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Green Destiny
2010-Nov-30, 05:45 AM
Calculating Rates in Magnetic Moments to Predict Period Interaction Between a Fermion and the Zero-Point Field

First we must identify the net magnetizationhttp://www.codecogs.com/eq.latex? \vec{M} as the sum of individual magnetic moments [1],

1. http://www.codecogs.com/eq.latex?\vec{M}= \sum^{N}_{i=1} \vec{\mu}_i

The magnetic moment http://www.codecogs.com/eq.latex?\vec{\mu}_i is a magnitude of the constant expression of http://www.codecogs.com/eq.latex?\frac{e \hbar}{2Mc}. If these equations where pertaining to a spinor field \psi (or simply an electron) then the electron in a certain magnitude of magnetic moments act like a clock to potentially absorb a photon at an energy of http://www.codecogs.com/eq.latex?\frac{e^2 h}{4Mc} [2] from the ZPF (zero-point field) which would imply an interaction term. Using the algebra of a limiting vector on the classical vector component http://www.codecogs.com/eq.latex?\vec{\mu}_i then we would find the upper bound at the squared value of http://www.codecogs.com/eq.latex?\vec{\mu}^2.

Green Destiny
2010-Nov-30, 05:46 AM
Taking the dot product of Equation 1. we have:

http://www.codecogs.com/eq.latex?F= \nabla(\vec{M} \cdot B)= \nabla(\sum^{N}_{i=1} \vec{\mu}_i \cdot B)

which implies also that

http://www.codecogs.com/eq.latex?\nabla(\sum^{N}_{i=1} \vec{\mu}_i \cdot B)= \nabla(MBcos \theta)

The force here can now be viewed in terms of a Lorentz force where the charge is seen in terms of magnetism:

http://www.codecogs.com/eq.latex?\nabla(\sum^{N}_{i=1} \vec{\mu}_i \cdot B)= (\oint \vec{\mu_i}M \cdot \partial \ell)v \times B

http://www.codecogs.com/eq.latex?= q_Mv \times B

Which would imply a force due to magnetism http://www.codecogs.com/eq.latex?F_M, where http://www.codecogs.com/eq.latex?q_M is the magnetic charge. Since the sum of magnetic moments calculate the exact absorption rate of photons when electrons have an energy of http://www.codecogs.com/eq.latex?\frac{e^2h}{4Mc} then the magnetization can be seen as giving rise to the interaction between the electron and the zero point field - remember, the magnetization is the sum of the magnetic moments, and the magnetic moment is a magnitude of one half less than that required for zero-point energy absorption. The rate in which an electron may obtain an upper bound of energy at http://www.codecogs.com/eq.latex?hf=15MeV and their respective Magnetizations should be investigated as possible co-roles.

Ref.

http://www.pma.caltech.edu/~ph77/labs/nmr.pdf

macaw
2010-Nov-30, 06:38 AM
Taking the dot product of Equation 1. we have:

http://www.codecogs.com/eq.latex?F= \nabla(\vec{M} \cdot B)= \nabla(\sum^{N}_{i=1} \vec{\mu}_i \cdot B)

Q1: How did you derive the above?
Q2: what sort of force is F?
Q3: Can you define the action of the operator http://www.codecogs.com/eq.latex? \nabla?

The force here can now be viewed in terms of a Lorentz force where the charge is seen in terms of magnetism:

http://www.codecogs.com/eq.latex?\nabla(\sum^{N}_{i=1} \vec{\mu}_i \cdot B)= (\oint \vec{\mu_i}M \cdot \partial \ell)v \times B

http://www.codecogs.com/eq.latex?= q_Mv \times B

Q4: Why would F be a Lorentz force?
Q5: How did you derive the above ?
Q6: Where is http://www.codecogs.com/eq.latex? \partial \ell coming from? What physical entity does it represent?
Q7: Where did the speed v show up in the formulas since it is not present anywhere else?
Q8: Where did the crossproduct come from?
Q9: What does the symbol http://www.codecogs.com/eq.latex? \oint and how did it show up in the expression?
Q10: You "lost" the http://www.codecogs.com/eq.latex? \nabla operator, is this a mistake or are you simply stringing arbitrary symbols to demonstrate your ability with latex?
Q11: Seems like you are making up electrical charge out of nothing. This can't be. You didn't know that?

Green Destiny
2010-Nov-30, 06:54 AM
Q1: How did you derive the above?
Q2: what sort of force is F?
Q3: Can you define the action of the operator http://www.codecogs.com/eq.latex? \nabla?

Q4: Why would F be a Lorentz force?
Q5: How did you derive the above ?
Q6: Where is http://www.codecogs.com/eq.latex? \partial \ell coming from? What physical entity does it represent?
Q7: Where did the speed v show up in the formulas since it is not present anywhere else?
Q8: Where did the crossproduct come from?
Q9: Seems like you are making up electrcal charge out of nothing. This can't be?
Q10: What is your knowledge of basic electromagnetism?

Some of these questions would be embarrasing normally.

Q1 - if you follow the links provided, spent a little time reading them, you will see all the relevent information on the equation. As for Q2 and Q3 here is where you will find more relevent information http://en.wikipedia.org/wiki/Magnetic_field - are you sure you weren't aware of a Lorentz force?

Q4 is has an obvious answer if you have followed so far. Deriving this is simple if you make a substitution on the charge for being the magnetic charge as you will see using the loop integral. Q6 that is distance. Q7 again, how come you do not know about the Lorentz force qv x B = F? Q8 from years of mathematics of course. Q9 Em no, if you can follow this, you would not have said this. If you cannot see how F equals the sum of the magnetization, or simply the dot product of the magnetic moment and the magnetic field, then I don't know what to say. The gradient simply measures the strength between \mu_i and B. Q10 I was about to ask you the same thing funnily enough.

Green Destiny
2010-Nov-30, 07:03 AM
Q10 Yes that is a mistake.

(The rest of the questions are all the same and I have answered them.)

macaw
2010-Nov-30, 07:05 AM
Some of these questions would be embarrasing normally.

Q1 - if you follow the links provided, spent a little time reading them, you will see all the relevent information on the equation.

As for Q2 and Q3 here is where you will find more relevent information http://en.wikipedia.org/wiki/Magnetic_field - are you sure you weren't aware of a Lorentz force?

I am fully aware of the Lorentz force, what you put down as a "derivation" is another form of nonsense, this is why I asked you step by step questions, so would you please answer them as asked?

Q4 is has an obvious answer if you have followed so far. Deriving this is simple if you make a substitution on the charge for being the magnetic charge

How could that be? Electric charge and magnetic momentum are totally unrelated entities, are you rewriting basic electrodynamics?

as you will see using the loop integral. Q6 that is distance.

Distance? Why the partial derivative sign? Partial derivative with respect to what? Is it a vector or a scalar? Where is it coming from?

Q7 again, how come you do not know about the Lorentz force qv x B = F?

I know very well, the question is about your bogus "derivation". How do you arrive to the formula?

Q8 from years of mathematics of course.

But your expression are all wrong. So what "years of mathematics" do you have?

Q9 Em no, if you can follow this, you would not have said this. If you cannot see how F equals the sum of the magnetization, or simply the dot product of the magnetic moment and the magnetic field,

Well, F is a vector, so it cannot be equal to a dot product. Try again.

. Q10 I was about to ask you the same thing funnily enough.

Q12: Are you copying stuff from this (http://en.wikipedia.org/wiki/Magnetic_field) wiki page while introducing your own mistakes? It would appear that this is what you do in all your ATMs (copy a random page from a mainstream site, introduce aseries of hilarious mistakes and call it your ATM).

macaw
2010-Nov-30, 07:06 AM
Q10 Yes that is a mistake.

(The rest of the questions are all the same and I have answered them.)

I don't think so, you are simply making up bogus formulas riddled with mistakes.

Green Destiny
2010-Nov-30, 07:08 AM

I am fully aware of the Lorentz force, what you put down as a "derivation" is another form of nonsense, this is why I asked you step by step questions, so would you please answer them as asked?

How could that be? Electric charge and magnetic momentum are totally unrelated entities, are you rewriting basic electrodynamics?

Distance? Why the partial derivative sign? Partial derivative with respect to what? Is it a vector or a scalar? Where is it coming from?

I know very well, the question is about your bogus "derivation". How do you arrive to the formula?

But your expression are all wrong. So what "years of mathematics" do you have?

Well, F is a vector, so it cannot be equal to a dot product. Try again.

You will find all relevent equations in the links provided. And I am not rewriting electrodynamics, thank you.

macaw
2010-Nov-30, 07:13 AM
You will find all relevent equations in the links provided.

And I am not rewriting electrodynamics, thank you.

Sure you are. While inserting some very amusing mistakes that show that you don't understand what you are cutting and pasting.

Green Destiny
2010-Nov-30, 07:20 AM

Sure you are. While inserting some very amusing mistakes that show that you don't understand what you are cutting and pasting.

One question at a time then.

Let us concentrate first on equation 1. Have you found its source yet? bothered to read the links like I respectfully asked you?

macaw
2010-Nov-30, 07:24 AM
One question at a time then.

Let us concentrate first on equation 1. Have you found its source yet? bothered to read the links like I respectfully asked you?

The equation is bogus, you don't have any links . Please stop the diversions and answer the questions

Green Destiny
2010-Nov-30, 07:26 AM
In the OP there are references at the very end. You are starting to annoy me.

Green Destiny
2010-Nov-30, 07:26 AM
Well, OP part 2 since the message cannot contain any more than 15 images

Green Destiny
2010-Nov-30, 07:28 AM
I mean questions like this, make me not even want to continue

'' Electric charge and magnetic momentum are totally unrelated entities, are you rewriting basic electrodynamics?''

Who did that? I said a substitution on q for q_M which gives you the necessery dimensions to rewrite the charge in terms of the loop integral. Who said anything about a magnetic momentum as being the same as an electric charge? It's a magnetic charge, not momentum.

I am going to report you - you have a basic duty to ask coherent questions with the added effort to understand why work is presented the way it is. If you cannot do this, you shouldn't be allowed to spam peoples threads like this.

macaw
2010-Nov-30, 07:30 AM
[b]

If these equations where pertaining to a spinor field \psi (or simply an electron)

Q13: Spinor fields and electron? In the same sentence? What is the connection?

then the electron in a certain magnitude of magnetic moments act like a clock

Q14: How can a magnetic momentum "act as a clock"?

to potentially absorb a photon at an energy

Q15: What is absorbing the photon? The magnetic momentum or the clock?

of http://www.codecogs.com/eq.latex?\frac{e^2 h}{4Mc} [2] from the ZPF (zero-point field)

Q16: Did you notice that the unts don't match?

Green Destiny
2010-Nov-30, 07:34 AM
Q13: The units don't match , how do you explain that?

Q14: Spinor fields and electron? In the same sentence? What is the connection?

Q15: How can a magnetic momentum "act as a clock"?

Q16: What is absorbing the photon? The magnetic momentum or the clock?

Q17: Did you notice that the unts don't match? Again.

From one question to another, and not even the balls to address your obvious contempt for misinformation concerning links, misidentification of variables. And you expect me to entertain these walls of words?

If by units don't match, do you mean in the link it was extracted from? Alpha is simply the fine structure constant, it is dimensionless.

Green Destiny
2010-Nov-30, 07:37 AM
You don't know how spinors relate to electrons?

I don't know if you any quantum field theory, but some of us know of an equation:

Dψ=dψ+ieAψ

if ψ is some spinor field (an electron) then ieAψ is an interaction term on the gauge field A where e is the presence of the charge of the particle. It is an inherent property of a moving mass where e is not equal to zero, and it is an intrinsic property for matter, taking values of either -1 or +1.

macaw
2010-Nov-30, 07:38 AM
Ref.

http://www.pma.caltech.edu/~ph77/labs/nmr.pdf

Q17: So, you are now cutting and pasting stuff that you don't understand from a homework page? Is this the starting point of your new ATM?

macaw
2010-Nov-30, 07:40 AM
You don't know how spinors relate to electrons?

I don't know if you any quantum field theory, but some of us know of an equation:

Dψ=dψ+ieAψ

if ψ is some spinor field (an electron) then ieAψ is an interaction term on the gauge field A where e is the presence of the charge of the particle. It is an inherent property of a moving mass where e is not equal to zero, and it is an intrinsic property for matter, taking values of either -1 or +1.

Q18; What does the above have to do with your botched attempt to derive the Lorentz force? You appear to think that electrons and spinor fields are interchangeable. At least, you use them the same way.

Green Destiny
2010-Nov-30, 07:40 AM
Q17: So, you are now cutting and pasting stuff that you don't understand from a homework page? Is this the starting point of your new ATM?

I think its a little more complicated than simply cutting and pasting. You've failed to see any relevence in anything. Not even connections in field algebra to spinors and electrons. You fail in all aspects, so I don't expect you for one minute to understand how the term in the link relates at all to the sum magnetization of a collection of magnetic moments.

Green Destiny
2010-Nov-30, 07:43 AM
Q18; What does the above have to do with your botched attempt to derive the Lorentz force? You appear to think that electrons and spinor fields are interchangeable. At least, you use them the same way.

That is what it says in quantum theory, so lump it. You asked me how a spinor is related to an electron, not how it botches a derivation of a Lorentz force, so stop molesting the two together.

macaw
2010-Nov-30, 07:45 AM
I think its a little more complicated than simply cutting and pasting.

True, as we have seen it is cutting and pasting and inserting your own errors.

Green Destiny
2010-Nov-30, 07:50 AM
True, as we have seen is cutting and pasting and inserting your own errors.

Any respectful reader will see the work above and see it is not a case of inserting errors. Simple mathematical substitutions do not even come close to a mathematical error. They are derivations performed all the time. Here is a perfect example

F=Ma

F=(E/c^2)a

Also I haven't ''rewritten'' the value of e^2h/4Mc (that is exactly how the value is given for the energy gained by an electron).

macaw
2010-Nov-30, 07:55 AM
Any respectful reader will see the work above and see it is not a case of inserting errors. Simple mathematical substitutions do not even come close to a mathematical error. They are derivations performed all the time. Here is a perfect example

F=Ma

F is not equal to Ma.

Q19: What is the correct relativistic definition of force?

F=(E/c^2)a

Q20: When you start improvising you get a lot worse a lot faster. I know how you derived the above and it is plain wrong. Do you know any basic relativity?

Also I haven't ''rewritten'' the value of e^2h/4Mc (that is exactly how the value is given for the energy gained by an electron).

First you tried rewriting classical electrodynamics, now you are trying to rewrite SR.

Green Destiny
2010-Nov-30, 08:00 AM
F is not equal to Ma.

Why are you being allowed to get away with assinine statements like this? So you are saying, as a true classical, meaning Newtonian approximation which makes quite accurate predictions, F =\= Ma?

:doh:

Newton will be rolling in his grave. Next you will be telling me P =\= Mv.

macaw
2010-Nov-30, 08:04 AM
Why are you being allowed to get away with assinine statements like this? So you are saying, as a true classical, meaning Newtonian approximation which makes quite accurate predictions, F =\= Ma?

:doh:

Newton will be rolling in his grave. Next you will be telling me P =\= Mv.

Q21: So what is the correct answer ? The ones you provided when you went off-script (off cutting and pasting from homework pages) are not correct. Please answer the last two questions, they are much easier than the ones you have been dodging throughout the thread.

macaw
2010-Nov-30, 08:11 AM
Newton will be rolling in his grave. Next you will be telling me P =\= Mv.

Q22: Newton? Weren't you pretending to be using SR and QM? Do you think that the homework page you've been copying from is using Newton?

Green Destiny
2010-Nov-30, 08:14 AM
So what is the correct answer ? The ones you provided when you went off-script (off cutting and pasting from homework pages) are not correct. Please answer the last two questions, they are much easier than the nnes you have been dodging throughout the thread.

What have they got to do with the OP? Totally off-topic, worth an infraction i'd say, Atleast, I would have got one.

Relativistic force

F=dP/dt

for relativistic speeds. F=Ma is valid for non-relativistic speeds as a low approximation, does not mean it is explicitely incorrect.

Q20 I know lots of basic relativity. Not that it has to do with the OP. Also you say my equations are incorrectly expressed, wrong. You haven't mathematically shown why. A very common occurrance with you.

Green Destiny
2010-Nov-30, 08:14 AM
The Zero Point Field is a classical limit. Did you not know that?

macaw
2010-Nov-30, 08:17 AM
What have they got to do with the OP? Totally off-topic, worth an infraction i'd say, Atleast, I would have got one.

You introduced this hilarious mis-step when you went off script.

Relativistic force

F=dP/dt

Good, you are googling fast. How about the hilarious expression of energy as a function of force and acceleration?

for relativistic speeds. F=Ma is valid for non-relativistic speeds as a low approximation, does not mean it is explicitely incorrect.

How about the hilarious F=aE/c^2 that you just "derived"? It is a fantastic combination of SR with Newtonian mechanics.

Green Destiny
2010-Nov-30, 08:20 AM
You introduced this hilarious mis-step when you went off script.

Good, you are googling fast. How about the hilarious expression of energy as a function of force and acceleration?

How about the hilarious F=aE/c^2 that you just "derived"?

Yes, hilarious.

F=Ma

M= E/c^2

I said in a substitution one could give F=(E/c^2)a as a very very trivial example. I wasn't making some grandiose claim. I was trying to massively simplify for you how the operations in the OP where working.

macaw
2010-Nov-30, 08:21 AM
The Zero Point Field is a classical limit. Did you not know that?

So, does this entitle you to combine Newtonian mechanics with particle physics? You just gave us a wonderful demonstration of combining SR and Newtonian mechanics.

macaw
2010-Nov-30, 08:23 AM
Yes, hilarious.

F=Ma

M= E/c^2

I said in a substitution one could give F=(E/c^2)a as a very very trivial example. I wasn't making some grandiose claim. I was trying to massively simplify for you how the operations in the OP where working.

You just combined SR with Newtonian mechanics. You tend to get really bad when you go off script.

Green Destiny
2010-Nov-30, 08:24 AM
Mind you I don't find it as half as hilarious as you claiming:

'' Electric charge and magnetic momentum are totally unrelated entities, are you rewriting basic electrodynamics?''
when...

Who did that? I said a substitution on q for q_M which gives you the necessery dimensions to rewrite the charge in terms of the loop integral. Who said anything about a magnetic momentum as being the same as an electric charge? It's a magnetic charge, not momentum.

and then asking how a spinor relates to an electron... but hey, we must be funny guys.

Green Destiny
2010-Nov-30, 08:25 AM
You just combined SR with Newtonian mechanics. You tend to get really bad when you go off script.

Obviously now I wish I hadn't used it as an example, as you have taken the example superfluously seriously.

macaw
2010-Nov-30, 08:27 AM
Obviously now I wish I hadn't used it as an example, as you have taken the example superfluously seriously.

No, it is good that you used it, it gives a gauge of your knowledge. Now, how about going back to question 1 and answering it?

Green Destiny
2010-Nov-30, 08:30 AM
No, it is good that you used it, it gives a gauge of your knowledge. Now, how about going back to question 1 and answering it?

Your gauge must be leaking then considering:

Mind you I don't find it as half as hilarious as you claiming:

'' Electric charge and magnetic momentum are totally unrelated entities, are you rewriting basic electrodynamics?''
when...

Who did that? I said a substitution on q for q_M which gives you the necessery dimensions to rewrite the charge in terms of the loop integral. Who said anything about a magnetic momentum as being the same as an electric charge? It's a magnetic charge, not momentum.

and then asking how a spinor relates to an electron... but hey, we must be funny guys.

macaw
2010-Nov-30, 08:33 AM
Your gauge must be leaking then considering:

Mind you I don't find it as half as hilarious as you claiming:

'' Electric charge and magnetic momentum are totally unrelated entities, are you rewriting basic electrodynamics?''
when...

Well it was you who claimed that "charge is seen in terms of magnetism" wasn't it?

Who did that? I said a substitution on q for q_M which gives you the necessery dimensions to rewrite the charge in terms of the loop integral.

http://www.codecogs.com/eq.latex?\nabla(\sum^{N}_{i=1} \vec{\mu}_i \cdot B)= (\oint \vec{\mu_i}M \cdot \partial \ell)v \times B

is riddled with typos and ridiculous mistakes. What homework did you try copying it from? Did you go off script like the E=(F/a)c^2 "discovery"?

Green Destiny
2010-Nov-30, 08:34 AM
So, does this entitle you to combine Newtonian mechanics with particle physics? You just gave us a wonderful demonstration of combining SR and Newtonian mechanics.

Well, actually yes studying the mathematics of the OP. The math clearly shows that on a certain ''collection'' of magnetic moments you can work out an absorption rate since the Newtonian limit is always conserved as a viable solution since the ZPF is a classical field theory. Magnetic moments are simply properties of a local system while the ZPF is a global system acting on a localized particle.

Green Destiny
2010-Nov-30, 08:36 AM
Well it was you who claimed that "charge can be viewed as magnetism" wasn't it?

But your "formula" is riddled with typos and ridiculous mistakes. Where did you try copying it from?

There are no typos. Apart from accidently not putting in the gradient in one of the right hand expressions, but there are not ''typos''. You've not even identified ''typos'' - you found one typo, and is not masked with ridiculous mistakes. If there was you should show some of these wonderful obvious examples.

Green Destiny
2010-Nov-30, 08:37 AM
Well it was you who claimed that "charge is seen in terms of magnetism" wasn't it?

What had that statement got to do with momentum?

macaw
2010-Nov-30, 08:40 AM
Well, actually yes studying the mathematics of the OP. The math clearly shows that on a certain ''collection'' of magnetic moments you can work out an absorption rate since the Newtonian limit is always conserved as a viable solution since the ZPF is a classical field theory. Magnetic moments are simply properties of a local system while the ZPF is a global system acting on a localized particle.

Green Destiny
2010-Nov-30, 08:42 AM
Well it was you who claimed that "charge is seen in terms of magnetism" wasn't it?

http://www.codecogs.com/eq.latex?\nabla(\sum^{N}_{i=1} \vec{\mu}_i \cdot B)= \nabla(MBcos \theta) is riddled with typos and ridiculous mistakes. What homework did you try copying it from? Did you go off script like the E=(F/a)c^2 "discovery"?

That equation is not wrong. (MBcos \theta) is M.B where M is the magnetic moment. The gradient is just the same gradient in F=\nabla(M.B) the force relation equation I told you about. [img]http://www.codecogs.com/eq.latex?\nabla(\sum^{N}_{i=1} \vec{\mu}_i \cdot B)[img] This is nothing but the sum of a collection of magnetic moments which is a dot product on B. Both sides are equivalent and I do not see any problem, other than you ''saying'' there is a problem, but never defining one.

Green Destiny
2010-Nov-30, 08:43 AM

Green Destiny
2010-Nov-30, 08:44 AM
I mean, afterall

Originally Posted by macaw
Well it was you who claimed that "charge is seen in terms of magnetism" wasn't it?

......

What had that statement got to do with momentum?

macaw
2010-Nov-30, 08:44 AM

That equation is not wrong. (MBcos \theta) is M.B where M is the magnetic moment. The gradient is just the same gradient in F=\nabla(M.B) the force relation equation I told you about. [img]http://www.codecogs.com/eq.latex?\nabla(\sum^{N}_{i=1} \vec{\mu}_i \cdot B)[img] This is nothing but the sum of a collection of magnetic moments which is a dot product on B. Both sides are equivalent and I do not see any problem, other than you ''saying'' there is a problem, but never defining one.

It is not even wrong. Let's try again:

Q5: How did you derive the above ?
Q6: Where is http://www.codecogs.com/eq.latex? \partial \ell coming from? What physical entity does it represent? How can it be a partial derivative with respect to...nothing?
Q7: Where did the speed v show up in the formulas since it is not present anywhere else?
Q8: Where did the crossproduct come from? There is nothing in the originating expression to make it pop up.
Q9: What does the symbol http://www.codecogs.com/eq.latex? \oint and how did it show up in the expression?

Green Destiny
2010-Nov-30, 08:46 AM
''where the gradient ∇ is the change of the quantity m · B per unit distance and the direction is that of maximum increase of m · B. To understand this equation, note that the dot product m · B = mBcos(θ), where m and B represent the magnitude of the m and B vectors and θ is the angle between them. ''

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

macaw
2010-Nov-30, 08:47 AM
I mean, afterall

Originally Posted by macaw
Well it was you who claimed that "charge is seen in terms of magnetism" wasn't it?

Don't you recognize your own words:

"The force here can now be viewed in terms of a Lorentz force where the charge is seen in terms of magnetism:"

So, let's see, your ATM is :

1. Rewriting (incorrectly) basic electrodynamics
2. Combining SR with Newtonian mechanics (this is a first!)
3. Making inroads in particle physics by confusing spinor fields with electrons

I must have missed some other "discoveries"

Green Destiny
2010-Nov-30, 08:48 AM
Look at the ''Magnetization'' section as well.

I can't believe after numerous attempts at getting you to read the links, you still fail to see where the equations come from.

Green Destiny
2010-Nov-30, 08:50 AM
Don't you recognize your own words:

"The force here can now be viewed in terms of a Lorentz force where the charge is seen in terms of magnetism:"

I know my own words. The qv x B is the lorentz force part, which you asked me about too, very strange question. I also explained that the q will be seen in terms of the magnetic charge q_M. I know my words sir when I say charge is seen in terms of magnetism.

macaw
2010-Nov-30, 08:55 AM
I know my own words. The qv x B is the lorentz force part, which you asked me about too, very strange question.

It isn't strange, you simply pasted in the expression for the Lorentz force (incomplete) and the LHS expression
http://www.codecogs.com/eq.latex? \oint \vec{\mu_i}M \cdot \partial \ell)v \times B leading to it was riddled with mistakes. See the unanswered questions Q5-Q9.

Q5: How did you derive the above ?
Q6: Where is http://www.codecogs.com/eq.latex? \partial \ell coming from? What physical entity does it represent? How can it be a partial derivative with respect to...nothing?
Q7: Where did the speed v show up in the formulas since it is not present anywhere else?
Q8: Where did the crossproduct come from? There is nothing in the originating expression to make it pop up.
Q9: What does the symbol http://www.codecogs.com/eq.latex? \oint and how did it show up in the expression?

tusenfem
2010-Nov-30, 10:11 AM
Oh for crying out loud, you guys get your own bulletin board to fight.

Green Destiny you will have to explain yourself much more clearly, you cannot use a vector capital M for magnetization and then just a scalar capital M for mass, don't you see how confusing that is? Similarly saying "taking the dot product of eq. 1" don't you think you should tell the people the dot product of WHAT????? Please if you start up another thread, than work on your presentation skills, because the are abominable.

macaw your agressive attitude is not appreciated, questions 1 though 6 are just totally ridiculous, deluging people with bezillion of question is not the way to go. I already complained in another thread that if you really want this ATM to improve than YOU should do some work too. Not just Q1Q2Q3Q4Q5Q6Q7Q8Q9Q10, if there is something significantly wrong than you show that on the board, don't play a grammar school teacher.

So enough, thread closed. One more of these kind of fights and I will infract you both off the board.