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How does a Faraday disk really work?



 
 
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  #1  
Old August 2nd 08 posted to sci.physics.electromag
Salmon Egg
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Posts: 412
Default How does a Faraday disk really work?

After all these these years dealing with electromagnetism, I still do
not fully understand a Faraday disk based upon Maxwell's equations. To
be specific let me describe a particular configuration.

1. There is a uniform magnetic field in the vertical direction
extending over the region that will contain all the apparatus.

2. The axle mounting a circular disk is vertical so that the disk is in
the horizontal plane. This axis and disk, both of conducting material is
what spins.

3. To make electrical connection, there is a conducting circular
cylindrical skirt attached the disk's periphery. This skirt dips into a
circular trough containing mercury. A similar but smaller skirt and
trough arrangement is used to connect to the axle shaft. Connections to
measure what happens are made to the trough bodies. These wires are not
moved during the testing.

4. These wires are connected to a variable dc source through a
galvanometer. When running, the source is adjusted so that the current
flow is zero. By using this potentiometric technique, there will be no
current to confuse what is happening. Maxwell's equations will also be
simplified.

Maxwell's equations for this case a

div B = 0. Always true anyway.

div D = 0. There is no free charge. There is the possible
- exception at the disk edge and the like, but I do
- not think this charge is significant.

curl E = 0. There is no time variation of the magnetic field.

curl H = 0. There is no conducted or displacement current.

When the disk turns, how do you explain the potential difference
measured by the external potentiometer?

Something has to be added to the Maxwell equations. For example, I think
that you have to add an equation representing the definition of the
electric field. Charge is a concept that is not explicitly defined but
which can be measured in terms of forces between charges. Given a charge
of size q, the electrical field is determined by F = q E once the
mechanical force is measured. Once you know and measure charge, you know
what current is.

I think that the other thing that needs to be known is equivalent to
knowing how the electromagnetic tensor in 4-space transforms with
motion. that is, relativistically, when a point moves through a magnetic
field, it sees an electric field introduced by the motion at the point.

This electric field may be called the Lorentz force or something
similar, but it seems to be something that must be added to the
equations I have already listed. This arises simply out of relativity
where the EM field has to be a tensor and has to transform properly with
a Lorentz transformation.

Questions:

1. Can this Lorentz force be derived without invoking relativity or
another law of nature? That is, are Maxwell's equations sufficient?

2. Given Maxwell's equations and F = q E, can the forces between wires
carrying current be derived?
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  #2  
Old August 2nd 08 posted to sci.physics.electromag
Benj
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Posts: 2,214
Default How does a Faraday disk really work?

On Aug 1, 8:41*pm, Salmon Egg wrote:

This electric field may be called the Lorentz force or something
similar, but it seems to be something that must be added to *the
equations I have already listed. This arises simply out of relativity
where the EM field has to be a tensor and has to transform properly with
a Lorentz transformation.

Questions:

1. *Can this Lorentz force be derived without invoking relativity or
another law of nature? That is, are Maxwell's equations sufficient?


I see that you've come to realize that all the faith-based physics
dogma as spewed by PBS and others simply doesn't cut it. The dogma
that says, "All electromagnetic phenomena can be "explained" by
Maxwell's Equations" is wrong. They can't even explain a simple
Faraday Generator! So, No, Maxwell's equations alone are not
sufficient. And, Yes, this is an excellent place to poke your nose!

It is the Lorentz relation that is missing. It is related to Maxwell's
equations but is not one of them! It is often taken to be the
defining relation for B. The Lorentz relation is written as:

F = q(E +(v X B)) And the law does not depend upon the inertial
reference frame in which the various quantities are measured. .

For more information I urge you to examine Jefimenko's expositions on
the subject, especially his derivation of the transformations of the
Lorentz force. "Electromagnetic Retardation and the Theory of
Relativity" by Oleg D. Jefimenko available from Amazon.com.

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

"Some people say I'm a dreamer,
But I'm not the only one!
I hope one day you'll join us,
And Electromagnetics can live as one! "

  #3  
Old August 2nd 08 posted to sci.physics.electromag
Salmon Egg
external usenet poster
 
Posts: 412
Default How does a Faraday disk really work?

In article
,
Benj wrote:

Questions:

1. *Can this Lorentz force be derived without invoking relativity or
another law of nature? That is, are Maxwell's equations sufficient?


I see that you've come to realize that all the faith-based physics
dogma as spewed by PBS and others simply doesn't cut it. The dogma
that says, "All electromagnetic phenomena can be "explained" by
Maxwell's Equations" is wrong. They can't even explain a simple
Faraday Generator! So, No, Maxwell's equations alone are not
sufficient. And, Yes, this is an excellent place to poke your nose!

It is the Lorentz relation that is missing. It is related to Maxwell's
equations but is not one of them! It is often taken to be the
defining relation for B. The Lorentz relation is written as:

F = q(E +(v X B)) And the law does not depend upon the inertial
reference frame in which the various quantities are measured. .

For more information I urge you to examine Jefimenko's expositions on
the subject, especially his derivation of the transformations of the
Lorentz force. "Electromagnetic Retardation and the Theory of
Relativity" by Oleg D. Jefimenko available from Amazon.com.


Wow! You really have a chip on your shoulder.

I never doubted that Maxwell's equations are correct. Whether or not
they are, the theory of relativity will hold. That is sufficient in my
mind to explain how the Faraday disk works. The Lorentz force you
describe is a way to understand relativity on the cheap without having
to invoke four-dimensional tensors. Engineers have come up with such
devices over the years. One example is the Mohr circle for understanding
stress and strain without resorting to tensors.

I have always heard that the Maxwell equations are relativistically
invariant but have never seen a proof--certainly not a simple proof. If
so, do these equations somehow hide the transformation of the
electromagnetic tensor that combines electric and magnetic fields into
one unified field?

I already realize that I need F = q E to define the electric field for
use in Maxwell's equations. What is the equivalent definition for B? Is
it F = q(E +(v X B)) ? What is the history of that?

Bill
  #4  
Old August 3rd 08 posted to sci.physics.electromag
Timo A. Nieminen
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Posts: 1,216
Default How does a Faraday disk really work?

On Fri, 1 Aug 2008, Salmon Egg wrote:

After all these these years dealing with electromagnetism, I still do
not fully understand a Faraday disk based upon Maxwell's equations.

[cut]
Something has to be added to the Maxwell equations.

[cut]
This electric field may be called the Lorentz force or something
similar, but it seems to be something that must be added to the
equations I have already listed.


Sure, the Maxwell equations alone are not sufficient. They tell you how
the fields depend on the sources. How do the sources depend on the fields?
For that, you need the consitutive relations D = epsilon E, B = mu H,
J = conductivity E.

This arises simply out of relativity
where the EM field has to be a tensor and has to transform properly with
a Lorentz transformation.


.... or how epsilon and mu transform. Epsilon and mu contain the effects of
electric polarisation and magnetisation. The question concerns a rotating
magnetised disk.

Questions:

1. Can this Lorentz force be derived without invoking relativity or
another law of nature? That is, are Maxwell's equations sufficient?


A qualified "no". In the sense that the Lorentz force provides an
operational definition of E and B, the Maxwell equations don't make any
sense without it, so it's already there. In the sense that the Maxwell
equations don't tell you anything about how fields act on the sources,
it's an extra add-on.

Maxwell included the effects of fields on sources in _his_ set of
equations. The 4-equation Hertz-Heaviside distillation is what omits them
in the usual telling of the story. Neither Hertz nor Heaviside ignored
them though, but considering that the constitutive relations and Lorentz
force are not PDEs, why should they be lumped in with the 4 PDEs that tell
us about the behaviour of the fields?

2. Given Maxwell's equations and F = q E, can the forces between wires
carrying current be derived?


No. Given these and special relativity, yes (and I've seen students
discover that it's harder than they first thought, but educational). You
don't need the entire Maxwell equations, all you need is Coulomb's law +
SR.

--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/...,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html

  #5  
Old August 3rd 08 posted to sci.physics.electromag
Timo A. Nieminen
external usenet poster
 
Posts: 1,216
Default How does a Faraday disk really work?

On Sat, 2 Aug 2008, Salmon Egg wrote:

I have always heard that the Maxwell equations are relativistically
invariant but have never seen a proof--certainly not a simple proof. If
so, do these equations somehow hide the transformation of the
electromagnetic tensor that combines electric and magnetic fields into
one unified field?


No. The Maxwell equations are just a bunch of PDEs. They're
relativistically invariant under the Lorentz transformations, but also
under the Galilei transformations. Again, along the lines of your original
post, the Maxwell equations alone are not the whole story.

What makes classical EM "relativistic" is epsilon_0 and mu_0 being the
same in all inertial reference frames. Note that Maxwell and Hertz assumed
that D = e0 E and B = u0 H only held in the rest frame of the ether. See
Maxwell's "G", the velocity of a point (relative to the ether) in his
Treatise, and Hertz's paper on the electrodynamics of moving media. Until
Lorentz, the theory of classical electrodynamics was _not_ relativistic
(in the modern sense of "relativistic"). Of course, until Lorentz, it also
failed to describe electromagnetism in moving media properly ...

Given a field equation, describing how a field A depends on source B,

C(A) = B,

where C is some linear differential operator, and an equation giving the
force (density?) acting on the source,

F = D(B,E)

where D is a function of B and some field E,

and a consitutive relation telling us how E depends on A,

E = a A,

if we assume a to be a Lorentz-invariant scalar (i.e., the same in all
inertial reference frames), what restrictions are there on C and D for the
whole theory to be Lorentz invariant?

I don't know the answer to this, but I do know that the key element in the
"relativistic" part is "a" being Lorentz-invariant. What C and D will tell
you is the relativistic transformation of the fields.

Which is why the Lorentz-relativistic transformation of EM fields differs
from the Galilei-relativistic transformation of EM fields, even though the
EM fields in both cases satisfy the Maxwell equations.

I already realize that I need F = q E to define the electric field for
use in Maxwell's equations. What is the equivalent definition for B? Is
it F = q(E +(v X B)) ?


Yes. In modern terms.

What is the history of that?


Good question. Perhaps one needs to go back and read Ampere for this. I've
read that Ampere never wrote down what we call "Ampere's law". But perhaps
it's in there. It's there by Maxwell's time.

--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/...,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html

  #6  
Old August 3rd 08 posted to sci.physics.electromag
Benj
external usenet poster
 
Posts: 2,214
Default How does a Faraday disk really work?

On Aug 2, 6:28*pm, "Timo A. Nieminen" wrote:

2. *Given Maxwell's equations and F = q E, can the forces between wires
carrying current be derived?


No. Given these and special relativity, yes (and I've seen students
discover that it's harder than they first thought, but educational). You
don't need the entire Maxwell equations, all you need is Coulomb's law +
SR.


Actually the whole issue of electromagnetic forces is a problem. It is
discussed in some detail in appendix 3 of the reference I just gave
you. He calculates the "force" between two dielectric plates 4 ways
and then between two parallel wires 4 ways. Each gives the same VALUE
of the "force", but in each case the force is acting in a different
place. This leads to a conclusion that the action of "force" from a
field actually is only a mathematical device and does not exist in the
mechanical sense. Further investigation shows that it is actually
MOMENTUM transferred from the fields that is the essence of what
appears to be electromagnetic force. It's all a very interesting
exposition.

Personally I try to never get involved in the on-going USENET "who is
smarter than Einstein" relativity debates.
I have my own dead horses to beat!

  #7  
Old August 7th 08 posted to sci.physics.electromag
blackhead
external usenet poster
 
Posts: 209
Default How does a Faraday disk really work?

On 2 Aug, 16:59, Benj wrote:
On Aug 1, 8:41*pm, Salmon Egg wrote:

This electric field may be called the Lorentz force or something
similar, but it seems to be something that must be added to *the
equations I have already listed. This arises simply out of relativity
where the EM field has to be a tensor and has to transform properly with
a Lorentz transformation.


Questions:


1. *Can this Lorentz force be derived without invoking relativity or
another law of nature? That is, are Maxwell's equations sufficient?


I see that you've come to realize that all the faith-based physics
dogma as spewed by PBS and others simply doesn't cut it. The dogma
that says, "All electromagnetic phenomena can be "explained" by
Maxwell's Equations" is wrong. They can't even explain a simple
Faraday Generator! *So, No, Maxwell's equations alone are not
sufficient. And, Yes, this is an excellent place to poke your nose!

It is the Lorentz relation that is missing. It is related to Maxwell's
equations but is not one of them! *It is often taken to be the
defining relation for B. *The Lorentz relation is written as:

F = q(E +(v X B)) *And the law does not depend upon the inertial
reference frame in which the various quantities are measured. .

For more information I urge you to examine Jefimenko's expositions on
the subject, especially his derivation of the transformations of the
Lorentz force. *"Electromagnetic Retardation and the Theory of
Relativity" by Oleg D. Jefimenko available from Amazon.com.


But the Lorentz Law is a special case of Faraday's Law in differential
form when B is time independent. I don't understand where you're
getting this idea from that Maxwell's equations aren't sufficient to
explain the Faraday generator.

Title:
A Simple Proof that the Lorentz Force, Law Implied Faraday's Law of
Induction, when B is Time Independent

http://adsabs.harvard.edu/abs/1973AmJPh..41..713D


Deriving the Lorentz force equation from Maxwellapos;s equations
http://ieeexplore.ieee.org/Xplore/lo...32..pdf?temp=x




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

"Some people say I'm a dreamer,
But I'm not the only one!
I hope one day you'll join us,
And Electromagnetics can live as one! "


  #8  
Old August 7th 08 posted to sci.physics.electromag
Salmon Egg
external usenet poster
 
Posts: 412
Default How does a Faraday disk really work?

In article
,
blackhead wrote:


But the Lorentz Law is a special case of Faraday's Law in differential
form when B is time independent. I don't understand where you're
getting this idea from that Maxwell's equations aren't sufficient to
explain the Faraday generator.

Title:
A Simple Proof that the Lorentz Force, Law Implied Faraday's Law of
Induction, when B is Time Independent

http://adsabs.harvard.edu/abs/1973AmJPh..41..713D


Deriving the Lorentz force equation from Maxwellapos;s equations
http://ieeexplore.ieee.org/Xplore/lo.../00995632.pdf?
temp=x


The only possible Maxwell equation in differential form that describes
Faraday induction must be:

curl E = -B/t {partial of B wrt t if symbols don't show}

This seems useless if B does not vary with time.

The references also seem of little value. I might be able to dig up the
Harvard reference at a local university library. Its title is so
ungrammatical as to be of no value.

Even though I am an IEEE member, I am not about to spend the money for a
subscription.

It seems clear to me that special relativity is the easy approach to
flux "cutting." I just do not see flux cutting arising from the Maxwell
equation. I am beginning to believe that using the Lorentz force as a
law by itself is a back road to describe relativity.

Where do we go from here?

Bill
  #9  
Old August 7th 08 posted to sci.physics.electromag
blackhead
external usenet poster
 
Posts: 209
Default How does a Faraday disk really work?

On 7 Aug, 06:23, Salmon Egg wrote:
In article
,





*blackhead wrote:

But the Lorentz Law is a special case of Faraday's Law in differential
form when B is time independent. I don't understand where you're
getting this idea from that Maxwell's equations aren't sufficient to
explain the Faraday generator.


Title:
*A Simple Proof that the Lorentz Force, Law Implied Faraday's Law of
Induction, when B is Time Independent


http://adsabs.harvard.edu/abs/1973AmJPh..41..713D


Deriving the Lorentz force equation from Maxwellapos;s equations
http://ieeexplore.ieee.org/Xplore/lo...814/21481/0099....
temp=x


The only possible Maxwell equation in differential form that describes
Faraday induction must be:

curl E = - B/ t *{partial of B wrt t if symbols don't show}

This seems useless if B does not vary with time.


huh?

(@Ez/@y - @Ey/@z)i = 0
(@Ex/@z - @Ez/@x)j = 0
(@Ey/@x - @Ex/@y)k = 0

I don't see how these equation are useless just because there's a 0 on
the RHS

The references also seem of little value. I might be able to dig up th
Harvard reference at a local university library. Its title is so
ungrammatical as to be of no value.

Even though I am an IEEE member, I am not about to spend the money for a
subscription.

It seems clear to me that special relativity is the easy approach to
flux "cutting." I just do not see flux cutting arising from the Maxwell
equation. I am beginning to believe that using the Lorentz force as a
law by itself is a back road to describe relativity.

Where do we go from here?


Maxwell's equations can be derived from Coulomb's Law, Special
Relativity and Hamilton's principle if you want to look at
fundamentals:

http://arxiv.org/abs/physics/0501130

"Flux cutting" and "tubes of flux" are pictures that were created by
Faraday around 200 years ago and are no longer a cutting edge
explanation of electromagnetism nowadays, even though still widely
practised by engineers.


Bill- Hide quoted text -

- Show quoted text -


  #10  
Old August 8th 08 posted to sci.physics.electromag
Benj
external usenet poster
 
Posts: 2,214
Default How does a Faraday disk really work?

On Aug 6, 7:34*pm, blackhead wrote:
On 2 Aug, 16:59, Benj wrote:


For more information I urge you to examine Jefimenko's expositions on
the subject, especially his derivation of the transformations of the
Lorentz force. *"Electromagnetic Retardation and the Theory of
Relativity" by Oleg D. Jefimenko available from Amazon.com.


But the Lorentz Law is a special case of Faraday's Law in differential
form when B is time independent. I don't understand where you're
getting this idea from that Maxwell's equations aren't sufficient to
explain the Faraday generator.


Actually the Lorentz Law is a special case of Faraday Induction where
the moving magnet represents elementary currents. These currents
produce a force upon a charge as given by the Lorentz equation. But
historically the Lorentz equation is taken as a definition of B which
places it outside the Maxwell's equations. But even if we take the
view above that it is derivable as a coordinate transformation the
forces themselves call into question whether the equations involved
are "fundamentaL" or part of Maxwell's set.

I quote from the above reference: P 322. "Although Eq.( A3.45)
(momentum transfer between mechanical and electromagnetic momentum) is
usually considered a derived equation subordinate to Lorentz force
equation, our analysis shows that Eq. (A3.45) is a fundamental
electromagnetic equation, and that it is quite correct to regard
Loretnz force equation as a consequence of Eq. (A3.45)"

Hence it follows that if the Lorentz eq is a consequence of a
fundamental momentum equation outside the Maxwellian set, it follows
that Maxwell's equations alone are not sufficient to describe all
electromagnetic phenomena as is widely asserted. OK?

 




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