Natural Science Forum / Physics / Particle Physics / April 2005

No one is, and since you claim to be a researcher
you should know that.

We haven't discussed QM yet.

Ok, just GR...

A_u = A*(x^u/s) is a positional potential.

B_u = A*(dx^u/ds) is a velocity potential.

Yes, but it's not physical.

I'm not sure I understand that?

The partial derivative,

a*F_uv,w =/=0 ,

and that's what Maxwell uses.

But, I find the covariant derivative,

a*F_uv;w=0,

from GR.

I argue all force is relative, there is
no invariant force, thus f=0, f=f_u U^u.

The Relativity of Force is because it
requires an interaction between two
distinguishable particles, (aka points/
systems/FoR's.

Let's fall back on Newton's Laws for a
breather, as these have been modified
by GR.

Newton's 1st law leads to the concept
of a geodesic in a g-field as the particle
is in free-fall.

Newton's 3rd, involves "action" and "reaction",
and certainly needs 2 particles.

But action is invariant and quantized in GR,
by "h", that action CANNOT be transformed
away, but no decision can be arrived at
to presumed one or the other particle is
accelerating in the interaction.

For example, shoot off a cannon, you have
a complete right to say the cannonball is
stationary or the cannon is stationary as
you want. Choosing the cannonball to be a
stationary frame still leaves intact the
action, that's the invariant.

I'll stop here pending the usual spit-balls
Regards
Ken S. Tucker

Notation convolution.

Your solution is imaginary, very much so.

Why not just write down the appropriate asymmetrical
tensor.

You missed the point.
My daughter is a P.Geo/P.Eng. In HS she was
having problems with chemistry, specifically
electrolysis. So I studied up on the subject
and took her for a boat ride out to the middle
of our lake, shut-off the motor and sat in the
sun and talked awhile. It's was very quiet and
peaceful, the occasional lap of a wave on the
side of the boat.
 She is by nature a high achiever, and I was
quite concerned with how to explain electrolysis,
to alleviate her frustration.
 I chose an analogy of dating, something akin
to her thinking, and I spoke about how positive
ions attract negative ions on the side.
 After a couple of hours she was peaceful and
felt she really understood the problem.
 We did that sort of thing quite often and her
marks took an upswing. She went on to obtain a
BSc honors, cum something or another, then I
think a Master of Geo and now a P Geo, she has
lots of alphabet behind her name, it's like a
hobby for her.

Probably, because I'm one of them.

You've never given birth.

You seem to be setting a narrow agenda,
SR bores me to tears, and QM is a sleeper.

Photonic relation ((Platonic too)).

You could have merely explained what you meant,
in the context of your arguement,( no need to type
out a dictionary).

I hear Australia is going to legalize nudity.

Is it true Australia is using sharks to
control over-population?

Are you threw? or are you through?

Of course they commute, how else would they
get there.

So you have spent the better portion of an hour
to explain the speed of light is "c", and
"through" a few spit-balls.
Would you like to tell us why you think "h"
is invariant?

Regards
Ken S. Tucker

Rubbish.  The notation E_x(r,t) is a STANDARD notation to denote the x
component of the (classical or quantum) electric field evaluated at
position r and time t.  The notation B_y(r',t') is a STANDARD notation to
denote the y component of the (classical or quantum) magnetic field
evaluated at position r' and time t'.  E_x(r,t) B_y(r',t') is the product
of E_x(r,t) and B_y(r',t'), in that order.  Similarly, B_y(r',t') E_x(r,t)
is the product of B_y(r',t') and E_x(r,t), in that order.  There is
absolutely nothing here that should be in any way unfamiliar to you.  
There is no "notation convolution".  In classical electromagnetic theory,
E_x(r,t) B_y(r',t') and B_y(r',t') E_x(r,t) are equal.  But in the case
of the quantum electromagnetic field, E_x(r,t) B_y(r',t') and
B_y(r',t') E_x(r,t) are not equal if the events (r,t) and (r',t') are
separated by a light-like interval.

What is evidently making your brain explode at this point is the
possibility that in the case of the quantum electromagnetic field,
E_x(r,t) B_y(r',t') and B_y(r',t') E_x(r,t) may not be equal.  But this
is a concept which is familiar to anybody who has more than just a
superficial grounding in quantum mechanics.  It is an example of the most
important difference between physical observables in the non-quantum case
and physical observables in the quantum case.

In classical mechanics, if A and B are physical observables, then AB = BA.

In quantum mechanics, if A and B are physical observables, then AB is not
necessarily equal to BA.

The canonical commutation relation in quantum mechanics states that
qp - pq = i hbar, where q is a coordinate, and p is the conjugate
momentum.  The usual representation in the Schroedinger picture for q and
p is determined by

    (q psi)(x) = x psi(x),

    (p psi)(x) = - i hbar psi'(x).

It is left as an exercise for the reader to prove that this is indeed a
representation of the commutation relation qp - pq = i hbar.  It is
directly from the canonical commutation relation, qp - pq = i hbar, that
the Heisenberg Uncertainty Principle is based.  The Heisenberg Uncertainty
Principle is the statement that the product of the indeterminacy (or
uncertainty) of q and the indeterminacy (or uncertainty) of p is no less
than hbar/2, i.e. (Delta q) (Delta p) >=  (1/2) hbar, where a >= b denotes
that a is greater than or equal to b, and for a physical observable A,
Delta A denotes its uncertainty or indeterminacy.  The proof of this is
given in Appendix A below.

The canonical commutation relation, qp - pq = i hbar, should be familiar
to anybody who has anything more than a basic superficial familiarity with
quantum mechanics.  And so, the concept that, if A and B are physical
observables in quantum mechanics, then AB and BA may not be equal, should
also be familiar to anybody with anything more than a basic superficial
understanding.

Let us take a look at the case of angular momentum, as another example.  
In three dimensions, the components of the orbital angular momentum are
given by

    L_x = y p_z - z p_y,

    L_y = z p_x - x p_z,

    L_z = x p_y - y p_x,

where x, y and z are the three coordinates, and p_x, p_y, p_z, are their
conjugate momenta.  The coordinates and momenta satisfy

    q_i q_j = q_j q_i, i, j = x, y, z,

    p_i p_j = p_j p_i, i, j = x, y, z,

    q_i p_j - p_j q_i = i hbar delta_{ij}, i, j = x, y, z,

i.e.

    x y = y x, x z = z x, y z = z y, p_x p_y = p_y p_x,

    p_x p_z = p_z p_x, p_y p_z = p_z p_y, x p_x - p_x x = i hbar,

    x p_y = p_y x, x p_z = p_z x, y p_x = p_x y,

    y p_y - p_y y = i hbar, y p_z = p_z y, z p_x = p_x z,

    z p_y = p_y z, z p_z - p_z z = i hbar.

So

 L_x x - x L_x = y p_z x - z p_y x - x y p_z + x z p_y = 0,

 L_x y - y L_x = y p_z y - z p_y y - y^2 p_z + y z p_y = i hbar z,

 L_x z - z L_x = y p_z z - z p_y z - z y p_z + z^2 p_y = - i hbar y,

 L_x p_x - p_x L_x = y p_z p_x - z p_y p_x - p_x y p_z + p_x z p_y = 0,

 L_x p_y - p_y L_x = y p_z p_y - z p_y^2 - p_y y p_z + p_y z p_y

= i hbar p_z,

 L_x p_z - p_z L_x = y p_z^2 - z p_y p_z - p_z y p_z + p_z z p_y

= - i hbar p_y.

In a similar manner, the following can be proven:

    L_y x - x L_y = - i hbar z,

    L_y y - y L_y = 0,

    L_y z - z L_y = i hbar x,

    L_y p_x - p_x L_y = - i hbar p_z,

    L_y p_y - p_y L_y = 0,

    L_y p_z - p_z L_y = i hbar p_x,

    L_z x - x L_z = i hbar y,

    L_z y - y L_z = - i hbar x,

    L_z z - z L_z = 0,

    L_z p_x - p_x L_z = i hbar p_y,

    L_z p_y - p_y L_z = - i hbar p_x,

    L_z p_z - p_z L_z = 0.

It follows that

     L_x L_y

    = L_x z p_x - L_x x p_z

    = z L_x p_x - i hbar y p_x - x L_x p_z

    = z p_x L_x - i hbar y p_x - x p_z L_x + i hbar x p_y

    = L_y L_x + i hbar L_z,

and so

    L_x L_y - L_y L_x = i hbar L_z.

Similarly,

    L_y L_z - L_z L_y = i hbar L_x,

    L_z L_x - L_x L_z = i hbar L_y.

It follows that if the system is in a state in which both L_x and L_y have
definite values, then the system is in a state psi such that

    L_x psi = a psi,

    L_y psi = b psi,

for some real a and b.  Therefore

    L_x L_y psi = L_x (b psi) = b L_x psi = a b psi,

    L_y L_x psi = L_y (a psi) = a L_y psi = a b psi,

and so

   i hbar L_z psi = L_x L_y psi - L_y L_x psi = a b psi - a b psi = 0.

It follows that L_z psi = 0.  Therefore

    L_y L_z psi = L_x (0) = 0,

    L_z L_y psi = L_z (b psi) = b L_z psi = 0,

and so

   i hbar L_x psi = L_y L_z psi - L_z L_y psi = 0 - 0 = 0.

It follows that L_x psi = 0 (i.e. a = 0).  Similarly, L_y psi = 0
(i.e. b = 0).  Therefore, if two components of the angular momentum are
simultaneously determinate, then all three are determinate, and all three
are equal to zero in that case.

The equations,

    L_x L_y - L_y L_x = i hbar L_z,

    L_y L_z - L_z L_y = i hbar L_x,

    L_z L_x - L_x L_z = i hbar L_y,

form the basis of the quantum theory of angular momentum.

In the quantum theory of angular momentum, for any three physical
observables J_x, J_y, J_z, such that

    J_x J_y - J_y J_x = i hbar J_z,

    J_y J_z - J_z J_y = i hbar J_x,

    J_z J_x - J_x J_z = i hbar J_y,

J^2 = J_x^2 + J_y^2 + J_z^2 can only take values of the form j(j+1) hbar^2,
where 2j is a nonnegative integer (i.e. J^2 can only take the values 0,
3 hbar^2/4, 2 hbar^2. 15 hbar^2/4, ...), and when J^2 has the determinate
value j(j+1) hbar^2, J_z can only take values j hbar, (j-1) hbar, ...,
(-j+1) hbar, -j hbar, so that when J^2 has value 3 hbar^2/4, J_z can only
take the values hbar/2, -hbar/2, and when J^2 has value 2 hbar^2, J_z can
only take values hbar, 0, -hbar.  This is proven in Appendix B below, and
it is why we say that angular momentum is quantized.

In the case of orbital angular momentum, L^2 can only take values
l(l+1) hbar^2 where l is a nonnegative integer (it follows that L_z can
only take values m hbar, where m is an integer).

An example of a quantum system is the simple harmonic oscillator, with
Hamiltonian,

    H = p^2/(2m) + (1/2) m w^2 q^2.

The energy levels of the harmonic oscillator are given by

    E_n = (n + 1/2) hbar w,

for n = 0, 1, 2, 3, ...  Note that w is the angular frequency of the
corresponding classical harmonic oscillator.  It follows that the energy
of a simple harmonic oscillator is quantized.  This is proven in Appendix
C, below.

Another example of a quantum system is the hydrogen atom, with
Hamiltonian,

    H = p^2/(2m) - e^2/r,

the Hamiltonian (or energy) can only take values given by

    E_n = - m e^4/(2 n^2 hbar^2),

and all nonnegative real values (i.e. any nonnegative real energy can be
taken as a value by the Hamiltonian - these are the so-called ionization
states).  The values of E_n are determined by certain integrability
requirements for the corresponding wave-function.  If the Hamiltonian has
a nonnegative value, then L^2 can take values l(l+1) hbar^2 for all
nonnegative integers l.  If the Hamiltonian has value E_n, then L^2 can
take values l(l+1) hbar^2 for l = 0, 1, ..., n-1, and for no other values.
Specifically, if H has the value E_1 = - m e^4/(2 hbar^2), then the
orbital angular momentum has the determinate value of 0.  This is in
contrast to the classical case where the orbital angular momentum of an
electron in orbit must be nonzero.  The fact that the angular momentum of
an electron in the ground state (with energy E_1) must have angular
momentum zero is against all classical expectations.

This is an example of how quantum mechanics works to quantize energy.  If
you are given a Hamiltonian, then you may be able to use the explicit form
of the Hamiltonian to work out what values it can take, and those values
are the legitimate values for the energy, i.e. the actual energy levels
are imposed by the form of the Hamiltonian, and not via any external
agency.

With regard to some of your other claims, we have already seen several
examples of your reading books without understanding.  One example is in
Section 9 (and Equation (20c)) of Einstein's 1916 General Relativity
paper, where you placed an inordinate importance on the minutiae within a
proof or derivation, and in which you failed to recognize that the only
significance of kappa_sigma in (20c) is that

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

 - 1/(2w) dg_{mu nu}/dx_sigma dx_mu/d lambda dx_nu/d lambda

 = 0

for a geodetic line, and that once you have this result, the symbol
kappa_sigma is no longer of any significance.  In other words, the ONLY
significance of kappa_sigma is that it is shorthand notation for the right
hand side of (20b) with the indices corrected, and that the significance
of (20c) is that the right hand side of (20b) (with the indices
corrected) is equal to zero for a geodetic.  You have suggested that this
way of regarding kappa_sigma in Section 9 is indicative of treating the
paper in a fragmented manner.  Your suggestion failed to take into account
what is actually important in the logical development of a consistent
theory.  The fact that the right hand side of (20b) is equal to zero for a
geodetic is important.  The fact that the right hand side was denoted by
kappa_sigma is not important.  The fact that the right hand side of (20b)
is equal to zero for geodetics is one step in the development of the
theory.  A self-consistent theory is not just a set of equations.  A
self-consistent theory is a (hopefully small) set of assumptions, and all
the assertions which follow from them.

Another example is your misreading of Sections 2.7 and 13.4 in Weinberg's
"Gravitation and Cosmology" (your misinterpretation of the passage
following (2.7.9) and your misinterpretation of (13.4.11), both of which
have been the subject of much discussion between us, and which do not need
elaboration here).

And in the more recent past, and until now, still unremarked upon, there
is your misinterpretation of Hermann Weyl's paper, "Gravitation and
Electricity".  Specifically, you have explicitly made the false statement
that, when Weyl introduced the infinitesimal d phi before Equation (6),
and then expressed d phi as d phi = sum phi_mu dx_mu, he was introducing
the gradient of a scalar function (presumably denoted by phi).  Presumably
it was Weyl's usage of the notation d phi which threw you, but Weyl *never*
introduced any such scalar function, nor did he ever make reference to any
such scalar function.  This scalar function is yet another figment of your
imagination.  Weyl was merely using d phi to denote an infinitesimal
quantity.  And it is not unprecedented that a notation like d phi should
be used for an inexact differential.  The line element ds is an inexact
differential, otherwise there would be no variational principle.  In
thermodynamics, infinitesimal quantities like dW (work) and dQ (heat)
are inexact differentials.  One of the implications of the Second Law of
Thermodynamics is that dQ/T is exact, where T is the temperature (for
proof, see Zemansky, for example - in fact, dQ/T = dS, where S is the
entropy).  This statement would have been meaningless if all
differentials had been exact.  The fact that Weyl at no time ever made any
mention of this scalar function which you accused him of constructing, and
that he introduced d phi specifically as an infinitesimal quantity alone,
means that he is completely undeserving of the scorn which you poured on
him.  If anybody deserves your scorn, it is yourself, since the concoction
that you were pouring scorn on is your own concoction, not Weyl's.

Also, I daresay, that when I pointed out earlier that certain expressions
for the charge density and current density are gauge-invariant, you were
probably not even aware what a gauge transformation is, or why the
gauge-invariance of these quantities is so important (for example, why we
need a quantity to be gauge-invariant before it can be regarded as
physical, and why we need a set of equations to be gauge-covariant
before they can be regarded as physical).

The commutator of E_x(r,t) and B_y(r',t') is certainly an imaginary
q-number, and it is therefore equal to the square root of -1 multiplied
by a real q-number.  After all, that is the mathematical meaning of the
word "imaginary".

I have already written down the Lorentz-covariant form for

    F_{uv}(x) F_{u'v'}(x') - F_{u'v'}(x') F_{uv}(x),