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Feb 4, 2000, 3:00:00 AM2/4/00

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"Bergervoet J.R.M." wrote:

> You started by calling electromagnetism as we know it "garbage". I

> replied by using the same terminology about your postings. Don't

> you think that's appropriate?

Jos, I tend not to argue with a perfect logical argument. You're right!

Hey sometimes I overreact to criticism but that's not your fault so I

apologize.

>

>

> > the force between two current

> > elements is a Coulomb force and if you want proof I will gladly

supply it.

>

> Then it is not the usual Biot-Savart-Lorentz force. It should still

> give the same total force on one segment (from all the other segments

in a

> closed current loop).

This is exactly what gives the idea at least some credibility.

> Even if you do, it will be impossible to say whether your force, or

the

> Biot-Savart-Lorentz force is the"correct" one, precisely because they

must give

> the same observable results. Of course we don't know whether your

force does

> that, since you don't give equations.

> -- Jos

I thought, when I began doing the math, that I would discover precisely

the above to be true.

As it turns out the two approaches are not entirely in agreement.

So, yes, there should be an observable difference.

I have seen in a few physics texts, the idea proposed that the magnetic

interaction is due to electrostatic forces, although it is presented as

being due to the relativistic length contraction of the lines of charges

due to their

motions relative to one another. i.e. Parallel currents of electrons

don't attract one another, they

attract the stationary protons in the opposing conductor.

I tested this idea years ago and couldn't get it to work. Maybe it does

work, I don't know. I'm not asking for an answer to this (it's

irrelevant). The drift rate of electrons in normal conductors is

extremely slow and any length contraction would be insignificant. Anyway

I can see no way to apply this concept to the Biot-Savart-Lorentz

force between current elements when we let the lengths of the elements

approach

zero.

However, the idea that the force is the vector sum of

"induced" Coulomb forces --whether the force is due to length

contraction or not-- seems to be the only sensible approach.

Argument: The classical equation, in the form F/l=uQeVeQe'Ve'/2(pi)d

for the force

acting between infinitely long parallel conductors, reveals the

following: Since u is a constant, if d is held constant and the line

densities

of charge, Qe

and Qe', are held to be constant, then a change in the measured

force on the conductors can only be produced by a change in Ve and/or

Ve' (the drift rates

of the electrons in conductors 1 and 2). We can only conclude that the

force

is induced by the velocities of the charges.

Please note*** this force is necessarily a function of the velocity of

the

charges

"relative to one another" rather than relative to some arbitrary

observer.

For equal drift rates in the same direction the following arguments

apply:

If we assume the frame of reference of the electrons, then no motion or

even relative motion, of the electrons exist. From this frame of

reference the only motion is that of the protons. We can mathematically

attribute the force to the motion of the protons, if we wish, by

postulating that the motion of positive charge generates a B Field which

is opposite in polarity to that generated by the same motion of negative

charge.

This argument leads to the following question "Is the magnetic

field a property of

the electrons or a property of the protons, or both?" It seems to depend

upon your frame of reference. (Don't

misquote me here, this is just the logical conclusion of the preceding

argument). Two arguments against the same preceding argument:

1. The line density of positive charge must exactly equal the line

density of negative charge.

2. If we consider electron beams we lose the convenience of transferring

the field to the protons since there are none present.

This leads to the only logically valid conclusion: The force is due to

the motion of the electrons relative to the protons.

Note* (For unequal currents or currents in

opposite directions the force is a function of all observed relative

velocities.)

It is very easy to derive an equation to mirror this concept.

It has long been known that when both Ve and Ve' are equal to c (the

speed

of light) that the Magnetic force is numerically equal to the Coulomb

force acting between the electrons alone. The equivalence goes even

further

than this; When both Ve and Ve' are equal to c, the Magnetic equation

reduces "exactly" to the Coulomb equation for the electrons alone. With

this in mind we can

postulate that the electrostatic magnetic-equation, for the force

between equal

parallel currents, will take on this form:

-F/l = (kQeQp'(Ve-Vp')^2/dc^2) + (kQe'Qp(Ve'-Vp)^2/dc^2)

Where k is Coulomb's constant Qe and Qp are the line densities of charge

of the electrons and protons in Conductor 1 respectively.

Qe' and Qp' are the line densities of charge in Conductor 2

respectively.

All charges are expressed as vectors.

All velocities (drift rates) are expressed as vectors.

The absolute value of (Ve-Vp') gives the "speed" of the electrons in

conductor 1 relative to the protons in conductor 2, likewise for

(Ve'-Vp) and these speeds are

invariant with respect to Galilean frames of reference.

Since we are taking the square of the speeds it is not

necessary to indicate their absolute values.

The minus sign at the left is supplied to produce the same convention of

force indicated by the classical equation.

For any two parallel conductors the full equation is:

-F/l = (kQeQp'(Ve-Vp')^2/dc^2) + (kQe'Qp(Ve'-Vp)^2/dc^2) +

(kQeQe'(Ve-Ve')^2/dc^2) + (kQpQp'(Vp-Vp')^2/dc^2)

If you play around with this equation you will soon find that it can be

reduced to the classical equation, although not very easily.

Reciprocally it can be derived from the classical equation. Therein lies

my proof, for now. If you see any typos or faults in the equation or in

my logical arguments please feel free to point them out to me.

Richard Perry rp...@cswnet.com

Feb 7, 2000, 3:00:00 AM2/7/00

to

Richard Perry wrote:

: I have seen in a few physics texts, the idea proposed that the magnetic

: interaction is due to electrostatic forces, although it is presented as

: being due to the relativistic length contraction of the lines of charges

: due to their motions relative to one another. i.e. Parallel currents of

: electrons don't attract one another, they attract the stationary protons

: in the opposing conductor.

:

: I tested this idea years ago and couldn't get it to work. Maybe it

: does work, I don't know. I'm not asking for an answer to this (it's

: irrelevant).

Do you know whether it is possible to detect a Hall voltage in

mutually attracting conductors?

: The drift rate of electrons in normal conductors is extremely slow and

: any length contraction would be insignificant.

I'm not sure whether your argument is valid, but there is an even more

serious problem. An effect resulting from length contraction would be

proportional to the square of the velocity v of the electrons.

1 / gamma = sqrt(1-v^2/c^2) = 1 - 0.5 v^2/c^2 v << c

The attraction between two currents remains however unchanged if

we increase the speed of the electrons by the same factor we reduce

their line density. So the SR explanation of this effect cannot

be correct.

: However, the idea that the force is the vector sum of

: "induced" Coulomb forces --whether the force is due to length

: contraction or not-- seems to be the only sensible approach.

It doesn't seem a good approach to me. An electric current has a

clearly defined direction, and this direction cannot be explained

by relatively moving charges. A vector in direction of the current

must somehow be involved (a possible solution: "pre-magnetic"

dipoles).

You can easily recognize this if you try to generate currents in

the same and in the opposite direction by relative motion of

charged rods:

rod 1 --------------------- e.g. -->

rod 2 ---------------------

rod 3 +++++++++++++++++++++ e.g. <--

So Maxwell's fourth equation (Ampčre's law) must be as wrong

within SR as it is in an original Maxwellian framework.

Regards, Wolfgang

http://members.lol.li/twostone/E/physics1.html

Feb 7, 2000, 3:00:00 AM2/7/00

to

Richard Perry writes:

>As it turns out the two approaches are not entirely in agreement.

>So, yes, there should be an observable difference.

To solve the question, you then should try to think of a simple experiment

that would show the difference. For instance: do you get the same tensile

force on a circular current-carrying wire as with the Lorentz force?

-- Jos

Feb 7, 2000, 3:00:00 AM2/7/00

to

"Bergervoet J.R.M." wrote:

> Richard Perry writes:

>

> >As it turns out the two approaches are not entirely in agreement.

> >So, yes, there should be an observable difference.

>

> To solve the question, you then should try to think of a simple experiment

> that would show the difference. For instance: do you get the same tensile

> force on a circular current-carrying wire as with the Lorentz force?

>

> -- Jos

Jos, I don't have time for a lengthy response but I just wanted to share

a few

quick ideas with you. If you cross examined my equation you no doubt

found it

to be in perfect agreement with the classical equation. As you should

have,

since it basically "is" the classical equation.

There is a very distinct difference between two however.

1. My equation is valid for non neutral conductors, whereas the

classical

equation is valid for neutral conductors only.

2. The terms may be selectively eliminated from my equation.

It may be used, for instance, to determine the influence of a current

carrying

conductor on an electron beam directed parallel to it, by eliminating

the

terms containing p'. Classical theory would predict the same

Magnetostatic

force on this beam whether it is contained in a conductor or not. If

you

calculate this force with my equation you will find that when Ve'

(velocity of

the electron beam) is much greater than Ve (drift rate of electrons in

the

conductor) the two equations are in approximate agreement. However as

Ve'

approaches Ve the equations begin to diverge, reaching a maximum

divergence at

Ve' =Ve/2 at which point,using my equation, the force reduces to zero.

(Assuming currents in the same direction). From which it follows that my

equation requires a force to be apparent on a Coulomb test charge at

rest

relative to the conductor, although a feeble one, and no force when the

velocity of the test charge is equal to Ve/2.

If my equation is correct, and if the "one current" theory( very

important) is

correct we can use either of these conclusions to precisely determine

the

drift rate of electrons through any material. Do you know anyone with

the facilities to conduct these experiments?

One final note, I also have an equation for the Biot-Savart-Lorentz

force

which is similar in form to my previous equation. It was on the basis of

this

equation that I posted the arguments you first replied to. It provides

for the net magnetostatic force between parallel conductors, but more

importantly, it also provides for magnetic inductance. More on this when

I have

more time.

Regards,

Richard Perry rp...@cswnet.com

Feb 8, 2000, 3:00:00 AM2/8/00

to

Richard Perry writes:

> Richard Perry wrote earlier:

> >

> >As it turns out the two approaches are not entirely in agreement.

> >So, yes, there should be an observable difference.

>Jos, I don't have time for a lengthy response but I just wanted to share

>a few quick ideas with you. If you cross examined my equation you no doubt

>found it to be in perfect agreement with the classical equation.

I'm affraid you contradicted yourself. Observable difference or not?

That's the question. It would be interesting if you can either:

1) Replace the Lorentz force between current elements with a central

force (directed along the line connecting the segments) and still

obtain the same observable results as classical electrodynamics.

or:

2) Replace something and give a clear example showing how observable

results are different from classical electrodynamics.

Of course case 1 would be more interesting (since case 2 has little chance

of passing experimental tests) but until now we saw neither 1 nor 2.

-- Jos

Feb 8, 2000, 3:00:00 AM2/8/00

to

"z@z" wrote:

>

> Richard Perry wrote:

>

> : I have seen in a few physics texts, the idea proposed that the magnetic

> : interaction is due to electrostatic forces, although it is presented as

> : being due to the relativistic length contraction of the lines of charges

> : due to their motions relative to one another. i.e. Parallel currents of

> : electrons don't attract one another, they attract the stationary protons

> : in the opposing conductor.

>

> Richard Perry wrote:

>

> : I have seen in a few physics texts, the idea proposed that the magnetic

> : interaction is due to electrostatic forces, although it is presented as

> : being due to the relativistic length contraction of the lines of charges

> : due to their motions relative to one another. i.e. Parallel currents of

> : electrons don't attract one another, they attract the stationary protons

> : in the opposing conductor.

> I'm not sure whether your argument is valid, but there is an even more

> serious problem. An effect resulting from length contraction would be

> proportional to the square of the velocity v of the electrons.

>

> 1 / gamma = sqrt(1-v^2/c^2) = 1 - 0.5 v^2/c^2 v << c

>

> The attraction between two currents remains however unchanged if

> we increase the speed of the electrons by the same factor we reduce

> their line density. So the SR explanation of this effect cannot

> be correct.

Assuming that we are talking about wires that

have neutral charge in the laboratory frame of

reference, you have missed a factor. If we keep

the same same current by doubling the velocity

and halving the number of electrons per unit length,

it is true that each proton (and electron) in the

wire would see twice the force, but there would only

be half as many of them (neutrality) so the net

force on the wire would be unchanged.

Opinions expressed herein are my own and may not represent those of my employer.

Feb 9, 2000, 3:00:00 AM2/9/00

to

: = Roy McCammon

:: = Wolfgang G. (on attraction between parallel currents):

:: = Wolfgang G. (on attraction between parallel currents):

:: An effect resulting from length contraction would be

:: proportional to the square of the velocity v of the electrons.

::

:: 1 / gamma = sqrt(1-v^2/c^2) = 1 - 0.5 v^2/c^2 v << c

::

:: The attraction between two currents remains however unchanged if

:: we increase the speed of the electrons by the same factor we reduce

:: their line density. So the SR explanation of this effect cannot

:: be correct.

:

: Assuming that we are talking about wires that

: have neutral charge in the laboratory frame of

: reference, you have missed a factor.

What kind of factor?

Take the case of two identical parallel wires with exactly the

same current.

-------------------- --->

electron speed v

-------------------- --->

Because v is the only involved velocity, an attraction resulting

from different charge densities because of length contraction

can only depend on v. The difference in charge must therefore

be proportional to rho * [L - L/gamma]

where rho is the line density of electrons (and protons) and L is

the length of a given segment of the wire in the rest frame.

Diff_in_charge_1 = const * rho * L - L/gamma =

= const * rho * 0.5 v^2/c^2

When doubling the line density and halving the electron speed

==================== ->

electron speed v/2

==================== ->

we get:

Diff_in_charge_2 = const * 2 rho * 0.5 [v/2]^2 / c^2

= 0.5 Diff_in_charge_1

So an effect resulting from a difference in charge because of

length contraction would in the second case yield an attractive

force between the currents which is only a fourth of the force

in the first case. We know however that the force is (more or

less) the same in both cases!

: If we keep the same current by doubling the velocity

: and halving the number of electrons per unit length,

: it is true that each proton (and electron) in the

: wire would see twice the force, but there would only

: be half as many of them (neutrality) so the net

: force on the wire would be unchanged.

This classical reasoning is out of question.

'Essential Relativity', Wolfgang Rindler, 1977 (1986), p. 104:

"Consider now a test charge moving with velocity u parallel to

the wire. It experiences a force in the direction u x H, i.e.

radially towards or away from the wire. In its rest frame, where

it can be affected only by E fields, it sees two moving lines

of positive and negative charge, respectively, but with sightly

different line densities, numerically. FOR LENGTH CONTRACTION

SEES TO IT THAT if the line densities are equal and opposite

in the lab frame, they cannot be so in any other frame moving

parallelly to the wire. And it is this difference which provides

the net E field that causes the charge to accelerate in its

rest frame." (emphasis mine)

Even the understandable part

" if the line densities are equal and opposite

in the lab frame, they cannot be so in any other frame moving

parallelly to the wire "

of the ununderstandable sentence is rather dubious. If the line

densities (of positive and negative charge) are equal and

opposite in the frame of the positive charges (why?), they

should also be equal and opposite in the frame of the (moving)

electrons. We can move the the whole wire in such a way that the

electrons are at rest in the laboratory frame and the positive

charges are moving in the opposite direction.

In any case, I've not yet seen a convincing (physical) presentation

of the relativistic identity between electric and magnetic fields.

Special relativity was (primarily) developed to remove obvious

contradictions from Maxwell's theory. Interestingly it turns out

that relativity as a space-time theory and a generalization of

classical mechanics could be more consistent than Maxwell's

theory itself.

Wolfgang Gottfried G.

My previous post of this thread:

http://www.deja.com/=dnc/getdoc.xp?AN=582450140

Feb 9, 2000, 3:00:00 AM2/9/00

to

"z@z" wrote:

I would like to add an important ingredient.

It was Einstein who said that: The perfect form of the laws of physics

must be invariant with respect to rectilinear frames of reference. It was

for this reason that Einstein adopted the Lorentz transformation, in the

hopes that when it was applied to Maxwell's equations, it would render

them invariant with respect to all rectilinear frames.

Although SR worked for the value c, which turns up as a constant in

Maxwell's work, it does not work for other velocities. Your arguments

above are just a few out of thousands posted on the web, that show

conclusively that SR fails in its designed task.

I fully agree with Einstein's statement of invariance however , and it was

for this reason that I sought a form of the equation for parallel

conductors that was invariant. I feel that I have obeyed Einstein's

mandate in a much more literal way than he did. If the electromagnetic

equations had been cast in an invariant form to start with SR would not

exist today.

For the sake of accuracy a velocity should always be related to some

reference point. In the case of the classical equation for the force

between infinitely parallel conductors we find only the terms Ve and Ve'.

These two velocities are properly (Ve-Vp) and (Ve'-Vp'), where all of

these velocities are vectors. This follows from the definition of the

ampere: The passage of one Coulomb of charge through a cross sectional

area of a conductor (ie. relative to the conductor) per second.

If you substitute these terms into the classical equation, and take the

conductors to be neutral, you should be able to derive the equation in my

original post, which I arrived at through very different means. If you

can't figure it out or don't have the time or the desire, check back on

this thread, I'll post it when I have time and desire.

Hint: begin by multiplying the speeds and paying close attention to the

negative signs all the way through.

Regards,

Richard Perry rp...@cswnet.com

Feb 9, 2000, 3:00:00 AM2/9/00

to

"Bergervoet J.R.M." wrote:

> Richard Perry writes:

>

> > Richard Perry wrote earlier:

> > >

> > >As it turns out the two approaches are not entirely in agreement.

> > >So, yes, there should be an observable difference.

>

> >Jos, I don't have time for a lengthy response but I just wanted to share

> >a few quick ideas with you. If you cross examined my equation you no doubt

> >found it to be in perfect agreement with the classical equation.

>

> I'm affraid you contradicted yourself. Observable difference or not?

> That's the question. It would be interesting if you can either:

>

> 1) Replace the Lorentz force between current elements with a central

> force (directed along the line connecting the segments) and still

> obtain the same observable results as classical electrodynamics.

> or:

>

> 2) Replace something and give a clear example showing how observable

> results are different from classical electrodynamics.

>

> Of course case 1 would be more interesting (since case 2 has little chance

> of passing experimental tests) but until now we saw neither 1 nor 2.

>

> -- Jos

In agreement is synonymous with the phrase "not contradictory". In case 1 I was

refering to the outcomes(contradictory ie not in agreement), in case 2 was

referring to the equations (complimentary when the following statement is

considered)

Regarding nuetral conductors the two equations are in complete agreement.

However if the conductors are not neutral the equations disagree. The classical

equation is a special case of my equation, and again, this only holds true if

the one-current model is true. If it is not, then the two equations will always

agree.

Do you think my equation is valid or not?

That is the question.

A simple yes or no will do.

Feb 10, 2000, 3:00:00 AM2/10/00

to

Richard Perry wrote:

> Your arguments above are just a few out of thousands

> posted on the web, that show conclusively that SR

> fails in its designed task.

> Your arguments above are just a few out of thousands

> posted on the web, that show conclusively that SR

> fails in its designed task.

Without taking a position on the validity of SR,

almost everyone of these "arguments" merely

demonstrate that the person advocating the

argument does not understand the theory being

debated. You cannot misapply a theory to show

an incorrect result and use that to debunk the

theory.

To be a little more specific, in every one of

these counter arguments that I have looked

at, the author fails to apply SR transformations

to the forces. If you don't do that, you

won't get the correct results. You can

neglect it at low speeds and get an answer that

has the right behavior (predicts the direction

of the forces for example). But don't expect

to get he numerically correct answer. And don't

be surprised if it gets significantly wrong

answers at high speeds.

To be even more specific regarding forces of

a wire. You cannot calculate the forces on

the stationary protons in their FOR (frame of

reference), calculate the forces on the moving

electrons in their FOE and merely add those

two forces to get the total force on the wire.

You have to transform the forces into the

reference frame in which the force is being

measured.

Most of the posters here don't know this.

Most of those who do know it, don't know

how to to do it (but could learn it). Most

of those that know how would probably make a

algebraic mistake somewhere along the way.

It is hard to apply SR and coulomb force to

derive the magnetic field effects. I certainly

haven't done it. But I can read Feynman and Purcel

who say it has been done. It may be tricky math,

but it is objective math. Its been verified by

many competent minds, including physicists and

mathematicians. And count on it, if there was an

error, some of the holders of these competent

minds would injure themselves in the rush to be

the first one to say it aint so.

So next time you try this and get an

answer that contradicts many competent

minds consider the following:

1. maybe you don't know the theory

2. maybe you misapplied the theory

3. maybe you made a math mistake.

Feb 10, 2000, 3:00:00 AM2/10/00

to

In <38A24E8D...@cswnet.com> Richard Perry <rp...@cswnet.com> writes:

>"Bergervoet J.R.M." wrote:

>Regarding nuetral conductors the two equations are in complete agreement.

>However if the conductors are not neutral the equations disagree. The

>classical equation is a special case of my equation, ...

I meant that to explain your theory, you could have given an explicit

example where it disagrees with electrodynamics. But OK, I'll do it

for you, using your formula:

>

> -F/l = (kQeQp'(Ve-Vp')^2/dc^2) + (kQe'Qp(Ve'-Vp)^2/dc^2) +

> (kQeQe'(Ve-Ve')^2/dc^2) + (kQpQp'(Vp-Vp')^2/dc^2)

On the first wire I take:

Qp = -Qe > 0, Vp=0, Ve > 0. (simple neutral metalic conductor)

On the second wire I take a pure negative charge without current:

Qe' < 0, Qp'=0, Vp'=Ve'=0. (non-moving line of pure negative charge)

Your formula reduces to the 3rd term only, and gives:

-F/l = k/d Qe Qe' Ve^2/c^2

But classical electrodynamics (CED) tells us the force is 0 (since

uncharged current and current-free charge don't attract or repel).

The CED answer (0) just means that a magnetic field does not attract

a stationary charge. As far as I know this is well-tested, so to

answer your last question: I think the CED result is true.

-- Jos

Feb 10, 2000, 3:00:00 AM2/10/00

to

"Bergervoet J.R.M." wrote:

> In <38A24E8D...@cswnet.com> Richard Perry <rp...@cswnet.com> writes:

>

> >"Bergervoet J.R.M." wrote:

>

> >Regarding nuetral conductors the two equations are in complete agreement.

> >However if the conductors are not neutral the equations disagree. The

> >classical equation is a special case of my equation, ...

>

> I meant that to explain your theory, you could have given an explicit

> example where it disagrees with electrodynamics. But OK, I'll do it

> for you, using your formula:

> >

> > -F/l = (kQeQp'(Ve-Vp')^2/dc^2) + (kQe'Qp(Ve'-Vp)^2/dc^2) +

> > (kQeQe'(Ve-Ve')^2/dc^2) + (kQpQp'(Vp-Vp')^2/dc^2)

>

> On the first wire I take:

>

> Qp = -Qe > 0, Vp=0, Ve > 0. (simple neutral metalic conductor)

>

> On the second wire I take a pure negative charge without current:

>

> Qe' < 0, Qp'=0, Vp'=Ve'=0. (non-moving line of pure negative charge)

>

> Your formula reduces to the 3rd term only, and gives:

>

> -F/l = k/d Qe Qe' Ve^2/c^2

>

> But classical electrodynamics (CED) tells us the force is 0 (since

> uncharged current and current-free charge don't attract or repel).

I was hoping that you would say this. Although I was quite aware of these

facts I can now use your statement in another proof with out having you

discrediting it. Since I can algebraically derive my equation, with a simple

and valid progression, from the classical equation, then it must be valid

even within the context of CED.

The result that you just arrived at proves that the one-current model is

invalid. I never believed that it was. The two-current model must be accepted

before I present you with the modified Biot-Savart equation.

Feb 11, 2000, 3:00:00 AM2/11/00

to

: = Roy McCammon

:: = Richard Perry

::: = z@z http://www.deja.com/=dnc/getdoc.xp?AN=583845550

:: = Richard Perry

::: = z@z http://www.deja.com/=dnc/getdoc.xp?AN=583845550

:: Your arguments above are just a few out of thousands

:: posted on the web, that show conclusively that SR

:: fails in its designed task.

I'd like to see the simplest and most convincing of those

arguments. How can I find them?

: To be a little more specific, in every one of

: these counter arguments that I have looked

: at, the author fails to apply SR transformations

: to the forces.

If you believe in currently accepted physics, then you

unconsciously conclude from any counter argument to the

incompetence of its author.

: To be even more specific regarding forces of

: a wire. You cannot calculate the forces on

: the stationary protons in their FOR (frame of

: reference), calculate the forces on the moving

: electrons in their FOR and merely add those

: two forces to get the total force on the wire.

: You have to transform the forces into the

: reference frame in which the force is being

: measured.

:

: Most of the posters here don't know this.

: Most of those who do know it, don't know

: how to to do it (but could learn it). Most

: of those that know how would probably make a

: algebraic mistake somewhere along the way.

If relativistic Maxwellian theory were correct, the

identity between magnetic and electric fields would

be a very fundamental principle. Really fundamental

principles however, must finally turn out to be simple

and clear. In 1905 Einstein had good reasons to assume

that this case and similar open questions will be

solved IN A TRANSPARENT WAY.

: It is hard to apply SR and coulomb force to

: derive the magnetic field effects. I certainly

: haven't done it. But I can read Feynman and Purcel

: who say it has been done.

So you believe in the authority of high priests such

as Richard Feynman. I'm highly skeptical about

Feynman's "science". Many of his ideas bear too much

resemblance with medieval theological concepts. QED

photons "travel all possible distances in all possible

times with all possible velocities". So they are not

only able to be everywhere at the same time, they even

know what will happen in the future!

http://www.deja.com/=dnc/getdoc.xp?AN=547437435

: It may be tricky math, but it is objective math.

I do not believe in your "objective math". In many

so-called mathematical proofs (such as e.g. Gödel's

incompleteness theorem) the complicated math is rather

unimportant, whereas non-mathematical (e.g. semantic)

assumptions and conclusions are crucial.

: Its been verified by

: many competent minds, including physicists and

: mathematicians. And count on it, if there was an

: error, some of the holders of these competent

: minds would injure themselves in the rush to be

: the first one to say it aint so.

History clearly shows that you are wrong. The most

obvious contradictions have always been defended by all

means by the leading scientists and their disciples.

One of the most obvious contradictions of modern science

is Maxwell's claim that all electromagnetic effects

propagate at the speed of light.

Apart from prejudice there is no convincing argument at

all that the relations described by Maxwell's equations

propagate at the same velocity as the wave solutions

which can be derived from them.

http://members.lol.li/twostone/E/physics1.html

Or assume we have two identical circular circuit loops in

two neighbouring parallel planes. "Simple" logic tells us

that it is impossible the explain the magnetic attraction

between the loops (in the case of currents in the same

direction) by E fields resulting somehow from different

line densities of charge (resulting somehow from SR length

contraction).

: So next time you try this and get an

: answer that contradicts many competent

: minds consider the following:

: 1. maybe you don't know the theory

: 2. maybe you misapplied the theory

: 3. maybe you made a math mistake.

Maybe too many people have taken these three considerations

too seriously.

Wolfgang Gottfried G.

Feb 11, 2000, 3:00:00 AM2/11/00

to

In <38A37C0E...@cswnet.com> Richard Perry <rp...@cswnet.com> writes:

>"Bergervoet J.R.M." wrote:

>>

>> Your formula reduces to the 3rd term only, and gives:

>>

>> -F/l = k/d Qe Qe' Ve^2/c^2

>>

>> But classical electrodynamics (CED) tells us the force is 0 (since

>I was hoping that you would say this. Although I was quite aware of these

>facts ...

Of course. You knew it, but just didn't want to tell us :-)

> ...I can now use your statement in another proof ...

Oh, you gave any proofs until now?

> ... with out having you

>discrediting it. Since I can algebraically derive my equation, with a simple

>and valid progression, ...

It doesn't matter how clever you are in deriving your results (as the

other Richard said,) it's useless if they don't match experiment.

Your task is now to show EXPERIMENTALLY that electrons start moving in a

magnetic field. If you do that, and after it is independently reproduced,

I'll believe you.

Success,

Jos

Feb 12, 2000, 3:00:00 AM2/12/00

to Bergervoet J.R.M.

> >"Bergervoet J.R.M." wrote:

> >>

> >> Your formula reduces to the 3rd term only, and gives:

> >>

> >> -F/l = k/d Qe Qe' Ve^2/c^2

> >>

> >> But classical electrodynamics (CED) tells us the force is 0 (since

> >>

> >> Your formula reduces to the 3rd term only, and gives:

> >>

> >> -F/l = k/d Qe Qe' Ve^2/c^2

> >>

> >> But classical electrodynamics (CED) tells us the force is 0 (since

Note to all other readers. The above result is derived by evaluating the

force on a test charge "at

rest"

relative to a conductor, assuming a "one-current" model of electrical

flow, using an equation which I derived.

> Your task is now to show EXPERIMENTALLY that electrons start moving in a

> magnetic field. If you do that, and after it is independently reproduced,

> I'll believe you.

>

> Success,

> Jos

The two-current model will predict a net force of zero on the test

charge, which

is in agreement with the empirical data. (Assuming a negative test

charge, and a

two-current model, the attraction due to the positive current will

cancel the

repulsion due to the negative current leaving a net force of zero.)

You just proved that the

one-current model is invalid, but apparently you didn't see it. The

empirical data and the mathematical

argument

presented by you "are" the proof. I thought that you would be able to

work out

the mathematical equivalence of my equation to the classical equation,

but

apparently you haven't. I will do this for you now.

Once having formulated a mathematical

proof of my equation's validity, the validity of the two current-model

will rest in

your own mathematical argument.

Here is the derivation:

Beginning with the classical equation for the force of magnetism acting

on two infinitely long

parallel conductors:

F/l=µQeQe'VeVe'/2(pi)d

The velocities in this equation are properly "speeds" and should read

(Ve-Vp')

and (Ve'-Vp) respectively, where the terms within parentheses are

vectors as measured in any Galilean frame of reference.

This follows from the definition of the ampere: The passage of one

Coulomb of

charge through a cross sectional area of the conductor (*ie. relative to

the

conductor) per second. You may argue that this only applies if we assume

a

one-current model (setting the positive charge at rest relative to the

conductor), but this is not the case.

If the current in the conductor is

due to the algebraic sum of the negative and positive sub-currents, then

the

following mathematical argument applies:

I(net)= Ie+Ip

Ie=Qe(Ve-Vconductor) and Ip=Qp(Vp-Vconductor)

I(net)=Qe(Ve-Vconductor)+Qp(Vp-Vconductor)

Qe=-Qp by substitution we get,

I(net)=Qe(Ve-Vconductor)-Qe(Vp-Vconductor)

I(net)=Qe[(Ve-Vconductor)-(Vp-Vconductor)]

I(net)=Qe(Ve-Vconductor-Vp+Vconductor)

I(net)=Qe(Ve-Vp) for the two-current model

Note that in the case of the one-current model Vp=Vconductor

and by substitution into the third equation above we get,

I(net)=Qe(Ve-Vp)+Qp(Vp-Vp)=Qe(Ve-Vp)

The classical equation in either case, is therefore properly:

F/l = µQeQe'(Ve-Vp)(Ve'-Vp')/2(pi)d

µ/2(pi)=2k/c^2 by substitution,

F/l = 2kQeQe'(Ve-Vp)(Ve'-Vp')/dc^2

F/l = 2kQeQe'(VeVe'+VpVp'-VeVp'-Ve'Vp)/dc^2

F/l = kQeQe'(2VeVe'+2VpVp'-2VeVp'-2Ve'Vp)/dc^2

-F/l = kQeQe'(2VeVp'+2Ve'Vp-2VeVe'-2VpVp')/dc^2

-F/l =

kQeQe'[(2VeVp'+2Ve'Vp-2VeVe'-2VpVp')+(Ve^2-Ve^2+Ve'^2-Ve'^2+Vp^2-Vp^2+Vp'^2-Vp'^2)]/dc^2

-F/l =

kQeQe'(-Ve^2+2VeVp'-Vp'^2-Ve'^2+2Ve'Vp-Vp^2+Ve^2-2VeVe'+Ve'^2+Vp^2-2VpVp'+Vp'^2)/dc^2

-F/l =

kQeQe'[-(+Ve^2-2VeVp'+Vp'^2)-(+Ve'^2-2Ve'Vp+Vp^2)+(Ve^2-2VeVe'+Ve'^2)+(Vp^2-2VpVp'+Vp'^2)]/dc^2

-F/l = kQeQe'[-(Ve-Vp')^2-(Ve'-Vp)^2+(Ve-Ve')^2+(Vp-Vp')^2]/dc^2

-F/l = -(kQeQe'(Ve-Vp')^2/dc^2) - (kQeQe'(Ve'-Vp)^2/dc^2) +

(kQeQe'(Ve-Ve')^2/dc^2) + (kQeQe'(Vp-Vp')^2/dc^2)

Qe=-Qp and Qe'=-Qp' by substitution,

-F/l = (kQeQp'(Ve-Vp')^2/dc^2) + (kQe'Qp(Ve'-Vp)^2/dc^2) +

(kQeQe'(Ve-Ve')^2/dc^2) + (kQpQp'(Vp-Vp')^2/dc^2)

You may recognize this last form of the "classical equation" as the one

I first presented to you.

I may have derived this equation but it is truly the the property of

CED.

I originally arrived at this equation with the logical arguments in the

original post to this thread, but the derivation came afterwards, and I

might add, did not come easily. It wasn't easy to type out either.

The equation clearly states that the force apparent between parallel

conductors is due to the Coulomb interactions of the charges, and not to

some other force, ie. magnetism.(So long to Dirac's monopole)

Having shown that the relative motion of charges produces a Coulomb

interaction, it is a simple hop skip and jump to the understanding

that the Coulomb force proper (ie. Force between stationary test

charges) is also due to the relative motions of the charged particles

deposited on them. (Unless of course you hold the conviction that these

particles are at rest relative to one another.)

Regards,

Richard Perry rp...@cswnet.com

*(The material above is the intellectual property of the author, but may be

freely

distributed so long as its authorship is clearly indicated.)

Feb 13, 2000, 3:00:00 AM2/13/00

to

"z@z" wrote:

> Take the case of two identical parallel wires with exactly the

> same current.

> [... diagram and some incomplete equations ...]> Take the case of two identical parallel wires with exactly the

> same current.

> So an effect resulting from a difference in charge because of

> length contraction would in the second case yield an attractive

> force between the currents which is only a fourth of the force

> in the first case. We know however that the force is (more or

> less) the same in both cases!

In the rest frame of the wires there is a net current in each wire,

but no net charge. The force on the lower wire is proportional to

ixB where i is the current in the lower wire and B is the magnetic

field at the location of the lower wire due to the upper wire;

similarly for the force on the upper wire. For both wires this is

attractive. Double the charge density and halve the velocity and

you leave both i and B unchanged, and therefore the force is

unchanged.

Your attempt to compute this in terms of length-contracted charge

density is incomplete -- you forgot the positive charges in the

wire: in the rest-frame of the electrons they feel a magnetic

force which you ignored, and they affect the net charge density

of the wires in this frame -- there is indeed a net electrostatic

force, but it is not as large as you think (the positive charges

cancel some of it).

So there are two components to the force in the electrons' frame:

the electrostatic force due to the net charge density in the

wires, and the magnetic force due to the currents of the positive

charges. If you double the charge density and halve the velocity

(in the rest frame of the wires), then these two components

vary such that the total force is unchanged.

> If the line

> densities (of positive and negative charge) are equal and

> opposite in the frame of the positive charges (why?),

The frame of the positive charges is the rest frame of the wires,

and we _observe_ that wires carrying a net current remain uncharged

themselves.

> they

> should also be equal and opposite in the frame of the (moving)

> electrons.

Not true. If you took a snapshot of the charges in the rest frame

of the wire you would find more electrons per meter than protons

per meter (because of the length contraction of the moving electron

frame and the requirement that the net charge density in this frame

be 0). Boost to any other frame and the positive and negative

charge densities will not cancel exactly. Basically the battery

which sources the current in the wires had to pump extra electrons

into the wires to get the current started; it is at rest in the

rest frame of the wires.

> We can move the the whole wire in such a way that the

> electrons are at rest in the laboratory frame and the positive

> charges are moving in the opposite direction.

Yes you could, but that is then a different physical situation when

observed in the lab. Note you need to move the battery as well and

_that_ is why this is a different situation in the lab.

> In any case, I've not yet seen a convincing (physical) presentation

> of the relativistic identity between electric and magnetic fields.

They aren't identical. But they are related. And if they weren't

related as predicted by SR and Maxwell's equations, then no

synchrotron in the world could possibly work. They do.

Tom Roberts tjro...@lucent.com

Feb 13, 2000, 3:00:00 AM2/13/00

to

Richard Perry wrote:

> It was Einstein who said that: The perfect form of the laws of physics

> must be invariant with respect to rectilinear frames of reference. It was

> for this reason that Einstein adopted the Lorentz transformation, in the

> hopes that when it was applied to Maxwell's equations, it would render

> them invariant with respect to all rectilinear frames.

> It was Einstein who said that: The perfect form of the laws of physics

> must be invariant with respect to rectilinear frames of reference. It was

> for this reason that Einstein adopted the Lorentz transformation, in the

> hopes that when it was applied to Maxwell's equations, it would render

> them invariant with respect to all rectilinear frames.

This is more than a mere hope, it is a fact (theorem).

> Although SR worked for the value c, which turns up as a constant in

> Maxwell's work, it does not work for other velocities.

That's plain not true. Look at any particle experiment....

Hmmm. Unless you mean that the invariant velocity of the

Lorentz transform (c) must be the same as the speed of

light (also c), then you are right. maxwell's equations

are only invariant over the Lorentz group if this is so.

There are experiments which show this is valid to within

2 parts in 10^7.

> Your arguments

> above are just a few out of thousands posted on the web, that show

> conclusively that SR fails in its designed task.

Not true. His argument is flawed. While I haven't seen the "thousands

posted on the web", I strongly suspect they are flawed too if they claim

to show that SR is wrong. The web is no paragon of good information, and

you have to already be knowledgeable about the subject you research (so

you can distinguish valid information from nonsense).

> I fully agree with Einstein's statement of invariance however , and it was

> for this reason that I sought a form of the equation for parallel

> conductors that was invariant.

Try Maxwell's equations and the Lorentz force law:

dF = 0

*d*F = J

f = F.J

They are manifestly invariant, even in a curved manifold.

> I feel that I have obeyed Einstein's

> mandate in a much more literal way than he did. If the electromagnetic

> equations had been cast in an invariant form to start with SR would not

> exist today.

Not true. If they had been cast in a manifestly-invariant form then

SR would have been known since the 1860's, not merely since 1905.

Tom Roberts tjro...@lucent.com

Feb 13, 2000, 3:00:00 AM2/13/00

to

"z@z" wrote:

> QED

> photons "travel all possible distances in all possible

> times with all possible velocities". So they are not

> only able to be everywhere at the same time, they even

> know what will happen in the future!

> http://www.deja.com/=dnc/getdoc.xp?AN=547437435

> QED

> photons "travel all possible distances in all possible

> times with all possible velocities". So they are not

> only able to be everywhere at the same time, they even

> know what will happen in the future!

> http://www.deja.com/=dnc/getdoc.xp?AN=547437435

You really should understand QED before attempting to summarize it.

Your statement is not true -- those photons don't _KNOW_ what will

happen in the future, they take account of _anything_ and

_everything_ which could possibly happen in the future! But note

that the contributions from outside the past light cone of any

given event will all cancel out when considering that event, and

only field values within its past light cone actually contribute.

> One of the most obvious contradictions of modern science

> is Maxwell's claim that all electromagnetic effects

> propagate at the speed of light.

>

> Apart from prejudice there is no convincing argument at

> all that the relations described by Maxwell's equations

> propagate at the same velocity as the wave solutions

> which can be derived from them.

> http://members.lol.li/twostone/E/physics1.html

This is merely the near-field of a moving charged particle and

a measurement of essentially 0 aberration or time delay in its

value. It is well known and predicted by Maxwell's equations (the

velocity-dependent terms cancel the aberration due to the finite

speed of light). See Feynman's lectures, vol 2.

> Or assume we have two identical circular circuit loops in

> two neighbouring parallel planes. "Simple" logic tells us

> that it is impossible the explain the magnetic attraction

> between the loops (in the case of currents in the same

> direction) by E fields resulting somehow from different

> line densities of charge (resulting somehow from SR length

> contraction).

I don't know what "simple logic" you are using, but Maxwell's

equations work quite well to predict this. Note that for loops

you cannot easily use SR to boost into the frame of the

electrons, as they are not moving inertially. So this is not

at all a simple problem to which one could apply length

contraction. And note my other response to you about your

errors in attempting such a "simple" analysis of linear wires.

> Maybe too many people have taken these three considerations

> too seriously.

Maybe too many people haven't actually applied Maxwell's theory

(as embodied in his famous equations). Yourself, for instance.

Tom Roberts tjro...@lucent.com

Feb 13, 2000, 3:00:00 AM2/13/00

to

In sci.physics.electromag Richard Perry <rp...@cswnet.com> wrote:

> The two-current model will predict a net force of zero on the test

> charge, which is in agreement with the empirical data. (Assuming a

> negative test charge, and ...

Aha, you need to restrict everything to "two-current" situations. So

you need a different behavior of nature before your new theory works.

That's bad for your market share in _this_ universe (but say hallo to

Alice, if you see her!)

-- Jos

Feb 13, 2000, 3:00:00 AM2/13/00

to

Roy, Tom and others:

I'm one of those who admits to NOT understanding a lot of this. So I

promise not to prpound any new theories. I'm an amateur just tying to

understand some of this. Please forgive the very amateur tone of the

comments that follow.

I while back I came across a basic electronics text that said that

the magnetic force can be completely explained by the relativistic

effects of the charges in motion in the wires. I really liked this

book -- I liked the idea that SR could provide the "link" between the

electric force and the magnetic force.

We started discussing this (I think in the COMPUSERVE Science group -

way back when) and my bubble was quickly burst by posters who

discounted this beautiful link. Their position seemed to be that:

1) There are relativistic effects that acount for some of the force

between the parallel wires, but...

2) Charges in motion also create a separate magnetic field that is

distinct from the relativistic effect of the charges in motion.

Someone pointed out that a single electron in motion would have a

magnetic field and this could not be explained by increased charge

density produced by SR effects.

Someone else said that the SR effects don't provide an explanation of

the self inductance (indictive reactance) in a coil.

Also, if the magnetic field is really just the electric field

undergoing length contraction due to SR effects, why do loop antennas

that are supposed to respond only to the magnetic field component of

an EM wave work?

Thanks, Bill

http://www.erols.com/wmeara

Feb 13, 2000, 3:00:00 AM2/13/00

to

In sci.physics.electromag Bill Meara <n2...@erols.com> wrote:

> ... Their position seemed to be that:

> 1) There are relativistic effects that acount for some of the force

> between the parallel wires, but...

> 2) Charges in motion also create a separate magnetic field that is

> distinct from the relativistic effect of the charges in motion.

I think what they mean is: SR effects require something like the

magnetic field to exist, but its shape is not completely determined

by SR alone.

After all, in a stationary case the electric force is a lot like the

gravitational force (albeit with another proportionality constant). In

both cases we can apply relativity, but gravitation does not have the

same magnetic part as electromagnetism. It does have a magnetic part

(called the gravito-magnetic force) but that is 4 times stronger (in

relation to the static part) than in the electromagnetic case.

-- Jos

Feb 13, 2000, 3:00:00 AM2/13/00

to

Jos Bergervoet wrote:

This is not at all Wonderland Jos.

Set an electron at the end of a positively charged segment.

Now give the electron motion towards the other end.

You will note that the center of positive charge moves in the opposite

direction to the electron, producing a very real drift of positive charge

relative to a charge at rest outside this segment, even though the

particles of positive charge do not move. Read silicon diode theory; the

part about hole conduction.

Richard

Feb 14, 2000, 3:00:00 AM2/14/00

to

> I'm one of those who admits to NOT understanding a lot of this. So I

> promise not to prpound any new theories. I'm an amateur just tying to

> understand some of this. Please forgive the very amateur tone of the

> comments that follow.

Unfortunately I don't know of any qualitative argument which will

give a convinving non-mathematical explanation. It really needs

some acquaintance with 4-vectors and vector calculus. So this won't

be a very satisfactory reply.

> I while back I came across a basic electronics text that said that

> the magnetic force can be completely explained by the relativistic

> effects of the charges in motion in the wires. I really liked this

> book -- I liked the idea that SR could provide the "link" between the

> electric force and the magnetic force.

So far, so good.

> We started discussing this (I think in the COMPUSERVE Science group -

> way back when) and my bubble was quickly burst by posters who

> discounted this beautiful link. Their position seemed to be that:

> 1) There are relativistic effects that acount for some of the force

> between the parallel wires, but...

> 2) Charges in motion also create a separate magnetic field that is

> distinct from the relativistic effect of the charges in motion.

Did they offer any *proof* of this statement?

The magnetic field *is* one of the relativistic effects.

> Someone pointed out that a single electron in motion would have a

> magnetic field and this could not be explained by increased charge

> density produced by SR effects.

Then they didn't understand what SR actually says. It doesn't just

predict increased charge density, but a velocity-dependent force

which is the magnetic field.

> Someone else said that the SR effects don't provide an explanation of

> the self inductance (indictive reactance) in a coil.

They do.

> Also, if the magnetic field is really just the electric field

> undergoing length contraction due to SR effects,

It isn't. Time comes into it as well.

> why do loop antennas

> that are supposed to respond only to the magnetic field component of

> an EM wave work?

As above.

--

Richard Herring | <richard...@gecm.com>

Feb 14, 2000, 3:00:00 AM2/14/00

to

Bill Meara wrote:

> 1) There are relativistic effects that acount for some of the force

> between the parallel wires, but...

> 1) There are relativistic effects that acount for some of the force

> between the parallel wires, but...

Hmmm. A Lorentz transform from the wire rest frame to any other inertial

frame will yield the same result as obtained in the wire rest frame.

Is that a "relativistic effect"? The force between the wires is well

described by Maxwell's equations; that is not a relativistic effect....

> 2) Charges in motion also create a separate magnetic field that is

> distinct from the relativistic effect of the charges in motion.

No, the magnetic field _is_ the relativistic effect of charges in

motion (of course there is also an electric field from them). In

its rest frame a charge has a purely electric field, but in a frame

in which it is moving there are both electric and magnetic fields.

This is due to the way the electromagnetic fields transform under

a Lorentz transform.

> Someone pointed out that a single electron in motion would have a

> magnetic field and this could not be explained by increased charge

> density produced by SR effects.

Yes it has a magnetic field. But yes it is predicted by SR, and that

prediction is in excellent agreement with actual measurements. This

is not due merely to "increased charge density", but is due to the

way the electromagnetic fields transform under a Lorentz transform.

If SR and Maxwell's equations did not agree with the way the world

actually works, no synchrotron in the world could possibly work

(they are designed using SR and the ME). They work.

> Someone else said that the SR effects don't provide an explanation of

> the self inductance (indictive reactance) in a coil.

They do.

> Also, if the magnetic field is really just the electric field

> undergoing length contraction due to SR effects, why do loop antennas

> that are supposed to respond only to the magnetic field component of

> an EM wave work?

1) the magnetic field is more than you say. 2) loop antennas are more

complicated than you say -- the changing electric field of a passing

E&M wave also induces a signal in one.

Tom Roberts tjro...@lucent.com

Feb 14, 2000, 3:00:00 AM2/14/00

to

Bill Meara wrote:

> Someone else said that the SR effects don't provide an explanation of

> the self inductance (indictive reactance) in a coil.

Since the motion is not in a straight line,

you would have to use GR instead of SR. That is

much more difficult.

It looks like you've already gotten some good answers.

I'll just mention that when the magnetic force is

derived from the electric force using the infinite

parallel wires, what you usually are shown is an

incomplete derivation. SR is applied selectively

in order to show you approximately how it works.

For example, Lorenz contraction is used, but the

forces aren't transformed. You won't get exactly

the correct answer.

Rather than add to the confusion, I'll see if I

can dig up an exact quote from Feynman or Purcell.

Feb 14, 2000, 3:00:00 AM2/14/00

to

Tom Roberts writes:

>> 2) Charges in motion also create a separate magnetic field that is

>> distinct from the relativistic effect of the charges in motion.

>> 2) Charges in motion also create a separate magnetic field that is

>> distinct from the relativistic effect of the charges in motion.

> No, the magnetic field _is_ the relativistic effect of charges in

> motion (of course there is also an electric field from them). In

> its rest frame a charge has a purely electric field, but in a frame

> in which it is moving there are both electric and magnetic fields.

> This is due to the way the electromagnetic fields transform under

> a Lorentz transform.

Hmm. Is this really a useful way to think of the magnetic field,

though? It never seemed so to me. Magnetic fields have their own

existence, at least macroscopically. Since E . B is a Lorentz

invariant, there usually is no frame in which the field is entirely an

E-field.

--

Johann Hibschman joh...@physics.berkeley.edu

Feb 14, 2000, 3:00:00 AM2/14/00

to

Bill Meara wrote:

> I while back I came across a basic electronics text that said that

> the magnetic force can be completely explained by the relativistic

> effects of the charges in motion in the wires. I really liked this

> book -- I liked the idea that SR could provide the "link" between the

> electric force and the magnetic force.

Here's what Feynman says (Lectures on Physics,

Vol 2, page 13-9) "If we take into account the

fact that forces also transform when we go from

one system to the other, we find that the two

ways of looking at what happens do indeed give

the same physical result for any velocity."

This is in the context of two infinite parallel

wires carrying constant current.

Then on page 26-2 "It is sometimes said, by

people who are careless, that all of electro-

dynamics can be deduced solely from the Lorentz

transformation and the Coulomb force. Of course,

that is completely false. ..." He goes on to

note that there are many tactic assumptions

that don't apply in the general case. This

quote is too long to reproduce entirely, you

need to read that entire section yourself to

see what he meant.

> Also, if the magnetic field is really just the electric field

> undergoing length contraction due to SR effects, why do loop antennas

> that are supposed to respond only to the magnetic field component of

> an EM wave work?

I've been working with loop antenna's in near field

situations and very small B fields, and I assure you

that they do respond also to E-field.

Feb 14, 2000, 3:00:00 AM2/14/00

to

In article <38A8567F...@mmm.com>, rbmcc...@mmm.com (Roy McCammon) wrote:

> Then on page 26-2 "It is sometimes said, by

> people who are careless, that all of electro-

> dynamics can be deduced solely from the Lorentz

> transformation and the Coulomb force. Of course,

> that is completely false. ..."

I always wanted to leave that up on the blackboard for my prof to find

when he walked in the room, since he was a big fan of this approach

(and wasn't quite fully rigorous). But I never had the guts.

Feb 15, 2000, 3:00:00 AM2/15/00

to

In article <38A83734...@mmm.com>, Roy McCammon (rbmcc...@mmm.com) wrote:

> Bill Meara wrote:

>

> > Someone else said that the SR effects don't provide an explanation of

> > the self inductance (indictive reactance) in a coil.

> Since the motion is not in a straight line,

> you would have to use GR instead of SR. That is

> much more difficult.

BZZZZZZT!!!!! Fallacy! Fallacy!

Not so. SR is perfectly capable of dealing with accelerated

frames; it merely becomes mathematically messy.

The one thing it can't handle which GR can is *gravity*,

which is not an issue here.

--

Richard Herring | <richard...@gecm.com>

Feb 15, 2000, 3:00:00 AM2/15/00

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Thanks to all who are helping me with my very amateur efforts to get

some understanding of the relationship between the electric and the

magentic fields. I realize that I won't really understand this until

I get a much better grasp of the math, but permit me to throw out a

few non-mathematical ideas -- please tell me if I'm missing the

(basic) point:

some understanding of the relationship between the electric and the

magentic fields. I realize that I won't really understand this until

I get a much better grasp of the math, but permit me to throw out a

few non-mathematical ideas -- please tell me if I'm missing the

(basic) point:

-- We start out with charged particles. At rest they have electrical

fields only.

-- When we put them in motion relative to a fixed observer, the motion

causes them to undego a Lorentz transform.

-- This is -- in effect -- a change in "charge density" and it is this

changed charge density that we perceive as the magnetic field.

-- That -- in a very basic way -- is how SR "unifies" the electric and

the magnetic fields.

Thanks, Bill

http://www.erols.com/wmeara

Feb 15, 2000, 3:00:00 AM2/15/00

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In article <38a9394f....@news.erols.com>, Bill Meara (n2...@erols.com) wrote:

> Thanks to all who are helping me with my very amateur efforts to get

> some understanding of the relationship between the electric and the

> magentic fields. I realize that I won't really understand this until

> I get a much better grasp of the math, but permit me to throw out a

> few non-mathematical ideas -- please tell me if I'm missing the

> (basic) point:

> Thanks to all who are helping me with my very amateur efforts to get

> some understanding of the relationship between the electric and the

> magentic fields. I realize that I won't really understand this until

> I get a much better grasp of the math, but permit me to throw out a

> few non-mathematical ideas -- please tell me if I'm missing the

> (basic) point:

> -- We start out with charged particles. At rest they have electrical

> fields only.

Yes.

> -- When we put them in motion relative to a fixed observer, the motion

> causes them to undego a Lorentz transform.

Causes *the fields* to be L. transformed.

> -- This is -- in effect -- a change in "charge density" and it is this

> changed charge density that we perceive as the magnetic field.

No, it's a rotation in Minkowski space, which mixes some space into

time, and vice versa. It does the same thing to the magnetic and

electric potentials which in turn cause some E to rotate into B.

--

Richard Herring | <richard...@gecm.com>

Feb 15, 2000, 3:00:00 AM2/15/00

to

Bill Meara wrote:

>

> Thanks to all who are helping me with my very amateur efforts to get

> some understanding of the relationship between the electric and the

> magentic fields. I realize that I won't really understand this until

> I get a much better grasp of the math, but permit me to throw out a

> few non-mathematical ideas -- please tell me if I'm missing the

> (basic) point:

>

> -- We start out with charged particles. At rest they have electrical

> fields only.

>

> Thanks to all who are helping me with my very amateur efforts to get

> some understanding of the relationship between the electric and the

> magentic fields. I realize that I won't really understand this until

> I get a much better grasp of the math, but permit me to throw out a

> few non-mathematical ideas -- please tell me if I'm missing the

> (basic) point:

>

> -- We start out with charged particles. At rest they have electrical

> fields only.

In their own inertial frame of reference, they

produce and feel only electric fields.

> -- When we put them in motion relative to a fixed observer, the motion

> causes them to undego a Lorentz transform.

> -- This is -- in effect -- a change in "charge density" and it is this

> changed charge density that we perceive as the magnetic field.

Yes, that would produce magnetic like effects.

But Lorentz contraction is not always present

(but SR is). For example, I could have a charged

disk moving perdedicular to its flat side. As it

went past you, you not see any Lorentz contraction

(the flat disk is already at zero thickness), but it

would see a changing B field.

> -- That -- in a very basic way -- is how SR "unifies" the electric and

> the magnetic fields.

yes. Or perhaps better said that SR allows us

to examine and understand the unity between

E & M.

Feb 15, 2000, 3:00:00 AM2/15/00

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Richard Herring wrote:

> Not so. SR is perfectly capable of dealing with accelerated

> frames; it merely becomes mathematically messy.

Perhaps you could elaborate.

Feb 15, 2000, 3:00:00 AM2/15/00

to

Be fore you smirk too much, you better read

the section.

He says you can deduce electrodynamics from

three facts.

1. The formula for the Coulomb potential for

a stationary charge.

2. The knowledge that the electromagnetic

potentials are a four vector.

3. The potential produced by a moving

charge depend only on the velocity

and position at the retarded time.

Feb 15, 2000, 3:00:00 AM2/15/00

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"z@z" wrote:

> : To be a little more specific, in every one of

> : these counter arguments that I have looked

> : at, the author fails to apply SR transformations

> : to the forces.

>

> If you believe in currently accepted physics, then you

> unconsciously conclude from any counter argument to the

> incompetence of its author.

My bias is not relevant. If you want to

show a theory is incorrect, you have to

show that when the theory is properly

applied it yields an incorrect result.

You cannot prove anything by misapplying

a theory, except your own lack of knowledge

about the theory. This is basic logic.

> If relativistic Maxwellian theory were correct, the

> identity between magnetic and electric fields would

> be a very fundamental principle.

It is.

> Really fundamental

> principles however, must finally turn out to be simple

> and clear. In 1905 Einstein had good reasons to assume

> that this case and similar open questions will be

> solved IN A TRANSPARENT WAY.

Einstien certanly preferred simple and clear,

but he never insisted that it was a requirment.

> So you believe in the authority of high priests such

> as Richard Feynman. I'm highly skeptical about

> Feynman's "science".

So your position is that you cannot show that it

works and you don't believe others who say it

does. That is at least a defensible position.

Nothing wrong with being skeptical.

> Many of his ideas bear too much

> resemblance with medieval theological concepts. QED

> photons "travel all possible distances in all possible

> times with all possible velocities". So they are not

> only able to be everywhere at the same time, they even

> know what will happen in the future!

You are reading too much into it. The math

says the result is "as if" the photons did

all those things in a probabilistic sense.

But we use that kind of math all the time.

For instance when we characterize the

"frequency response" for an amplifier

we use the mathematical fiction that

the stimulating sine wave has been present

since the infinite past and will be present

through the infinite future. Its not,

but it doesn't stop us from using the math.

> I do not believe in your "objective math".

In the case of two infinite wires with steady

current, the math is high school algebra level.

There are lots of reciprocal of square roots of

quadratics and many terms, but it is just

high school algebra. The SR knowledge is usually

acquired in the second year of a physics education.

If you don't do it a lot, its easy to make mistakes,

but, go to any first rate physics institution, and

you will find third year undergraduates that are

capable of verifying that case.

This is not inaccesible math and theory that must

be taken on faith.

> One of the most obvious contradictions of modern science

> is Maxwell's claim that all electromagnetic effects

> propagate at the speed of light.

Since light and electromagnetic effects are

the same thing, its hardly contradictory

that light travels at the speed of light.

> Apart from prejudice there is no convincing argument at

> all that the relations described by Maxwell's equations

> propagate at the same velocity as the wave solutions

> which can be derived from them.

Relations don't propagte. Prehaps you refer

to unobservable mathematical terms which

may be constructed. For example, if the

potential everywhere were 0 and I decided

to express that as +1 - 1, the -1 would

propagate instantly everywhere in the universe

but would be unobservable because the +1 would

also propagate and cancel it out. There are

certainly things like that which can be

construted from Maxwell.

> Or assume we have two identical circular circuit loops in

> two neighbouring parallel planes. "Simple" logic tells us

> that it is impossible the explain the magnetic attraction

> between the loops (in the case of currents in the same

> direction) by E fields resulting somehow from different

> line densities of charge (resulting somehow from SR length

> contraction).

Seems that Feynman agrees with you that it

takes more in general than Lorentz contraction.

Feb 15, 2000, 3:00:00 AM2/15/00

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Johann Hibschman wrote:

> [about magnetic fields as Lorentz transforms of electric fields]

> Hmm. Is this really a useful way to think of the magnetic field,

> though?

> [about magnetic fields as Lorentz transforms of electric fields]

> Hmm. Is this really a useful way to think of the magnetic field,

> though?

Sometimes.

> It never seemed so to me. Magnetic fields have their own

> existence, at least macroscopically. Since E . B is a Lorentz

> invariant, there usually is no frame in which the field is entirely an

> E-field.

Except in the important case of electrostatics.

Note also that ths usual interpretation of the Maxwell's equations is

that the charges cause both electric and magnetic fields. As there

are no magnetic monopoles, the only way to have a B field is due to

the Lorentz transform of the native electric field of some moving

charges.

An equally-valid interpretation is that the fields create the

charges. Then the best way to imagine this is to consider the

electromagnetic field as a two-form on spacetime, and that

inherently robs the magnetic and electric fields of any

individual identities.

Tom Roberts tjro...@lucent.com

Feb 15, 2000, 3:00:00 AM2/15/00

to

> > Not so. SR is perfectly capable of dealing with accelerated

> > frames; it merely becomes mathematically messy.

> Perhaps you could elaborate.

See the sci.relativity FAQ at

http://hepweb.rl.ac.uk/ppUK/PhysFAQ/acceleration.html

and various mirrors.

Really, there's not much to be said. 4-acceleration can be defined

and you can solve for the equation of motion of accelerating particles.

This isn't too surprising, since SR is consistent with Maxwellian

electrodynamics, and any theory involving forces is bound to

produce acceleration.

--

Richard Herring | <richard...@gecm.com>

Feb 15, 2000, 3:00:00 AM2/15/00

to

Bill Meara wrote:

> -- We start out with charged particles. At rest they have electrical

> fields only.

> -- We start out with charged particles. At rest they have electrical

> fields only.

When at rest wrt the observer, yes.

> -- When we put them in motion relative to a fixed observer, the motion

> causes them to undego a Lorentz transform.

Hmmm. Say rather that since they are moving wrt the observer the

observer sees a Lorentz transform of the properties they have in

their restframe. The charges themselves remain unchanged; all that

changes is the observer's perspective of them.

> -- This is -- in effect -- a change in "charge density" and it is this

> changed charge density that we perceive as the magnetic field.

Yes it changes the charge density, but that does not become a

magnetic field, that merely means that the electric field is

different from when the charges were at rest. In any inertial

frame (e.g. that of the observer here) Maxwell's equations

show that charge density _measured_in_that_frame_ generates

an electric field. It is the _motion_ of the charges which

generate the magnetic field, not any change in their charge

density. This occurs because the Lorentz transform of their

electric field in their rest frame becomes a combination of

electric and magnetic fields in the observer's frame.

Note that for a line of charges moving along their line (e.g. a

current in a wire) the magnetic field is perpendicular to their

motion wrt the observer -- a simple change in charge density

cannot do that, but a Lorentz transform of their rest-frame

fields _can_.

> -- That -- in a very basic way -- is how SR "unifies" the electric and

> the magnetic fields.

The electric and magnetic fields in any inertial frame are

related by the appropriate Lorentz transform to the electric

and magnetic fields in any other inertial frame. That transform

intermixes the components of the electric and magnetic fields

in such a complicated way that it is better to think of them as

the 6 components of the electromagnetic field (rather than as

individual fields with 3 components each).

You may have seen Maxwell's equations written in terms of

electric and magnetic fields -- rather complicated aren't

they? Written in their natural 4-dimensional language in

terms of the electromagnetic field they are quite simple

(and the 4 original equations become 2).

Tom Roberts tjro...@lucent.com

Feb 17, 2000, 3:00:00 AM2/17/00

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Hello Richard.

I am following E=B, which I comprehend. I dont agree however. My problem is

I dont believe that the b field exists at all, I refer to the conventional

magnetic field as the b field. For what its worth, I reference

http://www.terraworld.net/leepugh/

Please read the last half first.

I am following E=B, which I comprehend. I dont agree however. My problem is

I dont believe that the b field exists at all, I refer to the conventional

magnetic field as the b field. For what its worth, I reference

http://www.terraworld.net/leepugh/

Please read the last half first.

If you can simply convince me that the magnetic field is not an illusion, I

would like for you to do that. Believe me, it would save me a lot of grief.

I do not like my position any more than you do, however I seem to be picking

up a following, and If I am in error, I would like to try to fix it now

rather than later.

Regards, Lee Pugh

Feb 17, 2000, 3:00:00 AM2/17/00

to

Lee Pugh wrote:

> I dont believe that the b field exists at all,

It is real in the sense that you can use it

to make accurate predictions.

Per Feynman, it is not a real physical field

in the sense that it is not local.

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