# B=E

1 view

### Richard Perry

Feb 4, 2000, 3:00:00 AM2/4/00
to
"Bergervoet J.R.M." wrote:

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

### z@z

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

### Bergervoet J.R.M.

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

### Richard Perry

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

### Bergervoet J.R.M.

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

### Roy McCammon

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.

> 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.

### z@z

Feb 9, 2000, 3:00:00 AM2/9/00
to
: = Roy McCammon
:: = 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

### Richard Perry

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
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

### Richard Perry

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.

### Roy McCammon

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.

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

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
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.

### Bergervoet J.R.M.

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

>
> -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

-- Jos

### Richard Perry

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.

### z@z

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

:: 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

: 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.

### Bergervoet J.R.M.

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

### Richard Perry

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

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

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.)

### Tom Roberts

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 ...]

> 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

### Tom Roberts

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.

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.

> 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
> 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

### Tom Roberts

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

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

### Jos Bergervoet

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

### Bill Meara

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

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

### Jos Bergervoet

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

### Richard Perry

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

Richard

### Richard Herring

Feb 14, 2000, 3:00:00 AM2/14/00
to
In article <38a696be...@news.erols.com>, Bill Meara (n2...@erols.com) wrote:
> 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

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

> 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>

### Tom Roberts

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...

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

### Roy McCammon

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.

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

Rather than add to the confusion, I'll see if I
can dig up an exact quote from Feynman or Purcell.

### Johann Hibschman

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.

> 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

### Roy McCammon

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.

### Nathan Urban

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.

### Richard Herring

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>

### Bill Meara

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

-- 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

### Richard Herring

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

> -- 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>

### Roy McCammon

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.

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.

### Roy McCammon

Feb 15, 2000, 3:00:00 AM2/15/00
to
Richard Herring wrote:

> Not so. SR is perfectly capable of dealing with accelerated
> frames; it merely becomes mathematically messy.

Perhaps you could elaborate.

### Roy McCammon

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.

### Roy McCammon

Feb 15, 2000, 3:00:00 AM2/15/00
to
"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.

### Tom Roberts

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

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

### Richard Herring

Feb 15, 2000, 3:00:00 AM2/15/00
to
In article <38A95C75...@mmm.com>, Roy McCammon (rbmcc...@mmm.com) wrote:
> Richard Herring wrote:

> > 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>

### Tom Roberts

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.

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

### Lee Pugh

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

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

### Roy McCammon

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.