Non ame

From: Ken S. Tucker (dynamics_at_vianet.on.ca)
Date: 03/19/05 

Date: 19 Mar 2005 08:36:33 -0800

"Ken S. Tucker" <dynam...@vianet.on.ca> writes:

>Non Ame wrote:
>> "Ken S. Tucker" <dynam...@vianet.on.ca> writes:

>> >Non Ame wrote:
>> >> "Ken S. Tucker" writes:

>> >> >Non Ame wrote:
>> >> >> "Ken S. Tucker" writo:

>> >> >> >QT = requires a quantized variation of energy.

>> >> >> This looks like one of the ad hoc rules of the Old Quantum
>Theory,
>> >> >like
>> >> >> the ad hoc rule in the Old Quantum Theory that the orbital
>angular
>> >> >> momentum must be an integral multiple of hbar. This rule for
>the
>> >> >angular
>> >> >> momentum is actually true in Quantum Mechanics, but is derived
>> >from
>> >> >the
>> >> >> laws of Quantum Mechanics, rather than being an assumption of
>the
>> >> >theory.

>> >> >"laws" are they generally covariant?

>> >> Not the laws of Quantum Mechanics as such. That is why there is
a
>> >great
>> >> desire to attempt to unify the two theories.

>> >Ok good, that's done.

>> No, it has not been done.
>It evidentally has been in nature.

But humanity does not have such a theory. That is what some of the
physicists are working on.

>> >> >no, what you call "laws" are to others
>> >> >"guidelines".

>> >> Quantum Mechanics has certain postulates, just as Special
>Relativity
>> >and
>> >> General Relativity also have certain postulates.

>> >Ok, how does General Covariance fit in with
>> >your idea of postulates?

>> That question is irrelevant.
>WHAT??, We regard GC as the Foundation of the
>LAws of Nature.

That was not true before 1915. General Covariance of the laws of
nature
was not taken into account in any laws of nature before that time. The
Galilean statement of the Principle of Relativity was an observation
based
on what was known of the laws of mechanics at that time. Einstein had
to
explicitly state his version of the Principle of Relativity in his
paper
of 1905. When Quantum Mechaniocs was formulated in the mid-1920s,
there
was no thought to determine a theory which satisfied a General
Covariamce
of physical laws.

>Take a review of Dovers
>"Principle of Relativity", pg 121 in AE's GR1916,
>about half way down the page.
>Read, "If therefore a law of nature ... a tensor
>to zero, it is GC".

This is a paper on General Relativity. General Covariance of physical
laws is one of the postulates of General Relativity. This means that a
General Covariance of physical laws is important for any discussion on
General Relativity. I have already pointed that out. This was in
response to my pointing out that

        Quantum Mechanics has certain postulates, just as Special
        Relativity and General Relativity also have certain postulates.

As I have remarked, a General Covariance of physical laws was
introduced
as one of the postulates of General Relativity, and so it is only in
theories that subsume General Relativity in some way that have a
requirement of a General Covariance of physical laws. Quantum
Mechanics
is not such a theory. The present efforts to unite Quantum Mechanics
and
General Relativity are attempts to introduce a theory of Quantum
Mechanics
(built on many of the same principles as present day Quantum Mechanics)
which do satisfy a General Covariance of physical laws.

Nobody denies the importance of General Covariance. That is why there
has
been such a effort to unify the two theories.

When Quantum Mechanics was formulated in the mid 1920s, it was not even
Lorentz covariant. Time was placed on a completely different footing
to
the coordinates x, y and z. If the fact that the wave function psi is
a
function of t, x, y, z, gives the impression that t, x, y, z, are on a
similar footing, then that is an illusion. The time t is a real-valued
parameter. On the other hand, x, y and z are operators. This means
that
the theory is not Lorentz covariant.

Second quantization has the advantage that it does put t, x, y and z on
a
similar footing: as coordinates in a spacetime manifold, and therefore
as
real-valued parameters.

> If you intend to fiddle with GC and laws of nature
>start a new thread.

It was *you* who tried to connect the postulates of Quantum Mechanics
with
a General Covariance of physical laws. This is a discussion about what
postulates belong to what theory. A General Covariance of physical
laws
is a postulate of General Relativity. It is not a postulate of Quantum
Mechanics.

>>Each theory has its own postulates - i.e.
>> each theory has axioms on which the theory is based. A general
>covariance
>> of laws is a postulate of General Relativity.
>OK, How would you create an exception?

As I have already pointed out, a General Covariance of physical laws
was
not even a postulate of any theory until 1915.

>>General Covariance is not a
>> postulate of Special Relativity (the special case of the Principle
of
>> Relativity is the appropriate postulate of Special Relativity).
>General
>> Covariance holds no place in the postulates of Quantum Mechanics.
>"holds no place" is a bit strong.

It is not. As I remarked above, the postulates of Quantum Mechanics as
formulated in the mid 1920s were not even Lorentz covariant. It has
been
a long slog to unify Quantum Mechanics first with Special Relativity,
then
with General Relativity, and the task is not yet finished.

>> >> And you still haven't specified the axioms and postulates of what
>you
>> >are
>> >> calling Quantum Theory. Is there a reason for this reluctance of
>> >yours to
>> >> commit to a specific set of postulates?

>> >No.

>> Then please provide the axiomatic basis for what you are calling
>Quantum
>> Theory.
>A,u =0 , A= norm of 4-potential.

That is not an axiomatic basis. Do you understand what is meant by
"axiomatic basis"?

Also, the condition that you have written above is not gauge-covariant,
which means that it is not physically meaningful unless you impose a
specific gauge.

>> >Let's define the EM-4 potential by A_u and
>> >it's norm by either,

>> >A^2 = A_u A^u = g_uv A^u A^v.

>> >Wait and look see the problem?

>> Not at all. The fact that this quantity is unphysical has nothing
to
>do
>> with whether or not it is an invariant scalar. Obviously gauge
>> transformations will affect the quantity, but gauge transformations
>are not
>> Lorentz Transformations. In fact, gauge transformations form an
>entirely
>> different group.

>> Any formula in which the above expression appears for A^2 is gauge-
>> covariant (possibly with the additional requirement of a canonical
>> transformation in the non-quantum case). This means that any such
>> formula has a physical meaning.

>> So what problem do you find with the above expression?

No answer, I see, from you with regard to this.

>> >The Unified Field Theory is in accord with,

>> >A,u =0 .

>> Why would you want to impose a condition like A,u = 0? Also, u is
>not an
>> observable in the quantum case (in fact, u is meaningless in the
>context
>> of Quantum Mechanics).

>> Or do you mean dA/dx^u?
>Yes.

The four-potential A_u cannot be expressed as dA/dx^u unless the
electromagnetic field is zero at all events in spacetime. This is
equally
true in both Special Relativity and General Relativity. A simple
substitution of this form of the four-potential into the formula for
the
electromagnetic field proves the necessity of the vanishing of the
electromagnetic field at all events in spacetime.

Remember that the curl of a gradient is equal to zero. This is just as
true in General Relativity as it is in Special Relativity.

>>If so, what is your scalar quantity A, and why
>> is it relevant to any potential?
>A=q/s , where q and s are invariant,

And what are q and s? Electric charge and proper time? If q is the
electric charge and s is the proper time, then A is not a scalar field
and
it makes no sense to define the four-potential as its gradient.

>{s*/\s = x*/\x} == [s*ds = x*dx]
>where "/\" stand for finite increment.

You should use standard notation. And why are you discussing "finite
increments"?

>The {} == [], shows how to transform the
>"quantum derivative" on the LHS, to the
>continous derivative on the RHS.

Are {} and [] supposed to mean "Poisson bracket" (non-qauantum) and
"commutator" (quantum) respectively? If so, then the connection
between
the two has been known for decades.

>>From the perspective of GR the Energy Density,
>is the product of two E-fields from charges "a"
>and "b" i.e.,
>T_00 = E(a)*E(b)

If you ignore the magnetic field, and if you set epsilon_0 = 2, which
is
inconsistent with a convention that you adopted previously. Under the
convention that you are presently taking, the magnitude force between
two
point charges is given by Q_1 Q_2/(8 pi r^2), where Q_1 and Q_2 are the
charges and r is the distance between them.

Also, you have ignored other terms in the energy density. If you
ignore
the magnetic field, then T_00 = E.E, for the *full* electric field E.

Under your previous convention in which the magnitude of the force
between
the charges is Q_1 Q_2/r^2, the energy density is T_00 = E.E/(8 pi).

>T_00 = - E^2 ,(attractive a=-b)
>T_00 = + E^2 ,(repulsive a= +b)
>These in turn set-up the G_00, (G_uv=T_uv)

The *correct* expressions for T_00 contribute to the nett T_00, which
is
what appears in the equation. Note that, under your convention, the
magnitude of the gravitational force between two masses in the
Newtonian
approximation is given by M_1 M_2/(8 pi r^2) , where M_1 and M_2 are
the
masses and r is the distance between them. Your convention imposes
that
Newton's gravitational constant have value 1/(8 pi). In the absence of
other matter, the nett energy density is given by E.E + B.B, under your
present convention, and this is the energy density that is used in the
equation G_uv = T_uv.

>and
>the resulting g_uv and from them the geodesics.

Once you determine the *correct* formula for T_uv.

>Energy and therefore Energy Density varies dis-
>continuosly, hence the E-fields (E^2 above) also
>vary discontinuously, that would be by photons
>emitted by the pairing of protons and
>anti-protons.

This is nothing to with photons. Also, you have continued to misapply
what Quantum Theory actually states to a situation where you seem to be
imagining a classical behaviour except for quantum jumps in energy. If
that is how you see it, then you have completely the wrong idea of what
Quantum Theory says.

A photon is a quantum of energy in a mode in the quantum radiation
electromagnetic field.

And why do you specify protons and anti-protons?

The mutual annihilation of an electron positron leads to gamma rays.

> Virtual photon exchange is a conceptualiztion
>of mathematics. If you want you are entitled
>to think in terms of the annilihation of
>virtual antiphotons and photons, to create the
>-E^2 above, and by doing so emit a photon.

An antiphoton is exactly the same thing as a photon. There is no
separate
particle that can be called an antiphoton.

> I wouldn't argue with that.
>> And where are your long-promised postulates for Quantum Theory?
>You getting there!

What you meant was "You're getting there!"

What is stopping you from just stating the postulates?

>You
>> stated that there is no reason for your not telling us what your
>> postulates are, so please enlighten us.

>> >>Or perhaps you are
>> >> too timid to commit to whatever postulates you have for Quantum
>> >Theory.

>> >> It would also be interesting if you show us how your version of
>> >Quantum
>> >> Theory explains the result of the double-slit experiment, i.e.
the
>> >> outcome when the slit through which the particle passes is not
>> >measured,
>> >> and also the outcome when the slit through which the particle
>passes
>> >is
>> >> measured.
>Do you mean "wave-mechanics" I gotta guessing
>what you mean, QT, QM, and WM are different
>things.

No. Wave Mechanics was Schroedinger's formulation of Quantum
Mechanics.
Matrix Mechanics was Heisenberg's formulation of Quantum Mechanics.
Dirac
showed the the two are equivalent. So Wave Mechanics, Matrix Mechanics
and Quantum Mechanics are the same theory, up to equivalence.

Quantum Mechanics explains the result of the double-slit experiment.
If
whatever it is that you are calling Quantum Theory has any merit, then
it
has to be able to do the same.

>> >We should be discrete, if the answer to you're

>> The word you actually mean is "your", not "you're" (which means
>> "you are"). For some reason you keep making this type of error.

>> >question was posted, we'd ruin a prefectly good
>> >problem, is that what you want?

>> Yes. Your theory should be capable of explaining the phenomena
>related to
>> the double-slit experiment. Quantum Mechanics can provide an
>explanation.
>No it can't.

Yes, it can.

I don't know where you got the idea that Schroedinger's Wave Mechanics,
heisenberg's Matrix Mechanics and Quantum Mechanics were different
theories. They are not.

>> I cannot think of any possible reason why you would not want to
>provide an
>> explanation for the results of the double-slit experiment, except
for
>the
>> possibility that you actually have no explanation.
>Let's do the basics then we'll do applications.

But you point blank refuse to even discuss the basics, in spite of
numerous attempts to get you to do so.

>> And *STILL* no detailed line of reasoning. Until further evidence
>comes
>> along, I must conclude that you can't derive Quantum Theory from
>General
>> Relativity, otherwise you would have done it by now.
>A critical mind is a good mind.

Is this to justify your refusal to supply a detailed line of argument
to
get from General Relativity to whatever it is that you are calling
Quantum
Theory (which is still a state secret as a consequence of your
persistent
refusal to divulge any information about it)?

>> >> >BTW, recall f^0=0.

>> >> Only in "The Universe according to Ken S. Tucker".

>> >Well my dear your bulb is dim, Methinks I'm wasting
>> >time, see Weinberg's Eq((2.7.9), then down read f^0=0.

>> Let me quote that equation from that book in context:

>> "That this (i.e. (2.7.9)) is correct may be seen by repeating
the
>> arguments of Section 3. Equation (2.7.9) is correct in a
>> reference system in which the particle is at rest because in
this
>> frame it gives f = eE, f^0 = 0, and it transforms like a
>> four-vector, so it (i.e. (2.7.9)) is correct for all
velocities."

>> Note that Weinberg stated that f^0 = 0 in the special case when the
>> charged particle is *stationary* - i.e. Weinberg stated that f^0 = 0
>in
>> the very special reference frame in which the particle is stationary
>> (incidentally, in my derivation of my formula for f^2, I used the
>fact
>> that f^0 = 0 in the frame in which the charge is stationary). At no
>point
>> did Weinberg state that f^0 = 0 in all reference frames.

>> What Weinberg did state was that because (2.7.9) holds in one frame,
>then
>> it holds in all frames. Weinberg also states that (f^0,f)
transforms
>like
>> a four-vector. It follows from this observation that if I take a
>primed
>> reference frame which, relative to the reference frame in which the
>charge
>> is stationary, moves in a direction which is not orthogonal to E,
>then
>> f'^0 is nonzero (i.e. f^0 is not equal to zero in the primed
>reference
>> frame).

>> It looks like your bulb is the one which is dim. You read that
>Weinberg
>> stated that f^0 = 0 in the reference frame in which the charged
>particle
>> is stationary, and you took what was written in the section to mean
>that
>> f^0 = 0 in all frames, and this in spite of the fact that Weinberg
>never
>> made such a claim.
>Did you read "so is correct for all velocities"

I did, and I also pointed out above that when Weinberg wrote that
phrase,
he was referring to Equation (2.7.9). He was not referring to the
equation "f^0 = 0", which he only stated was specifically true for the
reference frame in which the particle is stationary. Let's take a very
close look at what Weinberg was doing in that passage,

(1) Weinberg notes that in a reference system in which the particle is
at
rest, f^0 = 0 and f = eE.

(2) Weinberg notes that it follows that in a reference system in which
the
particle is at rest, f^alpha = e F^alpha_gamma dx^gamma/dt (i.e.
(2.7.9)).
This is seen by noting that in such a reference system, the components
of
this equation are f^0 = 0, f = eE.

(3) Weinberg notes that (f^0,f) transforms as a four-vector.

(4) It follows that (2.7.9) is a tensor equality.

(5) (2.7.9) is a tensor equality which is true in one frame, therefore
(2.7.9) is true is all frames.

And that is the end of what Weinberg is doing in this passage.

f^0 = 0 is not a tensor equation. The fact that it is true in one
frame
does not make it true in all frames. All that Weinberg used the
equation
f^0 = 0 for in the passage is

        (1) to note that it is true in the reference frame in which the
            charge is stationary;

        (2) to note that in the frame in which the charge is
stationary,
            f^0 = 0 and f = eE together are equivalent to (2.7.9).

That is the *full* extent of Weinberg's interest in the equation f^0 =
0
in that discourse, and he never refers to f^0 = 0 in any frame of
reference in which the charged particle is not at rest. I note that
you
aren't trying to claim that he states that f = eE holds in all frames
of
reference, and I can only presume that this is because he explicitly
gives
the formula dp/dt = e [E + v x B] at the end of the paragraph.

>(lead a donkey to water but you can't make it
>drink), read to the end of sentences, your
>credibility is suffering.

No. Your credibility is suffering. When Weinberg wrote that phrase
that
you quoted, he was actually referring to Equation (2.7.9), and only to
that equation. The fact that you read that passage and failed to pick
that up betrays the shallowness of your comprehension of the subject.

>> >> If we were to take your insistence that f^u = 0 seriously, that
>would
>> >mean
>> >> that we would require that J and E be orthogonal (since J.E = 0)
>and
>> >that
>> >> rho E + J x B = 0 at all events in spacetime.

>> >Charge is quantized, your "rho" is useless,

>> Rubbish. Firstly, in classical electromagnetic theory, it does not
>matter
>> that the charge is quantized, and my observations are apposite.
>What does "apposite" mean, the discussion involves
>a peak at ways in which QT and GR relate.

It is apposite to the classical theory of electromagnetism, before
there
was a Theory of Relativity or Quantum Mechanics.

>Also,
>> Maxwell's Equations have an integral form which is capable of
>handling
>> point charges:

>> int_S E.dS = 4 pi Q,

>> int_S B.dS = 0,

>> where S is a closed surface, Q is the nett charge enclosed within
the
>> surface, and the convention epsilon_0 = 1/(4 pi), mu_0 = 4 pi, has
>been
>> adopted, and

>> int_C E.ds = - d(int_S B.dS)/dt,

>> int_C B.ds = 4 pi I + d(int_S E.dS)/dt,

>> where S is a surface, C is its boundary, and I is the current
flowing
>> through S.

[snip examples of charge density and current density in quantum
mechanics]

>> You may have to reinstate it, since not even quantization of charge
>has
>> stopped people from defining a charge density and current density
for
>> various cases in Quantum Mechanics.
>Good ahead, keep those equations, sheesh,
>it's none of my business.

So I take it that all that discussion on charge density and current
density was beyond your comprehension.

>>You should learn more before you
>> consign things to the junk heap. You may actually learn something
if
>you
>> take the time (and make the effort) to do so, instead of just
>'throwing
>> things away' for no better reason than that they do not make sense
to
>you.
>Discard prejudice and move on.

Well, it is good to see that you are discarding your own prejudices.

>> >>Furthermore, considering
>> >> that neither E nor B should have any discontinuities if the
charge
>is
>> >not
>> >> concentrated in point charges, and considering the level of
>> >arbitrariness
>> >> available for rho and J, then your requirement of f^u = 0 would
>> >require
>> >> that J.E = 0 for all possible values of J, and rho E + J x B = 0
>for
>> >all
>> >> possible values of rho and J, and therefore that E = B = 0. This
>is
>> >a
>> >> statement which contradicts Maxwell's Equations unless rho = 0
and
>J
>> >= 0.

>> >:)

>> So why the smiley?
>Well, I'm a natural flirt, and you're a girl,
>see even spelledulated "you're" right.

No relevance to the topic at hand, then. You should keep your mind on
the
subject. It is this type of distraction which has kept you from
comprehending. The fact that I am a girl should not make any
difference
whatsoever. This is not a battle of the sexes, and female of the
species
is not inferior to the male of the species. Females are just as
intelligent as males.

>> >> It is also a statement which contradicts the nature of light as
an
>> >> electromagnetic wave. It also contradicts the laws of Quantum
>> >> Electrodynamics, since neither E nor B is zero in that theory,
>since
>> >> neither is a c-number (for example, in the case of radiation,
>> >> [E_x(r,t),B_y(r',t)] = - 4 pi i hbar d(delta(r-r'))/dz, where
>> >delta(r) is
>> >> the Dirac delta function, and epsilon_0 = 1/(4 pi), mu_0 = 4 pi).

>> >You're over you head, your
>> >trying to

>> Where you have written 'your' should also be spelled as "you're"
>(which is
>> short for 'you are'). I have just realized that you are lazy and
>can't be
>> bothered getting the spelling correct, otherwise you would not have
>used
>> both "you're" and "your" when you meant the same in both instances.
>bla-bla

Maybe if you took greater care, there would not be any mistakes for me
to
point out.

>> >mix radiation (like F_uv,w) with E and B, lets
>> >stay with tensors.

>> The word you mean here is "let's" (short for 'let us'), not "lets".

>> Since E and B are components of F, then both usages are correct. I
>> reiterate that:

>> E_x(r,t) B_y(r',t) - B_y(r',t) E_x(r,t)

>> = - 4 pi i hbar d(delta(r-r'))/dz.
>You have written 0=0, is that what you meant?

Neither the left hand side, nor the right hand side, of the above
equation
is equal to zero. What makes you think that the left hand side of the
equation is equal to zero? Is it because you believe that

        E_x(r,t) B_y(r',t) = B_y(r',t) E_x(r,t)

in Quantum Electrodynamics, and if so, why would you believe such a
statement?

The fact that the right hand side is nonzero can be seen from the fact
that for any test function f : R^3 -> R,

        int f(r') d(delta(r-r'))/dz dx' dy' dz' = df(r)/dz,

        int f(r) d(delta(r-r'))/dz dx dy dz = - df(r')/dz'.

Since

        int f(r') 0 dx' dy' dz' = 0,

        int f(r) 0 dx dy dz = 0,

then d(delta(r-r'))/dz is most definitely not equal to zero, and so
neither is the right hand side above.

It is true that if r and r' are not equal, then

        E_x(r,t) B_y(r',t) = B_y(r',t) E_x(r,t)

in Quantum Electrodynamics. This is not because the equation is true
as
a matter of course. The reason why the equality holds if r and r' are
not equal is because the two events (r,t) and (r',t) are separated by a
space-like interval, and so an inequality of E_x(r,t) B_y(r',t) and
B_y(r',t) E_x(r,t) would violate causality.

>> >> I have ALREADY pointed out that (20c) is completely irrelevant to
>the

>> >> portion of the paper in question. This is because the definition
>of
>> >> kappa_sigma in (20c) has no relevance outside of Section 9 of
that
>> >paper.

>> No comment, I see. You should withdraw the reference to (20c) since
>it is
>> irrelevant to the present discussion.
>If you want to consider AE's GR1916 paper as
>disjointed go ahead, but you have presented
>no reason to discount GR1916, and I rather
>doubt you'll ever have that ability.

I do not consider the paper to be disjointed. I have taken the trouble
to actually understand what is being written in the paper.

I am becoming increasingly of the view that you are reading these
texts,
but that you are not taking the trouble to actually comprehend what
they
say, so that you are getting a lot of facts without any depth of
comprehension. Reading is easy. Actually comprehending and taking in
the
significance of what is written is much harder work, and it looks like
you
are not prepared to do that work, in spite of the necessity of that
work.

A proper comprehension of the paper would lead to knowing what is of
greater significance, and what is of more transient importance (as a
stepping stone to another more significant statement). One example of
this is Weinberg's usage of f^0 = 0 in Section 2.7 of "Gravitation and
Cosmology". Weinberg introduced f^0 = 0 as an equation relevant to a
reference frame in which the charged body is at rest. He then used the
equation as a stepping stone to move on to Equation (2.7.9), and after
he
had used the equation f^0 = 0 as a stepping stone, it was no longer of
any
interest to him, since it had served its purpose.

Section 9 of Albert Einstein's 1916 paper starts with a variational
principle, which eventually leads to the equation

int_{lambda_1}^{lambda_2} kappa_sigma delta x_sigma d lambda = 0,

where Einstein defines kappa_sigma in (20b) (except for the typo where
sigma should have been used as an index in the first term, rather than
nu), after which he used the arbitrariness of delta x_sigma to conclude
that kappa_sigma = 0, i.e. (20c). But he did not have to use the
notation
kappa_sigma at this point. Einstein could have used an entirely
different
symbol (e.g. tau_sigma) without affecting anything else in the paper,
including later references to other quantities denoted by kappa_sigma.
The
*only* importance that kappa_sigma has in Section 9 is as a short hand
method of denoting the right hand side of (20b), and the significance
of
(20c) is not that kappa_sigma is equal to zero. The significance of
(20c)
is that the right hand side of (20b) is equal to zero, i.e. the
significance of (20c) is that

(d/d lambda) {(g_{mu sigma}/w) d x_mu/d lambda}

- (1/(2w)) (dg_{mu nu}/d x_sigma) (d x_mu/d lambda) (d x_nu/d lambda)

= 0,

as a consequence of the variational principle. Once we have the
equation
as written here, which is what (20c) actually means, kappa_sigma is no
longer of any importance, since it has served its purpose. The next
time
that the notation kappa_sigma appears in the paper is in Equation (65),
which forms a definition for a new vector which is denoted by
kappa_sigma,
and which is unrelated to the symbol kappa_sigma in Section 9.

>> >> I will give some more thought to the passages about (65a) and
>(66).

>> >> >HP, you're barking up the wrong tree,
>> >> >blaming others, like Einstein, to be
>> >> >"sloppy" when you don't understand GR.

>> >> I *do* understand GR. The equation f^u = 0 for a charged body is
>NOT
>> >a
>> >> part of GR.

>> >You're circular, I'm bored.

>> And you have not proven that f^u = 0 for a charged body is a part of
>GR.
>> That's just your wishful thinking. If you believe that it is a part
>of
>> GR, then PROVE it.

>> And you get bored easily. Is that because you keep facing the same
>> challenge time after time, even though you have never met the
>challenge?
>I did, you simply ignored my proofs.

No. You pointed to two portions of Weinberg, neither of which support
your view. You have never given a proof constructed from General
Relativity either. All we have is your unsupported assertions.

>> >>In Section 5.2 of "Gravitation and Cosmology", Weinberg gives
>> >> the formula for f^u without ever specifying that f^u = 0. Are
you
>> >saying
>> >> that Weinberg deliberately left out f^u = 0, or that he was
sloppy
>> >and
>> >> left it out accidentally?

>> >Still listening...

>> You are condescending, and you have no right to be, considering your
>lack
>> of any sort of superiority.
>See above, (yawn).

This is referring to the part above where you tried to pretend that
Weinberg's phrase, "so it is correct for all velocities", was referring
to
the equation f^0 = 0, when it was not referring to that equation, but
to
(2.7.9).

>> >>Similarly, Tolman gives the force law for a
>> >> charge in Section 103 of "Relativity Thermodynamics and
Cosmology"
>> >without
>> >> specifying that the force must be zero. Can you name even one
>other
>> >text
>> >> on General Relativity which appears to take the line that f^u =
0?

>> >MAXIMALLY SYMMETRICAL SUBSPACES,
>> >Weinberg (13.4.11)
>> >f,u =0.

>> That has nothing to do with the four-force on a charged body in an
>> electromagnetic field.
>If you prounouce that then no intelligent
>rebuttal is possible.

You can't give an intelligent rebuttal because you have obviously read
Section 13.4 without understanding it. It's another example of what I
pointed out earlier. You are willing to do the easy task of reading
and
taking in facts, but when it comes to the hard yakka (hard work) of
actually trying to understand what is being written and the
significance
of what has been written, you can't be bothered.

Again, let us analyze what Weinberg was doing in Section 13.4. Let us
start with Equation (13.4.10), i.e. B_{sigma nu} = fg_{sigma nu}. Do
you see that bit immediately underneath where he wrote

        "where f = (1/N) B_mu^mu",

except that there are three lines in the equals sign, rather than just
two? Well, that was where he DEFINED f. Furthermore, he defined f
without reference to electromagnetic fields or four-forces or anything
like that. I know this because, not only does he make absolutely no
reference to either in the Section (recall also that B_{mu nu} was
iintroduced for the first time, immediately after (13.4.7) as an
example
of the most general second rank maximally form-invariant tensor), but
also
because such considerations are irrelevant to the study of maximally
symmetric spaces. Weinberg then uses the fact that B_{mu nu} is a
maximally form-invariant tensor to conclude that df/dx^lambda = 0,
still
without any consideration of anything to do with the electromagnetic
field
or with four-forces.

Also, let us take the fiction that the four-force on a particle is
equal
to the gradient of some scalar field at each event on its trajectory,
and
see where that leads us. Since g_{uv} p^u p^v = m^2 is a constant,
then
the four-force and the four-momentum are orthogonal, and so f_u U^u =
0,
where U is the four-velocity. For the trajectory of the particle,
dx^u = U^u ds, where s is the proper time, and so f_u dx^u = f_u U^u =
0
at all events on the trajectory. But we have assumed that the
four-force
is equal to the gradient of some scalar field f (i.e. f_u = f,u), and
so,
for the particle, df = f,u dx^u = 0, and so f is a constant. It
follows
that the trajectory remains within a level surface of f.

What happens if we extend the fiction to assuming the existence of a
scalar field f such that the force law for the particle requires that
the
four-force on the particle be equal to the gradient of f? We know that
f
must be the same at all events on the trajectory (from the previous
paragraph), and by the arbitrariness available for possible
trajectories,
it follows that f must be a constant through all of spacetime, and the
four-force is equal to zero in all of spacetime. This is unduly
restrictive, and it eliminates the idea of a scalar potential such that
the force law for the body is determined by the rule that the
four-force
on the body is equal to the gradient of a scalar field.

>>Where does Weinberg make any connection between
>> that formula and anything to do with electromagnetism?
>You *pronounced* he doesn't, why would I want
>to confuse you with facts?

Translation: you can't point to where Weinberg made the connection,
otherwise you would have done so. The reason why you can't is because
he never did make such a connection.

You use the "why would I want to confuse you with facts?" gambit when
you
know that you can't supply the evidence that is required of you, but
you
still want to pretend that it's there, so that you can save face.

Also, you do not answer such a challenge by "why would I want to
confuse
you with facts?" That as much as claims that you expect to be believed
purely on the basis that you are you, and that I am me. That is not
how
scientific discourse is conducted.

>> I get it. He used the notation f, and you instantly jumped to the
>> conclusion that he was discussing the four-force on a charged body
in
>an
>> electromagnetic field.

>> >That Weinberg loves a snappy ending.

>> But let's look at your very own contribution to reaching the
>conclusion
>> that the four-force on a charged body in an electromagnetic field is
>equal
>> to zero. Your contribution was very important to reaching this
>> conclusion, since your contribution was to see the expression f,u
and
>to
>> misinterpret it as the four-force on a charged body in an
>electromagnetic
>> field.
>How was that a misinterpretation?

See above. f was not even introduced into the text until (13.4.10),
and its introduction bore no connection to electromagnetic fields or
four-forces.

>>Weinberg, himself, gave NO indication of any connection between
>> f,u and electromagnetism whatsoever.
>See (13.1.5) => (13.4.11)

(13.1.5) is the equation of a Killing vector. Again, there is no
connection either with the electromagnetic field or with four-forces.
The equation for Killing vectors was derived using only geometric
considerations, without any regard to dynamics.

>> >I guess he knows how to sort out the
>> >ding-bats.

>> And the ding-bats that his equation (13.4.11) sorts out are not the
>people
>> you think, since the equation bears absolutely no resemblance to
what
>you
>> think it says. Whatever made you point to an equation so completely
>> unrelated to the discussion at hand as (13.4.11)?
>hmmm...

No answer. There is no relevance. f is only introduced for the first
time in (13.4.10), and it is a purely geometric object which has no
connection with the electromagnetic field or with four-forces.

>> >You have a choice, accept GR as it is
>> >or you do the work and recognize some edges.

>> It looks like YOU will have to accept GR as it is, and not as you
>would
>> like it to be. (13.4.11) had absolutely nothing to do with the
>four-force
>> on a charged body in an electromagnetic field. You still have to
>give a
>> valid and correct answer to my question:

>> Can you name even one other text on General Relativity which
>> appears to take the line that f^u = 0?
>Yes, you got two, AE and Weinberg, what's it worth
>to you to get more?

A proper analysis of what was actually happening in Weinberg's Section
2.7
and Section 13.4 shows that neither of these support your hypothesis.
The one phrase in Section 2.7 that you are relying on ("so it is
correct
for all velocities") does not apply to f^0 = 0 as you had hoped, but to
(2.7.9). The scalar function f in Section 13.4 was introduced FOR THE
FIRST TIME in Equation (13.4.10) and it was defined at that point as
well,
and its definition was completely unrelated to any considerations of
the
electromagnetic field, charges or four-forces. The passage from
(13.4.10)
to (13.4.11) was also completely unrelated to any considerations of the
electromagnetic field, charges or four-forces. All of these
conclusions
are actually there in the book for anyone who is actually willing to do
the hard work of comprehending what is written, rather than just
reading
the work on a superficial level.

>> >AE himself had issues with GR, and I
>> >agree, it can be improved.
>> >...

>> You have to understand it first. After your laughable error on
>Weinberg's
>> Equation (13.4.11), I'd say that you are still a long way off.
>The SAVAGE has SPoKEN, keep laughing, good
>you get a laugh.

Perhaps now that you have had an analysis of the context of Equation
(13.4.11) laid out before your eyes, you might reconsider your
judgement.
It is not a good idea to use pejorative words to describe those with
whom
you disagree. In this case, I happened to be right and you happened to
be
wrong, and your epithet was inappropriate. You betrayed the
shallowness
of your understanding because you could not be bothered to actually do
the
work of properly understanding what was written in Section 13.4.

>> >> >He was good at series (McClaren Taylor stuff)
>> >> >but weak in advanced geometry.
>> >> > So to explain space warps (Curl, Div and Grad)

>> >> Those are NOT space warps. They are exterior derivatives. Even
>in
>> >flat
>> >> Euclidean space, every differentiable scalar function has a
>gradient,
>> >and
>> >> every differentiable vector function has both a curl and a
>> >divergence.

>> >Your rebuttal does not deserve my answer.

>> So why do you think that the gradient, curl and divergence are space
>> warps? They are well-defined operators in flat Euclidean space.
>Well for my part, it has to do with embedding
>dimensions. You know, take a globe and examine
>it's 2D surface in 3D.
> Suppose you're restricted from that embededding,
>so that you must use a 2D surface only, like a
>rubber membrane, and demo Grad, Div and Curl in
>those rules, that excludes embedding, and then
>requires the inhabited geometry to be sufficient,
>if it is eleastic, as our 4D demos.

More detail required. That is too vague to get any useful information.

I should also like to point out that, in two dimensions, the curl of a
vector field is a scalar, not a vector. In fact, for a two-dimensional
manifold, the curl of a vector field is just the divergence of the
vector
field after being rotated through 90 degrees (i.e. for each point in
the
manifold, we take the value of the vector field and rotate it through
90
degrees - after we have done that at all points in the manifold, we
take
the divergence of the resulting vector field, and the result that we
get
is the curl of the original vector field).

>> >> > It was Dr. Sawyer who gave the Lawden book
>> >> >I wrote about, and Dr. Sawyer was having
>> >> >some problems with tensors,

>> >> I bet he wasn't. I would say that that is more a faulty memory
on
>> >your
>> >> part.

>> >I think you are very stupid.

>> You must be looking in a mirror. We only have you word for it that
>> W.W. Sawyer was having some problems with tensors. I find that you
>lack
>> credibility.
>Join the club.

Certainly. I will join the club of Dirk van de Moortel, Tom Roberts,
and
all the others who also find that you lack credibility.

>> If W.W. Sawyer was having any problems with tensors (and I'm not
>saying
>> that he wasn't), then those problems would have been far beyond
>anything
>> that you are capable of comprehending.
>Then I suggest you terminate posting to me.

You mean because you are indeed incapable of comprehending what I am
trying to explain to you? Anyway, I am posting to the newsgroup. I am
not posting directly to you - nor do I intend to ever do so.

>> >I read the math you wrote and
>> >you failed to use basis vectors.

>> I DID use basis vectors. At each event in Minkowski space, {d/dt,
>d/dx,
>> d/dy, d/dz} is a set of basis vectors for the vector space of
tangent
>> vectors (contravariant vectors), and {dt, dx, dy, dz} is a set of
>basis
>> vectors for the vector space of cotangent vectors (one-forms,
>covariant
>> vectors). I explicitly made reference to these particular basis
>vectors.
>> If they are in a notation that you don't recognize, then why don't
>you
>> say so? How are you supposed to answer my question if you can't
even
>> recognize a vector basis when you see one?

>> If you prefer, I could introduce a basis {e_u : u = 0, 1, 2, 3} of
>> contravariant vectors, whose components are (e_u)^v = delta^v_u, and
>> the dual basis {e^0, e^1, e^2, e^3} of covariant vectors (so that
the
>> components of the covariant vector e^u are (e^u)_v = delta^u_v).
>> Specifically, e_0 = d/dt, e_1 = d/dx, e_2 = d/dy, e_3 = d/dz, e^0 =
>dt,
>> e^1 = dx, e^2 = dy, e^3 = dz. The Minkowski metric is determined by

>> g(e_0,e_0) = 1, g(e_0,e_1) = g(e_1,e_0) = 0,

>> g(e_0,e_2) = g(e_2,e_0) = 0, g(e_0,e_3) = g(e_3,e_0) = 0,

>> g(e_1,e_1) = - 1, g(e_1,e_2) = g(e_2,e_1) = 0,

>> g(e_1,e_3) = g(e_3,e_1) = 0, g(e_2,e_2) = - 1,

>> g(e_2,e_3) = g(e_3,e_2) = 0, g(e_3,e_3) = - 1.
>Evidentally you are one to pay attention to the basics,
>((I do, so that's a definite compliment)).
>What would be the Kronecker Delta using
>those relations above?

Yes it is the Kronecker Delta. So please fulfil the promise that you
made
regarding pointing out an error about assuming a signature of (+---).

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