Relativity and Aharonov-Bohm
Lawrence Foard
be performed without an electron crossing an area of changing vector
potential (if only in direction). Which from a relativistic view point
translates to a changing electric potential (electric field). Obviously
altering its motion. So first thought is "whats the big deal".
But this doesn't seem right because one observer sees a field energy
and another does not.
I'm also assuming if Aharonov-Bohm was so simply resolved it would not
be of any interest. Does anyone understand why its supposed to show no
effect on electrons unless quantum mechanics is used? What am I missing
with the relativistic transformation of the vector potential?
--
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Rave: Immanentization of the Eschaton in a Temporary Autonomous Zone.
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-Benjamin Franklin
Pmb
"Lawrence Foard" <ent...@farviolet.com> wrote in message
news:b81533$fle$1...@farviolet.com...
> What am I missing here. I can't picture a way this experiment could
> be performed without an electron crossing an area of changing vector
> potential (if only in direction).
Hmmm! Interesting!
You know something Lawrence? Every great once in a while there comes along a
really intersting post. And this one sure is interesting!!!
Thank you *very* much for mentioning this.I assume that you are referring to
the fact that
E = -grad Phi + &A/&t
So that if A has a spatial dependence than even though curl A vanishes, in
another frame A is time dependant and thus there is an electric field?
Cool! I had never thought of that before. But bravo for seeing it!
Pmb
Bilge
>be performed without an electron crossing an area of changing vector
>potential (if only in direction). Which from a relativistic view point
>translates to a changing electric potential (electric field). Obviously
>altering its motion. So first thought is "whats the big deal".
solenoid are both zero. Changing frames can change magnetic fields
inot electric fields or electric fields into magnetic fields, but
can't change fields which are both zero into non-zero fields.
E&M is gauge invariant, i.e., for any potential, A^{u}, you can
add the gradient of a scalar function with F = F(x^{u}) a function
of the coordinates, such that the result:
A^{u} -> A^{u}' = A^{u} - d^{u}F
doesn't give different fields. Classically, the vector potential has
no meaning whatsoever. It's a mathematical artifice.
>But this doesn't seem right because one observer sees a field energy
>and another does not.
And, it isn't right. No observers see any fields.
>I'm also assuming if Aharonov-Bohm was so simply resolved it would not
>be of any interest. Does anyone understand why its supposed to show no
>effect on electrons unless quantum mechanics is used? What am I missing
>with the relativistic transformation of the vector potential?
Gauge invariance. There are no E or B fields quantum mechanically,
either. The aharanov-bohm effect is due to the local gauge invariaance
of the electron wavefunction. The solution to the schroedinger equation
for an electron in the presence of a vector potential, is:
\Psi(x,t) = \phi(x,t)exp(-iS/hbar)
where the phase, S = e\integral dx.A, where the integral is over the
path, dx.
Write the electron wavefunction in two parts:
\Psi(x,t) = \phi(x,t)exp(-iS1/hbar) + \chi(x,t)exp(-iS2/hbar)
You then have two path integrals and the difference in the paths is
the closed loop:
S1 - S2 = e\integral dx.A(x)
This is just a phase difference between the two wavefuntions passing
to either side above. For a fully relativistic expression, substitute
dx_{u}A^{u} for dx.A(x).
Lawrence Foard
Thanks.
Now for the fly in the soup :)
In the electron frame there is a field energy 1/2E^2 (hope I'm remembering
that right). In the lab frame there is supposedly no energy because there are
no fields. It also gives rise to a non zero E&M field tensor in one reference
frame which is completely 0 in another.
Which makes me think I must be missing something.
Bilge
It's also not true. E&M is gauge invariant. Use the expression you
wrote above, include the restrictions imposed by the magnetic field
(which is also zero) and make the result covariant.
Hypermars
> But this doesn't seem right because one observer sees a field energy
> and another does not.
I agree with pmb, this is very interesting. As my knowledge of relativity is
very limited, would you please show me what electric field is present in the
following experimental setup, if we are in a reference system which travels
with the electron?
The simplest AB-experiment I can think of is an infinite solenoid of
negligible radius, carrying a magnetic flux \phi, lying along the y-axis of
a cartesian reference system. Its magnetic field can be written as
B = \phi \delta(x) \delta(z) {0,1,0}
and its vector potential can be chosen as
A = \phi/(2 \pi) 1/(x^2+z^2) {-z,0,x}
so that the observable quantity "electron-optical phase shift" (observable
by electron holography for example) associated to the circular integral of
the vector potential (it is a gauge-invariant quantity, of course) is
\varphi(x) = \pi/2 Sign(x)
What happens in the electron reference system? What is the new B of the
solenoid, the new A and E (or electrostatic potential)? the electron is
traveling nearly at the speed of light (we can assume that its accelerating
voltage is 300kV, and it should be possible to calculate \gamma just from
knowing that) and it propagates perpendicular to the (x,y) plane (it should
be considered a plane wave, but maybe this is already going out of
relativity towards QM).
Thanks
Bye
Hyper
Pmb
"Lawrence Foard" <ent...@farviolet.com> wrote
> Thanks.
>
> Now for the fly in the soup :)
Yum! :-)
> In the electron frame there is a field energy 1/2E^2 (hope I'm remembering
> that right).
Good enough. I think there are units in which you can write it like that but
I've never been good at memorizing systems of units etc.
> In the lab frame there is supposedly no energy because there are
> no fields. It also gives rise to a non zero E&M field tensor in one
reference
> frame which is completely 0 in another.
First off let me see if I get this straight. Is the vector potential really
non-uniform in the lab frame? Is the Aharonov-Bohm effect dependant on the
field being non-uniform?
> Which makes me think I must be missing something.
I think that you have to take a literal look at it. In a real life situation
a real magnetic field will not have a vector potential in one area and
vanish everywhere else. What is required is that in each frame the total
energy must be constant. Energy conservation does not imply that if energy
is conserved one frame then its conserved in all frames. And energy
conservation does not mean that if energy is zero in one frame then it has
to be non-zero in another frame.
Pmb
Lawrence Foard
Bilge <cra...@fghfgigtu.com> wrote:
> Lawrence Foard:
> >What am I missing here. I can't picture a way this experiment could
> >be performed without an electron crossing an area of changing vector
> >potential (if only in direction). Which from a relativistic view point
> >translates to a changing electric potential (electric field). Obviously
> >altering its motion. So first thought is "whats the big deal".
>
> The electric and magnetic fields everywhere outside of a long
>solenoid are both zero. Changing frames can change magnetic fields
>inot electric fields or electric fields into magnetic fields, but
>can't change fields which are both zero into non-zero fields.
>E&M is gauge invariant, i.e., for any potential, A^{u}, you can
>add the gradient of a scalar function with F = F(x^{u}) a function
>of the coordinates, such that the result:
>
> A^{u} -> A^{u}' = A^{u} - d^{u}F
>
>doesn't give different fields. Classically, the vector potential has
>no meaning whatsoever. It's a mathematical artifice.
I'm thinking rather of a 4-vector potential:
(0,a,0,0)
Where a has a non zero derivative with respect to x.
As seen by an observer moving in the x direction. You now
have something like:
(b,c,0,0)
Where b and c now have non zero derivatives with respect to x.
Doesn't matter if you change the gauge you still have an electric
field in the reference frame of the electron.
> >But this doesn't seem right because one observer sees a field energy
> >and another does not.
>
> And, it isn't right. No observers see any fields.
Yet relativistic transformation of a vector potential (0,a,0,0) where
a varies with position appears to show an electric field in the moving
frame. I know this presents problems that why I asked about it.
> >I'm also assuming if Aharonov-Bohm was so simply resolved it would not
> >be of any interest. Does anyone understand why its supposed to show no
> >effect on electrons unless quantum mechanics is used? What am I missing
> >with the relativistic transformation of the vector potential?
>
> Gauge invariance. There are no E or B fields quantum mechanically,
>either. The aharanov-bohm effect is due to the local gauge invariaance
>of the electron wavefunction.
It certainly works out quantum mechanically. But there seems to be a
contradiction in the classical description of the experiment. In one
reference frame I expect to see an effect, in another I do not. It
seems highly unlikely but possible that what we are seeing
is not a superiority of quantum mechanics, but rather an internal
inconsistency in classical relativistic E&M.
This may be terribly wrong but,
Using STA notation, let y0, y1, y2, y3 represent the unit vectors for
t, x, y, z.
The standard E&M field
F = E + iB
F is composed of 6 bivector quantities.
F is related to the vector potential A by:
F = grad A
To simplify for a region of space let A be of the form, where X is a
location in space time.
A = f(X)*y1
The gradient appears to give three bivectors and a scalar, something
like:
F = (df(X)/dt)*y1*y0 + (df(X)/dx) + (df(X)/dy)*y2*y0 + (df(X)/dz)*y3*y0
Notice that there is now a scalar component thats not normally represented
in the E or B fields which together represent only 6 bivectors. df(X)/dx
is non zero in a region where a component of A varies with space. Such
as that outside of the solenoid.
Under the lorentz transform, this scalar component will mix with the
bivector components. If we manage to contrive an experiment where the
bivector components of F are zero but the scalar component is not.
The lorentz transform will create a bivector component which contains
y0 and the direction of motion, exactly what we observe, an electric
field either assisting or opposing the electron in its direction of
motion.
What about the field energy?
1/2Fy0F
A scalar component of F will end up contributing to y0 component
of energy/momentum. So this space varying A would appear to contain
an energy which is not accounted for in the E + B fields.
I assume a similar argument can be made using the 16 element E&M
field tensor. I presume resulting in some of the 0 components along
the diagonal becoming non zero. Why are they 0? Are they defined as
zero simply by definition because the E+B fields contain only
6 components believed to represent everything?
There is only one reason I believe this must be all wrong. Thats because its
so obvious that it couldn't have gone unnoticed. What am I doing wrong?
Lawrence Foard
Lawrence Foard <ent...@farviolet.com> wrote:
> I assume a similar argument can be made using the 16 element E&M
> field tensor. I presume resulting in some of the 0 components along
> the diagonal becoming non zero. Why are they 0? Are they defined as
> zero simply by definition because the E+B fields contain only
> 6 components believed to represent everything?
Actually it appears that the scalar mentioned in the last message
is excluded from the normal 16 element tensor by definition. It would
be interesting if geometric algebra actually provides something which
the normal relativistic definition of fields does not. So maybe it
really was never noticed.
Lawrence Foard
Pmb <peter....@verizon.net> wrote:
>
>> frame which is completely 0 in another.
>
>First off let me see if I get this straight. Is the vector potential really
>non-uniform in the lab frame? Is the Aharonov-Bohm effect dependant on the
>field being non-uniform?
That would be the obvious flaw. But I believe the vector potential magnitude
has to fall off with distance? The double slit experiment shows a particle
coming from some distance, going around both directions, and then hitting
a screen at some distance. From the particles frame I believe (but
am probably wrong) that one has fought against a field in approaching the
solenoid, and one was helped along by it.
>> Which makes me think I must be missing something.
>
>I think that you have to take a literal look at it. In a real life situation
>a real magnetic field will not have a vector potential in one area and
>vanish everywhere else. What is required is that in each frame the total
>energy must be constant. Energy conservation does not imply that if energy
>is conserved one frame then its conserved in all frames. And energy
>conservation does not mean that if energy is zero in one frame then it has
>to be non-zero in another frame.
Or conversily if energy is non zero in one frame its non zero in all. I believe
that is true. Yet conventional E&M applied to this situation appears to give a
contradictory answer. I posted another message where Hestene's STA appears to
provide an answer, a field like component with energy which mixes with the E
field under lorentz transform yet is not normally counted in the E + B fields.
It appears to fall out of the STA notation trivially when applied to this situation.
Bilge
>I'm thinking rather of a 4-vector potential:
>
>(0,a,0,0)
>
>Where a has a non zero derivative with respect to x.
>
>As seen by an observer moving in the x direction. You now
>have something like:
>
>(b,c,0,0)
>
>Where b and c now have non zero derivatives with respect to x.
>
>Doesn't matter if you change the gauge you still have an electric
>field in the reference frame of the electron.
[...]
>> And, it isn't right. No observers see any fields.
>
>Yet relativistic transformation of a vector potential (0,a,0,0) where
>a varies with position appears to show an electric field in the moving
>frame. I know this presents problems that why I asked about it.
No, it doesn't. Look up gauge transformation.
[...]
>
>Using STA notation, let y0, y1, y2, y3 represent the unit vectors for
> t, x, y, z.
What is "STA notation"?
[...]
> Under the lorentz transform, this scalar component will mix with the
> bivector components.
I suggest using the standard mathematics. If you want to use some
alternative formulation and get the wrong results, consult the reference
from which you obtained the method.
[...]
You're ignoring gauge invariance and probably getting confused by
whatever method you are trying to employ to address the question.
Pmb
"Lawrence Foard" <ent...@farviolet.com> wrote in message
news:b81fu3$goq$1...@farviolet.com...
> In article <%TWoa.20339$ot1....@nwrdny02.gnilink.net>,
> Pmb <peter....@verizon.net> wrote:
> >
> >> frame which is completely 0 in another.
> >
> >First off let me see if I get this straight. Is the vector potential
really
> >non-uniform in the lab frame? Is the Aharonov-Bohm effect dependant on
the
> >field being non-uniform?
>
> That would be the obvious flaw. But I believe the vector potential
magnitude
> has to fall off with distance?
Why? I thought that the effect depended in the vector potential being
non-zero in a specific region. It doesn't have to be zero everywhere from
what I understand. Only the region in which the particle is moving.
> The double slit experiment shows a particle
> coming from some distance, going around both directions, and then hitting
> a screen at some distance.
Maybe the B field is non-zero in the regions the electron doesn't travel in?
> >I think that you have to take a literal look at it. In a real life
situation
> >a real magnetic field will not have a vector potential in one area and
> >vanish everywhere else. What is required is that in each frame the total
> >energy must be constant. Energy conservation does not imply that if
energy
> >is conserved one frame then its conserved in all frames. And energy
> >conservation does not mean that if energy is zero in one frame then it
has
> >to be non-zero in another frame.
>
> Or conversily if energy is non zero in one frame its non zero in all.
All cases I know *in relativity* are like that. If there's a particle in one
frame then there's a particle in all frames and all particles have energy.
However this is not the case in classical physics.
> I believe
> that is true. Yet conventional E&M applied to this situation appears to
give a
> contradictory answer. I posted another message where Hestene's STA
What is "STA"?
> appears to
> provide an answer, a field like component with energy which mixes with the
E
> field under lorentz transform yet is not normally counted in the E + B
fields.
> It appears to fall out of the STA notation trivially when applied to this
situation.
Pmb
Lawrence Foard
Pmb <peter....@verizon.net> wrote:
>
>
>"Lawrence Foard" <ent...@farviolet.com> wrote in message
>news:b81fu3$goq$1...@farviolet.com...
>
>> In article <%TWoa.20339$ot1....@nwrdny02.gnilink.net>,
>> Pmb <peter....@verizon.net> wrote:
>> >
>> >> frame which is completely 0 in another.
>> >
>> >First off let me see if I get this straight. Is the vector potential
>really
>> >non-uniform in the lab frame? Is the Aharonov-Bohm effect dependant on
>the
>> >field being non-uniform?
>>
>> That would be the obvious flaw. But I believe the vector potential
>magnitude
>> has to fall off with distance?
>
>Why? I thought that the effect depended in the vector potential being
>non-zero in a specific region. It doesn't have to be zero everywhere from
>what I understand. Only the region in which the particle is moving.
The vector potential is non zero in the region containing the particle,
but the magnetic and electric fields are zero.
>> The double slit experiment shows a particle
>> coming from some distance, going around both directions, and then hitting
>> a screen at some distance.
>
>Maybe the B field is non-zero in the regions the electron doesn't travel in?
Yes its non zero in the solenoid.
>> Or conversily if energy is non zero in one frame its non zero in all.
>
>All cases I know *in relativity* are like that. If there's a particle in one
>frame then there's a particle in all frames and all particles have energy.
>
>However this is not the case in classical physics.
>
>> I believe
>> that is true. Yet conventional E&M applied to this situation appears to
>give a
>> contradictory answer. I posted another message where Hestene's STA
>
>What is "STA"?
Space Time Algebra, geometric algebra approach to relativity.
http://modelingnts.la.asu.edu/html/overview.html
This isn't an alternative theory just an alternative notation,
and an extremely elegant one at that. Its certainly possible that
I goofed, but it certainly seems that there is an extra field component
whose actions in a region lacking E and B fields could resist or assist
the motion of a charge in its direction of motion.
Pmb
"Lawrence Foard" <ent...@farviolet.com> wrote
> Space Time Algebra, geometric algebra approach to relativity.
>
> http://modelingnts.la.asu.edu/html/overview.html
Looks like an interesting paper. I know a lot about spacetime physics etc -
never heard the term "Space Time Algebra" per se.
I see this person won the Oerstead medal! A friend of mine won that a few
years back. I think he'll enjoy this paper. Who wrote it?
>
> This isn't an alternative theory just an alternative notation,
> and an extremely elegant one at that. Its certainly possible that
> I goofed, but it certainly seems that there is an extra field component
> whose actions in a region lacking E and B fields could resist or assist
> the motion of a charge in its direction of motion.
Still - interesting topic!
Pete
Ken S. Tucker
an be written as
>"Lawrence Foard" <ent...@farviolet.com> wrote in message
>news:b81533$fle$1...@farviolet.com...
>
>> But this doesn't seem right because one observer sees a field energy
>> and another does not.
My understanding of this experiment had to do with the difference
inside an enclosed sphere with a high potential energy on it's exterior,
but no gradient or electric field in the interior of the sphere. (This
was a Laplace thingly, I'll ref if you need that).
My knowledge on this is tenuous, but I'll share. The thought has
occurred that an atomic clock could be placed into a sphere of very
high electrical potential to measure GR space-time density increase
following Einsteins Law, G_uv =k*T_uv on the basis of electrical
potential. Well the experimental physicts said can't measure it.
So what I understand is A-B decided to drill two holes in the
sphere, and use X-ray frequency as an inferometer, as the X-rays
propagated threw the higher electric potential in the sphere,
then as MM did compare the changes wrt voltage on the sphere.
In the refinement, it was decided that powerful magnetic fields
would be more effective than mere electrostatics, hence leading
to soleniods to sub for the spheres.
The equipment can be calibrated to use X-ray inferometry
prior to application of field so everything is alligned and
defined at neutral. Issues do arise as power is applied to
this appartus, and only a good understanding of paramagnetic
effects that may twist the apparatus.This is a function of the
experimenters care. So I'm a bit irritated because I do NOT
know if this has been generally confirmed (GRRRR).
Regards
Ken S. Tucker
Lawrence Foard
Bilge <cra...@fghfgigtu.com> wrote:
> Lawrence Foard:
> >
> >I'm thinking rather of a 4-vector potential:
> >
> >(0,a,0,0)
>
> You don't get to invent one. Calculate A outside of a long solenoid.
Its been done before already. I should have thrown in a Y component as
well. However its irrelevant as it does not mix with the time component
under transformation.
> >Where a has a non zero derivative with respect to x.
> >
> >As seen by an observer moving in the x direction. You now
> >have something like:
> >
> >(b,c,0,0)
> >
> >Where b and c now have non zero derivatives with respect to x.
> >
> >Doesn't matter if you change the gauge you still have an electric
> >field in the reference frame of the electron.
>
> No, you don't. Make the gauge transformation.
Assuming that you could gauge it away there is another issue to consider.
A gauge transformation is based on the assumption that E + B are the only
observables. Any argument that asserts there can be no other observables
based on a gauge transformation of that sort is circular. Obviously there
is another observable currently considered only in quantum mechanics. Is
it possible that a classical experiment with two charged particles would
simply show one arriving at a different time than the other? Is so there
is another classical observable that is possibly not considered in gauge
transforms.
> >Using STA notation, let y0, y1, y2, y3 represent the unit vectors for
> > t, x, y, z.
>
> What is "STA notation"?
See other post.
> > Under the lorentz transform, this scalar component will mix with the
> > bivector components.
>
> I suggest using the standard mathematics. If you want to use some
>alternative formulation and get the wrong results, consult the reference
>from which you obtained the method.
Actually unless I find a problem in this real soon I'm going to bring
it up with David Hestene's, his equations appear to differ from the
standard E&M field tensor in this case. Need to double check and make
sure I'm not misapplying space time calculus here.
Here is what it comes down to:
In geometric algebra the E&M tensor is formed by vector multiplication
of a vector of differentials with the A vector. In this form of multiplication
both parallel and perpendicular components are considered. Perpendicular
anti commute and parallel commute. Its immediately obvious that the
conventional 16 element tensor represents only perpendicular multiplies,
it fails to properly deal with the parallel components:
Normally its: F{uv} = d{u}A{v} - d{v}A{u}
Obviously giving 0 when u and v are identical. However when u and v
are the same we are no longer dealing with perpendicular vectors, and
multiplication doesn't anti commute, so instead of a subtraction there
is an addition:
F{uu} = 2 * d{u}A{u}
In the area outside the solenoid all terms except for these would be
zero. However under transformation they are going to mix with the other
components, providing an electric field visible to a moving particle.
Perhaps this is an observable quantity in classical mechanics if you
look really carefully. It may be that its normally overwhelmed to an
extent where it can be easily missed.
However there might be other places to look for it. For example a slight
extra energy storage in a solenoid.
Lawrence Foard
Lawrence Foard <ent...@farviolet.com> wrote:
>Actually unless I find a problem in this real soon I'm going to bring
>it up with David Hestene's, his equations appear to differ from the
>standard E&M field tensor in this case. Need to double check and make
>sure I'm not misapplying space time calculus here.
Oops. It appears that it only uses the outer product (perpendicular) in STA,
with the inner (parallel) assigned to 0. Need to investigate that requirement
for the inner product a bit.
Bilge
[...]
>> No, you don't. Make the gauge transformation.
>
>Assuming that you could gauge it away there is another issue to consider.
>A gauge transformation is based on the assumption that E + B are the only
>observables.
No, it isn't. It's based on the assumption that you can write E and B
in terms of potentials and that maxwell's equations are lorentz covariant.
In classical mechanics, the potential is a mathematical artifice. In
quantum mechanics, the potentials have significance.
>Any argument that asserts there can be no other observables based on a
>gauge transformation of that sort is circular.
See chapter 6 in "Classical Exlectrodynamics", Jackson, J.D.
>Obviously there is another observable currently considered only in
>quantum mechanics.
You are using the term "observable", in a weird way.
[...]
>> I suggest using the standard mathematics. If you want to use some
>>alternative formulation and get the wrong results, consult the reference
>>from which you obtained the method.
>
>Actually unless I find a problem in this real soon I'm going to bring
>it up with David Hestene's, his equations appear to differ from the
>standard E&M field tensor in this case. Need to double check and make
>sure I'm not misapplying space time calculus here.
Contrary to what hestenes claims, I really find his claims of clarity
to be highly exaggerated.
>
>Here is what it comes down to:
>
>In geometric algebra the E&M tensor is formed by vector multiplication
>of a vector of differentials with the A vector. In this form of
>multiplication both parallel and perpendicular components are considered.
>Perpendicular anti commute and parallel commute. Its immediately obvious
>that the conventional 16 element tensor represents only perpendicular
>multiplies, it fails to properly deal with the parallel components:
>
>Normally its: F{uv} = d{u}A{v} - d{v}A{u}
At least indicate super and subscripts.
>Obviously giving 0 when u and v are identical.
And?
>However when u and v
>are the same we are no longer dealing with perpendicular vectors, and
>multiplication doesn't anti commute, so instead of a subtraction there
>is an addition:
Just use the conventional math. It's much more straight forward, despite
the hype about "spacetime algebra".
[...]
>However there might be other places to look for it. For example a slight
>extra energy storage in a solenoid.
The vector potetential doesn't change the momentum of the electron
passing outside the solenoid.
Lawrence Foard
Bilge <cra...@fghfgigtu.com> wrote:
> Lawrence Foard:
> >In article <slrnba8khv....@radioactivex.lebesque-al.net>,
> >Bilge <cra...@fghfgigtu.com> wrote:
>
>[...]
> >> No, you don't. Make the gauge transformation.
> >
> >Assuming that you could gauge it away there is another issue to consider.
> >A gauge transformation is based on the assumption that E + B are the only
> >observables.
>
> No, it isn't. It's based on the assumption that you can write E and B
>in terms of potentials and that maxwell's equations are lorentz covariant.
>In classical mechanics, the potential is a mathematical artifice. In
>quantum mechanics, the potentials have significance.
It appears your probably right about the gauge invariance issue. However
it doesn't provide a direct answer to the original question. The solution
to Aharonov-Bohm under the Lorentz transform appears to be the following.
The gradient of the time component of A, appears to be balanced by the
time derivative of A which also contributes to the electric field.
Still there is the interesting question of the nature of a field which
in addition to the E + B fields could completely reconstruct A locally,
including the part needed by Aharonov-Bohm. Is there a way to contrive
to observe this classically?
Bilge
>Bilge <cra...@fghfgigtu.com> wrote:
>> Lawrence Foard:
>> >In article <slrnba8khv....@radioactivex.lebesque-al.net>,
>> >Bilge <cra...@fghfgigtu.com> wrote:
>>
>>[...]
>> >> No, you don't. Make the gauge transformation.
>> >
>> >Assuming that you could gauge it away there is another issue to consider.
>> >A gauge transformation is based on the assumption that E + B are the only
>> >observables.
>>
>> No, it isn't. It's based on the assumption that you can write E and B
>>in terms of potentials and that maxwell's equations are lorentz covariant.
>>In classical mechanics, the potential is a mathematical artifice. In
>>quantum mechanics, the potentials have significance.
>
>It appears your probably right about the gauge invariance issue. However
>it doesn't provide a direct answer to the original question. The solution
>to Aharonov-Bohm under the Lorentz transform appears to be the following.
>The gradient of the time component of A, appears to be balanced by the
>time derivative of A which also contributes to the electric field.
discover why your "solution" to the aharanov-bohm effect is a
non-sequiter.
>Still there is the interesting question of the nature of a field which
>in addition to the E + B fields could completely reconstruct A locally,
>including the part needed by Aharonov-Bohm. Is there a way to contrive
>to observe this classically?
You won't understand how meaningless your question is until you
go read about gauge invariance.
Lawrence Foard
Bilge <cra...@fghfgigtu.com> wrote:
> Lawrence Foard:
> >In article <slrnba97v6....@radioactivex.lebesque-al.net>,
> >Bilge <cra...@fghfgigtu.com> wrote:
> >> Lawrence Foard:
> >> >In article <slrnba8khv....@radioactivex.lebesque-al.net>,
> >> >Bilge <cra...@fghfgigtu.com> wrote:
> >>
> >>[...]
> >> >> No, you don't. Make the gauge transformation.
> >> >
> >> >Assuming that you could gauge it away there is another issue to consider.
> >> >A gauge transformation is based on the assumption that E + B are the only
> >> >observables.
> >>
> >> No, it isn't. It's based on the assumption that you can write E and B
> >>in terms of potentials and that maxwell's equations are lorentz covariant.
> >>In classical mechanics, the potential is a mathematical artifice. In
> >>quantum mechanics, the potentials have significance.
> >
> >It appears your probably right about the gauge invariance issue. However
> >it doesn't provide a direct answer to the original question. The solution
> >to Aharonov-Bohm under the Lorentz transform appears to be the following.
> >The gradient of the time component of A, appears to be balanced by the
> >time derivative of A which also contributes to the electric field.
>
> Once you figure out that I'm right about gauge invariance, you'll
>discover why your "solution" to the aharanov-bohm effect is a
>non-sequiter.
The first equation on this page is what I'm refering to.
http://farside.ph.utexas.edu/~rfitzp/teaching/em1/lectures/node42.html
The reason a moving observer (relative to solenoid) will see no fields
is that the two components on the right side of this equation while non
zero are equal and opposite.
Bilge
>The first equation on this page is what I'm refering to.
>
>http://farside.ph.utexas.edu/~rfitzp/teaching/em1/lectures/node42.html
>
>The reason a moving observer (relative to solenoid) will see no fields
>is that the two components on the right side of this equation while non
>zero are equal and opposite.
Stop trying to invent fields that don't exist just for the sake of
cancelling them. That webpage does not say what you are trying to
read into it. Either go argue with hestenes, or learn E&M the way
everyone else does. Gauge invariance doesn't have anything to do with
"cancelling fields". It's about the existence of the fields in the first
place. If start with no magnetic or electric field, a lorentz transform
does not create them. I'll say it once more: The potentials in classical
E&M are a mathematical artifice. They have no physical meaning whatsoever.
Pmb
"Lawrence Foard" <ent...@farviolet.com> wrote
> > >It appears your probably right about the gauge invariance issue.
However
> > >it doesn't provide a direct answer to the original question. The
solution
> > >to Aharonov-Bohm under the Lorentz transform appears to be the
following.
> > >The gradient of the time component of A, appears to be balanced by the
> > >time derivative of A which also contributes to the electric field.
> >
> > Once you figure out that I'm right about gauge invariance, you'll
> >discover why your "solution" to the aharanov-bohm effect is a
> >non-sequiter.
>
> The first equation on this page is what I'm refering to.
>
> http://farside.ph.utexas.edu/~rfitzp/teaching/em1/lectures/node42.html
>
> The reason a moving observer (relative to solenoid) will see no fields
> is that the two components on the right side of this equation while non
> zero are equal and opposite.
Lawrence - can you bring me up to speed on this? What is all this about
gauge invariance? What is the disagreement about?
Pete
Pmb
"Lawrence Foard" <ent...@farviolet.com> wrote in message
news:b81533$fle$1...@farviolet.com...
> What am I missing here. I can't picture a way this experiment could
> be performed without an electron crossing an area of changing vector
> potential (if only in direction). Which from a relativistic view point
> translates to a changing electric potential (electric field). Obviously
> altering its motion. So first thought is "whats the big deal".
>
> But this doesn't seem right because one observer sees a field energy
> and another does not.
>
> I'm also assuming if Aharonov-Bohm was so simply resolved it would not
> be of any interest. Does anyone understand why its supposed to show no
> effect on electrons unless quantum mechanics is used? What am I missing
> with the relativistic transformation of the vector potential?
Let me take a crack at this from scratch.
In frame O there is a solenoid. Around that solenoid there is a
non-vanishing vector potential and that is not a constant (the vector
potential kind of circles the solenoid like the B-field of a line of current
does). In O the electric field is zero by design.
In O
E = -grad Phi - &A/&t
Since, in O, there is no E or B field then the Faraday tensor must vanish.
This also implies there is no EM energy. Now change to O'. The Faraday
tensor vanishes there too. Therefore E' must vanish in O' as well. This
implies that that quantity
E' = -grad Phi' - &A'/&t
vanishes giving
grad Phi' = &A'/&t
So what must hold true is this equation. Maybe you were assuming that Phi
vanished in O'? I know that was the mistake I made ... assuming this is the
answer to the dilemma.
Pmb
Lawrence Foard
Bilge <cra...@fghfgigtu.com> wrote:
> Lawrence Foard:
>
> >The first equation on this page is what I'm refering to.
> >
> >http://farside.ph.utexas.edu/~rfitzp/teaching/em1/lectures/node42.html
> >
> >The reason a moving observer (relative to solenoid) will see no fields
> >is that the two components on the right side of this equation while non
> >zero are equal and opposite.
>
> Stop trying to invent fields that don't exist just for the sake of
>cancelling them. That webpage does not say what you are trying to
>read into it.
That equation resolved the original question. There is no E or B field
in any reference frame outside the solenoid. Originally I had counted only
the dphi/dx component and forgotten that there is a second contribution to
the field.
>Either go argue with hestenes, or learn E&M the way
>everyone else does.
Where was Hestene's mentioned on that page? As I already pointed out
Hestene's and olde fashioned E&M give the same answer. The problem
was the missing dAx/dt term.
>Gauge invariance doesn't have anything to do with
>"cancelling fields".
Yes it does. Gauge invariance works because those two parts of the equation
cancel in all reference frames.
>It's about the existence of the fields in the first
>place. If start with no magnetic or electric field, a lorentz transform
>does not create them.
Thats the point.
>I'll say it once more: The potentials in classical
>E&M are a mathematical artifice. They have no physical meaning whatsoever.
However clearly there is something missing from the classical fields that
in the vector potential. Otherwise this experiment would show a null result.
Gauge
Yup. Noticed it when I thought about it more.
Lawrence Foard
Pmb <peter....@verizon.net> wrote:
>Let me take a crack at this from scratch.
>
>In frame O there is a solenoid. Around that solenoid there is a
>non-vanishing vector potential and that is not a constant (the vector
>potential kind of circles the solenoid like the B-field of a line of current
>does). In O the electric field is zero by design.
>
>In O
>
>E = -grad Phi - &A/&t
>
>Since, in O, there is no E or B field then the Faraday tensor must vanish.
>This also implies there is no EM energy. Now change to O'. The Faraday
>tensor vanishes there too. Therefore E' must vanish in O' as well. This
>implies that that quantity
>
>E' = -grad Phi' - &A'/&t
>
>vanishes giving
>
>grad Phi' = &A'/&t
>
>So what must hold true is this equation. Maybe you were assuming that Phi
>vanished in O'? I know that was the mistake I made ... assuming this is the
>answer to the dilemma.
Actually I think phi can vanish in O. Its just the &A/&t part I had been
missing before. Any change in grad Phi in a reference frame will be
mirrored by an equal and opposite change in &A/&t.
The Lorentz transform to the electron frame does appear to provide a more
intuitive look at the experiment however. If grad Phi is non vanishing, in
the electron frame, then the electron will see an electrical potential for
part of the trip, electrons on opposite sides of the solenoid will see
an opposite. This will altered the frequency of its wavefunction,
yet due to E=0 it will experience no alteration in its path.
While grad A is 0, I'm still curious about the individual components of
grad A. They seem to be important to this argument.
Pmb
> >I'll say it once more: The potentials in classical
> >E&M are a mathematical artifice. They have no physical meaning
whatsoever.
>
> However clearly there is something missing from the classical fields that
> in the vector potential. Otherwise this experiment would show a null
result.
What I said earlied (in another post) holds true, i.e.
-------------------------------------------------------------------
In frame O there is a solenoid. Around that solenoid there is a
non-vanishing vector potential and that is not a constant (the vector
potential kind of circles the solenoid like the B-field of a line of current
does). In O the electric field is zero by design.
In O:
E = -grad Phi - &A/&t
Since, in O, there is no E or B field then the Faraday tensor must vanish.
This also implies there is no EM energy. Now change to O'. The Faraday
tensor vanishes there too. Therefore E' must vanish in O' as well. This
implies that that quantity
E' = -grad Phi' - &A'/&t
vanishes in O'
-------------------------------------------------------------------
That is certainly true. So here's what I haven't mentioned yet (just dawned
on me - cobwebs are clearing now) -
If you want to work, not with E and B, but with Phi and A then these are not
uniquely defined (I think bilge said this) Phi and A are gauge-dependent,
therefore questions of their existence are meaningless. One must first pin
them down with gauge conditions.
Now if you want to write A and Phi you have to choose a gauge. After that
you can then talk about how these things transform and then you can show
that
E = -grad Phi - &A/&t
E' = -grad Phi' - &A'/&t
which, of course, must be true. How does that sound? Have you tried this out
for yourself explicitly?
Pmb
AntiCrank
"Pmb" <peter....@verizon.net> wrote in message
news:9Oapa.11497$xR4....@nwrdny03.gnilink.net...
Crank.
Ken S. Tucker
>I'll say it once more: The potentials in classical
>E&M are a mathematical artifice.
>They have no physical meaning whatsoever.
That's a pretty sweeping statement. You could
apply that to everthing that is conceptual.
Since photons are the only valid comparisons of
scientific measurements, that leaves us with only
photons left. Everything else is imaginary concepts
explaining photon's rendering measurements.
Even electrostatic potential, ie. voltage, could be
explained when you're shocked, by photonics
communicating energy between molecules.
So using General Relativity in an otherwise classical
way, one can store electrostatic potential between two
repelling disks and expect an effect on photons passing
between the discs. This is due to the increased energy
density from G_uv = kT_uv.
This effect should alter the photon path length and
produce a relative phase shift, with an attendant
interference.
Aharnov and Bohm have found a similiar prediction
using Quantum Theory, IMO this is an example of
Quantum Gravity.
Regards
Ken S. Tucker
Pmb
news:2202379a.03042...@posting.google.com...
> dub...@radioactivex.lebesque-al.net (Bilge) wrote in message
news:<slrnbaa1r3....@radioactivex.lebesque-al.net>...
> >I'll say it once more: The potentials in classical
> >E&M are a mathematical artifice.
> >They have no physical meaning whatsoever.
>
> That's a pretty sweeping statement.
But universally accepted as true. What you measure is in the lab is the
position of the particle as a function of time from which one can deduce the
force. The force then tells you what the fields are given that you know the
rest mass and the charge etc. However it is then an abstraction to find the
electric scalar potential and magnetic vector potential. Especially since
they are not uniquely defined. You have to make a choice as to how they are
defined and that is something that can't be measured.
Pmb
Bilge
>message news:<slrnbaa1r3....@radioactivex.lebesque-al.net>...
>>I'll say it once more: The potentials in classical
>>E&M are a mathematical artifice.
>>They have no physical meaning whatsoever.
>
>That's a pretty sweeping statement.
Nevertheless, it's true.
>You could apply that to everthing that is conceptual.
Only those concepts that don't lead to physical effects imposed
strictly by the concept itself rather than as a secondary effect
of the concept. In classical E&M, the potentials give you only
F^{uv}. F^{uv} alone is not sufficient to account for all electro-
magnetic effects.
>Since photons are the only valid comparisons of
>scientific measurements, that leaves us with only
>photons left. Everything else is imaginary concepts
>explaining photon's rendering measurements.
You're going off into la-la land again.
>Even electrostatic potential, ie. voltage, could be
>explained when you're shocked, by photonics
>communicating energy between molecules.
The electrostatic potential has no physical meaning of its own.
all that matters are differences in the potential, i.e., an electric
field. If I add 1,000,000,000 volts to the universe, nothing changes.
>So using General Relativity in an otherwise classical
>way, one can store electrostatic potential between two
>repelling disks and expect an effect on photons passing
>between the discs. This is due to the increased energy
>density from G_uv = kT_uv.
No, it's due to the increased energy needed to keep the disks
from moving and from falling apart. That's all electrostatic.
> This effect should alter the photon path length and
>produce a relative phase shift, with an attendant
>interference.
>Aharnov and Bohm have found a similiar prediction
>using Quantum Theory, IMO this is an example of
>Quantum Gravity.
It's not. An effect from quantum gravity would be interference from
objects whose phase is affected by a gravitational potential (not a
gravitational field).
Bilge
>Bilge <cra...@fghfgigtu.com> wrote:
>> Lawrence Foard:
>>
>> >The first equation on this page is what I'm refering to.
>> >
>> >http://farside.ph.utexas.edu/~rfitzp/teaching/em1/lectures/node42.html
>> >
>> >The reason a moving observer (relative to solenoid) will see no fields
>> >is that the two components on the right side of this equation while non
>> >zero are equal and opposite.
>>
>> Stop trying to invent fields that don't exist just for the sake of
>>cancelling them. That webpage does not say what you are trying to
>>read into it.
>
>That equation resolved the original question. There is no E or B field
>in any reference frame outside the solenoid. Originally I had counted only
>the dphi/dx component and forgotten that there is a second contribution to
>the field.
>
>>Either go argue with hestenes, or learn E&M the way
>>everyone else does.
>
>Where was Hestene's mentioned on that page? As I already pointed out
>Hestene's and olde fashioned E&M give the same answer. The problem
>was the missing dAx/dt term.
the standard description of what gauge invariance means.
>
>>Gauge invariance doesn't have anything to do with
>>"cancelling fields".
>
>Yes it does.
No, it doesn't.
>Gauge invariance works because those two parts of the equation
>cancel in all reference frames.
You are confuing mathematics and equations with physics. Gauge invariance
"works" classically, because the potentials are not physical quantities
and are themselves, meaningless. ANY potentials which give the same fields
are valid potentials.
You can see right away that if the fields had to cancel, then your
idea about the potentials would be acausal. The scalar potential
propagates instantly over all space in the coulomb gauge. Its gradient
doesn't represent a physical field. I could equally well add a lot of
terms which always add up to zero, but that doesn't mean that individually,
those terms ever represented anything physical unless I can manipulate
them independently.
>>It's about the existence of the fields in the first
>>place. If start with no magnetic or electric field, a lorentz transform
>>does not create them.
>
>Thats the point.
And that's what I've been saying.
>
>>I'll say it once more: The potentials in classical
>>E&M are a mathematical artifice. They have no physical meaning whatsoever.
>
>However clearly there is something missing from the classical fields that
>in the vector potential.
Obviously. I also showed you what it was in my first response where I
obtained the interference for the aharanov-bohm effect. It's called a
phase. By the way, this phase is what leads directly to the existence
of the field itself through the dirac equation, conserved (and quantized)
charge and ultimately to the qed lagrangian through noether's theorem.
You can start with nothing but the dirac equation for a particle and
show that it (1) must have a field and (2) lead to a conserved current
just by requiring local gauge invariance (classical E&M is strictly a
global invariance). Charge is only conserved globally.
Pmb
relativity? Is this the introduction of this concept or has this been around
for a while (excluding 4-tensor stuff that I know has been around etc)?
And nice! I see they introduce mass in the appropriate way!
"The spacetime split of p into energy (or relative mass) E and relative
momentum p is given by.
[...]
where m is the proper mass of the particle."
Now *that* is the way mass ought to be introduced/used/defined! Bravo for
the author!
Pmb
"Lawrence Foard" <ent...@farviolet.com> wrote in message
news:b81lhl$hb1$1...@farviolet.com...
Lawrence Foard
> >Hestene's and olde fashioned E&M give the same answer. The problem
> >was the missing dAx/dt term.
>
> He wasn't, but your ideas about gauge invariance didn't originate with
>the standard description of what gauge invariance means.
What ideas? I've already pointed out that I found the mistake in my
original view of the effect of a Lorentz transform of the apparatus.
However you seem stuck on trying to disagree with Hestene's where there
is nothing to disagree with. The statement that F is a bivector, implies
the lorentz condition. Quite a bit more elegant than requiring
an independant equation to represent it I must say. The fact that I
got a non zero scalar part hinted directly at an error on my part,
in using conventional math a mistake of that sort would not have been
as immediately apparent. It was obvious as soon as I went back and
discovered that Hestene's had addressed the issue of parallel components,
stating that they must sum to zero due to the Lorentz condition.
When I've got some time I'll post in response to PMB's question about
the equations for A and A'.
Pmb
> In article <slrnbab9ur....@radioactivex.lebesque-al.net>,
> Bilge <cra...@fghfgigtu.com> wrote:
> > Lawrence Foard:
> > >
> > >Where was Hestene's mentioned on that page? As I already pointed out
> > >Hestene's and olde fashioned E&M give the same answer. The problem
> > >was the missing dAx/dt term.
> >
> > He wasn't, but your ideas about gauge invariance didn't originate with
> >the standard description of what gauge invariance means.
>
> What ideas? I've already pointed out that I found the mistake in my
> original view of the effect of a Lorentz transform of the apparatus.
> However you seem stuck on trying to disagree with Hestene's where there
> is nothing to disagree with. The statement that F is a bivector, implies
> the lorentz condition. Quite a bit more elegant than requiring
> an independant equation to represent it I must say. The fact that I
> got a non zero scalar part hinted directly at an error on my part,
> in using conventional math a mistake of that sort would not have been
> as immediately apparent. It was obvious as soon as I went back and
> discovered that Hestene's had addressed the issue of parallel components,
> stating that they must sum to zero due to the Lorentz condition.
>
> When I've got some time I'll post in response to PMB's question about
> the equations for A and A'.
What do you mean by "Lorentz condition"? I think you mean "Lorenz condition"
as in "Lorenz Gauge".
Pmb
Bilge
>Bilge <cra...@fghfgigtu.com> wrote:
>> Lawrence Foard:
>> >
>> >Where was Hestene's mentioned on that page? As I already pointed out
>> >Hestene's and olde fashioned E&M give the same answer. The problem
>> >was the missing dAx/dt term.
>>
>> He wasn't, but your ideas about gauge invariance didn't originate with
>>the standard description of what gauge invariance means.
>
>What ideas?
The idea that there are fields which cancel.
Bilge
d_u A^u = 0. It's only a gauge condition for a massless field.
For a massive field (or particle) it's a condition necessary for
lorentz invariance. In particular, there is still gauge freedom
after applying the lorentz condition to the electromagnetic field,
which is removed by choosing the coulomb gauge.
Gauge
Then it's spelled "Lorenz" not "Lorentz" - that's all I was saying.
Pmb
Bilge
>message news:<slrnbablv5....@radioactivex.lebesque-al.net>...
>> Pmb:
>> >
>> >What do you mean by "Lorentz condition"? I think you mean
>> >"Lorenz condition" as in "Lorenz Gauge".
>> d_u A^u = 0. It's only a gauge condition for a massless field.
>> For a massive field (or particle) it's a condition necessary for
>> lorentz invariance. In particular, there is still gauge freedom
>> after applying the lorentz condition to the electromagnetic field,
>> which is removed by choosing the coulomb gauge.
>
>Then it's spelled "Lorenz" not "Lorentz" - that's all I was saying.
Uh, not quite. For example:
Classical Electrodynamics, Jackson, J.D., Lorentz Condition, p220
Lorentz Gauge, p211
Quarks and Leptons, Halzen & Martin, Lorentz Condition, p133
Quantum Mechanics, Schiff, L.I., Lorentz Condition, p399
Lawrence Foard
Oops. It appears your correct:
http://scienceworld.wolfram.com/biography/Lorenz.html
This must be the most common misspelling in all of physics. Not only
does David Hestene's misspell it, but "Lorentz condition" gives 900
google hits, "Lorenz condition" only 80.
Lawrence Foard
Pmb <peter....@verizon.net> wrote:
>I was finally able to download that STA paper. Cool! Something new in
>relativity? Is this the introduction of this concept or has this been around
>for a while (excluding 4-tensor stuff that I know has been around etc)?
>
>And nice! I see they introduce mass in the appropriate way!
>
>"The spacetime split of p into energy (or relative mass) E and relative
>momentum p is given by.
>
>[...]
>
>where m is the proper mass of the particle."
>
>Now *that* is the way mass ought to be introduced/used/defined! Bravo for
He has been writting papers for decades. I believe he published something
called space time algebra back in the 60's (?), but I assume this is newly
updated.
I'm not sure how much of the application to relativity is his
invention, and how much is just making existing geometric algebra fit
into a consistent elegant framework.
Lawrence Foard
Pmb <peter....@verizon.net> wrote:
>
>If you want to work, not with E and B, but with Phi and A then these are not
>uniquely defined (I think bilge said this) Phi and A are gauge-dependent,
>therefore questions of their existence are meaningless. One must first pin
>them down with gauge conditions.
>
>Now if you want to write A and Phi you have to choose a gauge. After that
>you can then talk about how these things transform and then you can show
>that
>
>E = -grad Phi - &A/&t
>
>E' = -grad Phi' - &A'/&t
>
>which, of course, must be true. How does that sound? Have you tried this out
>for yourself explicitly?
Let there be a solenoid at the origin in the unprimed frame, with its axis
in the z direction.
Let C be a function of the magnetic flux in the solenoid, a constant for our
purposes.
For points outside the solenoid:
A(x,y) = (0, C * y/(x^2 + y^2), -C * x/(x^2 + y^2), 0)
A circulating vector whose magnitude falls off as 1/r.
Which obviously produces an electric field E=0
Does A(x,y) obey the Lorenz (?) condition?
&Ax/&x + &Ay/&y = 0 ?
-2*C*x*y/(y^2 + x^2)^2 + 2*C*x*y/(y^2 + x^2)^2 = 0
Thats convient :)
Transform to a reference frame moving at speed v in the x direction.
A'=(-v * C * y/((x^2 + y^2)*gamma), C * y/((x^2 + y^2) * gamma),
-C * x/(x^2 + y^2), 0)
Where y = y', and x = (x' + vt')/gamma
x component of E
E'x = -&(-v * C * y/((x^2 + y^2)*gamma))/&x'
-&(C * y/((x^2 + y^2) * gamma))/&t'
x is a function of x' and t', so both components will be non zero in
everyplace where y != 0. However notice that v = &x'/&t' after cancellation
of &x' the two components -grad Phi and -&A/&t are equal and opposite under
the lorentz transform.
Pmb
"Lawrence Foard" <ent...@farviolet.com> wrote
> Oops. It appears your correct:
>
> http://scienceworld.wolfram.com/biography/Lorenz.html
>
> This must be the most common misspelling in all of physics. Not only
> does David Hestene's misspell it, but "Lorentz condition" gives 900
> google hits, "Lorenz condition" only 80.
Let me rephrase that. I still disagree with bilge on this since. People can
call things anything they want I guess but what's in a name? I would
therefore say that it is "Lorenz" and not "Lorentz" who should be credited
for that condition. Therefore I believe those text books bilge quoted are
all in error ... at least according to J.D. Jackson himself. I have
Jackson's 2nd Ed but I think Jackson's 3rd Ed has this correction in it.
See "Historical roots of gauge invariance," J.D. Jackson, L.B. Okun,
Rev.Mod.Phys. 73 (2001) 663-680
http://xxx.lanl.gov/abs/hep-ph/0012061
"Abstract: [...] 4-div A = 0 was proposed L. V. Lorenz in the middle of
1860's . In most of the modern texts the latter condition is attributed to
H. A. Lorentz, who half a century later was one of the key figures in the
final formulation of classical electrodynamics. [...]"
E.g.
---------------------------------------------------------------
This equation, now almost universally called the "Lorentz condition," is
seen to originate with Lorenz more than 25 years before Lorentz. In
discussing the quasi-static limit, Lorenz remarks (p. 292) that the retarded
potentials (in modern, corrected terms, the "Lorenz" gauge potentials) give
the same fields as the instantaneous scalar potential and a vector potential
that is "a mean between Weber's and Neumann's theories," namely, (17),
appropriate for Maxwell's choice of เฐ A = 0. Without explicit reference,
Lorenz was
apparently aware of and made use of what we call gauge transformations.
[.....]
It is amusing how little the authors of text-books know about the history of
physics. For a further reading the Resource Letter (Cheng and Li, 1988) is
recommended.
---------------------------------------------------------------
Texts are now being published with this in mind.
For example: "Relativity: Special, General, and Cosmological," Wolfgang
Rindler, Oxford University Press, 2001, page 145
"We can therefore assume, without loss of generality, that the potential
satisfies the 'Lorenz (not Lorentz!) gauge condition (7.445) ..."
The "(not Lorentz!)" is Rindler's comment. Not mine.
I see that in Sean Carroll's lecture notes he uses "Lortenz gauge" too. I
forwarded this to him also since he's publishing his notes into a text which
I'm sure we are all eager to see in hard cover in a full fledged text rather
than online. Have you see those nice set of notes?
Pmb
Ilja Schmelzer
> What am I missing here. I can't picture a way this experiment could
> be performed without an electron crossing an area of changing vector
> translates to a changing electric potential (electric field). Obviously
> altering its motion. So first thought is "whats the big deal".
The big deal is that you need the potential A_i. Assume a description
with forces E_i, H_i alone would be sufficient. Then they would be
zero outside the solenoid. Thus, they could not influence electrons
which do not reach the inside.
Ilja
--
I. Schmelzer, <il...@ilja-schmelzer.net>, http://ilja-schmelzer.net
Lawrence Foard
always deal only with a potential difference rather than worrying about
gauge. However this is only the electric potential part of A. Is a
mathmatical treatment of A where the influence of A is expressed only as
a different in A between two points unique under gauge transformation?
Pmb
"Lawrence Foard" <ent...@farviolet.com> wrote in message
news:b867k8$vt6$1...@farviolet.com...
> So another dumb question dealing with gauge issues. In electronics you
> always deal only with a potential difference rather than worrying about
> gauge. However this is only the electric potential part of A. Is a
> mathmatical treatment of A where the influence of A is expressed only as
> a different in A between two points unique under gauge transformation?
Hmmm! Good question! Not sure of the answer and overly rusty on this.
However let me toss this your way since I'm at a point where I want to lube
this are of my mind
In electronics what you measure is an EMF right? As such you have
EMF(between A and B) =- integral of E*dl from point A to point B.
For a static field E = -grad Phi
EMF = Phi(B)- Phi(A)
right? But if the field is not stating then there is that other term -&A/&t.
That gives you a -d(Phi)/dt term
So if I understand you correctly you're looking for the equivalent of this
in magnetism?
Pmb
Pmb
"Lawrence Foard" <ent...@farviolet.com> wrote in message
news:b867k8$vt6$1...@farviolet.com...
> So another dumb question dealing with gauge issues. In electronics you
> always deal only with a potential difference rather than worrying about
> gauge. However this is only the electric potential part of A. Is a
> mathmatical treatment of A where the influence of A is expressed only as
> a different in A between two points unique under gauge transformation?
Since EMF applies to electric circuits perhaps what you're interested in
applies to magnetic circuits?
Pmb
Lawrence Foard
Actually thinking of it that way would seem to imply that it is not, since
energy gained or lost by a particle is path dependant in the case of something
like a transformer.
Lawrence Foard
Lawrence Foard <ent...@farviolet.com> wrote:
>In article <9pxpa.10186$vs2...@nwrdny01.gnilink.net>,
>Pmb <peter....@verizon.net> wrote:
>>
>>
>>"Lawrence Foard" <ent...@farviolet.com> wrote in message
>>news:b867k8$vt6$1...@farviolet.com...
>>
>>> So another dumb question dealing with gauge issues. In electronics you
>>
>>> always deal only with a potential difference rather than worrying about
>>
>>> gauge. However this is only the electric potential part of A. Is a
>>
>>> mathmatical treatment of A where the influence of A is expressed only as
>>
>>> a different in A between two points unique under gauge transformation?
>>
>>Hmmm! Good question! Not sure of the answer and overly rusty on this.
>>However let me toss this your way since I'm at a point where I want to lube
>>this are of my mind
>>
>>In electronics what you measure is an EMF right? As such you have
>>
>>EMF(between A and B) =- integral of E*dl from point A to point B.
>>
>>For a static field E = -grad Phi
>>
>>EMF = Phi(B)- Phi(A)
>>
>>right? But if the field is not stating then there is that other term -&A/&t.
>>That gives you a -d(Phi)/dt term
>>
>>So if I understand you correctly you're looking for the equivalent of this
>>in magnetism?
>
>Actually thinking of it that way would seem to imply that it is not, since
>energy gained or lost by a particle is path dependant in the case of something
>like a transformer.
Now I'm no so sure again...
Consider the following using the electrical analogy:
Your standing under a high tension power line. You choose the ground
to be electrical potential zero. You assign a potential to the power
line, phi, and an energy per charge E = phi * q2.
From your view point a local charge q1 has an energy/momentum vector
E1=(m0,0,0,0), and a vector potential A1=(0,0,0,0) and a charge q2 on the
power line has an energy/momentum vector E2=(m0 + phi*q2,0,0,0) and a
vector potential A2=(phi,0,0,0). Of course the energy is really in the
fields, and an observer on the power line could choose to group it
with your charge rather than a charge on the power line, but we can
assemble a system out of any group of objects we want.
An observer on the ground moves at a speed v with respect to the two
test charges.
E1'=(gamma*m0, -gamma*v*m0, 0, 0)
A1'=(0,0,0,0)
E2'=(gamma*(m0 + phi*q2), -gamma*v*(m0 + phi*q2), 0, 0)
A2'=(gamma*phi, -gamma*v*phi, 0, 0)
A difference in A between two points seems to correspond to a difference between the
energy and momentum vectors of two charges. There may be something I'm missing for
the dynamic case, but at first glance it seems that since any valid vector potential
can be represented as the sum of the vector potentials of charges of assorted velocities
then this should be valid in general.
When viewed from 'ground' (0,0,0,0) A contributes to the energy momentum vector of a
particle, such that E = A * q
Warning shameless speculation follows :)
Viewed from 'ground' (where ever you choose to assign it), Aharonov-Bohm seems to be
saying that the momentum of a particle going around the left hand path is different
than the momentum going around the right hand path, yet the energy remains constant.
This disposes of the issue of requiring field energy to accomplish this change,
as there is no energy change, there is no field energy required.
Quantum mechanically the wave equation seems to say the same thing, an A.ds
dependant spatial frequency (momentum) with no alteration in energy. Bilge
however said there is no momentum difference between the two paths, I'd be
curious how this can be when the spatial frequency is different. The fact
that an interference pattern exists as the result of two paths of identical
length itself clearly indicates that the paths differ in spatial frequency.
Ken S. Tucker
>Ken S. Tucker:
> >dub...@radioactivex.lebesque-al.net (Bilge) wrote in
> >message news:<slrnbaa1r3....@radioactivex.lebesque-al.net>...
> >>I'll say it once more: The potentials in classical
> >>E&M are a mathematical artifice.
> >>They have no physical meaning whatsoever.
> >
> >That's a pretty sweeping statement.
>
> Nevertheless, it's true.
Sure, based on the meaning of "physical".
> >You could apply that to everthing that is conceptual.
>
> Only those concepts that don't lead to physical effects imposed
>strictly by the concept itself rather than as a secondary effect
>of the concept. In classical E&M, the potentials give you only
>F^{uv}. F^{uv} alone is not sufficient to account for all electro-
>magnetic effects.
Please name one E&M effect F^{uv} doesn't explain.
> >Since photons are the only valid comparisons of
> >scientific measurements, that leaves us with only
> >photons left. Everything else is imaginary concepts
> >explaining photon's rendering measurements.
>
> You're going off into la-la land again.
No, the whole point of Relativity is our connection
with reality by photons. Think about that.
(It ain't idle philopophy, I'm agreeing with you).
> >Even electrostatic potential, ie. voltage, could be
> >explained when you're shocked, by photonics
> >communicating energy between molecules.
>
> The electrostatic potential has no physical meaning of its own.
>all that matters are differences in the potential,
This is the same old argument that speed is not real.
>i.e., an electric field.
An electric field is not a potential difference,
>If I add 1,000,000,000 volts to the universe, nothing changes.
That's wrong, that would change the metric, recall g_00 has
an electrical component dependant on (e/r)^2.
> >So using General Relativity in an otherwise classical
> >way, one can store electrostatic potential between two
> >repelling disks and expect an effect on photons passing
> >between the discs. This is due to the increased energy
> >density from G_uv = kT_uv.
>
> No, it's due to the increased energy needed to keep the disks
>from moving and from falling apart. That's all electrostatic.
The stored energy increment varies the gravity.
> > This effect should alter the photon path length and
> >produce a relative phase shift, with an attendant
> >interference.
> >Aharnov and Bohm have found a similiar prediction
> >using Quantum Theory, IMO this is an example of
> >Quantum Gravity.
>
> It's not. An effect from quantum gravity would be interference from
>objects whose phase is affected by a gravitational potential (not a
>gravitational field).
Recall a shell of mass has no internal field, but a constant
interior potential. A photon passing threw this sphere will
have a phase shift proportional to potential.
Regards Ken S. Tucker
Bilge
>"Lawrence Foard" <ent...@farviolet.com> wrote
>
>> Oops. It appears your correct:
>>
>> http://scienceworld.wolfram.com/biography/Lorenz.html
>>
>> This must be the most common misspelling in all of physics. Not only
>> does David Hestene's misspell it, but "Lorentz condition" gives 900
>> google hits, "Lorenz condition" only 80.
>
>Let me rephrase that. I still disagree with bilge on this since. People can
>call things anything they want I guess but what's in a name? I would
>therefore say that it is "Lorenz" and not "Lorentz" who should be credited
>for that condition. Therefore I believe those text books bilge quoted are
>all in error ... at least according to J.D. Jackson himself. I have
>Jackson's 2nd Ed but I think Jackson's 3rd Ed has this correction in it.
>
I'm willing to believe the attribution to lorentz is an historical
oversight, but I'd like to see what stephen speicher has on this,
since he probably has the best collection of texts which could
clarify this.
Bilge
>dub...@radioactivex.lebesque-al.net (Bilge) wrote in message
>news:<slrnbab8ll....@radioactivex.lebesque-al.net>...
>>Ken S. Tucker:
>> >dub...@radioactivex.lebesque-al.net (Bilge) wrote in
>> >message news:<slrnbaa1r3....@radioactivex.lebesque-al.net>...
>> >>I'll say it once more: The potentials in classical
>> >>E&M are a mathematical artifice.
>> >>They have no physical meaning whatsoever.
>> >
>> >That's a pretty sweeping statement.
>>
>> Nevertheless, it's true.
>
>Sure, based on the meaning of "physical".
classical E&M, which happens to be E and B, not A and \phi.
>> >You could apply that to everthing that is conceptual.
>>
>> Only those concepts that don't lead to physical effects imposed
>>strictly by the concept itself rather than as a secondary effect
>>of the concept. In classical E&M, the potentials give you only
>>F^{uv}. F^{uv} alone is not sufficient to account for all electro-
>>magnetic effects.
>
>Please name one E&M effect F^{uv} doesn't explain.
The aharanov-bohm effect being discussed.
>> >Since photons are the only valid comparisons of
>> >scientific measurements, that leaves us with only
>> >photons left. Everything else is imaginary concepts
>> >explaining photon's rendering measurements.
>>
>> You're going off into la-la land again.
>
>No, the whole point of Relativity is our connection with reality by
>photons. Think about that. (It ain't idle philopophy, I'm agreeing
>with you).
But relativity does not have anything to do with light, per se. The
only connection with the speed of light could be stated more accurately
as the speed of a massless object.
>> The electrostatic potential has no physical meaning of its own.
>>all that matters are differences in the potential,
>
>This is the same old argument that speed is not real.
>
>>i.e., an electric field.
>
>An electric field is not a potential difference,
>
>>If I add 1,000,000,000 volts to the universe, nothing changes.
>
>That's wrong, that would change the metric, recall g_00 has
>an electrical component dependant on (e/r)^2.
That (e/r)^2 is also not anything like grad\phi. Your argument doesn't
apply for (at least) three reasons. (1) g_00 is not c\phi, which is the
time component of the electromagnetic potential, (c\phi, A), (2) The
reissner-nordstrom metric for a spherical mass is:
ds^2 = [1 - (2m/r) -(q/r)^2]dt^2 - [1 - (2m/r) -(q/r)^2]^-1 dx^2
- r^2 d\Omega^2
It does not reduce to anything if m = 0. No charged, massless particles
are known to exist and any electromagnetic theory of which I am aware,
precludes the existence of massless charges by virtue of charge conser-
vation, which I believe is intrinsic to obtaining the reissner-nordstrom
metric in the first place. Even for a rotating charged object, the
reisner-nordstrom metric requires the charge to be zero if the mass
is zero, (3) applying a potential to the entire universe doesn't change
e^2/r any more than floating a 1 kV power supply to 20 kV changes the
output of the supply to something other than 1 kV (I've done this
personally with an entire cabinet full of power supplies powered by
a standard 220 VAC line fed into an isolation transformer, in order to
not have to use a cabinet full of 25 kV supplies, so I know for a fact
this works as advertised - you just have to use insulating rods to
turn the controls unless you are dumb enough to stand on the insulating
platform.)
If the entire universe were "floated" to 1,000,000,000 volts, you would
not have an external reference, there would be no preferred direction
for \grad\phi to point and the result would be unobservable. This
is what is meant by gauge invariance in classical E&M.
>> >So using General Relativity in an otherwise classical
>> >way, one can store electrostatic potential between two
>> >repelling disks and expect an effect on photons passing
>> >between the discs. This is due to the increased energy
>> >density from G_uv = kT_uv.
>>
>> No, it's due to the increased energy needed to keep the disks
>>from moving and from falling apart. That's all electrostatic.
>
>The stored energy increment varies the gravity.
Are you saying that the energy would be identical if the plates
were not held fixed?
>> > This effect should alter the photon path length and
>> >produce a relative phase shift, with an attendant
>> >interference.
>> >Aharnov and Bohm have found a similiar prediction
>> >using Quantum Theory, IMO this is an example of
>> >Quantum Gravity.
>>
>> It's not. An effect from quantum gravity would be interference from
>>objects whose phase is affected by a gravitational potential (not a
>>gravitational field).
>
>Recall a shell of mass has no internal field, but a constant
>interior potential. A photon passing threw this sphere will
>have a phase shift proportional to potential.
Not so. The phase shift is given by the difference of two paths which
enclose a flux. Inside a shell with a mass M, none of the paths within the
shell enclose any "gravitational flux". The potential is constant, all of
the paths are geodesics so no net difference occurs from enclosing any
region inside the shell by different paths through the shell.
The aharanov-bohm effect is not due to any forces acting on the charges.
Pmb
"Bilge" <dub...@radioactivex.lebesque-al.net> wrote
> I'm willing to believe the attribution to lorentz is an historical
> oversight, but I'd like to see what stephen speicher has on this,
> since he probably has the best collection of texts which could
> clarify this.
It seems that I was right. Jackson did make this change in his new edition
i.e. of "Classical Electrodynamics - Third Edition"
The gentlemen who teaches this at the University of Cambridge notes this in
his lecture notes
http://www.tcm.phy.cam.ac.uk/~nrc25/red/index.html
See "Electrodynamics and Radiation," Dr. Nigel Cooper, footnote on page 12
"Often referred to as the \Lorentz condition", incorrectly attributing this
to Dutch physicist H.A. Lorentz rather than Danish physicist L.V. Lorenz.
See Jackson, page 294."
where he is referring to Jackson's 3rd edition (as listed in "Special
Relativity")
Pmb
Patrick Reany
> In article <jdipa.7338$vs2....@nwrdny01.gnilink.net>,
> Pmb <peter....@verizon.net> wrote:
> >I was finally able to download that STA paper. Cool! Something new in
> >relativity? Is this the introduction of this concept or has this been around
> >for a while (excluding 4-tensor stuff that I know has been around etc)?
> >
> >And nice! I see they introduce mass in the appropriate way!
> >
> >"The spacetime split of p into energy (or relative mass) E and relative
> >momentum p is given by.
> >
> >[...]
> >
> >where m is the proper mass of the particle."
> >
> >Now *that* is the way mass ought to be introduced/used/defined! Bravo for
> >the author.
>
> He has been writting papers for decades. I believe he published something
> called space time algebra back in the 60's (?), but I assume this is newly
> updated.
>
> I'm not sure how much of the application to relativity is his
> invention, and how much is just making existing geometric algebra fit
> into a consistent elegant framework.
See
http://modelingNTS.la.asu.edu/html/Evolution.html
Patrick
Ken S. Tucker
>Ken S. Tucker:
> >dub...@radioactivex.lebesque-al.net (Bilge) wrote in message
> >news:<slrnbab8ll....@radioactivex.lebesque-al.net>...
> >>Ken S. Tucker:
> In classical E&M, physical refers to what is measurable as derived from
>classical E&M, which happens to be E and B, not A and \phi.
Well you need a charge "q" to measure E by q*E. But even this
isn't really measureable, what you need to displace a needle on a
meter is some heat (photons) given by, q(\phi_2 - \phi_1).
> >> Only those concepts that don't lead to physical effects imposed
> >>strictly by the concept itself rather than as a secondary effect
> >>of the concept. In classical E&M, the potentials give you only
> >>F^{uv}. F^{uv} alone is not sufficient to account for all electro-
> >>magnetic effects.
> >
> >Please name one E&M effect F^{uv} doesn't explain.
>
> The aharanov-bohm effect being discussed.
Is aharanov-bohm effect, "electromagnetic"?
> >> >Since photons are the only valid comparisons of
> >> >scientific measurements, that leaves us with only
> >> >photons left. Everything else is imaginary concepts
> >> >explaining photon's rendering measurements.
> >>
> >> You're going off into la-la land again.
> >
> >No, the whole point of Relativity is our connection with reality by
> >photons. Think about that. (It ain't idle philopophy, I'm agreeing
> >with you).
>
> But relativity does not have anything to do with light, per se. The
>only connection with the speed of light could be stated more accurately
>as the speed of a massless object.
Ok, it's a matter of sematics...
(I've done some electrical engineering too), and your example
provides a thought experiment.
Suppose two conducting boxes, A and B are isolated in space,
and are equal in size, material etc.
Charge up box A to1kV and box B to 26kV, (diff =25kV).
OR
Only charge box B to 25kV.
Box B will require an input of electrical energy against
the force of repulsion in the charging process.
It follows that Box B has a greater mass (m = E/c^2).
Would experimental measurements performed in box A,
yield the same numerical result as measurements done in
box B?
> >> >So using General Relativity in an otherwise classical
> >> >way, one can store electrostatic potential between two
> >> >repelling disks and expect an effect on photons passing
> >> >between the discs. This is due to the increased energy
> >> >density from G_uv = kT_uv.
> >>
> >> No, it's due to the increased energy needed to keep the disks
> >>from moving and from falling apart. That's all electrostatic.
> >
> >The stored energy increment varies the gravity.
>
> Are you saying that the energy would be identical if the plates
>were not held fixed?
Fix the discs, and maintain the potential.
> >> > This effect should alter the photon path length and
> >> >produce a relative phase shift, with an attendant
> >> >interference.
> >> >Aharnov and Bohm have found a similiar prediction
> >> >using Quantum Theory, IMO this is an example of
> >> >Quantum Gravity.
> >>
> >> It's not. An effect from quantum gravity would be interference from
> >>objects whose phase is affected by a gravitational potential (not a
> >>gravitational field).
> >
> >Recall a shell of mass has no internal field, but a constant
> >interior potential. A photon passing threw this sphere will
> >have a phase shift proportional to potential.
>
> Not so. The phase shift is given by the difference of two paths which
>enclose a flux. Inside a shell with a mass M, none of the paths within the
>shell enclose any "gravitational flux". The potential is constant, all of
>the paths are geodesics so no net difference occurs from enclosing any
>region inside the shell by different paths through the shell.
Well one can imagine the effect on the Michelson Morley
experiment if a thin piece of glass was placed in the light path of
one of the arms. The light would have two cancelling potential
differences (as Lawrence alluded to) that cancel by entering
and leaving the glass. In your thinking, the interior of the glass
is gauge invariant, but it still accumulates a difference of time
travel for the photon and results in interferometric fringes,
when recombined.
> The aharanov-bohm effect is not due to any forces acting on the charges.
Agreed and Regards
Ken S. Tucker
Pmb
I was a bit confused about some things in your derivation, not to mention
that reading equations in text is hard enough!
So I made a web page to show you how I did this. See
http://www.geocities.com/physics_world/vector_potential.htm
Pmb
Pmb
> I was a bit confused about some things in your derivation, not to mention
> that reading equations in text is hard enough!
>
> So I made a web page to show you how I did this. See
>
> http://www.geocities.com/physics_world/vector_potential.htm
Note: There are some errors in this page and am working on correcting them -
but please take a look and let me know what you think.
Pmb
AntiCrank
"Pmb" <peter....@verizon.net> wrote in message
news:6rZpa.32789$ot1...@nwrdny02.gnilink.net...
Ok...........
I think there are errors on your page Brown.
Stephen Speicher
days, and may soon do the same.
On Wed, 23 Apr 2003, Bilge wrote:
>
> I'm willing to believe the attribution to lorentz is an historical
> oversight, but I'd like to see what stephen speicher has on this,
> since he probably has the best collection of texts which could
> clarify this.
>
Yes, there is no question that it was Ludvig Valentin Lorenz, not
Hendrik Antoon Lorentz, who first specified what is now commonly
known as the Lorentz condition. It was in 1867 that Lorenz
showed the condition as a mathematical consequence of the
retarded potentials. At that early time, however, the retarded
potentials were looked upon in a suspicious light by most
theorists in the field (with the notable exception of Maxwell)
and Lorenz' work on the gauge was pretty much lost by the time
that Lorentz got around to the issue in 1892.
The circumstances regarding priority with Lorentz/Lorenz are
different from the Lorentz/Voigt priority issue over the Lorentz
transformation. It was clear that Lorentz did not know about
Voigt's earlier work when he developed his transformation. But,
some time later, when Voigt sent to Lorentz the paper he had
published earlier, Lorentz publicly acknowledged Voigt in
Lorentz' 1909 book "The Theory of Electrons."
However, Lorentz never acknowledged Lorenz' work on this from
some 25 years earlier than his own, nor did he correct several
references from notable authors of the time on the priority. It
is almost certain that Lorentz knew of Lorenz' work, from several
different sources. For instance, Lorentz most certainly knew of
it from Maxwell's "Treatise" which we know that Lorentz devoured.
In the "Treatise" Maxwell references Lorenz, and even notes that
the conclusions of Lorenz are quite similar to his own work two
years earlier -- Maxwell is claiming some priority, though based
on an entirely different method! The other authors which we know
that Lorentz read were, at least, Bucherer, Abraham, and
Heavyside, all of whom either identified Lorentz alone, or with
others, but not Lorenz, for priority. Lorentz, however, never
disputed this, which I find peculiar since Lorentz was always
known as a stalwart purveyor of the truth.
So the early books (early 1900s) gave credit to Lorentz, yet what
we now call the Lorentz condition was frequently just referred to
as "the condition." It isn't until the 1940s to the 1950s that
the term "Lorentz condition" began to appear in earnest, and
eventually that term was firmly established. For the past
half-century, in perhaps one book in ten or twenty there will be
a reference to Lorenz, not Lorentz, and the same average occurs
in papers as well. Not too long ago I computer searched more than
20 million papers going back some 60 years for occurrences of
"Lorentz condition"/"Lorenz condition" and "Lorentz
gauge"/"Lorenz gauge" and something like 95% of the papers
referenced Lorentz, not Lorenz.
Over the years there have been several books and papers which
point out the Lorentz/Lorenz discrepancy -- some going back to
the early 1900s, and two which appeared just a couple of years
ago -- and the suggestion was made to acknowledge Lorenz as the
designation. However, despite the occasional inclusion of Lorenz,
I suspect, just as the "Lorentz transformation" has stuck all
these years, the overwhelming use of "Lorentz condition in books
and papers will also continue, since Lorentz has remained so
strongly associated with that early work (Lorentz force law,
Lorentz invariance, etc.).
--
Stephen
s...@speicher.com
Ignorance is just a placeholder for knowledge.
Printed using 100% recycled electrons.
-----------------------------------------------------------
Lawrence Foard
Pmb <peter....@verizon.net> wrote:
>
>I was a bit confused about some things in your derivation, not to mention
>that reading equations in text is hard enough!
>
>So I made a web page to show you how I did this. See
>
>http://www.geocities.com/physics_world/vector_potential.htm
That looks right, I didn't do out the differentiation as much
since I was just looking for equality.
What software did you use to make the web page?
Pmb
> In article <3YXpa.15497$vs2....@nwrdny01.gnilink.net>,
> Pmb <peter....@verizon.net> wrote:
> >
> >I was a bit confused about some things in your derivation, not to mention
> >that reading equations in text is hard enough!
> >
> >So I made a web page to show you how I did this. See
> >
> >http://www.geocities.com/physics_world/vector_potential.htm
>
> That looks right, I didn't do out the differentiation as much
> since I was just looking for equality.
I'm still not getting zero though. I can't figure out why.
> What software did you use to make the web page?
I wrote it first with MS-word. I then copied it to FrontPage. I didn't like
the way MS Word did the equation pictures so I then I used PaintShop Pro
(PSP) to do a screen capture of the equations. I used PSP to do the figures.
Pmb
Bilge
>dub...@radioactivex.lebesque-al.net (Bilge) wrote in message
>news:<slrnbaeavq....@radioactivex.lebesque-al.net>...
>>Ken S. Tucker:
>> >dub...@radioactivex.lebesque-al.net (Bilge) wrote in message
>> >news:<slrnbab8ll....@radioactivex.lebesque-al.net>...
>> >>Ken S. Tucker:
>
>> In classical E&M, physical refers to what is measurable as derived from
>>classical E&M, which happens to be E and B, not A and \phi.
>
>Well you need a charge "q" to measure E by q*E. But even this
>isn't really measureable, what you need to displace a needle on a
>meter is some heat (photons) given by, q(\phi_2 - \phi_1).
Knowing the potential difference doesn't help unless you know the
detailed geometry over which the potential difference occurs and
the charge on the surfaces:
q = CV
The capacitance, C is a geometric quantity.
[...]
>> >Please name one E&M effect F^{uv} doesn't explain.
>>
>> The aharanov-bohm effect being discussed.
>
>Is aharanov-bohm effect, "electromagnetic"?
Yup.
[...]
>> But relativity does not have anything to do with light, per se. The
>>only connection with the speed of light could be stated more accurately
>>as the speed of a massless object.
>
>Ok, it's a matter of sematics...
Not really. The issue of whether or not the photon is massless (and
thherefore, propagates a `c') is directly tied to conservation of
electric charge through noether's theorem.
[...]
>> If the entire universe were "floated" to 1,000,000,000 volts, you would
>>not have an external reference, there would be no preferred direction
>>for \grad\phi to point and the result would be unobservable. This
>>is what is meant by gauge invariance in classical E&M.
>
>(I've done some electrical engineering too), and your example
>provides a thought experiment.
> Suppose two conducting boxes, A and B are isolated in space,
>and are equal in size, material etc.
>
>Charge up box A to1kV and box B to 26kV, (diff =25kV).
>OR
>Only charge box B to 25kV.
>
>Box B will require an input of electrical energy against
>the force of repulsion in the charging process.
> It follows that Box B has a greater mass (m = E/c^2).
>
>Would experimental measurements performed in box A,
>yield the same numerical result as measurements done in
>box B?
Uhuhhhhhh. Cease and desist. You do not have "only" two boxes.
You have at least 3 other wise you couldn't reference boxes A
and B to a ground potential (0 volts) [which is where you got
the charge]. If you have only two boxes, you can only express their
potential difference and all of the work goes into moving charge
from one box to another.
....furthermore, giving the mass as m = E/c^2 where E is the
electrostatic energy is manifestly incorrect. Try the following.
For an electron, which has an upper limit of 10^-18 meters for
it's radius, as determinied by experiment, calculate the self-energy
required to assemble the charge into a spherical volume 4pi r^3/3.
(feel free to choose any shape you want so long as all of the
charge is contained within a radius no larger than 10^-18 meters.
Calculate m. How many hundreds of times larger is this than the
electron mass?
[...]
>> >The stored energy increment varies the gravity.
>>
>> Are you saying that the energy would be identical if the plates
>>were not held fixed?
>
>Fix the discs, and maintain the potential.
Then wouldn't the forces holding the discs fixed have to contribute
to the stress-energy tensor?
[...]
>> Not so. The phase shift is given by the difference of two paths which
>>enclose a flux. Inside a shell with a mass M, none of the paths within the
>>shell enclose any "gravitational flux". The potential is constant, all of
>>the paths are geodesics so no net difference occurs from enclosing any
>>region inside the shell by different paths through the shell.
>
>Well one can imagine the effect on the Michelson Morley
>experiment if a thin piece of glass was placed in the light path of
>one of the arms. The light would have two cancelling potential
>differences (as Lawrence alluded to) that cancel by entering
>and leaving the glass. In your thinking, the interior of the glass
>is gauge invariant, but it still accumulates a difference of time
>travel for the photon and results in interferometric fringes,
>when recombined.
No, it's just that there's more to interference than it superficially
appears.
FrediFizzx
news:slrnbai52b...@radioactivex.lebesque-al.net...
<snip>
| ....furthermore, giving the mass as m = E/c^2 where E is the
| electrostatic energy is manifestly incorrect. Try the following.
| For an electron, which has an upper limit of 10^-18 meters for
| it's radius, as determinied by experiment, calculate the self-energy
| required to assemble the charge into a spherical volume 4pi r^3/3.
| (feel free to choose any shape you want so long as all of the
| charge is contained within a radius no larger than 10^-18 meters.
| Calculate m. How many hundreds of times larger is this than the
| electron mass?
I don't think it is exactly true that the upper limit for "electron" radius
is 10^-18 meters for "observed" charge distribution. It maybe is more on
the order of the Compton wavelength divided by 2*pi. See Milonni "The
Quantum Vacuum" section 11.5 "How Big is an Electron?". Yes, back to the
issue of bare elementary point-like charge vs. "electron". Did you give up
on your proposal that bare elementary charge might be infinitesimal? I
still have not found anything that proposes anything other than bare
elementary charge up close being greater than observed charge from a
distance. At least please give me a clue as to why you might think that.
FrediFizzx
Ken S. Tucker
[snip]
> >Charge up box A to1kV and box B to 26kV, (diff =25kV).
> >OR Only charge box B to 25kV.
> >Box B will require an input of electrical energy against
> >the force of repulsion in the charging process.
> > It follows that Box B has a greater mass (m = E/c^2).
> >Would experimental measurements performed in box A,
> >yield the same numerical result as measurements done in
> >box B?
>
> Uhuhhhhhh. Cease and desist.
I see Fred wants to speak with you so I shall be brief.....
>You do not have "only" two boxes.
>You have at least 3 other wise you couldn't reference boxes A
>and B to a ground potential (0 volts) [which is where you got
>the charge]. If you have only two boxes, you can only express their
>potential difference and all of the work goes into moving charge
>from one box to another.
Please answer my question.
> ....furthermore, giving the mass as m = E/c^2 where E is the
>electrostatic energy is manifestly incorrect. Try the following.
>For an electron, which has an upper limit of 10^-18 meters for
>it's radius, as determinied by experiment, calculate the self-energy
>required to assemble the charge into a spherical volume 4pi r^3/3.
>(feel free to choose any shape you want so long as all of the
>charge is contained within a radius no larger than 10^-18 meters.
>Calculate m. How many hundreds of times larger is this than the
>electron mass?
We did this before...
>[...]
> >> >The stored energy increment varies the gravity.
> >>
> >> Are you saying that the energy would be identical if the plates
> >>were not held fixed?
> >
> >Fix the discs, and maintain the potential.
>
> Then wouldn't the forces holding the discs fixed have to contribute
>to the stress-energy tensor?
>
>[...]
> >> Not so. The phase shift is given by the difference of two paths which
> >>enclose a flux. Inside a shell with a mass M, none of the paths within the
> >>shell enclose any "gravitational flux". The potential is constant, all of
> >>the paths are geodesics so no net difference occurs from enclosing any
> >>region inside the shell by different paths through the shell.
> >
> >Well one can imagine the effect on the Michelson Morley
> >experiment if a thin piece of glass was placed in the light path of
> >one of the arms. The light would have two cancelling potential
> >differences (as Lawrence alluded to) that cancel by entering
> >and leaving the glass. In your thinking, the interior of the glass
> >is gauge invariant, but it still accumulates a difference of time
> >travel for the photon and results in interferometric fringes,
> >when recombined.
>
> No, it's just that there's more to interference than it superficially
>appears.
Did my wife just walk by ??? or do I sniff a Red Herring.
With Pleasure Bilge...Ken S. Tucker
Tom Roberts
> What am I missing here. I can't picture a way this experiment could
> be performed without an electron crossing an area of changing vector
> potential (if only in direction).
Right. That's the point. In the rest frame of the solenoid, the vector
potential outside a perfect infinitely-long solenoid is circular. So the
integral of A_u dx^u over a path that goes past the solenoid on one side
or the other (in a plane perpendicular to the axis) depends on which
side of the solenoid it passes (in this frame A_0 = 0, so this is really
a spatial integral of a 3-vector field).
> Which from a relativistic view point
> translates to a changing electric potential (electric field).
Not necessarily, and specifically not in this case (see below).
> But this doesn't seem right because one observer sees a field energy
> and another does not.
Not true. Outside a perfect infinitely-long solenoid F = 0. As F is a
tensor (a 2-form encompasing both E and B), it is zero in all frames. So
no observer sees any E&M field energy outside the solenoid.
> I'm also assuming if Aharonov-Bohm was so simply resolved it would not
> be of any interest. Does anyone understand why its supposed to show no
> effect on electrons unless quantum mechanics is used?
In classical E&M, only E and B can affect the path of an electron, and
the vector potential is merely a mathematical artifice that is sometimes
useful in computing them; classically the vector potential has no
physical effects, only E and B do. But in quantum mechanics the effect
of E&M fields on the phase of a particle is proportional to the integral
of A_u dx^u over the path (sum over all possible paths...). The vector
potential has physical effects in QM, and they are observed experimentally.
> What am I missing
> with the relativistic transformation of the vector potential?
I dunno. It seems more like you goofed in the transformation of the
electromagnetic field components F_uv. Note that A_u dx^u is an
invariant, so if its integral over a given path is nonzero in one frame,
it is nonzero in all frames.
Tom Roberts tjro...@lucent.com
Bilge
>dub...@radioactivex.lebesque-al.net (Bilge) wrote in message
>news:<slrnbai52b...@radioactivex.lebesque-al.net>...
>>Ken S. Tucker:
>> >dub...@radioactivex.lebesque-al.net (Bilge) wrote in message
>> >news:<slrnbaeavq....@radioactivex.lebesque-al.net>...
>[snip]
>
>> >Charge up box A to1kV and box B to 26kV, (diff =25kV).
>> >OR Only charge box B to 25kV.
>> >Box B will require an input of electrical energy against
>> >the force of repulsion in the charging process.
>> > It follows that Box B has a greater mass (m = E/c^2).
>> >Would experimental measurements performed in box A,
>> >yield the same numerical result as measurements done in
>> >box B?
>>
>> Uhuhhhhhh. Cease and desist.
>
>I see Fred wants to speak with you so I shall be brief.....
>
>>You do not have "only" two boxes.
>>You have at least 3 other wise you couldn't reference boxes A
>>and B to a ground potential (0 volts) [which is where you got
>>the charge]. If you have only two boxes, you can only express their
>>potential difference and all of the work goes into moving charge
>>from one box to another.
>
>Please answer my question.
didn't expect and probably didn't want.
I'll charge one box by _removing_ negative charge so that a positive
charge remains. I then charge the other box by removing positive charge
so that only negative charge remains. Now, which do you think is
lighter, the charged or uncharged boxes?
>> ....furthermore, giving the mass as m = E/c^2 where E is the
>>electrostatic energy is manifestly incorrect. Try the following.
>>For an electron, which has an upper limit of 10^-18 meters for
>>it's radius, as determinied by experiment, calculate the self-energy
>>required to assemble the charge into a spherical volume 4pi r^3/3.
>>(feel free to choose any shape you want so long as all of the
>>charge is contained within a radius no larger than 10^-18 meters.
>>Calculate m. How many hundreds of times larger is this than the
>>electron mass?
>
>We did this before...
Then you must have missed the point.
>> No, it's just that there's more to interference than it superficially
>>appears.
>
>Did my wife just walk by ??? or do I sniff a Red Herring.
>
OK. If you think there is not, lets ignore interference that everyone
thinks is familiar and venture into the regime in which you can't fall
back on E&M to try and conjure up an answer.
First the preliminaries. Special relativity and the dirac theory
allows for 5 and only 5 types of potentials. Scalars (S), Pseudo-
scalars (P), Vectors (V), Axial vectors (A) and pseudo-Tensors (T).
These are all of the bilinear forms which can be constructed. All
others reduce to one of the forms. (You should prove this. Take the
dirac matrices (including \gamma^5) and figure out how many combinations
there are with unique transformation properties).
Now, the decay of 60Co, is parity violating, and for a relativistically
correct theory, we need a potential which is parity violating. None
of the forms listed are themselves parity violating, so we need a
combination of them. I claim the following: (1) The sum of any combination
of the above will _not_ give a parity violating potential without
interference, (2) that a parity violating potential is not unique
and in fact, one can show that the same potential may be constructed
in a number of ways through fierz reordering, for example the form
V-A could also be written in terms of S,P,T, (3) that it is possible
to differentiate between those forms through (fierz) interference.
Does any of this remotely resemble what you've come to think of as
interference?
Bilge
>news:slrnbai52b...@radioactivex.lebesque-al.net...
><snip>
>| ....furthermore, giving the mass as m = E/c^2 where E is the
>| electrostatic energy is manifestly incorrect. Try the following.
>| For an electron, which has an upper limit of 10^-18 meters for
>| it's radius, as determinied by experiment, calculate the self-energy
>| required to assemble the charge into a spherical volume 4pi r^3/3.
>| (feel free to choose any shape you want so long as all of the
>| charge is contained within a radius no larger than 10^-18 meters.
>| Calculate m. How many hundreds of times larger is this than the
>| electron mass?
>
>I don't think it is exactly true that the upper limit for "electron" radius
>is 10^-18 meters for "observed" charge distribution. It maybe is more on
>the order of the Compton wavelength divided by 2*pi. See Milonni "The
>Quantum Vacuum" section 11.5 "How Big is an Electron?".
What defines the size of an electron other than the radius one
determines from the force it exerts scattering a particle? If you look at
the data for the proton radius, you'll find that it's quoted as the rms
charge radius. That is because the radius might appear to be somewhat
different if you scatter pi_0's from it. Now, as for the electron. It can
interact through two different forces. The electromagnetic and weak force.
That leaves out giving the electron a radius any different than can be
measured using those two forces. The "charge radius" is known to be <
10^-18 m. The weak force on has a range of about 10^-18 m, there isn't
much hope of making that radius any larger. That places the "real" electron
radius at < 10^-18 m. The compton wavelength is something entirely
different. It represents the probability of the electron being found
in a region such that the probability with that region is larger than
1/e (where e is 2.71...), just like any gaussian.
>Yes, back to the
>issue of bare elementary point-like charge vs. "electron". Did you give up
>on your proposal that bare elementary charge might be infinitesimal?
No and it isn't my proposal.
>I still have not found anything that proposes anything other than bare
>elementary charge up close being greater than observed charge from a
>distance. At least please give me a clue as to why you might think that.
You won't be able to renormalize the charge otherwise.
Pmb
> > Which from a relativistic view point
> > translates to a changing electric potential (electric field).
>
"Tom Roberts" wrote
> Not necessarily, and specifically not in this case (see below).
Why! I'd have to disagree here. In the rest frame of the solenoid, S', the
4-potential is
A^u = (0, A)
in a frame moving perpendicular to the length of the solenoid, S', The
4-potential becomes
A'^u = (Phi'/c, A')
The electric field is zero but the vector potential A' in this frame is a
function of time t', i.e.
E' = -grad Phi' - (1/c) &A'/&t' = 0
Or
grad Phi' = (1/c) &A'/&t'
&A'/&t' is not zero it follows that grad Phi' is not zero and that means
that Phi's is not zero and is a not a constant.
In S' the total energy is given by (kinetic energy - rest energy - potential
energy)
E' = K' + E_o + q*Phi'
Which implies that the total energy in S' depends on the charge of the
particle - seems that its not constant from I can see. But I'm not certain
as to why this is so and what it means.
Feel free to rip it all apart. This is as far as I got --
www.geocities.com/physics_world/relativistic_charge.htm
www.geocities.com/physics_world/vector_potential.htm
Pmb
Ken S. Tucker
>Lawrence Foard wrote:
>> What am I missing here. I can't picture a way this experiment could
>> be performed without an electron crossing an area of changing vector
>> potential (if only in direction).
>
>Right. That's the point. In the rest frame of the solenoid, the vector
>potential outside a perfect infinitely-long solenoid is circular. So the
>integral of A_u dx^u over a path that goes past the solenoid on one side
>or the other (in a plane perpendicular to the axis) depends on which
>side of the solenoid it passes (in this frame A_0 = 0, so this is really
>a spatial integral of a 3-vector field).
This can be explained classically by induction, if I
understand your apparatus correctly...
Viewing the solenoid from either end...
A moving electron is a small current.
------> top electron
==>
+ Solenoid currents (+ is center)
<==
------> bottom electron
Think of the path of the top and bottom electrons
as the outer windings around an inner winding (solenoid)
of a transformer. (Each electron is of course sweeping
out an arc relative to "+").
The top electron is equivalent to a current in a
winding increasing the voltage in the solenoid,
the bottom electron is opposing the voltage, and
therefore will experience a higher impedeance
(resistance) along that path.
These two different paths will create a phase
difference.
If this is a correct interpretation, the Aharonov-Bohm
effect discovered the QM equivalent of Induction.
I'm sure this must have been accounted for though?
Regards
Ken S. Tucker
[snip]
>Tom Roberts tjro...@lucent.com
Tom Roberts
>>Lawrence Foard wrote:
>>>Which from a relativistic view point
>>>translates to a changing electric potential (electric field).
> "Tom Roberts" wrote
>>Not necessarily, and specifically not in this case (see below).
>
Yes, I misspoke. And I did not resolve the misstatements made by the
original poster.
The main point is: outside the solenoid, F=0. This implies that in any
frame F_uv=0 and therefore E=0 and B=0 -- classically that implies that
the solenoid does not affect the motion of the electrons.
The actual expressions for A_u in a moving frame are quite
complicated.... And ultimately irrelevant to the conclusion
(A_u dx^u is invariant, so compute in the original frame).
Tom Roberts tjro...@lucent.com
Pmb
Thank you for responding. What about the energy? I recall that it was one of
the main points on this thread?
The energy of a charge in this field will, of course, depend of the frame in
which you measure it and only the Coulomb potential is part of the energy -
I.e. Energy depends on the time component of the 4-potential. In the
stationary frame the 4-potential has no time component. But in the moving
frame it does and it contributes to the energy. The potential energy is then
proportional to the charge.
A charged particle has more/less energy than a non-charged particle
Is that a measurable thing?
Pmb
FrediFizzx
Doesn't the charge distribution radius depend upon the energies involved in
the scattering? At "rest" energies, the probability of finding another
electron much closer than the Compton wavelength divided by 2*pi to the
original electron is very small. As we probe closer to the bare charge with
higher energy, its charge value seems to be increasing. Although bare
charge seems to be point-like and has no structure, the bare charge along
with vacuum and relativistic effects seems to give the "electron" some
"structure" on the order of the Compton wavelength/2*pi. At low energies we
can't really remove bare charge from the vacuum, so the electron always has
an effective "spread" about this size. It seems to me that since the
quantum vacuum should be fairly homogeneous, the electron may have some
fairly regular spread out structure involving its "jiggling" about. Which
maybe gives us magnet moment and spin. IOW, if it were possible to take
bare charge out of the quantum vacuum and test it in a "pure" vacuum at
total complete rest, would it still have magnet moment and spin? I don't
think so. But probably just me still thinking too classical again.
| >Yes, back to the
| >issue of bare elementary point-like charge vs. "electron". Did you give
up
| >on your proposal that bare elementary charge might be infinitesimal?
|
| No and it isn't my proposal.
|
| >I still have not found anything that proposes anything other than bare
| >elementary charge up close being greater than observed charge from a
| >distance. At least please give me a clue as to why you might think
that.
|
| You won't be able to renormalize the charge otherwise.
OK, thanks for the clue.
FrediFizzx
Ken S. Tucker
>> Lawrence Foard wrote:
>> > Which from a relativistic view point
>> > translates to a changing electric potential (electric field).
>"Tom Roberts" wrote
>
>> Not necessarily, and specifically not in this case (see below).
Pete must you over use your space bar so much?
>Why! I'd have to disagree here. In the rest frame of the solenoid, S', the
>4-potential is
>A^u = (0, A)
>in a frame moving perpendicular to the length of the solenoid, S', The
>4-potential becomes
>A'^u = (Phi'/c, A')
>The electric field is zero but the vector potential A' in this frame is a
>function of time t', i.e.
>E' = -grad Phi' - (1/c) &A'/&t' = 0
>Or
>grad Phi' = (1/c) &A'/&t'
>&A'/&t' is not zero it follows that grad Phi' is not zero and that means
>that Phi's is not zero and is a not a constant.
>In S' the total energy is given by (kinetic energy - rest energy - potential
>energy)
>E' = K' + E_o + q*Phi'
>Which implies that the total energy in S' depends on the charge of the
>particle - seems that its not constant from I can see. But I'm not certain
>as to why this is so and what it means.
Pete, collect a GOLD star, (IMO) because you've determined
the law of induction, gutsy, my post supports this.
>Feel free to rip it all apart. This is as far as I got --
>www.geocities.com/physics_world/relativistic_charge.htm
>www.geocities.com/physics_world/vector_potential.htm
>Pmb
Nope, we're either both wrong or both right. You have
arrived at the same place I have but by sophisicated and patiently
applied mathematics, however I wouldn't critic your equations
until you have had an opportunity to check them several times.
Two Thumbs Up
Ken S. Tucker
Pmb
"Ken S. Tucker" <dyna...@vianet.on.ca> wrote
> Pete must you over use your space bar so much?
There's a problem with my computer (or MS's software). Too many times I
would take the time to write a carefully thought out response and then I'd
hit a particular key ("Enter" I think) and the program would crash. I would
then have lost the entire thing I took all that time to type in. So I now
copy the post into a "Text" file in MS-Word. I think type it in there. I'd
save it and then I'd copy it back to the post. It looks perfect when I send
it. But later when I read it there are sometimes a bunch of spaces. I don't
know why there are there nor do I know how they got there.
> Pete, collect a GOLD star, (IMO) because you've determined
> the law of induction, gutsy, my post supports this.
What induction? That requires the presence of a B field in the region of
interest and there is none.
>
> >Feel free to rip it all apart. This is as far as I got --
> >www.geocities.com/physics_world/relativistic_charge.htm
> >www.geocities.com/physics_world/vector_potential.htm
> >Pmb
>
> Nope, we're either both wrong or both right. You have
> arrived at the same place I have but by sophisicated and patiently
> applied mathematics, however I wouldn't critic your equations
> until you have had an opportunity to check them several times.
> Two Thumbs Up
I've checked them but I put them down for now. The page at
www.geocities.com/physics_world/relativistic_charge.htm
is 100% correct. I know that much. I had someone else check it (we touched
base in e-mail) too but there was some confusion that remains unresolved.
Since that person posts here I'll let them take it from here. Otherwise I'm
going to wait until I get some more feedback from some friends of mine.
Pete
ps - If you can tell me how to get "MS Outlook Express" to stop crashing
then I'll be able to solve the "space" thing
Bilge
attributed to impedence. The potentials responsible for the effect are
purely those which satisfy the gauge condition, which classically can
have no effect, by the very way in which classical E&M defines them.
> These two different paths will create a phase difference.
That part is certainly true, but classically, the phase is precisely
that which is not-physically meaningful. Classical E&M stops with F^{uv}
and F^{uv} cannot contain the phases by definition:
F^{uv} = d^{u}A^{v} - d^{v}A^{u}
Substitute A'^{u} = A^{u} - d^{u}Q
where Q is a scalar function of the coordinates, and you'll see that
F^{uv} -> F^{uv}
>If this is a correct interpretation, the Aharonov-Bohm effect discovered
>the QM equivalent of Induction. I'm sure this must have been accounted
>for though?
While it might initially appear that way, the aharanov-bohm effect is
entirely due that which induces nothing. There is _no_ force exerted
on the electron by the solenoid (or vice-versa since F_12 = -F_21).
Bilge
>> Lawrence Foard wrote:
>> > Which from a relativistic view point
>> > translates to a changing electric potential (electric field).
>"Tom Roberts" wrote
>> Not necessarily, and specifically not in this case (see below).
>Why! I'd have to disagree here. In the rest frame of the solenoid, S', the
>4-potential is
>A^u = (0, A)
>in a frame moving perpendicular to the length of the solenoid, S', The
>4-potential becomes
>A'^u = (Phi'/c, A')
>The electric field is zero but the vector potential A' in this frame is a
>function of time t', i.e.
>E' = -grad Phi' - (1/c) &A'/&t' = 0
>Or
>grad Phi' = (1/c) &A'/&t'
>&A'/&t' is not zero it follows that grad Phi' is not zero and that means
>that Phi's is not zero and is a not a constant.
But that is precisely the gauge condition. That potential is purely
a result of the gauge freedom in classical E&M and cannot contribute
to any effect classically.
>In S' the total energy is given by (kinetic energy - rest energy - potential
>energy)
>E' = K' + E_o + q*Phi'
The lagrangian is not the total energy. The hamiltonian is. Your
assertion about the potential energy in relativity is going to come
back to haunt you. The hamiltonian for a charged particle in an
electromagenetic field is:
E = (1/2m)(p - (e/c)A)^2 + q\phi not E = p^2/2m - e\phi
You can generalize that for the relativistic case.
Pmb
"Bilge" <dub...@radioactivex.lebesque-al.net> wrote
[pmb wrote]
> >grad Phi' = (1/c) &A'/&t'
>
> >&A'/&t' is not zero it follows that grad Phi' is not zero and that means
> >that Phi's is not zero and is a not a constant.
[bilge wrote]
> But that is precisely the gauge condition.
Please define "the gauge condition," in particular the "the" part? This
doesn't have the form either the Lorenz gauge nor the Coulomb gauge. The
gauge was chosen as part of the definition of A when the relation was
derived. The Lorenz gauge is
div A + (1/c) &Phi/&t = 0
You can call it the Coulomb gauge as well since Phi = constant and thus
&Phi/&t = 0. In any case div A = in frame O.
> That potential is purely
> a result of the gauge freedom in classical E&M and cannot contribute
> to any effect classically.
Why? In what sense does it make to say "purely a result of the gauge
freedom" and how do you know it can't contribute to any effect.
Energy is simply an integral of motion. Phi is part of it.
>
> >In S' the total energy is given by (kinetic energy - rest energy -
potential
> >energy)
>
> >E' = K' + E_o + q*Phi'
>
> The lagrangian is not the total energy.
Who said it was??? Not I. Had you looked at the derivation you would have
seen that the relation
E' = K' + E_o + q*Phi'
is exact. And had you read the derivation that I gave a link to then you
would have seen that. It now appears that you're criticizing a post and an
equation without having seen the derivation. This *is* the energy.
www.geocities.com/physics_world/relativistic_charge.htm
> The hamiltonian is.
I never said that it wasn't. You made an error in asserting that what I gave
was a lagrangian.
> Your
> assertion about the potential energy in relativity is going to come
> back to haunt you.
Actually you have it the other way around bilge. It's biting you in the butt
at this very moment.
> The hamiltonian for a charged particle in an
> electromagenetic field is:
>
> E = (1/2m)(p - (e/c)A)^2 + q\phi not E = p^2/2m - e\phi
>
> You can generalize that for the relativistic case.
I did! I keep telling to read that derivation and you continue to refuse.
Equation #6 in --- www.geocities.com/physics_world/relativistic_charge.htm
gives the definition of "energy function" as
h == v_i &L&v_i - L
This is **by definition** the total energy of the system E. The value, in
frame O', is
E' = K' + E_o + q*Phi'
and that is relativistically the exact *correct* value!!!
**IF** you wish to comment further then before you criticize read the
derivation so you know what you're talking about.
The error you just made is in not recognizing (or not asking - try it
sometime) the relation.
h = energy function *** I S *** the Hamiltonian in value. All the
Hamiltonian is is the energy function expressed in terms of the canonical
momentum - **period**. And the fact that you didn't notice that quite
interesting - This is undergrad physics!
You should have paid attention when I was trying to explain this to you.
Pmb
Bilge
Learn to use a newsreader.
>"Bilge" <dub...@radioactivex.lebesque-al.net> wrote
>[pmb wrote]
>
>> >grad Phi' = (1/c) &A'/&t'
>> >&A'/&t' is not zero it follows that grad Phi' is not zero and that
>> >means that Phi's is not zero and is a not a constant.
>[bilge wrote]
>
>> But that is precisely the gauge condition.
>Please define "the gauge condition," in particular the "the" part?
The gauge substitutions are:
A -> A' = A + \grad Q
\phi -> \phi' = \phi - (1/c)dQ/dt
Where Q is any scalar function.
\grad\phi' -> \grad\phi - (1/c)(d/dt)\grad Q
\grad\phi' = \grad\phi - (1/c)(d/dt)(A' - A)
\grad\phi' - (1/c)(dA'/dt) = \grad\phi - dA/dt
If the potentials ever satisfy the condition, \grad\phi = (1/c) dA/dt,
they will always satisfy it.
More compactly:
F^{uv} = d^{u}A^{v} - d^{v}A^{u}
Substitute A' = A + d^{u}\Q
F^{uv}' = d^{u}[ A^{v} - d^{v}Q ] - d^{v}[ A^{u} - d^{u}Q ]
= d^{u}A^{v} - d^{v}A^{u} - [d^{u}d^{v}Q - d^{u}d^{v}Q]
= d^{u}A^{v} - d^{v}A^{u} - [d^{u}d^{v}Q - d^{u}d^{v}Q]
= d^{u}A^{v} - d^{v}A^{u}
= F^{uv}
>> That potential is purely a result of the gauge freedom in
>> classical E&M and cannot contribute to any effect classically.
>Why? In what sense does it make to say "purely a result of the gauge
>freedom" and how do you know it can't contribute to any effect.
For the reason I just gave above. See jackson, chapter 6, p220-221
2nd ed.
>> >In S' the total energy is given by (kinetic energy - rest energy -
>> >potential energy)
>> >
>> >E' = K' + E_o + q*Phi'
>> The lagrangian is not the total energy.
>Who said it was???
You did above where you wrote "In S' the total energy is given by
(Kinetic energy - rest energy - potential energy)". The lagrangian
is (classically) L = T - U. That's what you wrote. That's what I
assumed you meant.
>Not I. Had you looked at the derivation you would have seen that the
>relation
>E' = K' + E_o + q*Phi'
Why did you write "kinetic energy - rest energy - potential energy"?
>is exact. And had you read the derivation that I gave a link to then you
>would have seen that. It now appears that you're criticizing a post and an
>equation without having seen the derivation. This *is* the energy.
I'm not particularly interested in reading your web pages. You can't
post a consistent news article nor format your posts, so I'm not going
to go to any additional effort to read more of the same. The newsgroups
are for news articles.
>> The hamiltonian is.
>I never said that it wasn't. You made an error in asserting that what
>I gave was a lagrangian.
Last I checked, T - U is a lagrangian. If that's not what you meant,
then try correcting the problem at the source rather than blame me for
reading what you wrote.
>> The hamiltonian for a charged particle in an electromagenetic field is:
>> E = (1/2m)(p - (e/c)A)^2 + q\phi not E = p^2/2m - e\phi
>> You can generalize that for the relativistic case.
>I did! I keep telling to read that derivation and you continue to refuse.
I'm responding to what you posted.
[...]
>
>**IF** you wish to comment further then before you criticize read the
>derivation so you know what you're talking about.
>The error you just made is in not recognizing (or not asking - try it
>sometime) the relation.
Try writing two consecutive lines which are self-consistent.
>h = energy function *** I S *** the Hamiltonian in value. All the
>Hamiltonian is is the energy function expressed in terms of the canonical
>momentum - **period**. And the fact that you didn't notice that quite
>interesting - This is undergrad physics!
Then why are you arguing about it instead of simply correccting what you
wrote above?
>You should have paid attention when I was trying to explain this to you.
Which part was that? The part about gauge freedom or the hamiltonian is
"(kinetic energy - rest energy - potential energy)"? Which do you think
that is - a lagrangian or hamiltonian? No thanks. I don't want to unlearn
what I spent effort learning.
Bilge
>"Bilge" <dub...@radioactivex.lebesque-al.net> wrote in message
>Doesn't the charge distribution radius depend upon the energies involved in
>the scattering?
Yes. For example, that's how we know there are quarks within the
proton.
>At "rest" energies, the probability of finding another electron much
>closer than the Compton wavelength divided by 2*pi to the original
>electron is very small. As we probe closer to the bare charge with
>higher energy, its charge value seems to be increasing. Although bare
>charge seems to be point-like and has no structure, the bare charge along
>with vacuum and relativistic effects seems to give the "electron" some
>"structure" on the order of the Compton wavelength/2*pi.
In a manner of speaking. However at low energy one is not measuring
"the electron".
>At low energies we can't really remove bare charge from the vacuum,
>so the electron always has an effective "spread" about this size.
Something along those lines. It sounds like you've done some
reading.
>It seems to me that since the quantum vacuum should be fairly homogeneous,
It has to be. The fluctuations are entirely random. If that wasn't
the case, I'm not sure how it could be homogeneous.
>the electron may have some fairly regular spread out structure involving
>its "jiggling" about. Which maybe gives us magnet moment and spin. IOW,
>if it were possible to take bare charge out of the quantum vacuum and
>test it in a "pure" vacuum at total complete rest, would it still have
>magnet moment and spin? I don't think so. But probably just me still
>thinking too classical again.
What you are doing is assuming you can do something that is physically
impossible, which might be called "classical" in the sense that one
often idealizes being able to examine something independently of its
environment. This isn't always true. For example, consider a bubble
of air in water tank. You can't "remove" the bubble to examine its
properties independent of the water.
If you want to think about the fundamental building blocks of
nature, you have to stop considering the ability to measure some-
thing independent of what you are measuring. And I don't mean
that in a way that suggests the measurement disturbs the system,
and that you could, in principle, measure everything, if only
you could make your probe inobtrusive enough. You have to consider
the possibility that the reason you can't do that also has something
to do with the reason that nature is what it is.
Pmb
"Bilge" <dub...@radioactivex.lebesque-al.net> wrote
> You did above where you wrote "In S' the total energy is given by
> (Kinetic energy - rest energy - potential energy)".
I made a 'slight' error. I wrote
---------------------------------------------------------------------
In S' the total energy is given by (kinetic energy - rest energy - potential
energy)
E' = K' + E_o + q*Phi'
---------------------------------------------------------------------
You mean to tell me that you think these two expressions are consistent???
You mean to tell me that you didn't recognize a small mistake like a sign???
Tell me something bilge. You thought I made an error so instead of saying
that "-" signs in
"kinetic energy - rest energy - potential energy"
should be "+" why did you, instead, conclude that what I claimed was "the
total energy" was a Lagrangian? There's absolutely no reason to think that
is a Lagrangian! I told you it was the total energy! Rest energy doesn't
appear in a Lagrangian that way!!!!
You've never seen the expression E' = K' + E_o + q*Phi' before?? If not why?
This is the total relativistic energy for a charged particle in an EM field.
Since E' is obviously the total energy I'd like to ask you why you simply
didn't explain to me that the error was in the signs?????
> The lagrangian
> is (classically) L = T - U. That's what you wrote.
Nope. That's not what I wrote. I made an error in signs
I never stated that anything that I posted was a Lagrangian. In fact L = T -
U is not a relativistic Lagrangian. And I don't see why you would assume
that it was. Having the rest energy in an expression is a sure sign that
we're talking relativity. This is classical (i.e. non-quantum) relativistic
electrodynamics we're talking about after all.
> I'm not particularly interested in reading your web pages.
You took the time to read *all* of my post on **physics** and then you typed
in all of the above yet you refuse to read the complete and correct
derivation I've done on the **physics**? Why would you read what I write in
a post but refuse to read the derivation? I only put it in web page to make
the math clear. That kind of math can't be typed in and still be readable.
> You can't
> post a consistent news article nor format your posts, ..
Why is it you always find the need to start posting insults? It's uncalled
for.
> Last I checked, T - U is a lagrangian.
That is true only under *certain conditions* and then only in
non-relativistic cases. This is relativity. Or didn't you know that when
"rest energy" appears in a post that it just might be on relativity???
The only place I gave a Lagrangian was in a derivation that you refuse to
follow.
The correct Lagrangian for a relativistic charged particle in an EM field is
derived in Jackson's text in Section 12.1 - "Lagrangian and Hamiltonian for
a Relativistic Charged Particle in an External Electromagnetic Field" and is
given in subsection (a) "Elementary Approach to a Relativistic Lagrangian"
(Jackson - Eq 12.9) L = -m_o*c^2*sqrt[1 - (v/c)^2] + (q/c)A*v - q*Phi
When this *relativistically correct* Lagrangian is substituted into
Lagrange's equations (not the covariant one - the other one) it yields the
*relativistically correct* equations of motion (Note: 12.9 is not the
covariant Lagrangian but it is a relativistic Lagrangian). But both the
covariant Lagrangian and this Lagrangian yield the same equations of motion
when used appropriately. And by 'appropriately' I mean that there are two
forms of Lagrange's equations - The covariant form and the non-covariant
form. Each is still correct but each uses a different approach.
If you take the covariant equation
dU^dT = (q/c*m_o)F^uv*U_v
And write out its components then you'd see that it's this (i.e. Eq. 12.1 in
Jackson)
dp/dt = q[E + (v/c)xB]
which is the *relativistically correct* equation of motion and that is the
equation of motion that one gets with the *relativistically correct*
Lagrangian that I gave and used.
> If that's not what you meant,
> then try correcting the problem at the source rather than blame me for
> reading what you wrote.
When you refuse to read entire post then I'll keep correcting you. The web
page had the derivations and the correct results. And that was the topic of
my post. If you refuse to read the topic of my post (i.e. the physics in
that web page) then please do not respond to my post.
> I'm responding to what you posted.
Nope. You responded to an argument that you didn't fully read. The topic of
my post was the physics in the web pages I gave. Instead you spent all this
time on a slight error I made in signs. However I know that can happen. And
it is for THAT reason I gave the URLs. Please do not respond to my posts
when you refuse to read the entire post - web page and all.
Part of the **content of my post** was the mathematical derivations leading
to what I posted. As such I gave the URL to that derivations so there would
be no uncertainty in this respect. That kind of lengthy derivation can't be
posted in text format and still be clear - too lengthy!!!
> >h = energy function *** I S *** the Hamiltonian in value. All the
> >Hamiltonian is is the energy function expressed in terms of the
canonical
> >momentum - **period**. And the fact that you didn't notice that quite
> >interesting - This is undergrad physics!
>
> Then why are you arguing about it instead of simply correcting what you
> wrote above?
Because you're wrong. I never mentioned a Lagrangian and I never mentioned a
Hamiltonian. What I did mention was the total energy. However you didn't
recognize that
E' = K' + E_o + q*Phi'
was the total energy in for a particle. Had you recognize this equation then
you would have realize that the negative signs were wrong and that E' = K' +
E_o + q*Phi' really was the correct expression for the energy.
See what happens when you refuse to read an entire post and only comment on
a portion of an argument????
> >You should have paid attention when I was trying to explain this to
you.
>
> Which part was that?
When I posted the argument itself. You simply refuse to read the argument -
you just wanted to post on half of an argument. The part where you didn't
have to "click" on a URL. I know that errors come up when I type. THAT is
why I created the web page.
And your excuse for not reading it is pretty lame. You're the one who choose
to read what I wrote here. So don't give me that BS about not reading
something else I wrote in support of it - just because its on a web page!!
> or the hamiltonian is
> "(kinetic energy - rest energy - potential energy)"?
I posted the correct physics in the page the URL points to. I've also stated
this result a huge number of times in the "potential" thread *****on the
exact same problem - I.e. the Lagrangian for a relativistic charged in an EM
field. I can't even count how many times that I told you that in such a case
the total energy = Kinetic energy + rest energy + potential energy so you
should have know that I made a typo especially since I posted the link to
the physics.
When you *try* to correct a post when you have not read the entire content
then you should not comment on it.
When you find that you are able to read a web page let me know. Until then
please do not respond - I'm not interested in your comments on only a piece
of what you read. If you want to read an entire argument - web page and
all - let me know.
Until then please leave me alone. I'm not interested in anything you have to
say until you know exactly the physics I'm referring to and to do that you'd
have to read a web page - seems that you're not capable of that.
Pmb
Pmb
"Bilge" <dub...@radioactivex.lebesque-al.net>
> Learn to use a newsreader.
(sigh!) Condescending as always huh? When are you going to stop behaving in
this manner? I know how to use it just fine and had you read the response I
wrote to Ken on this (or had **asked** which is something that you do not
have a talent for) then you'd know that I have a software problem.
> >"Bilge" <dub...@radioactivex.lebesque-al.net> wrote
> >[pmb wrote]
> >
> >> >grad Phi' = (1/c) &A'/&t'
> >> >&A'/&t' is not zero it follows that grad Phi' is not zero and that
> >> >means that Phi's is not zero and is a not a constant.
>
> >[bilge wrote]
> >
> >> But that is precisely the gauge condition.
>
> >Please define "the gauge condition," in particular the "the" part?
>
> The gauge substitutions are:
>
> A -> A' = A + \grad Q
>
> \phi -> \phi' = \phi - (1/c)dQ/dt
>
> Where Q is any scalar function.
>
> \grad\phi' -> \grad\phi - (1/c)(d/dt)\grad Q
>
> \grad\phi' = \grad\phi - (1/c)(d/dt)(A' - A)
>
> \grad\phi' - (1/c)(dA'/dt) = \grad\phi - dA/dt
(sigh!) Seems like I made a slight error a sign again. I wrote
----------------------------------------
E' = -grad Phi' - (1/c) &A'/&t' = 0
Or
grad Phi' = (1/c) &A'/&t'
----------------------------------------
I missed the "-" sign in going from one to the other. It should have read
grad Phi' = - (1/c) &A'/&t'
However that was obvious anyway!
> If the potentials ever satisfy the condition, \grad\phi = (1/c) dA/dt,
> they will always satisfy it.
So what? Who said anything about the relation "\grad\phi = (1/c)
dA/dt"???????
The only relation that I posted was
-grad Phi' - (1/c) &A'/&t' = 0
or if you like
grad Phi' = - (1/c) &A'/&t'
which is a very different relation than what you wrote. "&" represents the
partial derivative. That's obvious from the use of "&" in the relation for
the electric field
"E' = -grad Phi' - (1/c) &A'/&t'"
So "&" is a *partial* derivative. d/dt is a *total* derivative.
So how is "grad Phi' = - (1/c) &A'/&t'" a "gauge condition????
Pmb
Ken S. Tucker
Can you simplify this....
>The effect cannot be attributed to impedence.
>The potentials responsible for the effect are
>purely those which satisfy the gauge condition, which classically can
>have no effect, by the very way in which classical E&M defines them.
I'll disagree, please see below....
> > These two different paths will create a phase difference.
>
> That part is certainly true, but classically, the phase is precisely
>that which is not-physically meaningful. Classical E&M stops with F^{uv}
>and F^{uv} cannot contain the phases by definition:
> F^{uv} = d^{u}A^{v} - d^{v}A^{u}
> Substitute A'^{u} = A^{u} - d^{u}Q
> where Q is a scalar function of the coordinates, and you'll see that
>F^{uv} -> F^{uv}
Ok, Curl x Grad=0, that's agreed. A phase shift will appear
if electron energy, velocity, or paths are relatively different.
> >If this is a correct interpretation, the Aharonov-Bohm effect discovered
> >the QM equivalent of Induction. I'm sure this must have been accounted
> >for though?
>
> While it might initially appear that way, the aharanov-bohm effect is
>entirely due that which induces nothing. There is _no_ force exerted
>on the electron by the solenoid (or vice-versa since F_12 = -F_21).
What?? A solenoid is a conductor. A current pulse, such as an
electron moving by it, is subject to inductance because charges
are relatively moving, the electron is creating a changing magnetic
field relative to a conductor - producing radiation, and this
will happen if solenoid current is zero.
This is very basic electricity of transformers and inductors,
have a look at the theory of *air-core* transformers.
Let's cover the basics before jumping to conclusions, no
amount of complex math, inappropriately applied, will
change the experimental result.
Regards Ken S. Tucker
AntiCrank
"Pmb" <physic...@yahoo.coom> wrote in message
>
>
> "Bilge" <dub...@radioactivex.lebesque-al.net> wrote
>
>
>
> > You did above where you wrote "In S' the total energy is given by
>
> > (Kinetic energy - rest energy - potential energy)".
>
>
>
> I made a 'slight' error. I wrote
*yawn* So what else is new, loser?
> > Then why are you arguing about it instead of simply correcting what
you
> > wrote above?
>
>
>
> Because you're wrong.
You are the one who made an error, you even said as much.
>
>
> Until then please leave me alone. I'm not interested in anything you have
to
> say until you know exactly the physics I'm referring to and to do that
you'd
> have to read a web page - seems that you're not capable of that.
Seems like bilge knows the physics, and was quite capable of pointing out
your errors, Brown.
AntiCrank
"Pmb" <physic...@yahoo.coom> wrote in message
>
>
> "Bilge" <dub...@radioactivex.lebesque-al.net>
>
>
>
> > Learn to use a newsreader.
>
>
>
> (sigh!) Condescending as always huh? When are you going to stop behaving
in
> this manner? I know how to use it just fine and had you read the response
I
> wrote to Ken on this (or had **asked** which is something that you do not
> have a talent for) then you'd know that I have a software problem.
Learn to use a newsreader Brown, there is such a function as save file.
Just hit Ctrl. c and you will save a copy of the message you are editing.
That way if your newsreader crashes, you will not have to start over again
and you can get rid of your multiple spaces.
You are welcome Brown, just like you are welcome for the spam block
information.
Ken S. Tucker
>Ken S. Tucker:
> >dub...@radioactivex.lebesque-al.net (Bilge) wrote in message
> >news:<slrnbai52b...@radioactivex.lebesque-al.net>...
> >>Ken S. Tucker:
> >>You do not have "only" two boxes.
> >>You have at least 3 other wise you couldn't reference boxes A
> >>and B to a ground potential (0 volts) [which is where you got
> >>the charge]. If you have only two boxes, you can only express their
> >>potential difference and all of the work goes into moving charge
> >>from one box to another.
> >
> >Please answer my question.
>
> OK. If you insist, I'm going to answer it precisely in a way you
>didn't expect and probably didn't want.
>
> I'll charge one box by _removing_ negative charge so that a positive
>charge remains. I then charge the other box by removing positive charge
>so that only negative charge remains. Now, which do you think is
>lighter, the charged or uncharged boxes?
While unsure this question has a point, uncharged boxes
are lighter.
(It was your suggestion to charge up one box (the universe) to
a gazillion volts, and say it makes no diff).
> >> ....furthermore, giving the mass as m = E/c^2 where E is the
> >>electrostatic energy is manifestly incorrect. Try the following.
> >>For an electron, which has an upper limit of 10^-18 meters for
> >>it's radius, as determinied by experiment, calculate the self-energy
> >>required to assemble the charge into a spherical volume 4pi r^3/3.
> >>(feel free to choose any shape you want so long as all of the
> >>charge is contained within a radius no larger than 10^-18 meters.
> >>Calculate m. How many hundreds of times larger is this than the
> >>electron mass?
> >
> >We did this before...
>
> Then you must have missed the point.
No, I concluded a single charge does not self energize.
> >> No, it's just that there's more to interference than it superficially
> >>appears.
> >
> >Did my wife just walk by ??? or do I sniff a Red Herring.
> >
> OK. If you think there is not, lets ignore interference that everyone
>thinks is familiar and venture into the regime in which you can't fall
>back on E&M to try and conjure up an answer.
Ok, I'll use GR....
> First the preliminaries. Special relativity and the dirac theory
>allows for 5 and only 5 types of potentials. Scalars (S), Pseudo-
>scalars (P), Vectors (V), Axial vectors (A) and pseudo-Tensors (T).
>These are all of the bilinear forms which can be constructed. All
>others reduce to one of the forms. (You should prove this. Take the
>dirac matrices (including \gamma^5) and figure out how many combinations
>there are with unique transformation properties).
Yes, one can go in circles by reconfiguring mathematics without
any physical reason....otherwise known as mathematical torment.
> Now, the decay of 60Co, is parity violating, and for a relativistically
>correct theory, we need a potential which is parity violating. None
>of the forms listed are themselves parity violating, so we need a
>combination of them.
>I claim the following: (1) The sum of any combination
>of the above will _not_ give a parity violating potential without
>interference,
>(2) that a parity violating potential is not unique
>and in fact, one can show that the same potential may be constructed
>in a number of ways through fierz reordering, for example the form
>V-A could also be written in terms of S,P,T, (3) that it is possible
>to differentiate between those forms through (fierz) interference.
> Does any of this remotely resemble what you've come to think of as
>interference?
Yes, the terminology you employ is sometimes unfamiliar,
phase is (IMO) a means of translating phenomenological
action (mathematically) in finite space-time intervals to
predict effects. For example...
Although it is interesting that one may
use Newtonian "instanteous action" with a GR correction
to get "the right answer" in celestrial mechanics, we do
know we are violating GR in this calculation, that
requires a finite gravitation speed, yet we do it.
If I were to solve celestrial mechanical problems
using GR strictly, phase (or equivalent) would be
necessary.Why not try to demonstrate a 2 body
gravitational solution with limited g-speed using
phases. This mathematical technique should be
available from QM, evidently, you're an expert!
(I'm just guessing, but I think Lagrange points
would end up being standing waves?)...
Interesting and Regards
Ken S. Tucker
Ken S. Tucker
>> Learn to use a newsreader.
>(sigh!) Condescending as always huh? When are you going to stop behaving in
>this manner? I know how to use it just fine and had you read the response I
>wrote to Ken on this (or had **asked** which is something that you do not
>have a talent for) then you'd know that I have a software problem.
Know what you are saying Pete, I've posted and reviewed ascii
diagrams, and they turn up screwed anyhow, ok, I understand,
it's not deliberate....no big deal, sorry if I offended you, it was
not intentional.
>> >"Bilge" <dub...@radioactivex.lebesque-al.net> wrote
>> >[pmb wrote]
>> >
>> >> >grad Phi' = (1/c) &A'/&t'
>> >> >&A'/&t' is not zero it follows that grad Phi' is not zero and that
>> >> >means that Phi's is not zero and is a not a constant.
>> >[bilge wrote]
>> >
>> >> But that is precisely the gauge condition.
>> >Please define "the gauge condition," in particular the "the" part?
>>
>> The gauge substitutions are:
>>
>> A -> A' = A + \grad Q
>>
>> \phi -> \phi' = \phi - (1/c)dQ/dt
>>
>> Where Q is any scalar function.
>>
>> \grad\phi' -> \grad\phi - (1/c)(d/dt)\grad Q
>>
>> \grad\phi' = \grad\phi - (1/c)(d/dt)(A' - A)
>>
>> \grad\phi' - (1/c)(dA'/dt) = \grad\phi - dA/dt
But isn't this just Curl x Grad =0, Bilge, isn't it easier to begin
your essay on common vector differentition rules, that are
basic and common too those you are communicating to,
McAnally, is certainly better than you at detail, as you said,
but our NG holds an interest in understanding.
Math equations alone is not good enough. Everyone in this
NG is much more interested in how to acquire "common
thoughts" that are rationally concieved.
In spite of the fact you have an exceptional intelligence,
can you assist, the unification of our NG, in view of your
ideas, by anologies that may assist the average American
to be more educated in physics?
Regards
Ken S. Tucker
>(sigh!) Seems like I made a slight error a sign again. I wrote
>
>----------------------------------------
>
>E' = -grad Phi' - (1/c) &A'/&t' = 0
>
>Or
>
>grad Phi' = (1/c) &A'/&t'
>----------------------------------------
>I missed the "-" sign in going from one to the other. It should have read
>grad Phi' = - (1/c) &A'/&t'
>
>However that was obvious anyway!
>
>> If the potentials ever satisfy the condition, \grad\phi = (1/c) dA/dt,
>> they will always satisfy it.
>So what? Who said anything about the relation "\grad\phi = (1/c)
>dA/dt"???????
>
>The only relation that I posted was
>
>-grad Phi' - (1/c) &A'/&t' = 0
>
>or if you like
>
>grad Phi' = - (1/c) &A'/&t'
>
>which is a very different relation than what you wrote. "&" represents the
>partial derivative. That's obvious from the use of "&" in the relation for
>the electric field
>
>"E' = -grad Phi' - (1/c) &A'/&t'"
>
>So "&" is a *partial* derivative. d/dt is a *total* derivative.
>
>So how is "grad Phi' = - (1/c) &A'/&t'" a "gauge condition????
>Pmb
Our good friend Bilge is on the ropes here. In a few hours,
my post will show up, and in geometric terms, produce a
classical explanation for Tom Robert's apparatus, contrary
to Bilges complex math.
Ken S. Tucker
Pmb
"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
news:2202379a.03042...@posting.google.com...
He's trying very hard to stay off the ropes. Hence his refusal to look at
the derivation that I posted (through a web page)
> In a few hours,
> my post will show up, and in geometric terms, produce a
> classical explanation for Tom Robert's apparatus, contrary
> to Bilges complex math.
>
> Ken S. Tucker
We've been assuming that whatever we put in the field was not changing the
field. If I have a charged particle at rest in the moving frame S' then it's
moving relative to the coil. It will therefore generate a magnetic field
thus altering the current in the coil. The changing current will then
generate a field which will exert a force on the charge. If I want to keep
it moving at constant velocity then I have to do work on it. Hence the
functional dependence of the total energy on the charge. If I want to move a
charge into the field I have to do work
How's that sound
Pete
Lawrence Foard
I've been doing some more thinking about this problem. Decided to think
about it only in terms of fields rather than potentials. Fields to me are
feel much more real and concrete since they have properties of 'real'
objects, like conserved energy and momentum.
Locality by proxy (for slow electrons):
Lets suppose that the following is true. The forces on a charge(s)
in an E&M field is the result of a position dependant variation in
the total energy stored in the E+B fields. This seems like an innocent
enough statement.
In general two particles fields have an interaction that looks like this
(ignoring constants):
Let E1 and B1 be the electric and magnetic fields of object 1
Let E2 and B2 be the electric and magnetic fields of object 2
energy = integral Et(X)^2 + Bt(X)^2 =
integral (E1(X) + E2(X))^2 + (B1(X) + B2(X))^2 =
integral E1(X)^2 + E2(X)^2 + 2E1(X)E2(X) + B1(X)^2 + B2(X)^2 + 2B1(X)*B2(X)
Notice that the difference between the particles interacting, vs
being infinitly far apart is the energy:
integral 2E1(X)E2(X) + 2B1(X)B2(X)
The spatial derivatives of this energy should reduce approximately to the
standard E&M force equations. Due to the fact that the vast majority of
a charges field energy is within a microscopic radius of the charge. Due to
this fact, the vast majority of the forces from the E and B fields are
simply proportional to the E & B fields local to the charge.
However this isn't the whole story, the vast majority of the field energy
is 'local' to the charge, but some extends further. It should be possible
given the right setup and a sufficiently sensitive detector to measure
the effects of remote field interactions on the motion of a charge.
The Aharonov-Bohm apparatus is such an experiment, the phase of an electron
is an incredibly sensitive detector of forces on the particle, forces which
would be very hard to detect in any classical experiment.
Notice that in the AB apparatus the sign of integral 2B1(X)*B2(X) is going
to depend on the side of the solenoid which a particle goes around in the
apparatus. Going one way the electron has to 'contribute' integral 2B1(X)*B2(X)
energy to the fields the only energy it has to spare is kinetic energy, so it
seems that its velocity must slow. The other way, it is given an energy
integral 2B1(X)*B2(X), it seems its velocity must increase.
Also notice that there is an energy flow in the solenoid due to the poynting
vector, which results from the electrons electric field crossed with the
solenoids magnetic field.
This behaviour may seem odd, normally a particles velocity is unchanged
by a magnetic field, only its direction. But thats because when dealing
with a B field local to the particle, the energy difference
integral 2B1(X)*B2(X) = 0,
on one side of the electron 2B1(X)*B2(X) is positive, on the otherside its
equally negative. However a magnetic field gradient should case some effect,
but given its tiny and transient nature (no lasting velocity change) it will
be very difficult to detect.
Prediction:
Given the right apparatus there will be classically detectable changes
in particle velocity which are not predicted by the classical E&M force
equations.
Issues:
This considers only the slow moving charge case. In reality remote fields
can't communicate with there charges instantly, rather communication is
delayed by the speed of light. Its hard to imagine the mechanism for
communication back to the charge, I assume it would take the form of
directed E&M radiation?
What happens when magnetic shielding is added to AB? What happens to
the poynting vector in the solenoid due to the shielding effects of the
wire? My first thought is that this effect must be such that shielding
cannot negate it.
--
Be a counter terrorist perpetrate random senseless acts of kindness
Rave: Immanentization of the Eschaton in a Temporary Autonomous Zone.
"Anyone who trades liberty for security deserves neither liberty nor security"
-Benjamin Franklin
Pmb
"Lawrence Foard" <ent...@farviolet.com> wrote in message
news:b8k6uf$kvd$1...@farviolet.com...
>
> I've been doing some more thinking about this problem. Decided to think
> about it only in terms of fields rather than potentials. Fields to me are
> feel much more real and concrete since they have properties of 'real'
> objects, like conserved energy and momentum.
Same here
I was assuming that whatever I was doing was not changing the field. If I
have a charged particle at rest in the moving frame then it's moving
relative to the coil. It will therefore generate a magnetic field thus
altering the current in the coil. The changing current will then generate a
field which will exert a force on the charge. If I want to keep it moving at
constant velocity then I have to do work on it. Hence the functional
dependence of the total energy on the charge. If I want to move a charge
into the field I have to do work
Pmb
Bilge
>news:<hc6ra.16783$xw4....@nwrdny01.gnilink.net>...
>>"Bilge" <dub...@radioactivex.lebesque-al.net>
>>> >
>>> >> >grad Phi' = (1/c) &A'/&t'
>>> >> >&A'/&t' is not zero it follows that grad Phi' is not zero and that
>>> >> >means that Phi's is not zero and is a not a constant.
>>> >[bilge wrote]
>>> >
>>> >> But that is precisely the gauge condition.
>>> >Please define "the gauge condition," in particular the "the" part?
>>>
>>> The gauge substitutions are:
>>>
>>> A -> A' = A + \grad Q
>>>
>>> \phi -> \phi' = \phi - (1/c)dQ/dt
>>>
>>> Where Q is any scalar function.
>>>
>>> \grad\phi' -> \grad\phi - (1/c)(d/dt)\grad Q
>>>
>>> \grad\phi' = \grad\phi - (1/c)(d/dt)(A' - A)
>>>
>>> \grad\phi' - (1/c)(dA'/dt) = \grad\phi - dA/dt
>
>But isn't this just Curl x Grad =0, Bilge, isn't it easier to begin
>your essay on common vector differentition rules, that are
>basic and common too those you are communicating to,
rules or else they shouldn't be arguing in the first place. Anyone
else who has a question, is free to ask. If you indicate you under-
stand the material enough to argue about it, then I shouldn't have
to explain the basics of E&M.
> McAnally, is certainly better than you at detail, as you said,
>but our NG holds an interest in understanding.
I didn't feel that it was necessary to post an introduction to
vector calculus and E&M, since anyone who claims to know enough
physics to be arguing about this should already be familiar with
what I wrote, not to mention the basics. Asking a question is
different from arguing a point. The latter presumes a knowledge
of the point in questionn while the former indicates a desire
to understand the point.
> Math equations alone is not good enough. Everyone in this
>NG is much more interested in how to acquire "common
>thoughts" that are rationally concieved.
Sorry, but in some cases you just have to struggle with the
mathematics before the concept sinks in. It's a mistake to
think equations express no physical meaning or lack the ability
to convey fundamental concepts. Equations only fail to convey
their content when one is used to using "formulae" to obtain
numerical solutions rather than to convey ideas. I'm not going
to write a book every time I respond to someone who clearly
insists he/she understands the point and wishes to argue about
it. There are lots of textbooks that already do that.
> In spite of the fact you have an exceptional intelligence,
>can you assist, the unification of our NG, in view of your
> ideas, by anologies that may assist the average American
> to be more educated in physics?
I try to do that by responding at the level I believe the person
insists he/she be addressed. If someone asks what gauge freedom
means, I'll explain it. If someone argues about what gauge freedom
means, I'll assume I don't have to explain what it is first.
The way one can get a response at the appropriate level is by
phrasing one's post to indicate that. If someone includes a
lot of jargon, then I can only assume they know what it means.
Bilge
>> Learn to use a newsreader.
>(sigh!) Condescending as always huh? When are you going to stop behaving
>in this manner?
insert 3 or 4 blank lines between every paragraph and turn a 30 line
post into one which is 120 lines, which I have to reformat.
>I know how to use it just fine and had you read the response I
in your posts.
>wrote to Ken on this (or had **asked** which is something that you do not
>have a talent for) then you'd know that I have a software problem.
>>
>> >Please define "the gauge condition," in particular the "the" part?
>>
>> The gauge substitutions are:
>>
>> A -> A' = A + \grad Q
>>
>> \phi -> \phi' = \phi - (1/c)dQ/dt
>>
>> Where Q is any scalar function.
>>
>> \grad\phi' -> \grad\phi - (1/c)(d/dt)\grad Q
>>
>> \grad\phi' = \grad\phi - (1/c)(d/dt)(A' - A)
>>
>> \grad\phi' - (1/c)(dA'/dt) = \grad\phi - dA/dt
>
>
>
>(sigh!) Seems like I made a slight error a sign again. I wrote
That is not my fault.
>----------------------------------------
>
>E' = -grad Phi' - (1/c) &A'/&t' = 0
>
>Or
>
>grad Phi' = (1/c) &A'/&t'
>----------------------------------------
>I missed the "-" sign in going from one to the other. It should have read
>grad Phi' = - (1/c) &A'/&t' However that was obvious anyway!
So what? I responded to your question: "please define the gauge
condition...". I didn't really even bother to look at whether you
had any signs correct, ambiguous or otherwise.
>> If the potentials ever satisfy the condition, \grad\phi = (1/c) dA/dt,
>> they will always satisfy it.
>So what? Who said anything about the relation "\grad\phi = (1/c)
>dA/dt"???????
Why, you did. Just above.
>The only relation that I posted was
>-grad Phi' - (1/c) &A'/&t' = 0
And?
>or if you like
>grad Phi' = - (1/c) &A'/&t'
And?
>which is a very different relation than what you wrote. "&" represents the
>partial derivative.
So what? Do you see any equation I wrote which contains both partial
and total derivites so that it's necessary to differentiate between
them using cryptic symbols? Did I say anywhere that d meant either
partial or total?
>That's obvious from the use of "&" in the relation for the electric field
So what? I read that as a partial.
>"E' = -grad Phi' - (1/c) &A'/&t'"
>So "&" is a *partial* derivative. d/dt is a *total* derivative.
Did I say anywhere that d/dt meant a total derivative? You didn't
bother to say what &A/&t means, yet I managed to figure that out
for myself. How do I know you didn't mean total derivative? The
only "standard" way of expressing a partial on the science newsgroups
would be the same way one does in a journal article, using the
standard TeX notation: \partial .
>So how is "grad Phi' = - (1/c) &A'/&t'" a "gauge condition????
Surely you can figure that one out. It's simple. I started with the
expression jackson uses for A and \phi, so that I could reference that
section in jackson. If I had written A -> A - \grad Q, I'd get the
opposite sign. If I use A^{u} rather than \phi and A, then I get
the same signs you have written: d_u A^u = 0. That happens to
be a gauge condition for a massless vector field.
Bilge
>>The potentials responsible for the effect are
>>purely those which satisfy the gauge condition, which classically can
>>have no effect, by the very way in which classical E&M defines them.
>
>I'll disagree, please see below....
>> That part is certainly true, but classically, the phase is precisely
>>that which is not-physically meaningful. Classical E&M stops with F^{uv}
>>and F^{uv} cannot contain the phases by definition:
>> F^{uv} = d^{u}A^{v} - d^{v}A^{u}
>>
>> Substitute A'^{u} = A^{u} - d^{u}Q
>.
>> where Q is a scalar function of the coordinates, and you'll see that
>>F^{uv} -> F^{uv}
>
>Ok, Curl x Grad=0, that's agreed. A phase shift will appear
>if electron energy, velocity, or paths are relatively different.
I don't know what you mean by that second sentence, but let me
point out that "gauge invariance" means "phase invariance". The
entire concept of phase invariance is the basis of qed and quantum
field theories, so you can't simply be cavalier in associating to
concepts likee energy, velocity, etc. In the aharanov-bohm effect,
there is _NO_ force exerted on the electrons passing by the solenoid.
>>
>> While it might initially appear that way, the aharanov-bohm effect is
>>entirely due that which induces nothing. There is _no_ force exerted
>>on the electron by the solenoid (or vice-versa since F_12 = -F_21).
>
>What?? A solenoid is a conductor.
So what? There is no B-field or E-field outside a solenoid.
>A current pulse, such as an electron moving by it, is subject to
>inductance because charges
Any effect due to something like that is over and above the
aharanov-bohm effect. A force acting on an electron changes
it's direction and/or momentum. The aharanov-bohm effect
does neither. It results in an interference pattern.
[...]
> This is very basic electricity of transformers and inductors,
>have a look at the theory of *air-core* transformers.
I know "basic electricity". I worked on missile guidance
systems for a summer after my first year in college. The
aharanov-bohm effect is not basic electricity.
> Let's cover the basics before jumping to conclusions, no
>amount of complex math, inappropriately applied, will
>change the experimental result.
Ken, in another article you complained about my explanations not being
detailed enough. The reason is that you presume that I don't know what you
are talking about and so you try to argue as if you knew what I was
talking about and so you get an answer that assumes you do. The aharanov-
bohm effect is not "basic electricity". Believe it or not, "basic
electricity" is something physicists study, especially experimentalists,
since one often has to design one's own electronics. So, let's get on
the same page here. Either decide you know what the aharanov-bohm effect
is and argue about that and not things like inductance, energy shifts,
or whatever or decide you don't know what it is and ask questions or
read something about it that will help you understand it.
Bilge
>dub...@radioactivex.lebesque-al.net (Bilge) wrote in message
>news:<slrnbane9d....@radioactivex.lebesque-al.net>...
>>Ken S. Tucker:
I'll charge one box by _removing_ negative charge so that a positive
>>charge remains. I then charge the other box by removing positive charge
>>so that only negative charge remains. Now, which do you think is
>>lighter, the charged or uncharged boxes?
>
>While unsure this question has a point, uncharged boxes are lighter.
Since it was in response to your insistance that I answer your
question about charging boxes, and some tangent about a reissner-
nordstrom black hole, if you don't see the point, then I certainly
don't.
>(It was your suggestion to charge up one box (the universe) to
>a gazillion volts, and say it makes no diff).
The universe as a box is not two boxes. And changing the potential
on something has nothing to do with "charging it up". Only changing
a potential difference between two objects does. This is a rather
crucial concept.
[...]
>> >>required to assemble the charge into a spherical volume 4pi r^3/3.
>> >>(feel free to choose any shape you want so long as all of the
>> >>charge is contained within a radius no larger than 10^-18 meters.
>> >>Calculate m. How many hundreds of times larger is this than the
>> >>electron mass?
>> >
>> >We did this before...
>>
>> Then you must have missed the point.
>
>No, I concluded a single charge does not self energize.
But that's manifestly false. What do you think holds the charge
together? Do you think the conncept of electrostatic energy is invalid?
[...]
>
>Ok, I'll use GR....
>
>> First the preliminaries. Special relativity and the dirac theory
>>allows for 5 and only 5 types of potentials. Scalars (S), Pseudo-
>>scalars (P), Vectors (V), Axial vectors (A) and pseudo-Tensors (T).
>>These are all of the bilinear forms which can be constructed. All
>>others reduce to one of the forms. (You should prove this. Take the
>>dirac matrices (including \gamma^5) and figure out how many combinations
>>there are with unique transformation properties).
>
>Yes, one can go in circles by reconfiguring mathematics without
>any physical reason....otherwise known as mathematical torment.
One is not going in circles. The entire result is a consequence
of the symmetries of minkowski space.
[...]
>Yes, the terminology you employ is sometimes unfamiliar, phase is
>(IMO) a means of translating phenomenological action (mathematically)
>in finite space-time intervals to predict effects. For example...
> Although it is interesting that one may use Newtonian "instanteous
>action" with a GR correction to get "the right answer" in celestrial
>mechanics, we do know we are violating GR in this calculation, that
>requires a finite gravitation speed, yet we do it.
>
> If I were to solve celestrial mechanical problems using GR strictly,
>phase (or equivalent) would be necessary.Why not try to demonstrate a 2
>body gravitational solution with limited g-speed using phases. This
>mathematical technique should be available from QM, evidently, you're an
>expert!
Your task is to start with the dirac lagrangian and make it ivariant
under a transformation of the electron wavefunction: \psi->\psi\exp(iS).
Once you've done that, you will (almost) have the qed lagrangian. Then
you may proceed to explain how this is related to what you just said.
If you aren't sure about how to do this, why are telling me what
to compare phase invariance to in general relativity?
Bilge
> The Aharonov-Bohm apparatus is such an experiment, the phase of an electron
> is an incredibly sensitive detector of forces on the particle, forces which
> would be very hard to detect in any classical experiment.
The phase of the electron is not an observable. It's a measure of nothing.
That is the _entire_ reason the aharanov-bohm effect occurs. The reason
that electrons are electrons is precisely because such phases are
meaningless.
[...]
>
> What happens when magnetic shielding is added to AB? What happens to
> the poynting vector in the solenoid due to the shielding effects of the
> wire? My first thought is that this effect must be such that shielding
> cannot negate it.
HUH? There is no B-field to shield.
Lawrence Foard
Bilge <cra...@fghfgigtu.com> wrote:
> Lawrence Foard:
>
> > The Aharonov-Bohm apparatus is such an experiment, the phase of an electron
> > is an incredibly sensitive detector of forces on the particle, forces which
> > would be very hard to detect in any classical experiment.
>
> The phase of the electron is not an observable. It's a measure of nothing.
>That is the _entire_ reason the aharanov-bohm effect occurs. The reason
>that electrons are electrons is precisely because such phases are
>meaningless.
The experiment measures the relative phase shift resulting from the
two paths. This makes it incredibly sensitive to miniscule changes.
>[...]
> >
> > What happens when magnetic shielding is added to AB? What happens to
> > the poynting vector in the solenoid due to the shielding effects of the
> > wire? My first thought is that this effect must be such that shielding
> > cannot negate it.
>
> HUH? There is no B-field to shield.
Yes there is, the B-field of the electron, that was the entire point of the
message.
Pmb
"Bilge" <dub...@radioactivex.lebesque-al.net> wrote
> When you learn to use your newsreader or get a newsreader that doesn't
> insert 3 or 4 blank lines between every paragraph and turn a 30 line
> post into one which is 120 lines, which I have to reformat.
Then don't respond!! Nobody is forcing you to. In fact if you're going tto
have this attitude because I'm having a software problem then it's highly
undesireable - So in the future please just don't respond until you check
the attitude at the door!
> >I missed the "-" sign in going from one to the other. It should have
read
> >grad Phi' = - (1/c) &A'/&t' However that was obvious anyway!
>
> So what? I responded to your question: "please define the gauge
> condition...". I didn't really even bother to look at whether you
> had any signs correct, ambiguous or otherwise.
My mistake. I thought that since I was using &A/&t for partial and since
everyone uses dQ/dt for total derivative then you meant total derivative.
> >So how is "grad Phi' = - (1/c) &A'/&t'" a "gauge condition????
>
> Surely you can figure that one out. It's simple. I started with the
> expression jackson uses for A and \phi, so that I could reference that
> section in jackson. If I had written A -> A - \grad Q, I'd get the
> opposite sign. If I use A^{u} rather than \phi and A, then I get
> the same signs you have written: d_u A^u = 0. That happens to
> be a gauge condition for a massless vector field.
Actually this point is moot. The vector potential is fully defined in frame
O. The fact that it satisfies the "grad Phi' = - (1/c) &A'/&t'" is simply a
matter of fact. It has nothing to do with the physical meaning of Phi
Pmb
AntiCrank
"Pmb" <physic...@yahoo.coom> wrote in message
>
> "Bilge" <dub...@radioactivex.lebesque-al.net> wrote
>
> > When you learn to use your newsreader or get a newsreader that
doesn't
> > insert 3 or 4 blank lines between every paragraph and turn a 30 line
> > post into one which is 120 lines, which I have to reformat.
>
> Then don't respond!! Nobody is forcing you to. In fact if you're going tto
> have this attitude because I'm having a software problem then it's highly
> undesireable - So in the future please just don't respond until you check
> the attitude at the door!
*sigh*
So once again, to summarize Peter M. Brown.
Brown is a 40 something unemployed physics failure.
He has Acute Myeloid Leukemia, a cancer with a 5 year survival rate of
around 20%, which means that people might not have to put up with him for
very long. In any event, Brown likes to whine about his AML in a bid for
sympathy, forgetting that the sci.physics hierarchy is not like his LLS
board, where he can whine and "vent" and have people fawn over him. Brown
has attempted suicide at least twice, but was unable to pull it off
successfully. It is likely he did this as a bid for attention, and for
sympathy points with his "friends" at the LLS board. Even though this might
earn him sympathy points with his "friends", it might hurt him in any future
employment opportunities.
Even though he failed out of not only his PhD. program, but his Ms. program
as well, Brown considers himself an expert in physics. Granted, he can sure
juggle those indices, but has trouble understanding physics at its basic
levels. In other words, he gets lost in the math, and has trouble
connecting the equations with reality, or even observation.
As a result, he often feels the need to fight with people over definitions
(many of which he has made up to suit his current needs), and quote
excessively from various journals or texts. This spew of quotes usually
serves only to further highlight the fact that Brown has a poor
understanding of the physics involved.
Brown has had no published papers what so ever, not even as a co-author, and
thus when a person on the sci.physics hierarchy publishes, he feels the need
to "pooh-pooh" the paper in an attempt to boost his ego. He is quick to
point out that he is acknowledged in a few books for helping with the proof
reading, suggesting cover art, or concatenating a glossary of terms. While
these are useful accomplishments, on the order of what the usual undergrad
or grad student does when their professor writes a book, they hardly signify
any significant scientific progress. This does not stop Brown from claiming
that his most current work has made it through "pier" review, and that the
authors could find no error. However he fails to grasp why his current work
remains unpublished.
Brown is prone to argument, and has great difficulty in admitting error.
When someone points out an error, or even disagrees with Brown, they are
attacked violently! A simple 5 word comment can cause Brown to go off the
deep end, resulting in his attacking the person at their place of
employment. These attacks involve creating web sites that attack his
detractor (Which are usually removed by his ISP), harassing the persons
co-workers, attempting to cause trouble at the "highest levels", writing his
congressman or the FBI (not to mention CNN), and calling the local
authorities.
At the end of these attacks, Brown withdraws to his LLS board to commiserate
with his "friends", who cheer him on, and boost his ego for his next foray.
At the end of each day, Brown likes to recap his exploits and battles, make
a fool of himself once again, and leave with a parting shot in order to
continue the battle the next day. Brown finishes his evenings with soup,
his favorite food and something which he hoards from everyone else.
"No soup for you!" is his retreat bugle.
Pmb
on
bilge - do not respond if you need to post more flames. I'm not going(or
rather my system is inapable of ) changing my posting habits until the
software problem is resolved. The program crashes every time I type a
response of more than a page or two. It is simply impossible to do. And I
have no other word processing program that does spell checmking. So don't
whine about me not knowing how to use a newsreader. If you're not mature
enough to post without flaming then I don't care if you respond at all! When
Microsoft fixes the problem the I won't have to post like this.
However - If you can respond in a civilized manner then I welcome your
contribution and please continue...
"Bilge" <dub...@radioactivex.lebesque-al.net> wrote
<Snipped bilge's flame>
[Pmb wrote]
> >> >grad Phi' = (1/c) &A'/&t'
> >> >&A'/&t' is not zero it follows that grad Phi' is not zero and that
> >> >means that Phi's is not zero and is a not a constant.
>
[Bilge wrote]
> >
> >> But that is precisely the gauge condition.
[Pmb wrote]
> >Please define "the gauge condition," in particular the "the" part?
[Bilge wrote]
> The gauge substitutions are:
>
> A -> A' = A + \grad Q
>
> \phi -> \phi' = \phi - (1/c)dQ/dt
>
> Where Q is any scalar function.
>
> \grad\phi' -> \grad\phi - (1/c)(d/dt)\grad Q
>
> \grad\phi' = \grad\phi - (1/c)(d/dt)(A' - A)
>
> \grad\phi' - (1/c)(dA'/dt) = \grad\phi - dA/dt
>
> If the potentials ever satisfy the condition, \grad\phi = (1/c) dA/dt,
> they will always satisfy it.
Why are you calling this "the gauge condition"?? A gauge condition is when
the divergence of a vector field is given. Imposing the condition div A =
[something] means that, since you curl is already defined on A {i.e. B =
curl A} then, according to Helmholtz's theorem, the field is uniquely
defined up to an arbitrary constant. That's why the Lorenz gauge and the
Coulomb gauge are of the form "div A = [something]"
However I still don't see what you did above, i.e. given A and Phi I can
define A' and Phi' as
[1] A' = A + grad Q
[2] Phi' = Phi - (1/c)&Q/&t
Taking the partial derivative of (1) and multiplying through by "1/c" gives
[3] (1/c)&A'/&t = (1/c)&A/&t + (1/c)&(grad Q)/&t
Taking the grad of [2] gives
[4] grad Phi' = grad Phi - (1/c)&(grad Q)/&t
Add [3] to [4] to give
[5] grad Phi' + (1/c)&A'/&t = { grad Phi - (1/c)&(grad Q)/&t } + {
(1/c)&A/&t + (1/c)&(grad Q)/&t }
Canceling the "(1/c)&(grad Q)/&t" term in [5] gives
[6] grad Phi' + (1/c)&A'/&t = grad Phi + (1/c)&A/&t
However, as I said in my post to which this was a response to, I see no
reason why [5] has to vanish. In response to that statement you stated "But
that is precisely the gauge condition"
*what* is the gauge condition? I see no gauge condition since you never gave
a value to div A. All I know at this point is that
[7] grad Phi' + (1/c)&A'/&t
is gauge invariant. But so what?
Why/how did you get this -- "\grad\phi' = \grad\phi - (1/c)(d/dt)(A' - A)"
???
You went from this
> \grad\phi' -> \grad\phi - (1/c)(d/dt)\grad Q
to this
> \grad\phi' = \grad\phi - (1/c)(d/dt)(A' - A)
This seems to me that you're putting a condition on A which it may not
satisfy. I.e. for what reason do you hold this to be true?
Pmb
AntiCrank
"Pmb" <physic...@yahoo.coom> wrote in message
> I'm going to address this so-called "gauge condition" that bilge commented
> on
>
>
>
> bilge - do not respond if you need to post more flames. I'm not going(or
> rather my system is inapable of ) changing my posting habits until the
> software problem is resolved. The program crashes every time I type a
> response of more than a page or two. It is simply impossible to do. And I
> have no other word processing program that does spell checmking.
Not that it seems to help with your spell "Checmking." *smirk*
Why don't you do the Ctrl S combo every so often?