cyrek1 wrote:
But if the HA radiates when in a stationary quantum state, you have to account for the energy. This was one reason the classical view of an atom with an 'orbiting' electron didn't work.

Thats actually a high frequency (small wavelength) photon.

Hmm, cyrek1 you seem to be trying diligently to portray electrons as a classical particle. Are you suggesting that an electron is a particle that physically oscillates?

.

Sorry if I suggested that your comments were imaginary (that was never my intent).
I also believe in the EoL without a genuine expansion

cyrek reply
page 1

fortis
Regarding the classical electron, I believe I mentioned this before:

The electron will return to its ground state orbit because of the interaction between the electrons magnetic field from its orbital motion and the protons magnetic field from its spin. In the ground state, both magnetic fields are strong enough to prevent any further electron approach or collapse of the atom.
Also, in the ground state, the standing sinusoidal wave is radiated by all the other HA’s. That also prevents any further erosion of the electrons orbital velocity and subsequent collapsing of the atoms.

Where the standing waves of the HA’s are in a higher energy state, this would be due to its environments spatial temperature.

Proton spin:
Yes, the magnetic fields act like two bar magnets that are in a repulsive state.
If the proton was to flip to reverse its MF, it would quickly flip again because of the strong interaction between the coulomb forces and the electrons velocity around the proton that causes the proton to spin.

Ground state energy radiations:
Ground state atoms are all radiating the same energy so there is no absorption or radiation energy loss.
Other atoms in higher states do not absorb any energy either unless a photon of the proper frequency or wavelength (quanta) bounces an electron to a higher state. These SW’s cannot do that because they are sinusoidal and continuous rather than quanta pulses.

Further refinements of Bohr model:
The Schroedinger equations do improve some miniscule changes and the electron spin is non-existent in the real sense. It is the proton that spins.
I figure the electron does not spin because of its distorted shape charge just like the major satellites in our solar system with liquid centers that gravity has also distorted to eliminate their spin.

21 cm radiation:
There is no need to explain this when the Cosmological redshifts are concerned.
My main reason is to promote the ‘expansion of the light waves to replace the EoS invention (no empirical evidence).

EoS:
Your speculation about the origin of the BB has no scientific basis.


The Shroedinger orbitals:
My opinion of these orbitals is that they are time extended approximations of the different electron positions in close proximity to other electrons that create interactions to make the electrons keep changing positions.
As I have said before, in any instant in time, the electron would reveal itself in one position only within its current orbital state.

The classical model:
The comparison between the atomic model and planetary models is that the only difference is the vast EM strength in comparison to the gravitational strength.
But these particles are extremely dense compared to their sizes and have momentum so they would form a binary like ant other two attractive bodies.
This, to me, is basic physics.

Even gravity seems to be quantized with Bodes law?

Ampere:
Your definition of the ampere states that the infinite length wire quantized the ampere to ‘PER METER’ in this length. So, how does that make much of any difference. Like I said, the measured distance of this Newton force is for one meter of this infinite wire.

I do not see any difference. An ampere is a flow of electrons past that one meter length.
This measured flow creates a magnetic field to generate one Newton per wire.
True, the electrons in the wires are the fringe electrons in the metal atoms. So, I said my calculations are a start in computing this intrinsic force of the photo. Another thing I did not mention is that the electron trajectory is not like a straight wire. These modifications and comparisons to the ampere definition requires an expert mathematician which I am not. Regardless, I think I have proven that an electron in transition with a variable velocity and reducing orbit does generate a ‘magnetic pulse’ that has an intrinsic force to expand it similar the ‘expanded field patterns’ in magnetic fields.

__________________
aka Michael Cyrek

How is this the ground state if the flipping of the proton spin would allow the electron to approach more closely? The system, as you describe it, exists in an unstable state. Any disturbance would send the spin of the proton flipping.

I'm guessing that you have played with bar magnets before. Take two of them. Hang them by strings from their mid-points and bring them together. What happens?

One thing that I didn't mention before, but which seems pertinent, is that the ground state of the hydrogen atom possesses zero angular momentum, with the obvious implications for orbital velocity, currents, and interactions between the proton spin and electron orbit.

I've no problem with the notion that higher energy states become increasingly populated as the temperature increases. (Have a look at the notion of partion functions. ) Can you explain, within your theory, what energies these higher energy states have? How does quantisation occur?

How does this Coulomb interaction do the flipping? Are you modelling the proton as a positive charge distribution, in some way being dragged around, or spun-up by the electron, i.e. so the angular momentum of the proton is of the same sign as that of the electron orbital angular momentum? As the charge on the proton has the opposite sign to the charge on the electron, you may want to revise your picture here.

What you seem to be postulating here is an electromagnetic equivalent of thermal equilibrium. First of all, you have to ask yourself if there is any experimental evidence for this radiation? It should have a characteristic wavelength that you could calculate within the framework of your theory. Secondly, how does this work in the context of other material being present? Take metals, which are pretty good at blocking a wide range of wavelengths. If you place hydrogen atoms on one side of a piece of metal, does it lead to a "shadow" on the other side, with a hydrogen atom their radiating to nothingness?

How do you calculate these "proper frequencies" within your model? (By the way, I think that you may have misunderstood what a standing wave is.)

The 21 cm emission being one of these "minsicule" changes.
I could go on, but hopefully you can see that the Bohr model is incapable of explaining many microscopic phenomena, such as multi-electron atoms, and the H2+ ion, etc. It is an approximation of certain features of a far more complete theory.

what evidence is there for the "distorted shape charge"? Current experiments have failed to resolve any structure within the electron, and so far it appears to be point-like. (If you know different please say. )

On the bit about liquid centres and spin elimination, I'm sure that there are folks with more astrophysical knowldge than I who could say say one way or the other if this is correct. Anyone?

Actually, it occurs to me that you might be referring to tidal locking, where the satellite has a rotational period equal to its orbital period. If this is what you are referring to, first of all I have not come across any evidence for this sort of behaviour with electrons, and second, the satellites still have a "spin" angular momentum, so it doesn't seem to lead to your conclusion anyway.

But your model appears to be built on faulty physics, and whilst being touted as a better model than QM, is unable to reproduce some of the basic successes of QM. Can you predict even 1 wavelength within your model? (And I'm not just meaning that you simply use the Bohr model. I mean the full on calculation of the ground state by balancing electrostatic forces with magnetic forces.)

The QM picture has a finite probability of the electron (in a single electron atom) being found close to the nucleus. There are no other electrons required, and not wanting to go on about this too much, this is an important component in the calculation of the 21 cm band. You like experiment, well this is an experimental result.

If you look for an electron you will find an electron. If you look for a stationary state, then you won't be able to pin down the position of the electron.

A major difference between the gravitational interaction and the EM interaction is that gravitational "charge" only comes in one sign, whereas EM charges can be positive and negative. This has implications that relate to the formation of dipoles. (There are others, but you should be aware of the dangers of naively going from a model of the solar system to a model of the atom.)

You should also be able to notice that you can treat the solar system using classical mechanics. (The only notable deviations being due to GR.) This is very different to what is seen with the microscopic H atom. Why should you be surprised to find that classical physics breaks down in a realm that you have no direct experience of? (Just as classical physics breaks down when we have objects moving at velocities approaching that of light.)

If your model does not represent the phenomena (which it doesn't appear to), then it isn't physics.

Not true, or else I wouldn't be able to put a satellite in an arbitrary orbit around the earth. (By the way, Bode's Law also fails for Neptune and beyond...)

Re-read what I said about the way that fields generated by lines of charge differ from those generated by point charges. You are using an equation in a regime for which it isn't valid.

Please read my earlier link about the Biot-Savart law and you will see why what you are doing here doesn't make physical sense.

If you want to work with the effects of accelerating charges (i.e. those with variable velocities) you really should try using Maxwell's equations, as they will finally allow you to see how the electric and magnetic fields interplay. You say that you are not an expert mathematician, and obviously this cannot be held against you. Unfortunately this will severely limit your ability to make sense of what you are doing, which is the equivalent of applying Newtonian mechanics to a problem involving particles travelling at 0.9c.

Further "refinements" of the Bohr model are a complete waste of time because better models based on actual physics exist. In particular the case of a one-electron atom can be solved exactly with the Dirac equation and yields precisely the energy levels that are observed. (There are some small shifts I will mention below.)

The electron has a spin of one-half h-bar, no more, no less. It has no shape or structure, at least down to about 10^-18m. The various energy levels in the Dirac solution mentioned above are labelled by a principal quantum number n, an orbital angular momentum l (indicated by s, p, d, f, g, ... for l = 0, 1, 2, 3, 4, ... respectively) and total angular momentum j which can be either l+1/2 or l-1/2. The first few energy levels are (please correct me if I'm wrong, I don't have my references in front of me) :

1s_1/2
2s_1/2 = 2p_1/2
2p_3/2
3s_1/2 = 3p_1/2
3p_3/2 = 3d_3/2 ?
4s_1/2
3d_5/2

The paired energy levels above are degenerate under the Dirac equation. When additional QED effects are taken into account, the energy levels are split. This resultant split is called the "Lamb-Retherford Shift". I challenge you to find any refinement of the Bohr atom that gives these energy levels.

As for proton spin, it too can be added or subtracted from the angular momentum of the electron, resulting in a further splitting of the levels. For example, the proton spin can align with the electron angular momentum giving a total angular momentum of one h-bar, so-called "parahydrogen" or "triplet" hydrogen, or the proton spin can align against the angular momentum of the electron giving a total angular momentum of zero, for "orthohydrogen" or "singlet" hydrogen.

All of this should be covered in a good introductory course on atomic spectra, except for the Dirac equation and Lamb-Retherford shift, but a good introductory text should at least mention these things.

It's time to hit the books again!

__________________
Microsoft is over if you want it.

The bar has been lowered for the promotion of ATM ideas; the bar for the acceptance of ATM ideas must remain high.

Today I hit the books. (I do take the advice that I give!) I am looking at Volume I of Relativistic Quantum Mechanics by Bjorken and Drell, published in 1964. It gives the solution of the Dirac equation for hydrogen-like atoms and ions in great detail and gives a general formula for the energy levels. I'm not going to quote it here, but I might post it tomorrow. It does give the values for the first few in a table which I will excerpt here.

1s_1/2: mc^2*sqrt(1-Z^2*alpha^2)

2s_1/2 = 2p_1/2: mc^2*sqrt((1+sqrt(1-Z^2*alpha^2))/2)

2p_3/2: (1/2)*mc^2*sqrt(4-Z^2*alpha^2)

I visited a library tonight and borrowed Atomic Spectra by H. G. Kuhn, published in 1962. It mentions Dirac but does not do a full derivation of the Dirac formula but opts for starting from the Bohr-Sommerfeld model and applying corrections to it. I need to read it more closely before I say any more about the book.

Edited to add: In the above formulae m is the mass of the electron, c is the speed of light, Z is the atomic number, so we are considering atomic H (Z=1), He+ (Z=2), Li++ (Z=3), etc., and alpha is the fine structure constant, approximately 137.036.

__________________
Microsoft is over if you want it.

The bar has been lowered for the promotion of ATM ideas; the bar for the acceptance of ATM ideas must remain high.

cyrek reply

Fortis and CM
All the details you suply are really irrelevant to the main discussion.
I saw illustrations of some pf the Schroedinger orbitals and they look like they are out of this world. Truly wierd.

Rather than reply to those details, I wrote the article below.

The ‘expansion of the light waves has real evidence for its support. Examples:

The magnetic field patterns where the central portion is expanded by an ‘intrinsic force’.
This is also true of the electric field.
The photons are primarily a magnetic pulse of radiation.
The electric component is primarily in the field surrounding the electron.
The photon energy uses this field for its transmission.

The electric motor makes use of these intrinsic forces within the EM fields

The ampere rating experimental setup is used to measure this intrinsic force and uses the relative characteristics for establishing its strength such as the amount of electrons passing through two parallel wires for a measured distance and spaced a measured distance apart. All measured distances are ‘one meter’. The result of this setup is based on a resultant force of ‘one Newton’ with a current rated at one ampere.

I reduced the above experimental components to the microscopic level of the hydrogen atom and the result was posted above. This is the final proof that this intrinsic force is real.

The Arp redshift anomalies show that these RS’s are temperature related and therefore intrinsic to the emitting objects and their radiation.

ON THE OTHER HAND
There is NO empirical evidence to prove the expansion of space. This modified hypothesis of the Doppler observations is based on false analogies such as ‘two dimensional no center space.

Using this expansion to justify the cosmological redshift is false because the EM fields are not intertwined with space but entirely separate components.

The Hubble Doppler observations is the only evidence used to fabricate the BB concept.
All subsequent corroborating evidence is purely circumstantial.

I still have not had an answer to the questuion I posted for 'empirical evidence' for the EoS?
Care to answer that?

__________________
aka Michael Cyrek

You really ought to stop going "La-la-la-la-la" with your fingers in your ears. It may aid in your self-education.

You are the one who goes on and on about the Bohr atom when in fact it has been completely overtaken and replaced by sophisticated quantum mechanical models that, guess what cyrek1, actually predict the energy levels that are measured, something that the Bohr-Sommerfeld model could not do even after extensive patching. It is the Bohr atom that is irrelevant to the main discussion and to modern science.

I will let others comment on the errors and fallacies in the mish-mash of word salad that is your "article".

__________________
Microsoft is over if you want it.

The bar has been lowered for the promotion of ATM ideas; the bar for the acceptance of ATM ideas must remain high.

In other words, instead of answering you're questions, I'll just ramble the same things I know you don't understand, hence the questions.

On a macroscopic scale, yes...

This is just more 'proof' you don't understand math or models of physics at all. What has been discovered (50 years ago) is that nothing about the micro-verse has any bearing on the macro-verse.

No evidence? None eh? Okay Mike. That analogy is only an approximation to help those who are 'undereducated' grasp the reality of the math that is involved.
How did you go about finding out that EM fields can exist without space? =D> Bravo. You found "proof" that no one else could have...

My guess is, you'll have to address the questions posed to your article first. Is it a two way street in your universe as well Mike?

There is plenty of 'evidence' for the expansion of space. You just dismiss it out of hand.
Evidence is only as good as the detective utilizing it. When you throw out evidence out of hand, for no reason other than "it's wierd," "it's not reality", and my favorite from you "[...any words...] BB [...any other words...]"

Humbly yours,

Trav

__________________

Feynman
>~~~~<
Science is a way of trying not to fool yourself. The first principle is that you must not fool yourself, and you are the easiest person to fool.

Religion is a culture of faith; science is a culture of doubt.

Oooh, another Gem:
EoLW and BB are not mutually exclusive...
That means that they don't exclude one another.
That means that even if your EoLW (not my reality, too wierd for me...)

__________________

Feynman
>~~~~<
Science is a way of trying not to fool yourself. The first principle is that you must not fool yourself, and you are the easiest person to fool.

Religion is a culture of faith; science is a culture of doubt.

They are not irrelevant if you are using an invalid model to try to prove your theory.

But more real than the planetary model.

I think that the best way to address this is by reference to your original post, so I'll do that in a following post. (If I've missed anything that is in addition to the original post please let me know.)

And as I have said many times, this is not an appropriate use of the definition of the Ampere. You need, at the very least, to use the Biot-Savart law.

See above.

As this is where your proof is contained I thought it simplest to address problems with it here.
As you know, this relationship only applies to wires of infinite length (which is a lot longer than the microscopic distances that you go on to discuss.)

Quite true.

As I have said, this is only true for two infinite, parallel, current carrying wires.

In the non-relativistic regime, the equation below gives you the magnitude magnetic field due to a point charge (with charge q) moving with a constant velocity, v.

B = (mu_0 /4Pi)*(q*vxr/r^3)
(If you don't want to derive it, it can be found in a number of places, such as, http://cpt.phys.utk.edu/~th/Physics231/Lecture08.pdf ) The variables in bold are vectors, and the "x" is the vector cross product.

Now if we have another charge roaming around, then the force exerted on it due to the magnetic field of the first charge is given by

F=q_2*v_2xB

These (along with the electrostatic forces) would be more appropriate for you to use.

Correct.

No.
What exactly do you mean by "The force between 1 wire"? This doesn't make a lot of physical sense.

The force generated by an electron on what?

I know that you're going to try to explain this later, but do you notice how, in my previous comments, these forces are acting on charged particles, not photons?

The above text has a whole bunch of problems with it, but let us, for the moment do a bit of dimensional analysis (which is always fun )

k_m (i.e. the 10^-17) has units of N A^-2 (You can see that this must be the case from your first equation.) which could, alternatively be written as N.C^-2.s^2 (where the C is Coulombs)

The unit of charge is, of course, the Coulomb, C.

Finally, the SI unit of length is the metre

So the right hand of your equation has units of N.C^-1.s^2

This is quite obviously not a unit of force. (Even with a bit of help from a velocity term, which has mysteriously disappeared, you can't salvage this one.)

This gets stranger. The Newton is (in your notation) kg.m/s/s. How did you lose the unit of mass?

How did you derive this value of three?

How have you coupled the force to the photon? You have calculated a number that doesn't have the units claimed for it. In addition you haven't shown how this number corresponds to some stretching of the photon. It's akin to saying that if I drop a ball, it will accelerate at 9.81 m/s^2, therefore my car will accelerate at 9.81 m/s^2.

Where did the kilogram come into this? I wouldn't worry about the figure being to low as the derivation of it doesn't appear to make any sense.

Nope.

You should try and sort out the previous bit of your model before you move on this.

Nope.

These exist, but have no bearing on your claim.

The electric motor is not evidence for your claim, and if you read my comments you shoudl see that your proof isn't either.

cyrek reply

fortis wrote
Definition of ampere:
The ampere is that constant current which, if maintained in two
straight parallel conductors of infinite length, of negligible
circular cross section, and placed 1 meter apart in vacuum, would
produce between these conductors a force equal to 2 x 10-7 newton
per meter of length.

reply
Since you have analyzed my work, I want to show you how to analyze
the definition above.

'of infinite length'
Practical? Mathmaticians frown on 'infinity' though here it really
does not have any application to what I used.

'of negligible circular cross section'
What does this mean? This statement means that the resistance of the
wire is not included? Apparently it could be very thin and is still
excluded? Regardless, this is also not applicable to my use of the
formula.

'per meter of length'
This is applicable! The force is based on ONE meter of lebgth of this
infinite wire.

So my use of limiting the data to 'one meter' is correct.

fortis, your whole argument is based on this 'infinite length' which
does not apply here because the 'one meter length' that is specified
for determining the force is the only criteria that is needed to solve
this problem. This statement excludes the 'infinite length' for
determining the result.

Regarding the use of one wire:
This data uses two wires to define the ampere. This is an interaction
between two magnetic fields generated by the currents.
The removal of one wire does not mean that the other wire does not
exist with its magnetic field. It is still there. It is still
generating ONE HALF of the interacting force for the distance
specified. In this case, it would be one half meter since the space
between the wires is one meter.

fortis wrote:
'the force generated by an electron on what?

reply
The force is in the magnetic field that would be acting on the 'virtual
negatively charged paticles' to condense them to create a photon of
condensed VNCP's that would constitute the photon.
This, of course, is my hypothesis of the nature of a photon pulse
(quanta).
You are right about the force acting on the charged particles.
The photon is the 'condensed congregate' of these charged particles
that would then be repelling each other to return to their normally
distributed state of tranquility.
My explanation for the 'intrinsic expansion' that light would have.

You made a 'typo' there of the 'magnetic constant' which is 10^-7.

Your interpretations of my formula are not accurate.
The coulomb is equivalent to an electrical force even though it is
stated as an electric unit of charge. It is surrounded by the field
particles that create the reactions and that constitute the force.

My ommission in leaving out the kilogram in the definition of the
Newton is not that important since it was defined in the text.

fortis wrote:
How did you derive the value of three?

reply
the value of 3 that divided the final result was an estimate of the
ratio of the electrons accelerated velocity to its standing orbital
velocities. It varies with wavelength.

My final result is the distance this photon would move one kilogram
in meters per s/s. This constitutes a ratio of this distance to the
length of the photon in meters to determine the time required to equal
one half wavelength for a redshift of one.
Remember, this result is calculated in Newtons which includes the
kilogram in its delinition.

Since you have not provided any evidence for justifying the EoS
concept, than I am sure my alternative view is correct.

__________________
aka Michael Cyrek

I'm guessing by this that you haven't read about the Biot-Savart law, which would explain why the definition is only exact in the context of infinitely long wires. For straight wires of finite length, the definition would only be an approximation. For wires that are really short it doesn't hold at all. As I gave you a formula for the field due to a moving point charge (which has a 1/r^2 dependance, rather than the 1/r dependance of the "long" wire), why don't you use that instead? It isn't really any more complicated.

The wires have to be of negligible thickness, or else the field produced by the first wire will be interacting with a current distrbution that has significant extent in all 3 dimensions. Look at what happens with Newtonian gravity when you include bodies with a non-negligible spatial extent. The bodies start noticing tidal forces.

Nope. It is the force per unit length exerted on a pair of very long wires. Let's look at the magnetic field due to a wire of finite length. From the Biot-Savart law, it is apparent that it is,
a) A function of the length of the wire, and
b) A function of your position both relative to your distance from the wire and your distance away from the ends of the wire.
As it is this field that exerts a force on the second wire, what do you think this variability means in terms of the forces?

Are you saying that the force per unit length is independant of the length of the wires?

No, for the reasons that I provide above.

I'm hoping that my earlier comments explain why this is incorrect. On the other hand, just in case they aren't, have you wondered why the Ampere is defined in terms of an infinite wire? Surely if you are correct, then they could just use wires that are only one metre in length? Is it perhaps that there is some conspiracy involving an international cabal of wire-makers?

By the way, given your apparent distaste for the SI definition of the Ampere, I find it curious that you attempt to adapt it to form the basis of your proof. Surely if you believe it to be wrong, then your proof doesn;t work either?

It is an interaction between a magnetic field generated by one wire, and the current flowing in the other wire. This should have been apparent from my earlier post which gave the equation for the force exerted on moving charge by a magnetic field.

Agreed.

No. The magnetic field is there. There is no force unless we have a moving charge. Notice that I said moving. If the charge is at rest, then there is no force. Take a wire with a current running along it. Place a charge as near to it as you like, and as long as the charge has no velocity, there will be no force exerted on it. (You should have been able to see that in from the equation in my previous post, F = q vxB, where F is the force on charge, q, that is moving with velocity v in a magnetic field, B. "x" is just the usual vector product operator.)

Let's take gravity as another example. What is the force of gravity at the surface of the earth? You can't answer that, because I haven't told you the mass of the object that gravity is acting on. (Mass being the equivalent of charge in Newtonian gravity.) So, what is the force due to a magnetic field?

Find a decent book on electromagnetism (one with at least a few equations) and have a read of it. Hopefully what I have been saying wil lthen make a bit more sense to you.

I'm afraid that this makes no sense, both from the standpoint of EM, and also from the standpoint of Newtonian mechanics. If the field due to the wire is exerting a force, then from Newton's laws, there should be an equal and opposite force exerted on the wire. Where is it?

Apart from issues regarding exactly what these VNCPs are, you run into the same problem that I've already mentioned. If they are static, then they feel no force. (And where are the positively charge particles, or are you claiming that electrical charge isn't conserved?)

It's fine to have a hypothesis, but can you demonstrate that it is consistent with experiment? (I'll explain this later.)

You claim that your quanta of light consist of negatively charged particles. This means that a source of EM radiation must, over a period of time, build up a positive charge. Is there any evidence of this? I'm not counting thermionic emission, as the negative particles (in this instance electrons) remain in the vicinity of the hot filament unless an electric field is applied. If I shine a light on a blackened plate, does it gain a negative charge?

Hopefully you can see that there are a whole bunch of difficulties with your model.

Where? Perhaps you're confused about mu_0, the permeability of free-space, which has a value of 4.Pi*10^-7 N.A^-2 ?
http://scienceworld.wolfram.com/phys...FreeSpace.html

So the Coulomb is a unit of force? If we assume that to be correct (which I don't believe that it is) then the dimensions of the right hand side of your equation is now in units of s^2. This still doesn't look anything like a force. Try doing the dimensional analysis yourself and show me how it is supposed to work, otherwise your equation is meaningless.

It is still important as you do this twice. At the end you talk about the force producing an acceleration on a kg mass (which makes a bit more sense), but why choose a kilogram? Is this the mass of a photon? What mass should you be using here to give you an acceleration? (And why do you appear to have your acceleration acting over the lifetime of the universe even though your force doesn't? I may be misinterpreting you here. If so, it would greatly help understanding if you could clarify what is going on here.)

This still doesn't explain the value of 3. Wouldn't a division by 2 be more likely to give you the average value of the velocity of particle under constant acceleration?

Why use 1 kg as the mass that this force is supposed to be acting on. Why not a mas of 10^-10 kg, or 10^10 kg? If you don't know the mass, then you can't get an acceleration.

Your model of the photon is not consistent with observation (e.g. charge conservation, among other problems.)

You cannot calculate a force unless you know the velocity (and charge) of your hypothetical negatively charged particles.

Even if you could calculate a force, you can't calculate an acceleration if you don't know the mass of the thing that it is supposed to be acting on.

The equations that you employ are incorrect (and I have already shown you what the correct equations are).

The final equation that you derive is dimensionally incorrect, so it is incapable of giving you the quantities that you want. One of the first things that you are taught, in a physics course, is to do a dimensional analysis of any equation that you derive. It is a great way to catch a lot of the errors that you might have made. It is something that you should have done to perform a sanity check on your equation. (Note that if the equation is correct, the dimensions should balance. The converse is not true. Just because the dimensions balance doesn't mean that the equation is correct. )

Even if the EoS was completely false as a theory, that doesn't mean that your theory is true. This is a logical fallacy. I might tell you that the electrons in the atom are actually little demons flying around. Just because the planetary theory is false, wouldn't mean that there really are little demons flying around in there.

cyrek reply

To fortis
'of infinite length'
The formula does not use infinite langth in its format. So why bother?

'of negligible circular cross section'
This term is also not used.

'per meter of length'
This is applicable! The force is based on ONE meter of length of this
infinite wire.
The length of the wire has nothng to do with the forces generated.
The main factor is the number of electrons passing a unit distance per
second that determine the strength of the forces.
In other words, the number of electrons that add up to one ampere.
So, I still say my use of limiting the data to 'one meter' is correct
because this unit of length is used to define the Newton. Using longer
lengths means adding more Newtons or one Newton per meter of length.

Regarding the use of one wire:
This data uses two wires to define the forces generated by one ampere
of current. This is an interaction between two magnetic fields
generated by the currents.
The removal of one wire does not mean that the other wire does not
exist with its magnetic field. It is still there.

fortis wrote
Agreed.

cyrek
It is still generating ONE HALF of the interacting force (one Newton)
for the distance specified.

fortis (excerpts)
No. The magnetic field is there. There is no force unless we have a
moving charge. Notice that I said moving. If the charge is at rest,
then there is no force. Take a wire with a current running along it.
Place a charge as near to it as you like, and as long as the charge
has no velocity, there will be no force exerted on it.

cyrek
Are you tryong to trick me? Where did the charge get into the act? This
is an action between two magnetice fields. The charges are moving in
the wires, remember?
Put a bar magnet near that remaining field and you will feel the force.
It is still there.
In this case, it would be one half meter from the wire that equals one
Newton since the space between the wires is one meter.

fortis
I'm afraid that this makes no sense, both from the standpoint of EM,
and also from the standpoint of Newtonian mechanics. If the field due
to the wire is exerting a force, then from Newton's laws, there should
be an equal and opposite force exerted on the wire. Where is it?

cyrek
What you say above does not make any sense since there is only one wire
and one field.

fortis wrote: 'the force generated by an electron on what?

cyrek (in the HA)
The force is in the magnetic field that would be acting on the 'virtual
negatively charged paticles' to condense them to create a photon of
condensed VNCP's that would constitute the photon.
This, of course, is my hypothesis of the nature of a photon pulse
(quanta).

fortis
Apart from issues regarding exactly what these VNCPs are, you run into
the same problem that I've already mentioned. If they are static, then
they feel no force. (And where are the positively charge particles, or
are you claiming that electrical charge isn't conserved?)

cyrek
These are field particles, not the self destructing space pairs. These
VNCP are responsive to magnetic forces just like electrons are.
The photon is the 'condensed congregate' of these charged particles
that would then be repelling each other to return to their normally
distributed state of tranquility.
There is no comservation problem since charged particles are just being
condensed, not added or subtracted.

fortis
You claim that your quanta of light consist of negatively charged
particles. This means that a source of EM radiation must, over a
period of time, build up a positive charge. Is there any evidence of
this? I'm not counting thermionic emission, as the negative particles
(in this instance electrons) remain in the vicinity of the hot
filament unless an electric field is applied. If I shine a light on
a blackened plate, does it gain a negative charge? Hopefully you can
see that there are a whole bunch of difficulties with your model.

cyrek
build up a positive charge? I do not understand what you are saying
here.
Shining a light on a blackened plate means that the plate would absorb
the energy. The photon particles would just dissipate.

cyrek
Your interpretations of my formula are not accurate. The coulomb is
equivalent to an electrical force even though it is stated as an
electric unit of charge. It is surrounded by the field particles that
create the reactions and that constitute the force.

fortis
So the Coulomb is a unit of force? If we assume that to be correct
(which I don't believe that it is) then the dimensions of the right
hand side of your equation is now in units of s^2. This still doesn't
look anything like a force. Try doing the dimensional analysis
yourself and show me how it is supposed to work, otherwise your
equation is meaningless.

cyrek
When the electron is in motion, it radiates a magnetic field as in a
wire. That is a force.

fortis
At the end you mention the force producing an acceleration on a kg
mass (which makes a bit more sense), but why choose a kilogram?
Is this the mass of a photon?

cyrek
These are dumb questions. The kilogram mass is used in defining the
Newton which is the force that the formula is determining.

fortis
What mass should you be using here to give you an acceleration?
(And why do you appear to have your acceleration acting over the
lifetime of the universe even though your force doesn't? I may be
misinterpreting you here. If so, it would greatly help understanding
if you could clarify what is going on here.)

cyrek
The accelerated electron from the outer to the inner orbit is what
generates the magnetic pulse that compacts the VCFP's that compose the
photon. These compacted particles than slowly repel each other to
return to their normally spaced tranquil state (redshift) and this
would take tens of billions of years for them to reach that state.

fortis
How did you derive the value of three?

cyrek
The value of 3 that divided the final result was an estimate of the
ratio of the electrons accelerated velocity (from outer to inner orbit)
to its standing averaged orbital velocities between the outer and inner
orbits. It varies with wavelength.

fortis
This still doesn't explain the value of 3. Wouldn't a division by 2 be
more likely to give you the average value of the velocity of particle
under constant acceleration?

cyrek
The change in the accelerated electron velocity that generates the
photon pulse is one third of the average of the velocities of the outer
and inner orbits (standing velocities), The electron stops accelerating
after settling back into the inner orbit and resumes a steady standing
velocity.

My final result is the distance this photon would move one kilogram in
meters per s/s. Because this is a photon with a extremely minute force,
it would take hundreds of billions of years, as I have calculated, to
move this weight foor a meaningful redshift.
But in 'free' space, this problem does not exist.

fortis
Why use 1 kg as the mass that this force is supposed to be acting on.
Why not a mass of 10^-10 kg, or 10^10 kg? If you don't know the mass,
then you can't get an acceleration.

cyrek
This result constitutes a ratio of this distance to the length of the
photon in meters to determine the time required to equal one half
wavelength for a redshift of one.
Remember, this result is calculated in Newtons which includes the
kilogram in its definition.

fortis
Your model of the photon is not consistent with observation (e.g.
charge conservation, among other problems.) You cannot calculate a
force unless you know the velocity (and charge) of your hypothetical
negatively charged particles. Even if you could calculate a force,
you can't calculate an acceleration if you don't know the mass of the
thing that it is supposed to be acting on. The equations that you
employ are incorrect (and I have already shown you what the correct
equations are). The final equation that you derive is dimensionally
incorrect, so it is incapable of giving you the quantities that you
want. One of the first things that you are taught, in a physics course,
is to do a dimensional analysis of any equation that you derive. It is
a great way to catch a lot of the errors that you might have made.
It is something that you should have done to perform a sanity check
on your equation. (Note that if the equation is correct, the dimensions
should balance. The converse is not true. Just because the dimensions
balance doesn't mean that the equation is correct.)

cyrek
True, the densities and charges of these VNCP's is unknown. That is a
problem for science to solve.

Since you have not provided any evidence for justifying the EoS concept,
than I am sure my alternative view is correct.

fortis
Even if the EoS was completely false as a theory, that doesn't mean
that your theory is true. This is a logical fallacy. I might tell you
that the electrons in the atom are actually little demons flying
around. Just because the planetary theory is false, wouldn't mean that
there really are little demons flying around in there.

cyrek
Ha Ha Ha. Is that the best you can do in closing your argument?

__________________
aka Michael Cyrek

I'll address the rest of your post in a later post, but I think it is really important to draw one particular point out.
Dimensional analysis isn't just a "nice to have". If your equation doesn't withstand dimensional analysis then it is like writing "2+2=Red". (I was going to write 2+2=5, but failing to balance the dimensions in an equation is far far worse than that.)

If you would like a primer on dimensional analysis, have a look at
http://www.physics.uoguelph.ca/tutorials/dimanaly/
or
http://en.wikipedia.org/wiki/Dimensional_analysis

Let's take a simple example, the formula for the pressure at the bottom of a pool of water. This is (ignoring atmospheric pressure)

Pressure = Rho.g.h,

where Rho is the density of water, g is the acceleration due to gravity, and h is the depth of water. We'll stick with SI units for the following.

Pressure is measured in Pascals, which is just N.m^-2 (a force per unit area), and expanding the Newton, we get the unit of pressure as

kg.m.s^-2.m^-2 or simplifying, kg.m^-1.s^-2 (you might prefer kg/m/s^2.)

Let's look at the right hand side of the equation at the units. We have

Rho is in kg.m^-3
g is in m.s^-2
h is in m

This means that the the right hand side of the equation is in units of

kg.m^-3.m.s^-2.m

which simplifies to

kg.m^-1.s^-2, or kg/m/s^2, which is a unit if pressure.

What happens if we get the equation wrong, for example by claiming
Pressure = Rho.g.h^2 ?

Now you can see that the units of the right hand side don't match the units of the left hand side, and hence the equation is nonsense. (As I said at the start, it is like saying "2+2=Red". )

Apart from the underlying physics behind your proof being wrong (which I shall explain in a later post), your equation makes as much sense as "2+2=Red" because the dimensions don't balance.

Unless, of course, you, can show me how the dimensions of your equation balance.

Please look at Biot-Savart and see why this is important. The formula is based on the assumption that the wire is of infinite length.

Please ask yourself why it is defined in terms of an infinite wire if the length of the wire makes no difference. (I mentioned this in my earlier post...)

Take a wire 1 cm in length. Calculate the magnetic field 1 m away from it. Now make the wire 100 m long. Are you telling me that the magnetic field at that point doesn't change? (Hint. It does. )

Increasing the length of the wire increases the field strength at a point, so this is wrong.

Actually, in my post I only agreed to the last two sentences. I disagreed strongly with the stuff before them. (The way that you've written this implies that I agree with everything in that quote, which is mis-leading.)

No. It is an interaction between a magnetic field and a moving charge. The bar magnet possesses a magnetic dipole due to lots of tiny little current loops, i.e. charges running around in circles. (This is a classical picture to help you picture what is happening. To do it properly requires QM). What you have is the magnetic field exerting a force on these little current loops. This is the origin of what you are interpreting as a force between two magnetic fields. It is not.

Is a force experienced by a static charge in a magnetic field. No. Please read an introductory book on EM. It'll help you to understand how electric motors really work.

What I said, made sense. What you said, did not.

Unless these particles possess spin, then they will not interact with a magnetic field unless they are moving. (Same as your "spinless" electrons.)

I still don't understand why you're focussing on the magnetic field when the electrostatic interactions in these hypothetical systems would probably be far stronger.

What happens to the charge? Let's say I have a hydrogen atom and it absorbs one of your photons. You claim that your photon possesses some (unspecified) quantity of negative charge. What happens to this negative charge? It can't just disappear.

You are taking a force and calculating an acceleration. A 1 N force will cause a 1 kg mass to accelerate at 1 ms^-2, and a 0.1 kg mass to accelerate at 10 ms^-2. If you take the mass of a proton, then a 1 N force will accelerate this at roughly 6*10^26 ms^-2. Don't you think that it is important to know what the mass is?

There is also the interesting question raised by the notion that your VNCPs might have a rest mass, i.e. they have a mass when measured in their rest frame. (You haven't explicitly stated that they do have a finite rest mass, but it seems to be implied by your description of them.) If true, then your photons would not travel at the speed of light.

So the correct force should be the electrostatic interaction (which causes the mutual repulsion) between these particles. Perhaps you should have been trying to calculate this instead.

Can you tell me what the starting and finishing electron velocities are? (Perhaps I'm being a bit dumb here?) How have you derived these quantities?

You need to have a mass term (physically motivated) otherwise your acceleration could take any value, and hence your equation would have no predictive capability.

Any guesses? What would be required in order for your model to give the observed red shift?

You have failed to show me that your model is any better than the flying demons, and until you do you are unlikely to persuade people that it is a viable alternative to GR and the BB.

Well. I won't say anything.

__________________

Feynman
>~~~~<
Science is a way of trying not to fool yourself. The first principle is that you must not fool yourself, and you are the easiest person to fool.

Religion is a culture of faith; science is a culture of doubt.

cyrek reply

fortis wrote
I'll address the rest of your post in a later post, but I think it is
really important to draw one particular point out.

cyrek1 quote:
Your interpretations of my formula are not accurate. The coulomb is
equivalent to an electrical force even though it is stated as an
electric unit of charge. It is surrounded by the field particles that
create the reactions and that constitute the force.

fortis quote
So the Coulomb is a unit of force? If we assume that to be correct
(which I don't believe that it is) then the dimensions of the right
hand side of your equation is now in units of s^2. This still doesn't
look anything like a force. Try doing the dimensional analysis
yourself and show me how it is supposed to work, otherwise your
equation is meaningless.

yrek reply
What balance? This equation is for solving a force. What are you
trying to balance?

cyrek1 quote:
The formula does not use infinite langth in its format. So why bother?

fortis quote
Please look at Biot-Savart and see why this is important. The formula
is based on the assumption that the wire is of infinite length.

cyrek reply
Why? Is not this somewhat virtual or abstract?

cyrek Quote:
'per meter of length'
This is applicable! The force is based on ONE meter of length of this
infinite wire.

fortis quote
Please ask yourself why it is defined in terms of an infinite wire if
the length of the wire makes no difference. (I mentioned this in my
earlier post...)

cyrek Quote:
The length of the wire has nothng to do with the forces generated.

fortis quote
Take a wire 1 cm in length. Calculate the magnetic field 1 m away
from it. Now make the wire 100 m long. Are you telling me that the
magnetic field at that point doesn't change? (Hint. It does. )

cyrek reply
The factors that determine the strength of the magnetic force is
the current passing at that point of magnetic radiation.

cyrek Quote:
Regarding the use of one wire:
This data uses two wires to define the forces generated by one ampere
of current. This is an interaction between two magnetic fields
generated by the currents.
The removal of one wire does not mean that the other wire does not
exist with its magnetic field. It is still there.

fortis quote
Agreed.

fortis quote
Actually, in my post I only agreed to the last two sentences. I
disagreed strongly with the stuff before them. (The way that you've
written this implies that I agree with everything in that quote,
which is mis-leading.)

cyrek Quote:
Where did the charge get into the act?
This is an action between two magnetice fields. The charges are moving
in the wires, remember?
Put a bar magnet near that remaining field and you will feel the force.
It is still there.
In this case, it would be one half meter from the wire that equals one
Newton since the space between the wires is one meter.

fortis quote
No. It is an interaction between a magnetic field and a moving charge.
The bar magnet possesses a magnetic dipole due to lots of tiny little
current loops, i.e. charges running around in circles. (This is a
classical picture to help you picture what is happening. To do it
properly requires QM). What you have is the magnetic field exerting
a force on these little current loops. This is the origin of what you
are interpreting as a force between two magnetic fields. It is not.

cyrek reply
Tiny loops? Are you using monopoles which are not even confirmed to
exist?

fortis quote
Apart from issues regarding exactly what these VNCPs are, you run into
the same problem that I've already mentioned. If they are static, then
they feel no force. (And where are the positively charge particles, or
are you claiming that electrical charge isn't conserved?)

cyrek quote
These are field particles, not the self destructing space pairs. These
VNCP are responsive to magnetic forces just like electrons are.
The photon is the 'condensed congregate' of these charged particles
that would then be repelling each other to return to their normally
distributed state of tranquility.
There is no comservation problem since charged particles are just
being condensed, not added or subtracted.

fortis quote
Unless these particles possess spin, then they will not interact with
a magnetic field unless they are moving. (Same as your "spinless"
electrons.)

cyrek reply
A moving magnetic field will move charged particles. The magnetic field
generated by the electron transition is a moving field increasing in
magnitude and then decreasing as a pulse.

fortis quote
I still don't understand why you're focussing on the magnetic field
when the electrostatic interactions in these hypothetical systems
would probably be far stronger.

cyrek reply
Not true. These 'virtual charged particles' have very weak charges
because of their virtual nature. The interaction now is very weak
electric repulsion, not magnetic.

fortis quote
What happens to the charge? Let's say I have a hydrogen atom and it
absorbs one of your photons. You claim that your photon possesses some
(unspecified) quantity of negative charge. What happens to this
negative charge? It can't just disappear.

cyrek reply
The electron has absorbed the photon energy. The photons concentrated
charged particles are dissipated in bumping the electron.

fortis quote
You are taking a force and calculating an acceleration. A 1 N force
will cause a 1 kg mass to accelerate at 1 ms^-2, and a 0.1 kg mass to
accelerate at 10 ms^-2. If you take the mass of a proton, then a 1 N
force will accelerate this at roughly 6*10^26 ms^-2. Don't you think
that it is important to know what the mass is?

cyrek reply
You are altering the equation requirement that uses the one kilogram in
defining the Newton.

fortis quote
So the correct force should be the electrostatic interaction (which
causes the mutual repulsion) between these particles. Perhaps you
should have been trying to calculate this instead.

cyrek reply
Agreed. But there is no data on these VNCP's as to their masses or
indivudual negative charges. I do not expect there will be any either
in the near future.

fortis quote
Can you tell me what the starting and finishing electron velocities
are? (Perhaps I'm being a bit dumb here?) How have you derived these
quantities?

cyrek reply
The Bohr HA has the fillowing characteristics:
Ground state radius is represented by the lower case Greek a with
sub 0. It equals .53x10^-10 m. Its orbital velocity is 2.19x10^6 m/s.
Higher orbital radiuses are squared. a sub 0 equals n^2 or R sub 2
equals 4 x R sub 1.
Higher orbital velocities are determined by dividing the ground state
velocity by orbital number. V sub n = V sub 1 / n. V sub 2 velocity
would be 1/2 of V sub 1.

fortis quote
Any guesses? What would be required in order for your model to give
the observed red shift?

cyrek reply
I said in the previous post that because the nature of these virtual
particles are unknown, that Euclidean geometry be used by correlating
the redshifts with the 'angular diameters' of the distant objects and
their luminosities reduced as inverely squared with the distance
because 'space is FLAT'.
Their spectral types could be done for classification prrposes.

fortis quote
You have failed to show me that your model is any better than the
flying demons, and until you do you are unlikely to persuade people
that it is a viable alternative to GR and the BB.

cyrek reply
Since the EoS is erroneously modified to make use of the observed
Doppler data, I think it should replace the EoS as the Cosmologocal
redshift.


TravisM
Well. I won't say anything.

cyrek reply
Ha ha ha. I do not blame you Travis

__________________
aka Michael Cyrek

Michael,
I notice that you appear to have almost completely ignored my earlier post, where I ask if you can demonstrate that your equation is dimensionally correct. (Though you do quote the very first line of it) As I state in the post, if the dimensions (i.e. the metres, seconds, kilograms, etc.) on either side of the equation don't balance then the equation is nonsense. It really is as bad as claiming that 2+2=Red. (OR for the surrealists out there, 2+2=Fish. )

When I carried out the analysis of your equation I found that while the left hand side is in Newtons, the right hand side of the equation was not, so the equation is wrong.

Can you demonstrate that the equation is dimensionally correct?
I even explained how to do it. It really is pretty simple, just write down the units of all the bits on the right hand side, and see what you're left with when you start cancelling out terms. If you can't do this, then it doesn't matter how strong or weak (and some of them are pretty weak) your underlying arguments are, it is still wrong.

What do you think?

Well, now I'll chime in. What's wrong with this equation Mike:

1 sec + 4 feet = 10 lbs.

If you can spot the "flaws" I will shut up.

I have a dollar and some pocket lint that says he'll ignore your post fortis and jump all over me with hahahahaha... 8)

And Mike, it's not a matter of blame, so no fretting you're exqusite head over me. I have not been posting lately because it's like trying to jump into a conversation about Poke' Mon... No matter what you ask the 'kid' trying to explain those little guys to you, you're wrong. So, Mike, I give up and won't try to 'help' you any. Because, well you've expressed your ignorance several times on this board, and I am content to let you amble about in la-la land for as long as you want...

Another thing:
Your dismissal of QM and now you have virtual particles? =D>

__________________

Feynman
>~~~~<
Science is a way of trying not to fool yourself. The first principle is that you must not fool yourself, and you are the easiest person to fool.

Religion is a culture of faith; science is a culture of doubt.

A side note:

I was playing connect 4 with my buddies Jacob's little girl, Adrianna. She's only 4 and the box says 7 and up, but Jacob wants his little girl to be a smarty, so we played anyway. We explained the rules to her but she still will put a few peices in and randomly she'll yell out "I win!"

Talkin with Mike reminds me alot of this. I even mentioned this board and this thread specifically to Jacob. He got a good kick out of it.

__________________

Feynman
>~~~~<
Science is a way of trying not to fool yourself. The first principle is that you must not fool yourself, and you are the easiest person to fool.

Religion is a culture of faith; science is a culture of doubt.

TravisM,
Something tells me that you're likely to be right.

I'm currently holding off on addressing the other issues in his post so that he won't be able to sidestep the problem with the units on the right hand side of his equation not matching the units on the left hand side. (It's tough though, as I desperately want to explain to him how a changing magnetic field generates an electric field which can act on stationary particles....)

I took a few years of electronics. I can build digital circuitry rather easily. It starts stressing my abilities when I build an amplifier though, all that analog whackery to deal with. But, I know a bit of how this works myself. 8) Once I understood those theories that are established and how well they are put into practice I stopped questioning the need for such 'complication'. :wink:

__________________

Feynman
>~~~~<
Science is a way of trying not to fool yourself. The first principle is that you must not fool yourself, and you are the easiest person to fool.

Religion is a culture of faith; science is a culture of doubt.

I agree completely. It appears easier and more beautiful to set a new and strange theory than studying the old theories.

The example of the CREIL is striking: using only the old semi-classical theory of the light-matter interactions, one finds that H* (atomic hydrogen in its states 2P or 2S) transfers energy from the electromagnetic beams having a high planck's temperature (generally high frequency) to the colder beams (often thermal radiation).

Searching where it may be some H* in the space, one explains:

- the blueshift of the radio from Pioneers 10 and 11 (the coolong of the solar wind at the limits of the solar system allows a combination of protons and electrons into some H*; thus, energy is transferred from the solar light to the radio).

- the extremely complicated spectrum of the quasars, which appear being accreting neutron stars (in 1980, Varshni wrote that the quasars are in galaxies, the micro-quasars are in our galaxy...)

- The proximity effects

- A lot of other effects : Arp's alignments,...

There is a general rule:
Why do think or make simple while it is easier to do complicated ?

__________________
JMB

It's an interesting approach, and has nothing to do with Michael Cyrek's theories of what is going on (just thought I'd better point that out for MC's benefit ), but is the underlying assumption that the universe is static, or is it conceivable that both CREIL and a BB expansion are happening?

The reason that I ask this is that GR has met pretty much every test that has been performed. (It is even claimed that frame-dragging has now been observed, and hopefully Gravity Probe-B will provide us with unambiguous evidence of this effect.) If we accept that GR is currently our best theory of gravity, and it doesn't appear to allow for any static solutions, then it seems natural to assume that that there may be an expansion or contraction contributing to a cosmological red or blue shift. This suggests a purely CREIL explanation for the redshifts requires new physics (cosmological constant, et al) in order to explain a static universe.

I may be completely missing the point, and I'll take this up on one of the CREIL related threads.
(I don't want to let MC off the hook on explaining how the dimensional analysis of his equation works out. )

The largest part of the redshifts of the quasars happens in the Lyman forest, or for higher z, therefore is intrinsic. The remainder may be produced by some H* in the space, or by a small expansion.
I do not know, but I am a little reluctant about GR because a lot of patches have been added to the initial GR, and they seem more ad hoc than serious. An other reason for my reluctance is the behaviour of many defenders of the big bang who criticize a priori, without any trial to understand.

__________________
JMB

cyrek reply

fortis
regarding the Ampere formula, Yes I did ignore you because I had to
study the formula again.
I then noticed that I made a mistake in assuming that the left side of
the equation was equal to one Newton rather than the permeability
factor 2x10^-7 N/A^2 which the formula equals when everything is
equal to one such as the distance between the wires, length of field
and the current size of one Ampere.
The two sides than balance as you say which I did not quite ubderstand.

But, This does not refute my downsizing the various components to the
atomic level.
It just shows that the force I figured is a lot smallar than the one
Newton I used at first that was being computed. The result would be
based on 2x10^-7 N/A^2. Wow, this is 7 millionths of a Newton which
makes the result seem more miniscule.

Well, its only one photon.

TravisM quote
Well, now I'll chime in. What's wrong with this equation Mike:

1 sec + 4 feet = 10 lbs.

reply
time(1s) + space(4 F) = ST (spacetime gravity)? Ha Ha.

Another thing:
Your dismissal of QM and now you have virtual particles?

reply
I do not dismiss QM. In the Bohr model, like I said, a picture is worth
a thousand words.
The virtual particles are a theory of electric field formations. They
were introduced to me by a PhD while explaining tha electric field
patterns between the electric charges using the Maxwell equations.

So I think this is the accepted theory of these fields.
They are still looking for monopoles.

__________________
aka Michael Cyrek

Not if the picture is wrong.

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"It's all about context!" - Vince Noir (The Mighty Boosh)

"I've never heard of such a brutal and shocking injustice that I cared so little about!" - Zapp Brannigan (Futurama)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

And fortis, I owe you a dollar and some pocket lint.

While Mike, you still owe me the errors. Nice way to re-state the equation, but you still haven't pointed out the flaw(s).

__________________

Feynman
>~~~~<
Science is a way of trying not to fool yourself. The first principle is that you must not fool yourself, and you are the easiest person to fool.

Religion is a culture of faith; science is a culture of doubt.

I'm glad that you had a look at it. As you can see the equation didn't make any sense because it didn't give an answer in Newtons. The N/A^2 should have been a bit of a clue that it describes a force between two currents.

As you only have one current in your model you can never, physically, extract a force.

Well, even if you have an equation with dimensions that balance (left and right), it doesn't mean that the equation is correct. Dimensional balance is a necessary, but not sufficient, condition for an equation being correct. In my earlier post I actually give you the correct equations relating to a moving, non-relativistic, point charge, i.e.
If you want to put your theory on anything like solid foundations, you should start with this.

For example, if we have two electrons moving in parallel with identical velocities, v, and seperated by a distance r, then the magnitude of the force between them is given by

F=e^2.v^2.(mu_0/(4.Pi))./r^2

where e is the charge on an electron. Notice how the force goes as 1/r^2, not 1/r as is implied by the equation that you are trying to use.

How did you go from your first incorrect result to a force in Newtons? What value did you use for the second current?
If you're talking about virtual particles and the EM field then I'm pretty sure that he would have been referring to virtual photons being the mediator of the EM field. These photons are neutral, i.e. they aren't the source of the EM field, that role is reserved for particles with electric charges. If photons had charge (as you claim), then the EM field would look a lot different to the one we currently know and love. Look, for example, at Quantum Chromodynamics (QCD), which is the theory of the strong nuclear force. In this theory the gluons (that bind the quarks together) also carry colour charge (the QCD equivalent of electric charge). This leads to very different behaviour to that which we observe for the EM field. I think that it is highly unlikely that we would get as good a correspondance between QED and the real world if what you suggest was in fact true.

Yup, QED is the accepted theory of the EM field. Unfortunately it doesn't correspond to your model.