as above.

Here we divide the distance into four sub-distances, taking us from A to C in 4 steps.

This new path, drawn by the sub-distances, is equivalent in length to AD + DC, as can be seen by any cursory analysis. It is nearer the path of the arc AC, but it retains the original distance of AD + DC.

If we divide the distance into eight sub-distances instead of four, we approach the path of the arc AC even nearer; but, again, we retain the original distance of AD + DC.

If we take this process to its limit, we take our path to the path of the arc AC.

This must mean that the length of the arc AC is equal to the distance AD + DC.

By substitution, this means that

arcAC = AB

Since 8 such arcs make up the circle, the circumference of the circle is 8AB.

Since AB = AO, and AO is the radius, we have found that

C = 8r

π does not exist in the circle equation. It is extinct.

The above is Mathis’s “proof” that Pi = 4, at first glance it seems plausible, the little ‘steps’ do visually look more and more like the arc of the circle as they are made ever smaller. All computer graphics are done in the same spirit; the pixels an LCD monitors are tiny squares, so any time anything curved is displayed on an LCD monitor it is being approximated visually by these squares. However a visual approximation of a curve is not the same as a curve. The error I mentioned where his explanation does not match the diagram involves his letter C. He uses it once as a point on the graph and then later as the circumference of the circle. This is not a major mistake and does not in any way change his argument or my proof that his method is flawed. I will now refer to Mathis’ method as the ‘step method’ because when drawn; the approximation looks like steps in a staircase. My proof that his method is invalid will now follow.

Definition and Labels

  1. Let a curve be the locus of points satisfying the equation y = f(x) where f is a continuous function or equivalently the set {(x,y) : xεR, y=f(x), f is continuous}. (think about drawing a curve above the x-axis where you never have to lift your pencil and above each x there is only one point on the curve and it never intersects itself)
  2. Let l(Z) denote the length of the curve Z
  3. In my diagram above, let R and B denote the red curve from (0,0) to (1,1) and the straight blue line from (0,0) to (1,1) respectively.
  4. A curve representing the function f(x) is said to be monotonic and strictly increasing if x<y implies f(x) < f(y) (If y is the right of x, then the point on the curve above y is higher than the point on the curve above x)
  5. A curve representing the function f(x) is said to be monotonic and strictly decreasing if x f(y) (If y is the right of x, then the point on the curve above y is lower than the point on the curve above x)
  6. A straight line is the curve between two points with the shortest distance. So if A is a straight line connecting any two points, and B is any curve distinct from A connecting the same two points we have l(A) < l(B) where l(Z) is the function that gives the length of the curve.

Assume that Mathis is correct and the length of a curve can be approximated by considering the lengths of the horizontal and vertical line segments that make up the ‘steps’ in the diagrams. Mathis is correct, no matter how few or how many of these steps we subdivide the interval [0,1] into, the total of their lengths will remain constant. They steps can be made small, medium, large, evenly sized, or unevenly sized yet their total length will remain the same. The below equation demonstrates this for evenly sized steps, which I will point out can only be drawn when the curve you are approximating is an arc of a circle or a straight line.

\lim_{n\to\infty} n\cdot(\frac{1}{n}+\frac{1}{n}) = 2

Now consider the curve R and B in the right side of the diagram. They can be approximated by ‘steps’ as in the Mathis proof. Notice that the big triangle in black has two sides, each of length 1 giving a combined length of 2. Thus the total length of the steps approximating B must be 2 implying the length of B is 2. The steps that approximate R must have total length 2 implying R has total length 2. We have shown that l(B) = l(R) or that the length of B and the length of R are in fact equal. However by assumption since B is a straight line it has the shortest possible length, and since R is a curve distinct from B it must have length greater than B. We have two contradictory statements: l(B) = l(R) and l(B) < l(R). The assumption that the ‘step method’ of calculating length is accurate leads to a contradiction and is therefore invalid.
\Box QED

One can extend the above line of reasoning to show a result of the Mathis method is the absurd idea that if m and n are any two distinct points, and X & Y are two distinct monotonic strictly increasing/decreasing curves joining m to n, then X and Y must have the same length. This would mean the length of the curves defined by the following functions are identical on [0,1]: f(x) = sin(Pi*x/2), g(x) = x, h(x) = x^2, j(x) = x^3, or even k(x) = x^n for any integer n>0.

Disproof #2

Assume that the Mathis’ “step method” can be used to calculate the length of a curve. In this situation our curve is the segment AC.
Let ABC be a right triangle with side lengths |AB| = 1, |AC| = 2, and |BC| unknown.
If we accept Mathis’ “step method” then we can approximate line segment AC by the red steps, and approximate the length |AC| by the lengths of the steps. As we make the steps smaller and smaller and pass to the limit, we have the length of the segment equal to the limit of the sum of the lengths of the steps, so as Mathis has demonstrated we have |AC| = |AB| + |BC|. Plugging in our numbers for |AC| and |AB| we get the equation 3 = 1 + |BC| which gives |BC| = 2.
From basic trigonometry we have sin(angle BAC) = opposite/hypotenuse = |BC|/|AC| = 2/2 = 1.

Now again from basic properties of triangles we can deduce that triangle ABC is “half” of an equilateral triangle. That is, if we reflected the triangle along the line segment BC and let A’ be the mirror image of A, then triangle ACA’ would be equilateral, as |AA’| = |AC| = |CA’|. This allows us to calculate angle BAC to be Pi/3 radians or 60 degrees.
Now given that we have found angle BAC to be Pi/3 radians we can calculate the sin of that angle in radians as sin(Pi/3) = sqrt(3)/2, or sin(60 degrees) = sqrt(3)/2)
So we have calculated sin(angle BAC) two different ways, and they should be the same.
Equating sin(BAC) = sin(BAC) and doing a few simple manipulations:

sin(BAC) = sin(BAC)
1=sqrt(3)/2
1^2 = [sqrt(3)/2]^2
1 = 3/4
4 = 3
4 – 3 = 3 – 3
1 = 0

Thus the assumption that the Mathis “step method” is an accurate way to calculate the length of a curve leads to the absurd conclusion that 1 = 0.
\Box QED

Afterthoughts

When I first set out to write this out I wasn’t entirely sure how to go about it. I found myself going on lengthy rants about Mathis’ ideas, methods, and conclusions. I could write a long piece attacking some of his ideas and explaining the folly in his way of thinking, but I had set out to discredit one specific paper. So I edited out my own rants and tried to make this as straight forward as possible and attack the method, not the man. I however must point out that the number Pi is so entrenched and intricately related to so many areas of mathematics and real world applications that if the value of Pi were to change, then everything in the following list would fall apart or at the very least need a significant rethink. So by Mathis attacking the value of Pi he is indirectly attacking so much more.
Here is a VERY short list of subjects and real world applications that Pi plays role in (it is more ‘important’ in some area than others):

  • complex analysis (contour integration in the complex plane)
  • probability, simulation, and testing
  • plane geometry
  • the GPS navigation system
  • signal processing (TV, radio, satellite)
  • physics (it shows up everywhere!)
  • real analysis (estimation see Stirling’s formula)

In closing I would like to quote Mathis from his long version of the Extinction of Pi, make a few short observations, and leave a question for Mathis and another for his supporters (I would really love to read a response!)

I show that in kinematic situations, Pi is 4. For all those going ballistic over my title, I repeat and stress that this paper applies to kinematic situations, not to static situations. I am analyzing an orbit, which is caused by motion and includes the time variable. In that situation, Pi becomes 4. When measuring your waistline, you are not creating an orbit, and you can keep Pi for that. So quit writing me nasty, uninformed letters.
-Miles Mathis

Pi is a constant so it never changes value based on the situation you are using it. I can only assume what he means is that in a kinematic situation he feels the constant Pi = 3.14159265… is an inappropriate constant to use and the number 4 should be used in its place. I promise you that that is not that case, but a proof of that will have to come at another time. I also acknowledge that I have not proved that Pi = 3.14159265, rather I have proved his method is flawed and any results derived from it are meaningless. There are many algorithms to calculate the exact value of Pi, one involves summation of the power series representation of the tangent function, some involve continued fractions, and some involve summation of other infinite series. Whatever the method used, the tools required are well outside the scope of this paper. To be completely honest, if you really believe that Pi equals anything other than the standard decimal representation, then the knowledge needed to understand methods of computing Pi is also well outside your grasp. Of the Mathis supporters I ask: what do you think is more likely; that in the history of scientific thought everyone else has been mistaken and the only person to truly understand the material is Mathis, or rather that Mathis misunderstands the very fundamentals and thus misunderstands anything built upon them? Take a minute to think about this and answer yourself honestly. My question to Mathis is if this notion Pi=4 only applies to kinematic situations, and if we assume by measuring your waistline you mean Pi = 3.14159265… for the purposes of plane geometry, then why did you bother with this little “proof” if you knew it was wrong?

– (by Dan)

Posted in Uncategorized | 32 Comments

A reply to: Is time travel possible?

(Lectures have started again at universities across europe and thus our time left for writing articles for Ex Falso … has been shortened significantly. I did however find the time to look at one of Mathis’ most recent papers about time travel.)

In one of his papers (Is time travel possible?) Miles Mathis, inspired by a Star Trek episode, ponders the possibility of time travel, mainly in connection with the so-called Twin Paradox in special relativity.
Special relativity is a delicate subject and this post can neither give you a full explanation of how it works nor clear up all the misconceptions about it. However, we’d at least like to point out some of the more obvious mistakes and places where Mathis is being too vague.

Right at the beginning of his paper, Mathis writes that the Lorentz transformations used in calculating the twin paradox “relies on treating velocity as a directionless vector”. That is certainly not true. So called Lorentz boosts (a special kind of transformations into a coordinate system of an observer moving with a certain velocity) depend both on the modulus and the direction of the velocity (see here). However, when doing calculations, one often simplifies things as much as possible. Since for the twin paradox one is only interested in the proper time \tau of the observers, the calculations can be simplified greatly be only considering how \tau changes. Since furthermore it turns out that time dilation does not depend on whether you are moving away or towards another observer, in this case it suffices to simply know the modulus of the velocity.

Mathis writes “The current interpretation implies that time passes more slowly for some events”. Although not a mathematical error, Miles’ usage of the term “event” is problematic. He has certainly picked it up from the literature on special and general relativity, but there it has a precise meaning, namely it is a point on the space-time manifold. It has different coordinates in different coordinate systems or reference frames, but the physical point in space-time it describes is unique. Time does not “pass” for events, just like time does not pass for the event “Paris, November 11th 2010 22:02”. If Mathis were a little more careful when using terms that have precise meanings (or defining his own terms), he would quite certainly become less prone to the many mistakes in his papers.

Although Miles repeatedly claims to agree with many of the results of special relativity, he suffers from a grave misconception about one of the most astonishing of its properties. He suggests that there is no actual time travel happening in the Star Trek example because Kirk and Spock can always “synch up their ‘nows’“. If we ignore special relativity for a moment and think about this with our intuition gained from day-to-day experience, it is clear what Mathis is talking about: Kirk does not travel into the future because at every instant, he is in the same instant together with Spock – His brain and body are just working a lot faster!
In Special Relativity however, the simultaneity of events (that is, whether an observer assigns the same time coordinate to two distinct events) is relative to the observer’s frame of reference. Two observers do not agree on what is “happening now” (see this video on youtube for a really short illustration). Even though this seems very counter-intuitive, it is one of the fundamental results of Special Relativity and Miles’ argument about synching up “now” shows that he does not fully understand its implications. (Miles is of course free to believe that the Theory of Relativity is fundamentally wrong . We think, however, that had he really understood its results, he would have at least mentioned that his arguments contradict them.)

At the end of his paper, Miles discusses the situation of an observer circling another observer at a certain radius. He argues that because of the very short distance between the observers, it will take light only a very short time to travel between them. This in turn will not be enough time for the “data” carried by the photons to be “skewed“. As in one of his earlier papers, neither does he give an explanation of what “data” or “skewing” is supposed to mean in this context and the meaning thus remains elusive. Reading Miles paper it rather seems that for a very distant observer (circling the other observer in a very large circle), there should be effects similar to those in special relativity (like time dilation), independent from the circling observer’s speed. This again, is in contradiction not only with the theory of special relativity but also with experiments.
(A calculation of this situation in the framework of special relativity can be found in N.M.J. Woodhouse’s very good book on special relativity)

Although the Star Trek example is amusing and might illustrate the point that even in special relativity the words “time travel” must be interpreted with caution, Miles article does in no way shed light either on the nature of time and space (as does relativity) nor on Miles own theory of “data”.

Posted in physics | 3 Comments

Aside

Miles Mathis recently published articles from his website in a book titled ‘The Un-unified Field: And Other Problems’. We would like to direct your attention to a review by R. Dowdall on Amazon.co.uk, or rather her reply to a comment by Steven Oostdijk (scroll down on the review page). Her reply echoes our own sentiments about Mathis’ work and we’re glad R. Dowdall took the time to write it.

(You may read Miles’ counter-critique here)

(Edit: I incorrectly assumed R. Dowdall was male. Sorry for that!)

Posted in Uncategorized | 4 Comments

A reply to “Angular Velocity and Angular Momentum”

In Angular Velocity and Angular Momentum Miles claims the current definitions of angular velocity and angular momentum are wrong.

To start with, look again at the basic equations

p = mv

L = rmv

Where L is the angular momentum. This equation tells us we can multiply a linear momentum by a radius and achieve an angular momentum. Is that sensible? No. It implies a big problem of scaling, for example. If r is greater than 1, the effective angular velocity is greater than the effective linear velocity. If r is less than 1, the effective angular velocity is less than the effective linear velocity. How is that logical?

First of all momentum and angular momentum are vectors and the correct definitions should read \vec{p}=m \vec{v} and \vec{L}=\vec{r} \times \vec{p}. Miles took the absolute value of the angular momentum |\vec{L}|=|\vec{r}| |\vec{p}| \sin(\vec{r},\vec{p}) while tacitly assuming that \vec{r}\bot \vec{p} (as in circular motion). It escapes us how he constructs his “problem of scaling”, but it seems he didn’t notice that \vec{L} and \vec{p} are different quantities. We now go straight to the part of the paper where a serious error occurs.

The correction for all this is fairly simple, although it required me to study the Principia very closely. We need a new equation to go from tangential or linear velocity to ω. Newton does not give us that equation, and no one else has supplied it since then. We can find it by following Newton to his ultimate interval, which is the same as going to the limit. We use the Pythagorean Theorem. As t→0,

ω2 → v2 – Δv2
and, v2 + r2 = (Δv + r)2
So, by substitution, ω2 + Δv2 + r2 = Δv2 + 2Δvr + r2
Δv = √ v2 + r2) – r = ω2/2r

ω = √[2r√v2 + r2) – 2r2]
r = √[ω4/(4v2 – 4ω2)]

This is Miles’ replacement for the current defintion of angular velocity. Apparently he got \omega^2 \to v^2 - \Delta v^2 and v^2 + r^2 = (\Delta v + r)^2 from Newton’s principia. It suffices to say that the units in the last equation do not match and it is therefore wrong.

Update:
Some days ago Miles added a section where he is supposedly clarifying the problems with units. He argues that he simply redefined \omega to be an acceleration. Redefine he did (according to the highlighted text), but \omega is still not an acceleration. It is a quantity which has no physical meaning, because the squares of two different quantities are added.

Posted in Uncategorized | 11 Comments

A reply to “A FINAL ARGUMENT AGAINST x’ = x – vt”

In one of his short articles on special relativity, Miles Mathis discusses the equation x’ = x – vt.
x is a spatial coordinate in one coordinate system and x’ is a spatial coordinate in a different coordinate system moving with velocity v with respect to the coordinate system whose coordinate is labeled x. You could imagine two observers, one on a platform and one in a train moving with velocity v relative to the platform. Each observer considers his own position to be the origin of his own coordinate system. Miles writes:

But the origin of x’ is not moving. If the origins of the two coordinate systems were together at t0 , then they are still together, since origins don’t move, by definition. This is just to say that if the train started from the station at t0 , then after time t the train still started from the station, which has not moved. Train stations do not move, just as origins do not move: t0’ and x0’ are still back at the origin, which is still back at the train station.

How origins of two different coordinate systems cannot move with respect to each other “by definition” is unclear to us. Can the two observers described above not set up coordinate system in which they themselves are at the origin? It seems that Miles has misunderstood the galilean principle of relativity.

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A reply to: Why Non-Euclidean Geometry is a Cheat

In Why Non-Euclidean Geometry is a Cheat Miles tries to establish that non-Euclidean geometry is a fraud only used to hide physical inconsistencies behind an opaque mathematical formalism. The paper also includes a critique of the complex numbers. As usual we try not to comment much on his metaphysical ramblings and silly ad hominem attacks but rather concentrate on mathematical errors. Therefore we will ignore for time being the whole first part of the paper which is concerned with non-Euclidean geometry and consists only of his personal opinion. The second part concerned with complex numbers proves a lot more interesting because Miles actually does some mathematical derivations.

First of all Miles gives the definition of a complex number. Then he continues to argue that i is not a constant.

Wikipedia, the ultimate and nearly perfect mouthpiece of institutional propaganda, defines the absolute value of the complex number in this way:

Algebraically, if z = x + yi

Then |z| = √x2 + y2

Surely someone besides me has noticed a problem there. If i is a constant, there is no way to make that true. That equality can work if and only if i is a variable. But i is not a variable.

Let x = 1 and y = 2

i = .618

Let x = 2 and y = 3

i = .535

Let x = 3 and y = 4

i = .5

But i is a number. A number cannot vary in a set of equations. Letting i vary like this is like letting 5 vary. If someone told you that in a given problem, the number 5 was sometimes worth 5.618, sometimes 5.535 and sometimes 5.5, you would look at them very strangely. I don’t think you would trust them as a mathematician.

While the definition is in fact correct his calculations are not. He equates z = |z| and solves for i. This is clearly wrong because |z| \neq z (see definition of complex numbers). (1)

There is one metaphysical quote we want to comment on because it concerns the whole of mathematics.

The fact that i equals anything is a major axiomatic problem, since it can’t equal anything but √-1, and √-1 is nothing.

and a few lines below

“How can you define something that does not exist?” Defining something that does not exist as “something that is imaginary” and then claiming that is a tight definition is a bit strange, is it not?

Without delving too deep into philosophy of math note that mathematics is not concerned with existence in the strict ontological sense. For an object to exist it suffices to write it down and show that it obeys certain required relationships. Sloppily speaking one could say that the (consistent) definition brings an object into existence. Miles’ statement above is clearly vacuous.

In a paragraph below he even states that it was wrong to equate |z| = z . He blames this (as far as we understand) on the wrong notion of an algebra.

The real reason you cannot solve for i here is that z = x + yi is not algebraic. It is not analogous in form to |z| = √x2 + y2, so the whole “if/then” claim above is false and misleading. The second equation is algebraic, but the first equation is a vector addition. I will be told that vector addition is part of vector algebra, so it must be “algebraic.” But I don’t like that use of the word algebra. In algebra, the mathematical signs like “+” should be directly applicable, without any expansion. In algebra, you should be able to solve for unknowns. As I have just shown, you can’t do that here.

What Miles’ notion of an algebra is clearly escapes us. Also Miles doesn’t seem to like the construction of \mathbb{C} via ordered pairs. Finally we want to point out one obivous error in his critique of this construction.
He tends to confuse the notion of identity and equality and seems not very proficient in manipulating equations through exchange of symbols for variables.

And another problem: where did he get (0,y) = (y,0)(0,1)? That is just equation finessing. He claims to have gotten it from here

z1z2 = (x1,y1)(x2,y2) = (x1x2 – y1y2, x1y2 + x2y1)

But according to that equation, y can never be in the first position. Look again at the middle part of that triple equation: (x1,y1)(x2,y2). Do you see a “y” in the first position there? No. We need some explanation of (y,0)(0,1), but historically we don’t get it. Then look at the last part of that triple equation:

(x1x2 – y1y2, x1y2 + x2y1)

We need to ultimately find (0,y) there, but the only way you can get 0 in the first position is if x1x2 = y1y2. And the only way to get “y” in the second position is if x1y2 + x2y1 = y.

If the second point is (0,1), as given here, then x2 is zero, which means that

x1x2 = y1y2 = 0

Since y2 is given as 1, then y1 must be 0.

So the correct equation must be

(0,y) = (x,0)(0,1)

And, since x1y2 + x2y1 = y,

then x = y

While the definition of complex multiplication in the second line is indeed correct the errors occurs in the marked line. Miles fails to recognize that (x1,y1)=(y,0) and thereby x1=y. Instead for unkown reasons he identifies x with x1, which leads to an equation but not an identity.

From what we have said it should become clear that Miles’ “revolutionary” results stem from simple errors in his calculations and not inconsistencies in the axioms of complex numbers.

(1)Note: Miles goes on by argueing that he should be able to solve for i because wikipedia says “the equations are algebraic“. If we were to write down x+yi = \sqrt{x^2 + y^2} this would indeed be an algebraic equation in i similar to the equation i = C or x = C where C is a constant. This does not save his argument because he mistook the axioms of complex numbers anyway.

Posted in mathematics | 9 Comments

A reply to: Relativity as a concept

Relativity as a concept

Miles Mathis’ first paper on the theory of relativity serves as an introductory text to his further papers on the same subject. He states

Absolutely everyone, Einstein included, thought that the transforms were transforming variables in one coordinate system to variables in another coordinate system. But this is not what the transforms do, mathematically or operationally. In any given experiment, what the transforms do is transform incoming data to local data.

Miles is simply wrong about this. It is precisely what the ‘transforms’ (Miles is talking about the Lorentz transformation) do ‘mathematically’ in the theory of special relativity. It can certainly be argued that the Lorentz transformation might not describe a physical aspect of reality. However, one of the ideas of special relativity that makes it so astonishing is this counter-intuitive transformation law between intertial reference frames. (1)

He argues that data we receive through electromagnetic radiation from very far or very fast objects is somehow skewed by the finite speed of light. In Miles’ description, the Lorentz transforms are simply a mathematical operation that corrects this change in the data on its way to us.
However, neither is the concept of data clearly defined at any point in his paper, nor does Miles give any argument as to why this elusive data should change. (2)

He writes:

Two of the fundamental equations and assumptions of SR concern the movement of light in the two fields or coordinate systems.
x = ct
x’ = ct’
The first equation is how light travels relative to us here on earth. The x and t variables are our own local variables. I have no problem with this equation.
The second equation is how light travels in the other field. But there is no other analogous field, in a strict sense. What I mean is that x’ and t’ are how the spacecraft’s lengths and times look to us. How do we put c into that data, if it just data? In what sense is data a field that light can travel in?

The term local variables is not correct in this context. x and t are coordinates in a coordinate system that is valid throughout the whole of spacetime, even though it is assigned to an observer that we might call ‘local’. Again, Miles misunderstands a fundamental part of special relativity: The variables x' and t' denote coordinates in a different interial reference frame. They are numbers an observer would assign to a point in spacetime. (Although field is a term used in physics and mathematics, it does not have any significance here.)

The arguments that follow rely on these gross misunderstandings of the special theory of relativity. If the reader is still in doubt as to wether Miles arguments are believable, let us instead look somewhere where we can really pin Miles down: He contradicts experimental evidence.

Miles writes:

As for the believers, they have also strayed far a-field. They have pretended to an understanding they never had. They have tried to force upon us twin paradoxes and varying atomic clocks in airplanes and all manner of other mysteries and mystifications.

Miles is refering to the Hafele-Keating experiment. This famous and amusing experiment (where several atomic clocks were flown around the earth to see wether there would be any difference in the time measured on the lights) is in very good agreement with the predictions of the theory of relativity. (Note that in this experiment, no data travels from a faraway place or a fast moving object to any observer. The clocks are compared at rest.) You can find the original papers here and here.

Many of the predictions of special relativity are very counter-intuitive and it is fair to ask wether the theory correctly describes what we measure. Miles however grossly misunderstands the theory and certainly does not give any corrections to it.

(1): If you are unfamiliar with the concepts of special relativity, it might seem that Miles simply has a different but equally valid interpretation of the mathematics behind Einstein’s theory. Although many of the concepts can be illustrated in famous examples such as the twin paradox or the train driving through a tunnel, there is no easy way to explain the mathematical side of special relativity in detail. Let me however assure you that, by assuming that the Lorentz transformations are simply a correction to skewed data, special relativity looses many, if not all, of it’s defining features.

(2) We feel that this is really all there is to say about Miles’ lengthy part about data.

Posted in physics | 18 Comments