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Juan R." González-Álvarez
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 Posted: Tue Jul 08, 2008 1:11 pm Post subject: massless or massive photon? |
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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 |
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Albertito
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| Posted: Wed Jul 09, 2008 9:09 am Post subject: Re: massless or massive photon? |
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On Jul 8, 2:11 pm, "Juan R." González-Álvarez
<juanREM...@canonicalscience.com> wrote:
| Quote: |
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. 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 |
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Sue...
Guest
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| Posted: Wed Jul 09, 2008 10:12 am Post subject: Re: massless or massive photon? |
|
|
On Jul 8, 9:11 am, "Juan R." González-Álvarez
<juanREM...@canonicalscience.com> wrote:
| Quote: |
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.
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...
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Albertito
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| Posted: Wed Jul 09, 2008 7:48 pm Post subject: Re: massless or massive photon? |
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On Jul 9, 4:34 pm, The TimeLord <math-n-physics-...@att.com> wrote:
| Quote: |
Am Wed, 09 Jul 2008 02:09:56 -0700 schrieb Albertito
albertito1...@gmail.com> in
d4bbceaf-082f-4726-a2e2-3be2e07ec...@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!
|
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. |
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The TimeLord
Guest
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| Posted: Wed Jul 09, 2008 8:34 pm Post subject: Re: massless or massive photon? |
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Am Wed, 09 Jul 2008 02:09:56 -0700 schrieb Albertito
<albertito1992@gmail.com> in
d4bbceaf-082f-4726-a2e2-3be2e07ecfdd@f36g2000hsa.googlegroups.com:
| Quote: |
On Jul 8, 2:11 pm, "Juan R." González-Ãlvarez
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
| Quote: |
deduction. So, where in SR is exactly the flaw? Let's modify the
|
SR has no flaw in this regard.
| Quote: |
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.
| Quote: |
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.
| Quote: |
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! |
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Guest
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| Posted: Wed Jul 09, 2008 11:38 pm Post subject: Re: massless or massive photon? |
|
|
On Jul 9, 6:12 am, "Sue..." <suzysewns...@yahoo.com.au> wrote:
| Quote: |
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.
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...
--
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. |
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Sue...
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Igor
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| Posted: Thu Jul 10, 2008 12:21 am Post subject: Re: massless or massive photon? |
|
|
On Jul 8, 9:11 am, "Juan R." González-Álvarez
<juanREM...@canonicalscience.com> wrote:
| Quote: |
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 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. |
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Eric Gisse
Guest
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| Posted: Thu Jul 10, 2008 1:10 am Post subject: Re: massless or massive photon? |
|
|
On Jul 8, 5:11 am, "Juan R." González-Álvarez
<juanREM...@canonicalscience.com> wrote:
| Quote: |
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.
| Quote: |
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.
| Quote: |
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.
| Quote: |
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.
| Quote: |
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.
| Quote: |
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.
| Quote: |
[#]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.
| Quote: |
--
Center for CANONICAL |SCIENCE) http://canonicalscience.org |
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Eric Gisse
Guest
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| Posted: Thu Jul 10, 2008 1:16 am Post subject: Re: massless or massive photon? |
|
|
On Jul 9, 1:09 am, Albertito <albertito1...@gmail.com> wrote:
| Quote: |
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. 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.
|
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.
| Quote: |
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.
| Quote: |
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. |
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Eric Gisse
Guest
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| Posted: Thu Jul 10, 2008 1:18 am Post subject: Re: massless or massive photon? |
|
|
On Jul 9, 11:48 am, Albertito <albertito1...@gmail.com> wrote:
| Quote: |
On Jul 9, 4:34 pm, The TimeLord <math-n-physics-...@att.com> wrote:
Am Wed, 09 Jul 2008 02:09:56 -0700 schrieb Albertito
albertito1...@gmail.com> in
d4bbceaf-082f-4726-a2e2-3be2e07ec...@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!
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.
|
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. |
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The TimeLord
Guest
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| Posted: Thu Jul 10, 2008 4:21 am Post subject: Re: massless or massive photon? |
|
|
Am Wed, 09 Jul 2008 12:48:54 -0700 schrieb Albertito
<albertito1992@gmail.com> in
a58fe8d2-41b4-4322-acbc-2997782f2c3a@k37g2000hsf.googlegroups.com:
| Quote: |
On Jul 9, 4:34 pm, The TimeLord <math-n-physics-...@att.com> wrote:
Am Wed, 09 Jul 2008 02:09:56 -0700 schrieb Albertito
albertito1...@gmail.com> in
d4bbceaf-082f-4726-a2e2-3be2e07ec...@f36g2000hsa.googlegroups.com:
On Jul 8, 2:11 pm, "Juan R." González-Ãlvarez
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.
[...]
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.
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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.)
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// The TimeLord says:
// Pogo 2.0 = We have met the aliens, and they are us! |
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Juan R." González-Álvarez
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| Posted: Thu Jul 10, 2008 1:12 pm Post subject: Re: massless or massive photon? |
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Albertito wrote on Wed, 09 Jul 2008 02:09:56 -0700:
| Quote: |
On Jul 8, 2:11 pm, "Juan R." González-Ãlvarez
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.
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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!
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But,
the relativistic momentum p = mv/sqrt(1 - v^2/c^2) forbids that
deduction. So, where in SR is exactly the flaw?
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No, v = c is a deduction from H = pc.
| Quote: |
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
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All of above is not right.
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Juan R." González-Álvarez
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| Posted: Thu Jul 10, 2008 1:22 pm Post subject: Re: massless or massive photon? |
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Sue... wrote on Wed, 09 Jul 2008 03:12:30 -0700:
| Quote: |
On Jul 8, 9:11 am, "Juan R." González-Ãlvarez
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.
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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.
(...)
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Juan R." González-Álvarez
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| Posted: Thu Jul 10, 2008 1:38 pm Post subject: Re: massless or massive photon? |
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Igor wrote on Wed, 09 Jul 2008 17:21:20 -0700:
| Quote: |
On Jul 8, 9:11 am, "Juan R." González-Ãlvarez
juanREM...@canonicalscience.com> wrote:
The usual argument for massless photons uses the Hamiltonian
H = (\sqrt (m^2c^4 + p^2c^2) ) Â (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) Â 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.
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Don't sure about that argument but will think about. Probably your
argument was right and then (2) is just (1).
| Quote: |
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.
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Yes, would be different massive particles, that is because wrote
"somewhat as".
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http://canonicalscience.org |
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