In one of his papers on Goldbach’s conjecture, Miles Mathis claims to prove the conjecture in a formalized way. While the lack of definitions and explanations of some terminology used by Miles makes the paper harder to follow, we can still find some inconsistencies and flaws to point out.

Even though the method of the proof is itself flawed, let us jump right at an inconsistency in one of the definitions. Let be an even number greater than 2. Miles writes

Let y = the number of primes less than x.

[…]

Let b = the number of non-prime odds less than x, as a fraction of numbers less than x.

b = [(x/2) – y] /(x – 1)

When calculating , Miles correctly calculates the number of odd natural numbers less than as . But when substracting the number of primes less than x from this value, he wrongly assumes that all primes are odd numbers as well. Most of them are, but 2 is not. The formula can of course easily be corrected. However, upon sending him an email about this mistake Miles wrote

I don’t include 2 as a prime. The current definition of prime is muddled, in my opinion. If you include 2, you logically should include 1 and 0 as primes.

mm@milesmathis.com to the author of this blog, Jun 12 2010

While it is perfectly reasonable to make up one’s own definitions, stating them is certainly necessary when giving a proof! Excluding 2 from the prime numbers, the given formula for is correct – But now we find that the conjecture Miles is about to prove does not hold any longer for the number 4! (Again, this is not a very big problem because conjecture could still be true for all even integers larger than 4 and the problem at hand would just be one of definition. However, the authors felt that it is worthwhile to point out these inconsistencies to the readers.)

But there is a more serious problem with Mathis’ proof. At no point does Mathis use any properties of the prime numbers, except that they are odd. If the proof were valid, replacing the word prime in the proof with any word that signifies a set of odd numbers would again give a valid proof and thus every even integer would be representable as the sum of two such numbers. That is certainly not the case for all sets of odd numbers, showing that Mathis’ proof is not correct.

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What we will see is that the mechanism of Bremsstrahlung, though roughly correct, is not completely correct. The electron is not emitting a photon, it is becoming a photon.

Mathis takes offence with current theory’s lack of explanation as to *why* and especially *how* electrons can emit – that is create – photons. They are not there before, so how can they be there afterwards? He repeatedly expresses the view that scientific description of nature should be “mechanical”.

But notice how convenient it is for them that quantum mechanics has no mechanics. Although they claim to be physicists, the fact that QM and QED are not mechanical allows them to dodge all physical questions.

We do not profess to know what Mathis means when he uses the term “mechanical”, but picture a model of small colliding spheres, not unlike billard balls. In this picture, interaction and forces are transmitted via collisions. Although it seems intuitive for nature to be “mechanical” in this everyday sense, there is no deeper reason why this should be true. Common sense tells us for example that the earth is flat, which is cleary not the case. It remains to say that contempary “non-mechanical” theories like electrodynamics and quantum mechanics are not only useful but make incredibly precise and well verified predictions and are justified in this pragmatist sense.

Mathis clings to his mechanical view because he cannot imagine the universe to be otherwise. His side blow

I thought we were done with force at a distance

to modern theory is in vain, because since the advent of field theory, building so called “local” theories is not a problem anymore. (There, force is mediated through local fields (e.g. the electromagnetic field), which explains the mysterious action at a distance that troubled Newton.)

On a philosophical side note, science has many heuristics, which “explain” how electrons emit real photons in the bremsstrahlung process (for example a cloud of virtual photons surrounding the electron). But they aren’t to be taken at face value from a philosophical perspective. They only thing science can do is make predictions and verify their consistency with experiments. It is true that scientists tend to have a lot of faith in their models, especially if they are established beyond reasonable doubt like general relativity and the standard model. But in the end they will have to hold up to experiment.

This brings us to the most problematic point in Mathis’ theory: He claims that the electron turns into a photon. This would imply that the electron’s negative charge somehow gets lost in the process, leaving the universe with an overall excess of one positive charge. To the best of our knowledge, so far no processes have been observed that violate charge conservation.

(*) “model” is really too strong a word for Mathis’ thinking. He uses heuristics, rather than clearly defined terms, in order to shed light on the problem. His method however, only enables him to make vague descriptions of what *might* be the case, but fails to produce predictions comparable to experiments.

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Let us look at four positions of the two planets, in four diagrams, in a straight 3-body problem.

With gravity as a force of attraction, and no other force fields playing a part, we find that we would expect the gap between Jupiter and Saturn to be getting smaller. Because, while Jupiter is sometimes being pulled away from the Sun by Saturn and sometimes toward it, Saturn is always being pulled toward the Sun by Jupiter. I challenge you to find a relative position of the two planets where Saturn is not being pulled into a lower orbit by Jupiter.

*(Diagrams created by Miles Mathis.)*

It seems to us that this argument does not make any sense. Consider for example the following similar diagram where the big dot represents the Sun and the smaller dot represents the earth:

If Mathis’s arguments were valid, a two-body system could certainly not be stable because the earth could only fall into orbits closer to the sun. The (mathematical) two-body problem however exhibits stable solutions – we are, approximately, living on one!

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The *Blog-Health-o-Meter™* reads This blog is on fire!.

A Boeing 747-400 passenger jet can hold 416 passengers. This blog was viewed about **2,600** times in 2010. That’s about 6 full 747s.

In 2010, there were **11** new posts, not bad for the first year! There were **2** pictures uploaded, taking up a total of 36kb.

The busiest day of the year was November 18th with **135** views. The most popular post that day was A reply to “The Extinction of Pi: The short version”.

The top referring sites in 2010 were **scientopia.org**, **forum.objectivismonline.net**, **thunderbolts.info**, **shmups.system11.org**, and **iceinspace.com.au**.

Some visitors came searching, mostly for **miles mathis**, **ex falso mathis**, **miles mathis wikipedia**, **miles mathis ex falso**, and **“miles mathis”**.

These are the posts and pages that got the most views in 2010.

1

A reply to “The Extinction of Pi: The short version” November 2010

4 comments

2

A reply to: Relativity as a concept June 2010

12 comments

3

A reply to “Angular Velocity and Angular Momentum” July 2010

9 comments

4

A reply to: Why Non-Euclidean Geometry is a Cheat June 2010

7 comments

5

About June 2010

3 comments

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The online journal Physics has a nice summary of the results which should appeal to a wider audience. There is also a free download of the original paper at Physical Review Letters.

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We’ve all heard that if you had an infinite number of monkeys plugging away on an infinite number of typewriters for an infinite amount of time then eventually those monkeys would produce all the works of Shakespeare. In his “paper” Mathis sets his sights on disproving this statement, and fails miserably. This is fundamentally a problem of statistics, yet nowhere does Mathis ever use a single statistical argument.

Mathis asserts that since it is possible for a monkey to type an infinite string of “S”’s, that there is a non-zero probability that this monkey will not produce a work of Shakespeare. It is true that there is the possibility that no work of Shakespeare will be produced, but there is a greater probability one will be produced. Mathis seems to quibble over the statement of the monkey theorem given in layman’s terms, had he taken the time to see the precise formal statement of the theory he would realize his argument is flawed. There are a few ways this theorem can be stated, I will use a formulation that does not appeal to the notion of infinite monkeys. To make this formulation precise a few simplifying assumptions are made that in no way change the problem and the process is clearly defined.

**The Experiment**

Instead of all Shakespeare’s works, let us consider only Macbeth. Let us assume that to reproduce Macbeth it would require *n* keystrokes. On the given typewriter being used let us assume there are *m* keys. I stress not knowing the exact numbers in no way changes the results being presented.

The experiment will be conducted as follows. One monkey will randomly press one key at a time on the typewriter. Once the monkey has made *n* keystrokes we will examine the document produced. If the document matches the Macbeth, then we will call it a success. If it is not a success then the monkey will be given another try. The experiment is complete once the monkey produces a success.

**Statistical Analysis & Predictions**

We are interested in predicting how many tries the monkey will need until he reproduces Macbeth. Let P(S) be the probability the monkey succeeds on a given try. In order to produce Macbeth the monkey must press n characters in the perfect order. Since there are *m* possible keys and only one of them is right for the j^{th} character, the probability the monkey presses the right key for the j^{th} character is (1/*m*). Since he has to press *n *keys in total and by assumption the keystrokes are statistically independent, the probability of success is given by:

Let X be the random variable that represents the number of trials before the monkey succeeds. Let P(X = k) be the probability that the monkey fails the first (k -1) tries and succeeds on the k^{th} try. The probability of failure on a given try is 1 minus probability of success. Since each try is statistically independent we multiply the probabilities of the individual events to find the probability of the compound event. Thus we have the following:

This is an example of what is known as a geometric random variable. What we are interested in is the expected value of X. If you are unfamiliar with the language of statistics, the expected value of X means if we were to repeat the experiment over and over, how many trials it takes for the monkey to succeed on average.

**Expected Value of a Discrete Random Variable:**

** **

If X is a discrete random variable with probability function P(X), then the expected value of X is given by:

Fortunately since X is a geometric random variable there is a simple formula that gives the expected value of X.

**Expected Value of a Geometric Random Variable:**

** **

If X is a geometric random variable with probability P(S) of success on any given trial then the expected value of X is given by:

Now in our case we know P(S) and upon substituting the known value we have:

Now m^{n} is an admittedly vary large number in this situation, but it is a finite number. This means that if we were to repeat this experiment over and over then on average the monkey would need m^{n }tries before he successfully typed Macbeth. Note that if we assume that Macbeth has 100 000 characters and a typewriter has 50 keys, this gives us 50^{100 000} tries on average. Now this is staggeringly large number and shows that this is not a practical undertaking, but practical or not the result is valid.

**A Second Analysis**

Above we calculated the probability of a success on a given try to be P(S). Now instead of calculating the expected number of tries before success, we can look at it a different way and ask *what is the probability the monkey eventually succeeds?*

What exactly does it mean for the monkey to eventually succeed? I know this sounds like a pedantic question, but it is useful to properly formulating the problem. This means that on some attempt, say the j^{th} one he finally reproduces Macbeth, but fail on all previous attempts. So to find the probability that he eventually succeeds we need to add the probabilities of success on each try. We need to add the probability he succeeds on the first try, the second try, the third try, and so on. To succeed on the j^{th} try means on the first (j-1) tries he fails then finally gets it right on the j^{th} try. Since each attempt is assumed statistically independent, we simply multiply the probability of failing (j-1) times by the probability of succeeding once on the j^{th} try.

So now we add these probabilities up.

P(eventual success) = P(success on 1^{st}) + P(success on 2^{nd}) + P(success on 3^{rd})…

This is a geometric series with first term P(S) and ratio [1 – P(S)]. Since the common ratio [1 – P(S)] is less than one, we can apply a simple formula to calculate this sum.

**Sum of Geometric Series:**

We now apply this formula to our series:

This tells us that that as we give the monkey more and more tries, as the number of tries grows without bound; the probability of a success goes to 1. In the language of statistics this means the success occurs with probability 1, or the event occurs almost surely. Simply put, given enough tries the monkey will almost surely reproduce Macbeth.

Mathis has accomplished a few things with this “paper”. First he has provided a great example illustrating the need for theorems to be stated in precise mathematical language. Everyday language is far too cumbersome to properly attack a problem, the advantages of equations are clear. When a mathematical problem is formulated in everyday language there is the possibility of it being interpreted incorrectly, as Mathis did when he confused the word *must* with the notion *with probability 1*.* *Subtle as the difference is, it completely changes the analysis and final conclusion. Equations give us concrete representation with very precise meaning, a clear way to model the problem and move forward solving it. Second he has showed that no amount of folksy wisdom or “common sense” can ever be a suitable replacement for true mathematical or statistical analysis of a problem. There is a very good reason why mathematical statements are made precise; among other things it helps minimize possibility of what you think should be true interfering with what really is true.

Mathis has clearly demonstrated his lack of understanding of the most basic concepts of probability. I once again suggest he spend the money and buy a decent introductory statistics text, *An Introduction to Mathematical Statistics and Its Applications (3rd Edition)* by Larson and Marx might do. It is obvious that Mathis has a deep interest in mathematics and physics, so what I don’t understand is why he refuses to learn any of it properly.

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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

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**

- 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)
- Let l(Z) denote the length of the curve Z
- 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.
- 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)
- 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)
- 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.

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.

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.

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)*

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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 of the observers, the calculations can be simplified greatly be only considering how 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”.

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(You may read Miles’ counter-critique here)

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

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