Miles Mathis

[Doc Wattson]

First post here, please don't eat me!

I am a fan of physics as a hobby but by no means am an expert. I ran across some writing of Miles Mathis today and found what he said was interesting. This lead me to do some research on him since I've never heard of him and I have found people either think he is a loon or ahead of his time. In an effort to find out what the truth is I thought I might cross post in different forums to see what the response is when both sides of the Miles Mathis debate interact.

Here is one thread from a different website that bashes Miles Mathis and some quotes from it:

http://www.theflatearthsociety.org/foru ... ic=27538.0

"I too had a quick look over it and by quick I mean lightning. The first topic I chose to look at was on angular momentum and was quickly laughing quite loudly (which didn't go down too well at work).

"Seems very simple, but it is wrong. In the equation v = 2πr/t, the velocity is not a linear velocity. Linear velocity is linear, by the equation x/t. It is a straight-line vector. But 2πr/t curves; it is not linear. The value 2πr is the circumference of the circle, which is a curve."

Unfortunately for him, 2(pi)r is not a curve as the radius of a circle is constant. So the rest of his argument is pretty much useless. And if something as simple as a 3 constants producing a straight line screws with his mind, I dont even want to think what would happen if he tackled anything more complex..."

"A car hitting you will apply a force, whether or not it is accelerating." -From the kinetic energy page.

This guy makes me laugh!

At least from experience, when a car hits something, I think it does decelerate...

" Einstein was nice enough to provide us with this simple equation, but not nice enough to tell us why the energy depends on the square of the speed of light."

Ya, I'm sure that Einstein wrote down a bunch of random equations that "seemed to work", even though he deduced it all from a small set of assumptions...

This guy is hopeless.

Well if it didn't have a linear velocity, it wouldn't be accelerating, therefore you shouldn't feel an acceleration on turns. A simple car ride disproves him.

Another thread that even mentions these forums and calls Mathis a crackpot:

http://www.boards.ie/vbulletin/showthre ... 2055920504

Actually I did go to the trouble of reading it, but it is simply incorrect, as I tried to explain above. There is a bunch of quite fundamental errors, but the one that stuck out at me was that he didn't really get what energy was. The kinetic energy is by definition, the work required to bring a particle from its rest state up to a given velocity. On the page he questions how energy can be defined if we do not actually do this acceleration. This is absolute nonsense, and an extremely fundamental mistake. We only care about the amount of work needed in theory, not whether you actually did this. That's the mistake right there. He basically is not accepting any equation that comes from thinking about an abstraction, and has replaced it with nonsense.

He is making very basic mistakes about kinematics, so it seems pointless to me at least to pay any further attention to any of his claims.

but your right, as is my book, and he is trying to take the fact that acceleration is simply involved in the algebra and then tell us everything is wrong.

I believe this is similar to what he did in the paper on Relativity I linked to but I think he did it with a partial differential operator that time

The more I think about it, the more I realise there's no way somebody could make that mistake & how deluded it is. Notice how he goes on to use E=mc² to then validate how everything is wrong without actually addressing motion according to his formulation.

Do you guys as Mathis supports have any counter arguments to points made in that thread? That seem rather damning yet there are people and places out there, like this thread, that treat Miles Mathis as a genius. What am I missing?

Please don't take this as an attempt to bash Miles Mathis, I am simply trying to do my own research to figure out if he is someone to pay attention to instead of just taking one forums opinion on him as fact.

-Gary-

[Jarvamundo]

Hi and welcome to TB.

It seems their quotes have already answered your questions...

"I too had a quick look over it and by quick I mean lightning."
yup... kind of struck himself out there didn't he...

"So the rest of his argument is pretty much useless. And if something as simple as a 3 constants producing a straight line screws with his mind, I dont even want to think what would happen if he tackled anything more complex.."
Miles makes clear distinctions of this in his book. The pi of geometry Vs the pi of AM. again... might be revealing his experience (or lack of) with the material?

"it seems pointless to me at least to pay any further attention to any of his claims"
again... striking ones credibility out... ?

I guess were now stuck here... do we rely on opinion provided by self declared lack of credibility, or do we check if the door is locked yourself?

The best "research" is usually done by examining source material in full and not internet opinions. Unfortunately this is not the most common route. This is especially relevant to Mile's work as you may 'land on a paper' through the inter-mer-net in a state totally void of his large body of clear definitions, which are required to form a coherent view of his work. Fortunately he's published a walk through in softcover form.

I summarize the discussions here as not from "supporters" per se' (i still have some difficulties with some of it), but just from folk who study the material with an open mind, not just a "lightning quick look". Read this thread start-to finish and this becomes bleedingly obvious.

All the best.
PS: You'll now start to notice how many 1 star reviews incorporate "i never read the book". Ironically it seems to actually 'see red', you've got to actually taste the red pill. check for yourself brother.

[seasmith]

Image of a single ion that had been trapped in an optical potential. The colors encode the intensity of its emitted fluorescence light. (Image: Max Planck Institute of Quantum Optics)

A Micro-example of a smaller photonic b-field being entrained by a larger b-field ?

Because of their electric charge ions are also most easily to influence by external electromagnetic fields. Hence physicists have used the method of trapping ions with alternating radiofrequency fields by now for more than sixty years. Meanwhile storage times of up to several months are achieved. However, these systems suffer from a severe drawback: it is very difficult to scale them to larger architectures, which limits the possibilities of performing quantum simulations with sufficiently many ions. So, what is the reason that up to now optical lattices have not been used as an alternative for storing ions?

"Optical fields are disfavored because they don't allow for potential wells nearly as deep as they are guaranteed by radiofrequency fields," Schätz said. "At the same time ions react in a very sensitive way to external stray fields. This has caused the widely believed prejudice that optical potentials are too shallow and therefore unable to trap ions. But, as a matter of fact, we were able to experimentally demonstrate that ions can indeed be trapped by the interaction with light.

"In principle, both traps (the radio frequency as well as the optical) work the same way: they capture the particle by a fast changing electromagnetic field," said Schätz

[Their bias]

Magnetism of a solid state can occur when the individual elemental atoms carry an angular momentum, a so-call spin. Depending on external conditions the interaction of each two spins makes them to align either parallel or antiparallel, thus eventually leading to ferromagnetic or antiferromagnetic (in the latter for uneven spin numbers even "frustrated") states.

http://www.mpq.mpg.de/
http://www.photonics.com/Article.aspx?[ ... Newsletter[/url]


But as MM states, the EM [~and hence ES] field is inseparable from photonic fields and any 'particle' spins .

s

[Good_Science]

Doc Wattson get out while you still can. Miles Mathis is a crackpot and while his ideas may be interesting they are certainly not correct or well founded.

The idea that Pi can change depending on the situation is laughable. If the object you're dealing with doesn't have its diameter and circumference in the ration of 1:3.1415926535 etc then it is not a circle, it's as simple as that. Pi has been proved mathematically to be irrational. Mathematical proofs are unlike any other proofs in that they are completely correct and it is impossible for them to be wrong. Thay are based on logic and reasoning. If pi truely were changing or variable surely we would have heard about it. Contrary to what people think mathematicians don't hide discoveries. If pi were 4 in some cases it would be more widely known. This isn't even considering the physical implication of pi=4 that would be obvious to everyone.

Another piece of evidence on his lack of scientific knowledge is his understanding, or lack thereof, of density. He claims it makes more sense to use volume and density in NEwtons formula for gravitaional attraction because

Mass is not a fundamental characteristic, like density or volume is. To know a mass, you have to know both a density and a volume. But to know a volume, you only need to know lengths. Likewise with density. Density, like volume, can be measured only with a yardstick. You will say that if density and volume can be measured with a yardstick, so can mass, since mass is defined by density and volume. True. But mass is a step more abstract, since it requires both measurements. Mass requires density and volume. But density and volume do not require mass.

Really? Does he not know the definition of density? Density is the ammount of mass per unit volume, the clue is in the units kg/m^3. You cannot measure density with a yardstck. It is impossible. Density is defined my mass and volume, not the other way round. If he has made such a huge mistake with such a simple concept then what else has he got wrong?

I say again to Doc Wattson, get out while you still can

[RealtyBasedWorldview]

Of course you can measure density with a yard stick. You measure the distance between particles, and then take the reciprocal (1/x) of it. The only logical definition of density is "number of blah per unit volume" because mass is, as you have already seen, not fundamental and thus can't be used.

If mass is so fundamental, why are those people working at CERN (supposedly) looking for a particle that causes mass? Mainstream science seems to have contradicted itself with this one!

[Good_Science]

You can measure number density with a yardstick, but mass density cannot be measured this way, and if it's number density he's using in his equations then he's even more inccorect than I initially thought.

You seemed to have made the incorrect assumption that just because a particle gets its mass through the interaction with a field (not particle as you seem to think) that mass is not a fundamental property.

[RealtyBasedWorldview]

You can measure number density with a yardstick, but mass density cannot be measured this way, and if it's number density he's using in his equations then he's even more inccorect than I initially thought.

You admit you are wrong about density, then immediately expect me to accept your assertion that Mathis is 'inccorect' without providing supporting evidence? Cite your sources!

You seemed to have made the incorrect assumption that just because a particle gets its mass through the interaction with a field (not particle as you seem to think) that mass is not a fundamental property.

Wait, so you are saying that the billions of Euros that European taxpayers gave to CERN to find the Higgs *particle* was for nothing? That they ripped us off building their giant machine to search for something that isn't there?

Glad we agree on something

[Good_Science]

Wait, so you are saying that the billions of Euros that European taxpayers gave to CERN to find the Higgs *particle* was for nothing? That they ripped us off building their giant machine to search for something that isn't there?

Thats a non sequitur. Just because the particle may not be the thing that gives mass doesn't mean it doesn't exist or it's not worth looking for it.

You admit you are wrong about density, then immediately expect me to accept your assertion that Mathis is 'inccorect' without providing supporting evidence? Cite your sources!

I did provide evidence.

Density is the ammount of mass per unit volume, the clue is in the units kg/m^3

That is the definition of density, if he's using number density to replace mass then he has a flawed understanding.

[Jarvamundo]

If pi truely were changing or variable surely we would have heard about it.
....
Mathematical proofs are unlike any other proofs in that they are completely correct and it is impossible for them to be wrong.
....
If pi were 4 in some cases it would be more widely known.

These seem argumentative fallacies, well identified by members of this forum. near useless.

I fear you may be missing much of Mathis' direction, and the above peppering carries no weight, especially here.

Of-course mathematical proofs remain correct in their applicable domains and co-ordinate systems, this is well acknowledged by Mathis' writings, however without these proper and full definitions seemingly correct mathematics cannot be applied to all scenario's of reality and labeled as 'physics', this (for me) is what Mathis, in his full works to date, has addressed, from first principles.

Just a guy who read the book, who does not subscribe fully to M's thesis, but sees here a lack of experience with the material, jumping all over an equation out of context like this 25+ pages into a well contributed thread, reveals a psuedo-skeptic.

Miles Mathis is a crackpot

Welcome to TB. + 1 could we see sources please 'Good Science'.

[webolife]

Just for clarification for those who haven't "done the math":
2-dim pi is the circular ratio of circumference to diameter, C/D = 3.1415926...
3-dim pi is the spherical ratio of surface area to cross-sectional area, 4(pi)R2/(pi)R2 = 4

[sponsoredwalk]

Another thread that even mentions these forums and calls Mathis a crackpot:

http://www.boards.ie/vbulletin/showthre ... 2055920504

Actually I did go to the trouble of reading it, but it is simply incorrect, as I tried to explain above. There is a bunch of quite fundamental errors, but the one that stuck out at me was that he didn't really get what energy was. The kinetic energy is by definition, the work required to bring a particle from its rest state up to a given velocity. On the page he questions how energy can be defined if we do not actually do this acceleration. This is absolute nonsense, and an extremely fundamental mistake. We only care about the amount of work needed in theory, not whether you actually did this. That's the mistake right there. He basically is not accepting any equation that comes from thinking about an abstraction, and has replaced it with nonsense.

He is making very basic mistakes about kinematics, so it seems pointless to me at least to pay any further attention to any of his claims.

but your right, as is my book, and he is trying to take the fact that acceleration is simply involved in the algebra and then tell us everything is wrong.

I believe this is similar to what he did in the paper on Relativity I linked to but I think he did it with a partial differential operator that time

The more I think about it, the more I realise there's no way somebody could make that mistake & how deluded it is. Notice how he goes on to use E=mc² to then validate how everything is wrong without actually addressing motion according to his formulation.

Do you guys as Mathis supports have any counter arguments to points made in that thread? That seem rather damning yet there are people and places out there, like this thread, that treat Miles Mathis as a genius. What am I missing?

Please don't take this as an attempt to bash Miles Mathis, I am simply trying to do my own research to figure out if he is someone to pay attention to instead of just taking one forums opinion on him as fact.

-Gary-

Hi, thanks for quoting my thread on boards I see you've gotten interested in Mathis claims as well.
Back then when I wrote that message I was still struggling with basic physics & as you can clearly see
I was having difficulty understanding what Mathis was saying. If I said the word crackpot or quack
you'll have to forgive me because this is internet land, I mean you do get these claims all over the
internet if you look slightly hard Lets look at what he's said:

If we let the initial velocity equal zero, and define work as force through a distance, we get

W = E = Fd = ½ mvf2

Work is then defined as the change in kinetic energy, in the famous work-energy theorem.

This is incorrect, W = ΔE

It is very important & makes a big difference when we look at his next argument:

Let me put it another way. If vf = vi ,
then the postulate equation becomes 2ad = vf2 - vi2 = 0
2Fd/m = 0
E = Fd = 0

You cannot postulate an acceleration in order to develop an equation, and then dump the acceleration. The equations that come after the first equation depend on the first equation. You cannot have different assumptions in the postulate equation and the derived equations. You cannot have variable motion in the first equation, and then derive constant motion from it! We see once again how our textbooks are riddled with gloriously negligent math.

Again, See where he wrote E = Fd = 0, the Δ is missing again. If v = v₀ Then the equation says that
the final kinetic energy is equal to the initial kinetic energy so no work has been done because no
force has acted. Therefore there is no CHANGE in the mechanical energy. The body has energy, but
the work-energy theorem deals in the CHANGE in mechanical energy. I'll admit it took me a while to
realise that, because he left out the Δ it's easy to forget the theorem deals in ΔE's

Some quarters try to dodge this problem as Wikipedia does when it says, “Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes.” But this is absurd. The equation is developed from the acceleration, as I just showed.

Now, this is explained when we think about the Δ. The equation is developed from an acceleration, and as
wikipedia says “Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes”. If you think about it, once the body has gained the kinetic energy, it's moving along in
space unperturbed, one point in the journey it has a certain kinetic energy, a few moments later it has the same kinetic energy, and then we just pull out W = Fd = ΔK = E and see that K₁= K₂, (kinetic energy hasn't changed) written another way W = Fd = K₂ - K₁ = 0 = E E = 0??? But wait, the theorem does not say anything like this,
it says W = Fd = K₂ - K₁ = 0 = ΔE. Obviously the body has energy of K. That wont alter until a force
acts.

Also:

You cannot postulate an acceleration in order to develop an equation, and then dump the acceleration.

Nobody is dumping any acceleration. If a body is not accelerating then acceleration is a constant, i.e. a = 0.
Nobody has dumped anything. If you've ever looked at these gloriously negligent textbooks they develop this concept in one of two ways, from the rate of change of momentum equation of just from F = ma.
If F = ma, a = F/m, and using v² = v₀² + 2ad we get v² = v₀² + 2(F/m)d and from this Fd = ½mv² - ½mv₀².
Nobody has dropped acceleration. Now, Fd = ½mv² - ½mv₀² = ΔK.E. Notice, this is NOT energy. This is one part of the Total Mechanical Energy term.

But can we derive the kinetic energy equation without a force?
...

The work-energy theorem requires a change in velocity, which is an acceleration. You cannot get work without a force and you cannot get a force without an acceleration. But the current kinetic energy equation has no change in velocity.

The kinetic Energy DOES have a change in velocity, Fd = ½mv² - ½mv₀² = ΔK.E.
If you are thinking about the situation where ½mv² = ½mv₀² then yes there is no change in
velocity because no force acts, Fd = 0, ½mv² - ½mv₀² = 0, ΔE = 0, BUT E ≠ 0, E = ½mv².

A particle has kinetic energy with a constant velocity. If the kinetic energy equation is developed from an acceleration, it means the energy depends on the acceleration. The particle should have kinetic energy only while it is being accelerated.

The energy depends on the acceleration during the period the force has acted, Just as Lavoisier theorized that
the mass will remain constant, so too does the energy remain constant in the body until some force changes
this, as the body moves it has within it the energy gained while the force acted. In other words, you mixed up
basic physics and had to go off insulting textbook authors and display that characteristic arrogance in your
papers, it's unfortunate to read this from someone who makes big claims and could potentially influence people.

To sum up, if you neglect the fact that the work-energy theorem describes the change in energy
his paper makes sense. If you actually read those gloriously negligent textbooks they make it explicitly
clear that they define a term from the derived equation ΔK.E. ½mv² - ½mv₀² as E - E₀ and call the term
½mv² energy. E = ½mv². The work-energy theorem then says that Fd = ΔK.E. = ΔE. It deals in the
change in kinetic energy. When Mathis sets v = v₀ and then goes off to claim "But the current kinetic energy equation has no change in velocity" we see that E is not equal to zero, as you'd find from the work-energy
theorem if you thought it described actual energy, but if you realise it describes the change in energy
you see that it all makes sense.

[SleestackVII]

Hi Everyone, I have been studying Mathis' work for about a year now after reading about it in this thread. I find most of his work compelling, thought provoking, and ahead of its time. I am also encourage by the fact that Mathis' reworking of fundamental “standard” theory and the EU/PC theories are not mutually exclusive. Mathis' real spin of qaunta, the resulting recycling of a photon bombardment field, and the E/M field produced from these interactions may be the basic reason for plasma instabilities at a fundamental level, especially near a large coherent bombardment field like the Earth's.

What prompted me to post was the narrative of sponsoredwalk and the fact the he completely misses the point Mathis is trying to get across. I have to admit that I have read and re-read Mathis' stuff to get a better understanding of where he is coming from and sometimes it isn't clear at first. The subject matter is certainly challenging for many and I count myself among the challenged in this respect. I contend that perhaps you should also re-read the original kinetic energy paper one more time. If you do I would like you to pay attention to Mathis' main thesis. I will try to highlight the most important parts for your convenience.


He begins by showing the current derivation for the kinetic energy equation:


Mathis


“To find out, let us look at how the first equation is derived in textbooks. We start with the constant acceleration equation,
2ad = vf2 - vi2
Then substitute a = F/m into that
2Fd/m = vf2 - vi2
If we let the initial velocity equal zero, and define work as force through a distance, we get
W = E = Fd = ½ mvf2

Work is then defined as the change in kinetic energy, in the famous work-energy theorem."

"The hole here couldn’t be bigger, though we never see comment on it. Lots of things have kinetic energy that don’t have accelerations. A photon is a prime example, but there are millions of other easy examples. The equation itself makes this clear, since it doesn’t have an acceleration in it. You can plug any particle with any constant velocity into it, and achieve a kinetic energy. So this derivation is misdirection. It implies that we need an acceleration in order to have a force or kinetic energy, but we don’t. Any object with any velocity will have a force. A car hitting you will apply a force, whether or not it is accelerating.”



So what he is saying is that the equation has been pushed for reasons that are unexplained. He never says that the final equation is false or not at least heuristically valid. Plainly he is saying that although the last equation is true the steps taken to get to that equation are physically invalid. (Because an acceleration and a constant velocity are physically distinct.)
At this point you create a straw man argument and begin to fixate on the missing “Δ” E in his last work equation but Mathis never disputes that there is zero change in kinetic energy at constant velocities.
You seem to make it his argument:

sponsoredwalk said,

“To sum up, if you neglect the fact that the work-energy theorem describes the change in energy his paper makes sense.”


But the change in energy, or the lack thereof , is not disputed. In fact Miles spells this out very clearly.

Mathis


“The equation is developed from the acceleration, as I just showed. The work-energy theorem requires a change in velocity, which is an acceleration. You cannot get work without a force and you cannot get a force without an acceleration. But the current kinetic energy equation has no change in velocity. A particle has kinetic energy with a constant velocity. If the kinetic energy equation is developed from an acceleration, it means the energy depends on the acceleration. The particle should have kinetic energy only while it is being accelerated."


So again it is the way the equation was derived that Mathis takes issue with. He does not, “--neglect the fact that the work-energy theorem describes the change in energy.”

Mathis then begins his second theme, asking why the velocity is squared.

Mathis


"In another paper, I showed that we can develop the equation E = ½mv2 from the equation E = mc2, by reworking Einstein’s equations and making a few corrections. But this brings us back to explaining E = mc2. Einstein develops this equation mainly by assuming that E/c is the momentum of the photon. Then, by the equation E/c = mv, and substituting c for v, he gets his new equation. But E/c was found by experiment, not by theory, so the theory becomes circular at this point. We keep returning to the question, why c2? Einstein gives us the number but not the mechanical explanation. Why square the speed of light? Why should the energy depend on c2? Or, to extend the question, why should the energy of any moving particle, moving with a constant velocity, depend on the square of that velocity?”


Later he reiterates his main thesis, (the pushing of the equation for unexplained reasons), reconfirms his acceptance of the current work-energy equation and explains why his theory of real spinning photons/quanta account for the squared velocity.


Mathis


“Even if I didn’t have an answer for this question, I think it would be important to show the hole. These holes should not be covered up, they should be put on the front page: that is the only hope of an answer. But I do have an answer. I can tell you why the kinetic energy is dependent on the square velocity.“
“In my paper on photon motion, I showed that the measured wavelength and the real wavelength of the photon differ by a factor of c2. This is because the linear motion of the photon stretches the spin wavelength. The linear velocity is c, of course, and the circular velocity approaches 1/c. The difference between the two is c2. Energy, like velocity, is a relative measurement. A quantum with a certain energy has that energy only relative to us, since it has its velocity only relative to us. If the wavelength has to be multiplied by c2 in order to match it to our measurements, then the mass or mass equivalence will also. Hence the equation E = mc2. In this way, c2 is not a velocity or a velocity squared, it is a velocity transform. It tells us how much the wavelength is stretched, and therefore how much the mass and energy are stretched, due to the motion of the object.”
“The same analysis can be applied to any object. The energy of any object is determined by summing the energies of its constituent atomic and quantum particles, and all these particles also have spins. The quanta will impart this spin energy in collision, so this spin energy must be included in the total kinetic energy.”


He later clarifies,


Mathis


“So the short answer is that the kinetic energy equation, like the equation E = mc2, always included the spin energy; but no one recognized that. Just as with the photon, all matter has a wavelength (see de Broglie), and the wavelength is determined by spin. The spin has a radius, and this radius is the local wavelength. Any linear velocity of the spinning particle will stretch our measurement of this wavelength, in a simple mechanical manner, as I showed in the photon paper. As the linear velocity increases, the spin velocity relative to the linear velocity decreases, by a factor of 1/v. This makes the difference between the linear velocity and the spin velocity v2. The term v2 transforms the local wavelength into the measured wavelength. This is why we find the term in the energy equation.”
The only question remaining is why we have the term ½ in the kinetic energy equation. The reason is simple: we are basically multiplying a wavelength transform by a mass, in order to calculate an energy. So we have to look at how the mass and the wavelength interact. I have shown that the wavelength is caused by stacking several spins (at least two spins), so what we have is a material particle spinning end-over-end. If we look at this spin over any extended time interval, we find that half the time the material particle is moving in the reverse direction of the linear motion. Circular motion cannot follow linear motion, of course, and if we average the circular motion over time, only half the circular motion will match the linear vector. This means that half the effective mass will be lost, hence the equation we have.
This does not apply to the photon, or to the equation E=mc2, due only to the fact that c2 is so large. The photon has the same mechanics, but ½ is so small relative to c2 that it can be ignored. The equation E=mc2 can be applied to the photon, to find a kinetic energy (since the photon is almost all kinetic energy), so the equation is analogous in all ways to E = ½mv2. But since the mass is negligible compared to c2 , we can ignore the ½. It does not matter that the mass is moving opposite to the linear motion, since ½ of almost nothing is still almost nothing”



It should be clear that Mathis' argument is dependent on the accepted kinetic energy equation being correct. So I'm not sure what your argument is sponsoredwalk. It can't be that Mathis misunderstands the equation. Perhaps he's just a sloppy typist.

[sponsoredwalk]

"The hole here couldn’t be bigger, though we never see comment on it. Lots of things have kinetic energy that don’t have accelerations. A photon is a prime example, but there are millions of other easy examples. The equation itself makes this clear, since it doesn’t have an acceleration in it. You can plug any particle with any constant velocity into it, and achieve a kinetic energy. So this derivation is misdirection. It implies that we need an acceleration in order to have a force or kinetic energy, but we don’t. Any object with any velocity will have a force.A car hitting you will apply a force, whether or not it is accelerating.”

This is still just a basic misunderstanding of the whole concept though. Just so we're clear, Δ represents
change, W = ΔK.E. = ΔE is an equation describing the WORK. Work is DEFINED as F•d. (force times distance).
What this means is that there can only be WORK done when an object moves under the action of a force,
which in turn causes an acceleration. The term K.E. = ½mv² is just an identity, the fact that the WORK equation
depends on an acceleration is important, and my "strawman" Δ is pivotal to understanding the significance of
this, but K.E. = ½mv², is just an identity. The gloriously negligent textbooks say that they are just going to
call the term ½mv² as K.E., it just is.
W = F•d = = K.E.₂ - K.E.₁ = ΔK.E. = ½m₂v² - ½mv₁² = ΔE
See, it's just one term in a bigger equation that tells you how to find the CHANGE in the quantity this identity
represents when a force acts over a distance.
As for this argument that a photon has kinetic energy without an acceleration, I mean I think it's obvious a
photon will not start moving in the first place unless something caused it to move, photons do not do work in
the sense that W = F•d, it's to do with energy states and electron configurations and all this QM stuff.
Lets be more specific, a photon will not leave a light source unless an atom moves into a lower energy state,
e.g. the ground state, whereby energy is released where it once was stored up.
http://en.wikipedia.org/wiki/Spontaneous_emission
Notice there is an explanation for this that has nothing to do with the concept of W = F•d = ΔK.E. = ΔE
because ΔE is a term representing the change in MECHANICAL ENERGY. It's a specific term, and notice
that the term describing the kinetic energy of a photon is not K.E. = ½mv², it's actually very misleading to
imply these things are similar, and that a photon's kinetic energy could be described by an identity
whose origin comes from an equation using accelerations and forces in the classical sense. It's either
dishonesty or just confusion.
http://www.greatians.com/physics/wave/e ... photon.htm

But the change in energy, or the lack thereof , is not disputed. In fact Miles spells this out very clearly.

Mathis


“The equation is developed from the acceleration, as I just showed. The work-energy theorem requires a change in velocity, which is an acceleration. You cannot get work without a force and you cannot get a force without an acceleration. But the current kinetic energy equation has no change in velocity. A particle has kinetic energy with a constant velocity. If the kinetic energy equation is developed from an acceleration, it means the energy depends on the acceleration. The particle should have kinetic energy only while it is being accelerated."


So again it is the way the equation was derived that Mathis takes issue with. He does not, “--neglect the fact that the work-energy theorem describes the change in energy.”

Leaving aside the fact that there is a change of velocity in the formula:
W = F•d = = K.E.₂ - K.E.₁ = ΔK.E. = ½m₂v² - ½mv₁² = ΔE
even though Mathis says "But the current kinetic energy equation has no change in velocity":
"If the kinetic energy equation is developed from an acceleration, it means the energy depends on the acceleration. The particle should have kinetic energy only while it is being accelerated"
here is another reason why my Δ is important, I mean what kind of a conclusion is this to draw? The equation will
describe the change in kinetic energy while the body is being accelerated but once the acceleration (well, Force)
stops acting there is no reason to think the body should lose all kinetic energy according to any reasoning other
than the confused variety. To reiterate, W = F•d = ΔK.E. = ½m₂v² - ½mv₁² = ΔE describes the
change that undergoes when a force acts, once the force stops, or the acceleration if we use Mathis language
the equation gives it's answer for the change in energy, when Mathis says "it means the energy depends on the
acceleration" this is just another example of my strawman Δ being forgotten by Mathis, it means the change in energy depends on the acceleration.
Now, once the force stops acting W = F•d = 0 = ΔK.E. = ½m₂v² - ½mv₁² = ΔE whereby ½m₂v² - ½mv₁² = 0 and ½m₂v² = ½mv₁².This equation gives us zero. But the eq. is only talking the change, this says NOTHING about the identity K.E. = ½m₂v² in-and-of-itself. We just see there is no change in energy. So there is no reason to think that "The particle should have kinetic energy only while it is being accelerated", rather "The particle should have a change in kinetic energy only while it is being accelerated"

Finally, while Mathis sees:
You can plug any particle with any constant velocity into it, and achieve a kinetic energy. So this derivation is misdirection. It implies that we need an acceleration in order to have a force or kinetic energy, but we don’t.
this as some sort of bone of contention, it's just his conclusion that is the problem. The equations do not
imply that "we need an acceleration in order to have a force or kinetic energy", they explicitly state that
we need a force acting over a distance to cause a CHANGE in kinetic energy. Also, "You can plug any particle
with any constant velocity into it, and achieve a kinetic energy. So this derivation is misdirection", is not a
misdirection, you can plug K.E. = ½m₂v² in for any old object and achieve a kinetic energy, yes, but
as Mathis sees acceleration as so important he should apply it in it's proper domain only,
namely in W = F•d = ΔK.E., NOT in the identity K.E. = ½m₂v² which is just an identity giving
you the kinetic energy, one term of a larger equation
W = F•d = = K.E.₂ - K.E.₁ = ΔK.E. = ½m₂v² - ½mv₁² = ΔE
whose answer depends on the existence of an acceleration. ½mv² Does not depend on acceleration because
it's not in the identity...

[SleestackVII]

sponsoredwalk

I want to be fair to you even tho you are not exactly being fair to Mathis' work. In reading through your post I have noted several problems with your argument.

First of all you may have had an adversarial mindset when you skimmed the paper.

http://milesmathis.com/kinetic.html

I believe this may have tainted your first reading of it. Please I ask you to take another whack at it. This time assume that Miles isn't trying to sneak one by you. He seems to me to be very adept at mathematics and equation manipulation and understands these equations very well. Its important to pay attention to the arguments he makes in the actual paper and not make up phantom arguments for Mathis that can be easily cut down, hence, strawman arguments.

For example,

quoting sponsoredwalk:


Work is DEFINED as F•d. (force times distance).
What this means is that there can only be WORK done when an object moves under the action of a force,
which in turn causes an acceleration. The term K.E. = ½mv² is just an identity, the fact that the WORK equation
depends on an acceleration is important, and my "strawman" Δ is pivotal to understanding the significance of
this, but K.E. = ½mv², is just an identity. The gloriously negligent textbooks say that they are just going to
call the term ½mv² as K.E., it just is.


I think you have a good handle on the kinetic energy equation and the fact that to do Work the object in motion has to accelerate or decelerate, when the force is imparted to another object. This is when the actual Work is done. The only problem is that Mathis' paper is not about the Work theorm specifically. He is asking the question,
Why is the “v” (velocity) in the equation:

W = ΔE = Fd = ½ mvf2

Why is that v squared?

That means you can replace the “v” with a constant velocity and get the correct answer by squaring it. Making you think it acts like an acceleration. But it doesn't. Mathis main point is that the squared velocity is not due to an acceleration but due to the stretching of the wavelength instead.


Mathis writes,


“In my paper on photon motion, I showed that the measured wavelength and the real wavelength of the photon differ by a factor of c2. This is because the linear motion of the photon stretches the spin wavelength. The linear velocity is c, of course, and the circular velocity approaches 1/c. The difference between the two is c2.”

He later concludes,

Hence the equation E = mc2. In this way, c2 is not a velocity or a velocity squared, it is a velocity transform. It tells us how much the wavelength is stretched, and therefore how much the mass and energy are stretched, due to the motion of the object.”

At the end of the paper he generalizes the concept to include the kinetic energy equation.

Mathis

“So the short answer is that the kinetic energy equation, like the equation E = mc2, always included the spin energy; but no one recognized that. Just as with the photon, all matter has a wavelength."


So it is agreed by all, including Mathis, that an object must accelerate to perform Work but it is quite beside the point of Mathis' argument in this paper.

In the last part of the above quote you mention an “identity”.

Quoting spondoredwalk:

The term K.E. = ½mv² is just an identity, the fact that the WORK equation
depends on an acceleration is important, and my "strawman" Δ is pivotal to understanding the significance of
this, but K.E. = ½mv², is just an identity.
W = F•d = = K.E.₂ - K.E.₁ = ΔK.E. = ½m₂v² - ½mv₁² = ΔE
See, it's just one term in a bigger equation that tells you how to find the CHANGE in the quantity of this identity

Can we agree that identity in this context means “equals” as in the word equation? That means all the terms between all the equal signs are identical to all the others. Its not just an identity it is the backbone of mathematics. I understand that you already know this but how is it relevant to the discussion? Mathis is not focusing on the terms that you wish to focus on but you must admit that if the equation holds true then any of the terms may be set against any of the others using an equal sign and focused on exclusively.

ΔE = ½ mvf2

Do you see? The constant velocity is still squared in the above equation even though the Delta is properly positioned exactly where it should be.

I agree that it appears Mathis forgot to type the Δ in his paper but the error never enters his argument. His paper is dealing with the other side of the equation! Although Mathis understands there is a change in kinetic energy (ΔE) resulting from accelerations. He makes no specific claims about kinetic energy changes (ΔE) related to this equation. He is very specific when he talks about the Standard Theory's problem:

Mathis
“If it could derive the kinetic energy equation with good math, it wouldn’t need to derive it with bad math. If anyone could tell you why the kinetic energy is a function of the square velocity, they wouldn’t need to pin the equation to acceleration.”

So sponsoredwalk I give you an opportunity here to answer Mathis' question. Why is the total kinetic energy of an object a function of the square velocity instead of its acceleration? When answering please don't forget that Mathis agrees the object has an intrinsic kinetic energy even at constant velocity.

speaking about the kinetic energy equation,

Mathis

“You can plug any particle with any constant velocity into it, and achieve a kinetic energy.”

and this quote,

“Lots of things have kinetic energy that don’t have accelerations. A photon is a prime example, but there are millions of other easy examples.”



This brings us to the next point in your post that needs clarification.

Quoting sponsoredwalk:

“As for this argument that a photon has kinetic energy without an acceleration, I mean I think it's obvious a photon will not start moving in the first place unless something caused it to move,”


It is generally accepted by Standard Theory that light moves at a constant velocity c and has kinetic energy, The photoelectric effect describes this where a photon displaces an electron out of its orbit.

From Wikipedia:
“Study of the photoelectric effect led to important steps in understanding the quantum nature of light and electrons and influenced the formation of the concept of wave–particle duality.[1]”

http://en.wikipedia.org/wiki/Photoelectric_effect

Don't forget Mathis contends that there is no wave particle duality - that light is a real particle with stacked spins and is the basic building block of all matter. Several of his papers discuss this in detail.

http://milesmathis.com/photon4.html
http://milesmathis.com/super.html
http://milesmathis.com/super2.html
http://milesmathis.com/double.html

Finally let's take a look at something else you've posted above:

quoting sponsoredwalk:

“Leaving aside the fact that there is a change of velocity in the formula:
W = F•d = = K.E.₂ - K.E.₁ = ΔK.E. = ½m₂v² - ½mv₁² = ΔE
even though Mathis says ""But the current kinetic energy equation has no change in velocity":
"If the kinetic energy equation is developed from an acceleration, it means the energy depends on the acceleration. The particle should have kinetic energy only while it is being accelerated""
here is another reason why my Δ is important, I mean what kind of a conclusion is this to draw? The equation will
describe the change in kinetic energy while the body is being accelerated but once the acceleration (well, Force)
stops acting there is no reason to think the body should lose all kinetic energy.”


You should note that in the above quote Mathis is being facetious.

http://www.merriam-webster.com/dictionary/facetious

Mathis

"But the current kinetic energy equation has no change in velocity":
"If the kinetic energy equation is developed from an acceleration, it means the energy depends on the acceleration. The particle should have kinetic energy only while it is being accelerated"

The last statement only holds true if the kinetic energy is a result of an acceleration. You argue above that is not the case. Well Mathis agrees with you but you don't realize it.

quoting sponsoredwalk refering to Mathis' facetious statement:

"I mean what kind of a conclusion is this to draw? The equation will
describe the change in kinetic energy while the body is being accelerated but once the acceleration (well, Force)
stops acting there is no reason to think the body should lose all kinetic energy according to any reasoning other
than the confused variety."

You take this last part and run with it, misrepresenting Mathis when you do. Frankly I think you were the one that was confused.

One more thing before I conclude. sponsoredwalk I appretiate you having an interest in Mathis' work, EU/PC theories, and Physics in general like I do. I am going to assume you are trying to assess Mathis honestly just like I am. So, bottom line, its important to represent Mathis' arguments accurately when making that assessment. That is my goal and I trust it is yours too.

[sponsoredwalk]

sponsoredwalk

I want to be fair to you even tho you are not exactly being fair to Mathis' work. In reading through your post I have noted several problems with your argument.

First of all you may have had an adversarial mindset when you skimmed the paper.

...


One more thing before I conclude. sponsoredwalk I appretiate you having an interest in Mathis' work, EU/PC theories, and Physics in general like I do. I am going to assume you are trying to assess Mathis honestly just like I am. So, bottom line, its important to represent Mathis' arguments accurately when making that assessment. That is my goal and I trust it is yours too

I believe this may have tainted your first reading of it. Please I ask you to take another whack at it. This time assume that Miles isn't trying to sneak one by you. He seems to me to be very adept at mathematics and equation manipulation and understands these equations very well. Its important to pay attention to the arguments he makes in the actual paper and not make up phantom arguments for Mathis that can be easily cut down, hence, strawman arguments..

No offence, but you're completely wrong. Not only have I read Mathis paper with an open mind, I read this
paper 7 months ago, thought about it for a month and then started the thread quoted by Doc Watson above,
then I thought about it every so often for the past 6 months to make sure I haven't made a mistake, to make
sure the people in that thread who I asked about this paper were not just writing him off so
please don't accuse me of reading this adversarially and letting some phantom bias cloud my judgement.
Mathis' arguments are wrong, and notice I've focused on specific things he's said. I never went near his
argument about wavelengths, or why there is a v² in the equation, yet that's what you start your
response to me talking about. Who is creating phantom arguments? I've only focused on what I know I can
talk about, the basic stuff, because it's the basic stuff he's getting wrong. You can accuse me of not being
fair to him all you want but all I've done is analyse his words and talk about what I know about them. If being
skeptical of someone's work is not "fair" I don't think we agree on the definition of fairness. If I'm coming
across as challenging it's because I am in fact challenging something I see as wrong. It's nothing personal
to anyone, I'm just focusing on the material.

The substance of Mathis paper, for me, is the beginning. His claim is our textbooks are wrong. Well I was
only learning about what kinetic energy was 7 months ago properly. In fact it was a wave of relief to
learn the source of the equation I'd heard about in school for 6 years not knowing a thing about where it
came from etc... I can't tell you how happy I was to see a logical derivation of the equation that haunted me
as part of a wider "haunting" I experienced. So when someone comes along and claims this is actually wrong
of course I'm going to give it some proper thought. I definitely thought there was something to it until
thinking about it clearly. A lot of my confusion sprang from the missing Δ, hence why I emphasize it's importance.

Mathis:
W = E = Fd = ½ mvf2

Work is then defined as the change in kinetic energy, in the famous work-energy theorem.

The hole here couldn’t be bigger, though we never see comment on it. Lots of things have kinetic energy that don’t have accelerations. A photon is a prime example, but there are millions of other easy examples. The equation itself makes this clear, since it doesn’t have an acceleration in it. You can plug any particle with any constant velocity into it, and achieve a kinetic energy. So this derivation is misdirection. It implies that we need an acceleration in order to have a force or kinetic energy, but we don’t. Any object with any velocity will have a force. A car hitting you will apply a force, whether or not it is accelerating.

See the part in red, that is wrong. It's actually terrible language too, I mean you need a force to get an
acceleration. A force is the reason for an acceleration to occur. Things don't just accelerate, they accelerate
due to the action of a force. And yes, the reason I constantly rewrite the equation
W = F•d = ΔK.E. = ½mv₂² - ½mv₁² = ΔE
is to emphasize that a force causes an object to change it's kinetic energy. If a body is originally motionless,
a force will cause it to change it's kinetic energy. Then as it moves along and hits you, as Mathis mentions
in the blue text above, the body will apply a force. The reason a moving object can apply that force to
you as it moves is because it originally gained kinetic energy due to a force acting on it at that time.
Essentially, Mathis claims that because an object that is moving at constant velocity hits you, thereby
being a force that acts on you, this somehow shows the derivation of the equation to be a misdirection
because acceleration is used in the derivation as part of the algebraic manipulation. As I've already said,
my Δ "strawman" gives a logical answer to Mathis, the body that is moving at constant velocity has
gained it's kinetic energy of motion because a force (with acceleration coming along as an accomplice) caused
it to move in the first place. He totally neglects the fact a body is only moving because a force got it moving
to begin with. Also, when he mentions "You can plug any particle with any constant velocity into it, and achieve a kinetic energy.", his obvious assertion is that because you can do this it somehow invalidates the derivation of
the kinetic energy term because it was itself initially derived from an equation that had acceleration in it.
Read his words - "The hole here couldn’t be bigger, though we never see comment on it. Lots of things have kinetic energy that don’t have accelerations" ... "So this derivation is misdirection. It implies that we need an acceleration in order to have a force or kinetic energy, but we don’t."
It''s as if he's saying: Look! A body with constant velocity is being described by an equation with an acceleration in it! Obviously it's the work of gloriously negligent textbook authors because I don't see an acceleration, do you? In fact, I'm even more right because a photon doesn't have an acceleration either! See! Lots of evidence on my side!
How can you not see how dishonest mentioning a photon at this instant is? I know light moves at c,
that is totally irrelevant to us. It's only relevant if you think a body moving at constant velocity somehow disproves
a different equation that has nothing to do with that body moving at constant velocity...

In the last part of the above quote you mention an “identity”.

Can we agree that identity in this context means “equals” as in the word equation? That means all the terms between all the equal signs are identical to all the others. Its not just an identity it is the backbone of mathematics. I understand that you already know this but how is it relevant to the discussion? Mathis is not focusing on the terms that you wish to focus on but you must admit that if the equation holds true then any of the terms may be set against any of the others using an equal sign and focused on exclusively.

No, that is not what I meant by an identity. The equals sign has nothing to do with what I'm saying about this,
maybe it was a bad choice of words on my part. All I meant was that K.E. = ½mv² is just a term
that gives a certain number, in this case a scalar value, to some physical quantity. It is independent
of everything else other than velocity and mass. Acceleration has nothing to do with this value,
an acceleration will change it and you'll get new instantaneous values for this scalar quantity when
you plug in the new velocity that the acceleration has given the body. That said, an acceleration
does nothing but change the value. The whole derivation does indeed stumble upon this quantity
but the derivation in fact stumbles on the change in this quantity, it comes upon the change in this quantity.
This does not invalidate what that quantity represents.

Just to be crystal clear, I can't respond to most of your post seeing as you chose to talk about everything I
purposely neglected in his paper. I'm only focusing on what I can talk about and you haven't addressed any point
I made other than to say Mathis was being facetious at one point. I'll just have to go through the paper again
and show you where he is making mistakes and notice it will be in those points having nothing to do with
wavelengths etc... I can't speak about them.

The hole here couldn’t be bigger, though we never see comment on it. Lots of things have kinetic energy that
don’t have accelerations. A photon is a prime example, but there are millions of other easy examples. The
equation itself makes this clear, since it doesn’t have an acceleration in it. You can plug any particle with any
constant velocity into it, and achieve a kinetic energy. So this derivation is misdirection. It implies that we
need an acceleration in order to have a force or kinetic energy, but we don’t. Any object with any velocity will
have a force. A car hitting you will apply a force, whether or not it is accelerating.

How does a photon having constant velocity disprove an equation that explicitly deals with acceleration?
How does the fact that lots of bodies having kinetic energies without having accelerations imply there is a
huge hole we never comment on and imply the derivation is a misdirection?
How is the derivation of Work a misdirection, as implied by his comments seeing as he is deriving this quantity?
Why does Mathis imply that because K.E. = ½mv² is derived in an equation with acceleration in it
somehow invalidates the derivation?
Why does Mathis incorrectly tell us that this derivation implies we need an acceleration to have a force or
kinetic energy? By his logic here it's understandable to believe his next conclusion, that a car applying a force
whether or not it accelerates, somehow accords with a misdirected derivation, but the equation implies
nothing of the kind. This badly phrased sentence, i.e. focusing on acceleration as being the prime
cause when nobody else thinks that, is just confused. He is the only one who thinks this equation implies
we need an acceleration in order to have an force, I don't and in fact neither does the equation, seeing as
it emphasizes Force as causing a change in K.E.

But can we derive the kinetic energy equation without a force? Can we achieve a square velocity without assuming an acceleration? ...
Some quarters try to dodge this problem as Wikipedia does when it says, “Having gained this energy during its
acceleration, the body maintains this kinetic energy unless its speed changes.” But this is absurd. The equation
is developed from the acceleration, as I just showed. The work-energy theorem requires a change in velocity,
which is an acceleration. You cannot get work without a force and you cannot get a force without an acceleration.
But the current kinetic energy equation has no change in velocity. A particle has kinetic energy with a constant
velocity. If the kinetic energy equation is developed from an acceleration, it means the energy depends on the
acceleration. The particle should have kinetic energy only while it is being accelerated.

If you're going to argue this is facetiousness then it invalidates Mathis claims because there is no substance
to his arguements here. Here he explicitly states that wikipedia is absurd for it's conclusion seeing as the
equation is developed from an acceleration. Is this the fecetious statement? Also, "you cannot get a force without an acceleration", if this is true then it invalidates the whole area of physics known as statics
It seems again that he focuses on accelerations more than forces as being prime causers, you can get a force
without an acceleration, that is what the study of a simple free-body diagram showing a boy acted on by gravity
and a normal force clearly shows. So he must of been facetious here too. But his next conclusion, that
a particle has K.E. with constant velocity, this is true. But why does he say that K.E. is developed from an
acceleration? Facetiousness? K.E. is developed from acceleration, as he & wikipedia clearly said! But he said
this was absurd? Where is the facetiousness??? If this is supposed to be a revolutionary paper you'd think he'd
alleviate all chance for confusion! The point is he's not being facetious, he's saying wikipedia is absurd in
it's conclusion that the body maintains it's kinetic energy unless the speed changes. Proof? Because he not only
quotes this line and calls it absurd but then goes on to say that "the particle should have kinetic energy only while
it is being accelerated.". If he's being factious in the last line then quoting wikipedia and calling it absurd
means he was wrong about wikipedia, that arbiter of propaganda, and if he's right in calling wiki absurd
then his conclusion is not facetious but serious.

Also, when Mathis says:

You cannot postulate an acceleration in order to develop an equation, and then dump the acceleration. The
equations that come after the first equation depend on the first equation. You cannot have different assumptions
in the postulate equation and the derived equations. You cannot have variable motion in the first equation, and
then derive constant motion from it! We see once again how our textbooks are riddled with gloriously negligent
math.

it only further shows that he actually believes "the particle should have kinetic energy only while it is being accelerated" because he thinks we're dumping acceleration when it's equal to zero. Again, as I originally
said, the only conclusion here is that K.E. = ½mv² is somehow a misdirection because the term ½mv² is
derived in an equation with an acceleration that is not zero. But this equation is describing the change in energy, not the energy itself. K.E. = ½mv² is just "an identity", or rather a term, that tells you the energy of the
body. There is no change in any assumption at all.

Based off all of this, his reasoning to think that the derivation is a misdirection is fundamentally
flawed. There is nothing absurd about thinking a body maintians it's kinetic energy unless it's speed
changes, there is nothing misdirected in the derivation because a body that hits you will impart a force
whether it's accelerating or not and there is certainly nothing misdirected in the equation because a
photon travels at the constant speed of light. Mathis mentioning all of these points as motivating factors for
his conclusions are all ridiculous, I have nothing to say about the conclusions he draws from these observations
as you clearly see. The Δ explains "If the kinetic energy equation is developed from an acceleration, it means
the energy depends on the acceleration", it means he not only forgets the Δ, which indicates change, he also
forgot to include the word change in his descriptions using words. If the kinetic energy is developed from an
acceleration, it means the change in energy depends on the acceleration. This says nothing about the energy
the body had before the force acted to change the acceleration. This is described by a previous event, and also
explains Mathis example of a car hitting you imparting a force whether or not it was accelerating, because it
absurdly has the kinetic energy it accumulated when it first began moving.