massless or massive photon?
Juan R.
H = (\sqrt (m^2c^4 + p^2c^2) ) (1)
and set (m = 0) to yield
H = pc
However, the original Hamiltonian (1) was derived using the Legendre
transformation
H = pv - L = (\sqrt (m^2c^4 + p^2c^2) )
If alternatively we start assuming (v = c) in the transformation, the
result is
H = pc - L = pc (2)
where no assumption was taken about the mass.
Is this Hamiltonian (2) representing some kind of massive photon (somewhat
as in Proca theory [#]) or is really the same that (1)?
[#] http://en.wikipedia.org/wiki/Proca_action
--
Center for CANONICAL |SCIENCE)
http://canonicalscience.org
Albertito
Hi Juan,
IMHO, SR is flawed :-) You need an equation as
H^2 = m^2c^4 + p^2c^2 from which, if m=0, then
you could deduce that v=c, and vice versa. But,
the relativistic momentum p = mv/sqrt(1 - v^2/c^2)
forbids that deduction. So, where in SR is exactly
the flaw? Let's modify the relativistic momentum
to be
p = mc sinh(v/c),
allowing v to be even superluminal.
Now, we can see that for v=c, p does not diverge
for non-zero m.
Therefore, the Hamiltonian would read
H^2 = m^2c^4 + (mc sinh(v/c))^2 c^2,
that after some algebra, it yields
H = mc^2 cosh(v/c),
and now we can see no momentum is needed to compute H.
The interesting issue is that when you quantize that H,
you can attain all the masses of charged leptons in a
very natural and straightforward manner :-)
Sue...
So you then have an imaginary mass that won't couple
to a real gravito-inertial field.
Now you just need an imaginary shape for the
pseudo-particle to massage the nasty radiation
patterns that we measure for real light into
a point.
http://www.rp-photonics.com/gaussian_beams.html
There is surely a "Nobel" or a knighthood in it
for you.
<<The Nobel Committee avoids committing itself
to the particle concept. Light-quanta or with
modern terminology, photons, were explicitly
mentioned in the reports on which the prize
decision rested only in connection with emission
and absorption processes. The Committee says
that the most important application of Einstein's
photoelectric law and also its most convincing
confirmation has come from the use Bohr made of
it in his theory of atoms, which explains a vast
amount of spectroscopic data. >>
http://nobelprize.org/nobel_prizes/physics/articles/ekspong/index.html
<<"It must come sometimes to 'jam today,'" Alice objected.
"No, it can't" said the Queen. "It's jam any other day—today isn't any
other day, you know."
"I don't understand you," said Alice. "It's dreadfully confusing!"
"That's the effect of living backward," the Queen said kindly. "It
always makes one giddy at first—"
"Living backward!" Alice repeated in great astonishment. "I never
heard of such a thing!"
"—but there's one great advantage in it, that one's memory works both
ways."
"I'm sure mine only works one way," Alice remarked. "I can't remember
things before they happen."
"It's a poor sort of memory that only works backward," the Queen
remarked. >>
--Lewis Carroll
"Sir Juan" does have a nice ring to it.
http://www.wam.umd.edu/~david/images/doredonandwindmills.jpg
:o)
Sue...
The TimeLord
<albert...@gmail.com> in
d4bbceaf-082f-4726...@f36g2000hsa.googlegroups.com:
> On Jul 8, 2:11 pm, "Juan R." González-Álvarez
> <juanREM...@canonicalscience.com> wrote:
>> The usual argument for massless photons uses the Hamiltonian
>>
>> H = (\sqrt (m^2c^4 + p^2c^2) ) (1)
>>
>> and set (m = 0) to yield
>>
>> H = pc
[...]
> Hi Juan,
>
> IMHO, SR is flawed :-) You need an equation as H^2 = m^2c^4 + p^2c^2
> from which, if m=0, then you could deduce that v=c, and vice versa. But,
> the relativistic momentum p = mv/sqrt(1 - v^2/c^2) forbids that
However, that equation is valid only for particles with mass. The photon
doesn't have mass, so it is invalid for photons. For photons
p = E/c = h/lambda
> deduction. So, where in SR is exactly the flaw? Let's modify the
SR has no flaw in this regard.
> relativistic momentum to be
>
> p = mc sinh(v/c),
Which is wrong, since momentum doesn't depend on the "sinh" of speed.
Instead you should use
p = mv/sqrt[1-v^2/c^2]
which is consistent with experimental observations. Your equation is not.
> allowing v to be even superluminal.
If a particle has mass, then it can not be superluminal since it would
require more energy than exists in the universe to achieve such a speed.
>
> Now, we can see that for v=c, p does not diverge for non-zero m.
> Therefore, the Hamiltonian would read
>
> H^2 = m^2c^4 + (mc sinh(v/c))^2 c^2,
[...]
Non-sequitur.
--
// The TimeLord says:
// Pogo 2.0 = We have met the aliens, and they are us!
Albertito
> Am Wed, 09 Jul 2008 02:09:56 -0700 schrieb Albertito
> d4bbceaf-082f-4726-a2e2-3be2e07ec...@f36g2000hsa.googlegroups.com:
Dear The TimeLord,
As I'd pointed out to you elsewhere, if the relativistic
momentum p = mv/sqrt(1-v^2/c^2) passes a experimental
test, then p = mc sinh(v/c) passes it too. You should
be more careful with your quick claims.
The TimeLord
<albert...@gmail.com> in
a58fe8d2-41b4-4322...@k37g2000hsf.googlegroups.com:
[...]
> Dear The TimeLord,
>
> As I'd pointed out to you elsewhere, if the relativistic momentum p =
> mv/sqrt(1-v^2/c^2) passes a experimental test, then p = mc sinh(v/c)
> passes it too. You should be more careful with your quick claims.
However, the two are not equivalent. Besides, where is there experimental
evidence of your equation? (I have visited the website and it is in
error.)
xxe...@gmail.com
>
> :o)
>
> Sue...
>
>
>
>
>
> > --
> > Center for CANONICAL |SCIENCE) http://canonicalscience.org- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
xxein: Nice quote. But even the Nobel committee cannot declare a
physic.
In my earlier foundational research (circa 1985 and still valid) I
quickly found out that the photo-electric effect (as described by
Einstein) is NOT the description of the effect.
Of course it IS for those that haven't really thought about enough of
the physic.
I find myself in a dilemna in which I don't know everything, but can
sense when thing are wrongly accepted. I know too much without
knowing everything. Things cannot possibly work within our popular
and accepted theories. Not even that close for comfort.
That said, I know that I am open to criticism. So, first, I would say
that we live with what we got and seem to do well with the
understanding what we have that provides for it. But a transistor did
not invent itself. This is opposed by what we think about it. The
same for the photo-electric effect.
Transistors did not naturally occur in nature. We invented the
prospect of it having a usefulness by manipulating what we could
manipulate. Gravity (otoh) does exist without any of our
manipulations.
We have, of course, made progress in our understanding of how the
physic works, but haven't we sort of forgotten the old notions in
favor of the new? Will we think differently 10 yrs from now? Will it
depend on a 'transistor' or a gravity?
These are the things that need sorting out. Do we understand the
physic, or do we just manipulate its components into a 'physics' that
we use?
Come to think about, I want all the criticism and abuse for my
thinking about this. I want to learn like I suppose everyone else
does.
Sue...
[...]
>
> Come to think about, I want all the criticism and abuse for my
> thinking about this. I want to learn like I suppose everyone else
> does.
Well...
Try using a wave model for light. When the anomalous twin
vanishes you may have learned *better* that everyone else.
http://farside.ph.utexas.edu/teaching/em/lectures/lectures.html
http://web.mit.edu/8.02t/www/802TEAL3D/visualizations/light/index.htm
Sue...
Igor
As I've always understood it, the assignment of v = c as a special
case should only occur AFTER the Legendre tranform has been
performed. I think in this particular case, you get the correct
result coincidently. But in general that probably wouldn't happen.
Besides, the Lagrangian in the wiki link you've provided is a field
Lagrangian density, whereas the analysis you've provided entails a
mechanical Lagrangian. They're entirely different animals.
Eric Gisse
<juanREM...@canonicalscience.com> wrote:
> The usual argument for massless photons uses the Hamiltonian
No, the usual argument for the massless photon is the following, in
this order:
a) Experiments and observation don't support it.
b) Proca's equations with m--> 0 recovers Maxwell's equations and the
expected vacuum dispersion relation.
c) SR + Quantum field theory ---> photons travel along null paths --->
photons must be massless.
>
> H = (\sqrt (m^2c^4 + p^2c^2) ) (1)
No. The special relativistic Hamiltonian is H = L = -mc^2 * [1 - v^2/
c^2 ]. There is no reference to electromagnetism regardless, so any
argument about electromagnetism that _neglects_ the electromagnetic
contributions to the Lagrangian is destined to be rather silly.
>
> and set (m = 0) to yield
>
> H = pc
>
> However, the original Hamiltonian (1) was derived using the Legendre
> transformation
No, you derived the original Hamiltonian is by assuming it is the
energy in the magnitude of four-momentum.
>
> H = pv - L = (\sqrt (m^2c^4 + p^2c^2) )
>
> If alternatively we start assuming (v = c) in the transformation, the
> result is
>
> H = pc - L = pc (2)
How does pc - (\sqrt (m^2c^4 + p^2c^2) ) = pc in your world? You can
only obtain that by _assuming_ a massless particle.
>
> where no assumption was taken about the mass.
By equating v = c you have DEFINED the particle to be massless. Either
that or you have defined a contradiction by allowing a massive
particle to travel on a null path.
>
> Is this Hamiltonian (2) representing some kind of massive photon (somewhat
> as in Proca theory [#]) or is really the same that (1)?
No, the _only_ thing (2) represents is the energy of a particle that
travels along a null path. If the photon is massive it travels along a
time-like geodesic and NOT a null path. Its' energy will be different.
>
> [#]http://en.wikipedia.org/wiki/Proca_action
Notice there are terms corresponding to the vector field in the
Lagrange density?
Notice you have made no reference to anything but SR? Regardless -
particles traveling along a null path _cannot have mass_ in special
relativity.
Eric Gisse
The only flaw is that you don't know what the fuck you are talking
about. Arbitrarily tinkering with the equations derived from the
Lagrangian means you aren't doing special relativity - you are doing a
characture of relativity that has no bearing on the actual theory.
This is also known as "beating on a strawman", which you /excel/ at.
>
> Now, we can see that for v=c, p does not diverge
> for non-zero m.
> Therefore, the Hamiltonian would read
>
> H^2 = m^2c^4 + (mc sinh(v/c))^2 c^2,
>
> that after some algebra, it yields
>
> H = mc^2 cosh(v/c),
>
> and now we can see no momentum is needed to compute H.
You are stupid. You have no idea what a Hamiltonian or Lagrangian is,
or how the conjugate coordinates [position, velocity, momentum] relate
to any of them.
> The interesting issue is that when you quantize that H,
> you can attain all the masses of charged leptons in a
> very natural and straightforward manner :-)
Says someone who doesn't understand anything about classical or
quantum mechanics, or relativity.
Eric Gisse
Only because you suffer the delusion that you are doing physics by
playing games with Taylor expansions. Special relativity is routinely
verified in the range v/c ~ 1.
Juan R.
> On Jul 8, 2:11 pm, "Juan R." González-Álvarez
> <juanREM...@canonicalscience.com> wrote:
>> The usual argument for massless photons uses the Hamiltonian
>>
>> H = (\sqrt (m^2c^4 + p^2c^2) ) (1)
>>
>> and set (m = 0) to yield
>>
>> H = pc
>>
>> However, the original Hamiltonian (1) was derived using the Legendre
>> transformation
>>
>> H = pv - L = (\sqrt (m^2c^4 + p^2c^2) )
>>
>> If alternatively we start assuming (v = c) in the transformation, the
>> result is
>>
>> H = pc - L = pc (2)
>>
>> where no assumption was taken about the mass.
>>
>> Is this Hamiltonian (2) representing some kind of massive photon
>> (somewhat as in Proca theory [#]) or is really the same that (1)?
>>
>> [#]http://en.wikipedia.org/wiki/Proca_action
>>
>> --
>> Center for CANONICAL |SCIENCE) http://canonicalscience.org
>
> Hi Juan,
>
> IMHO, SR is flawed :-) You need an equation as H^2 = m^2c^4 + p^2c^2
> from which, if m=0, then you could deduce that v=c, and vice versa.
Hi, I didn't said had a flaw in that. I did not even say above H were the
SR Hamiltonian even if looks so close!
> But,
> the relativistic momentum p = mv/sqrt(1 - v^2/c^2) forbids that
> deduction. So, where in SR is exactly the flaw?
No, v = c is a deduction from H = pc.
> Let's modify the
> relativistic momentum to be
>
> p = mc sinh(v/c),
> allowing v to be even superluminal.
>
> Now, we can see that for v=c, p does not diverge for non-zero m.
> Therefore, the Hamiltonian would read
>
> H^2 = m^2c^4 + (mc sinh(v/c))^2 c^2,
>
> that after some algebra, it yields
>
> H = mc^2 cosh(v/c),
>
> and now we can see no momentum is needed to compute H. The interesting
> issue is that when you quantize that H, you can attain all the masses of
> charged leptons in a very natural and straightforward manner :-)
All of above is not right.
Juan R.
> On Jul 8, 9:11 am, "Juan R." González-Álvarez
> <juanREM...@canonicalscience.com> wrote:
>> The usual argument for massless photons uses the Hamiltonian
>>
>> H = (\sqrt (m^2c^4 + p^2c^2) ) (1)
>>
>> and set (m = 0) to yield
>>
>> H = pc
>>
>> However, the original Hamiltonian (1) was derived using the Legendre
>> transformation
>>
>> H = pv - L = (\sqrt (m^2c^4 + p^2c^2) )
>>
>> If alternatively we start assuming (v = c) in the transformation, the
>> result is
>>
>> H = pc - L = pc (2)
>>
>> where no assumption was taken about the mass.
>>
>> Is this Hamiltonian (2) representing some kind of massive photon
>> (somewhat as in Proca theory [#]) or is really the same that (1)?
>>
>> [#]http://en.wikipedia.org/wiki/Proca_action
>
> So you then have an imaginary mass that won't couple to a real
> gravito-inertial field.
And your proof that m is imaginary for H = pc is?
In sci.physics.foundations Charles said that mass would be zero for (2),
but neither he offered any proof.
(...)
Juan R.
> On Jul 8, 9:11 am, "Juan R." González-Álvarez
> <juanREM...@canonicalscience.com> wrote:
>> The usual argument for massless photons uses the Hamiltonian
>>
>> H = (\sqrt (m^2c^4 + p^2c^2) ) (1)
>>
>> and set (m = 0) to yield
>>
>> H = pc
>>
>> However, the original Hamiltonian (1) was derived using the Legendre
>> transformation
>>
>> H = pv - L = (\sqrt (m^2c^4 + p^2c^2) )
>>
>> If alternatively we start assuming (v = c) in the transformation, the
>> result is
>>
>> H = pc - L = pc (2)
>>
>> where no assumption was taken about the mass.
>>
>> Is this Hamiltonian (2) representing some kind of massive photon
>> (somewhat as in Proca theory [#]) or is really the same that (1)?
>>
>> [#]http://en.wikipedia.org/wiki/Proca_action
>>
>> --
>> Center for CANONICAL |SCIENCE) http://canonicalscience.org
>
> As I've always understood it, the assignment of v = c as a special case
> should only occur AFTER the Legendre transform has been performed.
Don't sure about that argument but will think about. Probably your
argument was right and then (2) is just (1).
> I
> think in this particular case, you get the correct result coincidently.
> But in general that probably wouldn't happen.
>
> Besides, the Lagrangian in the wiki link you've provided is a field
> Lagrangian density, whereas the analysis you've provided entails a
> mechanical Lagrangian. They're entirely different animals.
Yes, would be different massive particles, that is because wrote
"somewhat as".
Juan R.
> On Jul 8, 5:11 am, "Juan R." González-Álvarez
> <juanREM...@canonicalscience.com> wrote:
>> The usual argument for massless photons uses the Hamiltonian
>
> No, the usual argument for the massless photon is the following, in this
> order:
>
> a) Experiments and observation don't support it.
Don't true. Experiments only provide a upper limit for m_ph.
> b) Proca's equations
> with m--> 0 recovers Maxwell's equations and the expected vacuum
> dispersion relation.
And Maxwell equation with massless photons is a special case of a more
general theory with massive photons. Your comment is irrelevant.
> c) SR + Quantum field theory ---> photons travel along null paths --->
> photons must be massless.
And QED with massless photons is a special case of a more general quantum
theory with massive photons. Your comment is also irrelevant.
>
>> H = (\sqrt (m^2c^4 + p^2c^2) ) (1)
>
> No. The special relativistic Hamiltonian is H = L = -mc^2 * [1 - v^2/
> c^2 ].
Nonsense.
> There is no reference to electromagnetism regardless, so any
> argument about electromagnetism that _neglects_ the electromagnetic
> contributions to the Lagrangian is destined to be rather silly.
Misreading or off-topic or what else...
>
>> and set (m = 0) to yield
>>
>> H = pc
>>
>> However, the original Hamiltonian (1) was derived using the Legendre
>> transformation
>
> No, you derived the original Hamiltonian is by assuming it is the energy
> in the magnitude of four-momentum.
No.
>
>> H = pv - L = (\sqrt (m^2c^4 + p^2c^2) )
>>
>> If alternatively we start assuming (v = c) in the transformation, the
>> result is
>>
>> H = pc - L = pc (2)
>
> How does pc - (\sqrt (m^2c^4 + p^2c^2) ) = pc in your world? You can
> only obtain that by _assuming_ a massless particle.
No. Read.
>
>> where no assumption was taken about the mass.
>
> By equating v = c you have DEFINED the particle to be massless.
No, a well-known counterexample is the electron in Dirac theory. It is
massive and its |v| = +-c.
> Either
> that or you have defined a contradiction by allowing a massive particle
> to travel on a null path.
It is to be proved that (2) and (1) represent the same particle.
(...)
Eric Gisse
<juanREM...@canonicalscience.com> wrote:
> Eric Gisse wrote on Wed, 09 Jul 2008 18:10:55 -0700:
>
> > On Jul 8, 5:11 am, "Juan R." González-Álvarez
> > <juanREM...@canonicalscience.com> wrote:
> >> The usual argument for massless photons uses the Hamiltonian
>
> > No, the usual argument for the massless photon is the following, in this
> > order:
>
> > a) Experiments and observation don't support it.
>
> Don't true. Experiments only provide a upper limit for m_ph.
...and you know what that upper limit is? It's in the region of 10^-17
eV.
http://pdg.lbl.gov/2007/listings/s000.pdf
Again - massive photons are inconsistent with observation in the same
way the aether is inconsistent with observation. Any theory that
incorporates a massive photon has to be hiding in incredibly tight
error bars.
>
> > b) Proca's equations
> > with m--> 0 recovers Maxwell's equations and the expected vacuum
> > dispersion relation.
>
> And Maxwell equation with massless photons is a special case of a more
> general theory with massive photons. Your comment is irrelevant.
Your short term memory is rather bad when it gets to the point where
you repeat exactly what I said back to me like it is a stunning
observation.
>
> > c) SR + Quantum field theory ---> photons travel along null paths --->
> > photons must be massless.
>
> And QED with massless photons is a special case of a more general quantum
> theory with massive photons. Your comment is also irrelevant.
Sure - but the photons won't travel along null paths, will have a
dispersion relation that is a function of mass, etc, etc, etc.
Inconsistent with observation.
>
>
>
> >> H = (\sqrt (m^2c^4 + p^2c^2) ) (1)
>
> > No. The special relativistic Hamiltonian is H = L = -mc^2 * [1 - v^2/
> > c^2 ].
>
> Nonsense.
*scratches head*
http://canonicalscience.blogspot.com/2007/08/relativistic-lagrangian-and-limitations.html
Says it right there. Then again my math ability is always bad this
early in the morning. Maybe I'm mistaken?
>
> > There is no reference to electromagnetism regardless, so any
> > argument about electromagnetism that _neglects_ the electromagnetic
> > contributions to the Lagrangian is destined to be rather silly.
>
> Misreading or off-topic or what else..
How is it either misleading or off-topic? It is silly to attempt to
discuss things like this without actually discussing electromagnetism.
>
>
>
> >> and set (m = 0) to yield
>
> >> H = pc
>
> >> However, the original Hamiltonian (1) was derived using the Legendre
> >> transformation
>
> > No, you derived the original Hamiltonian is by assuming it is the energy
> > in the magnitude of four-momentum.
>
> No.
To be fair that was a guess. How about explaining how you derive that
Hamiltonian?
Please don't insult my intelligence while trying to explain.
>
>
>
> >> H = pv - L = (\sqrt (m^2c^4 + p^2c^2) )
>
> >> If alternatively we start assuming (v = c) in the transformation, the
> >> result is
>
> >> H = pc - L = pc (2)
>
> > How does pc - (\sqrt (m^2c^4 + p^2c^2) ) = pc in your world? You can
> > only obtain that by _assuming_ a massless particle.
>
> No. Read.
>
>
>
> >> where no assumption was taken about the mass.
>
> > By equating v = c you have DEFINED the particle to be massless.
>
> No, a well-known counterexample is the electron in Dirac theory. It is
> massive and its |v| = +-c.
I can't decide whether the hypocracy or dishonesty annoys me more, so
I'll leave this be.
>
> > Either
> > that or you have defined a contradiction by allowing a massive particle
> > to travel on a null path.
>
> It is to be proved that (2) and (1) represent the same particle.
There's only one null path.
Sue...
<juanREM...@canonicalscience.com> wrote:
> Sue... wrote on Wed, 09 Jul 2008 03:12:30 -0700:
>
>
>
> > On Jul 8, 9:11 am, "Juan R." González-Álvarez
> > <juanREM...@canonicalscience.com> wrote:
> >> The usual argument for massless photons uses the Hamiltonian
>
> >> H = (\sqrt (m^2c^4 + p^2c^2) ) (1)
>
> >> and set (m = 0) to yield
>
> >> H = pc
>
> >> However, the original Hamiltonian (1) was derived using the Legendre
> >> transformation
>
> >> H = pv - L = (\sqrt (m^2c^4 + p^2c^2) )
>
> >> If alternatively we start assuming (v = c) in the transformation, the
> >> result is
>
> >> H = pc - L = pc (2)
>
> >> where no assumption was taken about the mass.
>
> >> Is this Hamiltonian (2) representing some kind of massive photon
> >> (somewhat as in Proca theory [#]) or is really the same that (1)?
>
> >> [#]http://en.wikipedia.org/wiki/Proca_action
>
> > So you then have an imaginary mass that won't couple to a real
> > gravito-inertial field.
>
> And your proof that m is imaginary for H = pc is?
>
> In sci.physics.foundations Charles said that mass would be zero for (2),
> but neither he offered any proof.
When an atom emits light quanta it loose mass.
When an atom absorbs light quanta it gains mass.
But the quanta absorbed is seldom 100 percent
causaly related to the emission. You have
to consider all the other EM disturbances
that could contribute to absorbtion and
subtract them out. Longer paths involve
more participants to be subtracted so
the mass approaches zero quicly.
Zero is close enough for me. :o)
In the subatomic realm the certainty
of a causal link is much higher and
surely worth accounting for.
http://en.wikipedia.org/wiki/Feynman_diagram#Motivation_and_history
Sue...
Juan R.
>> > So you then have an imaginary mass that won't couple to a real
>> > gravito-inertial field.
>>
>> And your proof that m is imaginary for H = pc is?
>>
>> In sci.physics.foundations Charles said that mass would be zero for
>> (2), but neither he offered any proof.
>
> When an atom emits light quanta it loose mass.
>
> When an atom absorbs light quanta it gains mass.
>
> But the quanta absorbed is seldom 100 percent causaly related to the
> emission. You have to consider all the other EM disturbances that could
> contribute to absorbtion and subtract them out. Longer paths involve
> more participants to be subtracted so the mass approaches zero quicly.
> Zero is close enough for me. :o)
>
> In the subatomic realm the certainty
> of a causal link is much higher and
> surely worth accounting for.
>
> http://en.wikipedia.org/wiki/Feynman_diagram#Motivation_and_history
>
>
> Sue...
>
No proof presented.
Sue...
You will be my proof when you spend 10 years
avoiding electromagnetism when you could
learn it in three.
http://farside.ph.utexas.edu/teaching/em/lectures/lectures.html
http://web.mit.edu/8.02t/www/802TEAL3D/visualizations/light/index.htm
http://www.ee.surrey.ac.uk/Personal/D.Jefferies/antennas.html
Sue...
Juan R.
>> > a) Experiments and observation don't support it.
>>
>> Don't true. Experiments only provide a upper limit for m_ph.
>
> ...and you know what that upper limit is? It's in the region of 10^-17
> eV.
>
> http://pdg.lbl.gov/2007/listings/s000.pdf
>
> Again - massive photons are inconsistent with observation in the same
> way the aether is inconsistent with observation.
Don't true again.
>> > b) Proca's equations
>> > with m--> 0 recovers Maxwell's equations and the expected vacuum
>> > dispersion relation.
>>
>> And Maxwell equation with massless photons is a special case of a more
>> general theory with massive photons. Your comment is irrelevant.
(snip another irrelevant comment from you)
>> > c) SR + Quantum field theory ---> photons travel along null paths
>> > ---> photons must be massless.
>>
>> And QED with massless photons is a special case of a more general
>> quantum theory with massive photons. Your comment is also irrelevant.
>
> Sure - but the photons won't travel along null paths, will have a
> dispersion relation that is a function of mass, etc, etc, etc.
> Inconsistent with observation.
Again don't true.
>> >> H = (\sqrt (m^2c^4 + p^2c^2) ) (1)
>>
>> > No. The special relativistic Hamiltonian is H = L = -mc^2 * [1 - v^2/
>> > c^2 ].
>>
>> Nonsense.
>
> *scratches head*
>
> http://canonicalscience.blogspot.com/2007/08/relativistic-lagrangian-
and-limitations.html
>
> Says it right there.
You say is completely false. The link say *nothing* about Hamiltonians!
> Then again my math ability is always bad this early
> in the morning. Maybe I'm mistaken?
After writing above nonsense now you wait to confound readers with your
usual unfair techniques.
What I wrote in that blog is the relativistic Lagrangian L, denoted by
symbol L and named (I quote from above link) the "free Lagrangian".
The Hamiltonian H is that I wrote above (1) in this newsgroup. What you
wrote was nonsensical.
You show once more again you have no idea of elementary facts of physics
still you reply adding noise. Go away!
I will use this recent post from you as warning to USENET users in a
future version of the USENET guidelines.
(rest of nonsense, irrelevant comments, and /ad hominem/ sniped)
Dono
<juanREM...@canonicalscience.com> wrote:
>
> >> >> H = (\sqrt (m^2c^4 + p^2c^2) ) (1)
>
> >> > No. The special relativistic Hamiltonian is H = L = -mc^2 * [1 - v^2/
> >> > c^2 ].
>
> >> Nonsense.
>
> > *scratches head*
>
> >http://canonicalscience.blogspot.com/2007/08/relativistic-lagrangian-
>
> and-limitations.html
>
>
>
> > Says it right there.
>
> You say is completely false. The link say *nothing* about Hamiltonians!
>
> > Then again my math ability is always bad this early
> > in the morning. Maybe I'm mistaken?
>
> After writing above nonsense now you wait to confound readers with your
> usual unfair techniques.
>
> What I wrote in that blog is the relativistic Lagrangian L, denoted by
> symbol L and named (I quote from above link) the "free Lagrangian".
>
Juanshito,
Your Alzheimer is progressing at an alarming pace:
1. You can't write proper English anymore
2. You contradict your own web page
3. You contradict yourself from post to post.
I am not even going to ask you to derive the relativistic Lagrangian
anymore.
You are clearly unable to.
Juan R.
But you continue without presenting any proof...
Some authors are developing electrodynamic theories where the photon has
a small mass and still the theory is compatible with high-precision tests
of QED:
E.V. Stefanovich,
Quantum Field Theory without Infinities
Ann. Phys. 292 (2001), 139-156.
See also discussion on massive photons on "S9c. What about relativistic
QFT at finite times?" on
http://www.mat.univie.ac.at/~neum/physics-faq.txt
There is an excellent discussion of the issue of experimental bounds on
massive photons in
http://arxiv.org/abs/hep-ph/0306245
Therefore your links, whereas interesting, are not addressing the issue
of this thread: Is (2), in my original post, representing same particle
than (1) or some other massive particle?
Juan R.
(...)
Fortunately, except you and Eric, nobody more here is confusing the
Hamiltonian H with the Lagrangian L.
Eric Gisse
<juanREM...@canonicalscience.com> wrote:
[snip blather]
You raise intellectual dishonesty to an art. Congratulations.
Juan R.
(snip)
You would not comment on posts without studying physics first. Learning
the difference between H and L is a pre-requisite for replying a post
about H and L.
Sue...
>
> But you continue without presenting any proof...
>
> Some authors are developing electrodynamic theories where the photon has
> a small mass and still the theory is compatible with high-precision tests
> of QED:
>
> E.V. Stefanovich,
> Quantum Field Theory without Infinities
> Ann. Phys. 292 (2001), 139-156.
>
I explained earlier how that can work.
Disroving a theory is the author's burden,
certainly not mine.
"Cargo Cult Science"
http://wwwcdf.pd.infn.it/~loreti/science.html
> See also discussion on massive photons on "S9c. What about relativistic
> QFT at finite times?" on
>
> http://www.mat.univie.ac.at/~neum/physics-faq.txt
>
> There is an excellent discussion of the issue of experimental bounds on
> massive photons in
>
> http://arxiv.org/abs/hep-ph/0306245
>
> Therefore your links, whereas interesting, are not addressing the issue
> of this thread: Is (2), in my original post, representing same particle
> than (1) or some other massive particle?
It seldom qualifies as a particle at all.
http://nobelprize.org/physics/articles/ekspong/index.html
...So anybody that cares what its mass is has already
defined it by the interaction it participates in.
Perhaps the refugees from EM-101 need to know?
http://farside.ph.utexas.edu/teaching/em/lectures/lectures.html
http://web.mit.edu/8.02t/www/802TEAL3D/visualizations/light/index.htm
http://www.ee.surrey.ac.uk/Personal/D.Jefferies/antennas.html
Sue...
Dono
<juanREM...@canonicalscience.com> wrote:
> Dono wrote on Thu, 10 Jul 2008 08:32:09 -0700:
>
> (...)
>
> Fortunately, except you and Eric, nobody more here is confusing the
> Hamiltonian H with the Lagrangian L.
>
Juanshito,
No one confused the hamiltonian with the lagrangian (not even you, in
spite your very advanced Alzheimer!)
The issue is that you contradicted yourself in discussing L.
The issue is that you contradicted your own web page.
The issue is that you have no clue how to derive L.
PRETENDER.
Juan R.
??
> Perhaps the refugees from EM-101 need to know?
>
> http://farside.ph.utexas.edu/teaching/em/lectures/lectures.html
> http://web.mit.edu/8.02t/www/802TEAL3D/visualizations/light/index.htm
> http://www.ee.surrey.ac.uk/Personal/D.Jefferies/antennas.html
>
> Sue...
Well, I already remarked that theories with massive photon are giving the
same experimental answers that classical electrodynamics in the
corresponding limit.
And also pointed to one example of quantum theory with massive photon
gives the same scattering amplitudes that computed using Feynman diagrams
for QED
Your links continue to be interesting but of no help. You continue
avoiding to reply a simple question I did.
Juan R.
> No one confused the hamiltonian with the lagrangian (not even you,
(...)
Eric confounded both, wrote H = L in previous message and did incorrect
claims about a blog article. You also did :-)
Sue...
> > that cares what its mass is has already defined it by the interaction it
> > participates in.
>
> ??
>
> > Perhaps the refugees from EM-101 need to know?
>
> >http://farside.ph.utexas.edu/teaching/em/lectures/lectures.html
> >http://web.mit.edu/8.02t/www/802TEAL3D/visualizations/light/index.htm
> >http://www.ee.surrey.ac.uk/Personal/D.Jefferies/antennas.html
>
> > Sue...
>
> Well, I already remarked that theories with massive photon are giving the
> same experimental answers that classical electrodynamics in the
> corresponding limit.
>
> And also pointed to one example of quantum theory with massive photon
> gives the same scattering amplitudes that computed using Feynman diagrams
> for QED
If you have a keen eye you may notice that only
three of the four forces on this web page has
a summary explanation.
http://hyperphysics.phy-astr.gsu.edu/hbase/forces/funfor.html
*Mass* is very relevant to what is absent from the page.
Can you write the missing material in terms of
your massive photon?
Sue...
Pmb
"Juan R. González-Álvarez" <juanR...@canonicalscience.com> wrote in
message news:pan.2008.07...@canonicalscience.com...
It would be helpful if people would define their terms here. I.e. what
definition of "mass" is being used here? Proper mass or inertial mass?
Pete
Pmb
"Juan R. González-Álvarez" <juanR...@canonicalscience.com> wrote in
message news:pan.2008.07...@canonicalscience.com...
>
>> No one confused the hamiltonian with the lagrangian (not even you,
>
> (...)
>
> Eric confounded both, wrote H = L in previous message and did incorrect
> claims about a blog article. You also did :-)
Eric isn't a physicist so don't expect him to be precise as one.
The Hamiltonian is identical to Jacobi's integral (sometimes called the
"energy function") in value. The former is expressed in terms of position
and canonical momentum whereas the later is expressed in terms of position
(q) and its first time derivative (dq/dt)
Pete
Sue...
[...]
>
> It would be helpful if people would define their terms here. I.e. what
> definition of "mass" is being used here? Proper mass or inertial mass?
Indeed it would.
I am refering to
"The invariant mass, intrinsic mass, proper mass or just mass..."
http://en.wikipedia.org/wiki/Rest_mass
http://en.wikipedia.org/wiki/Mass_in_special_relativity#Controversy
Sue...
>
> Pete
Pmb
"Sue..." <suzyse...@yahoo.com.au> wrote in message
news:c46db903-ef7d-42a8...@d45g2000hsc.googlegroups.com...
Thanks for clarifying. Now, is that what Juan is referring to?
Pete
Eric Gisse
> "Juan R. González-Álvarez" <juanREM...@canonicalscience.com> wrote in
> messagenews:pan.2008.07...@canonicalscience.com...
>
> > Dono wrote on Thu, 10 Jul 2008 09:52:26 -0700:
>
> >> No one confused the hamiltonian with the lagrangian (not even you,
>
> > (...)
>
> > Eric confounded both, wrote H = L in previous message and did incorrect
> > claims about a blog article. You also did :-)
>
> Eric isn't a physicist so don't expect him to be precise as one.
Yea but I'm smarter than both of you. By such a wide margin it boggles
the mind.
Exactly _neither of you_ are physicists which makes the snipe that
much more lolful. No, pete - sending complaints to a graduate school
doesn't count. No, Juan - lying about going to a conference doesn't
count.
Hell, you guys aren't even very good scientists. In fact, neither of
you are really that good of persons now that I think about it. Pete
has his whining bitchfits, and Juan has his compulsive intellectual
dishonesty.
I was going to make a point before I remembered how much I don't like
the two of you. Oh yea - the Lagrangian and Hamiltonian are equal when
there are no potentials under consideration. I figured between the two
of you quotemining classical mechanics textbooks, one of you would
have figured that out.
>
> The Hamiltonian is identical to Jacobi's integral (sometimes called the
> "energy function") in value. The former is expressed in terms of position
> and canonical momentum whereas the later is expressed in terms of position
> (q) and its first time derivative (dq/dt)
>
> Pete
Congratulations Pete - you opened a classical mechanics textbook at
some point in your life. Skip to the chapter that discusses the
Hamiltonian and how it relates to the Lagrangian.
That way you won't freak the fuck out when I use a trivial fact that's
obviously true but not explicitly written in your textbooks.
Eric Gisse
<juanREM...@canonicalscience.com> wrote:
> Dono wrote on Thu, 10 Jul 2008 09:52:26 -0700:
>
> > No one confused the hamiltonian with the lagrangian (not even you,
>
> (...)
>
> Eric confounded both, wrote H = L in previous message and did incorrect
> claims about a blog article. You also did :-)
Oh I'm sorry, are the big words and complicated symbols confusing you?
Here - Eric will help you past those little stumbling blocks that make
science so difficult.
The Lagrangian for a classical system is defined to be T - V.
Are you following me? Good.
The Hamiltonian for a classical system is defined to be T + V.
Now take a minute to relax, because this is where the magic happens
and you don't want to miss it!
When V = 0 ... H and L are the same. Isn't that fuckin' amazing?
And where did I get the value for L? Well - it was right there on your
blog. Sure I could have referenced Goldstein, Symon, Jackson, Misner /
Thorne / Wheeler, Carroll, or even Wiki-fucking-pedia.
However those references just lacked the certain something extra I get
by shoving your micro thought back in your face.
Alert readers: Given Juan's recent history of deleting posts on google
groups, who wants to take a guess how long his micro thought [sooo
apt] will stay up for the world to see?
Pmb
"Eric Gisse" <jow...@gmail.com> wrote in message
news:41d28527-f39e-4ea6...@e53g2000hsa.googlegroups.com...
On Jul 10, 9:54 am, "Juan R." González-Álvarez
<juanREM...@canonicalscience.com> wrote:
> Dono wrote on Thu, 10 Jul 2008 09:52:26 -0700:
>
> > No one confused the hamiltonian with the lagrangian (not even you,
>
> (...)
>
> Eric confounded both, wrote H = L in previous message and did incorrect
Okay. Enough is enough. You've provided more than adequate reason to be
killfiled. Plonk!
Pmb
On Jul 10, 10:58 am, "Pmb" <physics_wo...@yahoo.com> wrote:
> "Juan R. González-Álvarez" <juanREM...@canonicalscience.com> wrote in
> messagenews:pan.2008.07...@canonicalscience.com...
>
> > Dono wrote on Thu, 10 Jul 2008 09:52:26 -0700:
>
> >> No one confused the hamiltonian with the lagrangian (not even you,
>
> > (...)
>
> > Eric confounded both, wrote H = L in previous message and did incorrect
> > claims about a blog article. You also did :-)
>
> Eric isn't a physicist so don't expect him to be precise as one.
>Yea but I'm smarter than both of you.
Yah! Right! LOL!! You sure haven't demonstrated that here in the many
years I've been posting.
>Exactly _neither of you_ are physicists which makes the snipe that
>much more lolful.
Well that just goes to show how little you know. I have a BA in physics and
did some graduate work at Northeastern University
> No, pete - sending complaints to a graduate school
>doesn't count.
You're one very silly boy Eric. LOL!!
>Hell, you guys aren't even very good scientists. In fact, neither of
>you are really that good of persons now that I think about it. Pete
>has his whining bitchfits, ..
Yah. Right! Actually you've just proven to us that its you who is more
likely to have a bitchfit.
> and Juan has his compulsive intellectual dishonesty.
According to you? Sorry Eric but you're not the sort of person I'd take
their word for.
>:I was going to make a point before I remembered how much I don't like
>the two of you.
That's another strike against you. You're making personal judgements on
people you've never met based on superficial impressions on what you read on
a newsgroup. You have demonstrated that you read into things which were
never there.
>Oh yea - the Lagrangian and Hamiltonian are equal when
>there are no potentials under consideration. I figured between the two
>of you quotemining classical mechanics textbooks, one of you would
>have figured that out.
Oh my! Now you've shown that you don't know the basics of analytical
mechanics!! LOL!!
>>
>> The Hamiltonian is identical to Jacobi's integral (sometimes called the
>> "energy function") in value. The former is expressed in terms of position
>> and canonical momentum whereas the later is expressed in terms of
>> position
>> (q) and its first time derivative (dq/dt)
>>
>> Pete
>
>Congratulations Pete - you opened a classical mechanics textbook at
>some point in your life.
Its called a "college education".
>Skip to the chapter that discusses the
>Hamiltonian and how it relates to the Lagrangian.
Oy! Eric you aer such a silly boy! Who in the world are you attempting to
impress? It sure isn't I.
>That way you won't freak the fuck out when I use a trivial fact that's
>obviously true but not explicitly written in your textbooks.
You've just proven that you don't know even the most simplest parts of
analytical mechanics by the claim you made above regarding potentials.
Sorry Eric but all you've done here is to demonstrate what an arrogant sob
you are.
Later
Pete
Eric Gisse
> "Eric Gisse" <jowr...@gmail.com> wrote in message
....and here Pete realizes I'm right. Whoops!
Pmb
"Pmb" <physic...@yahoo.com> wrote in message
news:IP6dnTwT9KFyFOvV...@comcast.com...
>
> "Eric Gisse" <jow...@gmail.com> wrote in message
> news:41d28527-f39e-4ea6...@e53g2000hsa.googlegroups.com...
> <juanREM...@canonicalscience.com> wrote:
>> Dono wrote on Thu, 10 Jul 2008 09:52:26 -0700:
>>
>> > No one confused the hamiltonian with the lagrangian (not even you,
>>
>> (...)
>>
>> Eric confounded both, wrote H = L in previous message and did incorrect
>> claims about a blog article. You also did :-)\
>
> Okay. Enough is enough. You've provided more than adequate reason to be
> killfiled. Plonk!
btw - that was directed towards Eric
Tom Roberts
> The usual argument for massless photons uses the Hamiltonian [...]
> H = pc
You notion of "usual" and mine differ ENORMOUSLY.
First, I cannot see how such a Lagrangian gives photons at all -- for
that one needs a QUANTUM Lagrangian, as photons are inherently quantum
objects. If your theory does not have any photons, how can you sensibly
ask questions about their mass?
Second, given a proper quantum Lagrangian (and the corresponding
Hamiltonian if you wish), there's no general theoretical argument that
requires photons to be massless.
The REAL argument for massless photons is that when one treats the mass
of the photon as a free parameter and fits the theory to experimental
observations, one finds that the mass must be an INCREDIBLY small value,
and is consistent with zero. The PDG lists the current upper bound on
the photon mass as 6*10^-17 eV/c^2.
To put that in perspective, the lightest other particle
known (except neutrinos whose masses are not known) is the
electron. If an electron were scaled up to have the mass
of the earth, then the upper bound on the photon mass would
be scaled up to the mass of a half cubic meter of water.
> If alternatively we start assuming (v = c) in the transformation, the
> result is
> H = pc - L = pc (2)
> where no assumption was taken about the mass.
> Is this Hamiltonian (2) representing some kind of massive photon (somewhat
> as in Proca theory [#]) or is really the same that (1)?
The same objections apply. But on the question of this being different
from (1), ask yourself: what changes in the meanings of symbols have
occurred? If none, then these are obviously the same. But that is a
subtle question, in particular: what is the meaning of "c"?
> Well, I already remarked that theories with massive photon are giving the
> same experimental answers that classical electrodynamics in the
> corresponding limit.
> And also pointed to one example of quantum theory with massive photon
> gives the same scattering amplitudes that computed using Feynman diagrams
> for QED
But that it not how one tests theories! To do that, one goes to the PDG
website, looks up the experiments that they used to establish their
limit on the photon mass, and one fits the theory with a massive photon
to the experimental data, leaving mass as a free parameter in the
theory. But, of course, that is essentially what those experimenters and
the PDG already did to obtain their limit.
Tom Roberts
xxe...@gmail.com
> On Jul 9, 7:38 pm, xxe...@gmail.com wrote:
> [...]
>
>
>
> > Come to think about, I want all the criticism and abuse for my
> > thinking about this. I want to learn like I suppose everyone else
> > does.
>
> Well...
> Try using a wave model for light. When the anomalous twin
> vanishes you may have learned *better* that everyone else.
>
> http://farside.ph.utexas.edu/teaching/em/lectures/lectures.htmlhttp://web.mit.edu/8.02t/www/802TEAL3D/visualizations/light/index.htm
>
> Sue...
xxein: Sorry to have made myself appear stupid to you. I am
lightyears beyond such twins.
It's just that it appears that everybody has an answer for everything
and none of them coincide wrt to the other.
It may be that nothing really has a mass. What does a lump of
entangled energy feel like? Can it have a velocity (even if only
relative)? Can it have a momentum-like effect?
Although individual photons are measured to have a velocity of c in
all FORs, it does not necessarily follow that a one-way speed of c is
present in that frame. You don't measure the same in a different FOR,
do you? The point here is that in order to measure a photon's
velocity the same in different FORs, you would have to either contain
all photons in your FOR (and the hell with anyone else's), or that we
measure only TWLS and not OWLS. The matter of consideration is
obvious.
It's like tuning an engine. There is a dependency on timing, advance,
fuel flow and let's not forget how an engine is constructed to work.
But there's a difference between an engine and a physic. One is
physics and the forgotten step-child is the underlying physic.
It is not possible to make an engine without the underlying physic.
Many of our theories tend to side-step this basic issue and just make
some kind of continuity that is not deserved in in the physic's
(itself) breath. Yeah. It happens. We interfere. We can do that on
a small extent. But that extent is chosen by us. We can make
anything we want to out of it for how we think.
We can make a measurement, but we cannot know if we have captured the
physic by any such measurement. Certainly a physics is a local
phenomena (ex: Relativity) but it has to have a physic to rule how it
can work. It seems that no one makes this distinction.
This is what I mean by separating subjective measurability into a
different class to that of an objective viewpoint. Nobody seems to
understand.
As long as we can make an engine, who cares? That is the state of our
science anymore.
In a way, I hope that our minds could be obliterated so that we are
forced into re-thinking.
Sue...
> On Jul 9, 8:03 pm, "Sue..." <suzysewns...@yahoo.com.au> wrote:
>
>
>
> > On Jul 9, 7:38 pm, xxe...@gmail.com wrote:
> > [...]
>
> > > Come to think about, I want all the criticism and abuse for my
> > > thinking about this. I want to learn like I suppose everyone else
> > > does.
>
> > Well...
> > Try using a wave model for light. When the anomalous twin
> > vanishes you may have learned *better* that everyone else.
>
http://farside.ph.utexas.edu/teaching/em/lectures/lectures.html
>
>
> xxein: Sorry to have made myself appear stupid to you. I am
> lightyears beyond such twins.
>
> It's just that it appears that everybody has an answer for everything
> and none of them coincide wrt to the other.
>
> It may be that nothing really has a mass. What does a lump of
> entangled energy feel like? Can it have a velocity (even if only
> relative)? Can it have a momentum-like effect?
Yes... I often wonder if electrons should "fall".
Until my local welding supply begins selling them
in containers that I can put on a scale I am
remaining skeptical... but also not loosing
much sleep over it.
>
> Although individual photons are measured to have a velocity of c in
> all FORs, it does not necessarily follow that a one-way speed of c is
> present in that frame. You don't measure the same in a different FOR,
> do you? The point here is that in order to measure a photon's
> velocity the same in different FORs, you would have to either contain
> all photons in your FOR (and the hell with anyone else's), or that we
> measure only TWLS and not OWLS. The matter of consideration is
> obvious.
Q. What makes a bullet move in a straight line.
A. It seldom does. Its mass is coupled to neighboring masses.
Q. What makes light move in a straight line?
A. Antenna gain and homogenous or favourable dielectric.
http://www.rp-photonics.com/gaussian_beams.html
>
> It's like tuning an engine. There is a dependency on timing, advance,
> fuel flow and let's not forget how an engine is constructed to work.
> But there's a difference between an engine and a physic. One is
> physics and the forgotten step-child is the underlying physic.
>
> It is not possible to make an engine without the underlying physic.
> Many of our theories tend to side-step this basic issue and just make
> some kind of continuity that is not deserved in in the physic's
> (itself) breath. Yeah. It happens. We interfere. We can do that on
> a small extent. But that extent is chosen by us. We can make
> anything we want to out of it for how we think.
For the particle we have never met, discription
frequently preceeds discovery.
<<I think equation guessing might be the best method to
proceed to obtain the laws for the part of physics which
is presently unknown. Yet, when I was much younger, I
tried this equation guessing and I have seen many students
try this, but it is very easy to go off in wildly incorrect
and impossible directions. I think the problem is not to
find the best or most efficient method to proceed to a
discovery, but to find any method at all. Physical
reasoning does help some people to generate suggestions
as to how the unknown may be related to the known.>>
--R.P.Feynman
http://nobelprize.org/nobel_prizes/physics/laureates/1965/feynman-lecture.html
>
> We can make a measurement, but we cannot know if we have captured the
> physic by any such measurement. Certainly a physics is a local
> phenomena (ex: Relativity) but it has to have a physic to rule how it
> can work. It seems that no one makes this distinction.
Are you a sailor? Only sailors have a god given right
to complain. :o)
The *mechanism* is certainly more interesting and informs
further advances but the *formalism* is what we must test.
Without that, it is not science.
>
> This is what I mean by separating subjective measurability into a
> different class to that of an objective viewpoint. Nobody seems to
> understand.
You are describing light in terms of Newton's laws.
Try describing Newton's laws in terms of light.
"Relativity and electromagnetism"
http://farside.ph.utexas.edu/teaching/em/lectures/node106.html
"The Origin of Gravity" [untestable mechanism ]
http://arxiv.org/abs/physics/0107015v6
>
> As long as we can make an engine, who cares? That is the state of our
> science anymore.
>
> In a way, I hope that our minds could be obliterated so that we are
> forced into re-thinking.
Ouch! :o)
Sue...
Juan R.
>> Well, I already remarked that theories with massive photon are giving
>> the same experimental answers that classical electrodynamics in the
>> corresponding limit.
>>
>> And also pointed to one example of quantum theory with massive photon
>> gives the same scattering amplitudes that computed using Feynman
>> diagrams for QED
(...)
> It would be helpful if people would define their terms here. I.e. what
> definition of "mass" is being used here? Proper mass or inertial mass?
I am referring to rest mass (usually denoted by m).
In standard EM theory rest mass for the photon is zero but in other
theories and models it is not.
Juan R.
>> Well, I already remarked that theories with massive photon are giving
>> the same experimental answers that classical electrodynamics in the
>> corresponding limit.
>>
>> And also pointed to one example of quantum theory with massive photon
>> gives the same scattering amplitudes that computed using Feynman
>> diagrams for QED
>
> If you have a keen eye you may notice that only three of the four forces
> on this web page has a summary explanation.
>
> http://hyperphysics.phy-astr.gsu.edu/hbase/forces/funfor.html
>
> *Mass* is very relevant to what is absent from the page. Can you write
> the missing material in terms of your massive photon?
Precisely, the author of the quantum theory with massive photon cited
above has reused the relativistic quantum model he is developing and
applied to gravitational interactions, (the changes needed are small
charges by masses, some changes in interaction coefficients and so on)
He has showed that one can obtain a quantum theory of gravity free of
divergences that gives the same classical tests as GR (this includes
light deflection by Sun)...
Once again, you avoid to reply my question and add more useless links.
Juan R.
> On Jul 10, 9:54 am, "Juan R." González-Álvarez
> <juanREM...@canonicalscience.com> wrote:
>> Dono wrote on Thu, 10 Jul 2008 09:52:26 -0700:
>>
>> > No one confused the hamiltonian with the lagrangian (not even you,
>>
>> (...)
>>
>> Eric confounded both, wrote H = L in previous message and did incorrect
>> claims about a blog article. You also did :-)
>
> Oh I'm sorry, are the big words and complicated symbols confusing you?
> Here - Eric will help you past those little stumbling blocks that make
> science so difficult.
>
> The Lagrangian for a classical system is defined to be T - V.
>
> Are you following me? Good.
>
> The Hamiltonian for a classical system is defined to be T + V.
>
> Now take a minute to relax, because this is where the magic happens and
> you don't want to miss it!
>
> When V = 0 ... H and L are the same. Isn't that fuckin' amazing?
Beni replies your *nonsense* about 04:20 seconds
http://www.youtube.com/watch?v=SSZp3C65vps
Of course, you have absolutely no idea of basic facts of physics.
The Hamiltonian is, of course, that I wrote in the original poster
> >> >> H = (\sqrt (m^2c^4 + p^2c^2) ) (1)
You incorrect reply to that was
> >> > No. The special relativistic Hamiltonian is
> >> > H = L = -mc^2 * [1 - v^2/ c^2 ].
Is the kind of nonsense characterizes most of your posts here.
(...)
> Alert readers: Given Juan's recent history of deleting posts on google
> groups, who wants to take a guess how long his micro thought [sooo apt]
> will stay up for the world to see?
Well, you continue with your paranoia...
Don't worry I already said that your above nonsense (H = L) will be cited
as example of bizarre posting in a future improvement of the guidelines.
Readers will warned of the existence of people as you.
After your insistence for years finally you will cited in one of my
works. Be a bit patient!
Juan R.
> On Jul 10, 2:14 pm, "Pmb" <physics_wo...@yahoo.com> wrote:
>> "Eric Gisse" <jowr...@gmail.com> wrote in message
>>
>> news:41d28527-f39e-4ea6-9977-
fcfa73...@e53g2000hsa.googlegroups.com...
>> On Jul 10, 9:54 am, "Juan R." González-Álvarez
>>
>> <juanREM...@canonicalscience.com> wrote:
>> > Dono wrote on Thu, 10 Jul 2008 09:52:26 -0700:
>>
>> > > No one confused the hamiltonian with the lagrangian (not even you,
>>
>> > (...)
>>
>> > Eric confounded both, wrote H = L in previous message and did
>> > incorrect claims about a blog article. You also did :-)\
>>
>> Okay. Enough is enough. You've provided more than adequate reason to be
>> killfiled. Plonk!
>
> ....and here Pete realizes I'm right. Whoops!
Look Beni about 04:20 seconds
http://www.youtube.com/watch?v=SSZp3C65vps
--
Juan R.
> On Jul 10, 10:58 am, "Pmb" <physics_wo...@yahoo.com> wrote:
>> "Juan R. González-Álvarez" <juanREM...@canonicalscience.com> wrote in
>> messagenews:pan.2008.07...@canonicalscience.com...
>>
>> > Dono wrote on Thu, 10 Jul 2008 09:52:26 -0700:
>>
>> >> No one confused the hamiltonian with the lagrangian (not even you,
>>
>> > (...)
>>
>> > Eric confounded both, wrote H = L in previous message and did
>> > incorrect claims about a blog article. You also did :-)
>>
>> Eric isn't a physicist so don't expect him to be precise as one.
>
> Yea but I'm smarter than both of you. By such a wide margin it boggles
> the mind.
Yes the margin is very wide but so negative as your Hamiltonian H :-)
> No. The special relativistic Hamiltonian is
> H = L = -mc^2 * [1 - v^2/ c^2 ].
ha ha ha ha ha ha ha ha ha ha ha ha.
I want to be so smart as you :-)
Look Beni reaction when I explained him how smart (by a wide margin) you
are (look about 04:20 seconds)
http://www.youtube.com/watch?v=SSZp3C65vps
Sue...
<juanREM...@canonicalscience.com> wrote:
> Sue... wrote on Thu, 10 Jul 2008 11:09:39 -0700:
>
> >> Well, I already remarked that theories with massive photon are giving
> >> the same experimental answers that classical electrodynamics in the
> >> corresponding limit.
>
> >> And also pointed to one example of quantum theory with massive photon
> >> gives the same scattering amplitudes that computed using Feynman
> >> diagrams for QED
>
> > If you have a keen eye you may notice that only three of the four forces
> > on this web page has a summary explanation.
>
> >http://hyperphysics.phy-astr.gsu.edu/hbase/forces/funfor.html
>
> > *Mass* is very relevant to what is absent from the page. Can you write
> > the missing material in terms of your massive photon?
>
> Precisely, the author of the quantum theory with massive photon cited
> above has reused the relativistic quantum model he is developing and
> applied to gravitational interactions, (the changes needed are small
> charges by masses, some changes in interaction coefficients and so on)
>
> He has showed that one can obtain a quantum theory of gravity free of
> divergences that gives the same classical tests as GR (this includes
> light deflection by Sun)...
We anxiously await its appearance in a peer reviewed publication.
Sue...
Juan R.
> Juan R. González-Álvarez wrote:
>> The usual argument for massless photons uses the Hamiltonian [...] H =
>> pc
>
> You notion of "usual" and mine differ ENORMOUSLY.
I am sure! See below.
> First, I cannot see how such a Lagrangian gives photons at all -- for
> that one needs a QUANTUM Lagrangian, as photons are inherently quantum
> objects. If your theory does not have any photons, how can you sensibly
> ask questions about their mass?
It is described in several places how one can obtain the energy for a
photon from the Hamiltonian
H = (\sqrt (m^2c^4 + p^2c^2) )
when setting (m = 0) to yield the Hamiltonian energy
H = pc
This is standard stuff. Take a textbook or visit
http://en.wikipedia.org/wiki/
Mass_in_special_relativity#The_relativistic_energy-momentum_equation
You may also find useful the discussion in
http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html
but it may be something confusing for a first reading and maybe you will
enter into trouble.
> Second, given a proper quantum Lagrangian (and the corresponding
> Hamiltonian if you wish), there's no general theoretical argument that
> requires photons to be massless.
That was the *point*. Is Hamiltonian (2) in my original message
representing the same massless particle than (1) or not?
> The REAL argument for massless photons is that when one treats the mass
> of the photon as a free parameter and fits the theory to experimental
> observations, one finds that the mass must be an INCREDIBLY small value,
> and is consistent with zero. The PDG lists the current upper bound on
> the photon mass as 6*10^-17 eV/c^2.
None of those measurements invalidate the idea of a massive photon. You
are giving no real argument.
As remarked before, quantum theories of massive photon give the same
precision for experimental data than QED using a massless photon. In the
former case the small photon mass is used to cure divergences.
See also references cited in previous messages about actual measurements
and interpretation of mass bounds for photons.
(...)
>> Well, I already remarked that theories with massive photon are giving
>> the same experimental answers that classical electrodynamics in the
>> corresponding limit.
>> And also pointed to one example of quantum theory with massive photon
>> gives the same scattering amplitudes that computed using Feynman
>> diagrams for QED
>
> But that it not how one tests theories! To do that, one goes to the PDG
> website, looks up the experiments that they used to establish their
> limit on the photon mass, and one fits the theory with a massive photon
> to the experimental data, leaving mass as a free parameter in the
> theory. But, of course, that is essentially what those experimenters and
> the PDG already did to obtain their limit.
Read posts before replying.
Juan R.
> We anxiously await its appearance in a peer reviewed publication.
Another useless comment (/ad hominem/ on his paper?) that again avoids to
reply the original poster question.
Nobody here and in sci.physics.foundations has clearly replied my
question of if Hamiltonian (2) represent or not the same particle that
Hamiltonian (1).
The only useful reply on the topic has been by Igor. Probably he is right
and (2) is just (1).
Since it seems you insist on claiming that the idea of massive photon has
been experimentally discharged or answered on EM-101 :-) I add some other
links on the question of photon mass including gravitational and
cosmological consequences:
http://www.iop.org/EJ/abstract/0034-4885/68/1/R02/
http://www.aip.org/pnu/2003/split/625-2.html
http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/961102.html
http://prola.aps.org/abstract/PRD/v69/i10/e107501
http://www.iop.org/EJ/abstract/0049-1748/22/8/A21
http://prola.aps.org/abstract/PRL/v89/i10/e101301
Juan R.
> "Juan R. Gonzlez-lvarez" <juanR...@canonicalscience.com> wrote in
> message news:pan.2008.07...@canonicalscience.com...
>> Dono wrote on Thu, 10 Jul 2008 09:52:26 -0700:
>>
>>> No one confused the hamiltonian with the lagrangian (not even you,
>>
>> (...)
>>
>> Eric confounded both, wrote H = L in previous message and did incorrect
>> claims about a blog article. You also did :-)
>
> Eric isn't a physicist so don't expect him to be precise as one.
I don't wait Eric to be precise. As noticed in previous days he makes
dozens of mistakes each week, I only point a small fraction of his
nonsenses and I usually do when he decides to replies to me claiming some
mistake from mine, that he imagines in his usual paranoia...
But one thing is being no precise and other is his nonsensical claim that
> >> > No. The special relativistic Hamiltonian is
> >> > H = L = -mc^2 * [1 - v^2/ c^2 ].
> The Hamiltonian is identical to Jacobi's integral (sometimes called the
> "energy function") in value. The former is expressed in terms of
> position and canonical momentum whereas the later is expressed in terms
> of position (q) and its first time derivative (dq/dt)
But expressing a Hamiltonian as a function of velocity was only one of
Eric mistakes. The second being his confusion that (H = L) from the
components T and V!
The third mistake being his confusion between the Lagrangian and the
energy in special relativity. Note that Eric *multiplies* by the factor
(1 - (v^2/ c^2))
instead dividing by it. According to *Eric* when a particle travel to
energies close to c its energies vanishes...
But I am still more perturbed by the fact Eric and Dono considered the
special relativity Hamiltonian (energy) to be a negative quantity!!!
*They* sure us that Hamiltonian is negative
H(Eric) = -mc^2 * [1 - v^2/ c^2 ].
What physicist, student, or aficionado you know that would think that
energy of a massive particle in special relativity is negative?
Apart from ignorant and arrogant pair, Eric and Dono, I know of nobody
else.
And after his nonsense still Eric has claimed he is smart and his
formulae is right whereas insulted you!!!!!!!!
> Pete
My recommendation to users is to avoid Eric and Dono as plague. Several
of their posts with discussion of their favorite tactics will be cited in
a new version of USENET guidelines
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Finally they will get worldwide fame!
Juan R.
> Pmb wrote on Thu, 10 Jul 2008 14:58:40 -0400:
>
>> "Juan R. Gonzlez-lvarez" <juanR...@canonicalscience.com> wrote in
>> message news:pan.2008.07...@canonicalscience.com...
>>> Dono wrote on Thu, 10 Jul 2008 09:52:26 -0700:
>>>
>>>> No one confused the hamiltonian with the lagrangian (not even you,
>>>
>>> (...)
>>>
>>> Eric confounded both, wrote H = L in previous message and did
>>> incorrect claims about a blog article. You also did :-)
>>
>> Eric isn't a physicist so don't expect him to be precise as one.
>
> I don't wait Eric to be precise. As noticed in previous days he makes
> dozens of mistakes each week, I only point a small fraction of his
> nonsenses and I usually do when he decides to replies to me claiming
> some mistake from mine, that he imagines in his usual paranoia...
I don't wait Eric to be precise. As noticed in previous days he makes
dozens of mistakes each week, I only point a small fraction of his
nonsenses and I usually do when he decides to reply me claiming
some mistake from mine; 'mistakes' that he imagines in his usual
paranoia...
Pmb
"Juan R. González-Álvarez" <juanR...@canonicalscience.com> wrote in
message news:pan.2008.07...@canonicalscience.com...
>
>> On Jul 10, 10:58 am, "Pmb" <physics_wo...@yahoo.com> wrote:
>>> "Juan R. González-Álvarez" <juanREM...@canonicalscience.com> wrote in
>>> messagenews:pan.2008.07...@canonicalscience.com...
>>>
>>> > Dono wrote on Thu, 10 Jul 2008 09:52:26 -0700:
>>>
>>> >> No one confused the hamiltonian with the lagrangian (not even you,
>>>
>>> > (...)
>>>
>>> > Eric confounded both, wrote H = L in previous message and did
>>> > incorrect claims about a blog article. You also did :-)
>>>
>>> Eric isn't a physicist so don't expect him to be precise as one.
>>
>> Yea but I'm smarter than both of you. By such a wide margin it boggles
>> the mind.
>
> Yes the margin is very wide but so negative as your Hamiltonian H :-)
>
>> No. The special relativistic Hamiltonian is
>> H = L = -mc^2 * [1 - v^2/ c^2 ].
That's only the case for a particle which is not subject to a force since it
doesn't contain the potential.
Pete
Pmb
"Juan R. González-Álvarez" <juanR...@canonicalscience.com> wrote in
The actual Lagrangian for a free particle with proper mass "m" is L = -mc^2
sqrt[1 - v^2/c^2]. If it is subject to a potential V then L = -mc^2 sqrt[1 -
v^2/c^2] - V
>
> But I am still more perturbed by the fact Eric and Dono considered the
> special relativity Hamiltonian (energy) to be a negative quantity!!!
As I said, don't expect too much from Eric.
>
> *They* sure us that Hamiltonian is negative
>
> H(Eric) = -mc^2 * [1 - v^2/ c^2 ].
>
> What physicist, student, or aficionado you know that would think that
> energy of a massive particle in special relativity is negative?
The value of the Hamiltonian can be negative. All that means is that the
kinetic energy is so small that the potential energy has a greater
magnitude. For a bound particle this could mean a negative value for H,
depending on what the zero of potential is chosen to be. But H(Eric) = -mc^2
* [1 - v^2/ c^2 ] is impossible since it would mean a negative kinetic
energy
>
> Apart from ignorant and arrogant pair, Eric and Dono, I know of nobody
> else.
>
> And after his nonsense still Eric has claimed he is smart and his
> formulae is right whereas insulted you!!!!!!!!
As I said... lol!
>
>> Pete
>
> My recommendation to users is to avoid Eric and Dono as plague.
I've already plonked them both.
Pete
Sue...
<juanREM...@canonicalscience.com> wrote:
> Sue... wrote on Fri, 11 Jul 2008 03:31:10 -0700:
>
> > We anxiously await its appearance in a peer reviewed publication.
>
> Another useless comment (/ad hominem/ on his paper?) that again avoids to
> reply the original poster question.
>
> Nobody here and in sci.physics.foundations has clearly replied my
> question of if Hamiltonian (2) represent or not the same particle that
> Hamiltonian (1).
>
> The only useful reply on the topic has been by Igor. Probably he is right
> and (2) is just (1).
>
> Since it seems you insist on claiming that the idea of massive photon has
> been experimentally discharged or answered on EM-101 :-) I add some other
> links on the question of photon mass including gravitational and
> cosmological consequences:
>
> http://www.iop.org/EJ/abstract/0034-4885/68/1/R02/
>
<< Photon mass is expected to be zero by most
physicists, but this is an assumption which must
be checked experimentally. A nonzero mass would
make trouble for special relativity, Maxwell's
equations, and for Coulomb's inverse-square law
for electrical attraction. >>
http://www.aip.org/pnu/2003/split/625-2.html
Do you read your own references?
Sue...
Pmb
"Juan R. González-Álvarez" <juanR...@canonicalscience.com> wrote in
message news:pan.2008.07...@canonicalscience.com...
>
>> On Jul 10, 2:14 pm, "Pmb" <physics_wo...@yahoo.com> wrote:
>>> "Eric Gisse" <jowr...@gmail.com> wrote in message
>>>
>>> news:41d28527-f39e-4ea6-9977-
> fcfa73...@e53g2000hsa.googlegroups.com...
>>> On Jul 10, 9:54 am, "Juan R." González-Álvarez
>>>
>>> <juanREM...@canonicalscience.com> wrote:
>>> > Dono wrote on Thu, 10 Jul 2008 09:52:26 -0700:
>>>
>>> > > No one confused the hamiltonian with the lagrangian (not even you,
>>>
>>> > (...)
>>>
>>> > Eric confounded both, wrote H = L in previous message and did
>>> > incorrect claims about a blog article. You also did :-)\
>>>
>>> Okay. Enough is enough. You've provided more than adequate reason to be
>>> killfiled. Plonk!
>>
>> ....and here Pete realizes I'm right. Whoops!
Obviously a ploy used by children to antagonize people who know that their
posts aren't worth reading.
Eric has proven two things (1) he resorts to flaming when he doesn't
understand the physics and (2) is ignorant about relativity and analytical
mechanics (as made evident by his bogus comments on the Hamiltonian and
potential). Thus he was plonked. His claim above that H = L shows us how
very little he knows about physics.
Pete
Pmb
"Juan R. González-Álvarez" <juanR...@canonicalscience.com> wrote in
message news:pan.2008.07...@canonicalscience.com...
>
>> On Jul 10, 9:54 am, "Juan R." González-Álvarez
>> <juanREM...@canonicalscience.com> wrote:
>>> Dono wrote on Thu, 10 Jul 2008 09:52:26 -0700:
>>>
>>> > No one confused the hamiltonian with the lagrangian (not even you,
>>>
>>> (...)
>>>
>>> Eric confounded both, wrote H = L in previous message and did incorrect
>>> claims about a blog article. You also did :-)
>>
>> Oh I'm sorry, are the big words and complicated symbols confusing you?
>> Here - Eric will help you past those little stumbling blocks that make
>> science so difficult.
>>
>> The Lagrangian for a classical system is defined to be T - V.
>>
>> Are you following me? Good.
>>
>> The Hamiltonian for a classical system is defined to be T + V.
>>
>> Now take a minute to relax, because this is where the magic happens and
>> you don't want to miss it!
>>
>> When V = 0 ... H and L are the same. Isn't that fuckin' amazing?
A typical respose from a flamer. He assumes that swearing will help his
case.
Eric - You are so wrong its scary. In the first place you need to do a
Legendre transformation to change variables from position and veloclity
variables to position and momentum variables. Even when that is done you
will, by your (invalid) assumption, get the wrong value. According to you
the energy has the value h (q,v) = -mc^2(1 - v^2/c^2) which is totally
wrong. It implies that the energy, h, is all kinetic energy (since you set V
= 0) leaving h to be all kinetic energy. hence you assign not only the wrong
value of the kinetic energy but the wrong sign. The correct value for h is
(gamm - 1)mc^2. Not what you gave.
> Beni replies your *nonsense* about 04:20 seconds
>
> http://www.youtube.com/watch?v=SSZp3C65vps
>
> Of course, you have absolutely no idea of basic facts of physics.
>
> The Hamiltonian is, of course, that I wrote in the original poster
>
>> >> >> H = (\sqrt (m^2c^4 + p^2c^2) ) (1)
>
> You incorrect reply to that was
>
>> >> > No. The special relativistic Hamiltonian is
>> >> > H = L = -mc^2 * [1 - v^2/ c^2 ].
That isn't even the correct Lagrangian!
>
> Is the kind of nonsense characterizes most of your posts here.
>
> (...)
>
>> Alert readers: Given Juan's recent history of deleting posts on google
>> groups, who wants to take a guess how long his micro thought [sooo apt]
>> will stay up for the world to see?
>
> Well, you continue with your paranoia...
>
> Don't worry I already said that your above nonsense (H = L) will be cited
> as example of bizarre posting in a future improvement of the guidelines.
> Readers will warned of the existence of people as you.
>
> After your insistence for years finally you will cited in one of my
> works. Be a bit patient!
Not one of Eric's attributes.
Pete
Pmb
"Juan R. González-Álvarez" <juanR...@canonicalscience.com> wrote in
message news:pan.2008.07...@canonicalscience.com...
>
>> On Jul 10, 9:54 am, "Juan R." González-Álvarez
>> <juanREM...@canonicalscience.com> wrote:
>>> Dono wrote on Thu, 10 Jul 2008 09:52:26 -0700:
>>>
>>> > No one confused the hamiltonian with the lagrangian (not even you,
>>>
>>> (...)
>>>
>>> Eric confounded both, wrote H = L in previous message and did incorrect
>>> claims about a blog article. You also did :-)
>>
>> Oh I'm sorry, are the big words and complicated symbols confusing you?
>> Here - Eric will help you past those little stumbling blocks that make
>> science so difficult.
>>
>> The Lagrangian for a classical system is defined to be T - V.
>>
>> Are you following me? Good.
>>
>> The Hamiltonian for a classical system is defined to be T + V.
>>
>> Now take a minute to relax, because this is where the magic happens and
>> you don't want to miss it!
>>
>> When V = 0 ... H and L are the same. Isn't that fuckin' amazing?
I can't believe that he just posted that bizzare reasoning.
Eric - Here is how you just screwed up. In **relativity** the Lagrangian
**is not** given by L = T - V. That is only in non-relativistic mechanics.
LOL!!! Once again you'd demonstrated your ignorance in analytical
mechanics.
The correct physics is at
http://www.geocities.com/physics_world/sr/relativistic_energy.htm
Read it for once in your life!
> Beni replies your *nonsense* about 04:20 seconds
>
> http://www.youtube.com/watch?v=SSZp3C65vps
>
> Of course, you have absolutely no idea of basic facts of physics.
>
> The Hamiltonian is, of course, that I wrote in the original poster
>
>> >> >> H = (\sqrt (m^2c^4 + p^2c^2) ) (1)
>
> You incorrect reply to that was
>
>> >> > No. The special relativistic Hamiltonian is
>> >> > H = L = -mc^2 * [1 - v^2/ c^2 ].
>
> Is the kind of nonsense characterizes most of your posts here.
>
> (...)
>
>> Alert readers: Given Juan's recent history of deleting posts on google
>> groups, who wants to take a guess how long his micro thought [sooo apt]
>> will stay up for the world to see?
>
> Well, you continue with your paranoia...
Deleting posts on google? What is he, nuts??
> Don't worry I already said that your above nonsense (H = L) will be cited
> as example of bizarre posting in a future improvement of the guidelines.
Better yet, post his bogus assumption that L = T - V in relativity since
that is the source of his error.
> Readers will warned of the existence of people as you.
Thanks. :)
Pete
John Kennaugh
>On Jul 8, 9:11 am, "Juan R." González-Álvarez
><juanREM...@canonicalscience.com> wrote:
>> The usual argument for massless photons uses the Hamiltonian
>>
>>
>>
>> H = pc
>>
>> However, the original Hamiltonian (1) was derived using the Legendre
>> transformation
>>
>> H = pv - L = (\sqrt (m^2c^4 + p^2c^2) )
>>
>> If alternatively we start assuming (v = c) in the transformation, the
>> result is
>>
>> H = pc - L = pc (2)
>>
>> where no assumption was taken about the mass.
>>
>> Is this Hamiltonian (2) representing some kind of massive photon (somewhat
>> as in Proca theory [#]) or is really the same that (1)?
>>
>
>So you then have an imaginary mass that won't couple
>to a real gravito-inertial field.
>
>Now you just need an imaginary shape for the
>pseudo-particle to massage the nasty radiation
>patterns that we measure for real light into
>a point.
>
>http://www.rp-photonics.com/gaussian_beams.html
>
>
>
>There is surely a "Nobel" or a knighthood in it
>for you.
>
><<The Nobel Committee avoids committing itself
>to the particle concept. Light-quanta or with
>modern terminology, photons, were explicitly
>mentioned in the reports on which the prize
>decision rested only in connection with emission
>and absorption processes. The Committee says
>that the most important application of Einstein's
>photoelectric law and also its most convincing
>confirmation has come from the use Bohr made of
>it in his theory of atoms, which explains a vast
>amount of spectroscopic data. >>
>http://nobelprize.org/nobel_prizes/physics/articles/ekspong/index.html
It is an interesting link which shows the desperate measures which
physics has adopted to *retain* the wave-partical duality and avoid the
obvious conclusion that light is fundamentally particulate. The
experiment it describes to demonstrate the duality does nothing of the
kind when properly interpreted.
"Experiments with beams of light or of electrons have been made such
that both aspects - waves and particles - are observed. For interference
to occur it is among other things also necessary for the beam to have
available more than one path from source to detector (e.g. a screen).
Interference is explained by the wave picture. When the beam intensity
is sufficiently low and the detector suitable the impact of particles
one by one can be observed. The energy quanta are then localised as if
particles in space and time."
By definition "interference" implies that two things of different phase
have a net amplitude which will vary from virtually cancelling to the
sum of the two. What is not happening is that photons are arriving at a
point of minimum intensity and then being 'cancelled' by subsequent
photons. Once a photon arrives at a point it stays. The minimum in the
low light experiment represents the end of a path which, for whatever
reason, photons have a very low probability of taking and where the
maximum is the most probable direction. The result may mathematically
conform to the wave mathematical model but physically it is not and
cannot be interference for the reasons stated. Photons do not check with
the equations to see which direction to travel in. There is some
physical mechanism involved and whatever it is, it will also explain the
normal intensity pattern.
What is needed is a model of a photon which can give rise to wavelike
phenomena.
"The most hopeful of these led to the concept of the wave-packet.
In certain circumstances, chief among which is that the physical medium
in which they travel must be dispersive - a technical term - a group of
water waves will propagate together across a pond and will remain
concentrated together in the form of a package. The energy represented
by the wave system travels at the speed of the group, which is not the
same as the speed of the individual waves. (The mathematics of this
situation is quite elegant). Hence it was suggested that the quantum,
the particle-like concentration of light energy which was deduced from
the experiments, might be merely a wave-packet of dispersive
electromagnetic waves. That was the view which Planck himself took of
the matter and maintained with some vehemence.
The trouble with this idea is that although a suitable wavepacket could
remain stable indefinitely in the longitudinal direction, no
configuration of linear (Maxwell) waves can be devised which would
prevent a wave packet from dissipating across the direction of the
propagation. Now a beam of light will dissipate laterally, exactly like
a wave system, but the individual quanta of which it to be composed do
not dissipate. The wave-packet concept was a non-starter, disproved by
the evidence, but it is still offered to physics students today as
though it were valid and relevant." Dr Scott Murray
OK so that approach doesn't work and it smacks anyway of an extreme
reluctance to accept that light is not fundamentally made up of waves.
The only other approach which I am aware of it that by Waldron. A photon
contains equal positive and negative charge and spins. Now the
predictive accuracy of Maxwell's equations must mean something even if
his theory as a whole is flawed. Those equations are based upon the
behaviour of charge and only charge. So there is a link between light
and charge. There is a link between light and photons so it is
reasonable to assume there is a link between charge and photons.
In Maxwell's theory a field is a 'stress' (altered state) pattern in the
aether caused by a charge. If we now no longer believe in the aether it
is reasonable to ask what a charge IS. One may ask but the answer one
gets is that physics no longer considers such questions useful and makes
no attempt to answer them. I'm not a physicist and I still ask. If there
is no aether and if a 'field' is not some sort of unlikely unidentified
matter then it is a metaphysical 'field of influence' and cannot exist
without there being a source of influence. If Photons have associated
with them electromagnetic fields then they must contain charge. If those
fields are changing the charge must be rotating or oscillating.
Waldron may not have all the answers but it is an approach which as far
as I know has not been studied by main line physics which declares that
photons have no internal structure ensuring the perpetuation of the wave
particle duality. The status quo is always a great comfort.
--
John Kennaugh
"George W Bush is the elected leader of the world's only super power,
Controller of the world's most influential economic superstructure, Supreme
commander of the largest nuclear military force on the planet - you couldn't
make it up could you? But then 3.7 million Americans believe they have been
abducted by aliens and interrogated. Presumably the aliens wanted to know why?"
- Humphrey Lyttelton
Juan R.
>> Since it seems you insist on claiming that the idea of massive photon
>> has been experimentally discharged or answered on EM-101 :-) I add some
>> other links on the question of photon mass including gravitational and
>> cosmological consequences:
>>
>> http://www.iop.org/EJ/abstract/0034-4885/68/1/R02/
>>
>>
> << Photon mass is expected to be zero by most physicists, but this is an
> assumption which must be checked experimentally. A nonzero mass would
> make trouble for special relativity, Maxwell's equations, and for
> Coulomb's inverse-square law for electrical attraction. >>
> http://www.aip.org/pnu/2003/split/625-2.html
>
> Do you read your own references?
Another irrelevant comment /a la/ EM-101 :-)
The mass for a photon may introduce changes in special relativity,
Maxwell equation, and the Coulomb law.
For special relativity this is usually related to very special relativity
theories with varying c. For Maxwell equation, I already cited the Proca
action previously, which reduces to Maxwellian theory in the photon mass
limit
(m --> 0)
It is also known that a small photon mass adds a small exponential decay
to the 1/r^2 dependence of a pure Coulomb law.
But and this is the point you fail to understand again very special and
Proca-like theories coincide with usual special relativity and Maxwellian
theory in the low energy regime, giving the same results that classical
electrodynamnics and special relativity for the current tests.
This is why your EM-101 comments were useless...
In an analogous way the correction to the Coulomb law due to massive
photon is only noticeable for ultra-short distances, i.e. high energies.
The generalized law being totally compatible with all current EM data on
Coulomb 1/r^2 law.
This is why your EM-101 comments were useless...
I also remarked that some authors are using a small photon mass to cure
divergences of field theory but still their theory is completely
compatible with QED and classical electrodynamics *data*.
Write the generalized Coulomb law with the decay component associated to
a massive photon and compute the (r --> 0) limit. What you obtain?
Infinite? sure no...
Dono
<juanREM...@canonicalscience.com> wrote:
>
> The third mistake being his confusion between the Lagrangian and the
> energy in special relativity. Note that Eric *multiplies* by the factor
>
> (1 - (v^2/ c^2))
>
> instead dividing by it. According to *Eric* when a particle travel to
> energies close to c its energies vanishes...
JuanShit,
You ARE an idiot, the relativistic Lagrangian is indeed:
L= - m_0 * c^2 * sqrt((1-(v/c)^2))
MULTIPLICATION, Shito. Not division.
Square root
PRETENDER.
Juan R.
Right.
> All that means is that the
> kinetic energy is so small that the potential energy has a greater
> magnitude. For a bound particle this could mean a negative value for H,
> depending on what the zero of potential is chosen to be. But H(Eric) =
> -mc^2 * [1 - v^2/ c^2 ] is impossible since it would mean a negative
> kinetic energy
Yes it was pure nonsense becoming from anyone (Eric) who claim himself to
be one of the more knowledgeable posters in relativity, insulting others
who actually know physics.
>
>> Apart from ignorant and arrogant pair, Eric and Dono, I know of nobody
>> else.
>>
>> And after his nonsense still Eric has claimed he is smart and his
>> formulae is right whereas insulted you!!!!!!!!
>
> As I said... lol!
>
>
>>> Pete
>>
>> My recommendation to users is to avoid Eric and Dono as plague.
>
> I've already plonked them both.
>
> Pete
--
Dono
>
> > My recommendation to users is to avoid Eric and Dono as plague.
>
> I've already plonked them both.
>
> Pete
Wow, after I tried to correct the idiocies on your "Cyclotrons" page
(http://www.geocities.com/physics_world/sr/cyclotron.htm)?
BTW, your "cyclotron frequency" and "particle radius" are STILL wrong.
You should not mix Newtonian mechanics (F=ma!?) in the derivation.
Because you get...shit. :-)
Juan R.
> On Jul 11, 4:49 am, "JuanShit R." González-Álvarez
> <juanREM...@canonicalscience.com> wrote:
>
>
>> The third mistake being his confusion between the Lagrangian and the
>> energy in special relativity. Note that Eric *multiplies* by the factor
>>
>> (1 - (v^2/ c^2))
>>
>> instead dividing by it. According to *Eric* when a particle travel to
>> energies close to c its energies vanishes...
>
>
> JuanShit,
>
> You ARE an idiot, the relativistic Lagrangian is indeed:
>
> L=-m_0*c^2 *(1-(v/c)^2)
>
> MULTIPLICATION, Shito. Not division. PRETENDER.
Ha ha ha ha ha ha ha
You copied the Lagrangian from my blog article cited *before*
http://canonicalscience.blogspot.com/2007/08/relativistic-lagrangian-
and-limitations.html
You continue to confound the energy and the Hmailtonian with the
Lagrnagian L. This is hilarious :-)
Well done guys, continue to submit more nonsense instead recognizing both
of you did a very, very strong mistake.
Don't worry Dono you will got international fame as troll of poor specie
you deserve, when you will be cited in the guidelines together your
friend Eric.
Juan R.
ha ha ha ha ha ha ha ha ha ha ha ha ha ha
ha ha ha ha ha ha ha
Pmb
"Juan R. González-Álvarez" <juanR...@canonicalscience.com> wrote in
Eric actually claims to be one of the more knowledgeable posters in
relativity? You've got to be kidding me? He sure doesn't demonstrate that
here, especially with his poor understanding of relativistic analytical
mechanics.
Pete
Dono
<juanREM...@canonicalscience.com> wrote:
<imbecilities snipped>
MULTIPLCATION, shithead.
NOT DIVISION.
Your Alzheimer is progressing, you contradict yourself quite often:
> The third mistake being his confusion between the Lagrangian and the
> energy in special relativity. Note that Eric *multiplies* by the factor
> (1 - (v^2/ c^2))
> instead dividing by it.
You wrote that, didn't you, Juanshito? A few posts ago. Or you have
already "forgotten"?
Juan R.
you repeated this mistake above ha ha ha ha. I am tired from so many laugh
Please stop I cannot laugh more ha ha ha ha ha ha
You are 100% bufon ha ha ha ha you still think (H = L) ha ha ha
Juan R.
you repeated this mistake by *third* time ha ha ha ha. I am tired from so
many laugh i cannot sopt
Dono
So, Juanshito, you want to divide, eh? You forgot to take your pills,
again.
Juan R.
>>> All that means is that the
>>> kinetic energy is so small that the potential energy has a greater
>>> magnitude. For a bound particle this could mean a negative value for
>>> H, depending on what the zero of potential is chosen to be. But
>>> H(Eric) = -mc^2 * [1 - v^2/ c^2 ] is impossible since it would mean a
>>> negative kinetic energy
>>
>> Yes it was pure nonsense becoming from anyone (Eric) who claim himself
>> to be one of the more knowledgeable posters in relativity, insulting
>> others who actually know physics.
>
> Eric actually claims to be one of the more knowledgeable posters in
> relativity? You've got to be kidding me? He sure doesn't demonstrate
> that here, especially with his poor understanding of relativistic
> analytical mechanics.
I am not kidding! Eric *believes* that and once said us in this group
that he was the second or third more knowledgeable poster.
Eric also has problems with dimensional analysis
http://groups.google.com/group/sci.physics.relativity/msg/6fe7633a0e8130f8
Juan R.
Ha ha ha ha. I am tired from so many laugh I cannot stop.
ha ha ha ha ha ha ha ha ha ha habha ha h ah ah ha ha ha ha hah ah
Dono
You got is correctly on your website (where you dutiully COPIED from a
textbook):
http://canonicalscience.blogspot.com/2007_08_01_archive.html
See, you MULTIPLY -m_0*c^2 by sqrt(1-(v/c)^2) and you get the
relativistic Lagarangian.
So, why are you CONTRADICTING yourself in this thread and you are
asking for DIVISION by (1-(v/c)^2)?
Ahh, the Alzheimer, of course, old fart. You forgot to take your daily
medication :-)
Juan R.
Ha ha ha ha ha ha ha ha ha...
Ha ha ha for obtaining the energy Dono "MULTIPLIES" by sqrt(1-(v/c)^2)
and finally give a total "negative" sign Ha ha ha ha ha ha ha.
What bunch of nonsenses you say ha ha ha ha ha you still did not noticed
the nonsense you are saying ha ha ha ha
This is an excellent thread guys. Both of you are fantastic buffoons ha
ha ha I have getting dolor from so many laughs I can't stoppp :-)
Juan R.
Please continue trying to hide your strong mistake...
ha ha ha ha ha ha ha :-)
Juan R.
> So, why are you CONTRADICTING yourself in this thread and you are asking
> for DIVISION by (1-(v/c)^2)?
>
> Ahh, the Alzheimer, of course, old fart. You forgot to take your daily
> medication :-)
Please Dono ask Eric to write again
(\blockquote
No. The special relativistic Hamiltonian is
H = L = -mc^2 * [1 - v^2/ c^2 ].
)
and you repeat again your comic post saying that the (H = L), that energy
in special relativity is negative and contains a *multiplicative* factor
(1-(v/c)^2)
ha ha ha ha ha ha ha ha what bunch of nonsense ha ha ha ha ha :-)
Pmb
"Juan R. González-Álvarez" <juanR...@canonicalscience.com> wrote in
message news:pan.2008.07...@canonicalscience.com...
>
>> On Jul 11, 4:49 am, "JuanShit R." González-Álvarez
>> <juanREM...@canonicalscience.com> wrote:
>>
>>
>>> The third mistake being his confusion between the Lagrangian and the
>>> energy in special relativity. Note that Eric *multiplies* by the factor
>>>
>>> (1 - (v^2/ c^2))
>>>
>>> instead dividing by it. According to *Eric* when a particle travel to
>>> energies close to c its energies vanishes...
>>
>>
>> JuanShit,
>>
>> You ARE an idiot, the relativistic Lagrangian is indeed:
>>
>> L=-m_0*c^2 *(1-(v/c)^2)
Wrong. That is not the Lagrangian. Seems that some people just can't figure
this out Juan! LOL!
>>
>> MULTIPLICATION, Shito. Not division. PRETENDER.
>
> Ha ha ha ha ha ha ha
>
> You copied the Lagrangian from my blog article cited *before*
>
> http://canonicalscience.blogspot.com/2007/08/relativistic-lagrangian-
> and-limitations.html
If so then he copied it incorrectly Juan. You gave the correct
(non-covariant) Lagrangian, i.e. L = -mc^2 sqrt(1 - v^2/c^2), but that is
not what Dono and Eric keep posting.
Pete
Pmb
"Juan R. González-Álvarez" <juanR...@canonicalscience.com> wrote in
message news:pan.2008.07...@canonicalscience.com...
Oh my my! That's really funny! Whoever this Dono is he clearly doesn't
understand what he was reading.
Seems that Dono can't read English, i.e. "Since the motion is transverse to
the acceleration ..." - That part seems to have slipped past him. Probably
because he either didn't or couldn't understand it.
F = ma is on that page because it applies for a charged particle moving in a
purely magnetic field. I.e. When the acceleration is transverse to the
motion then the force has the value F = m_t a where m_t = transverse mass.
Dono seems to not understand that m_t = m = inertial mass (aka relativistic
mass).
Even when Dono was shown the derivation it seems he couldn't follow it
because he doesn't understand that the relation p = qBr is the correct
expression for cyclotron motion and thus r = p/qB.
Dono - please learn basic relativity before you attempt to correct something
you clearly don't understand.
Pete
Pmb
"Juan R. González-Álvarez" <juanR...@canonicalscience.com> wrote in
message news:pan.2008.07...@canonicalscience.com...
A very old whine from a crank. First off everybody learns these formulas
from textbooks. Only an idiot thinks that all physicists have all the
physics formulas memorized. Assuming that is just plain stupidity. And
because someone knows a formula is no indication that he just copied it from
a text. Dono shows his insecurity here in that he can't understand this well
known fact about textbooks as sources of reference for formulas. Clearly he
is the farthest thing from being a physicist that one can get.
>>
>> http://canonicalscience.blogspot.com/2007_08_01_archive.html
>>
>> See, you MULTIPLY -m_0*c^2 by sqrt(1-(v/c)^2) and you get the
>> relativistic Lagarangian.
>>
>> So, why are you CONTRADICTING yourself in this thread and you are asking
>> for DIVISION by (1-(v/c)^2)?
>>
>> Ahh, the Alzheimer, of course, old fart. You forgot to take your daily
>> medication :-)
>
> Ha ha ha ha ha ha ha ha ha...
>
> Ha ha ha for obtaining the energy Dono "MULTIPLIES" by sqrt(1-(v/c)^2)
> and finally give a total "negative" sign Ha ha ha ha ha ha ha.
>
> What bunch of nonsenses you say ha ha ha ha ha you still did not noticed
> the nonsense you are saying ha ha ha ha
>
> This is an excellent thread guys. Both of you are fantastic buffoons ha
> ha ha I have getting dolor from so many laughs I can't stoppp :-)
I just get sad to see what nonsense is being posted by them. I do have to
laugh everyone in a while, especially when they can't seem to learn from
their mistakes, even when they are pointed out to them.
Pete
Pmb
"Juan R. González-Álvarez" <juanR...@canonicalscience.com> wrote in
message news:pan.2008.07...@canonicalscience.com...
Especially where Eric claims that L = T - V in relativity. Now *that* is
pretty darn funny for someone who claims to know physics. lol!
Pete
Dono
Juanshit,
This is the same formula you have on your website (COPIED from a
textbook)
http://canonicalscience.blogspot.com/2007_08_01_archive.html
So, what's the problem?
Besides, it is not the "energy", imbecile, it is the relativistic
Lagrangian (L). Your Alzheimer is progressing at an alarming pace, old
fart.
Dono
> from textbooks. Only an idiot thinks that all physicists have all the
> physics formulas memorized. Assuming that is just plain stupidity. And
> because someone knows a formula is no indication that he just copied it from
> a text. Dono shows his insecurity here in that he can't understand this well
> known fact about textbooks as sources of reference for formulas. Clearly he
> is the farthest thing from being a physicist that one can get.
No, Pete
I happen to derive formulas, I do not copy or memorize them.
This is how I figured the errors in your "Cyclotrons" page :-)
It is nice to see how you and Juanshito support each other :-)
Dono
> "Juan R. González-Álvarez" <juanREM...@canonicalscience.com> wrote in
> messagenews:pan.2008.07...@canonicalscience.com...
>
>
>
>
>
> > Dono wrote on Fri, 11 Jul 2008 08:07:15 -0700:
>
> >> On Jul 11, 4:49 am, "JuanShit R." González-Álvarez
> >> <juanREM...@canonicalscience.com> wrote:
>
> >>> The third mistake being his confusion between the Lagrangian and the
> >>> energy in special relativity. Note that Eric *multiplies* by the factor
>
> >>> (1 - (v^2/ c^2))
>
> >>> instead dividing by it. According to *Eric* when a particle travel to
> >>> energies close to c its energies vanishes...
>
> >> JuanShit,
>
> >> You ARE an idiot, the relativistic Lagrangian is indeed:
>
> >> L=-m_0*c^2 *(1-(v/c)^2)
>
> Wrong. That is not the Lagrangian. Seems that some people just can't figure
> this out Juan! LOL!
>
>
>
> >> MULTIPLICATION, Shito. Not division. PRETENDER.
>
> > Ha ha ha ha ha ha ha
>
> > You copied the Lagrangian from my blog article cited *before*
>
> >http://canonicalscience.blogspot.com/2007/08/relativistic-lagrangian-
> > and-limitations.html
>
> If so then he copied it incorrectly Juan. You gave the correct
> (non-covariant) Lagrangian, i.e. L = -mc^2 sqrt(1 - v^2/c^2),
Yes, Pete
Juanshito has the correct (COPIED from textbook) Lagrangian on his
webpage. The funny thing is that he gave the INCORRECT formula in this
thread. He's a PRETENDER and so are you.
Pmb
"Juan R. González-Álvarez" <juanR...@canonicalscience.com> wrote in
message news:pan.2008.07...@canonicalscience.com...
Here's some more for you to laugh at Juan; F = ma was not due to Newton as
Dono seems to think. Its due to Euler. :) Newton's relation was F = dp/dt
which is precisely correct. Especially since Newtonian physics is not
inconsistent with relativity as far as definitions and laws. They can be
modified to be the same in the worst case. E.g. Newton's third law can be
made to work if it is restricted to contact forces. F = dp/dt already works
fine. p = mv is a definition and thus does not need to be changed
(especially since it always works in any case in sr and gr). All definitions
from Newtonian mechanics also apply to relativistic mechanics. Thus this
nonsense regarding Newtonian mechanics is just that, nonsense! The only
thing that is different is the results of observations, i.e. while p = mv is
a definition the law/observation that m is invariant does not apply in
relativity. But that is a law, not a definition.
Pete
Dono
You mean "buffoon", Juanshito?
Yes, you are a buffoon and all the people that know physics have your
number already. What I don't understand is why you feel compelled to
reaffirm it several times a day, every day? When do you have time to
do your "scientific research", Bozo? :-)
Dono
wrote:
>
> Here's some more for you to laugh at Juan; F = ma was not due to Newton as
> Dono seems to think. Its due to Euler. :)
Hey Pete,
You are now Brown-nosing the number one crank on this forum, mr.
Juanshito Alvarez.
Now, the point was that by your MIXING SR with Newtonian mechanics you
FUCKED UP your results on your "Cyclotrons" web page. You are using
the same tactics as Juanshito, when you are caught with mistakes you
try to deflect. This makes you what? A crank? Yes
The point is after all this time, your "cyclotron frequency" and your
"particle radius" are still WRONG, even after "harry" gave you a
helping hand.
Do you want to learn the right results and the correct derivation, mr.
Brown-nose?
>Especially since Newtonian physics is not
> inconsistent with relativity as far as definitions and laws.
Good, so why do you mix SR with Newtonian mechanics on your webpage?
This is a rookie mistake. PRETENDER, just like Juanshito.
Dono
wrote:
>
> >>> Pete
>
> >> Wow, after I tried to correct the idiocies on your "Cyclotrons" page
> >> (http://www.geocities.com/physics_world/sr/cyclotron.htm)?BTW, your
> >> "cyclotron frequency" and "particle radius" are STILL wrong. You should
> >> not mix Newtonian mechanics (F=ma!?) in the derivation. Because you
> >> get...shit. :-)
>
> Seems that Dono can't read English, i.e. "Since the motion is transverse to
> the acceleration ..." - That part seems to have slipped past him. Probably
> because he either didn't or couldn't understand it.
>
> F = ma is on that page because it applies for a charged particle moving in a
> purely magnetic field. I.e. When the acceleration is transverse to the
> motion then the force has the value F = m_t a where m_t = transverse mass.
> Dono seems to not understand that m_t = m = inertial mass (aka relativistic
> mass).
>
> Even when Dono was shown the derivation it seems he couldn't follow it
> because he doesn't understand that the relation p = qBr is the correct
> expression for cyclotron motion and thus r = p/qB.
>
> - Show quoted text -
Brown-nose,
You got the WRONG results.
Because of the ROOKIE mistake of mixing SR with Newtonian mechanics.
Because of your idiotic insistance of using "relativistic mass",
"transverse mass" and the rest of the ANTIQUATED approach.
Now, you can go back and do "kiss-ass" to your crank pal,
Juanshito :-)
Eric Gisse
> "Juan R. González-Álvarez" <juanREM...@canonicalscience.com> wrote in
> messagenews:pan.2008.07...@canonicalscience.com...
>
>
>
> > Eric Gisse wrote on Thu, 10 Jul 2008 15:33:30 -0700:
>
> >> On Jul 10, 2:14 pm, "Pmb" <physics_wo...@yahoo.com> wrote:
> >>> "Eric Gisse" <jowr...@gmail.com> wrote in message
>
> >>> news:41d28527-f39e-4ea6-9977-
> > fcfa73c66...@e53g2000hsa.googlegroups.com...
> >>> On Jul 10, 9:54 am, "Juan R." González-Álvarez
>
> >>> <juanREM...@canonicalscience.com> wrote:
> >>> > Dono wrote on Thu, 10 Jul 2008 09:52:26 -0700:
>
> >>> > > No one confused the hamiltonian with the lagrangian (not even you,
>
> >>> > (...)
>
> >>> > Eric confounded both, wrote H = L in previous message and did
>
> >>> Okay. Enough is enough. You've provided more than adequate reason to be
> >>> killfiled. Plonk!
>
> >> ....and here Pete realizes I'm right. Whoops!
>
> Obviously a ploy used by children to antagonize people who know that their
> posts aren't worth reading.
I didn't honestly expect it to work. Thanks for falling for it, Pete.
>
> Eric has proven two things (1) he resorts to flaming when he doesn't
> understand the physics and (2) is ignorant about relativity and analytical
> mechanics (as made evident by his bogus comments on the Hamiltonian and
> potential). Thus he was plonked. His claim above that H = L shows us how
> very little he knows about physics.
>
> Pete
You want to know why I don't respect you or Juan R, Pete? This is why.
All I did was use a _trivial_ fact from classical mechanics, in that
the Lagrangian and Hamiltonian for a system are /equivalent/ under the
specific condition that there are no potentials in the system. I know
Juan doesn't see this because he has specifically cranked out about
relativistic potentials in the past, but I didn't think you were so
blind that you couldn't understand this basic concept even after I
explained it to you.
BURT
>
> - Show quoted text -
Light has no kinetic energy. If it did all light would be the same
mass/energy because of its constancy of speed. Clearly this is not the
case. Light is not like matter. It cannot accelerate and thus become
more massive. Light's energy comes from the frequency of its wave.
Light is massive in energy.
Mitch Raemsch
Eric Gisse
> > messagenews:pan.2008.07...@canonicalscience.com...
> >> Dono wrote on Thu, 10 Jul 2008 09:52:26 -0700:
>
> >>> No one confused the hamiltonian with the lagrangian (not even you,
>
> >> (...)
>
> >> Eric confounded both, wrote H = L in previous message and did incorrect
> >> claims about a blog article. You also did :-)
>
>
> I don't wait Eric to be precise. As noticed in previous days he makes
> dozens of mistakes each week, I only point a small fraction of his
> nonsenses and I usually do when he decides to replies to me claiming some
> mistake from mine, that he imagines in his usual paranoia...
All those big words are confusing. Please forgive the lack of
precision in my verbiage.
>
> But one thing is being no precise and other is his nonsensical claim that
>
> > >> > No. The special relativistic Hamiltonian is
> > >> > H = L = -mc^2 * [1 - v^2/ c^2 ].
I left out the square root, and you didn't notice?!
You even whine about the placement of gamma, but you don't say word
one about the square root.
> > The Hamiltonian is identical to Jacobi's integral (sometimes called the
> > "energy function") in value. The former is expressed in terms of
> > position and canonical momentum whereas the later is expressed in terms
> > of position (q) and its first time derivative (dq/dt)
>
> But expressing a Hamiltonian as a function of velocity was only one of
> Eric mistakes.
I didn't bother writing the Hamiltonian in the correct set of
generalized coordinates, instead I left it be because IT IS SO FUCKING
OBVIOUS and it was irrelevant to my point.
But since you are pitching a shitfit over it, here: v = p/m --> H = -
mc^2 [1 - p^2 / (mc)^2 ]^1/2
Does that substitution of the bleedingly obvious make you feel better?
Its' still the same quantity - just in a different set of generalized
coordinates.
> The second being his confusion that (H = L) from the
> components T and V!
Did you fail every classical mechanics course you ever took? How can
you possibly disagree that H = T + V and L = T - V? Regardless, it
does not matter since the Lagrangian is constructed from the metric.
>
> The third mistake being his confusion between the Lagrangian and the
> energy in special relativity. Note that Eric *multiplies* by the factor
>
> (1 - (v^2/ c^2))
>
> instead dividing by it. According to *Eric* when a particle travel to
> energies close to c its energies vanishes...
No its' supposed to be like that, though with a ^1/2. Nice job on not
noticing an actual mistake.
Regardless since you disagree about the sign choice, Taylor expand L
in in v/c and you'll see why the sign is as it is. The negative sign
on the Lagrangian offsets the negative sign when you expand it, so the
v << c limit is Newtonian.
Given how hard of a time you have understanding Newtonian limits, I
expect a fair bit of arguing about this.
>
> But I am still more perturbed by the fact Eric and Dono considered the
> special relativity Hamiltonian (energy) to be a negative quantity!!!
The dynamics do not matter as long as the choice is a consistent
choice.
Regardless - I never claimed that H was an energy in special
relativity. Sure its' conserved, but that doesn't mean it is energy.
>
> *They* sure us that Hamiltonian is negative
>
> H(Eric) = -mc^2 * [1 - v^2/ c^2 ].
With a ^1/2...
I'm glad nobody noticed [or at least said anything] in the few hours I
left this message half-constructed.
Open up any goddamn mechanics textbook and you will see the Lagrangian
for special relativity constructed. It will be the same.
>
> What physicist, student, or aficionado you know that would think that
> energy of a massive particle in special relativity is negative?
I never said it was the energy of motion, stupid. All I ever said is
that it was conserved - which is not the same as saying it is energy.
The energy you want to talk about is constructed a different way.
>
> Apart from ignorant and arrogant pair, Eric and Dono, I know of nobody
> else.
>
> And after his nonsense still Eric has claimed he is smart and his
> formulae is right whereas insulted you!!!!!!!!
That's because the both of you are as smart as bricks, and just as
stubborn.
>
> > Pete
>
> My recommendation to users is to avoid Eric and Dono as plague. Several
> of their posts with discussion of their favorite tactics will be cited in
> a new version of USENET guidelines
>
> http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
>
> Finally they will get worldwide fame!
Why do you suffer the delusion that your guidelines are relevant to
ANYONE other than yourself?
Eric Gisse
[...]
> Besides, it is not the "energy", imbecile, it is the relativistic
> Lagrangian (L). Your Alzheimer is progressing at an alarming pace, old
> fart.
That's what makes this fun. That is, of course, on top of Pete
responding despite me being killfiled and their general shrillness
about a trivial topic. But that's just gravy.
I don't recall either of us explicitly claiming that this is the
energy. I only ever claimed that it was conserved - if I said it was
_the_ energy I was mistaken.
The actual energy of motion is constructed explicitly through g_uv p^u
p^v as the energy relevant to kinematics is a part of the four-
momentum.
I wonder what Juan is going to say when he reads my blurb about the
choice of sign for the Lagrangian being based on wishing the v << c
expansion to match Newton.
Tom Roberts
> Tom Roberts wrote on Thu, 10 Jul 2008 20:54:28 -0500:
>> given a proper quantum Lagrangian (and the corresponding
>> Hamiltonian if you wish), there's no general theoretical argument that
>> requires photons to be massless.
>
> That was the *point*. Is Hamiltonian (2) in my original message
> representing the same massless particle than (1) or not?
Where are the "particles" or "photons" in (1) and in (2)? How is it you
think either one is a quantum theory at all?
You fantasize that Chubykalo and Vlaev are discussing "quantum
fields" in a paper with "classical electrodynamics" in its
title. Perhaps you should LEARN what a quantum theory really is
so you won't make silly mistakes like this.
>> The REAL argument for massless photons is that when one treats the mass
>> of the photon as a free parameter and fits the theory to experimental
>> observations, one finds that the mass must be an INCREDIBLY small value,
>> and is consistent with zero. The PDG lists the current upper bound on
>> the photon mass as 6*10^-17 eV/c^2.
>
> None of those measurements invalidate the idea of a massive photon.
No, they merely set an upper bound on the mass of the photon (in a
PROPER quantum theory). This is how physics is actually done.
> You
> are giving no real argument.
I'm giving the argument that physicists use. The one that we have.
You seem to want to argue about "massless" or "massive" in some abstract
sense. But physics is a QUANTITATIVE science, and "massive" is
inherently quantitative -- experiments simply cannot distinguish
"massless" from "massive" when the mass is below the experimental upper
bound. This is how science actually works, regardless of what you might
wish. <shrug>
> As remarked before, quantum theories of massive photon give the same
> precision for experimental data than QED using a massless photon.
Sure, for a ludicrously small value of the mass. This is "fine tuning"
at its worst!
The PDG limit is currently published as 6*10^-17 eV. I have
seen references to a new result that reduces the limit by
almost two orders of magnitude. How will that fine tuning
behave as the limit is further reduced?
> In the
> former case the small photon mass is used to cure divergences.
There are other approaches, that remain applicable for a rigorously zero
mass of the photon.
Tom Roberts
Dono
> On Jul 11, 9:28 am, Dono <sa...@comcast.net> wrote:
>
> [...]
>
> > Besides, it is not the "energy", imbecile, it is the relativistic
> > Lagrangian (L). Your Alzheimer is progressing at an alarming pace, old
> > fart.
>
> That's what makes this fun. That is, of course, on top of Pete
> responding despite me being killfiled and their general shrillness
> about a trivial topic. But that's just gravy.
>
Pete Brown-nose killfiled me as well :-)
After I showed him how he fucked up a simple thing like his
"Cyclotrons" page.
The two nutters share the interesting trait that they cannot admit to
being wrong.
When you pin them down they find all kinds of evasive techniques. They
are a lot of fun when they squirm.
> I don't recall either of us explicitly claiming that this is the
> energy. I only ever claimed that it was conserved - if I said it was
> _the_ energy I was mistaken.
>
You didn't say. But they had to find something wrong with what you
said. BTW, I told the two PRETENDERS several times about SQRT(1-(v/
c)^2) insetad of your (1-(v/c)^2), the Juanshito species got it
eventually but not Brown-nose :-)
> I wonder what Juan is going to say when he reads my blurb about the
> choice of sign for the Lagrangian being based on wishing the v << c
> expansion to match Newton.
Never underestimate a pure nutter. I wonder how old he is. I would put
him in the "old fart" category, about 55-65 , maybe more. I also think
that he has no job, he's a "mamma-boy" with papa money. Never worked
a day in his life.
Pmb
"BURT" <macro...@yahoo.com> wrote in message
news:1c83740a-a2e9-41ab...@e39g2000hsf.googlegroups.com...
Oh Lord! You screwed up big time. Not only did you not understand that L =
T - V does *not* hold in relativity but I explained that fact in this thread
and you still didn't get it! L *does not* equal T - V in relativity. That
expression is derived with the assumption that does not hold in relativity.
Now stop being so ignorant and read a darn text which explains relativistic
Lagrangians.
> I know
> Juan doesn't see this because he has specifically cranked out about
> relativistic potentials in the past, but I didn't think you were so
> blind that you couldn't understand this basic concept even after I
> explained it to you.
I didn't see that. I guess you don't pay attention when someone tells you
that you have been plonked. Oh well. What else is new?
In any case if that is what you did then that is yet another error on your
part.
Pete
Eric Gisse
[snippy]
>
> > You want to know why I don't respect you or Juan R, Pete? This is why.
>
> > All I did was use a _trivial_ fact from classical mechanics, in that
> > the Lagrangian and Hamiltonian for a system are /equivalent/ under the
> > specific condition that there are no potentials in the system.
>
> Oh Lord! You screwed up big time. Not only did you not understand that L =
> T - V does *not* hold in relativity but I explained that fact in this thread
> and you still didn't get it! L *does not* equal T - V in relativity. That
> expression is derived with the assumption that does not hold in relativity.
Yet again I have to explain the obvious. I'm noticing a pattern here.
Captain Obvious to the rescue!
DUH it is not true that it holds. It was simply the case that is far
simpler to point to T-V / T+V for the case of V = 0 to explain my
point than take the time to construct it explicitly from the Minkowski
metric.
In fact, it is so obvious I fail to see why you had to point it out
except as a cheap attempt to sound smart. The SR Lagrangian is -mc^2 *
sqrt[1-v^2/c^2], whereas kinetic energy is mc^2/sqrt[1-v^2/c^2] -
mc^2. Kinda hard to confuse them.
I even pointed it out before you posted this whine!
http://groups.google.com/group/sci.physics.relativity/msg/d8b3c42f450091d8?dmode=source
"Did you fail every classical mechanics course you ever took? How can
you possibly disagree that H = T + V and L = T - V? Regardless, it
does not matter since the Lagrangian is constructed from the metric. "
>
> Now stop being so ignorant and read a darn text which explains relativistic
> Lagrangians.
I would, but I have better things to do.
Why didn't you notice my gaping huge error of saying the Lagrangian
goes as [1-v^2/c^2] instead of sqrt[1-v^2/c^2] with the correct form
right next to the incorrect form?
Too bad you missed an easy shot to make me look stupid - better luck
next time.
Since you didn't notice my huge gaffe, maybe you should open that
textbook on relativistic Lagrangians. There's a pretty good section in
Goldstein, as I recall. Do you need help finding the page number?
>
> > I know
> > Juan doesn't see this because he has specifically cranked out about
> > relativistic potentials in the past, but I didn't think you were so
> > blind that you couldn't understand this basic concept even after I
> > explained it to you.
>
> I didn't see that. I guess you don't pay attention when someone tells you
> that you have been plonked. Oh well. What else is new?
If you actually plonked me we wouldn't be having this discussion.
Did you know that plonking someone means you don't read or respond to
their posts? You seem to have the relationship backwards - it isn't my
job to ignore you.
Why would I ignore you, anyway? Shit like this is entertaining. Not as
entertaining as assembling virtualization software on a cluster, but
still damn entertaining.
The shortest route to getting me to ignore you is either not talk, or
not make a load of stupid errors then shove your foot in your mouth as
you defend yourself.
>
> In any case if that is what you did then that is yet another error on your
> part.
>
> Pete
Help me out here, Pete.
Is this an actual error, or just another error you invented to make
yourself seem relevant? Methinks it is more the latter than the former.
Juan R.
(...)
>> Eric has proven two things (1) he resorts to flaming when he doesn't
>> understand the physics and (2) is ignorant about relativity and
>> analytical mechanics (as made evident by his bogus comments on the
>> Hamiltonian and potential). Thus he was plonked. His claim above that
>> H = L shows us how very little he knows about physics.
>>
>> Pete
>
> You want to know why I don't respect you or Juan R, Pete? This is why.
>
> All I did was use a _trivial_ fact from classical mechanics, in that the
> Lagrangian and Hamiltonian for a system are /equivalent/ under the
> specific condition that there are no potentials in the system. I know
> Juan doesn't see this because he has specifically cranked out about
> relativistic potentials in the past, but I didn't think you were so
> blind that you couldn't understand this basic concept even after I
> explained it to you.
I already said you that your (H = L) claim is pure nonsense and that your
relativistic Hamiltonian is plain wrong.
If your nonsense was said by someone with a degree or a PhD on physics
the person would automatically lost them. In my country if a student
write your *nonsense*
(\blockquote
No. The special relativistic Hamiltonian is
H = L = -mc^2 * [1 - v^2/ c^2 ].
)
wouldn't pass an elementary course of mechanics.
You were also informed that this thread and your messages (and those from
Dono aka Karandash) will be cited in a future version of USENET
guidelines.
Are you aware that everyone will be reading the nonsense you are
submitting and how you continue to insult people who point your OBVIOUS
nonsense?
Juan R.
> On Jul 11, 11:25 pm, "Pmb" <physics_wo...@yahoo.com> wrote:
(...)
> Why didn't you notice my gaping huge error of saying the Lagrangian goes
> as [1-v^2/c^2] instead of sqrt[1-v^2/c^2] with the correct form right
> next to the incorrect form?
>
> Too bad you missed an easy shot to make me look stupid - better luck
> next time.
ha ha ha ha ha. But Pmb did that in his old message of 11 Jul when said
you
(\blockquote
That isn't even the correct Lagrangian!
)
Even when copying the Lagrangian from my blog you copied it bad :-)
If you cannot even copy, how could you understand?
I repeat again, are you aware that this thread and your messages will be
cited in a new version of USENET guidelines?
> Since you didn't notice my huge gaffe, maybe you should open that
> textbook on relativistic Lagrangians. There's a pretty good section in
> Goldstein, as I recall. Do you need help finding the page number?
but Pmb already noticed!