Quanta of Momentum
vernonner3voltazim
I've worked out some speculations in reasonable mathematical
detail, located at this URL:
http://www.nemitz.net/vernon/MOMEQUAT.htm
David McAnally
Momentum has been quantized. It has a continuous spectrum consisting of
all real numbers. On a bounded interval, momentum is quantized, with
discrete eigenvalues.
David McAnally
---------
David Thomson
news:42336979.02092...@posting.google.com...
Hi Vernon,
I read your paper. It's a nice try. But the quantum momentum must be based
on real quantum values. There is no place in the subatomic realm where a
mass (devoid of length) moves at a velocity. If you're going to quantize
momentum, it must be angular momentum. Electrons, protons, neutrons, and
photons all possess angular momentum.
Dave
Paul Cardinale
> Why not?
>
Because there is no reason to suspect that it might be.
> Hasn't everything else in Physics been quantized?
>
No. Consider: Mass, energy, position, acceleration, velocity,
temperature, force, gravity, ...
Many things are not quantized.
Paul Cardinale
vernonner3voltazim
Thanks for the responses!
Dave, it seems that you did not quite get the picture I was
trying to describe. I was not talking about "mass (devoid of
length)" in any sense. I was talking about a unique entity
that is totally independent of "mass" altogether. As you
know, Eass and Energy are equivalent entities, Space and Time
are reasonably equivalent entities, and Momentum is another
Fundamental Thing that is distinct from those others. I wish
you to think about that distinctness to the extent that
"could some amount of momentum exist all by itself -- a blob
of PURE momentum?" THAT would be the "quantum of momentum"
that I was talking about. And, as I tried to show in that
paper, it might be a good candidate for the graviton, too.
(As a bonus thought, once gravitons enter the picture, one
has to contemplate "gravity waves" -- so what sort of wave
might a quantum of momentum be? Perhaps a solitary wave....)
dlzc@aol.com (formerly)
> Thanks for the responses!
> Dave, it seems that you did not quite get the picture I was
> trying to describe. I was not talking about "mass (devoid of
> length)" in any sense. I was talking about a unique entity
> that is totally independent of "mass" altogether. As you
> know, Eass and Energy are equivalent entities, Space and Time
> are reasonably equivalent entities, and Momentum is another
> Fundamental Thing that is distinct from those others.
Not true for the photon, where Energy and Momentum are equivalent.
> I wish
> you to think about that distinctness to the extent that
> "could some amount of momentum exist all by itself -- a blob
> of PURE momentum?" THAT would be the "quantum of momentum"
> that I was talking about. And, as I tried to show in that
> paper, it might be a good candidate for the graviton, too.
Graviton has not been found. Momentum without mass is a photon, and its
momentum is a function of the frame that absorbs it. No real quantum here,
but a packet certainly...
> (As a bonus thought, once gravitons enter the picture, one
> has to contemplate "gravity waves" -- so what sort of wave
> might a quantum of momentum be? Perhaps a solitary wave....)
Solitons have been thought of already...
David A. Smith
vernonner3voltazim
> > know, Eass and Energy are equivalent entities, Space and Time
> > are reasonably equivalent entities, and Momentum is another
> > Fundamental Thing that is distinct from those others.
>
> Not true for the photon, where Energy and Momentum are equivalent.
No for the photon, Energy and Momentum are merely proportional,
not equivalent. A photon is "energy in motion", and thus has
momentum as a consequence, since mass and energy are equivalent.
> > I wish
> > you to think about that distinctness to the extent that
> > "could some amount of momentum exist all by itself -- a blob
> > of PURE momentum?" THAT would be the "quantum of momentum"
> > that I was talking about. And, as I tried to show in that
> > paper, it might be a good candidate for the graviton, too.
>
> Graviton has not been found.
That doesn't mean we can't devise experiments based on the
hypothesis described in my "momenton" paper. Suggestion:
http://www.halfbakery.com/idea/Gravity_20Waves#1025890913
> Momentum without mass is a photon, and its
> momentum is a function of the frame that absorbs it.
You are repeating your prior misconception. Think about
the trouble people have, learning that "mass" is not the
same thing as "weight". I ask you to contemplate a new
thought, that momentum might be able to exist independently
of Mass or Energy.
> > (As a bonus thought, once gravitons enter the picture, one
> > has to contemplate "gravity waves" -- so what sort of wave
> > might a quantum of momentum be? Perhaps a solitary wave....)
>
> Solitons have been thought of already...
So? If I hadn't known solitons existed, I couldn't have
suggested that the type of wave that a graviton might be,
might be a solitary wave, could I? Now, if you are saying
that others have also suggested that gravitons might be
solitary waves, well, OK, but you sure didn't say it very
clearly!
David Thomson
based
> > on real quantum values. There is no place in the subatomic realm where
a
> > mass (devoid of length) moves at a velocity. If you're going to
quantize
> > momentum, it must be angular momentum. Electrons, protons, neutrons,
and
> > photons all possess angular momentum.
> Thanks for the responses!
> Dave, it seems that you did not quite get the picture I was
> trying to describe. I was not talking about "mass (devoid of
> length)" in any sense. I was talking about a unique entity
> that is totally independent of "mass" altogether. As you
> know, Eass and Energy are equivalent entities, Space and Time
> are reasonably equivalent entities, and Momentum is another
> Fundamental Thing that is distinct from those others.
Vernon, you cannot have momentum independent of mass. Momentum is mass
times velocity.
> I wish
> you to think about that distinctness to the extent that
> "could some amount of momentum exist all by itself -- a blob
> of PURE momentum?" THAT would be the "quantum of momentum"
What do you propose is the basis for this quantum momentum? Do you have a
value for it?
> And, as I tried to show in that
> paper, it might be a good candidate for the graviton, too.
I think you are making a good effort in your paper. I'm giving you the
opportunity to defend your paper against a fair criticism.
Dave
David Thomson
"dl...@aol.com (formerly)" <dee-ell-...@cox.net (use dlzc1 in
beginning)> wrote in message
news:yd9m9.139522$Pf7.5...@news1.west.cox.net...
> Not true for the photon, where Energy and Momentum are equivalent.
Energy = mass * velocity^2
Momentum = mass * velocity
How are energy and momentum equivalent?
> Momentum without mass is a photon, and its
> momentum is a function of the frame that absorbs it. No real quantum
here,
> but a packet certainly...
photon = h * c (units are angular momentum times velocity)
momentum
----------- = velocity
mass
How does momentum without a mass equal a photon?
Dave
Patrick Reany
David Thomson wrote:
You're making the mistake of thinking about mass
and momentum as real essences that can't be messed
about with. But in science this is not so. Theories get
to define their terms arbitrarily so long as those definitions
are consistent with conventional norms. After that,
those definitions are either instrumental for producing
workable theories or they are not.
The biggest mistake people make in trying to learn
physics is to put their own personal intuitions as the
ultimate arbiters of what is reasonable as a construct
of a physical theory. In the end, the best you can do
is to replace an arbitrary construct that you don't like
with an arbitrary construct that you do like, but it's
still an arbitrary construct. Nature does not have to
be in its essence the way the human mind prefers
to think about it.
Physical concepts invented by humans live only
to be place in theories, be they prescientific
or scientific theories. And, although I am not
about to say that Nature does NOT operate
under a set of "laws," I am quite willing to reject
the notion that Nature operates under our theories
or any other notion of theory.
Patrick
Stephen Speicher
>
> Vernon, you cannot have momentum independent of mass. Momentum is mass
> times velocity.
>
If Newton were to come alive and discover what knowledge has been
accumulated since his time, he would laugh at you, as do we.
--
Stephen
s...@speicher.com
Ignorance is just a placeholder for knowledge.
Printed using 100% recycled electrons.
-----------------------------------------------------------
Stephen Speicher
> Hi David,
>
> "dl...@aol.com (formerly)" <dee-ell-...@cox.net (use dlzc1 in
> beginning)> wrote in message
> news:yd9m9.139522$Pf7.5...@news1.west.cox.net...
> > Not true for the photon, where Energy and Momentum are equivalent.
>
> Energy = mass * velocity^2
E^2 = m^2c^4 + p^2c^2
> Momentum = mass * velocity
>
p^2c^2 = E^2 - m^2c^4
for a photon, m = 0,
p = E/c
> How are energy and momentum equivalent?
>
See above.
>
> photon = h * c (units are angular momentum times velocity)
>
> momentum
> ----------- = velocity
> mass
>
> How does momentum without a mass equal a photon?
>
1. Go to technical bookstore.
2. Buy book on quantum mechanics.
3. Read book.
David Thomson
news:3D99F5E4...@asu.edu...
> David Thomson wrote:
>
> > Hi David,
> >
> > "dl...@aol.com (formerly)" <dee-ell-...@cox.net (use dlzc1 in
> > beginning)> wrote in message
> > news:yd9m9.139522$Pf7.5...@news1.west.cox.net...
> > > Not true for the photon, where Energy and Momentum are equivalent.
> >
> > Energy = mass * velocity^2
> > Momentum = mass * velocity
> >
> > How are energy and momentum equivalent?
> >
> > > Momentum without mass is a photon, and its
> > > momentum is a function of the frame that absorbs it. No real quantum
> > here,
> > > but a packet certainly...
> >
> > photon = h * c (units are angular momentum times velocity)
> >
> > momentum
> > ----------- = velocity
> > mass
> >
> > How does momentum without a mass equal a photon?
> >
> > Dave
>
> You're making the mistake of thinking about mass
> and momentum as real essences that can't be messed
> about with. But in science this is not so.
Hi Patrick,
I can accept the criticism, but you need to provide a counter argument.
Momentum has a specific definition. Provide a definition that disagrees
with my statement that it is mass times velocity. Also, explain how you can
mess with mass.
We're discussing the quantum of momentum here, so it would seem the smallest
quantum of momentum will be the mass of the electron times the speed of
light. What are your suggestions?
Dave
Bilge
>
>I can accept the criticism, but you need to provide a counter argument.
That claim has been demonstrated to be false. Not just overstated.
>Momentum has a specific definition. Provide a definition that disagrees
>with my statement that it is mass times velocity. Also, explain how you can
>mess with mass.
What exactly is the scientific meaning of the verb "to mess"?
>
>We're discussing the quantum of momentum here, so it would seem the smallest
>quantum of momentum will be the mass of the electron times the speed of
>light. What are your suggestions?
Since the momentum of an object in its own restframe is zero, and
velocity is a relative quantity which is also not a quantum mechanical
observable, why should momentum be quantized? The phase space, p.x, is
quantized.
Paul Cardinale
> Hi David,
>
> "dl...@aol.com (formerly)" <dee-ell-...@cox.net (use dlzc1 in
> beginning)> wrote in message
> news:yd9m9.139522$Pf7.5...@news1.west.cox.net...
> > Not true for the photon, where Energy and Momentum are equivalent.
>
> Energy = mass * velocity^2
> Momentum = mass * velocity
>
E^2 = (p*c)^2 + (m*c^2)^2
For a photon, m=0 so:
E^2 = (p*c)^2
E = p*c
Thus, for a photon, E and p differ only by a constant factor;
and if you choose units such that c=1, then E = p.
Paul Cardinale
Robert Kolker
Patrick Reany wrote:
> Physical concepts invented by humans live only
> to be place in theories, be they prescientific
> or scientific theories. And, although I am not
> about to say that Nature does NOT operate
> under a set of "laws," I am quite willing to reject
> the notion that Nature operates under our theories
> or any other notion of theory.
Quite so. The correct way of describing the fit between our "laws of
nature" and nature is that our laws of nature up to now predict and
describe correctly. Hopefully the will continue to do so when applied
under new circumstance, but there is not guarantee.
BTW, When is the Worldwide Association of Pragmatists and
Instrumentalists having their next convention? I do so love a good party.
Bob Kolker
dlzc@aol.com (formerly)
> > Not true for the photon, where Energy and Momentum are equivalent.
>
> Energy = mass * velocity^2
> Momentum = mass * velocity
>
> How are energy and momentum equivalent?
I will name the book you can read. You are correct that the units don't
"add up", however you have seen the formulas the others have supplied...
momentum * velocity = energy. The velocity cleverly chosen for the case of
light (or anything else) will be... c!
Energy is equivalent to mass for a particle at rest, as much as energy is
equivalent to momentum for a photon.
"Spacetime Physics", 2nd edition
Edwin F. Taylor & John Archibald Wheeler
Currently available (at least) at Borders for $38.00 USD, unless I bought
the last copy.
David A. Smith
vernonner3voltazim
> Vernon, you cannot have momentum independent of mass. Momentum is mass
> times velocity.
That is the conventional definition, true. It need not become
dogmatic, however (more below).
> > I wish
> > you to think about that distinctness to the extent that
> > "could some amount of momentum exist all by itself -- a blob
> > of PURE momentum?" THAT would be the "quantum of momentum"
>
> What do you propose is the basis for this quantum momentum? Do you have a
> value for it?
Hmmm...I cannot claim to be expert on the mathematics of quanta.
But I do have a pretty good logical argument, which is presented
at the Web page previously listed:
http://www.halfbakery.com/idea/Gravity_20Waves#1025890913
Didn't you go look? It's OK if you didn't; I'm going to post
the key elements of the argument below.
> > And, as I tried to show in that
> > paper, it might be a good candidate for the graviton, too.
>
> I think you are making a good effort in your paper. I'm giving you the
> opportunity to defend your paper against a fair criticism.
> Dave
Thank you!
OK, Start with the assumption that JUST BECAUSE ordinary mass/energy
exists, then "negative mass/energy" might also exist. This is not
a horrid assumption, partly because others have proposed it before
(going at least as far back as an Isaac Asimov essay, "I'm Looking
Over a Four-Leaf Clover", first published in The Magazine of Fantasy
and Science Fiction, and reprinted in Asimov's nonfiction book
"Science, Numbers and I" (about 1965)).
SO..., take a unit of ordinary mass and let it collide with a unit of
negative mass. Nothing should remain behind, right? (Dr. Robert L.
Forward did some of his own speculating about negative mass/energy,
and labelled this type of expectable destruction as "nullification".)
But something WILL be left behind! Look:
(m)(v)----> <----(-m)(-v) Assume equal/opposite masses/velocities
Well, if we can say (m)(v) is "one momentum unit", then (-m)(-v) will
ALSO equal "one momentum unit". After nullification, there will be
NO leftover mass, NO leftover kinetic energy, but TWO momentum units!
(And you can use Relativity to change the reference frame for this
interaction all you want [example, set (v) to 2 and (-v) to 0], and
the result will ALWAYS be two momentum units. Such constancy,
independent of reference frame, suggests to me that "quantum of
momentum" is not a bad idea....)
So, to whatever extent that it may be possible for negative mass/energy
to exist, one must be immediately ready to start wrapping one's mind
around the logical consequence that momentum might be so Fundamental
a thing that it can indeed exist independently of mass or energy.
Q.E.D., and enjoy!
P.S. Doesn't General Relativity say somewhere, something to the
effect that "If Mass is utterly destroyed, such as by falling into
a black hole, a gravity wave will be produced"? (Another possible
link between quantized momentum and gravitons....)
P.P.S. One argument against the existence of negative mass/energy
is: "Shouldn't it be able to pop into existence at the quantum level,
accompanied by equal quantities of ordinary mass/energy? Well,
where is it?!?" ANSWER: Conservation of Momentum prevents it!
(except temporarily, at the virtual-particle level....)
Robert Kolker
vernonner3voltazim wrote:
> So, to whatever extent that it may be possible for negative mass/energy
> to exist, one must be immediately ready to start wrapping one's mind
> around the logical consequence that momentum might be so Fundamental
> a thing that it can indeed exist independently of mass or energy.
> Q.E.D., and enjoy!
I pride myself on having some mental flexibility, but I am really having
a hard time understanding what negative mass could possibly mean. Mass
is a measure of inertia. It tells us how hard it is to get the body with
the mass into motion and to increase/decrease its speed. A body with 0
mass does not even need a push. As soon as it exists, it is going at
light speed. Example: A bound electron drops a quantum level and emitts
a photon. The photon comes into existence by virtue of the energy jump,
and zowie!, it is off and running. That also means infinite
acceleration. Does negative mass mean that a body possessing it is even
easier to set into motion or easier to slow down? Does it mean it will
go faster than light? If such a body had a velocity greater than c, it
would have relativistic mass, oooopppps! energy that was imaginary (or
complex). That is not even negative. If zero mass => infinite
acceleration (applied just for an infinitesimal time interval) what is
negative mass. Even greater than infinite acceleration? That does not
make much sense. Imaginary mass might mean acceleration backward in
time. That is a possibility, but where is negative mass in all this.
Look at it from the gravitational side. The gravitational mass of a body
tells us how much it bends spacetime ( or in Newton-speak, how much
force it exerts on other bodies). A zero mass particle with momentum
gravitates because it has energy (equal to its momentum). A particle
with less than zero mass gravitates less? In terms of the relativistic 4
vector, negative mass really does not make much sense.
Hep' me! Hep' me!
Bob Kolker
David Thomson
news:64050551.02100...@posting.google.com...
> > Energy = mass * velocity^2
> > Momentum = mass * velocity
> >
> Wrong. The relationship between energy (E), momentum (p) and mass (m) is:
> E^2 = (p*c)^2 + (m*c^2)^2
Since we are talking specifically about quantizing momentum the value for p
would be the mass of the electron times the speed of light. That is the
smallest quanta for momentum that there can be. The value for m would be
the mass of the electron.
What you have here is an inequality. The energy on the right side of the
equation is twice the energy on the left side. And your equation does
nothing to disprove that momentum is equal to mass times velocity or that
energy is equal to mass time velocity squared. In fact, your equation
supports these definitions.
If you tell me that a photon can't have mass, then I can tell you that mass
and momentum of a photon does not exist apart from its angular momentum.
You can't have either equation, neither p*c nor m*c^2 because a photon is
always in the state of angular momentum times the speed of light.
> For a photon, m=0 so:
> E^2 = (p*c)^2
> E = p*c
>
> Thus, for a photon, E and p differ only by a constant factor;
> and if you choose units such that c=1, then E = p.
I realize this voodoo math is what physicists live and die by today. The
equation E^2 = (p*c)^2 + (m*c^2)^2 should also apply to particles since the
universe does not discriminate between photon physics and particle physics.
If you apply this equation to the mass and momentum of the electron you get
an inequality.
You've really got to be out of your mind to believe that for E = p * c means
that c (which has the dimensions of m/sec) can be replaced by 1. Why even
bother with math if you're going to make the rules up as you go along? c is
equal specifically to the speed of light, not 1.
But if you're going to change the dimensions of c to equal 1, then you also
have to change the dimensions of E as well. E = p * c then becomes p = p,
not E = p.
Dave
David Thomson
Hi David, I paid $5 for my copy on half.com. I just explained in a reply to
Paul why the present physics is wrong. What you are being taught is voodoo
math. People make up the rules as they go along. Energy does not equal
momentum no matter how the numbers are cooked.
Also, energy does not equal mass. Energy is equal to mass times the
velocity of light squared. Just as in the case of momentum times c, if
you're going to divide mass times c^2 by c^2 to make it equal to one, you
also have to do it on the other side of the equation. You end up with mass
= mass, not energy = mass.
energy mass * c^2
------- = ------------
c^2 c^2
Dave
David Thomson
> OK, Start with the assumption that JUST BECAUSE ordinary mass/energy
> exists, then "negative mass/energy" might also exist. This is not
> a horrid assumption, partly because others have proposed it before
> (going at least as far back as an Isaac Asimov essay, "I'm Looking
> Over a Four-Leaf Clover", first published in The Magazine of Fantasy
> and Science Fiction, and reprinted in Asimov's nonfiction book
> "Science, Numbers and I" (about 1965)).
Agreed that one can imagine a negative mass, but a negative velocity is
relative to a vector. There is no such thing as an absolute negative
velocity. If two objects are heading toward each other they both have
positive velocities because the velocity is relative to the
mass-with-the-velocity, not the other mass. A negative mass times a
negative velocity is a positive momentum. But you have a negative mass
times a positive velocity which is equal to a negative momentum.
> SO..., take a unit of ordinary mass and let it collide with a unit of
> negative mass. Nothing should remain behind, right? (Dr. Robert L.
> Forward did some of his own speculating about negative mass/energy,
> and labelled this type of expectable destruction as "nullification".)
> But something WILL be left behind! Look:
> (m)(v)----> <----(-m)(-v) Assume equal/opposite masses/velocities
> Well, if we can say (m)(v) is "one momentum unit", then (-m)(-v) will
> ALSO equal "one momentum unit". After nullification, there will be
> NO leftover mass, NO leftover kinetic energy, but TWO momentum units!
The momenta nullify each other as well. There is a positive momentum and a
negative momentum.
> (And you can use Relativity to change the reference frame for this
> interaction all you want [example, set (v) to 2 and (-v) to 0], and
> the result will ALWAYS be two momentum units. Such constancy,
> independent of reference frame, suggests to me that "quantum of
> momentum" is not a bad idea....)
I pointed out the error in your suggestion by explaining that the velocity
of the negative mass must be positive in order for the positive mass and
negative mass to move toward each other. If your view were correct then
driving down the Interstate you will be going negative 70 miles per hour
while the cars driving in the opposite direction would be going 70 miles per
hour. Yet we all know that our speedometers don't work that way. And even
if you put your car into reverse and drove 70 miles per hour backward while
cars behind you were driving 70 miles per hour forward, you would still have
a net velocity of 140 miles per hour when you collide. So all velocity is
positive, even when it applies to a hypothetical negative mass.
> So, to whatever extent that it may be possible for negative mass/energy
> to exist, one must be immediately ready to start wrapping one's mind
> around the logical consequence that momentum might be so Fundamental
> a thing that it can indeed exist independently of mass or energy.
> Q.E.D., and enjoy!
What does exist in place of mass and energy at the subatomic level is
angular momentum. An electron, proton, neutron, or photon cannot exist
except as angular momentum. If the velocity or length associated with the
mass is altered in anyway, the particle ceases to exist, as happens in a
particle collider. Angular momentum is indivisible and talking about mass
or momentum of the particle apart from angular momentum is meaningless.
Dave
Bilge
>
>You've really got to be out of your mind to believe that for E = p * c
>means that c (which has the dimensions of m/sec) can be replaced by 1.
In case the idea never occured to you, _people_ made up the definitions
of velocity, mass, energy, and so on initially.
>Why even bother with math if you're going to make the rules up as you
>go along? c is equal specifically to the speed of light, not 1.
Because the proper way to specify speeds is by \beta == v/c, not v.
>But if you're going to change the dimensions of c to equal 1, then you also
>have to change the dimensions of E as well. E = p * c then becomes p = p,
>not E = p.
E^2 = p^2 + m^2
MeV^2 MeV^2 MeV^2
Bilge
Re: Quanta of Momentum to usenet:
>
>
>vernonner3voltazim wrote:
>> So, to whatever extent that it may be possible for negative mass/energy
>> to exist, one must be immediately ready to start wrapping one's mind
>> around the logical consequence that momentum might be so Fundamental
>> a thing that it can indeed exist independently of mass or energy.
>> Q.E.D., and enjoy!
>
>I pride myself on having some mental flexibility, but I am really having
>a hard time understanding what negative mass could possibly mean. Mass
>is a measure of inertia. It tells us how hard it is to get the body with
>the mass into motion and to increase/decrease its speed.
If you want to picture a negative mass, picture the vacuum as you would
the fermi surface of a semiconductor. Below the fermi surface, all of the
states are filled, and no interaction is possible without supplying enough
energy to raise a particle across the gap (here, 1.022 MeV). That produces
a particle-hole pair (e.g., particle/hole == electron/positron or vice
versa).
0.511 ---------- Dirac initially solved the problem of electrons
MeV falling into the negative energies "sea" by
postulating the levels were filled, since no two
0 ----------- fermions can have the same quantum numbers. This
picture was eventually replaced by a multiparticle
interpretation. This is exactly how an LED works,
-0.511 ----------- (except that the energies are much lower and the
MeV zero level is only zero within the semiconductor,
not the rest of the universes, however, note that
inside the semiconductor, it really does look similar,
and the reason one can create a particle-hole pair which
propagates is because neither the particle, nor hole
interacts with the atoms beneath the fermi surface. In
addition, the particle/hole pair do not have the mass
of the electron. They have effective masses determined by the potential
in the crystal (and yes, the holes really have masses).
> A body with 0 mass does not even need a push. As soon as it exists,
>it is going at light speed. Example: A bound electron drops a quantum
>level and emitts a photon.
>The photon comes into existence by virtue of the energy jump, and zowie!,
Well,...
>it is off and running. That also means infinite acceleration.
You'll notice that it is impossible for a free charge to emit a photon.
Given an electron with an initial 4-momentum p^{u} = (E,p), a final
4-momentum p^{u}' = (E',p') and a photon with 4-momentum q^{u} = (w, k),
you can't conserve energy and momentum:
p^2 = (p' + q)^2 = p'^2 + q^2 + 2q.p'
m^2 = m^2 + 0 + \gamma m w - m\gamma\beta.k
0 = w - \beta.k
which only works if the final electron velocity is 1 (i.e., v = c), or
the frequency (energy) of the photon is zero. Which is convenient, since a
non-zero photon mass would allow the electron to radiate itself away. But,
this would also be true for an atom, if the emission involved only the
recoil of the center of momentum of the atom. So, this is part of why a
change in angular momentum is necessary. Then your problem of "infinite
acceleration" goes away. If the agular momentum of the atom changes by
\hbar and the photon has a spin of \hbar, then angular momentum is
conserved. Similarly, if the angular momentum of the electron changes by
\hbar, then the product mvr has to change by hbar. If you play around with
that just using ordinary classical mechanics and include relativity so
far as you'll need the mass-energy relation, you'll (re)-discover all sorts
of results from the "old"-quantum theory, circa the bohr model.
>Does negative mass mean that a body possessing it is even
>easier to set into motion or easier to slow down? Does it mean it will
>go faster than light?
Look at it this way. If you "add" a "negative mass" to a proton
and electron, the proton and electron are no longer on mass-shell and
can only exist as a hydrogen atom. A hydrogen atom has a mass less than
that of its constituents.
>If such a body had a velocity greater than c, it
>would have relativistic mass, oooopppps! energy that was imaginary (or
>complex).
>Look at it from the gravitational side. The gravitational mass of a body
>tells us how much it bends spacetime ( or in Newton-speak, how much
>force it exerts on other bodies). A zero mass particle with momentum
>gravitates because it has energy (equal to its momentum). A particle
>with less than zero mass gravitates less? In terms of the relativistic 4
>vector, negative mass really does not make much sense.
What happens in hawking radiation? The simplest picture is that of a
radiated photon and a negative energy photon "falling" back into the
black hole so that it loses mass. That reduces the curvature.
What's hard to picture is a negative energy density in free space.
Paul Cardinale
> "Paul Cardinale" <pcard...@volcanomail.com> wrote in message
> news:64050551.02100...@posting.google.com...
> > > Energy = mass * velocity^2
> > > Momentum = mass * velocity
> > >
> > Wrong. The relationship between energy (E), momentum (p) and mass (m) is:
> > E^2 = (p*c)^2 + (m*c^2)^2
>
> Since we are talking specifically about quantizing momentum the value for p
> would be the mass of the electron times the speed of light. That is the
> smallest quanta for momentum that there can be.
>
> The value for m would be
> the mass of the electron.
>
> What you have here is an inequality.
>
No. That equation is valid in all cases. The equations you're using:
> > > Energy = mass * velocity^2
> > > Momentum = mass * velocity
are valid only when v is much smaller than c.
<snip>
>
> > For a photon, m=0 so:
> > E^2 = (p*c)^2
> > E = p*c
> >
> > Thus, for a photon, E and p differ only by a constant factor;
> > and if you choose units such that c=1, then E = p.
>
> I realize this voodoo math is what physicists live and die by today.
>
That's ordinary algebra; but I suppose that might seem like voodoo to a moron.
> The
> equation E^2 = (p*c)^2 + (m*c^2)^2 should also apply to particles since the
> universe does not discriminate between photon physics and particle physics.
>
Right. Photons are particles too, but with m=0.
> If you apply this equation to the mass and momentum of the electron you get
> an inequality.
>
No. Its valid. The classical equations you're using are not valid.
> You've really got to be out of your mind to believe that for E = p * c means
> that c (which has the dimensions of m/sec) can be replaced by 1.
>
Reread what I wrote. Try to pay attention this time.
> Why even
> bother with math if you're going to make the rules up as you go along? c is
> equal specifically to the speed of light, not 1.
>
The numerical value of c depends on the units chosen.
If I choose to use distance units of light-years, and time units of years,
then c = 1 light-year/year.
> But if you're going to change the dimensions of c to equal 1, then you also
> have to change the dimensions of E as well.
>
I'm not changing the dimensions of anything.
I'm selecting units in order to get a convenient value for c.
> E = p * c then becomes p = p,
> not E = p.
>
Wow. You really don't understand math at all.
Let's try really slowly:
E = p * c
Now we choose units such that c = 1.
This means that wherever there is a "c" we can substitute "1". So:
E = p * 1
Since 1 is the multiplicitive indentity, p * 1 is equal to p.
This means that wherever there is "p * 1" we can substitute "p", So:
E = p
Paul Cardinale
David Thomson
news:64050551.02100...@posting.google.com...
> "David Thomson" <ne...@volantis.org> wrote in message
news:<jPvm9.4489$lV3.3...@newsread1.prod.itd.earthlink.net>...
> > "Paul Cardinale" <pcard...@volcanomail.com> wrote in message
> > news:64050551.02100...@posting.google.com...
> > > > Energy = mass * velocity^2
> > > > Momentum = mass * velocity
> > > >
> > > Wrong. The relationship between energy (E), momentum (p) and mass (m)
is:
> > > E^2 = (p*c)^2 + (m*c^2)^2
> >
> > Since we are talking specifically about quantizing momentum the value
for p
> > would be the mass of the electron times the speed of light. That is the
> > smallest quanta for momentum that there can be.
> >
> D.O.A. Momenta smaller than that are observed.
Technically, yes, neutrinos may have a smaller unit of momentum than
electromagnetic systems. But for the purpose of dealing with energy, where
energy is most often transfered by moving electrons, we must work with the
electron as the quantum mass.
The smallest unit of momentum is the smallest unit of the mass times the
smallest length it can travel divided by the smallest time interval it can
travel in. The smallest length the electron can travel divided by the
smallest time it can travel in is the speed of light. So the smallest unit
of momentum of the electron is the mass of the electron times the speed of
light, which is 2.731 x 10^-22 kg*m/sec. You can't get a smaller unit of
momentum for the electron.
> > The value for m would be
> > the mass of the electron.
> >
> > What you have here is an inequality.
> >
> No. That equation is valid in all cases. The equations you're using:
> > > > Energy = mass * velocity^2
> > > > Momentum = mass * velocity
> are valid only when v is much smaller than c.
We're talking about dimensions here. The dimensions are always correct
regardless of what the velocity is. You're trying to prove that energy is
the same thing as momentum, which it is not.
> > > For a photon, m=0 so:
> > > E^2 = (p*c)^2
> > > E = p*c
> > >
> > > Thus, for a photon, E and p differ only by a constant factor;
> > > and if you choose units such that c=1, then E = p.
> >
> > I realize this voodoo math is what physicists live and die by today.
> >
> That's ordinary algebra; but I suppose that might seem like voodoo to a
moron.
Ordinary algebra is valid math. Making up new math rules on the fly for
physics is unscientific and illogical. You cannot make energy equal
momentum. Period.
> > The
> > equation E^2 = (p*c)^2 + (m*c^2)^2 should also apply to particles since
the
> > universe does not discriminate between photon physics and particle
physics.
> >
> Right. Photons are particles too, but with m=0.
Electrons are particles. But they don't work with this equation. Electrons
have both mass and momentum.
> > If you apply this equation to the mass and momentum of the electron you
get
> > an inequality.
> >
> No. Its valid. The classical equations you're using are not valid.
E^2 = (p*c)^2 + (m*c^2)^2
In the case of the electron, what you have is
energy^2 = energy^2 + energy^2.
If energy has the value of 1, you get 1 = 1 + 1. That is not a valid
equality.
> > You've really got to be out of your mind to believe that for E = p * c
means
> > that c (which has the dimensions of m/sec) can be replaced by 1.
> >
> Reread what I wrote. Try to pay attention this time.
I had to read what your wrote very carefully in order to respond so
accurately and clearly.
> > Why even
> > bother with math if you're going to make the rules up as you go along?
c is
> > equal specifically to the speed of light, not 1.
> >
> The numerical value of c depends on the units chosen.
> If I choose to use distance units of light-years, and time units of years,
> then c = 1 light-year/year.
You could use boogers per elephant if you want. But you have to divide both
sides by the same values. You can't just arbitrarily divide c by a quantity
and not divide energy by the same quantity.
> > But if you're going to change the dimensions of c to equal 1, then you
also
> > have to change the dimensions of E as well.
> >
> I'm not changing the dimensions of anything.
> I'm selecting units in order to get a convenient value for c.
c has the same equivalent value and dimensions no matter what units you use.
The speed of light is a specific velocity. You can't change the outcome of
an equation by merely choosing a convenient value.
> > E = p * c then becomes p = p,
> > not E = p.
> >
> Wow. You really don't understand math at all.
> Let's try really slowly:
> E = p * c
> Now we choose units such that c = 1.
> This means that wherever there is a "c" we can substitute "1". So:
> E = p * 1
Did you flunk math? E is equal to p * c, you said so yourself in your
equation above. You have:
p * c = p * c
If you divide out c from one side of the equation, you have to divide it
from the other side as well. You end up with
p = p
You can't just arbitrarily divide c from just one side of the equation!
> Since 1 is the multiplicitive indentity, p * 1 is equal to p.
> This means that wherever there is "p * 1" we can substitute "p", So:
> E = p
Do you have a degree of some sort? Where did you learn to do math like
this? Did you learn this method from a book somewhere? If so, I would like
to know what the title is and write to the author and publisher. If this is
published material it is a complete disgrace to physics.
Dave
Stephen Speicher
> "Paul Cardinale" <pcard...@volcanomail.com> wrote in message
> news:64050551.02100...@posting.google.com...
> >
> > Thus, for a photon, E and p differ only by a constant factor;
> > and if you choose units such that c=1, then E = p.
>
> You've really got to be out of your mind to believe that for E = p * c means
> that c (which has the dimensions of m/sec) can be replaced by 1. Why even
> bother with math if you're going to make the rules up as you go along? c is
> equal specifically to the speed of light, not 1.
>
You have repeatedly been told that _before_ you are critical of
modern physics, and _before_ you develop a replacement theory,
that you should at least become familiar with what modern physics
is, so that you do not spend all your time arguing against straw
men, or chasing your tail. The above makes clear you have not
even cracked open a basic book on physics, and by continually
failing to do so you condemn yourself to a state of self-imposed
ignorance. Do not wonder why your nonsense is not taken
seriously, when you prefer to remain ignorant rather than gain
knowledge.
David McAnally
>"Paul Cardinale" <pcard...@volcanomail.com> wrote in message
>news:64050551.02100...@posting.google.com...
>> D.O.A. Momenta smaller than that are observed.
>Technically, yes, neutrinos may have a smaller unit of momentum than
>electromagnetic systems. But for the purpose of dealing with energy, where
>energy is most often transfered by moving electrons, we must work with the
>electron as the quantum mass.
>The smallest unit of momentum is the smallest unit of the mass times the
>smallest length it can travel divided by the smallest time interval it can
>travel in.
This contradicts even the most elementary mathematics. As a divisor
decreases, the quotient INcreases. So even if you are only going to use
naive mathematical principles, the smallest unit of momentum would be the
smallest unit of mass times the smallest length divided by the LARGEST
time interval. That is assuming that everything is quantized.
>The smallest length the electron can travel divided by the
>smallest time it can travel in is the speed of light.
This statement was made without justification. HOW do you know that it is
the speed of light?
>E^2 = (p*c)^2 + (m*c^2)^2
>In the case of the electron, what you have is
>energy^2 = energy^2 + energy^2.
This is a statement involving dimensions.
>If energy has the value of 1, you get 1 = 1 + 1. That is not a valid
>equality.
This logic is invalid. You are assumimg that p*c = E and m*c^2 = E
without any justification. Your statement above was statement about
DIMENSIONS and was not quantitative. You are assuming a quantitative
result from a non-quantitative statement. You might as well take the
classical equation E = 1/2*m*v^2 + m*g*h, determine that that means that
energy = energy + energy, and conclude that that means that 1 = 1 + 1,
because that uses *exactly* the same logic, and is just as invalid in that
case as it is in yours.
David McAnally
------
Stephen Speicher
> "dl...@aol.com (formerly)" <dee-ell-...@cox.net (use dlzc1 in
> beginning)> wrote in message
> news:marm9.143184$Pf7.6...@news1.west.cox.net...
> >
> > "Spacetime Physics", 2nd edition
> > Edwin F. Taylor & John Archibald Wheeler
> >
> > Currently available (at least) at Borders for $38.00 USD, unless I bought
> > the last copy.
>
> Hi David, I paid $5 for my copy on half.com.
Studies have shown that mere ownership of books has little effect
on one's overall knowledge. Apparently actually reading the book
seems to be more effective.
> I just explained in a reply to
> Paul why the present physics is wrong. What you are being taught is voodoo
> math.
No. You just explained to Paul that you yourself choose to remain
so ignorant.
Stephen Speicher
> "Paul Cardinale" <pcard...@volcanomail.com> wrote in message
> news:64050551.02100...@posting.google.com...
> > E = p
>
> Do you have a degree of some sort? Where did you learn to do math like
> this? Did you learn this method from a book somewhere? If so, I would like
> to know what the title is and write to the author and publisher. If this is
> published material it is a complete disgrace to physics.
>
In another posting you claim to own the book. However, as I
pointed out in a reply to that posting, studies have shown that
mere ownership of a book does not have too great an impact on
one's knowledge. Apparently you have to actually _read_ the book,
and even then you must be on the premise of trying to understand.
Unederstand?
dlzc@aol.com (formerly)
> > > > Not true for the photon, where Energy and Momentum are equivalent.
> Energy does not equal
> momentum no matter how the numbers are cooked.
>
> Also, energy does not equal mass.
Right. Your intial claim was that energy and mass were equivalent for a
particle at rest. I extended this, in like vein, to the photon.
I did not say "equal", I said "equivalent"... your exact choice of words.
So if you decide the energy is not equivalent to momentum, then energy is c
times less equivalent to mass.
David A. Smith
Patrick Reany
David Thomson wrote:
> "Paul Cardinale" <pcard...@volcanomail.com> wrote in message
> [snip]
>
>
> > No. That equation is valid in all cases. The equations you're using:
> > > > > Energy = mass * velocity^2
> > > > > Momentum = mass * velocity
> > are valid only when v is much smaller than c.
>
> We're talking about dimensions here. The dimensions are always correct
> regardless of what the velocity is. You're trying to prove that energy is
> the same thing as momentum, which it is not.
>
> > > > For a photon, m=0 so:
> > > > E^2 = (p*c)^2
> > > > E = p*c
> > > >
> > > > Thus, for a photon, E and p differ only by a constant factor;
> > > > and if you choose units such that c=1, then E = p.
> > >
> > > I realize this voodoo math is what physicists live and die by today.
> > >
> > That's ordinary algebra; but I suppose that might seem like voodoo to a
> moron.
>
> Ordinary algebra is valid math. Making up new math rules on the fly for
> physics is unscientific and illogical.
Translation: I am in "debunking" mode, so I
don't have to try to determine if the other guy
is logical or not: If it looks fishy prima facie, then
it must be fishy, right?
> You cannot make energy equal
> momentum. Period.
You cannot make cans of peas equal money,
right? You can if you have the right conversion
factor to use. Grocery stores do it all the time.
Dave, light speed is given frequently in
both meters/sec and miles/sec. The numbers
are different between them. Which is the
TRUE unit system for velocity?
> You can't change the outcome of
> an equation by merely choosing a convenient value.
>
> > > E = p * c then becomes p = p,
> > > not E = p.
> > >
> > Wow. You really don't understand math at all.
> > Let's try really slowly:
> > E = p * c
> > Now we choose units such that c = 1.
> > This means that wherever there is a "c" we can substitute "1". So:
> > E = p * 1
>
> Did you flunk math? E is equal to p * c, you said so yourself in your
> equation above. You have:
>
> p * c = p * c
>
> If you divide out c from one side of the equation, you have to divide it
> from the other side as well. You end up with
>
> p = p
>
> You can't just arbitrarily divide c from just one side of the equation!
You can if it is equal to unity in some appropriate
unit system.
> > Since 1 is the multiplicitive indentity, p * 1 is equal to p.
> > This means that wherever there is "p * 1" we can substitute "p", So:
> > E = p
>
> Do you have a degree of some sort? Where did you learn to do math like
> this? Did you learn this method from a book somewhere? If so, I would like
> to know what the title is and write to the author and publisher. If this is
> published material it is a complete disgrace to physics.
>
> Dave
Dave, what is wrong with you that you've
yet to get it that this is nothing more than
a short-hand convention used to reduce
writing in long computations. All it is, is
to adopt for the sake of simplification a unit
of distance and time such that the speed
of light is numerically unity in this system.
I am surprised that you haven't seen this
before, since you're so accomplished
a student of relativity.
There is a sort of precedence to this
from E&M, by choosing either MKS
or cgs units one can simplify the units
to one's preference.
As I said repeatedly, once a person adopts the
"I'm a debunker of this nonsense" mode, that
person is at risk of loosing the ability to think
rationally and objectively. So, don't try this
at home, ladies and gentlemen. Leave debunking
to the experts!
Patrick
vernonner3voltazim
> Agreed that one can imagine a negative mass, but a negative velocity is
> relative to a vector. There is no such thing as an absolute negative
> velocity. If two objects are heading toward each other they both have
> positive velocities because the velocity is relative to the
> mass-with-the-velocity, not the other mass. A negative mass times a
> negative velocity is a positive momentum. But you have a negative mass
> times a positive velocity which is equal to a negative momentum.
I never said anything about "absolute negative velocity"!!!
Velocities are indeed relative to each other, and, yes, velocities
are vector values. They are a combination of "speed" (which is always
positive) and "direction" (which may be positive or negative, depending
on the reference frame). Negative velocities are quite possible simply
because they are the easy way to specify "a speed in the opposite
direction, relative to some first-specified positive velocity".
> > SO..., take a unit of ordinary mass and let it collide with a unit of
> > negative mass. Nothing should remain behind, right? (Dr. Robert L.
> > Forward did some of his own speculating about negative mass/energy,
> > and labelled this type of expectable destruction as "nullification".)
> > But something WILL be left behind! Look:
> > (m)(v)----> <----(-m)(-v) Assume equal/opposite masses/velocities
> > Well, if we can say (m)(v) is "one momentum unit", then (-m)(-v) will
> > ALSO equal "one momentum unit". After nullification, there will be
> > NO leftover mass, NO leftover kinetic energy, but TWO momentum units!
>
> The momenta nullify each other as well. There is a positive momentum and a
> negative momentum.
Sorry, you're still wrong. Let me break the diagram into smaller pieces:
(m)[(distance)/(time)]----> <----(-m)[(-distance)/(time)]
We still get positive momentum on both sides of the collision,
and therefore a positive total moment survives nullification.
> > (And you can use Relativity to change the reference frame for this
> > interaction all you want [example, set (v) to 2 and (-v) to 0], and
> > the result will ALWAYS be two momentum units. Such constancy,
> > independent of reference frame, suggests to me that "quantum of
> > momentum" is not a bad idea....)
>
> I pointed out the error in your suggestion by explaining that the velocity
> of the negative mass must be positive in order for the positive mass and
> negative mass to move toward each other. If your view were correct then
> driving down the Interstate you will be going negative 70 miles per hour
> while the cars driving in the opposite direction would be going 70 miles per
> hour. Yet we all know that our speedometers don't work that way.
Well, since they measure SPEED and not VELOCITY, it figures that
speedometers are called "speedometers". And speed is indeed always
positive!
> And even
> if you put your car into reverse and drove 70 miles per hour backward while
> cars behind you were driving 70 miles per hour forward, you would still have
> a net velocity of 140 miles per hour when you collide. So all velocity is
> positive, even when it applies to a hypothetical negative mass.
Actually, it has been shown that for equal-mass cars, the effect upon
each car, during collision, is the same as if the cars had instead
run into an immovable wall, that suddenly popped into existance between
them, just before the collision. Remember that those speedometers measure
speed RELATIVE TO THE GROUND, which is considered "stationary" in such
situations. So, since after the collision, each car is moving at zero
velocity, the decelleration has been from 70 to 0 mph, and so the value
of 140 mph never enters the picture. To get the kind of collision that
you describe, both cars would have to end up, after the collision, moving
at 70mph in the direction opposite to their original motions (before the
collision). And since you know that inelastic collisions don't do that,
you are wrong, again.
> > So, to whatever extent that it may be possible for negative mass/energy
> > to exist, one must be immediately ready to start wrapping one's mind
> > around the logical consequence that momentum might be so Fundamental
> > a thing that it can indeed exist independently of mass or energy.
> > Q.E.D., and enjoy!
>
> What does exist in place of mass and energy at the subatomic level is
> angular momentum. An electron, proton, neutron, or photon cannot exist
> except as angular momentum. If the velocity or length associated with the
> mass is altered in anyway, the particle ceases to exist, as happens in a
> particle collider. Angular momentum is indivisible and talking about mass
> or momentum of the particle apart from angular momentum is meaningless.
Angular momentum is a whole 'nother kettle of fish, which is outside
the scope of the current discussion. But I wouldn't be one bit surprised
if it could exist totally independently of mass or energy also, as seems
possible for linear momentum.
----------------
I do need to clarify one thing about my prior message. When one changes
the reference frame in which most-simple nullification is described, then
there will be some leftover kinetic energy as well as momentum. This is
why, in my paper at http://www.nemitz.net/vernon/MOMEQUAT.htm
I defined a "momenton" as having both momentum and velocity. See, if the
"blob" of pure momentum (p) that I have been discussing here, moves at
some velocity (v), then it could easily encompass the leftover kinetic
energy just mentioned, as the product (p)(v). [Yes, I used (s) for
speed in that paper, but I was merely trying to keep two different
velocities (the other was that of the mass) from being confused, in
the equations.]
Paul Cardinale
> "Paul Cardinale" <pcard...@volcanomail.com> wrote in message
> news:64050551.02100...@posting.google.com...
> > "David Thomson" <ne...@volantis.org> wrote in message
> news:<jPvm9.4489$lV3.3...@newsread1.prod.itd.earthlink.net>...
> > > "Paul Cardinale" <pcard...@volcanomail.com> wrote in message
> > > news:64050551.02100...@posting.google.com...
> > > > > Energy = mass * velocity^2
> > > > > Momentum = mass * velocity
> > > > >
> > > > Wrong. The relationship between energy (E), momentum (p) and mass (m)
> is:
> > > > E^2 = (p*c)^2 + (m*c^2)^2
> > >
> > > Since we are talking specifically about quantizing momentum the value
> for p
> > > would be the mass of the electron times the speed of light. That is the
> > > smallest quanta for momentum that there can be.
> > >
> > D.O.A. Momenta smaller than that are observed.
>
> Technically, yes, neutrinos may have a smaller unit of momentum than
> electromagnetic systems. But for the purpose of dealing with energy, where
> energy is most often transfered by moving electrons, we must work with the
> electron as the quantum mass.
>
quantized.
> The smallest unit of momentum is the smallest unit of the mass times the
> smallest length it can travel divided by the smallest time interval it can
> travel in. The smallest length the electron can travel divided by the
> smallest time it can travel in is the speed of light. So the smallest unit
> of momentum of the electron is the mass of the electron times the speed of
> light, which is 2.731 x 10^-22 kg*m/sec. You can't get a smaller unit of
> momentum for the electron.
>
Pathetically wrong. Try learning something before you make a fool of
yourself publicly.
> > > The value for m would be
> > > the mass of the electron.
> > >
> > > What you have here is an inequality.
> > >
> No. That equation is valid in all cases. The equations you're using:
> > > > > Energy = mass * velocity^2
> > > > > Momentum = mass * velocity
> > are valid only when v is much smaller than c.
>
> We're talking about dimensions here. The dimensions are always correct
> regardless of what the velocity is.
>
Just because the dimensions match, doesn't mean that the equations are
valid.
(And in this case, they're not.)
> You're trying to prove that energy is
> the same thing as momentum, which it is not.
>
In the case of the photon, E = p * c.
i.e. energy is proportional to momentum, and with the right units,
the numerical values are the same. Thus the equivalence.
> > > > For a photon, m=0 so:
> > > > E^2 = (p*c)^2
> > > > E = p*c
> > > >
> > > > Thus, for a photon, E and p differ only by a constant factor;
> > > > and if you choose units such that c=1, then E = p.
> > >
> > > I realize this voodoo math is what physicists live and die by today.
> > >
> > That's ordinary algebra; but I suppose that might seem like voodoo to a
> moron.
>
> Ordinary algebra is valid math. Making up new math rules on the fly for
> physics is unscientific and illogical. You cannot make energy equal
> momentum. Period.
>
> > > The
> > > equation E^2 = (p*c)^2 + (m*c^2)^2 should also apply to particles since
> the
> > > universe does not discriminate between photon physics and particle
> physics.
> > >
> > Right. Photons are particles too, but with m=0.
>
> Electrons are particles. But they don't work with this equation. Electrons
> have both mass and momentum.
>
The equation E^2 = (p*c)^2 + (m*c^2)^2 is valid (empirically) for all
particles.
> > > If you apply this equation to the mass and momentum of the electron you
> get
> > > an inequality.
> > >
> > No. Its valid. The classical equations you're using are not valid.
>
> E^2 = (p*c)^2 + (m*c^2)^2
>
> In the case of the electron, what you have is
>
> energy^2 = energy^2 + energy^2.
>
> If energy has the value of 1, you get 1 = 1 + 1. That is not a valid
> equality.
>
Stupidly wrong. Here you assumed that E^2 = (p*c)^2 = (m*c^2)^2.
Which is invalid.
> > > You've really got to be out of your mind to believe that for E = p * c
> means
> > > that c (which has the dimensions of m/sec) can be replaced by 1.
> > >
> > Reread what I wrote. Try to pay attention this time.
>
> I had to read what your wrote very carefully in order to respond so
> accurately and clearly.
>
I wrote: "Thus, for a photon, E and p differ only by a constant
factor; and if you choose units such that c=1, then E = p."
Which is not even close to your assertion that I "believe that for E =
p * c means that c (which has the dimensions of m/sec) can be replaced
by 1."
1. The dimensions of c are distance/time. m/sec are units, not the
dimensions.
2. As I showed you, but you couldn't learn, choosing the appropriate
UNITS results in c having a numerical value of 1 (the dimensions
remain the same).
> > > Why even
> > > bother with math if you're going to make the rules up as you go along?
> c is
> > > equal specifically to the speed of light, not 1.
> > >
> > The numerical value of c depends on the units chosen.
> > If I choose to use distance units of light-years, and time units of years,
> > then c = 1 light-year/year.
>
> You could use boogers per elephant if you want. But you have to divide both
> sides by the same values. You can't just arbitrarily divide c by a quantity
> and not divide energy by the same quantity.
>
I didn't divide c by anything. I chose units such that c = 1.
Naturally, the units of everything else in the equation must match.
> > > But if you're going to change the dimensions of c to equal 1, then you
> also
> > > have to change the dimensions of E as well.
> > >
> > I'm not changing the dimensions of anything.
> > I'm selecting units in order to get a convenient value for c.
>
> c has the same equivalent value and dimensions no matter what units you use.
> The speed of light is a specific velocity. You can't change the outcome of
> an equation by merely choosing a convenient value.
>
Depends what you mean by "outcome".
Certainly the numerical values will come out differently depending on
choice of units, but the relationship defined by the equation doesn't
change.
> > > E = p * c then becomes p = p,
> > > not E = p.
> > >
> > Wow. You really don't understand math at all.
> > Let's try really slowly:
> > E = p * c
> > Now we choose units such that c = 1.
> > This means that wherever there is a "c" we can substitute "1". So:
> > E = p * 1
>
> Did you flunk math? E is equal to p * c, you said so yourself in your
> equation above. You have:
>
> p * c = p * c
>
> If you divide out c from one side of the equation, you have to divide it
> from the other side as well. You end up with
>
> p = p
>
> You can't just arbitrarily divide c from just one side of the equation!
>
When I was in 3rd grade, I learned that multiplying by 1 is the same
as not multiplying at all. Perhaps you dropped out before that grade?
> > Since 1 is the multiplicitive indentity, p * 1 is equal to p.
> > This means that wherever there is "p * 1" we can substitute "p", So:
> > E = p
>
> Do you have a degree of some sort? Where did you learn to do math like
> this?
>
Where did I learn that p * 1 = p? I learned the concept in 3rd grade,
but I didn't get to algebra until grade 9.
(Try this on a calculator: multiply any number by 1 and you get the
original number! Just like magic!)
> Did you learn this method from a book somewhere? If so, I would like
> to know what the title is and write to the author and publisher. If this is
> published material it is a complete disgrace to physics.
>
So what do you think p * 1 is if not p?
Paul Cardinale
David Thomson
news:anfuqb$q76$1...@bunyip.cc.uq.edu.au...
> >The smallest unit of momentum is the smallest unit of the mass times the
> >smallest length it can travel divided by the smallest time interval it
can
> >travel in.
>
> This contradicts even the most elementary mathematics. As a divisor
> decreases, the quotient INcreases. So even if you are only going to use
> naive mathematical principles, the smallest unit of momentum would be the
> smallest unit of mass times the smallest length divided by the LARGEST
> time interval. That is assuming that everything is quantized.
Yes, you are correct that a smaller value is produced by increasing the
time. I was talking about the smallest unit of momentum *in terms of
quanta.* The discussion in this thread was about devising a quantum of
momentum. The dimensions of momentum (and all other units) are what can be
quantized, not the derived units. Every unit that has a fraction will
encounter the same problem.
> >The smallest length the electron can travel divided by the
> >smallest time it can travel in is the speed of light.
>
> This statement was made without justification. HOW do you know that it is
> the speed of light?
I really don't know this. I deduce this.
The speed of light in a vacuum is accurately measured. The Compton
wavelength is widely accepted as the quantum wavelength. This leaves the
quotient of Compton wavelength divided by the speed of light as the quantum
time. It must be a quantum time if the speed of light is constant and the
Compton wavelength is quantum.
> >E^2 = (p*c)^2 + (m*c^2)^2
>
> >In the case of the electron, what you have is
>
> >energy^2 = energy^2 + energy^2.
>
> This is a statement involving dimensions.
And...??? Is that your scientific rebuttal? I not only make a statement of
dimensions (which proves I'm right) I also make a statement of values.
Since energy is the common dimension, the dimensions can be divided out of
the equation and leave us only with values.
1 = 1 + 1
You and the entire scientific establishment that believes this voodoo math
are simply wrong. The proof is clear. There is no scientific rebuttal to
my argument.
> >If energy has the value of 1, you get 1 = 1 + 1. That is not a valid
> >equality.
>
> This logic is invalid. You are assumimg that p*c = E and m*c^2 = E
> without any justification.
Excuse me? Hello? Is anybody home?
Tell us what p*c is equal to. Tell us what m*c^2 is equal to. It is
standard physics knowledge that both of these products are equal to energy.
I have plenty of justification for making this point. This is exactly why
modern physics is so full of crap. There are guys like you who change the
rules whenever it seems convenient.
> Your statement above was statement about
> DIMENSIONS and was not quantitative. You are assuming a quantitative
> result from a non-quantitative statement. You might as well take the
> classical equation E = 1/2*m*v^2 + m*g*h, determine that that means that
> energy = energy + energy, and conclude that that means that 1 = 1 + 1,
> because that uses *exactly* the same logic, and is just as invalid in that
> case as it is in yours.
Exactly. You are exactly right on that. That is another example of voodoo
math in physics. Another one is the equation for energy in a capacitor:
E = 1/2 CV^2
The equation for the energy stored in a capacitor only looks at the energy
stored in the electric field. It completely ignores the fact that there is
an equal amount of energy stored in the dielectric of the capacitor. Common
sense would dictate that a capacitor could not spontaneously discharge
energy without having a force behind it to push the energy out. But this
can be mathematically proven as well. Since the force is doing work, if the
only energy in the capacitor were the energy in the electric field, then a
capacitor would drain a circuit of energy due to the work of pushing the
charge off the plates.
There are many areas of physics where voodoo math is brought into play in
order to conform to the present model. Instead of designing a physics model
that fits the facts, the physicists have altered the math so the facts would
fit the model.
Dave
David Thomson
news:64050551.02100...@posting.google.com...
> > > > > > Energy = mass * velocity^2
> > > > > > Momentum = mass * velocity
> > > are valid only when v is much smaller than c.
> >
> > We're talking about dimensions here. The dimensions are always correct
> > regardless of what the velocity is.
> >
> Just because the dimensions match, doesn't mean that the equations are
> valid.
> (And in this case, they're not.)
Give a mathematical example in the real world where the equations are not
valid.
> > You're trying to prove that energy is
> > the same thing as momentum, which it is not.
> >
> In the case of the photon, E = p * c.
> i.e. energy is proportional to momentum, and with the right units,
> the numerical values are the same. Thus the equivalence.
I agree that energy is proportional to momentum. But that isn't what you
were saying. In the previous message you said:
> Let's try really slowly:
> E = p * c
> Now we choose units such that c = 1.
> This means that wherever there is a "c" we can substitute "1". So:
> E = p * 1
There is a world of a difference between saying something is proportional
and something is equal. You tried to show equivalence by violating the laws
of mathematics.
> The equation E^2 = (p*c)^2 + (m*c^2)^2 is valid (empirically) for all
particles.
This is your unproven, personal opinion.
> > E^2 = (p*c)^2 + (m*c^2)^2
> >
> > In the case of the electron, what you have is
> >
> > energy^2 = energy^2 + energy^2.
> >
> > If energy has the value of 1, you get 1 = 1 + 1. That is not a valid
> > equality.
> >
> Stupidly wrong. Here you assumed that E^2 = (p*c)^2 = (m*c^2)^2.
> Which is invalid.
Trying to have it both ways are you?
In a previous post you stated:
>Wrong. The relationship between energy (E), momentum (p) and mass (m) is:
>E^2 = (p*c)^2 + (m*c^2)^2
>
>For a photon, m=0 so:
>E^2 = (p*c)^2
>E = p*c
You clearly showed that YOU believe E^2 = (p*c)^2 when m = 0. But since the
momentum of the photon is equal to the mass of an electron times the speed
of light, you cannot give m = 0 and p <> 0 since they are both based on the
same source of mass. There are not two different masses associated with a
particle.
When you apply the equation E^2 = (p*c)^2 + (m*c^2)^2 to an electron, you
don't differentiate between two entirely different masses in the particle.
You assume the mass of the electron is what it is and that it also is the
component of its own momentum.
You're doing the standard physicist thing of violating the laws of
mathematics in order to make the data fit your model. You model doesn't
work. The equations don't work. It's a failure and you don't have the
ability to mathematically defend your argument nor the wherewithall to admit
defeat. All you have are personal opinions that are not rooted in the
scientific method.
> I wrote: "Thus, for a photon, E and p differ only by a constant
> factor; and if you choose units such that c=1, then E = p."
> Which is not even close to your assertion that I "believe that for E =
> p * c means that c (which has the dimensions of m/sec) can be replaced
> by 1."
> 1. The dimensions of c are distance/time. m/sec are units, not the
> dimensions.
> 2. As I showed you, but you couldn't learn, choosing the appropriate
> UNITS results in c having a numerical value of 1 (the dimensions
> remain the same).
I agree that the dimensions of c are distance divided by time. And velocity
is the unit. But don't tell me you didn't say that c could be replaced by
1.
You wrote:
> Let's try really slowly:
> E = p * c
> Now we choose units such that c = 1.
> This means that wherever there is a "c" we can substitute "1". So:
> E = p * 1
It's pretty clear that you DID say that. You're helplessly confused by the
brainwashing you received in school. Like a cult member, you have been told
so many times that 1 = 1 + 1 that you actually believe it. You have also
been convinced that you can mathematically replace c with 1 on one side of
an equation without performing the same operation on the other side of the
equation. You've been snookered. Admit it!
> I didn't divide c by anything. I chose units such that c = 1.
> Naturally, the units of everything else in the equation must match.
What right do you have for choosing units for c such that c = 1 without
altering the other side of the equation? What is the mathematical law that
allows for this? Don't give my your personal philosophy or personal
opinion. Let's talk math and science here. Where do you get the authority
to make c = 1 without altering the other side of the equation?
Let me put it this way. If you can arbitrarily assign c = 1, what is to
stop you from arbitrarily assigning c = 12 or p = 52.3 or E = -1? What
happens to the reliability of mathematics when you can just arbitrarily
assign a different unit value to a variable in an algebraic equation without
keeping the equation in balance?
E is not equivalent to p.
> > c has the same equivalent value and dimensions no matter what units you
use.
> > The speed of light is a specific velocity. You can't change the outcome
of
> > an equation by merely choosing a convenient value.
> >
> Depends what you mean by "outcome".
> Certainly the numerical values will come out differently depending on
> choice of units, but the relationship defined by the equation doesn't
> change.
It does change the relationship defined by the equation. In one moment you
have an equality that shows energy is equal to momentum times velocity. In
the next moment you have an equality that show energy is equal to just
momentum.
energy = momentum * velocity
energy = momentum
The relationship has been completely changed.
> > > > E = p * c then becomes p = p,
> > > > not E = p.
> > > >
> > > Wow. You really don't understand math at all.
> > > Let's try really slowly:
> > > E = p * c
> > > Now we choose units such that c = 1.
> > > This means that wherever there is a "c" we can substitute "1". So:
> > > E = p * 1
> >
> > Did you flunk math? E is equal to p * c, you said so yourself in your
> > equation above. You have:
> >
> > p * c = p * c
> >
> > If you divide out c from one side of the equation, you have to divide it
> > from the other side as well. You end up with
> >
> > p = p
> >
> > You can't just arbitrarily divide c from just one side of the equation!
> >
> When I was in 3rd grade, I learned that multiplying by 1 is the same
> as not multiplying at all. Perhaps you dropped out before that grade?
You seemed to be pretty smart in the third grade. Use your third grade
brains to follow this:
> > > Let's try really slowly:
> > > E = p * c
> > > Now we choose units such that c = 1.
> > > This means that wherever there is a "c" we can substitute "1". So:
> > > E = p * 1
Since c is not equal to 1, c * 1 equals c, not 1. Is that any easier to
understand? 1 is not a replacement for c unless c had dimension and value
of 1 to begin with.
> > > Since 1 is the multiplicitive indentity, p * 1 is equal to p.
> > > This means that wherever there is "p * 1" we can substitute "p", So:
> > > E = p
> >
> > Do you have a degree of some sort? Where did you learn to do math like
> > this?
> >
> Where did I learn that p * 1 = p?
No, don't try to hide by switching the argument. You said:
E = p * 1
which is the same as
E = p
Just read the messages you wrote to verify this.
Dave
David Thomson
<personal opinions snipped>
Nothing left to respond to. Patrick, if you are going to talk personal
opinions, I have no interest. All I'm interested in is verifiable science
using verifiable mathematics principles.
Dave
Patrick Reany
David Thomson wrote:
Then why did you bother to reply? (Rhetorical
question only) Is this a physics / relativity NG
or a mathematical NG, anyway? Maybe you're
on the wrong NG. Besides, there's more
to physics than facts. Where does meaning fit
into your sanitized notion of physics?
Patrick
David Thomson
> Dear "David Thomson":
>
> > > > > Not true for the photon, where Energy and Momentum are equivalent.
>
> > Energy does not equal
> > momentum no matter how the numbers are cooked.
> >
> > Also, energy does not equal mass.
>
> Right. Your intial claim was that energy and mass were equivalent for a
> particle at rest. I extended this, in like vein, to the photon.
You must be talking about someone else. I was saying that energy and
momentum were not equivalent. I don't claim that the subatomic particles or
the photon can ever be at rest. Further, energy is neither equivalent nor
equal to mass, at any time. The two are proportional.
> I did not say "equal", I said "equivalent"... your exact choice of words.
Like I said, I was talking about energy and momentum, not energy and mass.
Energy and mass are neither equal nor equivalent. They are proportional.
You are the one who said:
>Energy is equivalent to mass for a particle at rest, as much as energy is
>equivalent to momentum for a photon.
State this mathematically.
> So if you decide the energy is not equivalent to momentum, then energy is
c
> times less equivalent to mass.
Energy is c^2 times "less equivalent" than mass.
What pages are you suggesting that I read in Spacetime Physics to understand
your view?
Dave
David Thomson
> > Agreed that one can imagine a negative mass, but a negative velocity is
> > relative to a vector. There is no such thing as an absolute negative
> > velocity. If two objects are heading toward each other they both have
> > positive velocities because the velocity is relative to the
> > mass-with-the-velocity, not the other mass. A negative mass times a
> > negative velocity is a positive momentum. But you have a negative mass
> > times a positive velocity which is equal to a negative momentum.
>
> I never said anything about "absolute negative velocity"!!!
You had to imply absolute negative velocity or your who argument would
immediately fall apart.
You are clearly speaking about two separate entities. There is a positive
mass and a negative mass. These two entities cannot share the same
velocity, they must each have their own velocities, regardless of whether
they are heading toward each other, away from each other, or parallel to
each other. Further, you specifically stated:
(m)(v)----> <----(-m)(-v)
You're clearly showing the velocity of the negative particle as having a
negative velocity. If you maintain that the negative velocity of the
negative mass belongs to the positive mass, then the positive mass has a net
velocity of zero (assuming the two velocities have the same absolute value)
and so does the negative mass.
> Negative velocities are quite possible simply
> because they are the easy way to specify "a speed in the opposite
> direction, relative to some first-specified positive velocity".
The are two separate vectors. The positive or negative state of one vector
has no bearing whatsoever on the second vector unless there is an absolute
reference. Your relationship does not show a vector nor identify the
reference frame. If you're going to conclusively show there is a momentum
remaining, you need specify all the parameters.
> Sorry, you're still wrong. Let me break the diagram into smaller pieces:
> (m)[(distance)/(time)]----> <----(-m)[(-distance)/(time)]
> We still get positive momentum on both sides of the collision,
> and therefore a positive total moment survives nullification.
I'd like to think that I was wrong. But from the relationship you are
providing, there is no justification why the exact same distance between the
two masses should be positive for one and negative for the other. In other
words, do we live 500 miles apart from each other or do we live -500 miles
apart from each other? How could either of us live -500 miles apart from
the other?
> > I pointed out the error in your suggestion by explaining that the
velocity
> > of the negative mass must be positive in order for the positive mass and
> > negative mass to move toward each other. If your view were correct then
> > driving down the Interstate you will be going negative 70 miles per hour
> > while the cars driving in the opposite direction would be going 70 miles
per
> > hour. Yet we all know that our speedometers don't work that way.
>
> Well, since they measure SPEED and not VELOCITY, it figures that
> speedometers are called "speedometers". And speed is indeed always
> positive!
Is it your position then that someone could invent a velocitymeter and that
it would have both a positive and negative scale (depending on which side
the road you are driving on?)
> > And even
> > if you put your car into reverse and drove 70 miles per hour backward
while
> > cars behind you were driving 70 miles per hour forward, you would still
have
> > a net velocity of 140 miles per hour when you collide. So all velocity
is
> > positive, even when it applies to a hypothetical negative mass.
>
> Actually, it has been shown that for equal-mass cars, the effect upon
> each car, during collision, is the same as if the cars had instead
> run into an immovable wall, that suddenly popped into existance between
> them, just before the collision. Remember that those speedometers measure
> speed RELATIVE TO THE GROUND, which is considered "stationary" in such
> situations. So, since after the collision, each car is moving at zero
> velocity, the decelleration has been from 70 to 0 mph, and so the value
> of 140 mph never enters the picture.
Yes it does. Both cars are decelerating by 70 mph. The total deceleration
is 140 miles per hour. At the moment just before impact, both cars are
going a positive speed of 70 mph. Relatively speaking, if one had been
standing still, the other would be going 140 mph.
> To get the kind of collision that
> you describe, both cars would have to end up, after the collision, moving
> at 70mph in the direction opposite to their original motions (before the
> collision). And since you know that inelastic collisions don't do that,
> you are wrong, again.
A car is not inelastic. Sure, they are designed to absorb shock, but the
full shock of the other vehicle needs to be absorbed in addition to the
shock of the "wall". If you look at the wall as providing a 70 mph shock to
the vehicle striking it, there is half the shock. But since the other
vehicle is also moving 70 mph, there is an addition amount of force that has
to be absorbed and it is the force of a car going 70 mph. The total force
being absorbed is a force equivalent to a car traveling 140 mph into a wall.
> Angular momentum is a whole 'nother kettle of fish, which is outside
> the scope of the current discussion. But I wouldn't be one bit surprised
> if it could exist totally independently of mass or energy also, as seems
> possible for linear momentum.
Angular momentum is a unit composed of a dimension of mass, times a
dimension of length, times the dimensions of length divided by time. If you
take away any of the dimensions, the unit changes to something else. The
same principle holds true for all units including momentum. If there is no
mass left over, there is no momentum.
Dave
dlzc@aol.com (formerly)
> > Right. Your intial claim was that energy and mass were equivalent for a
> > particle at rest. I extended this, in like vein, to the photon.
>
> You must be talking about someone else. I was saying that energy and
> momentum were not equivalent. I don't claim that the subatomic particles
or
> the photon can ever be at rest. Further, energy is neither equivalent nor
> equal to mass, at any time. The two are proportional.
Yes, again you are correct...
vernonner3voltazim <vne...@pinn.net> said:
> As you
> know, [M]ass and Energy are equivalent entities, Space and Time
> are reasonably equivalent entities, and Momentum is another
> Fundamental Thing that is distinct from those others.
> You are the one who said:
>
> >Energy is equivalent to mass for a particle at rest, as much as energy is
> >equivalent to momentum for a photon.
>
> State this mathematically.
for a particle at rest: E = mc^2
for a photon: E=pc
with whatever value you choose for c, knowing the penalty for picking
something wierd.
> What pages are you suggesting that I read in Spacetime Physics to
understand
> your view?
I was responding erroneously to one of your correct statements. However you
said:
[DAS]
>> Momentum without mass is a photon, and its
>> momentum is a function of the frame that absorbs it. No real quantum
>> here, but a packet certainly...
[YOU]
>
> photon = h * c (units are angular momentum times velocity)
>
> momentum
> ----------- = velocity
> mass
>
>How does momentum without a mass equal a photon?
Division by zero is not a valid operation. A photon has been determined to
have a mass smaller than we can currently measure (very like zero), yet is a
carrier for measreuable momentum (other than angular). And yes that
momentum is proportional to the photon's energy divided by c.
The only particle I know of that carries momentum yet has no mass is a
photon. But my experience is very limited.
Better?
David A. Smith
David Thomson
> Besides, there's more to physics than facts.
LOL!!! Boy, you hit that right. That's why I posted in this thread; to held
dispel some of the fiction.
> Where does meaning fit into your sanitized notion of physics?
The proper math reveals the true Geometry of the Universe. The meaning is
in the Geometry.
Dave
Todd Desiato
"Bilge" <ro...@radioactivex.lebesque-al.net> wrote in message
news:slrnapljn...@radioactivex.lebesque-al.net...
> Robert Kolker said some stuff about
> Re: Quanta of Momentum to usenet:
> >
> What's hard to picture is a negative energy density in free space.
>
It's only hard to picture because you neglect the source of the field. To
"define" what you mean by negative energy for a vector field, such as an
electric field you must consider the source charge. If you have a positive
charge and a negative electric field, that defines a negative energy. I
believe that without consideration of the sign of the charge, the sign of
the energy density is unknown and indeterminate, not positive definite.
If you work in units with c=1, you would also choose mu0 =1 and eps0 = 1.
You then get for the energy density,
T^00 = (E^2 + B^2) / 2
and when shown like this you are completely ignoring the "ratios" between
the source strength and the field strength. A more accurate way to write
this would be the original, classical energy density in a medium;
T^00 = ((eps0*eps_rel)*E^2 + B^2/(mu0*mu_rel)) / 2
where the relative permeability and permittivity are taken into
consideration. These are simply the ratio between the field strength and the
source strength in the medium with relative properties eps_rel and mu_rel.
Look at the reflection of a charge or current in a mirror, these values are
negative. If you place a positive charge in front of a mirror you get a
negative electric field reflected in the mirror from the image of the
charge. So a negative energy density is just a reflection of a positive
energy density.
IMO the WEC and the SEC are a farce caused by a misconception that you can
neglect the ratio between the local vector field and the local field source.
The local field can be manipulated by a non-local field source, and then
locally the energy density can appear to be negative relative to the local
source. By non-local I mean a source that exists at a distant set of
coordinates. The effect of electromagnetism is week gravitationally, but to
a charge in an electromagnetic field, Lorentz forces are just geodesic
motion. So even a small relative change in the field strength is a curvature
to a charged particle. So while you may treat the gravitational space-time
as flat, and work with EM fields in flat space-time, you can also treat the
EM field like a geometric manifold with respect to the charged particle.
Then it is not so flat, and local isn't so local any more.
Best Regards,
Todd Desiato
vernonner3voltazim
> >
> > I never said anything about "absolute negative velocity"!!!
>
> You had to imply absolute negative velocity or your who argument would
> immediately fall apart.
Wrong again. The precise thing I originally wrote was:
(m)(v)----> <----(-m)(-v) Assume equal/opposite masses/velocities
(the last part of which you conveniently did not copy in what
you quoted below below) It plainly says, "Assume equal/opposite
masses/velocities". You can make that assumption inside ANY
reference frame at all!!! Because, inside any reference frame,
it is possible to specify some point as being stationary. (Often
enough, it is declared to be the "origin" of the reference frame.)
(m)(v)---->.<----(-m)(-v) Assume equal/opposite masses/velocities
OK, I have added a RELATIVELY stationary reference point. Are you
satisfied now?
> You are clearly speaking about two separate entities. There is a positive
> mass and a negative mass. These two entities cannot share the same
> velocity, they must each have their own velocities, regardless of whether
> they are heading toward each other, away from each other, or parallel to
> each other. Further, you specifically stated:
>
> (m)(v)----> <----(-m)(-v)
>
> You're clearly showing the velocity of the negative particle as having a
> negative velocity.
Yes, indeed! An object may have a velocity that is totally independent
of the essence of the object. Velocity is merely a description of how
the object's position changes (steadily) with time. (more below)
> If you maintain that the negative velocity of the
> negative mass belongs to the positive mass, then the positive mass has a net
> velocity of zero (assuming the two velocities have the same absolute value)
> and so does the negative mass.
Wrong again. I most certainly do NOT "maintain that the negative
velocity of the negative mass belongs to the positive mass" --
mostly because that statement makes no sense! However, if I
interpret that to mean "Suppose you chose a new reference frame
in which the positive mass has a zero velocity", well, it happens
that the above diagram is kind-of like an equation, in which
the invisible equals sign is located at the point of interaction.
Thus, if I subract 1 velocity-unit from the left side of the
diagram, I must also subtract 1 velocity-unit from the right side.
Therefore, if I started with:
(1m)(1v)---->.<----(-1m)(-1v)
I end up with:
(1m)(0v).<--------(-1m)(-2v)
(You do know that -1 - 1 = -2, don't you? If you have any doubts,
look up a "number line" in a grade-school arithmetic book, and see
for yourself.)
> > Negative velocities are quite possible simply
> > because they are the easy way to specify "a speed in the opposite
> > direction, relative to some first-specified positive velocity".
>
> They are two separate vectors. The positive or negative state of
> one vector has no bearing whatsoever on the second vector unless
> there is an absolute reference.
Wrong again; any arbitrary-declared stationary reference point can
be used for something this simple.
> Your relationship does not show a vector nor identify the
> reference frame. If you're going to conclusively show there is a momentum
> remaining, you need specify all the parameters.
The reference frame was implied by the statement "Assume equal/
opposite masses/velocities". And as for vectors, what do you
think those arrows were????
> > Sorry, you're still wrong. Let me break the diagram into smaller pieces:
> > (m)[(distance)/(time)]----> <----(-m)[(-distance)/(time)]
> > We still get positive momentum on both sides of the collision,
> > and therefore a positive total moment survives nullification.
>
> I'd like to think that I was wrong. But from the relationship you are
> providing, there is no justification why the exact same distance between the
> two masses should be positive for one and negative for the other. In other
> words, do we live 500 miles apart from each other or do we live -500 miles
> apart from each other? How could either of us live -500 miles apart from
> the other?
You reach that problem ONLY because you ignored the statement:
"Assume equal/opposite masses/velocities".
And regarding negative distances, once SOME reference frame is
specified, with a stationary point, it is just as valid to say
"An object has moved -500 miles South toward that point", as it
is to say, "An object has moved 500 miles North toward that
point".
> > Well, since they measure SPEED and not VELOCITY, it figures that
> > speedometers are called "speedometers". And speed is indeed always
> > positive!
>
> Is it your position then that someone could invent a velocitymeter
> and that it would have both a positive and negative scale (depending
> on which side the road you are driving on?)
Such is not impossible. All such instruments would have to be
calibrated with respect to a particular point, though (thus
establishing a reference frame, which obviously would be as
arbitrary as any other point-of-origin). But because there is
likely little practical need (that is not already met by
satellite GPS gadgets), velocitymeters are not likely to be
hitting the roads anytime soon.
> > Actually, it has been shown that for equal-mass cars, the effect upon
> > each car, during collision, is the same as if the cars had instead
> > run into an immovable wall, that suddenly popped into existance between
> > them, just before the collision. Remember that those speedometers measure
> > speed RELATIVE TO THE GROUND, which is considered "stationary" in such
> > situations. So, since after the collision, each car is moving at zero
> > velocity, the decelleration has been from 70 to 0 mph, and so the value
> > of 140 mph never enters the picture.
>
> Yes it does. Both cars are decelerating by 70 mph. The total deceleration
> is 140 miles per hour. At the moment just before impact, both cars are
> going a positive speed of 70 mph. Relatively speaking, if one had been
> standing still, the other would be going 140 mph.
I'll take you on (assuming equal-mass cars)!
Reference frame 1 BEFORE: (70mph)[car A]-->.<--[car B](-70mph)
<--[your car](-70mph)
Reference frame 1 AFTER: (0mph)[car A].[car B](0mph)
<--[your car](-70mph)
Adding 70mph to both sides:
Reference frame 2 BEFORE: (140mph)[car A]---->.[car B](0mph)
.[your car](0mph)
Reference frame 2 AFTER: (70mph)[car A][car B](70mph)-->
.[your car](0mph)
Removal of the vectors for Cars A and B should not be
construed as indicating changes in relative positions.
Think in terms of "instant before" and "instant after" the
impact. Sure, I'm fudging the duration of impact, but this
is a thought-experiment, after all. The duration of impact
is not a factor relevant to figuring FINAL velocities.
Anyway, that business of picking another reference frame, to
modify the stated velocities by 70mph, has to be consistent
with BOTH the "before" AND the "after" descriptions! I added
[your car] to the picture because you know that Reference
Frame 2 is arbitrary; it STILL has to show that [your car] and
the wreck will be separating, after the collision, at 70mph.
THEREFORE NEITHER WRECKED CAR, IN EITHER REFERENCE FRAME,
EXPERIENCES A CHANGE IN VELOCITY ASSOCIATED WITH 140MPH.
Each has done the equivalent of hitting a wall at 70mph.
Q.E.D.
You have been suffering from the "a little knowledge is a
dangerous thing" syndrome. Get well soon, please!
> > To get the kind of collision that
> > you describe, both cars would have to end up, after the collision, moving
> > at 70mph in the direction opposite to their original motions (before the
> > collision). And since you know that inelastic collisions don't do that,
> > you are wrong, again.
>
> A car is not inelastic. Sure, they are designed to absorb shock, but the
> full shock of the other vehicle needs to be absorbed in addition to the
> shock of the "wall". If you look at the wall as providing a 70 mph shock to
> the vehicle striking it, there is half the shock. But since the other
> vehicle is also moving 70 mph, there is an addition amount of force that has
> to be absorbed and it is the force of a car going 70 mph. The total force
> being absorbed is a force equivalent to a car traveling 140 mph into a wall.
Worthless argumentation. See above proof.
[paraphrased quote]
> If you take away any of the dimensions of momentum, the unit changes to
> something else. If there is no mass left over, there is no momentum.
>
> Dave
True enough. However, I never claimed that I was changing the unit!
The hypothetical momenton DOES have a constituent unit of (p), which
is a conventional shorthand for (mv) --momentum, in other words. All
I was claiming is that the essence of (p) can, in terms of the logical
consequences of most-simple nullification, exist totally independently
of both mass AND velocity (the result of most-simple nullification is
a "blob" of pure momentum that lacks mass and isn't moving!) YES, I
know it is a weird idea to wrap one's thoughts around, but one's
inabiltiy to make that stretch in no way invalidates the logic. (Do
you have trouble with "the square root of -1", also?)
David Thomson
> > >Energy is equivalent to mass for a particle at rest, as much as energy
is
> > >equivalent to momentum for a photon.
> >
> > State this mathematically.
>
> for a particle at rest: E = mc^2
> for a photon: E=pc
> with whatever value you choose for c, knowing the penalty for picking
> something wierd.
Once again, energy is proportional to mass or momentum, but energy is not
equal or equivalent to mass or momentum.
> [DAS]
> >> Momentum without mass is a photon, and its
> >> momentum is a function of the frame that absorbs it. No real quantum
> >> here, but a packet certainly...
> [YOU]
> >
> > photon = h * c (units are angular momentum times velocity)
> >
> > momentum
> > ----------- = velocity
> > mass
> >
> >How does momentum without a mass equal a photon?
>
> Division by zero is not a valid operation.
The discussion is about momentum-with-no-mass becoming a photon. The
purpose of showing that momentum divided by mass (momentum with mass
removed) is equal to velocity is to demonstrate that there are not enough
dimensions to call it a photon. Momentum with its mass removed is NOT a
photon.
It now appears what you were saying is that mass is not removed, but rather
that it has a value equal to zero or near zero. A momentum (mass times
velocity) with zero mass is a non-existent entity.
> A photon has been determined to
> have a mass smaller than we can currently measure (very like zero), yet is
a
> carrier for measreuable momentum (other than angular). And yes that
> momentum is proportional to the photon's energy divided by c.
The only place the mass can be, is in the angular momentum. If the photon
is seen as angular momentum of an electron times the speed of light and
viewed like ripples in a pond, it would be seen that the mass associated
with the angular momentum is rapidly dispersed over a very wide area. If
the mass of a photon were going to be measured at all, it would have to be
very close to the emitter (like within a few Compton wavelengths.) The
photon is a very light particle equal in mass to the electron (the electron
has a mass of 9.109 x 10^-31 kg) and is spread out over a very wide area.
So if we have difficulty measuring an electron, it will be that much more
difficult to measure a photon.
Others on these newsgroups have been pondering whether or not photons could
account for the "dark matter" seen in the center of galaxies. When viewing
photons in the manner just described, it presents a possible explanation for
dark matter. Even though the photons are spread out over a wide space and
are traveling at c, there is a lot of space and a concentration of photons
near the center of galaxies.
> The only particle I know of that carries momentum yet has no mass is a
> photon. But my experience is very limited.
>
> Better?
I think you were more accurate above when you suggested the photon may have
a near zero mass as opposed to a zero mass. If the photon is expanding, the
measuring apparatus will capture only a small portion of the area of the
expanding photon. Since the mass of the photon, is distributed equally over
the area of angular momentum, only a very small fraction of the photon's
mass is being measured.
Dave
Patrick Reany
David Thomson wrote:
What criteria do you use to decide this
"true Geometry of the Universe" from the
"false" ones.?
Patrick
David Thomson
> What criteria do you use to decide this
> "true Geometry of the Universe" from the
> "false" ones.?
It's not that there are "false" geometries in physics, it's that physics
hasn't identified any geometry (except perhaps Bohr's radial energy states.)
Naturally, I'm referring only to the subatomic realm.
I have found that angular momentum is exactly one half steradian of a
sphere, and that the sphere and steradian are essential dimensions to
subatomic physics just as much as mass, length, and time. (Charge appears
to be optional depending on the units system.) I have found three levels of
charge in the aether, all three are related to each other through 4pi. Just
this morning, I discovered (with the help of Hanson) that the photon and
distributed charge are similarly related. Photon = distributed charge
divided by 16 pi^2 in the cgs system of units.
While you're busy denying the existence of the aether, I'm showing how the
aether mathematically aligns with charge and photons in our Universe. What
does this mean? I'm not ready to draw a conclusion yet. But because there
is a precise and ordered geometry between the units I'm certain that I am
hot on the trail of making a significant discovery. I'll be making another
discovery missed for 100 years because the scientific aristocracy has lifted
its nose to anyone who dared question the voodoo math of the physicists.
Dave
Paul Cardinale
> "Paul Cardinale" <pcard...@volcanomail.com> wrote in message
> news:64050551.02100...@posting.google.com...
> (Unubstantiated opinions intentionally ignored)
>
> > > > > > > Energy = mass * velocity^2
> > > > > > > Momentum = mass * velocity
> > > > are valid only when v is much smaller than c.
> > >
> > > We're talking about dimensions here. The dimensions are always correct
> > > regardless of what the velocity is.
> > >
> > Just because the dimensions match, doesn't mean that the equations are
> > valid.
> > (And in this case, they're not.)
>
> Give a mathematical example in the real world where the equations are not
> valid.
>
1. Photons. They have energy and momentum, but no mass.
2. CRTs larger than about 25". In order to get them to work well,
designers must use relativistic mechanics, not classical mechanics.
> > > You're trying to prove that energy is
> > > the same thing as momentum, which it is not.
> > >
> > In the case of the photon, E = p * c.
> > i.e. energy is proportional to momentum, and with the right units,
> > the numerical values are the same. Thus the equivalence.
>
> I agree that energy is proportional to momentum. But that isn't what you
> were saying. In the previous message you said:
>
> > Let's try really slowly:
> > E = p * c
> > Now we choose units such that c = 1.
> > This means that wherever there is a "c" we can substitute "1". So:
> > E = p * 1
>
> There is a world of a difference between saying something is proportional
> and something is equal.
>
There is only a small difference: If the proportionality constant
happens to be 1, then they are equal.
> You tried to show equivalence by violating the laws
> of mathematics.
>
No.
> > The equation E^2 = (p*c)^2 + (m*c^2)^2 is valid (empirically) for all
> particles.
>
> This is your unproven, personal opinion.
>
No. This is a fact. Experimentally verified.
> > > E^2 = (p*c)^2 + (m*c^2)^2
> > >
> > > In the case of the electron, what you have is
> > >
> > > energy^2 = energy^2 + energy^2.
> > >
> > > If energy has the value of 1, you get 1 = 1 + 1. That is not a valid
> > > equality.
> > >
> > Stupidly wrong. Here you assumed that E^2 = (p*c)^2 = (m*c^2)^2.
> > Which is invalid.
>
> Trying to have it both ways are you?
>
No. You seem to have a reading comprehension problem.
> In a previous post you stated:
>
> >Wrong. The relationship between energy (E), momentum (p) and mass (m) is:
> >E^2 = (p*c)^2 + (m*c^2)^2
> >
> >For a photon, m=0 so:
> >E^2 = (p*c)^2
> >E = p*c
>
> You clearly showed that YOU believe E^2 = (p*c)^2 when m = 0.
>
That has been verified experimentally.
> But since the
> momentum of the photon is equal to the mass of an electron times the speed
> of light,
>
Pure nonsense. Where did you come up with that idiocy?
The mass of an electron has nothing to due with the momentum of a
photon.
Furthermore, "the mass of an electron times the speed of light" is a
constant,
whereas the momentum of a photon is different for different
wavelengths.
> you cannot give m = 0 and p <> 0 since they are both based on the
> same source of mass.
You're wrong. You are assuming that p = m * v. Which is an
approximation that is only valid when v << c.
> There are not two different masses associated with a
> particle.
>
Correct. But you are the one trying to do that.
The mass of the photon is 0, but you want to use the mass of an
electron when dealing with a photon.
<snip>
> > I wrote: "Thus, for a photon, E and p differ only by a constant
> > factor; and if you choose units such that c=1, then E = p."
> > Which is not even close to your assertion that I "believe that for E =
> > p * c means that c (which has the dimensions of m/sec) can be replaced
> > by 1."
> > 1. The dimensions of c are distance/time. m/sec are units, not the
> > dimensions.
> > 2. As I showed you, but you couldn't learn, choosing the appropriate
> > UNITS results in c having a numerical value of 1 (the dimensions
> > remain the same).
>
> I agree that the dimensions of c are distance divided by time.
>
Fine.
> And velocity is the unit.
>
Wrong. Velocity is not a unit. It is a compound dimension.
> But don't tell me you didn't say that c could be replaced by 1.
>
I Didn't tell you I didn't say that (you really do have a reading
comprehension problem).
c is a constant. What numerical value it has depends on what units
you use.
If you choose your units carefully, you can have c = 1.
<snip>
> What right do you have for choosing units for c such that c = 1 without
> altering the other side of the equation?
>
You really are confused. The choosing of units is not a mathematical
operation.
In the equation E = p * c, both sides have dimensions of energy, but
the equation is devoid of units.
In order to apply the equation, we must choose units. We may use any
unit system.
Once a system of units has been chosen, the numerical value of c can
be determined.
If we use FPS, then c = 983,568,960. If we use MKS, then c =
299,791,820.
If we use units of seconds for the dimension of time, and units of
light-seconds for the dimension of distance,
then c = 1. Since we are free to choose any units, we may freely use
a system that results in convenient values.
Thus depending on the units, we could have:
E = p * 983,568,960
or
E = p * 299,791,820
or
E = p * 1
or any other value we like (noting that these equations are no longer
unitless, but unit-specific and thus the numbers to be used for E and
p must be in the corresponding units).
> What is the mathematical law that
> allows for this? Don't give my your personal philosophy or personal
> opinion. Let's talk math and science here. Where do you get the authority
> to make c = 1 without altering the other side of the equation?
>
Substituting a numberical value for symbol representing a constant
does not violate any law of mathematics.
c is one only one side of the equation, so replacing it with its
numerical value happens on only one side.
> Let me put it this way. If you can arbitrarily assign c = 1, what is to
> stop you from arbitrarily assigning c = 12
>
Nothing, you may choose a unit system such that c = 12.
> or p = 52.3 or E = -1?
>
You can't do that because p and E are not constants like c.
If you restrict yourself to a specific case, for example where the
value of p is known,
and it won't change, then you could choose units which result in any
specific numerical value for p.
Same with E. (But you cannot arbitrarily choose numbers for all 3 E,
p, c; just any 2).
> What
> happens to the reliability of mathematics when you can just arbitrarily
> assign a different unit value to a variable in an algebraic equation without
> keeping the equation in balance?
>
I never said that the units don't have to match on both sides.
<snip>
> Since c is not equal to 1,
>
It can be 1. See above.
> c * 1 equals c, not 1. Is that any easier to
> understand? 1 is not a replacement for c unless c had dimension and value
> of 1 to begin with.
>
You are confused about the difference between dimensions and units.
"To begin with", c has the dimension of velocity, but no units and no
numerical value.
It acquires units and a numerical value when we choose the units.
The equation E = p * c is valid for massless paricles in any system of
units.
The equation E = p is valid for massless paricles in a system of units
wherein c = 1.
Paul Cardinale
David Thomson
news:64050551.02100...@posting.google.com...
not valid.
> >
> Here are two:
> 1. Photons. They have energy and momentum, but no mass.
> 2. CRTs larger than about 25". In order to get them to work well,
> designers must use relativistic mechanics, not classical mechanics.
Neither are mathematical examples. The first situation you listed is
patently wrong. If the mass in the momentum is zero, then there is no
momentum. You have no genuine mathematical proof for your claim.
> > > Let's try really slowly:
> > > E = p * c
> > > Now we choose units such that c = 1.
> > > This means that wherever there is a "c" we can substitute "1". So:
> > > E = p * 1
> >
> > There is a world of a difference between saying something is
proportional
> > and something is equal.
> >
> There is only a small difference: If the proportionality constant
> happens to be 1, then they are equal.
But the proportionality constant is not 1, it is c. You arbitrarily put 1
in place of c. That's not real math.
> > You tried to show equivalence by violating the laws of mathematics.
> >
> No.
Yes
> > > The equation E^2 = (p*c)^2 + (m*c^2)^2 is valid (empirically) for all
> > particles.
> >
> > This is your unproven, personal opinion.
> >
> No. This is a fact. Experimentally verified.
Direct me to a published paper that supports your claim. I will deal with
the author's of the papers directly if they are still alive.
> No. You seem to have a reading comprehension problem.
You have a physics and mathematics comprehension problem. And it's not your
fault, entirely. You were brainwashed at an early age and haven't been
deprogrammed yet.
> > >Wrong. The relationship between energy (E), momentum (p) and mass (m)
is:
> > >E^2 = (p*c)^2 + (m*c^2)^2
> > >
> > >For a photon, m=0 so:
> > >E^2 = (p*c)^2
> > >E = p*c
> >
> > You clearly showed that YOU believe E^2 = (p*c)^2 when m = 0.
> >
> That has been verified experimentally.
Voodoo math cannot be experimentally verified. Someone got tired of looking
at the numbers, couldn't come up with a better solution, and convinced
themselves that the rules of math must be playing a trick on them. So they
altered the rules of math in order to get passed the problem and go on to
something else.
> > But since the
> > momentum of the photon is equal to the mass of an electron times the
speed
> > of light,
> >
> Pure nonsense. Where did you come up with that idiocy?
> The mass of an electron has nothing to due with the momentum of a photon.
Are you now saying that photons appear out of nowhere? An atom just
magically gives birth to photons without regard to physical laws? Photons
have everything to do with electrons, including the angular momentum of the
electron. I have a complete theory with regard to the photoelectric effect,
Compton effect, and pair production. All of these processes are identical
except that the frequency of photon and electron absorption is different for
each situation.
> Furthermore, "the mass of an electron times the speed of light" is a
constant,
> whereas the momentum of a photon is different for different wavelengths.
A photon is a quantum unit of light. A quantum unit cannot have a
frequency. This is another one of the failures of modern physics. Light
has a frequency, but photons do not. Light is made from photons, photons
are the quantum unit of light.
AND ONCE AGAIN, you are having it both ways. You tell me in one breath that
a photon has no mass, and in the next breath you're saying the photon has
momentum. We all know that momentum is mass times velocity. So if you
multiply zero times any velocity, you're going to get zero momentum. Voodoo
math, my friend, you are practicing voodoo math. In one moment the rules of
algebra are one way, and in the next the rules are different. It all
depends on what you want to believe and has no basis in the rules of
mathematics.
> > you cannot give m = 0 and p <> 0 since they are both based on the
> > same source of mass.
> You're wrong. You are assuming that p = m * v. Which is an
> approximation that is only valid when v << c.
It is only valid when you want it to be valid. If I want it to be valid,
you won't allow it. But I don't care what you want, I'm going by the
established rules of mathematics. You cannot change the units by
arbitrarily removing them from an equation. You're simply wrong.
> > There are not two different masses associated with a
> > particle.
> >
> Correct. But you are the one trying to do that.
> The mass of the photon is 0, but you want to use the mass of an
> electron when dealing with a photon.
If the mass of the photon is zero, then the momentum of the photon is zero.
There is only one mass associated with the photon.
When a photon is absorbed by an atom, the angular momentum has to go
somewhere according to the conservations of mass, energy, and angular
momentum. This absorbed mass goes into the angular momentum of the valence
electron. I have shown mathematical proofs for this. When the valence
electron reaches twice its ground state angular momentum, the electron is
ejected to the next energy level due to accumulating extra electron strong
charge. The double sized electron has to move the quantum length of one
Compton wavelength in one quantum unit of time (speed of light). This
imparts the speed of light to the double sized electron. In the case of the
Compton effect, half the angular momentum of the double sized electron is
ejected into space at the speed of light (h * c). This is the photon. The
photon is created directly as a result of the electron and share's its mass
and strong charge characteristics.
The photon has the full angular momentum of the electron but also it is
propagating at the speed of light. The reason there is no discernable mass
associated with the photon has to do with the angular momentum being in an
expanding state. Soon I will be able to show mathematically that despite
the lack of a discernable mass, the photon still interacts with gravity.
This is already borne by the experimental evidence. When a photon moves
near a large mass, it changes direction. This will also explain at least
some of the dark matter in the Universe.
> > > I wrote: "Thus, for a photon, E and p differ only by a constant
> > > factor; and if you choose units such that c=1, then E = p."
> > > Which is not even close to your assertion that I "believe that for E =
> > > p * c means that c (which has the dimensions of m/sec) can be replaced
> > > by 1."
> > > 1. The dimensions of c are distance/time. m/sec are units, not the
> > > dimensions.
> > > 2. As I showed you, but you couldn't learn, choosing the appropriate
> > > UNITS results in c having a numerical value of 1 (the dimensions
> > > remain the same).
> >
> > I agree that the dimensions of c are distance divided by time.
> >
> Fine.
>
> > And velocity is the unit.
> >
> Wrong. Velocity is not a unit. It is a compound dimension.
Write to these people and correct them. Also, do a search on the Internet
and correct everyone else, too.
http://www.site.uottawa.ca:4321/astronomy/index.html#velocityunit
> > But don't tell me you didn't say that c could be replaced by 1.
> >
> I Didn't tell you I didn't say that (you really do have a reading
> comprehension problem).
> c is a constant. What numerical value it has depends on what units
> you use.
> If you choose your units carefully, you can have c = 1.
A person with a reading comprehension problem can't tell the difference
between a rhetorical statement and a factual statement. The point is, you
are stating and reaffirming that you are replacing c with 1. You are
admitting that you are arbitrarily changing the units to make the equation
fit your preconceived idea. This is not math, nor is it physics. It's
voodoo math and self-deception.
> > What right do you have for choosing units for c such that c = 1 without
> > altering the other side of the equation?
> >
> You really are confused. The choosing of units is not a mathematical
> operation.
We're getting somewhere. You are correct, the choosing of units is not a
mathematical operation. It's merely a whim. The units have already been
chosen based on the definition of E, p and c. The equation has already been
established to be true when E = p * c. At this point, choosing different
units for c is a mere whim. It is not based on mathematics. We are in
agreement.
> In the equation E = p * c, both sides have dimensions of energy, but
> the equation is devoid of units.
Excuse me? This is algebra. The variables are defined by the dimensions
and values of the units they represent. They have very specific units. You
have already admitted this by agreeing that momentum is equal to mass times
velocity and that energy is equal to mass times velocity squared. The units
have been defined. Choosing to ignore the units does not give you
permission to bend the rules of mathematics in a vain attempt at describing
physics in your wrong views.
> In order to apply the equation, we must choose units. We may use any
> unit system.
You have already done that.
> Once a system of units has been chosen, the numerical value of c can
> be determined.
> If we use FPS, then c = 983,568,960. If we use MKS, then c =
> 299,791,820.
Nonsense. There is nothing that ties c to 299,791,820 without also
including the dimensions of m/sec in the MKS system.
> If we use units of seconds for the dimension of time, and units of
> light-seconds for the dimension of distance,
> then c = 1.
Then you have to use the same units for defining E and p. You can't have
energy in SI units, momentum in MKS units, and c in a new set just because
that's what you want. It's poor science even if you took the time to
properly establish the correct relationships between the different unit
systems. All equations need to be in the same system of units throughout
the equation.
> Since we are free to choose any units, we may freely use
> a system that results in convenient values.
> Thus depending on the units, we could have:
> E = p * 983,568,960
> or
> E = p * 299,791,820
> or
> E = p * 1
You are a hopeless case. You sound like the Soviet Communists of the 60s
and 70s who came over during the Olympics, walked through our grocery stores
and shopping malls, and publicly declared it was all propaganda. You are
blindly following the party line. Even within your own mind you know you
are fighting for a losing cause. But you are so afraid of confronting the
establishment, that you think it is easier to deny the truth than defend it.
If our country is defeated by an oppressive foreign enemy, the scientists
will undoubtedly be the first to give in and turn their backs on freedom.
This is a very disturbing thought.
> or any other value we like (noting that these equations are no longer
> unitless, but unit-specific and thus the numbers to be used for E and
> p must be in the corresponding units).
Whatever you want. Just make it up and it will be so.
> > What is the mathematical law that
> > allows for this? Don't give my your personal philosophy or personal
> > opinion. Let's talk math and science here. Where do you get the
authority
> > to make c = 1 without altering the other side of the equation?
> >
> Substituting a numberical value for symbol representing a constant
> does not violate any law of mathematics.
> c is one only one side of the equation, so replacing it with its
> numerical value happens on only one side.
Do you have access to a university math department? How about a local high
school? Take this problem to a qualified math teacher and have them explain
to you why you can't do this.
> > Let me put it this way. If you can arbitrarily assign c = 1, what is to
> > stop you from arbitrarily assigning c = 12
> >
> Nothing, you may choose a unit system such that c = 12.
>
> > or p = 52.3 or E = -1?
> >
> You can't do that because p and E are not constants like c.
Eh? You just said c is not a constant. If it were a constant it would not
change. Yet you are saying you can change c for whatever you want. Are you
presenting a new definition of "constant" now? A constant is anything you
want it to be because you can choose it and its dimensions are not important
to the equation?
> If you restrict yourself to a specific case
Sorry, I can't take any more time for this. You have a serious impediment
to understanding physics until you get rid of the voodoo math.
Dave
Bilge
Re: Quanta of Momentum to usenet:
>
>"Bilge" <ro...@radioactivex.lebesque-al.net> wrote in message
>news:slrnapljn...@radioactivex.lebesque-al.net...
>> Robert Kolker said some stuff about
>> Re: Quanta of Momentum to usenet:
>> >
>Bilge wrote...
>> What's hard to picture is a negative energy density in free space.
>>
>
>It's only hard to picture because you neglect the source of the field.
>To
>"define" what you mean by negative energy for a vector field, such as an
>electric field you must consider the source charge. If you have a positive
>charge and a negative electric field, that defines a negative energy. I
No, it doesn't:
| E- E+ |
|<-- +q == +q <--|
|<-- <- <- <--|
| |
>believe that without consideration of the sign of the charge, the sign of
>the energy density is unknown and indeterminate, not positive definite.
Once you define it as positive or negative, your freedom evaporates
and you have to be consistent. It's like defining an intrinsic parity.
It doesn't matter whether you choose even or odd for the electron, but
after that, the positron is what the the electron isn't.
>
>If you work in units with c=1, you would also choose mu0 =1 and eps0 = 1.
>You then get for the energy density,
>
>T^00 = (E^2 + B^2) / 2
>
>and when shown like this you are completely ignoring the "ratios" between
>the source strength and the field strength.
No, I'm not.
>A more accurate way to write
>this would be the original, classical energy density in a medium;
>
>T^00 = ((eps0*eps_rel)*E^2 + B^2/(mu0*mu_rel)) / 2
>
>where the relative permeability and permittivity are taken into
>consideration.
You've replaced a parameter (`c') which has no physics associated with
it's value and replaced it with two constants which only exist to convert
units invented before anyone ever realized electricity and magnetism were
the same phenomenon.
>These are simply the ratio between the field strength and the
>source strength in the medium with relative properties eps_rel and mu_rel.
>Look at the reflection of a charge or current in a mirror, these values are
>negative. If you place a positive charge in front of a mirror you get a
>negative electric field reflected in the mirror from the image of the
>charge. So a negative energy density is just a reflection of a positive
>energy density.
Think that one through again. Draw the reflection of a field from
a charge.
>IMO the WEC and the SEC are a farce caused by a misconception that you
>can neglect the ratio between the local vector field and the local
>field source.
Huh? (I assume you mean "weak equivalence principle" and "strong
equivalence principle" - and please don't use acronyms).
[...]
>to a charged particle. So while you may treat the gravitational space-time
>as flat, and work with EM fields in flat space-time, you can also treat the
>EM field like a geometric manifold with respect to the charged particle.
>Then it is not so flat, and local isn't so local any more.
Yes, it is. The geometry is still flat. After all, E&M is derived
from exactly that assumption. The proper geometric jargon is "fiber
bundle". I'm not "fiber bundle conversant" to any great extent, so
you would be better looking it up than having me "wing it" and misusing
the terminology.
Stephen Speicher
> "Paul Cardinale" <pcard...@volcanomail.com> wrote in message
> news:64050551.02100...@posting.google.com...
>
> > > > The equation E^2 = (p*c)^2 + (m*c^2)^2 is valid (empirically) for all
> > > particles.
> > >
> > > This is your unproven, personal opinion.
> > >
> > No. This is a fact. Experimentally verified.
>
> Direct me to a published paper that supports your claim. I will deal with
> the author's of the papers directly if they are still alive.
>
Sure. The author's are just waiting around, twiddling their
thumbs, counting the minutes till you come along and "deal" with
them. Being so pathetically ignorant of physics is a very poor
trait to possess when you are also truly megalomaniacal.
[Snip the rest of Thomson's inanities, including the argument
with Paul against setting c = 1, which literally brought tears of
laughter to my eyes.]
dlzc@aol.com (formerly)
> The discussion is about momentum-with-no-mass becoming a photon. The
> purpose of showing that momentum divided by mass (momentum with mass
> removed) is equal to velocity is to demonstrate that there are not enough
> dimensions to call it a photon. Momentum with its mass removed is NOT a
> photon.
If it is not by definition a photon, then it is merely an unsupportable
statement. Withdrawn.
> It now appears what you were saying is that mass is not removed, but
rather
> that it has a value equal to zero or near zero. A momentum (mass times
> velocity) with zero mass is a non-existent entity.
Or it is a photon, which has momentum and zero mass.
> > A photon has been determined to
> > have a mass smaller than we can currently measure (very like zero), yet
is
> a
> > carrier for measreuable momentum (other than angular). And yes that
> > momentum is proportional to the photon's energy divided by c.
>
> The only place the mass can be, is in the angular momentum. If the photon
A photon will have _no_ mass. A _pair_ of particles with non-zero momentum
can have more than their rest mass' worth of mass. One or more of these
could be photons.
So maybe the mass is in the coupling between the "angular momenta"?
> very close to the emitter (like within a few Compton wavelengths.) The
> photon is a very light particle equal in mass to the electron (the
electron
> has a mass of 9.109 x 10^-31 kg) and is spread out over a very wide area.
> So if we have difficulty measuring an electron, it will be that much more
> difficult to measure a photon.
Why an electron. Photons are absorbed by protons too. The only requirement
is that charge be involved. So what is the "mass" of charge, et al?
> Others on these newsgroups have been pondering whether or not photons
could
> account for the "dark matter" seen in the center of galaxies. When
viewing
> photons in the manner just described, it presents a possible explanation
for
> dark matter. Even though the photons are spread out over a wide space and
> are traveling at c, there is a lot of space and a concentration of photons
> near the center of galaxies.
>
> > The only particle I know of that carries momentum yet has no mass is a
> > photon. But my experience is very limited.
> >
> > Better?
>
> I think you were more accurate above when you suggested the photon may
have
> a near zero mass as opposed to a zero mass. If the photon is expanding,
the
> measuring apparatus will capture only a small portion of the area of the
> expanding photon. Since the mass of the photon, is distributed equally
over
> the area of angular momentum, only a very small fraction of the photon's
> mass is being measured.
And measurements of two photons that strike (or attempt to strke) each other
head-on, that has determined that the photons's size is "very small", what
of these experiments?
I prefer to think of a photon as an impulse (in the Fourier transform
sense). Infinitely short in duration, but with a finite effect. But then I
am a macroscopic kind of guy...
David A. Smith
vernonner3voltazim
> vernonner3voltazim wrote:
> > So, to whatever extent that it may be possible for negative mass/energy
> > to exist, one must be immediately ready to start wrapping one's mind
> > around the logical consequence that momentum might be so Fundamental
> > a thing that it can indeed exist independently of mass or energy.
> > Q.E.D., and enjoy!
>
> a hard time understanding what negative mass could possibly mean. Mass
> is a measure of inertia. It tells us how hard it is to get the body with
> the mass into motion and to increase/decrease its speed. A body with 0
> mass does not even need a push. As soon as it exists, it is going at
> light speed. Example: A bound electron drops a quantum level and emitts
> a photon. The photon comes into existence by virtue of the energy jump,
> and zowie!, it is off and running. That also means infinite
> acceleration. Does negative mass mean that a body possessing it is even
> easier to set into motion or easier to slow down?
With respect to inertia, the simplest way to think about a quantity of
negative mass is imagine pushing on it -- it responds by moving TOWARD
you, instead of away from you (the way ordinary mass responds).
Since F = (m)(a), F also = (-m)(-a), see?
> Does it mean it will
> go faster than light? If such a body had a velocity greater than c, it
> would have relativistic mass, oooopppps! energy that was imaginary (or
> complex). That is not even negative. If zero mass => infinite
> acceleration (applied just for an infinitesimal time interval) what is
> negative mass. Even greater than infinite acceleration? That does not
> make much sense. Imaginary mass might mean acceleration backward in
> time. That is a possibility, but where is negative mass in all this.
Well, since acceleration (a) has units of (distance)/[(time)(time)],
then (-a) could have units of either (-d)/[(t)(t)] or (d)/{-[(t)(t)]}.
The former is what I described above; the latter means we have to
start talking about "imaginary Time", distinct from either ordinary
flow-of-Time, or ordinary backwards-through-Time. I suppose we won't
know for sure which way negative mass behaves, until we get our hands
on some, to play with. But by Occam's Razor, I vote for the former.
> Look at it from the gravitational side. The gravitational mass of a body
> tells us how much it bends spacetime ( or in Newton-speak, how much
> force it exerts on other bodies). A zero mass particle with momentum
> gravitates because it has energy (equal to its momentum). A particle
> with less than zero mass gravitates less? In terms of the relativistic 4
> vector, negative mass really does not make much sense.
From the gravitational perspective, let's start with simple
multidimensional geometry. Consider a one-dimensional line:
<------.------> A zero-dimensional object (a mathematical point)
that is restricted to that line only has two ways it can move,
right?
Next, consider a mathematical plane, like a flat monitor
screen. A one-dimensional line, that has its endpoints fixed on
the plane, say like this: | can most easily be bent in either
of only two directions, so as to produce shapes like: < and >
Next, consider a volume of ordinary Space, occupied by a
trampoline. That construct is basically planar, but allowed to
flex in either of only two possible directions, right?
Finally, consider 4-dimensional hyperspace (NOT Space-Time!!!), in
which it becomes possible to talk about "curved 3D-Space". Well,
by simple extension of what has been previously described, it
follows that if 3D-Space can be curved in any direction at all,
inside 4D-hyperspace, then it must be possible to curve in the
OPPOSITE direction as well (minimum of two possible directions)!
(Once could say something equivalent, while discussing "curvature
of Space-Time", but I'm trying to avoid complications here, by
sticking to geometry of distances only.)
The point of the preceding is that, insofar as the Gravitation of
ordinary mass has been associated with "curved Space", then it
logically follows that Gravitation of negative mass should be
associated with the opposite "direction" of curved Space!
Hope that helps!
Robert Kolker
vernonner3voltazim wrote:
> The point of the preceding is that, insofar as the Gravitation of
> ordinary mass has been associated with "curved Space", then it
> logically follows that Gravitation of negative mass should be
> associated with the opposite "direction" of curved Space!
>
> Hope that helps!
Interesting. Well, you sure got further than I did.
Thank you.
Bob Kolker
vernonner3voltazim
> > Why not?
> >
> Because there is no reason to suspect that it might be.
>
> > Hasn't everything else in Physics been quantized?
> >
> No. Consider: Mass, energy, position, acceleration, velocity,
> temperature, force, gravity, ...
>
> Many things are not quantized.
>
> Paul Cardinale
Actually, various subatomic particles are all considered to
be "quanta" of Mass. The photon is considered to be a
"quantum" of Energy. The Planck Length is considered to
be a "quantum" of distance, and there is also a Planck Time,
too.
One of the things to BEWARE (or, more accurately, BE AWARE)
of, is the seeming that "quanta" are not necessarily
themselves quantized. For example, how identical can two
photons be, yet still be different? In theory, one could
imagine Photon A, having energy of EXACTLY 1 electron-volt,
while Photon B has {1 + [10^(-100)]} electron-volts. In
practice, however, the Uncertainty Principle interferes
with our ability to measure that many significant figures
of accuracy, due to something known as "Planck's Constant",
the "quantum of Action". To the best of my knowledge,
two such different photons as A and B could really exist,
but we would simply never be able to tell them apart.
The key fact is that EACH photon is a "quantum", a packet;
the amount of energy either carries, in its packet, doesn't
really matter, with respect to the definition of "quanta".
Next, temperature is actually a measurement of the average
kinetic energy of molecules -- and I already said that
energy is quantized. As for gravity, well, plenty of
physicists are attempting to resolve incompatibilities
between Quantum Mechanics and General Relativity. It is
widely expected that as soon as those incompatibilites
are resolved, quantized Gravitation will be one consequence.
Finally, as for acceleration, velocity, and force: Since
those things all have "dimensions" (like space, time, and
mass) which are quantized, it figures that they should
be quantizeable, too.
Momentum would be in that last category, of course.
vernonner3voltazim
> > What criteria do you use to decide this
> > "true Geometry of the Universe" from the
> > "false" ones.?
>
> It's not that there are "false" geometries in physics, it's that physics
> hasn't identified any geometry (except perhaps Bohr's radial energy states.)
> Naturally, I'm referring only to the subatomic realm.
>
> aether mathematically aligns with charge and photons in our Universe.
>
Personally, I think that the totality of virtual-particles in the vacuum,
sometimes called "the vacuum self-energy", qualifies as an aether.
For more details, see this: http://www.nemitz.net/vernon/GHOSTLY.pdf
Hope you enjoy it!
vernonner3voltazim
> > > >Energy is equivalent to mass for a particle at rest, as
> > > >much as energy is equivalent to momentum for a photon.
> > >
> > > State this mathematically.
> >
> > for a particle at rest: E = mc^2
> > for a photon: E=pc
> > with whatever value you choose for c, knowing the penalty for picking
> > something wierd.
>
> Once again, energy is proportional to mass or momentum, but energy is not
> equal or equivalent to mass or momentum.
I'd like to try clearing some air, here.
As we all know, a photon consists of energy,
and E=mc^2
I shall modify that like this: (E)=(m)(c)(c)
because it's easier to manipulate algebraically.
The thing to keep in mind is that the energy
of a photon is EQUIVALENT to some amount of mass,
(m)=(E)/[(c)(c)].
So, since the energy of a photon is non-zero,
its mass-equivalent is ALSO non-zero!
SURE, for an average photon, its mass-equivalent
is an EXTREMELY small magnitude, very near zero.
NEVERTHELESS: While a photon has no actual mass,
it ALWAYS has some non-zero equivalent of mass.
That non-zero mass-equivalent is a perfectly
valid thing to use in other equations.
Thus, since the photon moves at the speed of light
(c), AND because it has a non-zero equivalent of
mass, it will possess a non-zero quantity of
momentum (multiply the last equation by (c)):
(m)(c)=(E)/(c)
That momentum will also usually be quite a small
magnitude, close to zero, BUT DEFINITELY > ZERO.
Does that help???
Todd Desiato
> Todd Desiato said some stuff about
> Re: Quanta of Momentum to usenet:
> >
> >"Bilge" <ro...@radioactivex.lebesque-al.net> wrote in message
> >news:slrnapljn...@radioactivex.lebesque-al.net...
> >> Robert Kolker said some stuff about
> >> Re: Quanta of Momentum to usenet:
> >> >
> >Bilge wrote...
> >> What's hard to picture is a negative energy density in free space.
> >>
> >
> >It's only hard to picture because you neglect the source of the field.
>
> By definition, there are no sources in "free space".
Then you cannot measure the sign of the energy density, you can only define
the abigous value you measure to be positive. So for the sake of argument
consider there to be a local source, otherwise there is no point in
continuing.
>
> >To
> >"define" what you mean by negative energy for a vector field, such as an
> >electric field you must consider the source charge. If you have a
positive
> >charge and a negative electric field, that defines a negative energy. I
>
> No, it doesn't:
>
>
> | E- E+ |
> |<-- +q == +q <--|
> |<-- <- <- <--|
> | |
What is this supposed to be?
I meant that if you have a positive charge and a negative electric field
where you should have a postivie one, that defines negative energy for an
electric field. There is no other way to define it! The energy of the vector
field depends on the source charge squared, or if you think in terms of a
capacitor it is the voltage squared. So you cannot get any negative energy
unless your capacitor has a negative value. A negative valued capacitor is
just a capacitor whose impedance has a 180 phase shift in the complex plane.
Then you have a battery, or in effect as you add more positive charge you
get a more negative electric field. The negative impedance has the effect of
reversing the direction of the vector field relative to the sign of the
source charge.
If you disagree, then please define for me what is negative energy in an
electric field? Your original statment that I replied to implies you do not
know, so how do you know I'm wrong? I think you just like to be disagreable.
> >believe that without consideration of the sign of the charge, the sign
of
> >the energy density is unknown and indeterminate, not positive definite.
>
> Once you define it as positive or negative, your freedom evaporates
> and you have to be consistent. It's like defining an intrinsic parity.
> It doesn't matter whether you choose even or odd for the electron, but
> after that, the positron is what the the electron isn't.
> >
> >If you work in units with c=1, you would also choose mu0 =1 and eps0 =
1.
> >You then get for the energy density,
> >
> >T^00 = (E^2 + B^2) / 2
> >
> >and when shown like this you are completely ignoring the "ratios"
between
> >the source strength and the field strength.
>
> No, I'm not.
>
> >A more accurate way to write
> >this would be the original, classical energy density in a medium;
> >
> >T^00 = ((eps0*eps_rel)*E^2 + B^2/(mu0*mu_rel)) / 2
> >
> >where the relative permeability and permittivity are taken into
> >consideration.
>
> You've replaced a parameter (`c') which has no physics associated with
> it's value and replaced it with two constants which only exist to convert
> units invented before anyone ever realized electricity and magnetism were
> the same phenomenon.
I am being consistent. Consistent with the fact that the original and more
accurate form of writing the energy density term includes the ratio between
the local source strength and the local field strenth, which includes the
sign of both. By eliminating those constants you are eliminating that ratio
from the equation for energy density which then results in the Weak Energy
Condition and Strong Energy Condition, the violation of which should not
even be an issue. (See Wald, General Relativity, pg 219. for a definition of
the Weak and Strong Energy Conditions.)
>
> [...]
> >to a charged particle. So while you may treat the gravitational
space-time
> >as flat, and work with EM fields in flat space-time, you can also treat
the
> >EM field like a geometric manifold with respect to the charged particle.
> >Then it is not so flat, and local isn't so local any more.
>
> Yes, it is. The geometry is still flat.
Only to an uncharged mass!
> After all, E&M is derived
> from exactly that assumption. The proper geometric jargon is "fiber
> bundle". I'm not "fiber bundle conversant" to any great extent, so
> you would be better looking it up than having me "wing it" and misusing
> the terminology.
Engineers don't need fiber bundles we need the equations of motion and ways
to control them. You can keep your fiber bundles, that level of abstraction
is useless to me. Maxwell's equations are the equations of motion for a
charged particle. That particle moves along a geodesic of the
electromagnetic field just as a mass particle moves along a geodesic of the
gravitational field. The curvature of space-time to a particle with only
mass can be perfectly flat in the same region where the curvature as seen by
a charged particle is strong. You must learn to work with both using the
same tools, but separating the descriptions of the two objects. (See arxiv:
physics/0205086 as a reference on how to treat a charged particle in a
electromagnetic-curved space-time. In this paper you will see that intertial
mass and the geodesic motion of a charged particle are the same thing!) You
should relate to it, in effect you treat the electomagnetic field as a
geometrical manifold relative to the charges it contains, just like you
treat space-time as a geometric manifold relative to the mass it contains.
Now think about what is local to a charged particle in a strong, time
varying, non-uniform EM field. If I have a local positive source charge and
nearby "local" there is a region where the electric field has the wrong
sign, then in that region I have a negative energy density relative to the
local source.
Best Regards,
Todd Desiato
vernonner3voltazim
You're welcome!
But that's just the tip of the iceberg. For more info, try this:
http://www.nemitz.net/vernon/BALANCD2.pdf
Dirk Van de moortel
"David Thomson" <ne...@volantis.org> wrote in message news:lz%m9.6805$OB5.6...@newsread2.prod.itd.earthlink.net...
> "David McAnally" <D.McAnally@i'm_a_gnu.uq.net.au> wrote in message
> news:anfuqb$q76$1...@bunyip.cc.uq.edu.au...
[snip]
> > >E^2 = (p*c)^2 + (m*c^2)^2
> >
> > >In the case of the electron, what you have is
> >
> > >energy^2 = energy^2 + energy^2.
> >
> > This is a statement involving dimensions.
>
> And...??? Is that your scientific rebuttal? I not only make a statement of
> dimensions (which proves I'm right) I also make a statement of values.
> Since energy is the common dimension, the dimensions can be divided out of
> the equation and leave us only with values.
>
> 1 = 1 + 1
>
> You and the entire scientific establishment that believes this voodoo math
> are simply wrong. The proof is clear. There is no scientific rebuttal to
> my argument.
>
> > >If energy has the value of 1, you get 1 = 1 + 1. That is not a valid
> > >equality.
> >
> > This logic is invalid. You are assumimg that p*c = E and m*c^2 = E
> > without any justification.
>
> Excuse me? Hello? Is anybody home?
Hmm!
http://users.pandora.be/vdmoortel/dirk/Physics/ImmortalFumbles.html#ExcuseMe
Title: "Excuse me? Hello? Is anybody home?"
Dirk Vdm
David Thomson
in message news:GdAn9.146734$8o4....@afrodite.telenet-ops.be...
>
> Hmm!
>
http://users.pandora.be/vdmoortel/dirk/Physics/ImmortalFumbles.html#ExcuseMe
> Title: "Excuse me? Hello? Is anybody home?"
That's fair by me. But could you spell my name correctly on your web site?
Thanks,
Dave
David Thomson
> As we all know, a photon consists of energy,
> and E=mc^2
> I shall modify that like this: (E)=(m)(c)(c)
> because it's easier to manipulate algebraically.
Wait a minute. Just because the photon consists of energy does not mean
that the photon IS energy. And in fact,
photon <> energy
photon = energy * length
> The thing to keep in mind is that the energy
> of a photon is EQUIVALENT to some amount of mass,
> (m)=(E)/[(c)(c)].
>
> So, since the energy of a photon is non-zero,
> its mass-equivalent is ALSO non-zero!
I fully agree with this.
> SURE, for an average photon, its mass-equivalent
> is an EXTREMELY small magnitude, very near zero.
>
> NEVERTHELESS: While a photon has no actual mass,
> it ALWAYS has some non-zero equivalent of mass.
I agree with this, too.
> That non-zero mass-equivalent is a perfectly
> valid thing to use in other equations.
>
> Thus, since the photon moves at the speed of light
Technically not correct. The angular momentum is moving at the speed of
light. The photon is equal to:
photon = angular momentum * c
To say that the photon is moving at c is equal to:
photon * c = ?
> AND because it has a non-zero equivalent of
> mass, it will possess a non-zero quantity of
> momentum (multiply the last equation by (c)):
> (m)(c)=(E)/(c)
>
> That momentum will also usually be quite a small
> magnitude, close to zero, BUT DEFINITELY > ZERO.
>
> Does that help???
I fully agree that there is a very small amount of mass and momentum in the
photon.
Dave
Bilge
field.
>I meant that if you have a positive charge and a negative electric field
>where you should have a postivie one, that defines negative energy for an
>electric field.
No, it doesn't.
>There is no other way to define it! The energy of the vector field depends
>on the source charge squared, or if you think in terms of a capacitor it
>is the voltage squared. So you cannot get any negative energy unless your
>capacitor has a negative value.
Since capaitance is a purely geometric quantity and lengths are positive,
you've answered your own question.
>A negative valued capacitor is
>just a capacitor whose impedance has a 180 phase shift in the complex
>plane. Then you have a battery, or in effect as you add more positive
>charge you get a more negative electric field.
Nice try, but batteries require chemical processes to produce that energy.
>If you disagree, then please define for me what is negative energy in
>an electric field?
First, you brought up electric fields. I'm just telling how the energy
is defined.
> Your original statment that I replied to implies you do not know, so
>how do you know I'm wrong?
Because I know how the energy in of an electric field is defined.
The energy desnity is given by w = |E|^2/8pi. If you want to look
it up you can find it in one or more of (1) jackson, (2) griffiths,
(3) lorrain & corson.
>I think you just like to be disagreable.
I think you just like to be wrong.
[...]
>> You've replaced a parameter (`c') which has no physics associated with
>> it's value and replaced it with two constants which only exist to convert
>> units invented before anyone ever realized electricity and magnetism were
>> the same phenomenon.
>
>I am being consistent. Consistent with the fact that the original and more
>accurate form of writing the energy density term includes the ratio between
>the local source strength and the local field strenth, which includes the
>sign of both. By eliminating those constants you are eliminating that ratio
nonsense.
>from the equation for energy density which then results in the Weak Energy
>Condition and Strong Energy Condition, the violation of which should not
>even be an issue. (See Wald, General Relativity, pg 219. for a definition of
>the Weak and Strong Energy Conditions.)
And?
>> Yes, it is. The geometry is still flat.
>
>Only to an uncharged mass!
I really don't care what you believe at this point.
>> After all, E&M is derived
>> from exactly that assumption. The proper geometric jargon is "fiber
>> bundle". I'm not "fiber bundle conversant" to any great extent, so
>> you would be better looking it up than having me "wing it" and misusing
>> the terminology.
>
>Engineers don't need fiber bundles we need the equations of motion and ways
>to control them.
Look, YOU are the one that mentioned geometry and curvature. I've told
you previously what the electromagnetic equivalent to the curvature tensor
is. If you don't like it, fine. If you want to believe your own bullshit,
fine.
>You can keep your fiber bundles, that level of abstraction
>is useless to me.
Fine. Physics is basically useless to you. Now go away.
Bilge
Re: Quanta of Momentum to usenet:
>news:64050551.02100...@posting.google.com...
>> > Give a mathematical example in the real world where the equations are
>not valid.
>> >
>> Here are two:
>> 1. Photons. They have energy and momentum, but no mass.
>> 2. CRTs larger than about 25". In order to get them to work well,
>> designers must use relativistic mechanics, not classical mechanics.
>
>Neither are mathematical examples. The first situation you listed is
>patently wrong. If the mass in the momentum is zero, then there is no
>momentum. You have no genuine mathematical proof for your claim.
If m = 0, E^2 = p^2.
[...]
>> There is only a small difference: If the proportionality constant
>> happens to be 1, then they are equal.
>
>But the proportionality constant is not 1, it is c. You arbitrarily put 1
>in place of c. That's not real math.
You don't know any real math.
[...]
>> > > The equation E^2 = (p*c)^2 + (m*c^2)^2 is valid (empirically) for all
>> > particles.
>> >
>> > This is your unproven, personal opinion.
>> >
>> No. This is a fact. Experimentally verified.
>
>Direct me to a published paper that supports your claim. I will deal with
>the author's of the papers directly if they are still alive.
See any publication on accelerator design, charged particle beam optics,
charged particle spectroscopu, gamma ray spectroscopy, scattering, etc.,
the list is endless.
>
>> No. You seem to have a reading comprehension problem.
>
>You have a physics and mathematics comprehension problem. And it's not your
>fault, entirely. You were brainwashed at an early age and haven't been
>deprogrammed yet.
No. YOU have an all around comprehension problem which involves delusions
of grandeur.
[...]
>
>If the mass of the photon is zero, then the momentum of the photon is zero.
That is a totally idiotic statement.
>There is only one mass associated with the photon.
zero.
[...]
>A person with a reading comprehension problem can't tell the difference
>between a rhetorical statement and a factual statement. The point is, you
>are stating and reaffirming that you are replacing c with 1. You are
>admitting that you are arbitrarily changing the units to make the equation
>fit your preconceived idea.
That's an asinine statement. The only "preconception" is on your part,
for not being able to fathom a definition of velocity that differs from
the one you grew up with.
[...]
You win the golden shovel award for september.
vernonner3voltazim
> > I'd like to try clearing some air, here.
> > As we all know, a photon consists of energy,
> > and E=mc^2
> > I shall modify that like this: (E)=(m)(c)(c)
> > because it's easier to manipulate algebraically.
>
> Wait a minute. Just because the photon consists of energy does not mean
> that the photon IS energy. And in fact,
>
> photon <> energy
>
> photon = energy * length
>
Hmmm...
The closest equations I've ever seen, to something like that,
is [first]: (E)=(h)(f)
where (E) is the energy of the photon,
(h) is Planck's Constant,
and (f) is the photon's frequency (cycles per second, or Hertz).
[second]: (L)=(c)/(f)
where (L) is the photon's wavelength (usually a Greek Lambda),
(c) is the speed of light,
and (f) is again the photon's frequency,
Therefore, to relate (L) and (E) in the same equation:
(f)=(c)/(L)
(E)=(h)(c)/(L)
(h)(c)=(E)(L)
This does NOT look like what you wrote!
What evidence do you have to support your equation???
Current physicists have tested the accuracy of the
equations I wrote, to something like 18 significant
figures of precision. (Quantum Electrodynamics is
the #1 theory in Physics, with respect to theory
agreeing with experiment. Therefore, if you come
up with any equations that yield different results,
they can't be more different than in the 19th decimal
place or so!)
> > That non-zero mass-equivalent is a perfectly
> > valid thing to use in other equations.
> >
> > Thus, since the photon moves at the speed of light
>
> Technically not correct. The angular momentum is moving at the speed of
> light. The photon is equal to:
>
> photon = angular momentum * c
>
> To say that the photon is moving at c is equal to:
>
> photon * c = ?
Hmmm...again.
The only time I've ever suspected that a photon might
have angular momentum is when it comes in the form
known as "circularly polarized". All the other
photons I've ever read about are purely linear.
So, on what evidential basis can you claim that all
photons have angular momentum???
Dirk Van de moortel
"David Thomson" <ne...@volantis.org> wrote in message news:WpNn9.10855$OB5.1...@newsread2.prod.itd.earthlink.net...
Corrected.
You know, you can always change your mind about things.
Fumbles can be erased too, "mortalized" so to speak ;-)
Dirk Vdm
David McAnally
>"David McAnally" <D.McAnally@i'm_a_gnu.uq.net.au> wrote in message
>news:anfuqb$q76$1...@bunyip.cc.uq.edu.au...
>> >The smallest unit of momentum is the smallest unit of the mass times the
>> >smallest length it can travel divided by the smallest time interval it
>can
>> >travel in.
>>
>> This contradicts even the most elementary mathematics. As a divisor
>> decreases, the quotient INcreases. So even if you are only going to use
>> naive mathematical principles, the smallest unit of momentum would be the
>> smallest unit of mass times the smallest length divided by the LARGEST
>> time interval. That is assuming that everything is quantized.
>Yes, you are correct that a smaller value is produced by increasing the
>time. I was talking about the smallest unit of momentum *in terms of
>quanta.*
And do you know what the momentum operator looks like????? Do you know
anything of quantum mechanics? Are you aware of the canonical commutation
relations? Do you not realize that one *necessary* consequence of the
canonical commutation relations is that momentum has a *continuous*
spectrum which consists of the *entire* real line? You seem to be wanting
to rewrite the laws of quantum mecahanics without understanding the
present theory. Can you tell me about your understanding of the
postulates of quantum mechanics.
>The discussion in this thread was about devising a quantum of
>momentum.
And the question is why do it, when the spectrum of the momentum operator
was already determined in the early days of quantum mechanics.
>The dimensions of momentum (and all other units) are what can be
>quantized, not the derived units. Every unit that has a fraction will
>encounter the same problem.
What possible justification can you give for such an outrageous comment?
>> >The smallest length the electron can travel divided by the
>> >smallest time it can travel in is the speed of light.
>>
>> This statement was made without justification. HOW do you know that it is
>> the speed of light?
>I really don't know this. I deduce this.
That last comment does not even make sense. If you can deduce the claim
that you made, then you certainly know it, so your statements here
directly contradict each other. Also, you have given no evidence that you
*can* deduce such a claim. Let us see the deduction.
>The speed of light in a vacuum is accurately measured. The Compton
>wavelength is widely accepted as the quantum wavelength. This leaves the
>quotient of Compton wavelength divided by the speed of light as the quantum
>time.
Why???? Justify. And what possible justification can you give fort
treating the Compton wavelength as a quantum of length, anyway?
>It must be a quantum time if the speed of light is constant and the
>Compton wavelength is quantum.
>> >E^2 = (p*c)^2 + (m*c^2)^2
>>
>> >In the case of the electron, what you have is
>>
>> >energy^2 = energy^2 + energy^2.
>>
>> This is a statement involving dimensions.
>And...??? Is that your scientific rebuttal?
Okay, it is a statement involving dimensions *****ONLY****. There was
absolutely quantitative content to your statement
"energy^2 = energy^2 + energy^2".
>I not only make a statement of
>dimensions (which proves I'm right)
In what way do you possibly believe that it proves that your point is
right?
>I also make a statement of values.
No, you wrote "energy^2 = energy^2 + energy^2". That is a statement about
dimensions ONLY. There is no quantitative content to your statement.
>Since energy is the common dimension, the dimensions can be divided out of
>the equation and leave us only with values.
So?
>1 = 1 + 1
No. You have made the statement that a quantity whose dimension is
energy^2 is the sum of two quantities each of whose dimensions are
energy^2. There was NO quantitative content. If I wrote that
3m = 1m + 2m,
then your logic would dictate that that was also wrong because it could be
noted that the equation looked like length = length + length, and then you
would get 1 = 1 + 1 when you set length equal to 1.
The argument looks crazy here but it carries EXACTLY the SAME amount of
rigour as your argument, because it uses EXACTLY the SAME logic as your
argument. The upshot is that the above argument is flawed, and so os
yours. Since the arguments are LOGICALLY EQUIVALENT to each other, you
cannot claim that one is flawed and the other is not.
The fact is that length in my example above and energy^2 in your argument
merely describe the DIMENSIONS of the quantities, and cannot be set equal
to 1.
>You and the entire scientific establishment that believes this voodoo math
>are simply wrong. The proof is clear. There is no scientific rebuttal to
>my argument.
There is no rebuttal to person deluded enough to believe that there is
quantitative content in a statement which is completely devoid of
quantitative content.
>> >If energy has the value of 1, you get 1 = 1 + 1. That is not a valid
>> >equality.
>>
>> This logic is invalid. You are assumimg that p*c = E and m*c^2 = E
>> without any justification.
>Excuse me? Hello? Is anybody home?
More so with me than with you obviously.
>Tell us what p*c is equal to.
It is the speed of light multiplied by the momentum. In the case of a
stationary clasiical particle, its value is zero.
>Tell us what m*c^2 is equal to.
It is c^2 multiplied by the REST MASS of the particle (i.e. c^2 multiplied
by the mass of the particle in its rest frame, i.e. the REST energy of the
particle).
>It is
>standard physics knowledge that both of these products are equal to energy.
Rubbish. You are possibly thinking of E = mc^2, where m describes
constant of proportionality between the momentum and the velocity, also
sometimes called the mass. This quantity m is NOT equal to the rest mass,
which is the mass that appears in the equation E^2 = (pc)^2 + (mc^2)^2,
UNLESS the particle is at rest. As far as E = pc is concerned, you would
only have seen that equation when it comes to photons, and you would
DEFINITELY not have seen the equation when it comes to massive particles.
>I have plenty of justification for making this point.
No you don't. The two equations E = mc^2 and E = pc arise under MUTUALLY
EXCLUSIVE circumsrtances (i.e. they are never simultaneously satisfied).
LEARN about a theory before you criticise it. Making uneducated comments
does not make you look good.
>This is exactly why
>modern physics is so full of crap. There are guys like you who change the
>rules whenever it seems convenient.
No. It's because you can't be bothered applying yourself to actually
bother learning it.
>> Your statement above was statement about
>> DIMENSIONS and was not quantitative. You are assuming a quantitative
>> result from a non-quantitative statement. You might as well take the
>> classical equation E = 1/2*m*v^2 + m*g*h, determine that that means that
>> energy = energy + energy, and conclude that that means that 1 = 1 + 1,
>> because that uses *exactly* the same logic, and is just as invalid in that
>> case as it is in yours.
>Exactly. You are exactly right on that.
This is the most staggering comment that I have seen yet. So now you
don't believe in Newtonian mechanics either. Do you want to go back to
Aristotle??????
The rest of your message is deleted as it is becoming obvious that you are
just another perverted little troll who will be going straight into my
kill file. *PLONK!*
David McAnally
---------
David McAnally
>David Thomson wrote:
>> "Patrick Reany" <re...@asu.edu> wrote in message
>> news:3D9B9802...@asu.edu...
>>
>> <personal opinions snipped>
>>
>> Nothing left to respond to. Patrick, if you are going to talk personal
>> opinions, I have no interest. All I'm interested in is verifiable science
>> using verifiable mathematics principles.
>>
>> Dave
>Then why did you bother to reply? (Rhetorical
>question only) Is this a physics / relativity NG
>or a mathematical NG, anyway? Maybe you're
>on the wrong NG. Besides, there's more
>to physics than facts. Where does meaning fit
>into your sanitized notion of physics?
>Patrick
Don't inflict Thomson on a mathematics group. He is a mathematical
ignoramus. It's bad enough that he's here as a physics ignoramus.
David McAnally
--------
David McAnally
>Hmmm...again.
>The only time I've ever suspected that a photon might
>have angular momentum is when it comes in the form
>known as "circularly polarized". All the other
>photons I've ever read about are purely linear.
>So, on what evidential basis can you claim that all
>photons have angular momentum???
Photons have spin 1. Circularly polarized photons have angular momentum
\hbar or -\hbar along their direction of motion. A linearly polarized
photon is a superposition of circularly polarized photons, so the angular
momentum operator is nonsigular even on linearly polarized photons.
David McAnally
---------
David McAnally
>David Thomson wrote:
>> photon = h * c (units are angular momentum times velocity)
>>
>> momentum
>> ----------- = velocity
>> mass
>>
>> How does momentum without a mass equal a photon?
>>
>> Dave
>You're making the mistake of thinking about mass
>and momentum as real essences that can't be messed
>about with. But in science this is not so. Theories get
>to define their terms arbitrarily so long as those definitions
>are consistent with conventional norms. After that,
>those definitions are either instrumental for producing
>workable theories or they are not.
Thomson appears to have also made the mistake os assuming that quantum
mechanics is just classical mecahics with quanta of quantities thrown in.
Presumably the full effect of the actual theory was too much for his mind
to cope with.
David McAnally
-----------
David McAnally
>Why not? Hasn't everything else in Physics been quantized?
>I've worked out some speculations in reasonable mathematical
>detail, located at this URL:
>http://www.nemitz.net/vernon/MOMEQUAT.htm
Simple and fast answer: Take space to be R^3 (so it is infinite in extent
in all directions), then the x-component of the momentum satisfies
[q_x,p_x] = i \hbar, where q_x denotes the x-coordinate operator (i.e.
multiplication by x). For any constant P with the dimensions of momentum,
it is an immediate consequence that
exp(-i P q_x/\hbar) p_x exp(i P q_x/\hbar) = p_x + P.
This means that if \lambda is in the spectrum of p_x, so is \lambda + P.
But this is true for *any* value of P, so the spectrum of p_x consists of
the *entire* real line, and is continuous. There is therefore no quantum
of momentum.
David McAnally
------------
Paul Cardinale
I see your vying with the kenseto for dimmest bulb on the internet.
(Hmmm... is your TV controlling your mind?)
Paul Cardinale
David Thomson
> Patrick Reany <re...@asu.edu> writes:
> Thomson appears to have also made the mistake os assuming that quantum
> mechanics is just classical mecahics with quanta of quantities thrown in.
> Presumably the full effect of the actual theory was too much for his mind
> to cope with.
Perhaps Thomson critically reviewed the QM literature and refused to accept
that reality is based on voodoo math? Perhaps the present day QM students
have neither the wherewithal to admit the math doesn't work nor the ability
to see the correct perception of subatomic structure?
Dave
David Thomson
> Corrected.
Thanks.
> You know, you can always change your mind about things.
> Fumbles can be erased too, "mortalized" so to speak ;-)
I find it interesting that you would post this particular message on your
website although you have not been a participant in the discussion. It's
kind of like shooting someone in the back.
Why don't you weigh in on this discussion and give me a mathematically
correct explanation of the equivalence between energy and momentum? I
wouldn't want to develop the opinion of you as a cowardly opportunist.
Exactly at what point does the proportionality of energy and mass (or energy
and momentum) become the equivalence of energy and mass (or energy and
momentum?)
Dave
Dirk Van de moortel
"David Thomson" <ne...@volantis.org> wrote in message news:gTYn9.12394$lV3.1...@newsread1.prod.itd.earthlink.net...
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
> in message news:zMSn9.148163$8o4....@afrodite.telenet-ops.be...
> > Corrected.
>
> Thanks.
>
> > You know, you can always change your mind about things.
> > Fumbles can be erased too, "mortalized" so to speak ;-)
>
> I find it interesting that you would post this particular message on your
> website although you have not been a participant in the discussion. It's
> kind of like shooting someone in the back.
As far as this discussion was concerned, you can call me an
observer.
>
> Why don't you weigh in on this discussion and give me a mathematically
> correct explanation of the equivalence between energy and momentum? I
> wouldn't want to develop the opinion of you as a cowardly opportunist.
I don't care about your opinion of me and I don't care about what
opinions you would or wouldn't want to develop
>
> Exactly at what point does the proportionality of energy and mass (or energy
> and momentum) become the equivalence of energy and mass (or energy and
> momentum?)
Perhaps you could try to understand the 114 attempts at explaining
first. If you can work your way through those in a way that makes
me decide that you are even remotely interested, then maybe I will
have a go at it.
It might help to ask questions with the intent to learn.
Meanwhile you are welcome to develop any opinion you fancy of me.
Dirk Vdm
Dirk Van de moortel
"David Thomson" <ne...@volantis.org> wrote in message news:nrYn9.12362$lV3.1...@newsread1.prod.itd.earthlink.net...
Perhaps Thomson's idea of the QM literature consists of the
pathetic set containing Gribbin's "In Search of Schroedinger's Cat",
and Penrose's "The Emperor's New Mind".
Dirk Vdm
David Thomson
> Don't inflict Thomson on a mathematics group. He is a mathematical
> ignoramus. It's bad enough that he's here as a physics ignoramus.
It's easy to throw words like a child. But let's see you throw a corrected
mathematical interpretation where E is correctly shown to be equivalent to
p. There is a simple way of proving this. Try defining E, p and c on a
MathCAD spreadsheet and see if it allows you to arbitrarily change c to 1.
There IS an ignoramus in this group, in fact there are many. They
arrogantly claim to be educated in the complexities of Special Relativity
and Quantum Mechanics and they can't even do basic algebra correctly. The
ignoramus' in physics circles would sooner allow themselves to be swooned by
the boondoggle of SR and QM than to seek a more rational and mathematically
correct explanation of subatomic behavior.
If anybody would have difficulty in a mathematics group, it would be a
modern day physicist.
Dave
David Thomson
> > > I'd like to try clearing some air, here.
> > > As we all know, a photon consists of energy,
> > > and E=mc^2
> > > I shall modify that like this: (E)=(m)(c)(c)
> > > because it's easier to manipulate algebraically.
> >
> > Wait a minute. Just because the photon consists of energy does not mean
> > that the photon IS energy. And in fact,
> >
> > photon <> energy
> >
> > photon = energy * length
> >
>
> Hmmm...
> The closest equations I've ever seen, to something like that,
> is [first]: (E)=(h)(f)
> where (E) is the energy of the photon,
> (h) is Planck's Constant,
> and (f) is the photon's frequency (cycles per second, or Hertz).
The electron also has angular momentum (h) and frequency. Everything with
mass has energy. So the equation you are using applies to everything
whether it is a photon or an electron, proton, or neutron. The point is
that energy is NOT a photon. A photon, like electrons, protons, and
neutrons, has a different set of dimensions from energy. The electron,
proton, and neutron have dimensions of h and the photon has dimensions of h
* c.
> [second]: (L)=(c)/(f)
> where (L) is the photon's wavelength (usually a Greek Lambda),
> (c) is the speed of light,
> and (f) is again the photon's frequency,
All photons have the same inherent wavelength (Compton wavelength,) just as
all electrons, protons, and neutrons do. A photon is a quantum amount of
electromagnetic radiation that is always inititiated with the same amount of
angular momentum and traveling always at the speed of light (in a vacuum.)
The confusion in physics is in assuming that the photon is light and or
energy. The photon is not light. The photon is a quantum of light. Light
is the result of photons being emitted at a constant frequency, and this is
the only frequency that varies. The frequency of the light varies, not the
frequency of the photon. The frequency of the photon is one of its defining
characteristics.
h = angular momentum
E = h * f (quantum frequency)
photon = h * c
light = h * c * f (light frequency)
Using the above dimensions gives the proper relationship between each of the
units.
> Therefore, to relate (L) and (E) in the same equation:
> (f)=(c)/(L)
> (E)=(h)(c)/(L)
> (h)(c)=(E)(L)
> This does NOT look like what you wrote!
It is the same balanced units as I wrote. The only difference between our
views is that you are insinuating that E is not only the energy of the
photon, but the photon itself. I am clearly identifying the photon and
light as separate units from energy.
> What evidence do you have to support your equation???
> Current physicists have tested the accuracy of the
> equations I wrote, to something like 18 significant
> figures of precision. (Quantum Electrodynamics is
> the #1 theory in Physics, with respect to theory
> agreeing with experiment. Therefore, if you come
> up with any equations that yield different results,
> they can't be more different than in the 19th decimal
> place or so!)
I come up with the same precision but I reorder the equations to fit the
proper units. I also use a particle model that exactly fits the dimensions
of the particle, not the balls in orbit models. Using my method there is no
wave/particle duality as all particles are neither balls of mass nor waves
of energy. All particles in my model are angular momentum. And it is
because physical particles and photons are all angular momentum that one can
become the other without defying logic. Presently, it defies all logic that
a ball of matter can become a wave or that a wave can become a ball of
matter.
The reason Einstein ended up with E = p is because Einstein (and everyone
after him) used the wrong units for the photon. The math had to be fudged
in order for the correct solutions to come out in the math. But if the
correct units are used from the beginning, there is no need to violate the
rules of mathematics in order to prove the laws of physics.
> > > That non-zero mass-equivalent is a perfectly
> > > valid thing to use in other equations.
> > >
> > > Thus, since the photon moves at the speed of light
> >
> > Technically not correct. The angular momentum is moving at the speed of
> > light. The photon is equal to:
> >
> > photon = angular momentum * c
> >
> > To say that the photon is moving at c is equal to:
> >
> > photon * c = ?
>
> Hmmm...again.
> The only time I've ever suspected that a photon might
> have angular momentum is when it comes in the form
> known as "circularly polarized". All the other
> photons I've ever read about are purely linear.
For your own edification, go over the equations for a linear photon again.
Just for fun, entertain the idea that you are critically analyzing the
method in the mathematics. Look carefully at all the mathematical tricks
being used. If you get to a point where the math follows unconventional
rules (such as E = p) then stop and see how far you've gotten. Also,
critically analyze any assumptions, such as the assumption that energy equal
photon. I have already posted in previous messages equations for the
transfer of angular momentum that explain the photoelectric effect, Compton
effect, and pair production. I'll do it again if you missed it.
> So, on what evidential basis can you claim that all
> photons have angular momentum???
I can use the exact same experiments that claim the photon is a linear
traveling particle. The equations that "predict" a linear traveling
particle are not based on seeing a photon flying through space. There is no
equipment that can see a photon at greater resolution than the materials
that compose the equipment.
Further, I can use experiments such as the Bose Einstein condensates and MRI
technology to support my theory. Further, my theory will explain the
mechanics of the Faraday motor. Further, my theory explains where the
magnetic charge is in Maxwell's 4 equations. My theory ties all these
things together and much more.
Other than revealing the voodoo math used by the Standard Model, my physics
does not prove the outcome of QM to be wrong. It agrees with the outcome of
QM and SR. In fact, if it were not for the accuracy of the QM data, I would
not have been able to develop my theory.
Dave
David Thomson
Dave,
I agree that the spin of a photon is 1. This supports my theory that a
double sized (2 * 1/2 spin) electron angular momentum produces the photon.
I realize physics today views hbar as a unit of spin. I have commented on
why I disagree with hbar in other threads. The angular momentum is merely h
and the spin is inherent to the angular momentum, it is not due to a factor
of either 4pi or 2pi.
What determines whether one can observe the circular polarization of the
photon is the distance and angle the photon is being measured from the
emitter. The further away from the emitter, the more the photon appears to
be a "wave packet" type event. The wave packet is nothing more than
observing a section of the angular momentum due to the angular momentum
having spread out radially.
Dave
Todd Desiato
Therefore if you reduce the capacitance by adding a negative capacitance you
make length shorter. Gee, I wonder what you can use that for? A long trip
perhaps?
> >A negative valued capacitor is
> >just a capacitor whose impedance has a 180 phase shift in the complex
> >plane. Then you have a battery, or in effect as you add more positive
> >charge you get a more negative electric field.
>
> Nice try, but batteries require chemical processes to produce that
energy.
The impedance of a capacitor is;
Z(w) = 1/(jwC) = (1/wC)exp(-j*pi/2)
w = freqnecy
j = sqrt(-1)
A negative capacitor is just a 180 phase shift in the complex plane. If you
can't see that your blind. A capacitor stores energy and "then" can act as a
source of energy.(QED)
Best Regards,
Todd Desiato
vernonner3voltazim
>
> >Hmmm...again.
> >The only time I've ever suspected that a photon might
> >have angular momentum is when it comes in the form
> >known as "circularly polarized". All the other
> >photons I've ever read about are purely linear.
>
> \hbar or -\hbar along their direction of motion. A linearly polarized
> photon is a superposition of circularly polarized photons, so the angular
> momentum operator is nonsigular even on linearly polarized photons.
>
> David McAnally
Thanks, ...but:
The "spin 1" of photons and the "spin 1/2" of various fermions
was, I had read, considered to be something rather distinct
from ordinary classical angular momentum. That is, while it
was CALLED "spin", it was actually just a convenient handle
to allow discussion of something more abstruse. Yes, very
recently I encountered something that said, well, it could be
that the various subatomic particles can also have something
like ordinary angular momentum, but I got the impression that
it was optional, and not a built-in feature of the particles.
Do we have good evidence in favor of, for example, the
superposition of counter-cirularly-polarized photons? While
I agree that because they are spin-1 bosons, there is nothing
to prevent such a possibility, I'd never before heard that
this was the rule, rather than the exception! Besides, if
photonic angular momentum was optional, then a great many
photons could indeed be, as I indicated above, purely linear.
vernonner3voltazim
>
> > > > I'd like to try clearing some air, here.
> > > > As we all know, a photon consists of energy,
> > > > and E=mc^2
> > > > I shall modify that like this: (E)=(m)(c)(c)
> > > > because it's easier to manipulate algebraically.
> > >
> > > Wait a minute. Just because the photon consists of energy does not mean
> > > that the photon IS energy. And in fact,
> > >
> > > photon <> energy
> > >
> > > photon = energy * length
> > >
> >
> > Hmmm...
> > The closest equations I've ever seen, to something like that,
> > is [first]: (E)=(h)(f)
> > where (E) is the energy of the photon,
> > (h) is Planck's Constant,
> > and (f) is the photon's frequency (cycles per second, or Hertz).
>
> The electron also has angular momentum (h) and frequency. Everything with
> mass has energy. So the equation you are using applies to everything
> whether it is a photon or an electron, proton, or neutron. The point is
> that energy is NOT a photon. A photon, like electrons, protons, and
> neutrons, has a different set of dimensions from energy. The electron,
> proton, and neutron have dimensions of h and the photon has dimensions of h
> * c.
I think you need to study Bohm's Interpretation of Quantum Mechanics.
Certainly a photon is an "entity", and, because it is a quantum, it
has a chance of being perpetually stable. Yes, such an entity has
various properties, such as being a container of some energy, with
zero actual mass. However, the form that energy takes here is as
an enmeshed set of electric and magnetic fields. Note that by
themselves (ignoring the subject of photons for the moment), even
static electric and magnetic fields are considered to be FORMS of
energy. It is thus quite easy to assume that a photon is also
nothing more than a form of energy -- to assume that a photon IS a
piece of energy. (ESPECIALLY after one learns that macroscopic
static electric and magnetic fields are described, in Quantum
Mechanics, as being "made" from hordes of virtual photons!)
Anyway, there is a standard description of a photon, first
recognized by James Clerk Maxwell, long before Q.M. came by. He
figured out the relationship between electricity and magnetism,
and discovered that whenever an electric field changes, it causes
a magnetic field to come into existence, and vice-versa. Thus,
as an electric field collapses toward zero, it causes a magetic
field to come into existence from zero. AT zero, the electric
field can no longer supply energy to the magnetic field, so the
magnetic field begins to collapse -- which pumps new life into
the electric field! When the magnetic field has collapsed to
zero, it can no longer feed the electric field... et cetera.
Maxwell even figured out the rate at which such fields would
change/propagate, and it turned out to be the speed of light.
So, he wrote that he suspected that light was an electromagnetic
wave, but proof didn't arrive for a few more decades.
Quantum Mechanics has never changed that description. The ONLY
beef I have concerns "standard portrayals" of that description.
One such is at http://lompado.uah.edu/EMWave.htm
Note that that drawing shows BOTH the electric and magnetic
fields reaching zero (and reaching maximum values) IN SYNC!!!
(A cycle from 0 Energy to 200% Energy is an average of 100%,
but that cycle is slow enough that Uncertainty does NOT allow
that much Violation of Energy Conservation!!!!!!!!!!!!!!!!!)
Instead, to actually correspond with the description, the
two drawn fields should be desynchronized by perhaps 90 degrees.
I can't draw it here, but I can try to describe how the "waving"
intensity levels should more-or-less be associated/portrayed:
(the vertical bars are to indicate equal time intervals)
| | | | | | | | | |
EF: 100% 71% 0% -71% -100% -71% 0% 71% 100%
MF: 0% 71% 100% 71% 0% -71% -100% -71% 0%
degs: 0 45 90 135 180 225 270 315 360
Since the sine and cosine of 45 degrees is .707, I rounded
the energy levels to 71% -- I'd rather see 50%, again due
to Energy-Conservation, but don't see how to get there,
without turning a sine wave into a triangular wave. However,
Uncertainty may allow this much temporary violation, and so
I'll stick with those 71% values. Or, perhaps Physics just
doesn't KNOW exactly what shapes those waving fields really
have, inside a single photon....
Having wandered far from the original topic, I now try
to recover, and say, "One of the standard descriptions of
a photon is 'Energy in Motion'. Einstein supposedly required
all forms of pure energy to move at the speed of light." So,
since a photon moves at the speed of light, the implication
is that it is pure energy.
(Ever since Einstein, there have been arguments about
whether or not gravity waves travel at light-speed. The
problem was, while certainly they carried energy, were
they a form of PURE energy? If not, such as I described
in my paper at http://www.nemitz.net/vernon/MOMEQUAT.htm
then gravity waves might move at quite a range of velocities.)
> > [second]: (L)=(c)/(f)
> > where (L) is the photon's wavelength (usually a Greek Lambda),
> > (c) is the speed of light,
> > and (f) is again the photon's frequency,
>
> All photons have the same inherent wavelength (Compton wavelength,) just as
> all electrons, protons, and neutrons do.
You have GOT to work on your phrasing! My first impression of that
statement is that it implies every photon has the same wavelength,
period. After reading what followed, however, I could tell you
were talking about how a wavelength is computed, and not talking
about a particular value.
> A photon is a quantum amount of
> electromagnetic radiation that is always inititiated with the same amount of
> angular momentum and traveling always at the speed of light (in a vacuum.)
> The confusion in physics is in assuming that the photon is light and or
> energy. The photon is not light. The photon is a quantum of light. Light
> is the result of photons being emitted at a constant frequency, and this is
> the only frequency that varies. The frequency of the light varies, not the
> frequency of the photon. The frequency of the photon is one of its defining
> characteristics.
Ahem. The photon may or may not be a particle of light. If not light,
it may be a particle of microwave, or a particle of infrared, or a
particle of ultraviolet, et cetera. ON THE OTHER HAND, the word "light"
is sometimes used to refer to ALL electromagnetic waves, regardless of
whether they are "radio" or "gamma". By that definition, a photon is,
indeed, always "light"! So, don't jump on somebody about "light" unless
you know exactly how they are trying to use it, AND you know that that
particular usage is incorrect.
> h = angular momentum
> E = h * f (quantum frequency)
> photon = h * c
> light = h * c * f (light frequency)
>
> Using the above dimensions gives the proper relationship between each of the
> units.
OK, the "h" of Planck's Constant does indeed have the same
dimensions as classical angular momentum. However, Quantum
Physics is not Classical Physics, and one aspect of the
distinction, particularly the difference between Planck's
Constant and angular momentum, is described pretty well at:
http://kims.ms.u-tokyo.ac.jp/time/199905/0037.html
Therefore, while your version of the equation (E)=(h)(f) is
mathematically correct, it seems you have managed to
misinterpret its actual meaning. The photon has a wavelength
simply because it has some particular amount of Energy.
That amount of Energy is ALWAYS going to be associated with
the same wavelength, regardless of whether or not that
Energy takes the form of a photon, a neutrino, or some
other particle. Thus, for you to claim:
> > > photon = energy * length
is redundant and insignificant! Furthermore, most "light"
actually consists of vast numbers of photons that happen to
be travelling together. Thus:
instead of your (light)=(photon)(f)
the truth is more like: (light)=(photon)(quantity)
After all, every photon ALREADY HAS a frequency -- you said
so yourself. To qualify as visible light, that frequencey
merely needs to be within a certain range. Also, it happens
that in actual experiments, measurements made in pitch-dark
rooms, the human eye is capable of detecting individual
visible-light photons, about 1/6 of the time.
The people on the receiving end of those individual
test-photons said that the photons looked like tiny flashes
of LIGHT.
> > What evidence do you have to support your equation???
> > Current physicists have tested the accuracy of the
> > equations I wrote, to something like 18 significant
> > figures of precision. (Quantum Electrodynamics is
> > the #1 theory in Physics, with respect to theory
> > agreeing with experiment. Therefore, if you come
> > up with any equations that yield different results,
> > they can't be more different than in the 19th decimal
> > place or so!)
>
> I come up with the same precision but I reorder the equations to fit the
> proper units. I also use a particle model that exactly fits the dimensions
> of the particle, not the balls in orbit models.
Ahem, again! The "balls in orbit" models died back in the 1930s.
"Standing waves" replaced it.
> Using my method there is no
> wave/particle duality as all particles are neither balls of mass nor waves
> of energy.
Too bad. I find the wave/particle duality to be quite useful.
For example, Gravitation is directly proportional to Mass/Energy,
REGARDLESS of the FORM of that Mass or Energy. So, when one
wishes to contruct a "model" of Quantum Gravitation, how can
one directly associate a RATE of emission of virtual-gravitons
to the mass/energy of ANY object, massless or otherwise? Via
the wave/particle duality, which directly associates Mass/Energy
(via Momentum) with Frequency! For more info, see these essays:
http://www.nemitz.net/vernon/STUBBED2.pdf
http://www.nemitz.net/vernon/BALANCD2.pdf
http://www.nemitz.net/vernon/IMAGNARY.pdf
> All particles in my model are angular momentum. And it is
> because physical particles and photons are all angular momentum that one can
> become the other without defying logic. Presently, it defies all logic that
> a ball of matter can become a wave or that a wave can become a ball of
> matter.
Yes, that defies logic, BUT that is not what is claimed by
Quantum Physics! What is claimed is that particles always
have characteristics of BOTH particles and waves, at the
same time. No particle ever "becomes" one or the other.
EVEN SO, there is a way to "mentally picture" the goings-on
at the quantum level, without really trying to accept at
face value bald words like the preceding. READ THIS:
http://www.nemitz.net/vernon/GHOSTLY.pdf
> The reason Einstein ended up with E = p is because Einstein (and everyone
> after him) used the wrong units for the photon. The math had to be fudged
> in order for the correct solutions to come out in the math. But if the
> correct units are used from the beginning, there is no need to violate the
> rules of mathematics in order to prove the laws of physics.
Now you are being unclear again. What is "p" in that context?
If it is momentum, then you are most absolutely WRONG about
Einstein (or any other top physicist) claiming that
Energy = momentum.
> > > > That non-zero mass-equivalent is a perfectly
> > > > valid thing to use in other equations.
> > > >
> > > > Thus, since the photon moves at the speed of light
> > >
> > > Technically not correct. The angular momentum is moving at the speed of
> > > light. The photon is equal to:
> > >
> > > photon = angular momentum * c
> > >
> > > To say that the photon is moving at c is equal to:
> > >
> > > photon * c = ?
> >
> > Hmmm...again.
> > The only time I've ever suspected that a photon might
> > have angular momentum is when it comes in the form
> > known as "circularly polarized". All the other
> > photons I've ever read about are purely linear.
>
> For your own edification, go over the equations for a linear photon again.
> Just for fun, entertain the idea that you are critically analyzing the
> method in the mathematics. Look carefully at all the mathematical tricks
> being used. If you get to a point where the math follows unconventional
> rules (such as E = p) then stop and see how far you've gotten.
Sorry, I have no idea what false physics you are talking about.
The equations for photons that I already posted look plenty fine
and sensible to me!
> Also,
> critically analyze any assumptions, such as the assumption that energy equal
> photon. I have already posted in previous messages equations for the
> transfer of angular momentum that explain the photoelectric effect, Compton
> effect, and pair production. I'll do it again if you missed it.
That must have been some other thread. In this thread, I'd like to
stick with quanta of pure momentum that have no mass and can move at
any velocity, including zero (and even including FTL). Thanks!
> > So, on what evidential basis can you claim that all
> > photons have angular momentum???
>
> I can use the exact same experiments that claim the photon is a linear
> traveling particle. The equations that "predict" a linear traveling
> particle are not based on seeing a photon flying through space. There is no
> equipment that can see a photon at greater resolution than the materials
> that compose the equipment.
>
> Further, I can use experiments such as the Bose Einstein condensates and MRI
> technology to support my theory. Further, my theory will explain the
> mechanics of the Faraday motor. Further, my theory explains where the
> magnetic charge is in Maxwell's 4 equations. My theory ties all these
> things together and much more.
>
> Other than revealing the voodoo math used by the Standard Model, my physics
> does not prove the outcome of QM to be wrong. It agrees with the outcome of
> QM and SR. In fact, if it were not for the accuracy of the QM data, I would
> not have been able to develop my theory.
>
> Dave
Whatever. That was not very specific.... (the guy who posted after you
did a better job!)
vernonner3voltazim
Well, there are quantities and then there are quanta. Sure, to the
extent that some quantity of momentum is allowed to be part of a
continuum, then obviously NO momentum-possessing object HAS to
acquire or lose momentum in discrete amounts.
HOWEVER, one could possibly say the exact same thing with respect
to Kinetic Energy. Nevertheless, quanta of Energy are known to
exist! So why not any stand-alone quanta of momentum?
You may have missed the behind-the-scenes logical argument for
such a thing as a stand-alone quantum of momentum. To wit:
(1) Pretend there can be such a thing as "negative" mass/energy.
(2) Consider a simple interaction between ordinary and negative mass:
(m)(v)----> <----(-m)(-v) Assume equal/opposite masses/velocities
(3) Compute resulting mass, kinetic energy and momentum, and you
get: Zero mass, Zero kinetic energy, TWO momentum-units!
(4) That sure looks like a stand-alone quantity of pure momentum
to me! Furthermore, if you alter the reference frame, you STILL
get 2 momentum-units, every time. Looks like a quantum to me!
Elsewhere in this overall Thread are additional descriptions
(As I write this, Messages 63 and 75) written by me, that you
might review. Or, for a more complete description, including 5 or 6
corroborating/similar hypotheses, and suggested experiments, see
this: http://www.halfbakery.com/idea/Gravity_20Waves#1025890913
Bilge
Re: Quanta of Momentum to usenet:
>>
>> >Hmmm...again.
>> >The only time I've ever suspected that a photon might
>> >have angular momentum is when it comes in the form
>> >known as "circularly polarized". All the other
>> >photons I've ever read about are purely linear.
>>
>> Photons have spin 1. Circularly polarized photons have angular momentum
>> \hbar or -\hbar along their direction of motion. A linearly polarized
>> photon is a superposition of circularly polarized photons, so the angular
>> momentum operator is nonsigular even on linearly polarized photons.
>>
>> David McAnally
>
>Thanks, ...but:
>The "spin 1" of photons and the "spin 1/2" of various fermions
>was, I had read, considered to be something rather distinct
>from ordinary classical angular momentum. That is, while it
>was CALLED "spin", it was actually just a convenient handle
>to allow discussion of something more abstruse. Yes, very
that results in the selection rules for level transitions in atoms
and nuclei, which is \Delta L = +/- 1. The reason "spin" is identified
with angular momentum is that when solving the dirac equation, you
end up with an additional degree of freedom which when added to the
mechanical angular momentum, L, commutes with the dirac hamiltonian.
Neither L nor S commute separately. However, while integral values
of spin may become orbital angular momenta and vice-versa, the spin
is strictly a relativistic feature.
>recently I encountered something that said, well, it could be
>that the various subatomic particles can also have something
>like ordinary angular momentum, but I got the impression that
>it was optional, and not a built-in feature of the particles.
>
>Do we have good evidence in favor of, for example, the
>superposition of counter-cirularly-polarized photons? While
>I agree that because they are spin-1 bosons, there is nothing
>to prevent such a possibility, I'd never before heard that
>this was the rule, rather than the exception! Besides, if
>photonic angular momentum was optional, then a great many
>photons could indeed be, as I indicated above, purely linear.
you wish. If you have a circular basis, in terms of plane
polarizations:
|+> = (1/sqrt(2))[|x> + i|y>]
|-> = (1/sqrt(2))[|x> - i|y>]
David McAnally
>D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) wrote/quoted:
>>
>> >Hmmm...again.
>> >The only time I've ever suspected that a photon might
>> >have angular momentum is when it comes in the form
>> >known as "circularly polarized". All the other
>> >photons I've ever read about are purely linear.
>>
>> Photons have spin 1. Circularly polarized photons have angular momentum
>> \hbar or -\hbar along their direction of motion. A linearly polarized
>> photon is a superposition of circularly polarized photons, so the angular
>> momentum operator is nonsigular even on linearly polarized photons.
>>
>> David McAnally
>Thanks, ...but:
>The "spin 1" of photons and the "spin 1/2" of various fermions
>was, I had read, considered to be something rather distinct
>from ordinary classical angular momentum.
That is correct, but orbital angular momentum in quantum mechanics
is also distinct from ordinary classical angular momentum.
>That is, while it
>was CALLED "spin", it was actually just a convenient handle
>to allow discussion of something more abstruse. Yes, very
>recently I encountered something that said, well, it could be
>that the various subatomic particles can also have something
>like ordinary angular momentum, but I got the impression that
>it was optional, and not a built-in feature of the particles.
Well, orbital angular momentum can only have integer magnitudes,
so it might be a bit difficult to generate a spin of 1/2 from
purely orbital considerations. On the other hand, elementary
particles can have orbital angular momentum as well as spin -
that's where Clebsch-Gordan coefficiants come in, relating
sets of common eigenvectors between distinct complete sets of
commuting operators.
>Do we have good evidence in favor of, for example, the
>superposition of counter-cirularly-polarized photons?
The internal Hilbert space for a photon is two-dimensional.
Left circular polarization and right circular polarization
are mutually orthogonal states, and therefore form a basis
for the internal Hilbert space. This means that all
internal states of the photon is a superposition of these
two states.
As far experimental evidence is concerned, I will leave that
for others to comment on, but I think that there is experimental
equipment which distinguishes between left circular polarization
and right circular polarization (such materials could be useful
in quantum cryptography).
>While
>I agree that because they are spin-1 bosons, there is nothing
>to prevent such a possibility, I'd never before heard that
>this was the rule, rather than the exception!
It's a question of how much you know of the principles of
quantum mechanics. All internal states of the photon are
linear superpositions of the circularly polarized states,
and in exactly the same manner, all states of the photon
are linear superpositions of the vertically polarized state
and the horizontally polarized state, purely as a consequence
of the fact that the internal Hilbert space is two-dimensional.
If the internal Hilbert space of a particle has dimension n,
then given any set of n mutually orthogonal states for the
particle, the internal state of the particle is a linear
superposition of these mutually orthogonal states in the
set. This is the ruke, and in the case of the photon, it
reduces to the comment that I made.
>Besides, if
>photonic angular momentum was optional, then a great many
>photons could indeed be, as I indicated above, purely linear.
A linearly polarized photon is a linear superposition of
circularly polarized photons.
David McAnally
--------
David McAnally
>D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) wrote in message news:<anpe3k$1fc$1...@bunyip.cc.uq.edu.au>...
>> vne...@pinn.net (vernonner3voltazim) writes:
>>
>> >Why not? Hasn't everything else in Physics been quantized?
>> >I've worked out some speculations in reasonable mathematical
>> >detail, located at this URL:
>> >http://www.nemitz.net/vernon/MOMEQUAT.htm
>>
>> Simple and fast answer: Take space to be R^3 (so it is infinite in extent
>> in all directions), then the x-component of the momentum satisfies
>> [q_x,p_x] = i \hbar, where q_x denotes the x-coordinate operator (i.e.
>> multiplication by x). For any constant P with the dimensions of momentum,
>> it is an immediate consequence that
>>
>> exp(-i P q_x/\hbar) p_x exp(i P q_x/\hbar) = p_x + P.
>>
>> This means that if \lambda is in the spectrum of p_x, so is \lambda + P.
>> But this is true for *any* value of P, so the spectrum of p_x consists of
>> the *entire* real line, and is continuous. There is therefore no quantum
>> of momentum.
>Well, there are quantities and then there are quanta. Sure, to the
>extent that some quantity of momentum is allowed to be part of a
>continuum, then obviously NO momentum-possessing object HAS to
>acquire or lose momentum in discrete amounts.
>HOWEVER, one could possibly say the exact same thing with respect
>to Kinetic Energy. Nevertheless, quanta of Energy are known to
>exist! So why not any stand-alone quanta of momentum?
That is because one is able to determine the nature of the energy
spectra from the form of the Hamiltonian operator.
>You may have missed the behind-the-scenes logical argument for
>such a thing as a stand-alone quantum of momentum. To wit:
>(1) Pretend there can be such a thing as "negative" mass/energy.
>(2) Consider a simple interaction between ordinary and negative mass:
> (m)(v)----> <----(-m)(-v) Assume equal/opposite masses/velocities
This does not describe a quantum system. What happens classically
after the collision?
>(3) Compute resulting mass, kinetic energy and momentum, and you
>get: Zero mass, Zero kinetic energy, TWO momentum-units!
Nett mass = 0, nett kinetic energy = 0, nett momentum = 2mv.
>(4) That sure looks like a stand-alone quantity of pure momentum
>to me! Furthermore, if you alter the reference frame, you STILL
>get 2 momentum-units, every time.
Under Galilean transformations.
>Looks like a quantum to me!
Why????
>Elsewhere in this overall Thread are additional descriptions
>(As I write this, Messages 63 and 75) written by me, that you
>might review. Or, for a more complete description, including 5 or 6
>corroborating/similar hypotheses, and suggested experiments, see
>this: http://www.halfbakery.com/idea/Gravity_20Waves#1025890913
David McAnally
-----------
Stephen Speicher
> vne...@pinn.net (vernonner3voltazim) writes:
>
> >Do we have good evidence in favor of, for example, the
> >superposition of counter-cirularly-polarized photons?
>
> The internal Hilbert space for a photon is two-dimensional.
> Left circular polarization and right circular polarization
> are mutually orthogonal states, and therefore form a basis
> for the internal Hilbert space. This means that all
> internal states of the photon is a superposition of these
> two states.
>
> As far experimental evidence is concerned, I will leave that
> for others to comment on, but I think that there is experimental
> equipment which distinguishes between left circular polarization
> and right circular polarization (such materials could be useful
> in quantum cryptography).
>
Sure, you can use a circular polarizer to analyze the handedness
of a circular wave. Just take a series of two linear polarizers
and their matching 90deg. retarder, and each pair forms a
circular polarizer. You can set these two pair up in series such
that, for instance, only left-circular light gets transmitted.
> >
> >Besides, if
> >photonic angular momentum was optional, then a great many
> >photons could indeed be, as I indicated above, purely linear.
>
> A linearly polarized photon is a linear superposition of
> circularly polarized photons.
>
And both are, in general, special cases of ellipticaly polarized
light. Denoting P-state as linearly polarized, R-state and
L-state as right and left circular, respectively, and E-state as
elliptical polarization, then P-state is a superposition of
R-state and L-state of equal amplitudes, while E-state is a
superposition of R-state and L-state without restricting the
amplitudes to be the same.
--
Stephen
s...@speicher.com
Ignorance is just a placeholder for knowledge.
Printed using 100% recycled electrons.
-----------------------------------------------------------
vernonner3voltazim
>
> A linearly polarized photon is a linear superposition of
> circularly polarized photons.
>
left to be said, because now I realize I made a possibly
unwarranted assumption about "superposition".
That is, when you first used it I thought you were
talking about two separate photons superimposed
upon each other (as spin-1 bosons are allowed to do),
and so some of my doubts arose from what I knew about the
reasonably simple physics of quantum leaps and photon
generation --SINGLE photon generation, that is.
(So how can there be two superimposed photons?) AHA!
What you were really saying was something like, "That
single photon exists in both circularly polarized states
at the same time." I do know what that means.
OK, and thanks again!
vernonner3voltazim
> >>
> >> This means that if \lambda is in the spectrum of p_x, so is \lambda + P.
> >> But this is true for *any* value of P, so the spectrum of p_x consists of
> >> the *entire* real line, and is continuous. There is therefore no quantum
> >> of momentum.
>
> >Well, there are quantities and then there are quanta. Sure, to the
> >extent that some quantity of momentum is allowed to be part of a
> >continuum, then obviously NO momentum-possessing object HAS to
> >acquire or lose momentum in discrete amounts.
>
> >HOWEVER, one could possibly say the exact same thing with respect
> >to Kinetic Energy. Nevertheless, quanta of Energy are known to
> >exist! So why not any stand-alone quanta of momentum?
>
> That is because one is able to determine the nature of the energy
> spectra from the form of the Hamiltonian operator.
Sorry, that was clear as mud. Yes, I do know that Energy as found
in photons is not really the same thing as ordinary kinetic energy,
even though photon-absorption can alter a particle's K.E. So,
simply because photons exist and are observed/computed to be quanta,
we know that Energy can be quantized. However, why should that mean
that just because we have not SO FAR observed anything like that
with respect to momentum, quantized momentum must therefore be
impossible?
> >You may have missed the behind-the-scenes logical argument for
> >such a thing as a stand-alone quantum of momentum. To wit:
> >(1) Pretend there can be such a thing as "negative" mass/energy.
> >(2) Consider a simple interaction between ordinary and negative mass:
> > (m)(v)----> <----(-m)(-v) Assume equal/opposite masses/velocities
>
> This does not describe a quantum system. What happens classically
> after the collision?
Well, true, that is not at first glance a quantum system per se.
But what if the (m) was an ordinary neutrino, and the (-m) was the
negative-mass equivalent of a neutrino? (I'd like to bring up
electrons, but I've no idea how electric charges might or might
not balance to zero, in such a scenario.)
> >(3) Compute resulting mass, kinetic energy and momentum, and you
> >get: Zero mass, Zero kinetic energy, TWO momentum-units!
>
> Nett mass = 0, nett kinetic energy = 0, nett momentum = 2mv.
Yes, that is the classical result. But what FORM does that
resulting momentum possess??? Which is why I said the rest:
> >(4) That sure looks like a stand-alone quantity of pure momentum
> >to me! Furthermore, if you alter the reference frame, you STILL
> >get 2 momentum-units, every time.
>
> Under Galilean transformations.
>
> >Looks like a quantum to me!
>
> Why????
IF it, a stand-alone quantity of pure momentum, can really
exist, then why COULDN'T be called a quantum? As a simile, recall
that photons come in a wide variety of magnitudes, yet they are
still quanta. I see no reason why blobs of pure momentum
couldn't exist in an equally wide variety of magnitudes -- yet
their identity as distinct blobs means that they deserve SOME
kind of label equivalent to quanta, if not actually quanta.
> > http://www.halfbakery.com/idea/Gravity_20Waves#1025890913
I do appreciate the feedback. Thanks again!
David McAnally
>D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) wrote/quoted:
>> >>
>> >> exp(-i P q_x/\hbar) p_x exp(i P q_x/\hbar) = p_x + P.
>> >>
>> >> This means that if \lambda is in the spectrum of p_x, so is \lambda + P.
>> >> But this is true for *any* value of P, so the spectrum of p_x consists of
>> >> the *entire* real line, and is continuous. There is therefore no quantum
>> >> of momentum.
>>
>> >Well, there are quantities and then there are quanta. Sure, to the
>> >extent that some quantity of momentum is allowed to be part of a
>> >continuum, then obviously NO momentum-possessing object HAS to
>> >acquire or lose momentum in discrete amounts.
>>
>> >HOWEVER, one could possibly say the exact same thing with respect
>> >to Kinetic Energy. Nevertheless, quanta of Energy are known to
>> >exist! So why not any stand-alone quanta of momentum?
>>
>> That is because one is able to determine the nature of the energy
>> spectra from the form of the Hamiltonian operator.
>Sorry, that was clear as mud.
Do you know what the Hamiltonian is? Do you know anything about its
eigenvalues?
>Yes, I do know that Energy as found
>in photons is not really the same thing as ordinary kinetic energy,
Whay not?
>even though photon-absorption can alter a particle's K.E. So,
>simply because photons exist and are observed/computed to be quanta,
>we know that Energy can be quantized.
We know that the Hamiltonian can have dicrete eigenvalues, and that
jumps between eigenstates of the Hamiltonian lead to absorption or
emission of specified quantities of energy.
>However, why should that mean
>that just because we have not SO FAR observed anything like that
>with respect to momentum, quantized momentum must therefore be
>impossible?
I reiterate: the Canonical Commutation Relations.
>> >You may have missed the behind-the-scenes logical argument for
>> >such a thing as a stand-alone quantum of momentum. To wit:
>> >(1) Pretend there can be such a thing as "negative" mass/energy.
>> >(2) Consider a simple interaction between ordinary and negative mass:
>> > (m)(v)----> <----(-m)(-v) Assume equal/opposite masses/velocities
>>
>> This does not describe a quantum system. What happens classically
>> after the collision?
>Well, true, that is not at first glance a quantum system per se.
>But what if the (m) was an ordinary neutrino, and the (-m) was the
>negative-mass equivalent of a neutrino? (I'd like to bring up
>electrons, but I've no idea how electric charges might or might
>not balance to zero, in such a scenario.)
>
>> >(3) Compute resulting mass, kinetic energy and momentum, and you
>> >get: Zero mass, Zero kinetic energy, TWO momentum-units!
>>
>> Nett mass = 0, nett kinetic energy = 0, nett momentum = 2mv.
>Yes, that is the classical result. But what FORM does that
>resulting momentum possess??? Which is why I said the rest:
You haven't specified with what happens after the collision. I don't see
any problem with what you have already written.
>
>> >(4) That sure looks like a stand-alone quantity of pure momentum
>> >to me! Furthermore, if you alter the reference frame, you STILL
>> >get 2 momentum-units, every time.
>>
>> Under Galilean transformations.
>>
>> >Looks like a quantum to me!
>>
>> Why????
>IF it, a stand-alone quantity of pure momentum, can really
>exist, then why COULDN'T be called a quantum? As a simile, recall
>that photons come in a wide variety of magnitudes, yet they are
>still quanta.
Photons are indivisible. You haven't shown that your 'pure momentum' is
indivisible.
>I see no reason why blobs of pure momentum
>couldn't exist in an equally wide variety of magnitudes -- yet
>their identity as distinct blobs means that they deserve SOME
>kind of label equivalent to quanta, if not actually quanta.
But why? I see no sense of discreteness or indivisibility.
>> > http://www.halfbakery.com/idea/Gravity_20Waves#1025890913
>I do appreciate the feedback. Thanks again!
David McAnally
------------
Randy Poe
> "David McAnally" <D.McAnally@i'm_a_gnu.uq.net.au> wrote in message
> news:anpc81$n8v$1...@bunyip.cc.uq.edu.au...
>
>>Don't inflict Thomson on a mathematics group. He is a mathematical
>>ignoramus. It's bad enough that he's here as a physics ignoramus.
>
>
> It's easy to throw words like a child. But let's see you throw a corrected
> mathematical interpretation where E is correctly shown to be equivalent to
> p. There is a simple way of proving this. Try defining E, p and c on a
> MathCAD spreadsheet and see if it allows you to arbitrarily change c to 1.
I can put write c=1 in a computer program.
I can label a cell as c in an Excel or Quattro-pro spreadsheet,
and put the value "1" into that cell.
What was that supposed to prove?
Your weight is 10, or 1, or 100, or pi, in suitably chosen
units. Why does the fact that the numerical value of a constant
changes with the units bother you so much?
Did you know that the velocity of a car can be simultaneously
60, 96, and 88? Is that really so difficult?
- Randy
Bilge
>
>
>Z(w) = 1/(jwC) = (1/wC)exp(-j*pi/2)
>w = freqnecy
>j = sqrt(-1)
>
>A negative capacitor is just a 180 phase shift in the complex plane. If you
>can't see that your blind. A capacitor stores energy and "then" can act as a
>source of energy.(QED)
A phase shift of 180 degrees means you have an inductor.
Z = jwL
That's why you get a resonance when:
Z = jwL + 1/jwC = 0
jwL = j/wC => w^2 = 1/LC
If you want to call an inductor a "negative capacitance", go ahead.
Every one will correct you, but so long as you don't mind...
David McAnally
>D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) wrote/quoted:
>> >> >>
>> >> >> exp(-i P q_x/\hbar) p_x exp(i P q_x/\hbar) = p_x + P.
>> >> >>
>> >> >> This means that if \lambda is in the spectrum of p_x, so is \lambda + P.
>> >> >> But this is true for *any* value of P, so the spectrum of p_x consists of
>> >> >> the *entire* real line, and is continuous. There is therefore no quantum
>> >> >> of momentum.
>>
>> >> >Well, there are quantities and then there are quanta. Sure, to the
>> >> >extent that some quantity of momentum is allowed to be part of a
>> >> >continuum, then obviously NO momentum-possessing object HAS to
>> >> >acquire or lose momentum in discrete amounts.
>>
>> >> >HOWEVER, one could possibly say the exact same thing with respect
>> >> >to Kinetic Energy. Nevertheless, quanta of Energy are known to
>> >> >exist! So why not any stand-alone quanta of momentum?
>> >>
>> >> That is because one is able to determine the nature of the energy
>> >> spectra from the form of the Hamiltonian operator.
>>
>> >Sorry, that was clear as mud.
>>
>> Do you know what the Hamiltonian is? Do you know anything about its
>> eigenvalues?
>I'm sorry, but I never got that far into the nitpicky details
>of quantum theory. If there is a simple way to explain the
>Hamiltonian, please do.
One of the postulates of quantum mechanics is that in the absence of
measurement, the state vector \psi evolves according to the equation
(called Schrodinger's Equation) is
i \hbar d\psi/dt = H \psi,
where H is the Hamiltonian operator. The classical equivalent is the
classical Hamiltonian, H, which is a function of coordinates, their
conjugate momenta, and time. In the classical case, if
H = H(q_1,...,q_N,p_1,...,p_N,t), then the equations of motion are given
by:
dq_i/dt = @H/@p_i, dp_i/dt = - @H/@p_i,
for all i = 1,...,N, where @f/@t denotes the partial derivative of f with
respect to t, etc. Note that a necessary consequence that
dH/dt = @H/@t,
so that if H has no explicit dependence on time, then H is a constant of
the motion. In other words, in BOTH the classical case and the quantum
case, the time-evolution is completely specified by the Hamiltonian.
Calssically, the Hamiltonian is often equal to the energy, and so the
eigenvalues of the Hamiltonian operators are often regarded as energy
levels, and the eigenstates of the Hamiltonian operator are regarded as
energy states.
>(Alas, without details, you might as
>well be spouting science-fictional gobbledygook/buzzwords, with
>the result being a claim of how to go faster than light.)
>> >Yes, I do know that Energy as found
>> >in photons is not really the same thing as ordinary kinetic energy,
>>
>> Whay not?
>Well, ordinary kinetic energy is not quantized (didn't we discuss
>that already?), and photons are.
Different Hamiltonians, and therefore different spectra.
>
>> >even though photon-absorption can alter a particle's K.E. So,
>> >simply because photons exist and are observed/computed to be quanta,
>> >we know that Energy can be quantized.
>>
>> We know that the Hamiltonian can have dicrete eigenvalues, and that
>> jumps between eigenstates of the Hamiltonian lead to absorption or
>> emission of specified quantities of energy.
>USUALLY, I thought, that refers to quantum jumps of electrons
>inside atoms (or protons inside nuclei).
Not necessarily. There are other examples.
>But freely-flying
>mass-possessing particles CAN transfer oddball amounts of
>kinetic energy to each other, without bringing quanta into the
>description of the interaction...can't they? Only when photons
>are absorbed/emitted, (as you just said) do we then associate
>kinetic energy with quanta.
>
>> >However, why should that mean
>> >that just because we have not SO FAR observed anything like that
>> >with respect to momentum, quantized momentum must therefore be
>> >impossible?
>>
>> I reiterate: the Canonical Commutation Relations.
>Sorry, that sounds like more buzzwords, even though I know you
>presented some of the math behind it in a prior message.
The CCR's are:
[q_j,q_k] = 0, [p_j,p_k] = 0,
[q_j,p_k] = i \hbar \delta_{jk},
where \delta_{jk} is the Kronecker delta, i.e. \delta_{jk} = 1 if j = k,
and \delta_{jk} = 0 if j and k are distinct.
>What
>I have been trying to get at is the idea that, OK so ordinary
>particulate momentum is unquantized -- and since the difference
>in Dimensional Units, between momentum and kinetic energy, is
>merely a factor of "velocity" (unquantized!), it seems logical
>that ordinary particulate kinetic energy can also be unquantized.
>Nevertheless, since Energy can indeed exist in a quantized form,
>how (in plain English) is anything similar to that automatically
>and totally impossible for momentum?
Not totally impossible. Momentum is quantized in a finite box (i.e.
a rectangular solid on which V = 0, such that V = infinity on the
boundary). Incidentally, in this case, the CCR's don't hold.
>> >> >You may have missed the behind-the-scenes logical argument for
>> >> >such a thing as a stand-alone quantum of momentum. To wit:
>> >> >(1) Pretend there can be such a thing as "negative" mass/energy.
>> >> >(2) Consider a simple interaction between ordinary and negative mass:
>> >> > (m)(v)----> <----(-m)(-v) Assume equal/opposite masses/velocities
>> >>
>> >> This does not describe a quantum system. What happens classically
>> >> after the collision?
>>
>>
>> >> >(3) Compute resulting mass, kinetic energy and momentum, and you
>> >> >get: Zero mass, Zero kinetic energy, TWO momentum-units!
>> >>
>> >> Nett mass = 0, nett kinetic energy = 0, nett momentum = 2mv.
>>
>> >Yes, that is the classical result. But what FORM does that
>> >resulting momentum possess??? Which is why I said the rest:
>>
>> You haven't specified with what happens after the collision. I don't see
>> any problem with what you have already written.
>I'm not sure what you are asking for, here. In the reference
>frame describing a collision of equal/opposite masses/velocities,
>we end up with a quantity of pure momentum that doesn't go
>anywhere.
That doen't answer the question of the physical situation after the
collision. Or are you thinking of annihilation?
David McAnally
----------
David McAnally
>The CCR's are:
> [q_j,q_k] = 0, [p_j,p_k] = 0,
> [q_j,p_k] = i \hbar \delta_{jk},
>where \delta_{jk} is the Kronecker delta, i.e. \delta_{jk} = 1 if j = k,
>and \delta_{jk} = 0 if j and k are distinct.
These relations can also be written as follows:
q_j q_k = q_k q_j for all j, k
p_j p_k = p_k p_j for all j, k
q_j p_k - p_k q_j = i \hbar \delta_{jk}
David McAnally
--------
vernonner3voltazim
> >> eigenvalues?
>
> >I'm sorry, but I never got that far into the nitpicky details
> >of quantum theory. If there is a simple way to explain the
> >Hamiltonian, please do.
>
> One of the postulates of quantum mechanics is that in the absence of
> measurement, the state vector \psi evolves according to the equation
> (called Schrodinger's Equation) is
>
> i \hbar d\psi/dt = H \psi,
>
> where H is the Hamiltonian operator. The classical equivalent is the
> classical Hamiltonian, H, which is a function of coordinates, their
> conjugate momenta, and time. In the classical case, if
> H = H(q_1,...,q_N,p_1,...,p_N,t), then the equations of motion are given
> by:
>
> dq_i/dt = @H/@p_i, dp_i/dt = - @H/@p_i,
>
> for all i = 1,...,N, where @f/@t denotes the partial derivative of f with
> respect to t, etc. Note that a necessary consequence that
>
> dH/dt = @H/@t,
>
> so that if H has no explicit dependence on time, then H is a constant of
> the motion. In other words, in BOTH the classical case and the quantum
> case, the time-evolution is completely specified by the Hamiltonian.
>
> levels, and the eigenstates of the Hamiltonian operator are regarded as
> energy states.
Thank you!
> >> >Yes, I do know that Energy as found in photons is not
> >> >really the same thing as ordinary kinetic energy,
> >>
> >> Whay not?
>
> >Well, ordinary kinetic energy is not quantized (didn't we
> >discuss that already?), and photons are.
>
> Different Hamiltonians, and therefore different spectra.
Which, I think, means "that Energy as found in photons is not really
the same thing as ordinary kinetic energy" --right? (Those different
Hamiltonians mean different properties, too -- such as quantized
versus unquantized.)
> >> We know that the Hamiltonian can have dicrete eigenvalues, and that
> >> jumps between eigenstates of the Hamiltonian lead to absorption or
> >> emission of specified quantities of energy.
>
> >USUALLY, I thought, that refers to quantum jumps of electrons
> >inside atoms (or protons inside nuclei).
>
> Not necessarily. There are other examples.
Well, I did say "USUALLY" ...
> >> >However, why should that mean
> >> >that just because we have not SO FAR observed anything like that
> >> >with respect to momentum, quantized momentum must therefore be
> >> >impossible?
> >>
> >> I reiterate: the Canonical Commutation Relations.
>
> >Sorry, that sounds like more buzzwords, even though I know you
> >presented some of the math behind it in a prior message.
>
> The CCR's are:
>
> [q_j,q_k] = 0, [p_j,p_k] = 0,
>
> [q_j,p_k] = i \hbar \delta_{jk},
>
> where \delta_{jk} is the Kronecker delta, i.e. \delta_{jk} = 1 if j = k,
> and \delta_{jk} = 0 if j and k are distinct.
Uh huh.....
I suppose I might bring up question of, "Just because those descriptions
are "canonized" :) how do we know for sure that there are no loopholes?"
> >What
> >I have been trying to get at is the idea that, OK so ordinary
> >particulate momentum is unquantized -- and since the difference
> >in Dimensional Units, between momentum and kinetic energy, is
> >merely a factor of "velocity" (unquantized!), it seems logical
> >that ordinary particulate kinetic energy can also be unquantized.
> >Nevertheless, since Energy can indeed exist in a quantized form,
> >how (in plain English) is anything similar to that automatically
> >and totally impossible for momentum?
>
> Not totally impossible. Momentum is quantized in a finite box (i.e.
> a rectangular solid on which V = 0, such that V = infinity on the
> boundary). Incidentally, in this case, the CCR's don't hold.
Yeah, if the container/box is indivisible, then the contents of the
box are untouchable and therefore also indivisible.
Do you think there might be any other possibilities?
> >> >> >You may have missed the behind-the-scenes logical argument for
> >> >> >such a thing as a stand-alone quantum of momentum. To wit:
> >> >> >(1) Pretend there can be such a thing as "negative" mass/energy.
> >> >> >(2) Consider a simple interaction between ordinary and negative mass:
> >> >> > (m)(v)----> <----(-m)(-v) Assume equal/opposite masses/velocities
> >> >>
> >> >> This does not describe a quantum system. What happens classically
> >> >> after the collision?
> >>
> >> >> >you get: Zero mass, Zero kinetic energy, TWO momentum-units!
> >> >>
> >> You haven't specified with what happens after the collision. I don't see
> >> any problem with what you have already written.
>
> >I'm not sure what you are asking for, here. In the reference
> >frame describing a collision of equal/opposite masses/velocities,
> >we end up with a quantity of pure momentum that doesn't go
> >anywhere.
>
> That doen't answer the question of the physical situation after the
> collision. Or are you thinking of annihilation?
Well, the late Dr. Robert L. Forward did a fair amount of hypothetical
work with negative mass/energy (wrote a good novel, too: "Time Master"),
and he coined the term "nullification" to describe mutual annihilation
of equal/opposite masses. I know you came late to this discussion,
but I did mention this in prior messages, and had specified that the
above example be thought of as "most-simple nullification". I assumed
you had reviewed prior messages in the thread.
Anyway, the only remaining physical picture/puzzle concerns a
description of the leftover blob of pure momentum. At this point,
it seems to me more important to get folks to accept the logic,
that IF negative mass/energy exists, then so will also exist such
momentum-blobs. Part of the reason why I wanted people to look at
this page, the one that started this whole thread:
http://www.nemitz.net/vernon/MOMEQUAT.htm
and this one, which presents more consequences:
http://www.halfbakery.com/idea/Gravity_20Waves#1025890913
is because one thing leads to another: If the blobs exist, can they
be absorbed by ordinary masses? OK, if masses can absorb those blobs,
can they also EMIT those blobs (Time-Reversal Symmetry)? If so, then
can we manufacture the blobs without needing negative mass/energy?
I did suggest particular experiments. IF the experiments provide any
supporting evidence, then perhaps that evidence will lead to a
physical picture of the blobs (I vote for gravitons!).
One of the neat things about the overall idea of leftover momentum
concerns the sea of "virtual particles in the vacuum". As you know,
the Energy-Time version of the Uncertainty Principle allows local
fluctuations in energy, which manifests as the spontaneous appearance
of pairs of opposite particles. Well, there is also that ORIGINAL
version of the Uncertainty Principle, a Momentum-Position relation,
that could allow the spontaneous appearance of pairs of particles
having ordinary and negative mass/energy. Since, as you know, if we
inject enough REAL energy into one of those ordinary virtual-particle
pairs, we get a pair of real particles, it logically follows that if
we could inject enough REAL pure momentum (a blob!) into one of
those ordinary/negative pairs, then we should obtain some genuine
real particles having negative mass/energy. Wouldn't that be fun!
(I'd especially like to generate anti-matter/negative-anti-matter
pairs, because we could solve the energy crisis with the anti-matter,
and the negative-anti-matter isn't opposite enough to ordinary matter,
for it to nullify!)
David McAnally
>One of the postulates of quantum mechanics is that in the absence of
>measurement, the state vector \psi evolves according to the equation
>(called Schrodinger's Equation) is
> i \hbar d\psi/dt = H \psi,
>where H is the Hamiltonian operator. The classical equivalent is the
>classical Hamiltonian, H, which is a function of coordinates, their
>conjugate momenta, and time. In the classical case, if
>H = H(q_1,...,q_N,p_1,...,p_N,t), then the equations of motion are given
>by:
> dq_i/dt = @H/@p_i, dp_i/dt = - @H/@p_i,
Of course, the second equation should read:
dp_i/dt = - @H/@q_i.
>for all i = 1,...,N, where @f/@t denotes the partial derivative of f with
>respect to t, etc. Note that a necessary consequence that
> dH/dt = @H/@t,
>so that if H has no explicit dependence on time, then H is a constant of
>the motion. In other words, in BOTH the classical case and the quantum
>case, the time-evolution is completely specified by the Hamiltonian.
>Calssically, the Hamiltonian is often equal to the energy, and so the
>eigenvalues of the Hamiltonian operators are often regarded as energy
>levels, and the eigenstates of the Hamiltonian operator are regarded as
>energy states.
David McAnally
----------
vernonner3voltazim
Thanks again! Question: Any idea what happened to one of my prior
messages in the thread? You replied to it, but the message itself
seems to be gone. I sure didn't delete it!
Anyway, the most recent other message I wrote is listed in the thread,
but what you have written above doesn't really seem to be a reply to
it. I hope the discussion will continue!
Thanks again!
Vernon