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Maxwell's electromagnetic theory of light.

Discussion in 'Mathematics and Physics' started by copernicus1234, Apr 15, 2014.

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  1. copernicus1234

    copernicus1234 Member

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    Equations 2 and 3 are transverse waves. Using the EM transverse equations (equ 2 and 3), in equations 1a,b, results in the equating of the y and z dimensions.
     
  2. steveB

    steveB Well-Known Member Most Helpful Member

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    Your answer to your own question is that Eqs. 2 and 3 describe transverse waves.

    Ey = Eo cos(kx - wt) ...............................Equ 2
    Bz = Bo cos(kx -wt) ................................Equ 3

    My answer is that these equations describe wave propagation in the x-direction.

    I would say you are incorrect when you say that these are necessarily transverse waves. To be sure they are transverse waves you need to further specify that Ex=0 and Bx=0. So we are now able to correct another misconception on your part. Of course, Prof. Susskind, as well as Jenkins and White, as well as every other electromagnetics book is clear on this point. The fact that free space only allows transverse waves falls out of the theory. However, if you solve the wave equation in waveguides, transmission lines, optical fibers etc. you will discover that there are various types of wave solutions including, transverse, transverse magnetic, transverse electric and hybrid modes.

    So, you keep saying this, despite all evidence to the contrary. You are incorrect as I already showed.

    Let's write one of those equations, and derive the consequences correctly.

    dEy/dx = - (1/c)(dBz/dt).............................equ 1

    Here we are equating derivatives of field components, and not y and z dimensions. Nowhere in that equation does "vector" information or "direction" information show up.

    Likewise, equations Eq 2 and Eq 3 do not have vector information, nor direction information in them. These equations are all relations between the scalar field components. The fact that these scalar values are the components of the vector fields is irrelevant to this part of the derivation/discussion. Since direction information is not in these equations, it is impossible for them to be "equating the y and z dimensions".

    Let's go step by step to see what these equations really tell us. They tell us something interesting and useful, not something contradictory to common sense, as you keep insisting.

    Differentiating eqs. 2 and 3 gives.

    dEy/dx = - k Eo sin(kx - wt) ...............................Equ 4
    dBz/dt = +w Bo sin(kx -wt) ................................Equ 5

    Then substituting in to eq. 1 we obtain.

    - k Eo sin(kx - wt) = - (w/c) Bo sin(kx -wt) .............................equ 6

    The sine dependence is then redundant and can be removed as follows.

    k Eo = (w/c) Bo .............................equ 7

    This leads to the relation.

    Bo = k c Eo /w .............................equ 8

    And in free space, the dispersion relation is well known to be w/k = c which then leads to the result that

    Bo = Eo

    Hence, this is not leading to an equating of the y and z dimensions, but is instead leads to an equating of the field component amplitudes Eo and Bo. In other words we are discovering the wave impedance of free space. Your approach was to force Bo=Eo and then conclude that j=k. However, the right approach is to discover that Bo=Eo, and realize that this equation says nothing about j=k. Indeed, the equations were formulated under the assumption that j and k are orthogonal.

    The result of "unity" or "one" for the ratio Eo/Bo is a result of the chosen units systems, and we (in this forum) would typically use SI units, as most engineers do. If we had started with SI units, we would have discovered that Eo/Bo = c. Also, typically we would look at impedance as E/H which gives Zo=Eo/Ho=uEo/Bo=uc which is about Zo=377 Ohms. So this all leads to the identification of the impedance of freespace, not the "collapse" of free space via a j=k, k=i, i=j implosion.

    So congratulations to you for completely screwing up the whole works. Instead of discovering the wave impedance of free space, as you were supposed to do, you instead discovered that our "space" is broken and j=k.
     
    Last edited: May 6, 2014
  3. copernicus1234

    copernicus1234 Member

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    steveB........"I would say you are incorrect when you say that these are necessarily transverse waves."


    Ey = Eo cos(kx - wt) ...............................Equ 2
    Bz = Bo cos(kx -wt) ................................Equ 3


    copernicus1234..........then what do the subscripts y an z, of equations 2 and 3, represent? In addition,

    kx - wt = 0..........................4

    which also proves the derivation of equations 2 and 3 are physically and mathematically invalid.
     
    Last edited: May 6, 2014
  4. dave

    Dave New Member

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  5. steveB

    steveB Well-Known Member Most Helpful Member

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    The subscripts are labels which makes it clear which component of the field vector you are talking about. Maxwell never used subscripts like this. He used variable names like X, Y and Z, or A, B and C. The variables are scalar quantities. Their values are used to construct the field vector. The fields are vectors as usually written, but the equations become scalar equations when expanded.

    Strictly, our notion that the E and B fields are vectors is not quite accurate. The field components actually are components to a rank 2 tensor in 4-D. That is, the electromagnetic field tensor has 6 independent scalar values that are used to make the field tensor F.

    So strictly, Maxwell's equations are tensor equations if you want to be more accurate. But, whether you formulate Maxwell's equations as a collection (2) of tensor equations, as physicists do; or as a collection (4) of vector equations as engineers do, or as a collection of scalar equations between the field components, as Maxwell did, it all amounts to the same relations once you get to solving a real problem.

    This just proves that you are a troll, not that anything is mathematically invalid. The only way for kx-wt to be zero always for all x and all t, is if k=w=0. For free space waves, k=w/c and kx-wt=w(x/c-t).

    OK, what other nonsense would you like to shout out to the world?
     
  6. copernicus1234

    copernicus1234 Member

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    steveB........."The subscripts are labels which makes it clear which component of the field vector you are talking about." and "Maxwell never used subscripts like this. He used variable names like X, Y and Z, or A, B and C. The variables are scalar quantities."


    -----------------------------------------------------------------

    Usually, a field vector has a direction which makes it a field vector and not a scalar since a scalar does not have a direction. Also, Maxwell describes polarization using transverse waves, in his electromagnetic theory of light. Also, I distinctly remember you stating that Maxwell's equations are vector equations yet are insisting that the EM wave equations are not transverse waves that a vector equations.

    -----------------------------------------------------------------
     
    Last edited: May 6, 2014
  7. copernicus1234

    copernicus1234 Member

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    steveB..............The only way for kx-wt to be zero always for all x and all t, is if k=w=0.


    -----------------------------------------------------------


    steveB what JC did you take your math and phyiscs classes from? Or did you get your math and physics from the McDonlds or Jack in the Box...Did you get a Big Mac or a taco Supreme with your onion skin. Coke or Pepsi? Paper or plastic? Ketchup or Mayo?

    k = w ????? (have you been smoking that funny green stuff?)

    more like kx = wt............................................................1

    using

    k = 2pi/l (where l is the wavelength)........................2

    and

    ct = x...............................................................................3

    in

    kx = (2pi/l) (ct)................................................................4

    Using

    f x l = c (where f is the frequency).................................5


    in equation 4

    kx = (2pi/l) (f x l) t...........................................................6

    Equation 6 becomes,

    kx = (2pi)f t........................................................................7

    Since wt = (2pi)t

    equation 7 becomes,

    kx = wt..........or.............kx - wt = 0..................................8

    Who's the one full of nonsense. Scalar vector fields, come on steveB wakeup! I'm doing circles around you.
     
    Last edited: May 6, 2014
  8. steveB

    steveB Well-Known Member Most Helpful Member

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    Always a field vector has a direction, unless it is the zero vector. It is not a scalar, and yes a scalar does not have a direction. The point you keep missing is that a vector E can be represented with components as Ex i + Ey j + Ez k. And it is the component values Ex, Ey and Ez that are scalar quantities. These have no direction. It's very simple, yet you want to keep pretending you don't understand it. Everyone understands this.

    Correct.

    Yes, I did say that, even though before you lied in the other thread and said that I said there are not vector equations. But, thank you for correcting your lie. They are vector equations, or tensor equations, depending on how you want to express them.

    Again, you are lying. Where did I insist that the EM wave equations are not transverse waves. Show me where I said that. Maxwells' equations require free space waves to be transverse waves. You know very well that I said that, but you enjoy the trolling process. No one is fooled by your nonsense, - least of all me. But, I will point out your lies as many times as you spout them.

    It is possible to have non-transverse waves in waveguides, but free space allows only transverse wave solutions.
     
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  9. steveB

    steveB Well-Known Member Most Helpful Member

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    The sad thing here is that you are not even a good troll. By k=w=0 I just mean a short-hand notation for k=0 and w=0, which is the only way for kx-wt to be equal to zero for all possible values of x and t. Very simple, and anyone can see it.

    Your claim is kx-wt equals zero. We know in free space k=w/c which leads to kx-wt=w(x/c-t). Take the example of w=5 rad/s. We know the value of c is a constant, so, you would say that 5(x/c-t)=0, which is obviously not possible for all values of x and t. But if w=0 (static case), then the relation holds. If w=0 then it follows that k=0.

    Very simple.

    Your comments about JC don't bother me, as I know many JC grads that would run circles around you. I could tell you about my credentials, which include a Ph. D. in technical studies from an accredited University, but then as a troll you would insult my school, professors and everyone else. Well, that's all meaningless because you reject Maxwell himself. So, please keep calling me an idiot. I'm happy to be included in your grouping of idiots along with Maxwell and Susskind, as well as Einstein, Dirac and a myriad of other famous people that were not smart enough to see your "unit vector catastrophe".

    OK, your turn ... what's next troll?
     
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  10. copernicus1234

    copernicus1234 Member

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    "Always a field vector has a direction, unless it is the zero vector. It is not a scalar, and yes a scalar does not have a direction. The point you keep missing is that a vector E can be represented with components as Ex i + Ey j + Ez k. And it is the component values Ex, Ey and Ez that are scalar quantities. These have no direction."

    Then how can equations 2 and 3 represent an EM transverse wave. Also, the derivation is based on Only Ey, for the electric field. You're getting your ass wopped.
     
  11. copernicus1234

    copernicus1234 Member

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    steveB.................."Your answer to your own question is that Eqs. 2 and 3 describe transverse waves.

    Ey = Eo cos(kx - wt) ...............................Equ 2

    Bz = Bo cos(kx -wt) ................................Equ 3

    My answer is that these equations describe wave propagation in the x-direction.
    I would say you are incorrect when you say that these are necessarily transverse waves." (post #42).

    ------------------------------------------------------

    Are equations 2 and 3, in free space EM transverse waves. Note, Maxwell's equations are free space equations not wave guide equations. You are avoiding the question since the EM transverse wave equations results in the equating the the y and z directions which is a catastrophe.
     
  12. steveB

    steveB Well-Known Member Most Helpful Member

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    I'm not avoiding any questions. I've answered them all, and correctly too. You have avoided every answer and continue to spout your nonsense that is contradictory to 150 years of established science. You keep saying that the y and z directions are equated and it results in a catastrophe despite my very clear descriptions that make sense to the rest of the world. Those equations related the derivatives of the field components, not the y and z direction. Keep saying your nonsense if you like, and keep saying I'm wrong, or avoiding the question, but it's all nonsense. Everyone sees it. You've posted on 4 forums that I know of, and probably more. No one agrees with you, and you contradict Maxwell, Einstein, and 1000 other famous physicists. How does it feel to be so smart and alone in the world?

    Since you can't make a reasonable technical argument you resort to nonsense statements like, "You're getting your ass wopped.". Do you think this kind of talk can make your stupid ideas right? It doesn't work that way, but it does reveal an adolescent mind.
     
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  13. copernicus1234

    copernicus1234 Member

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    "Maxwell's electrodynamics proceeds in the same unusual way already analysed in studying his electrostatics. Under the influence of hypotheses which remain vague and undefined in his mind, Maxwell sketches a theory which he never completes, he does not even bother to remove contradictions from it; then he starts changing this theory, he imposes on it essential modifications which he does not notify to his reader; the latter tries in vain to fix the fugitive and intangible thought of the author; just when he thinks he has got it, even the parts of the doctrine dealing with the best studied phenomena are seen to vanish. And yet this strange and disconcerting method led Maxwell to the electromagnetic theory of light!" (Duhem, 1902).

    Duhem, Pierre. Les theories electriques de J. C. Maxwell. Paris. 1902.






    Putin
     
  14. steveB

    steveB Well-Known Member Most Helpful Member

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    It's certainly interesting to study the scientific debates of both the past and present. You've highlighted a criticism from a contemporary of Maxwell. The presence of criticism and debate is nothing new in science, although I know you are presenting the information as though it means something unusual. Previously in this thread, I highlighted the more recent debate/criticisms between Hawking and Susskind with regards to the "Black Hole War". This is the way science is supposed to work. All work must be criticized and debated and then the best theories eventually survive.

    Many people including Maxwell himself, along with all others at that time tried to figure out the real meaning and implications of the field equations. Some of Maxwell's ideas on how to reconcile the new theory with the old were wrong (for example his idea of a "luminiferous aether" as a medium for wave transmission). But, the theory was right even though the implications of the theory took time to work out.

    This is just an old outdated criticism, and it does not diminish the great work of Maxwell. Maxwell's work has withstood the test of time, and only in more recent times has quantum electrodynamics provided a more complete theory. Of course, at that time (1902, which is pre-relativity times) there were contradictions between electromagnetic theory and other classical theories. Indeed the very presence of contradictions with a new theory provides the basis to see if it is right or wrong. And, it was Einstein who showed that contradictions were flaws in the other theories, not in electromagnetism. Einstein solved these riddles with the introduction of relativity theory. Also, Einstein's work was heavily criticized, by just about all physicists. But, Einstein's and Maxwell's theories have both withstood the test of time and it is their classical theories that are the accepted classical theories (i.e. electromagnetics, special relativity and general relativity). Of course, quantum mechanics and quantum field theory are the better theories overall, but the classical theories are useful approximations when quantum effects are negligible.

    Anyway, your quote is a nice reminder of how the scientific process works to lead to new and better theories, and it provides a great lesson about a truism that I'll highlight with the following wording ...

    "True pioneers should be concerned with the critics of tomorrow, not those of today. Critics use the past to judge what they see, while the pioneer sets new standards for the future."

    So, are you a pioneer or a charlatan? The difference between the nonsense you present and the insight a true pioneer provides is that your ideas are based on old theories that were disproved so very long ago, you make basic fundamental errors in your work and can't even see them when they are pointed out, and you ignore the existing known data that disproves your theories. A real pioneer gives a new theory that makes predictions that can be tested experimentally. For example, Maxwell's equations revealed the Lorentz transformation which disagreed with the accepted Galilean transformation, but it was proved right experimentally. Maxwell's equations predicted the constancy of the speed of light, which was proved right experimentally. Einstein made many predictions from his new General Relativity theory, and one by one each prediction was verified experimentally.
     
    Last edited: May 11, 2014
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  15. copernicus1234

    copernicus1234 Member

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    Leibniz is the real genius not Newton. Maxwell-Einstein are abusing Leibniz's derivative.
     
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  16. steveB

    steveB Well-Known Member Most Helpful Member

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    Leibniz was a real genius. So was Newton. Maxwell and Einstein were as well. You are not.

    Your mistakes I've corrected in this thread are proof that you are not qualified to make these kinds of judgments.
     
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  17. copernicus1234

    copernicus1234 Member

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    I didn't post this new version of my recent findings and I will be refering to parts of this post in regards to the unit vector catastrophe in the next post. The new version is essentially the same as the shorter version except I introduced a new and improve method in describing the expansion and gradiant methods using * and **. Also, I clarified the ending by streamlining the discussion regarding the concealment strategy by disgarding one of the two equation references but I may at some latter date re-include these equations, if there are any suggestion that make things clearer. Also, I was thinking a adding Poynting's energy equation of light since it is related to Faraday or more like Ampere. Someday there may not be any need for capacitors (Faraday), with chips but there will always be current (Ampere). Isn't Ampere-Faraday relationship interesting. It was tough being great, back than, as it is today. What do you think ronsimplson?




    Maxwell's electromagnetic theory of light is based on Faraday induction experiment that is not optical. Planck uses the blackbody radiation effect that emits the radio induction effect and light to structurally unify induction with light but the blackbody light emission is not an induction effect since electrons are released from the blackbody surface when light is emitted. Lenard's photoelectric effect proves light is composed of particles which contradicts the lateral continuity of Maxwell's EM induction field. In addition, the wave effects and velocity of the radio induction effect does not justify Maxwell's theory since induction is not optical. Furthermore, the derivation of Maxwell's EM wave equations of light, using Maxwell's equation, based on the expansion* and gradient** methods are patently incorrect.
    **In the gradient method, http://en.wikipedia.org/wiki/Electromagnetic_wave_equation , a vector identity, that produces a second order gradient, results in the derivation of the EM horizontal wave equations of light that contradicts Maxwell's transverse waves that is used to represent polarization.


    -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------


    *In the expansion method (Jenkins, Francis and White, Harvy. Fundamentals of Optics. 3rd ed. McGraw-Hill. 1957. p. 410), an EM transverse wave equations of light are derived. After Maxwell's equations are expanded, 16 of the first order differential field components are eliminated to produce,


    dEy/dx = - (1/c)(dBz/dt] ........... - dBy/dx = (1/c)(dEz/dt)..............Equ 1a,b


    that are used to derive


    (d"E/d"x) = c(d"E/d"t).......................................................................Equ 2


    Equation 2 is used in the derivation of the x-direction EM transverse wave equations of light.


    Ex = Eo cos(kx - wt)j..............Bz = Bo cos(kx -wt)k...........................Equ 3a,b


    Using the electromagnetic transverse wave equations, in equations 1a forms,


    d/dx (Eo cos(kx - wt)j) = - (1/c) d/dt(Bo cos(kx - wt)k).......................Equ 4


    Using Bo = Eo, in equation 4 forms,


    j = k (unit vectors). ........................................................................Equ 5


    Equation 1b also produces the unit vector of catastrophe (equ 5). In addition, Condon also uses the expansion method to derive the EM transverse wave equations of light but neglects the representation of equations 1a,b (Condon, Handbook of Physics. McGraw-Hill. 1958. 4-108). Furthermore, Hecht also uses the expansion method to derive the EM transverse wave equations of light and also neglects the representation of equations 1a,b (Hecht, Eugene. Optics. Addison-Wesley. 4th ed. p. 44). In general, physicists uses the expansion method or the divergence method but physicists are intensionally concealing an extremely important and critical fact of the unit vector catastrophe because the wave theory of light is the foundation of modern theoretical physics................................................................................................................................................................................---------------------------------------------------------------------------------------------------------------------.....
     
    Last edited: May 13, 2014
  18. copernicus1234

    copernicus1234 Member

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    Einstein's relativity is based on Maxwell's equations derived using Faraday's induction experiment that is not optical. Planck uses the blackbody radiation effect's radio induction effect and light emissions to structurally unify induction with light but the blackbody light emission is not an induction effect since electrons are released from the blackbody surface when light is emitted. Lenard's photoelectric effect proves light is composed of particles which contradicts the lateral continuity of Maxwell's EM induction field. Mr. Pierre Duhem agrees with my findings:


    "Maxwell's electrodynamics proceeds in the same unusual way already analysed in studying his electrostatics. Under the influence of hypotheses which remain vague and undefined in his mind, Maxwell sketches a theory which he never completes, he does not even bother to remove contradictions from it; then he starts changing this theory, he imposes on it essential modifications which he does not notify to his reader; the latter tries in vain to fix the fugitive and intangible thought of the author; just when he thinks he has got it, even the parts of the doctrine dealing with the best studied phenomena are seen to vanish. And yet this strange and disconcerting method led Maxwell to the electromagnetic theory of light!" (Duhem, 1902).






    Duhem, Pierre. Les theories electriques de J. C. Maxwell. Paris. 1902.


    Note: Einstein does not use the EM wave equations of light to avoid the unit vector castrophe.
    Putin
     
  19. steveB

    steveB Well-Known Member Most Helpful Member

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    Since all you have done is repeat your previous nonsense, I will just repeat my previous correction. Your conclusions are wrong because your analysis is wrong. You don't even understand that your statement that "induction is not optical" is devoid of any substance or meaning.

    I think we've reached a point where you are deserving of the lowest form of criticism we can give you about your response. "Mr Madison, everyone in this room is now dumber for having listened to it"

     
    Last edited: May 13, 2014
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  20. tvtech

    tvtech Well-Known Member Most Helpful Member

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    Hi Steve

    You keep it accurate, informative, factual and on track with these supposedly "informative discussions". Way out of my league...but good to see someone who knows how to deal with a Troll nicely on an intellectual basis.

    You know how to let a Troll hang itself.

    And that is an Art that cannot be taught :cool:

    I have lots to learn.

    Regards,
    tvtech
     
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  21. copernicus1234

    copernicus1234 Member

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    Your correction is a follows:


    Ey = Eo cos(kx - wt) j ...............................Equ 2

    Bz = Bo cos(kx -wt) k ................................Equ 3


    I'm not going to go into all your confusing comments and misinterpretations of what I said. I'm not having trouble with these derivations at all. I mastered them years ago. You are the one that is having trouble with the derivation, which is why you arrived at your nonsensical conclusion that there is a unit vector catastrophe, and this is why you have posted this question on 3 or more technical forums and cant get one person in the world to agree with you.

    I already explained, but I can try again. Let's put it in a nutshell. Here is your mistake in bold underlined font ...

    Equations 2 and 3 above are incorrect. (imagine the sound of trumpets as you read that)

    If you think they are correct, please provide a reference for them.

    The correct equations are as follows.

    Ey = Eo cos(kx - wt) ...............................Equ 2
    Bz = Bo cos(kx -wt) ................................Equ 3

    Notice that I left the j and k unit vectors out of the equations because including them is mathematical nonsense, in the context of this derivation.

    Now, use these correct equations and you will not bump into your unit vector catastrophe.

    Why is it nonsense? Well a vector that is represented as E is writen as E = Ex i + Ey j + Ez k and if we substitute your equations with your suggestion of Ey = Eo cos(kx - wt) j, Ex=0 and Ez=0, then E is ...

    E
    = Ey j = (Eo cos(kx - wt) j) j

    The presence of two j unit vector reveals the nonsense of your viewpoint. The presence of two unit vectors would normally indicated a dyad (http://en.wikipedia.org/wiki/Dyad_product) and would imply a tensor of higher rank than a vector. This is utter nonsense in the context of this derivation.

    So, I ask you to provide your reference for equations 2 and 3, and then we can track down your mistake. Did another author mislead you or did you misinterpret another author? It might be useful to identify which is the case, and we can do so, if you provide your reference. I pointed out that your eq. 2 and 3 do not appear in Jenkins & White, while the correct versions I showed do show up there.
     
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