How does light travel through a vacuum?
chuckles
in the air how does light travel through a vacuum?
Regards,
Cathal
donsto...@hotmail.com
Sam Wormley
Light, unlike mechanical waves, requires no medium to propagate.
See Physics FAQ
http://hermes.physics.adelaide.edu.au/~dkoks/Faq/
chuckles
FrediFizzx
news:1134873846.9...@g49g2000cwa.googlegroups.com...
| If sound travels through the compression and rarefaction of particles
| in the air how does light travel through a vacuum?
Why do you think light travels through a vacuum? And/or what is your
definition of a vacuum?
FrediFizzx
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
Androcles
"chuckles" <chuckleb...@gmail.com> wrote in message
news:1134873846.9...@g49g2000cwa.googlegroups.com...
Same way a planet does, same way a spacecraft does, same way a bullet does,
same way an electron does, as in your TV.
BTW, sound doesn't travel through a vacuum.
Androcles.
Androcles
> Sorry for the stupid question, I wanted to know why. That's all.
Questions are not stupid, but some answers are.
Androcles.
Androcles
"FrediFizzx" <fredi...@hotmail.com> wrote in message
news:40khriF...@individual.net...
> "chuckles" <chuckleb...@gmail.com> wrote in message
> news:1134873846.9...@g49g2000cwa.googlegroups.com...
> | If sound travels through the compression and rarefaction of particles
> | in the air how does light travel through a vacuum?
>
> Why do you think light travels through a vacuum?
Because it does.
And/or what is your
> definition of a vacuum?
Nothing.
What is your definition of stupidity?
Mine is "FrediFizzx"
Androcles
Jan Panteltje
<chuckleb...@gmail.com> wrote in
<1134873846.9...@g49g2000cwa.googlegroups.com>:
>If sound travels through the compression and rarefaction of particles
>in the air how does light travel through a vacuum?
Well, some people thought light was a particle, and called that particle 'photon'.
It was the Einstein who came up with that.
However recently it was discovered that the vacuum is not really empty,
but particles are popping in and out of existence all the time.
And light is a wave.
<flame throwers>
Use this space
</end flame throwers>
Gregory L. Hansen
The same way electrons or baseballs or anything else travels through a
vacuum.
Thinking that light is a wave, and therefore some medium must be waving,
is a classical and 19th century point of view. In modern theory there is
little theoretical difference between light and massive particles other
than the value of the mass that's plugged into the equations. Massive
particles like neutrons will refract and scatter and exhibit all manner of
wave properties. Light is just another particle, with "particle"
interpreted in the quantum sense and not to imply little billiard balls.
--
"'No user-serviceable parts inside.' I'll be the judge of that!"
chuckles
> On a sunny day (17 Dec 2005 18:44:07 -0800) it happened "chuckles"
> <chuckleb...@gmail.com> wrote in
> <1134873846.9...@g49g2000cwa.googlegroups.com>:
>
> >If sound travels through the compression and rarefaction of particles
> >in the air how does light travel through a vacuum?
> Well, some people thought light was a particle, and called that particle 'photon'.
> It was the Einstein who came up with that.
>
Thanks for the replies people. I'm afraid I have no physics experience
I only learned about sound during my final year project write up.
(Computer science).
> However recently it was discovered that the vacuum is not really empty,
> but particles are popping in and out of existence all the time.
Could you elaborate on that a little bit?
> And light is a wave.
Is it or isn't it?
Sam Wormley
> Is it [light] or isn't it [a wave] ?
>
Photons are considered to be point particles. Other properties
and measurement are summarized at
http://pdg.lbl.gov/2005/listings/s000.pdf
"In physics, wave-particle duality holds that light and matter can
exhibit properties of both waves and of particles. This concept is
a key part of quantum mechanics".
http://en.wikipedia.org/wiki/Wave-Particle_duality
This has been shown to hold true for any number of elementary
particles and large assemblages of atoms including buckyballs and
viruses.
Jan Panteltje
<chuckleb...@gmail.com> wrote in
<1134931121.5...@g44g2000cwa.googlegroups.com>:
Oh, I have to go and see the sci fi movie, you know where the alien princess
is rescued.. 60ties stuff.. but anyways (and they had Flash Gordon too), the
thing is: every particle has associated with it a wave..
In the so called 2 slit experiment the 'photon' light particle (as
proposed by Einstein to explain the photo electric effect) produces a wave
like interference pattern, this whole thing is referred to as the
'particle-wave-duality-.
You should really read some physics text.
Then there are the personal views of many posters here, and some strongly
disagree with the current consensus that the 'photon' is a particle.
I for example claim it is a wave excitation.
And that the photo electric effect can be explained from the wave point of view.
Since a lot of things can then NOT be explained from the particle point of view,
with that EVERYTHING could be explained from the wave point of view.
perhaps even those virtual particles popping in and out of existence in the
'quantum ether' are just small wave excitations (noise?) of an ether.
Anyways, have a look here:
http://lectureonline.cl.msu.edu/~mmp/kap28/PhotoEffect/photo.htm
This is the best I know, also look up the main index for a nice physics overview:
Jan Panteltje
<swor...@mchsi.com> wrote in <oiipf.629297$x96.136162@attbi_s72>:
>
> This has been shown to hold true for any number of elementary
> particles and large assemblages of atoms including buckyballs and
> viruses.
Refuted for some posters ;-)
FrediFizzx
This will most likely be hard for you to understand without more quantum
physics and cosmology "under your belt".
| > And light is a wave.
|
| Is it or isn't it?
Light and all electromagnetic radiation certainly has wave properties.
But at the very microscopic level, it also has particle properties. I
suggest if you are really interested that you read Feynman's "QED: The
Strange Theory of Light and Matter" to get your feet wet. It is an
inexpensive paperback that is written for laypersons.
Mike
Since every hypothesis, true or false, implies a true proposition, then
pick your choice from below or make up your own:
A. If there are blue unicorns, then light travels though vacuum
B. if New York is in China, then light travels through vacuum
C. If 1 = 2,. then light travels through vacuum
D. if vauum is plenum, then light travels through vacuum
A - D are true conditional (implications)
seriously now, the answer to your questions would be easire if:
1) we knew what light is
2) we knew what is the the stuff we call vacuum made of
3) what "travels through" means
4) whether spacetime geometry is just a matter of convention (just
kidding)(
5) whether spontaneous symmetry breaking is related to the time arrow
(not kidding)
Now, you must happy you chose marketing and sociology.
Mike
The Ghost In The Machine
<swor...@mchsi.com>
wrote
on Sun, 18 Dec 2005 18:57:24 GMT
<oiipf.629297$x96.136162@attbi_s72>:
A photon being a point particle does not appear to square with
the notion of a spherical wavefront. Of course this doesn't
mean Lorentz doesn't apply; it turns out the equation
x_O^2 - c^2t_O^2 = 0
transforms into
x_A^2 - c^2t_A^2 = 0
as it should.
If one sees a light pulse at (0,0)_O and one knows it's
coming from a moving source, then one can attempt to solve
0 = x_O = (x_A - v * t_A) / sqrt(1-v^2/c^2)
0 = t_O = (t_A - v * x_A/c^2) / sqrt(1-v^2/c^2)
with v = c -- and immediately run into a divide by zero.
However, one can take the limit, with some care, by
manipulating the Lorentz:
x_O * sqrt(1-v^2/c^2) = x_A - v * t_A
x_O * sqrt(1-v^2/c^2) + v * t_A = x_A
Since x_O = 0, the second reduces to v * t_A = x_A. Of course
it turns out sqrt(1-v^2/c^2) is 0 as well.
We also know that
t_O * sqrt(1-v^2/c^2) = (t_A - (v^2/c^2)*t_A)
= t_A * (1 - v^2/c^2)
or
t_O = t_A * sqrt(1-v^2/c^2)
which tells us absolutely nothing; t_A can be any value at all,
at the limit. If we assume x_A = 0 then t_A = 0 from the
first equation. If we assume x_A != 0, then t_A = -x_A/c.
If we observe a light pulse at (0,t)_O things get even wackier.
The best we can do there is solve
0 = t_O *sqrt(1-v^2/c^2) = (t_A - v * x_A/c^2)
-> t_A - c * x_A
but all that does is state that the photon is somewhere
on a light cone (or a line thereof), and it's the exact
same equation as the one derived from
0 * sqrt(1-v^2/c^2) = 0 = x_O = (x_A - v * t_A)
as one takes the limit. It's getting messy, and at this
point one might contemplate doing something ugly with
Maxwell's Equations, as well as the Lorentz. :-)
--
#191, ewi...@earthlink.net
It's still legal to go .sigless.
chuckles
knows what light is, or what "space/vacuum" is made of. I'd be
interested to know. Thanks very much for all the replies specially jan
(Great links) and Androcles; of course all replies are very much
appreciated, you have all been most illuminating. Being honest I've
more questions know than when I started. I think I may have to chalk it
all up to magic and be done with it.
Thanks for all your help,
Cathal
P.S can anyone tell me what happens to light when it enters a black
hole? ;-)
Sam Wormley
Try not to get too hooked on what the trolls and cranks tell you. ;-)
Autymn D. C.
> If sound travels through the compression and rarefaction of particles
> in the air how does light travel through a vacuum?
This is easy:
<http://groups.google.com/groups?q=liht+OR+fotons+Autymn&start=0&scoring=d&filter=0>.
Look for the two links I posted in one message to Ross for my
transactional interpretation of energic radiation. Real electrons are
almost infinitely big, so the vacuum is no problem. And there is a
vacuum between the air molecules, so sound does fare in a vacuum.
There /are/ dumb asknesses (questions), those that are too easily or
already answeren. Frosend (Dismiss) fake wisdom. Fotons are waves,
but waves are not things. Fotons do not ever will (act) as particles;
all things are particles, yet they will wave and will as waves. The
fotons that seem to part are but particle conjugates, such as e+/-,
about to come out of the field. But the particles were always there.
-Aut
donsto...@hotmail.com
**************
Huh. Can one fit inside the causal horizon?
she...@yahoo.com
Gregory L. Hansen wrote:
> In article <1134873846.9...@g49g2000cwa.googlegroups.com>,
> chuckles <chuckleb...@gmail.com> wrote:
> >If sound travels through the compression and rarefaction of particles
> >in the air how does light travel through a vacuum?
> >
> >Regards,
> >Cathal
> >
>
>
> The same way electrons or baseballs or anything else travels through a
> vacuum.
>
> Thinking that light is a wave, and therefore some medium must be waving,
> is a classical and 19th century point of view.
And is in many ways correct.
> In modern theory there is
> little theoretical difference between light and massive particles other
> than the value of the mass that's plugged into the equations.
A huge difference right there!
> Massive
> particles like neutrons will refract and scatter and exhibit all manner of
> wave properties. Light is just another particle, with "particle"
> interpreted in the quantum sense and not to imply little billiard balls.
>
How is a radio wave, emitted by alternating current in an antenna, like
a particle?
For arguments sake, assume the antenna is driven constantly for
infinite time.
she...@yahoo.com
chuckles wrote:
> If sound travels through the compression and rarefaction of particles
> in the air how does light travel through a vacuum?
>
>
Rather than compression and rarefaction of space-time constituents,
light is a wave of spin or rotation of said constituents.
FrediFizzx
news:1134968647....@g47g2000cwa.googlegroups.com...
This is a bugaboo in quantum theory that no one likes to talk about.
;-) The classical and quantum wave patterns are the same for EM
radiation with the quantum one being a probability wave where the
probability of finding the photon is greater at greater amplitude.
Sheesh, the wave pattern for a radio wave can be as big as a house or
bigger! Our explanation is at the link in the sig. We pretty much
follow Volovik's concept of the quantum "vacuum" as a medium.
Jan Panteltje
<swor...@mchsi.com> wrote in <Yompf.636655$_o.608168@attbi_s71>:
As for the ability to do real analysis, you -Sam- MUST read this!
I STRONGLY recommend you read this.
http://www.nytimes.com/2005/12/13/science/13essa.html
Not that I would expect to you see how it applies... other then
-REMEMBERING- what you read, and PAROTING.
But PLEASE read this Sam, PLEASE!
(And I love this sort of stuff... actualy did a similar tests, but I asked
the guy for a screwdriver and took the whole thing apart).
'increadible technical insight'.
The other test they were all bowing to me at the end of the day, never
understood why, until I got the job.'
I had wondered why I turned in all that stuff way before anyone else though.
So guys, this NEEDS reading specially the so called 'dissidents and trolls'
should read it, as they have perhaps that gen that supports has BOTH abilities
(well SOME do).
I LOVE that sort of stuff, it is sooooooooooooo good.
Hero.van...@gmx.de
Jan Panteltje wrote:
> On a sunny day (Sun, 18 Dec 2005 23:37:28 GMT) it happened Sam Wormley
> <swor...@mchsi.com> wrote in <Yompf.636655$_o.608168@attbi_s71>:
>
> >chuckles wrote:
> >> O-k, Mike I amn't sure if you're being serious when you say no one
> >> knows what light is, or what "space/vacuum" is made of. I'd be
> >> P.S can anyone tell me what happens to light when it enters a black
> >> hole? ;-)
> >>
> >
> > Try not to get too hooked on what the trolls and cranks tell you. ;-)
> As for the ability to do real analysis, you -Sam- MUST read this!
> I STRONGLY recommend you read this.
> http://www.nytimes.com/2005/12/13/science/13essa.html
>
> Not that I would expect to you see how it applies... other then
> -REMEMBERING- what you read, and PAROTING.
> But PLEASE read this Sam, PLEASE!
>
> (And I love this sort of stuff... actualy did a similar tests, but I asked
> the guy for a screwdriver and took the whole thing apart).
> 'increadible technical insight'.
and how do
these move through Vacuum or "vacuum"?
Hero
Jan Panteltje
Hero.van...@gmx.de wrote in
<1134993590.1...@z14g2000cwz.googlegroups.com>:
I like the idea that all particles are an exitation of some 'ether'.
Since in this case I cannot open the box with my screwdriver, I do not expect
to have the answer for you in this lifetime.
However lots of capable people seem to be at it with screwdrivers and any other
possible tool.
I just do not think the ones who use mathematics as tool will be first.
That is like tinking what is in the box and making an abstraction while not
having looked inside.
Although in 'cybernetics' by 'shaking' the box we can find out.... it still
requires physical interaction, not philosophy.
'I think so I am' is tricky, real peace is sometimes experienced when not thinking,
so 'I experience so I am' would be closer, but that would only apply to the
aware part of us, as obviously in sleep we are still there....
'I do math so I am' NO, 'I am and I am doing math?' Possible.
Something for Baez to calculate all possible combinations.
Nuf.. ;-)
LOL
PD
chuckles wrote:
>
> > And light is a wave.
>
> Is it or isn't it?
It exhibits wave properties. Then again, sound exhibits wave
properties, but it's actually a little amiss to say sound *is* a wave.
(Sound has particle properties, too.)
The difference between sound and light is that sound is the
manifestation of the propagation of a disturbance in a material medium,
that propagation being dictated both by the inertia and the stiffness
of the medium. Light is not a manifestation of the same sort of thing,
though what it *is* a manifestation of happens to obey mathematical
equations that look similar to sound's. This coincidence in the
mathematics often leads novices to conclude that the underlying physics
is the same, which is a dreadful mistake.
PD
Gregory L. Hansen
she...@yahoo.com <she...@yahoo.com> wrote:
>
>Gregory L. Hansen wrote:
>> In article <1134873846.9...@g49g2000cwa.googlegroups.com>,
>> chuckles <chuckleb...@gmail.com> wrote:
>> >If sound travels through the compression and rarefaction of particles
>> >in the air how does light travel through a vacuum?
>> >
>> >Regards,
>> >Cathal
>> >
>>
>>
>> The same way electrons or baseballs or anything else travels through a
>> vacuum.
>>
>> Thinking that light is a wave, and therefore some medium must be waving,
>> is a classical and 19th century point of view.
>
>And is in many ways correct.
There's no evidence of that.
>
>> In modern theory there is
>> little theoretical difference between light and massive particles other
>> than the value of the mass that's plugged into the equations.
>
>A huge difference right there!
Not really. A perfectly good theory of massive light exists, after all.
And calculations involving massive particles at high energies can be
simplified without losing validity by letting the mass go to zero if the
energies involved are much higher than mc^2. Theoretically there's just
that parameter in the equations and you need to plug in a number, m=?
>
>> Massive
>> particles like neutrons will refract and scatter and exhibit all manner of
>> wave properties. Light is just another particle, with "particle"
>> interpreted in the quantum sense and not to imply little billiard balls.
>>
>
>How is a radio wave, emitted by alternating current in an antenna, like
>a particle?
>For arguments sake, assume the antenna is driven constantly for
>infinite time.
You're not thinking of little billiard balls, are you? I just told you
not to do that!
A radio wave, emitted by alternating current in an antenna, is not like
"a" particle. But it's like a continuous field in the same sense that a
water wave on your hand is like a continuous fluid-- lots of particles and
the effect of each one is too small for you to feel them individually.
--
"No one need be surprised that the subject of contagion was not clear to
our ancestors."-- Heironymus Fracastorius, 1546
frank...@yahoo.com
no mystery. To see how this works, see my Theory of Everything at:
http://www.geocities.com/franklinu/theory.html
fhutoe
Gregory L. Hansen
chuckles <chuckleb...@gmail.com> wrote:
>O-k, Mike I amn't sure if you're being serious when you say no one
>knows what light is, or what "space/vacuum" is made of. I'd be
>interested to know. Thanks very much for all the replies specially jan
>(Great links) and Androcles; of course all replies are very much
>appreciated, you have all been most illuminating. Being honest I've
>more questions know than when I started. I think I may have to chalk it
>all up to magic and be done with it.
You know about sound waves. But what are these atoms that jiggle around
to create the sound wave? What is the force between them that causes them
to interact? Be like the child, answering every explanation with the
simple question "Why?", and eventually a phenomenon can't be reduced
further than a set of basic postulates that are simply assumed to be true.
Why would two molecules bounce from each other, interacting? We would say
something about electrostatic forces. What causes the electrostatic
force? Some folks here on sci.physics like a "mechanical" explanation--
they would want to explain the force in terms of little particles that
bounce off of each other like colliding billiard balls. But would these
force-transmitting particles bounce off of each other? We can't invoke
an electrostatic field because the electrostatic field is what those
particles were supposed to explain. So we've come around full circle,
now stop asking questions.
Any theory has these primitive assumptions, or postulates. They're not
derived from more basic facts-- that would make them conclusions, not
postulates. And it seems to be a characteristic of some of the, um,
alternative thinkers around here that they don't realize their own
assumptions, and consider their own postulates to be logical necessities.
A theory is a human-created model of nature. And even if we happen to
create a theory that matches the Cosmic Blueprints, we can never know.
All we can do is give the degree to which the theory is known to correctly
predict obsevable things.
In modern theory, the most basic physical entity is the field. It is a
quantum field and produces particle-like behavior. But the field is not
made of particles, it is not something waving around, it simply exists.
>
>Thanks for all your help,
>
>Cathal
>
>P.S can anyone tell me what happens to light when it enters a black
>hole? ;-)
It stops.
--
"We need to remember that when we are faced with an unstructured problem
it is we who create the model in the form of a quantitative metaphor;
there is no correct model waiting in the wings for us to call onstage." --
Thomas L. Saaty, "Mathematical Methods of Operations Research" (1988)
kvark
If sound travels through gaseous body, then,
perheps light travels through a ether (polarized vacuum)?
Kvark.
kvark
Sam Wormley
> Franklin, your link doesn't work.
>
You are lucky
Gregory L. Hansen
Helium balloons float through the air because they're less dense than the
air. So birds must be able to fly because they're less dense than air.
Or maybe birds aren't helium balloons.
It's not always a good idea to generalize a special case to all similar
phenomena.
--
Irony: "Small businesses want relief from the flood of spam clogging their
in-boxes, but they fear a proposed national 'Do Not Spam' registry will
make it impossible to use e-mail as a marketing tool."
http://www.bizjournals.com/houston/stories/2003/11/10/newscolumn6.html
frank...@yahoo.com
http://www.geocities.com/franklinhu/theory.html
In particular, you should look at my concept of the aether
http://www.geocities.com/franklinhu/aether.html
This is a geocities web site, so you might get some errors on the page.
If you click past them, you should be able to view the contents.
Jeff_Relf
Hi Cathal, a.k.a. Chuckles, You asked:
If sound travels through
the compression and rarefaction of particles in the air
how does light travel through a vacuum ?
The SI second is more virtual than real.
I posit that SI_Defined speed of light in a virtual vacuum, at absolute zero,
C, is how fast waves propagate in the ether known as _Space_Time_.
But tell me, where do you find no magnetic fields, no gravitational fields,
no Unruh radiation, absolute zero and a true vacuum ?
C probably changes as Space_Time thins
from a virtual black hole into a virtual vacuum.
C can only be measured locally, of course,
no one can measure the one way speed of light,
...photons don't make round trips.
Measured or not, I posit that C is always local, with local causes.
WikiPedia explains the SI second here:
In 1967 the Thirteenth General Conference on Weights and Measures
defined the second of atomic time in the International System of Units as:
the duration of 9,192,631,770 periods of the radiation
corresponding to the transition between
the two hyperfine levels of the ground state of the caesium-133 atom.
The ground state is defined at zero magnetic field.
The second thus defined is equivalent to the ephemeris second.
The definition of the second was later refined at
the 1997 meeting of the BIPM to include the statement
This definition refers to a caesium atom at rest at a temperature of 0 K.
In practice, this means that high-precision realizations of the second
should compensate for the effects of ambient radiation
to try to extrapolate to the value of the second as defined above.
__ WikiPedia.ORG/wiki/Second#Explanation
Jeff_Relf
Hi Chuckles, You asked about Pseudo_Vacuums and Wave_Particles,
In the famous Double_Slit experiment,
the more you can know about which slit a photon/molecule passes through,
even in theory, the more randomly it hits the screen.
For example, warmer molecules are, in theory, more detectable,
because they emit radiation as they pass through a slit,
...so they show up more randomly on the screen... acting less wavelike.
A more wavelike molecule, i.e. one that's colder,
interferes more with All_Possible_Histories of itself,
producing a stronger interference pattern... adding predicability.
Are molecules really Point_Like particles, having no volume, no density ?
I doubt it, I posit that the density is simply unknown, that's all.
I posit that physical processes determine absolutely everything.
Randomness is naught but the byproduct of unknowns,
making things seem like a casino... where the house always wins in the end.
Everything, including the universe, is like a match burning out, dissipating.
Gravity, photons, molecules and even humanity itself are
naught but leftover density from the start of the big bang,
...sometimes winning in the Short_Term... consuming... glomming on...
but always losing in the Long_Term... dissipating.
As for what a Pseudo_Vacuum is, see my other post you you at:
news:Jeff_Relf_200...@Cotse.NET
Androcles
err... it illuminates the bright green flying elephant's eggs, of course.
If you ever find a black hole, look inside for me and see if there
are any eggshells. I've been trying to prove they exist.
Androcles.
she...@yahoo.com
Gregory L. Hansen wrote:
> In article <1134968647....@g47g2000cwa.googlegroups.com>,
> she...@yahoo.com <she...@yahoo.com> wrote:
> >
> >Gregory L. Hansen wrote:
> >> In article <1134873846.9...@g49g2000cwa.googlegroups.com>,
> >> chuckles <chuckleb...@gmail.com> wrote:
> >> >If sound travels through the compression and rarefaction of particles
> >> >in the air how does light travel through a vacuum?
> >> >
> >> >Regards,
> >> >Cathal
> >> >
> >>
> >>
> >> The same way electrons or baseballs or anything else travels through a
> >> vacuum.
> >>
> >> Thinking that light is a wave, and therefore some medium must be waving,
> >> is a classical and 19th century point of view.
> >
> >And is in many ways correct.
>
> There's no evidence of that.
>
Hmm.. no evidence of light as a wave? What about diffraction, prisms,
lenses, etc.?
> >
> >> In modern theory there is
> >> little theoretical difference between light and massive particles other
> >> than the value of the mass that's plugged into the equations.
> >
> >A huge difference right there!
>
> Not really. A perfectly good theory of massive light exists, after all.
> And calculations involving massive particles at high energies can be
> simplified without losing validity by letting the mass go to zero if the
> energies involved are much higher than mc^2. Theoretically there's just
> that parameter in the equations and you need to plug in a number, m=?
>
Yeah, you have a point there. Light does carry stress-energy after
all.
Mass or no mass - not much of a difference, but exactly what is the
difference? That's the million dollar question over at Clay I think.
> >
> >> Massive
> >> particles like neutrons will refract and scatter and exhibit all manner of
> >> wave properties. Light is just another particle, with "particle"
> >> interpreted in the quantum sense and not to imply little billiard balls.
> >>
> >
> >How is a radio wave, emitted by alternating current in an antenna, like
> >a particle?
> >For arguments sake, assume the antenna is driven constantly for
> >infinite time.
>
> You're not thinking of little billiard balls, are you? I just told you
> not to do that!
>
> A radio wave, emitted by alternating current in an antenna, is not like
> "a" particle. But it's like a continuous field in the same sense that a
> water wave on your hand is like a continuous fluid-- lots of particles and
> the effect of each one is too small for you to feel them individually.
>
Yes.
That exact same affect occurs inside a UV photon - as the wave packet
passes the continuous fluid is disturbed.
Photons, as emitted by quantized energy transitions of elements, are
not the particles that make up the ocean (water wave on your hand).
Photons are wave packets in the ocean. The radio wave example merely
shows that not all electromagnetic radiation must be emitted as
photons.
However, the all-light-is-photons approach can still be saved because
every transmission must have a beginning and an end, making it a large
photon.
Cheers- shevek
Gregory L. Hansen
she...@yahoo.com <she...@yahoo.com> wrote:
>
>Gregory L. Hansen wrote:
>> In article <1134968647....@g47g2000cwa.googlegroups.com>,
>> she...@yahoo.com <she...@yahoo.com> wrote:
>> >
>> >Gregory L. Hansen wrote:
>> >> In article <1134873846.9...@g49g2000cwa.googlegroups.com>,
>> >> chuckles <chuckleb...@gmail.com> wrote:
>> >> >If sound travels through the compression and rarefaction of particles
>> >> >in the air how does light travel through a vacuum?
>> >> >
>> >> >Regards,
>> >> >Cathal
>> >> >
>> >>
>> >>
>> >> The same way electrons or baseballs or anything else travels through a
>> >> vacuum.
>> >>
>> >> Thinking that light is a wave, and therefore some medium must be waving,
>> >> is a classical and 19th century point of view.
>> >
>> >And is in many ways correct.
>>
>> There's no evidence of that.
>>
>
>Hmm.. no evidence of light as a wave? What about diffraction, prisms,
>lenses, etc.?
No evidence of the medium that must be waving.
>
>
>> >
>> >> In modern theory there is
>> >> little theoretical difference between light and massive particles other
>> >> than the value of the mass that's plugged into the equations.
>> >
>> >A huge difference right there!
>>
>> Not really. A perfectly good theory of massive light exists, after all.
>> And calculations involving massive particles at high energies can be
>> simplified without losing validity by letting the mass go to zero if the
>> energies involved are much higher than mc^2. Theoretically there's just
>> that parameter in the equations and you need to plug in a number, m=?
>>
>
>Yeah, you have a point there. Light does carry stress-energy after
>all.
>Mass or no mass - not much of a difference, but exactly what is the
>difference? That's the million dollar question over at Clay I think.
A big difference is the threshhold for particle creation. To create an
electron you need at least 2*mc^2 before you have a particle and
antiparticle that you can begin giving kinetic energy to. Photons are
created literally at the drop of a hat when you consider thermal
radiation from frictional heating with the air and heating from the impact
with the ground.
>
>
>> >
>> >> Massive
>> >> particles like neutrons will refract and scatter and exhibit all manner of
>> >> wave properties. Light is just another particle, with "particle"
>> >> interpreted in the quantum sense and not to imply little billiard balls.
>> >>
>> >
>> >How is a radio wave, emitted by alternating current in an antenna, like
>> >a particle?
>> >For arguments sake, assume the antenna is driven constantly for
>> >infinite time.
>>
>> You're not thinking of little billiard balls, are you? I just told you
>> not to do that!
>>
>> A radio wave, emitted by alternating current in an antenna, is not like
>> "a" particle. But it's like a continuous field in the same sense that a
>> water wave on your hand is like a continuous fluid-- lots of particles and
>> the effect of each one is too small for you to feel them individually.
>>
>
>Yes.
>That exact same affect occurs inside a UV photon - as the wave packet
>passes the continuous fluid is disturbed.
>
>Photons, as emitted by quantized energy transitions of elements, are
>not the particles that make up the ocean (water wave on your hand).
>Photons are wave packets in the ocean. The radio wave example merely
>shows that not all electromagnetic radiation must be emitted as
>photons.
Photons are not wave packets. There you go again, thinking maybe they're
not little billiard balls, but at least they're localized like little
billiard balls are-- a billiard ball smeared out into a wave packet.
Photon modes are momentum eigenstates of the field. They're plane waves.
A wave packet is a sum of many photon modes. We say "a photon has been
exchanged" when momentum is transfered to or from the field, which, being
a quantum field, follows DeBroglie's relation and the superposition
principle.
>
>However, the all-light-is-photons approach can still be saved because
>every transmission must have a beginning and an end, making it a large
>photon.
Suppose an L=0 to L=0 transition in a hydrogen atom. You get a spherical
wavefunction. But the photon isn't detected throughout that area, it's
detected in a particular detector at a particular location. Don't confuse
the wavefunction with an energy density.
--
"I'm giving you the chance to look fate in those pretty eyes of hers
and say, 'Step off, bitch. This is my party and you're not invited.'"
-- Chris Shugart, _Testosterone Magazine_
chuckles
ignorance if I seem to be over simplifying.
> I posit that physical processes determine absolutely everything.
>
> Randomness is naught but the byproduct of unknowns,
> making things seem like a casino... where the house always wins in the end.
> Everything, including the universe, is like a match burning out, dissipating.
So you agree with Einstein that there is one unifying theroy to explain
the universe.
God doesn't roll dice?????
I thought Einstein's version of events discounted quantum mechanics, an
aspect of science he inadvertently invented with the discovery of the
photon?
I went to the library today and found some books on physics, lay
person's introduction to quantum physics and a book on quantum
computing. I would very much like to keep discussing these issues I now
have quite a few questions about light (how can a photon's mass be
zero?????) but I think I would find your answers more useful if I had
some basic physics under my belt.
Hoping you'll all watch this space.
Cathal
Hero.van...@gmx.de
> >> >chuckles wrote:
> >> >> O-k, Mike I amn't sure if you're being serious when you say no one
> >> >> knows what light is, or what "space/vacuum" is made of. I'd be
> >> >> interested to know. .,,,,,,,,,,,,
Jan Panteltje wrote:
> >> -Sam- MUST read this!
> >> I STRONGLY recommend you read this.
> >> http://www.nytimes.com/2005/12/13/science/13essa.html
> >>
> >> (And I love this sort of stuff... actualy did a similar tests, but I asked
> >> the guy for a screwdriver and took the whole thing apart).
Jan Panteltje wrote:
> I like the idea that all particles are an exitation of some 'ether'.
> Since in this case I cannot open the box with my screwdriver, I do not expect
> to have the answer for you in this lifetime.
> However lots of capable people seem to be at it with screwdrivers and any other
> possible tool.
> I just do not think the ones who use mathematics as tool will be first.
> That is like tinking what is in the box and making an abstraction while not
> having looked inside.
> Although in 'cybernetics' by 'shaking' the box we can find out.... it still
> requires physical interaction, not philosophy.
> 'I think so I am' is tricky, real peace is sometimes experienced when not thinking,
> so 'I experience so I am' would be closer, but that would only apply to the
> aware part of us, as obviously in sleep we are still there....
> 'I do math so I am' NO, 'I am and I am doing math?' Possible.
> Something for Baez to calculate all possible combinations.
> Nuf.. ;-)
>
> LOL
Jan really thought my thoughts and gave me a clever answer. So i took
quite some time, to give something back.
I think it's a bit about identity. At a certain age a child can talk
about itself ( can say "I am...happy." or things like that) - the child
makes a difference between an "I.." and the rest of the universe, the
"it" (like in
"it is raining").
The question "How does light travel through a vacuum ?" is nearly on
the same level - there's light, there's vacuum , so what else is in the
universe? Particles like atoms or it's parts, not much more? And vacuum
is the absence of these particles.
Now the child will notice that it can see into the second box, which
was transparent. So how can light enter this massive box, and come out
again and tell about, what's inside the box? This my father once
explained to me with the wheel of a bicycle, the atoms are just a
little center and very little stuff around, and nearly everything could
pass through, if not - and he gave the wheel some speed - there is
movement. This gives us, as we are slow, the impression of something
massive, solid, but light is very quick and can pass.
Some philosophical thinking on this level of particles, vacuum, light
and movement is easily done,as not much can be said.
The electrons and the stuff, what is moving as light is all made up of
particles, very small, or even smaller - this results in a kind of
mechanical universe.
So it's may be all a kind of fluid, not in the sense that it is made up
from atoms, but there's continuity and not beeing massive, so You can
explain atoms as iceblocks and light as movements of masses of water.
But this implies different densities.
I even read about a somehow similar model, particles are just like air
bubbles in the universal sea and this explains the high speed of light,
as the vacuum, the absence of bubbles, is very very dense.
As this is nearly all, what can be said, the question arises, why we
are searching for just one stuff, like other people searching for one
formula for everything ?
Confront this with the immense knowledge we have about light, from the
first fire humans lit up to the scrutinizing experiments finding out
the cohaerence of light from one ordinary source of about one meter,
just to give one detail of a million.
Another one of identity can be observed with air, as a comparision of
light and sound was raised in the question from chuckles. Sound waves
already have some kind of identity and life, but easier it's seen with
rotating air-masses, just some differences in temperature and pressure
(which is both due to movements) and this results in a low, spinning
around, growing and becoming a hurricane, we give it a name, like
Katharina, and we get afraid of it. It's not massive, like our body,
just air. No skin around it, and we can no longer push it from it's
path. In some respect similar is observed when creating light of more
and more vibes or energy like ultraviolett and beyond - it' s
developing particle-like behaviour. Shape and form with a certain
lifespan.
So if we follow the physicist path one day we can explain a bit more
about particles and from this we might conclude that light is not a
moving particle or a bunch of it, but a movement of something and when
this movement changes into the movement of an particle in Your skin,
You say that light is warming You.
But by assigning the quality of beeing a physicist to oneself one
imposes a very heavy restriction. That's what makes the childs
behaviour so attractive. A child can see, that the emperor is wearing
no cloths (and talk about it without fear).(My son was asking me these
days, as in a wave one has parts moving slower and other quicker, the
quicker parts have more speed, than the wave is propogating with).
So let's throw a bit of math into it, what is the amplitude of a light
wave, as one can block or let light pass through a polarisation
filter? This is an important number or measurement
And when light is spreading out from a source in a sphere or a cone,
how can there be polarisation? Somehow there is a discontinuity
transversal to the direction of propagation of light.
Or the biologist can tell You something about light can be absorbed and
changed into organic stuff with chlorophyll. And about the octopus deep
in the dark talking to his girlfriend -
by producing light in fascinating colours.
So nowadays a physicist has to think like a child and has to "talk" to
the octopus to find out, what the identity of light travelling in
vacuum is.
I'm not a physicist, only a bit curious, so
have fun
Hero
Jeff_Relf
Hi Cathal, You asked: How can a photon's mass be zero ? ? ? ? ?
Mass_Energy is relativistic. A photon hitting your skin has momentum... mass.
Yet, in theory, it has no mass in it's local frame of refernce.
But only massless objects can travel at C,
where C is the speed of light in a virtual vacuum
at a virtual temperature of absolute zero... yea right !
Virtual means in effect, not in reality.
For example, a map is a virtual territory.
Observations often take decades or even millenniums to catch up with Reality.
How does one measure a photon's mass in it's local frame of refernce ?
What is it's density ? No one knows.
You asked me:
So you agree with Einstein that
there is one unifying theroy to explain the universe.
God doesn't roll dice ? ? ? ? ?
Yes, Einstein wasn't just ahead of his time... he was ahead of our time !
He added a fourth spatial dimension, time,
and I post yet another one, entory: Space_Time_Entropy.
As I said...
Randomness is naught but the byproduct of unknowns,
making things seem like a casino... where the house always wins in the end.
Everything, including the universe, is like a match burning out, dissipating.
Black_Holes, photons, molecules and even humanity itself are
naught but leftover density from the start of the big bang,
...sometimes winning in the Short_Term... consuming... glomming on...
but always losing in the Long_Term... dissipating.
These is no place with no magnetic fields, no gravitational fields,
no Unruh radiation, absolute zero and a true vacuum.
C probably slows down as Space_Time thins
from the virtual sigularity at the start of the big bag
into a virtual vacuum of the end of all virtual Mass_Energy.
The biggest joke of physics is how unphysical it is, how virtual.
Despite many unknowns, rest assured that
entropy, dissipation, devolution, consumption, etc. is spatial,
...the fifth spatial dimension, Space_Time_Entropy... static and parochial.
Liquid water is just at a certain density, a place in Space_Time_Entropy,
it's nothing special... the Anthropic_Principle is Human_Centric,
recycled hubris from the ancient days of Earth_Centric cosmology.
Meaning belongs to me only, it's what matters to me, here and now.
Yet it seems like Over_Glorified inanity to others.
God is relative. For example, I'm God to a rat in my maze,
and, likewise, I'm a rat in my God's maze... controlled by him/it.
Speaking of _Him_, it's hubris to assume sentience requires liquid water.
Hero.van...@gmx.de
Jeff_Relf wrote:
> Hi Cathal, You asked: How can a photon's mass be zero ? ? ? ? ?
Hi Jeff, why did You change the title of the thread ?
This makes it difficult to follow it, as google groups is listing the
titles only and in sci-physics are so many threads, one cannot read all
of them.
Hero
PS
>
> God is relative. For example, I'm God to a rat in my maze,
> and, likewise, I'm a rat in my God's maze... controlled by him/it.
> Speaking of _Him_, it's hubris to assume sentience requires liquid water.
-playing God to mice.
Total control?
One mouse is talking to the other: "Look here. I've got the man in the
white coat so manipulated by me, that everytime i press this buttom, he
will give me food."
Peter Köhlmann
>
> Jeff_Relf wrote:
>> Hi Cathal, You asked: How can a photon's mass be zero ? ? ? ? ?
> ..........
> Hi Jeff, why did You change the title of the thread ?
< snip more garbage >
Why do you post this off-topic dung to a linux group?
--
Failure is not an option. It comes bundled with your Microsoft product.
Jiozef Ruchlewicz
> I understand that I'm too old to be spanked. However when my mum hits
> me across the face and it leaves marks, bruises, or gives me a bloody
> nose, is that excessive?
Linønut
> begin virus.txt.scr Hero.van...@gmx.de wrote:
>
>>
>> Jeff_Relf wrote:
>>> Hi Cathal, You asked: How can a photon's mass be zero ? ? ? ? ?
>> ..........
>> Hi Jeff, why did You change the title of the thread ?
>
> < snip more garbage >
>
> Why do you post this off-topic dung to a linux group?
He's a net dung-beetle?
--
I love the smell of code compiling in the morning. It smells like... Freedom.
Jeff_Relf
Hi Hero_van_Jindelt, You asked me:
Why did You change the title of the thread ?
This makes it difficult to follow it,
as google groups is listing the titles only
and in sci-physics are so many threads, one cannot read all of them.
Changing a title does not alter the thread that it's in.
I think Google makes title changes more prominent
to encourage the use of more relavent titles... I love those guys.
Nevertheless, you don't have to use Google,
I wrote my own newsreader, called X, which lets me edit
the full text of a newsgroup using MicroSoft's Visual_Studio_2005.
Editing and searching Phy.TXT in MicroSoft_Visual_Studio
is an example of how I read Sci.Phyics. It's currently 5.2 megabytes:
http://www.Cotse.NET/users/jeffrelf/Phy.TXT
Here's a screenshot of how that looks to me these days:
http://www.Cotse.NET/users/jeffrelf/Phy_TXT.PNG
Phy.TXT is maintained by X.EXE, my custom newsreader, SMTP_AUTH/POP3 client,
handling my On_The_Fly e-mail aliases.
X also generates what I call: Sci.Physics' _Most_Discussed_,
i.e. the 24 articles with the most replies, in terms of lines, 5 levels deep,
weighted by level, as shown in the number to the left of the _news:_ prefix.
Lines beginning with _>_ don't count, deeper quotes, > >, count negatively.
Only posts who's titles didn't begin with _Re:_ are shown.
1. Laurent, EKPhm, _N8Ce3, D Att.NET, Microsoft, 20, _.36 P
Alt.Philosophy, Alt.Sci.Physics.New-Theories, Sci.Physics, Sci.Physics.Relativity
Liberal Dialectic Materialism
671 news:7XZpf.321616$zb5....@bgtnsc04-news.ops.worldnet.att.net
2. a_plutonium_hotmail, _D7T26G, B Googlegroups.COM, Mozilla_5_0, 19, 1.16 A
Sci.Math, Sci.Physics
Writing a textbook that consolidates all the Adics and getting rid of base dependency; call them Infinite Integers and do they form a Field
512 news:1134983800.1...@g14g2000cwa.googlegroups.com
3. x600y_yahoo_com, _BTmFvt, B Googlegroups.COM, Mozilla_5_0, 17, 6.35 A
Sci.Physics, Sci.Energy, Misc.Consumers.Frugal-Living
Most efficient way to boil water?
361 news:1134830154....@z14g2000cwz.googlegroups.com
4. chuckles, _C5H0uv, B Googlegroups.COM, Mozilla_5_0, 17, 6.44 P
Sci.Physics
How does light travel through a vacuum?
342 news:1134873846.9...@g49g2000cwa.googlegroups.com
5. Pentcho_Valev, _D8gSiG, D Googlegroups.COM, Mozilla_4_0, 17, 7.29 A
Sci.Physics.Relativity, Sci.Physics, Sci.Skeptic, Sci.Philosophy.Tech
Superluminal Signals Cause No Alarm in Einstein's Cult
328 news:1134833362....@g43g2000cwa.googlegroups.com
6. EagleEye, _BRsJ5q, B Googlegroups.COM, Mozilla_4_0, 20, 5.21 P
Alt.Politics.Bush, Alt.Conspiracy, Sci.Skeptic, Sci.Physics, Alt.Religion.Christian
9/11 Update: Missing Black Boxes in World Trade Center Attacks Found by Firefighters, Analyzed by NTSB, Concealed by FBI
293 news:1135128079.7...@g44g2000cwa.googlegroups.com
7. habshi_anony, E Clara.NET, Forte_Free, 19, 4.29 P
Sci.Energy, Sci.Physics, Soc.Culture.Indian
If there is no ether why does light lose its energy?
223 news:43a74ffa...@news.clara.net
8. kkurtstocklmeir_aol, _Dvk3oN, B Googlegroups.COM, Mozilla_4_0, 17, _.29 A
Sci.Physics
God and curses
216 news:1134808145....@g44g2000cwa.googlegroups.com
9. a_plutonium_hotmail, _D8Xf6D, B Googlegroups.COM, Mozilla_5_0, 20, 10.45 A
Sci.Math, Sci.Physics
Counterexamples to Riemann Hypothesis, Goldbach and FLT and proof of infinitude of Even primes Re: Infinite-Integers
189 news:1135104356.3...@g14g2000cwa.googlegroups.com
10. The_U_S, Alt.NET, Xnews_5_04_25, 17, 10.59 A
Alt.Politics.Bush, Alt.Conspiracy, Sci.Skeptic, Sci.Physics, Alt.Folklore.Urban, Alt.Journalism
Briefcase Nuke at the WTCs?
164 news:Xns972F83FEC...@207.14.113.17
11. MobyDikc, _BNx5QO, B Googlegroups.COM, Mozilla_5_0, 20, 11.26 A
Sci.Physics
Proper Frequency
128 news:1135106766.3...@f14g2000cwb.googlegroups.com
12. Ed_Conrad, QP5WL, _BTrcib, G Gnilink.NET, Forte_Agent_2_0, 19, 7.18 A
Sci.Anthropology, Sci.MED, Sci.Physics, Sci.Astro
THE BEST CHRISTMAS PRESENT OF ALL
119 news:q99cq1hmrdj2kfbhp...@4ax.com
13. jgreenfield_seol_net, _DtdBcN, B Googlegroups.COM, Mozilla_4_0, 19, 3.47 P
Sci.Physics
Replacing helium on earth
118 news:1135036035.7...@g47g2000cwa.googlegroups.com
14. Jeff_Relf, A Individual.NET, X, 17, 2._6 P
Sci.Physics, Comp.OS.Linux.Advocacy, Sci.Math, Sci.Physics.Relativity
Take_therefore_no_thought_for_the_morrow.
115 news:Jeff_Relf_200...@Cotse.NET
15. jaska, _bQxBS, B Googlegroups.COM, Mozilla_4_0, 18, _.11 P
Sci.Physics
Orbiting planet
114 news:1134936700.7...@f14g2000cwb.googlegroups.com
16. Jeff_Relf, A Individual.NET, X, 17, 11.47 A
Sci.Physics, Comp.OS.Linux.Advocacy, Sci.Math, Sci.Physics.Relativity
Randomness is a byproduct of unknowns, nothing more.
109 news:Jeff_Relf_200...@Cotse.NET
17. glhansen_steel_ucs, C Indiana.EDU, trn_4_0_test62, 17, 11.53 A
Sci.Physics
Classical mechanics as if QM
107 news:do1qc1$kq$1...@rainier.uits.indiana.edu
18. pierceb, _cHZQ4, B Googlegroups.COM, Mozilla_5_0, 18, 10.45 A
Sci.Physics
Please Help
105 news:1134931549.5...@g49g2000cwa.googlegroups.com
19. Brablo, _DxzisG, B Googlegroups.COM, Mozilla_5_0, 18, 9.29 P
Sci.Physics
Transparency?
104 news:1134970190.7...@g49g2000cwa.googlegroups.com
20. x600y_yahoo_com, _BTmFvt, B Googlegroups.COM, Mozilla_5_0, 18, 9.33 A
Sci.Energy, Misc.Consumers.Frugal-Living, Sci.Physics
Recycling CDs for reflective heating?
95 news:1134927235.1...@g47g2000cwa.googlegroups.com
21. OsherD, _byL2D, B Googlegroups.COM, Mozilla_4_0, 16, 10.42 P
Sci.Physics
Complex Multiplication Does Not Preserve Causation
92 news:1134801745....@g43g2000cwa.googlegroups.com
22. N_dlzc_D_aol_T_com, EKPhm, _BSQFVm, D Cox.NET, Microsoft, 18, 8.16 P
Sci.Physics.Relativity, Sci.Physics, Sci.Physics.Electromag
Welcome! to the Physics Newsgroups
87 news:0uqpf.489$TI6.120@fed1read03
23. kvark, _CctdUX, B Googlegroups.COM, Mozilla_4_0, 19, _.26 A
Sci.Physics
Sound pressure, calculating thrust from pressure, theory
86 news:1134980778....@f14g2000cwb.googlegroups.com
24. don_findlay, _Drm0Ng, B Googlegroups.COM, Mozilla_5_0, 18, 6.52 P
Sci.Geo.Geology, Sci.Physics
convection versus rotation - and plate tectonics
85 news:1134960742.1...@g44g2000cwa.googlegroups.com
-G means _No_ Google record.
Only the last 5 Message-IDs in the _References_ header
of the last 2,000 posts were considered replies.
_No_ other deletions or retentions biased this analysis.
Generated using X.EXE, below... X.TXT has the settings:
http://www.Cotse.NET/users/jeffrelf/X.TXT
http://www.Cotse.NET/users/jeffrelf/X.EXE
My Home Page: Cotse.NET/users/jeffrelf
PD
You're going to discover two things:
1. The universe is much simpler than you thought it was.
2. The universe is much different than you thought it was.
>
> Cathal
she...@yahoo.com
That is a big difference, but it seems there is something more than the
amount of energy - there is also inertia and local confinement of the
energy.
Are you claiming that normal eigenfunctions are "photon modes"? Yes,
any function can be decomposed into normal modes, or orthogonal
polynomials, in an infinite number of ways. What does that have to do
with emission of light?
Light emitted or absorbed by a quantized system such as an electron
changing energy levels in a nucleus is often referred to as a photon.
Such an object can also be called a wave packet.
> We say "a photon has been
> exchanged" when momentum is transfered to or from the field, which, being
> a quantum field, follows DeBroglie's relation and the superposition
> principle.
>
> >
> >However, the all-light-is-photons approach can still be saved because
> >every transmission must have a beginning and an end, making it a large
> >photon.
>
> Suppose an L=0 to L=0 transition in a hydrogen atom. You get a spherical
> wavefunction. But the photon isn't detected throughout that area, it's
> detected in a particular detector at a particular location. Don't confuse
> the wavefunction with an energy density.
When the photon is detected, e.g. in a SSD, another transition is
taking place in the elemental atoms in the SSD. That is the detected
photon - we cannot say that that energy driving the transition came
from a particular place. AFAIK, photons cannot be traced from emission
to absorption.
Cheers - shevek
Gregory L. Hansen
she...@yahoo.com <she...@yahoo.com> wrote:
>
>Gregory L. Hansen wrote:
>> In article <1135053728....@g49g2000cwa.googlegroups.com>,
>> she...@yahoo.com <she...@yahoo.com> wrote:
>> >
>> >Gregory L. Hansen wrote:
>> >> In article <1134968647....@g47g2000cwa.googlegroups.com>,
>> >> she...@yahoo.com <she...@yahoo.com> wrote:
>> >> >
>> >> >Gregory L. Hansen wrote:
>> >> >> In article <1134873846.9...@g49g2000cwa.googlegroups.com>,
>> >> >> chuckles <chuckleb...@gmail.com> wrote:
>> >> >> >If sound travels through the compression and rarefaction of particles
>> >> >> >in the air how does light travel through a vacuum?
>> >> >> >
>> >> >> >Regards,
>> >> >> >Cathal
>> >> >> >
>> >> >>
>> >> >>
>> >> >> The same way electrons or baseballs or anything else travels through a
>> >> >> vacuum.
>> >> >>
>> >> >> Thinking that light is a wave, and therefore some medium must be waving,
>> >> >> is a classical and 19th century point of view.
>> >> >
>> >> >And is in many ways correct.
>> >>
>> >> There's no evidence of that.
>> >>
>> >
>> >Hmm.. no evidence of light as a wave? What about diffraction, prisms,
>> >lenses, etc.?
>>
>> No evidence of the medium that must be waving.
>>
>
>What about diffraction, prisms, lenses, etc.?
No evidence that a medium is involved. No more so for light than for a
baseball.
>
>> >
>> >
>> >> >
>> >> >> In modern theory there is
>> >> >> little theoretical difference between light and massive particles other
>> >> >> than the value of the mass that's plugged into the equations.
>> >> >
>> >> >A huge difference right there!
>> >>
>> >> Not really. A perfectly good theory of massive light exists, after all.
>> >> And calculations involving massive particles at high energies can be
>> >> simplified without losing validity by letting the mass go to zero if the
>> >> energies involved are much higher than mc^2. Theoretically there's just
>> >> that parameter in the equations and you need to plug in a number, m=?
>> >>
>> >
>> >Yeah, you have a point there. Light does carry stress-energy after
>> >all.
>> >Mass or no mass - not much of a difference, but exactly what is the
>> >difference? That's the million dollar question over at Clay I think.
>>
>> A big difference is the threshhold for particle creation. To create an
>> electron you need at least 2*mc^2 before you have a particle and
>> antiparticle that you can begin giving kinetic energy to. Photons are
>> created literally at the drop of a hat when you consider thermal
>> radiation from frictional heating with the air and heating from the impact
>> with the ground.
>>
>
>That is a big difference, but it seems there is something more than the
>amount of energy - there is also inertia and local confinement of the
>energy.
The lighter a particle is, the more it acts like light in those respects.
What meaningful measure of inertia does a particle have if its mass is
basically too small to measure, as with a neutrino?
E=pc=hf. A particular frequency of light corresponds to a particular
energy and a particular momentum.
>
>Light emitted or absorbed by a quantized system such as an electron
>changing energy levels in a nucleus is often referred to as a photon.
Yes.
>Such an object can also be called a wave packet.
No. The emission process can be modeled as a wave packet, and the wave
packet then gives a probability distribution that goes to zero in the
future and goes to zero in the past. The wave packet also gives a
possible distribution of frequencies-- the narrower in position space the
wavepacket is, the wider in frequency space it must be. The photon
is the transfer of momentum to the field, not the packet in itself. And
that transfer will correspond to a particular frequency, not the sum of
all frequencies existing in the wave packet.
Remember that wave packet can be spherically symmetric-- it might be more
or less localized in r, but the photon could be detected to the north,
south, east, or west.
>
>> We say "a photon has been
>> exchanged" when momentum is transfered to or from the field, which, being
>> a quantum field, follows DeBroglie's relation and the superposition
>> principle.
>>
>> >
>> >However, the all-light-is-photons approach can still be saved because
>> >every transmission must have a beginning and an end, making it a large
>> >photon.
>>
>> Suppose an L=0 to L=0 transition in a hydrogen atom. You get a spherical
>> wavefunction. But the photon isn't detected throughout that area, it's
>> detected in a particular detector at a particular location. Don't confuse
>> the wavefunction with an energy density.
>
>When the photon is detected, e.g. in a SSD, another transition is
>taking place in the elemental atoms in the SSD. That is the detected
>photon - we cannot say that that energy driving the transition came
>from a particular place. AFAIK, photons cannot be traced from emission
>to absorption.
The momentum, including direction, can be measured.
--
"For every problem there is a solution which is simple, clean and wrong."
-- Henry Louis Mencken
she...@yahoo.com
Show me a baseball that gives a diffraction pattern, splits in
different frequencies in a prism, or is focused by a lens, and then
perhaps I'll agree.
I don't know. Flavor switching I suppose, but I think we are in
unchartered territory here.
OK, suppose you take for example the frequency of 300khz. Can you tell
me the exact energy of an antenna running on that frequency?
But not precisely..
Cheers - shevek
Gregory L. Hansen
Baseballs are a bit large. But all of that can be shown for electrons,
neutrons, and other small particles. There's a whole study of neutron
optics, for instance, although it's not very different from x-ray optics.
Neutrons diffract around edges, neutrons lenses are made of MgF, and so
on.
As far as I know, buckyballs are the largest particles that have been
diffracted.
--
"When the fool walks through the street, in his lack of understanding he
calls everything foolish." -- Ecclesiastes 10:3, New American Bible
Greg Neill
>
> Gregory L. Hansen wrote:
> > In article <1135178099.1...@g44g2000cwa.googlegroups.com>,
> > she...@yahoo.com <she...@yahoo.com> wrote:
> > >
> > >Gregory L. Hansen wrote:
[snip]
> > >
> > >What about diffraction, prisms, lenses, etc.?
> >
> > No evidence that a medium is involved. No more so for light than for a
> > baseball.
> >
>
> Show me a baseball that gives a diffraction pattern, splits in
> different frequencies in a prism, or is focused by a lens, and then
> perhaps I'll agree.
Well let's see. A baseball masses about 143 grams. A good
velocity for a baseball is 45 m/sec (about 100 miles per hour).
That yields a de Broglie wavelength of about 1x10^-34 meters;
Clearly too small to measure.
But don't give up hope! Diffraction has been achieved with
some pretty hefty molecules. A molecular baseball in the form
of a buckyball as an example:
http://www.quantum.univie.ac.at/research/matterwave/c60/
[snip]
> >
> > E=pc=hf. A particular frequency of light corresponds to a particular
> > energy and a particular momentum.
> >
>
> OK, suppose you take for example the frequency of 300khz. Can you tell
> me the exact energy of an antenna running on that frequency?
That doesn't seem to make sense. An antenna will deal with
whatever amount of energy is provided in the form of a signal,
depending upon such things as impedance matching. The formula
that G. Hansen provided gives the energy of an individual
photon for a given frequency.
[snip]
> >
> > The momentum, including direction, can be measured.
> >
>
> But not precisely..
It can be measured as accurately as you wish, provided you
don't particularly care about the position...
she...@yahoo.com
Greg Neill wrote:
> <she...@yahoo.com> wrote in message news:1135219901.6...@o13g2000cwo.googlegroups.com...
> >
> > Gregory L. Hansen wrote:
> > > In article <1135178099.1...@g44g2000cwa.googlegroups.com>,
> > > she...@yahoo.com <she...@yahoo.com> wrote:
> > > >
> > > >Gregory L. Hansen wrote:
>
> [snip]
>
> > > >
> > > >What about diffraction, prisms, lenses, etc.?
> > >
> > > No evidence that a medium is involved. No more so for light than for a
> > > baseball.
> > >
> >
> > Show me a baseball that gives a diffraction pattern, splits in
> > different frequencies in a prism, or is focused by a lens, and then
> > perhaps I'll agree.
>
> Well let's see. A baseball masses about 143 grams. A good
> velocity for a baseball is 45 m/sec (about 100 miles per hour).
> That yields a de Broglie wavelength of about 1x10^-34 meters;
> Clearly too small to measure.
Yes, that is definitely in the realm of science fiction at this
juncture.
>
> But don't give up hope! Diffraction has been achieved with
> some pretty hefty molecules. A molecular baseball in the form
> of a buckyball as an example:
>
> http://www.quantum.univie.ac.at/research/matterwave/c60/
>
>
> [snip]
>
> > >
> > > E=pc=hf. A particular frequency of light corresponds to a particular
> > > energy and a particular momentum.
> > >
> >
> > OK, suppose you take for example the frequency of 300khz. Can you tell
> > me the exact energy of an antenna running on that frequency?
>
> That doesn't seem to make sense. An antenna will deal with
> whatever amount of energy is provided in the form of a signal,
> depending upon such things as impedance matching. The formula
> that G. Hansen provided gives the energy of an individual
> photon for a given frequency.
>
Exactly. My point was that E=hv applies to elemental atomic emission
and absorption, e.g. blackbody radiation or the photoelectric effect,
but not to other kinds of radiation such as bremstrahlung, synchrotron
emission, dipole antennae, etc. In those situations you are better off
trying E=B^2+E^2.
> [snip]
>
> > >
> > > The momentum, including direction, can be measured.
> > >
> >
> > But not precisely..
>
> It can be measured as accurately as you wish, provided you
> don't particularly care about the position...
Just curious, how do you intend to measure the momentum of a particle -
if you have no idea where it is?
Cheers - shevek
Greg Neill
[snip]
>
> Exactly. My point was that E=hv applies to elemental atomic emission
> and absorption, e.g. blackbody radiation or the photoelectric effect,
> but not to other kinds of radiation such as bremstrahlung, synchrotron
> emission, dipole antennae, etc. In those situations you are better off
> trying E=B^2+E^2.
So you're saying that you think there are different kinds
of photons, and E = hf applies only to some? That would
seem to contradict an awful lot of empirical evidence.
>
>
> > [snip]
> >
> > > >
> > > > The momentum, including direction, can be measured.
> > > >
> > >
> > > But not precisely..
> >
> > It can be measured as accurately as you wish, provided you
> > don't particularly care about the position...
>
> Just curious, how do you intend to measure the momentum of a particle -
> if you have no idea where it is?
Use a big detector!
FrediFizzx
news:1135235228.9...@g47g2000cwa.googlegroups.com...
That is not energy. It is energy density. Should be E/vol = B^2 + E^2.
Or E = (B^2 + E^2)*volume. For the B^2 and E^2 you can use,
B^2 = E^2 = hbar*w^4*n/(4pi^2*c^3),
For free space identical photons where w is omega, angular frequency and
n is the number of photons.
FrediFizzx
Ken S. Tucker
Hi Fredi, I understand what you mean, but you appear to
have duplicated "E", to be energy in "E/vol" and Electric
field in E^2 on the RHS. So if that's true, I'd recommend
using "p" as energy, as in the time component of the
momentum 4-vector. I know that sucks, but we're in
cheap ascii here....ugh
Regards
Ken S. Tucker
srp
Bremsstrahlung, synchrotron emission and dipole antenae emission are all
made up at the source of individually emitted photons, so E=hv still
applies, even if at the macro level you can deal with total emission
with E=B^2+E^2, which represents energy density as FrediFizzx explained.
>
>
>>[snip]
>>
>>
>>>>The momentum, including direction, can be measured.
>>>>
>>>
>>>But not precisely..
>>
>>It can be measured as accurately as you wish, provided you
>>don't particularly care about the position...
>
>
> Just curious, how do you intend to measure the momentum of a particle -
> if you have no idea where it is?
If you detect it, you definitely know where it is as you intercept it,
and it is easy to have a detector setup so that the precise direction
it came is ascertained.
André Michaud
Gregory L. Hansen
she...@yahoo.com <she...@yahoo.com> wrote:
>
>Greg Neill wrote:
>> <she...@yahoo.com> wrote in message
>news:1135219901.6...@o13g2000cwo.googlegroups.com...
>> >
>> > Gregory L. Hansen wrote:
>> > > In article <1135178099.1...@g44g2000cwa.googlegroups.com>,
>> > > she...@yahoo.com <she...@yahoo.com> wrote:
>> > > >
>> > > >Gregory L. Hansen wrote:
>> > >
>> > > E=pc=hf. A particular frequency of light corresponds to a particular
>> > > energy and a particular momentum.
>> > >
>> >
>> > OK, suppose you take for example the frequency of 300khz. Can you tell
>> > me the exact energy of an antenna running on that frequency?
The power transmitted would by P=(dn/dt)hf, with dn/dt the number of
photons per second.
>>
>> That doesn't seem to make sense. An antenna will deal with
>> whatever amount of energy is provided in the form of a signal,
>> depending upon such things as impedance matching. The formula
>> that G. Hansen provided gives the energy of an individual
>> photon for a given frequency.
>>
>
>Exactly. My point was that E=hv applies to elemental atomic emission
>and absorption, e.g. blackbody radiation or the photoelectric effect,
>but not to other kinds of radiation such as bremstrahlung, synchrotron
>emission, dipole antennae, etc. In those situations you are better off
>trying E=B^2+E^2.
Bremstrahlung and synchrotron emission are best treated quantum
mechanically if they go into the x-ray regimes. The best classical
treatment you can do of x-rays is to divide the field by hf and call that
the number of photons per second. And that worked well enough for Warren
in his treatment of x-ray diffraction.
>
>
>> [snip]
>>
>> > >
>> > > The momentum, including direction, can be measured.
>> > >
>> >
>> > But not precisely..
>>
>> It can be measured as accurately as you wish, provided you
>> don't particularly care about the position...
>
>Just curious, how do you intend to measure the momentum of a particle -
>if you have no idea where it is?
Surround the source with detectors.
--
"We don't grow up hearing stories around the camp fire anymore about
cultural figures. Instead we get them from books, TV or movies, so the
characters that today provide us a common language are corporate
creatures" -- Rebecca Tushnet
Justin Ape
The honorable glha...@steel.ucs.indiana.edu (Gregory L. Hansen) wrote
>>> > OK, suppose you take for example the frequency of 300khz. Can you tell
>>> > me the exact energy of an antenna running on that frequency?
>
>The power transmitted would by P=(dn/dt)hf, with dn/dt the number of
>photons per second.
Hi, I have been following this discussion now for some time.
Professor Wormly mentions the photon as a 'point particle', and you like
bucky balls? Anyways I am a bit confused about how to solve this:
.......................
. .
-----| 0 |------
| . electron . |
| ....................... |
| vacuum tube |
| |
------------ ~ ---------------
1Hz sinewave
signal generator
Suppose we have a vacuum tube, with a metal plate at each end, connected
to a sine wave generator (see drawing).
We have (just for argument sake) ONE free electron in the tube.
The tube is 1 meter long, and the signal amplitude is such that the electron
wanders back and forth between the 2 electrodes, and just does not touch these.
So the electron speed is a nice sine wave too.
No what I am confused about is this:
The electron is clearly a moving charge, and as such has an changing
electromagnetic field of in this case 1 Hz frequency associated with it.
But WHEN (and WHERE) along the trajectory is the 'photon' emitted?
And if emitted, is there also a photons absorbed ?
And how many photons? Clearly these must have energy related to 1Hz and hf?
I would like to understand this better, could you shed some light on this?
(Pun intended).
Justin Ape
Gregory L. Hansen
This situation involves more than one photon. There's a gazillion photons
per second exchanged between the plates and the electron. It's an
application of Bohr's correspondence principle, which says that quantum
mechanics reduces to classical mechanics in the limit of large quantum
numbers. In this case, photons per second.
One Hz corresponds to a very low energy E=hf. But supposing we have the
patience and the instrumentation to handle that, if the amplitude of your
sine wave is brought down low enough (the maximum field amplitude is low
enough) then the electron *wouldn't* wander back and forth in a smooth
sine wave. It would receive a series of kicks that would, over time,
average out to a sine wave.
You couldn't see that at 1 Hz, but you could see it at higher energies.
The photoelectric effect was probably the first example. Blue light would
kick out electrons from a metal while red light would not, even if the red
light was very bright. If the blue light were made very dim it would kick
out fewer electrons per second, but would still kick out electrons. And
it would do so immediately, which dispelled theories of a resonance that
gradually built up enough energy in the electrons over time.
Another example is acceleration of charged particles in a magnetic field.
The design of particle accelerators is purely classical electrodynamics
with the exception of the effect of radiation in the bending magnets.
Classical theory says that the radiation would be emitted continuously
and that the particles would be accelerated smoothly in the magnetic
fields. In reality the particles are accelerated in a series of kicks
which increases the divergence of the beam versus the classical
predictions, and the increased divergence has to be accounted for in the
design.
--
"Then they placed the ark of the Lord on the cart; along with the box
containing the golden mice and the images of the hemorrhoids."
-- 1 Samuel 6:11
she...@yahoo.com
Gregory L. Hansen wrote:
> In article <1135235228.9...@g47g2000cwa.googlegroups.com>,
> she...@yahoo.com <she...@yahoo.com> wrote:
> >
> >Greg Neill wrote:
> >> <she...@yahoo.com> wrote in message
> >news:1135219901.6...@o13g2000cwo.googlegroups.com...
> >> >
> >> > Gregory L. Hansen wrote:
> >> > > In article <1135178099.1...@g44g2000cwa.googlegroups.com>,
> >> > > she...@yahoo.com <she...@yahoo.com> wrote:
> >> > > >
> >> > > >Gregory L. Hansen wrote:
>
> >> > >
> >> > > E=pc=hf. A particular frequency of light corresponds to a particular
> >> > > energy and a particular momentum.
> >> > >
> >> >
> >> > OK, suppose you take for example the frequency of 300khz. Can you tell
> >> > me the exact energy of an antenna running on that frequency?
>
> The power transmitted would by P=(dn/dt)hf, with dn/dt the number of
> photons per second.
>
The point is that accelerated charges generate electromagnetic
radiation. For the case of an electron moving between discrete energy
states, the generated pulse of electromagnetic radiation is called a
photon. For the case of free electrons in a radio antenna (or a plasma
if you prefer), there's no real use for that term.
> >>
> >> That doesn't seem to make sense. An antenna will deal with
> >> whatever amount of energy is provided in the form of a signal,
> >> depending upon such things as impedance matching. The formula
> >> that G. Hansen provided gives the energy of an individual
> >> photon for a given frequency.
> >>
> >
> >Exactly. My point was that E=hv applies to elemental atomic emission
> >and absorption, e.g. blackbody radiation or the photoelectric effect,
> >but not to other kinds of radiation such as bremstrahlung, synchrotron
> >emission, dipole antennae, etc. In those situations you are better off
> >trying E=B^2+E^2.
>
> Bremstrahlung and synchrotron emission are best treated quantum
> mechanically if they go into the x-ray regimes.
Synchrotron radiation is derived entirely classically. "The language
of photons can be used if desired, by dividing the intensity
distribution by hbar omega" (Jackson, Sect. 14.7)
> The best classical
> treatment you can do of x-rays is to divide the field by hf and call that
> the number of photons per second.
You mean that's the best -quantum- treatment.
>And that worked well enough for Warren
> in his treatment of x-ray diffraction.
>
Warren was considering quantized systems involving electrons trapped in
atoms. Free electrons in very strong magnetic fields can give off
x-ray synchrotron emission. There's no quantization involved there -
until you are use an SSD to measure the radiation.
> >
> >
> >> [snip]
> >>
> >> > >
> >> > > The momentum, including direction, can be measured.
> >> > >
> >> >
> >> > But not precisely..
> >>
> >> It can be measured as accurately as you wish, provided you
> >> don't particularly care about the position...
> >
> >Just curious, how do you intend to measure the momentum of a particle -
> >if you have no idea where it is?
>
> Surround the source with detectors.
>
It sounds like you do have some idea of the position of the particle...
so you're not going to be able to measure the momentum as accurately
as you wish.
Cheers - shevek
she...@yahoo.com
Greg Neill wrote:
> <she...@yahoo.com> wrote in message news:1135235228.9...@g47g2000cwa.googlegroups.com...
>
> [snip]
>
> >
> > Exactly. My point was that E=hv applies to elemental atomic emission
> > and absorption, e.g. blackbody radiation or the photoelectric effect,
> > but not to other kinds of radiation such as bremstrahlung, synchrotron
> > emission, dipole antennae, etc. In those situations you are better off
> > trying E=B^2+E^2.
>
> So you're saying that you think there are different kinds
> of photons, and E = hf applies only to some? That would
> seem to contradict an awful lot of empirical evidence.
>
No, I'm saying that there are different kinds of electromagnetic
radiation - some of which are quantized into photons due to the nature
of the emission and some of which are not.
she...@yahoo.com
How do you figure? Electromagnetic radiation is caused by acceleration
of charges
see e.g.
R. Tsien, Amer. J. Phys. 40 (1972) 46.
If these accelerations aren't quantized, neither is the emission. If
the breaking force, magnetic field, or current (for the three examples,
respectively) are not quantized - than neither is the radiation.
Greg Neill
So you think that the classical field exists independently
of the quantum description of light, and that there is
some way to distinguish field from photon? So what's an
EM field made of?
srp
Presently, it is caused by breaking (slowing down) accelerated particles
or charges. When a particle is forced to slow down by an external field,
the energy that it has in excess of that required to maintain its new
velocity evacuates as a photon, this is why it is called bremsstrahlung
("slowing down radiation" in German).
> see e.g.
> R. Tsien, Amer. J. Phys. 40 (1972) 46.
>
> If these accelerations aren't quantized, neither is the emission. If
> the breaking force, magnetic field, or current (for the three examples,
> respectively) are not quantized - than neither is the radiation.
>
If by "quantized" you mean that these emissions do not respect the
preset and fixed values of emission from de-energizing electrons
or quarks in atoms, then you are right. These acceleration related
emission can have just about any energy, but E=hf still applies since
the ultimate source always is discrete quantities of em energy
(photons) being released as the related particles are forced to
slow down (one photon emitted per particle per slowing down).
Bremsstrahlung is self explantory, obviously.
The cause of synchrotron radiation emission is now very well
understood in accelerator circles. It specifically depends on the
trajectories of a charged particle beam (typically electrons) being
forced to ondulate (in linear accelerators), or to wiggle about the
main orbit in (circular accelerators), which causes repeated
transverse acceleration of the particles in the beam followed by
breaking to a transverse stop (which is when a photon is emitted by
each electron) prior to transverse accelerating (to eventual
transverse stop and emission of another photon) in the reverse
direction and so forth, which makes synchrotron radiation also to
turn out to also be bremsstrahlung.
I refer you for example to "Principles of Charged Particle
Acceleration" by S. Humphries Jr.
As for longer wave radiation emitted through antenae, bremsstrahlung
is also the name of the game, since a photon is emitted each time the
alternating motion of the current stops moving in one direction prior
to re-accelerating in the reverse direction. To my knowledge, a
phenomenon of resonance is also involved.
André Michaud
FrediFizzx
news:kJZqf.4706$%N1.6...@news20.bellglobal.com...
Zertons of course. ;-) The quantum nature of photons is due to the
quantum nature of the so-called quantum "vacuum".
FrediFizzx
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
Gregory L. Hansen
she...@yahoo.com <she...@yahoo.com> wrote:
>
>Gregory L. Hansen wrote:
>> In article <1135235228.9...@g47g2000cwa.googlegroups.com>,
>> she...@yahoo.com <she...@yahoo.com> wrote:
>> >
>> >Greg Neill wrote:
>> >> <she...@yahoo.com> wrote in message
>> >news:1135219901.6...@o13g2000cwo.googlegroups.com...
>> >> >
>> >> > Gregory L. Hansen wrote:
>> >> > > In article <1135178099.1...@g44g2000cwa.googlegroups.com>,
>> >> > > she...@yahoo.com <she...@yahoo.com> wrote:
>> >> > > >
>> >> > > >Gregory L. Hansen wrote:
>>
>> >> > >
>> >> > > E=pc=hf. A particular frequency of light corresponds to a particular
>> >> > > energy and a particular momentum.
>> >> > >
>> >> >
>> >> > OK, suppose you take for example the frequency of 300khz. Can you tell
>> >> > me the exact energy of an antenna running on that frequency?
>>
>> The power transmitted would by P=(dn/dt)hf, with dn/dt the number of
>> photons per second.
>>
>
>The point is that accelerated charges generate electromagnetic
>radiation. For the case of an electron moving between discrete energy
>states, the generated pulse of electromagnetic radiation is called a
>photon. For the case of free electrons in a radio antenna (or a plasma
>if you prefer), there's no real use for that term.
Photons are not about discrete energy states. Photons are about the
quanntum description of the electromagnetic field. And quantum mechanics
applies perfectly well to free electrons in radio antennas or magnetic
fields or whatever else you like. You can use the classical description
when the "granularity" is too small to notice, as it is in a radio
antenna.
>
>> >>
>> >> That doesn't seem to make sense. An antenna will deal with
>> >> whatever amount of energy is provided in the form of a signal,
>> >> depending upon such things as impedance matching. The formula
>> >> that G. Hansen provided gives the energy of an individual
>> >> photon for a given frequency.
>> >>
>> >
>> >Exactly. My point was that E=hv applies to elemental atomic emission
>> >and absorption, e.g. blackbody radiation or the photoelectric effect,
>> >but not to other kinds of radiation such as bremstrahlung, synchrotron
>> >emission, dipole antennae, etc. In those situations you are better off
>> >trying E=B^2+E^2.
>>
>> Bremstrahlung and synchrotron emission are best treated quantum
>> mechanically if they go into the x-ray regimes.
>
>Synchrotron radiation is derived entirely classically. "The language
>of photons can be used if desired, by dividing the intensity
>distribution by hbar omega" (Jackson, Sect. 14.7)
And if x-rays are coming out you'll get electromagnetic radiation that can
kick a particle in your wire chamber. When bend a particle at high
energies in a magnetic field you don't get a continuous emission of
radiation as the classical theory predicts. You get discrete momentum
recoils, which increases the divergence of your beam versus the classical
prediction and must be accounted for in accelerator design.
Have you ever seen the electromagnetic Lagrangian in quantum field theory?
It's exactly the same as the classical one. Quantum field theory is a
theory of fields. It really shouldn't be surprising that a lot of these
results can be derived with the classical theory because they both deal
with the same field! And it should be derived classically when that's
possible because it's easier to work with, even if you do have to use the
kludge of dividing by hf to get a number per second of photons.
Now Compton scattering is an example of a problem that one would think the
classical theory could deal with perfectly well, but doesn't.
>
>
>> The best classical
>> treatment you can do of x-rays is to divide the field by hf and call that
>> the number of photons per second.
>
>You mean that's the best -quantum- treatment.
Nope. That's a kludge that allows you to use the classical theory on a
problem that's really quantum mechanical. You realize that x-rays aren't
detected with a radio antenna, right?
>
>>And that worked well enough for Warren
>> in his treatment of x-ray diffraction.
>>
>
>Warren was considering quantized systems involving electrons trapped in
>atoms. Free electrons in very strong magnetic fields can give off
>x-ray synchrotron emission. There's no quantization involved there -
>until you are use an SSD to measure the radiation.
You didn't read Warren. But as I'd mentioned, yes there is quantization
involved with free electrons in very strong electromagnetic fields.
--
"The hardest conviction to get into the mind of the beginner is that the
education he is receiving in college is not a medical course but a life
course for which the work of a few years under teachers is but a
preparation." -- Sir William Osler
she...@yahoo.com
Good question! One way to model an EM field (in a vacuum) is as a
kinetic distribution of space-time contituents which are free to move
and rotate. The bulk motion of these constituents is the magnetic
vector potential A, while the pressure is the electric potential V.
she...@yahoo.com
I disagree entirely.
> And quantum mechanics
> applies perfectly well to free electrons in radio antennas or magnetic
> fields or whatever else you like.
You can try to apply techniques of quantum mechanics - however you are
applying them to a system which isn't quantized.
> You can use the classical description
> when the "granularity" is too small to notice, as it is in a radio
> antenna.
>
I'm not going to promote "the classical description", when of course
modern techniques are far better tested and robust. However, I will
say that the emission of radiation by a free electron is not a
quantized system.
> >
> >> >>
> >> >> That doesn't seem to make sense. An antenna will deal with
> >> >> whatever amount of energy is provided in the form of a signal,
> >> >> depending upon such things as impedance matching. The formula
> >> >> that G. Hansen provided gives the energy of an individual
> >> >> photon for a given frequency.
> >> >>
> >> >
> >> >Exactly. My point was that E=hv applies to elemental atomic emission
> >> >and absorption, e.g. blackbody radiation or the photoelectric effect,
> >> >but not to other kinds of radiation such as bremstrahlung, synchrotron
> >> >emission, dipole antennae, etc. In those situations you are better off
> >> >trying E=B^2+E^2.
> >>
> >> Bremstrahlung and synchrotron emission are best treated quantum
> >> mechanically if they go into the x-ray regimes.
> >
> >Synchrotron radiation is derived entirely classically. "The language
> >of photons can be used if desired, by dividing the intensity
> >distribution by hbar omega" (Jackson, Sect. 14.7)
>
> And if x-rays are coming out you'll get electromagnetic radiation that can
> kick a particle in your wire chamber. When bend a particle at high
> energies in a magnetic field you don't get a continuous emission of
> radiation as the classical theory predicts. You get discrete momentum
> recoils, which increases the divergence of your beam versus the classical
> prediction and must be accounted for in accelerator design.
Really? Reference please? THere must be something in your
experimental setup I'm not understanding.
>
> Have you ever seen the electromagnetic Lagrangian in quantum field theory?
> It's exactly the same as the classical one. Quantum field theory is a
> theory of fields. It really shouldn't be surprising that a lot of these
> results can be derived with the classical theory because they both deal
> with the same field! And it should be derived classically when that's
> possible because it's easier to work with, even if you do have to use the
> kludge of dividing by hf to get a number per second of photons.
>
I agree.
> Now Compton scattering is an example of a problem that one would think the
> classical theory could deal with perfectly well, but doesn't.
>
The problem is that classical theory might model the electron as a
point particle - which won't capture the interactions well at low
wavelengths.
> >
> >
> >> The best classical
> >> treatment you can do of x-rays is to divide the field by hf and call that
> >> the number of photons per second.
> >
> >You mean that's the best -quantum- treatment.
>
> Nope. That's a kludge that allows you to use the classical theory on a
> problem that's really quantum mechanical. You realize that x-rays aren't
> detected with a radio antenna, right?
>
Yes! :)
> >
> >>And that worked well enough for Warren
> >> in his treatment of x-ray diffraction.
> >>
> >
> >Warren was considering quantized systems involving electrons trapped in
> >atoms. Free electrons in very strong magnetic fields can give off
> >x-ray synchrotron emission. There's no quantization involved there -
> >until you are use an SSD to measure the radiation.
>
> You didn't read Warren. But as I'd mentioned, yes there is quantization
> involved with free electrons in very strong electromagnetic fields.
>
I'm very curious to hear more, please educate me further.
Cheers - shevek
she...@yahoo.com
srp wrote:
> she...@yahoo.com a écrit :
> > srp wrote:
> >
> >>she...@yahoo.com a écrit :
> >>
> >>>Greg Neill wrote:
> >>>
> >>>
> >>>><she...@yahoo.com> wrote in message news:1135219901.6...@o13g2000cwo.googlegroups.com...
> >>>>
> >>>>
> >>>>>Gregory L. Hansen wrote:
> >>>>>
> >>>>>
> >>>>>>In article <1135178099.1...@g44g2000cwa.googlegroups.com>,
> >>>>>>she...@yahoo.com <she...@yahoo.com> wrote:
> >>>>>>
> >>>>>>
> >>>>>>>Gregory L. Hansen wrote:
> >>>>
> >>>>[snip]
> >
> >
Thanks for the reference, I'll take a look at that book.
> As for longer wave radiation emitted through antenae, bremsstrahlung
> is also the name of the game, since a photon is emitted each time the
> alternating motion of the current stops moving in one direction prior
> to re-accelerating in the reverse direction. To my knowledge, a
> phenomenon of resonance is also involved.
>
?? The radiation is emitted continuously during the acceleration of
the electron - the electron doesn't have to stop - in which frame would
that be? A sinusoidal current on top of a constant current will also
emit.
> André Michaud
Gregory L. Hansen
On what grounds?
>
>> And quantum mechanics
>> applies perfectly well to free electrons in radio antennas or magnetic
>> fields or whatever else you like.
>
>You can try to apply techniques of quantum mechanics - however you are
>applying them to a system which isn't quantized.
It turns out that quantum mechanics doesn't just apply to the electronic
transitions of atoms. You recall one of the postulates of quantum
mechanics-- DeBroglie's relation p=h/lambda. That means, for instance, if
you have a pure sinusoidal electromagnetic wave it can transfer a momentum
p=h/lambda to a particle. Not p=0.5*h/lambda, not p=1.127*h/lambda. Just
p=h/lambda. It can do it twice, p=2h/lambda, or three times, or more.
But always in steps of h/lambda. That's quantization. That's what makes
an interaction with a field like a collision between particles. You don't
need discrete energy states for that, it applies to all interactions.
Undergraduate quantum mechanics doesn't quantize the field. In that sense
it's really a semiclassical theory-- the quantum mechanics of particles in
classical fields. That's a good enough approximation for many purposes.
But not, e.g., for the Lamb shift.
The text we used in the class on accelerator physics was Wiedemann,
"Particle Accelerator Physics".
>
>>
>> Have you ever seen the electromagnetic Lagrangian in quantum field theory?
>> It's exactly the same as the classical one. Quantum field theory is a
>> theory of fields. It really shouldn't be surprising that a lot of these
>> results can be derived with the classical theory because they both deal
>> with the same field! And it should be derived classically when that's
>> possible because it's easier to work with, even if you do have to use the
>> kludge of dividing by hf to get a number per second of photons.
>>
>
>I agree.
>
>> Now Compton scattering is an example of a problem that one would think the
>> classical theory could deal with perfectly well, but doesn't.
>>
>
>The problem is that classical theory might model the electron as a
>point particle - which won't capture the interactions well at low
>wavelengths.
The classical theory will say that the electron oscillates with the
frequency of the field no matter how large you decide the electron is,
and therefore that the radiation it emits must be the same as that of the
field that made it accelerate.
>
>> >
>> >
>> >> The best classical
>> >> treatment you can do of x-rays is to divide the field by hf and call that
>> >> the number of photons per second.
>> >
>> >You mean that's the best -quantum- treatment.
>>
>> Nope. That's a kludge that allows you to use the classical theory on a
>> problem that's really quantum mechanical. You realize that x-rays aren't
>> detected with a radio antenna, right?
>>
>
>Yes! :)
What's the difference between an x-ray and a radio wave?
>
>> >
>> >>And that worked well enough for Warren
>> >> in his treatment of x-ray diffraction.
>> >>
>> >
>> >Warren was considering quantized systems involving electrons trapped in
>> >atoms. Free electrons in very strong magnetic fields can give off
>> >x-ray synchrotron emission. There's no quantization involved there -
>> >until you are use an SSD to measure the radiation.
>>
>> You didn't read Warren. But as I'd mentioned, yes there is quantization
>> involved with free electrons in very strong electromagnetic fields.
>>
>
>I'm very curious to hear more, please educate me further.
He considered a free electron. Since an x-ray frequency is much higher
than the visible light frequencies of typical electronic transitions, it's
reasonable to neglect the restoring force of the nucleus.
srp
Continuous emission during electron acceleration was a hypothesis
put forward by Larmor and Lienard (if I remember correctly) at the
end of the 1800's, before any real conclusive experiments could be
conducted.
In the 1930's however, the whole physics community, with Bohr leading
the way, had finally bowed to the evidence that this was not the case
for the electron in the ground state of the hydrogen atom.
Moreover, Even for electrons made to accelerate in a magnetic field,
it was observed that on perfectly circular orbits, no radiation,
continuous or otherwise, could be detected, in the only real conclusive
experiments that were conducted, that is, with the GE 100 MeV Betatron.
When numerous electrons were fed in, their mutual repulsion made it
difficult for all of them to perfectly follow the main trajectory (the
perfectly circular orbit that only betatrons ideally allow) and most of
them tend to oscillate about that perfect trajectory as they move along.
This oscillation amounts to repeated transverse accelerations of the
particles involved followed by braking to transverse stops, at which
points a photon is emitted by each stopping electron, which was named,
after the cause had been identified in the first synchrotron accelerator
"synchrotron radiation".
But when only a few electrons were fed in so that they minimally
interfere with each other, perfectly circular orbits can
be obtained in a Betatron, with electrons moving at very near light
speed. And this is precisely what happened in the first Betatron
experiments, before general interest quickly grew for the useful
applications of synchrotron radiation to the point of causing
interest for research on the characteristics of motion of charged
particles on perfectly circular orbits to wane considerably.
Ref: "Radiation Losses in the Induction Electron Accelerator",
John P. Blewett, (Phys. Rev. 69, 87 (1946))
Schwinger later (1949 I think) tried to explain away the loss,
but only on theoretical grounds. No one ever tried to
experimentally clear the case later on with Betatrons due
to lack of interest.
It is easy today to observe that when the beam is not ondulated
in linear accelerator, no acceleration related radiation is
emitted until the target is reached.
On the other hand, how could an electron verifiably accelerate
in exact proportion with the energy that is communicated to it
if it continuously radiated that energy?
I really don't understand why so-called continuous emission of
radiation during acceleration is still being taught. As far as
I can see, it was proven false way back in the 1930's for
electrons in atoms and for circularly (really circular orbits)
accelerated electrons in the 1940's and for non-ondulated beams
in linear accelerators ever since they came in service.
> A sinusoidal current on top of a constant current will also
> emit.
"Sinusoidal current" really is "to and fro motion" of the free
moving electrons in the conductor. To and fro motion involves
repeated acceleration followed by stopping and reaccelerating
in the reverse direction, etc.
André Michaud
she...@yahoo.com
An atom emits a photon.. by dropping from one discrete energy state to
another. That contradicts your statement that photons are not about
discrete energy states.
> >
> >> And quantum mechanics
> >> applies perfectly well to free electrons in radio antennas or magnetic
> >> fields or whatever else you like.
> >
> >You can try to apply techniques of quantum mechanics - however you are
> >applying them to a system which isn't quantized.
>
> It turns out that quantum mechanics doesn't just apply to the electronic
> transitions of atoms. You recall one of the postulates of quantum
> mechanics-- DeBroglie's relation p=h/lambda. That means, for instance, if
> you have a pure sinusoidal electromagnetic wave it can transfer a momentum
> p=h/lambda to a particle. Not p=0.5*h/lambda, not p=1.127*h/lambda. Just
> p=h/lambda. It can do it twice, p=2h/lambda, or three times, or more.
> But always in steps of h/lambda. That's quantization. That's what makes
> an interaction with a field like a collision between particles. You don't
> need discrete energy states for that, it applies to all interactions.
>
> Undergraduate quantum mechanics doesn't quantize the field. In that sense
> it's really a semiclassical theory-- the quantum mechanics of particles in
> classical fields. That's a good enough approximation for many purposes.
> But not, e.g., for the Lamb shift.
>
I don't know much about the Lamb shift, but I do know it involves bound
electrons - quantized due to the central potential of the nucleus.
It's not about free electrons.
In general, just the frequency/wavelength/energy. However there are
lots of other differences, such as x-rays are often produced as
products of nuclear decays - those discrete energy states again.
> >
> >> >
> >> >>And that worked well enough for Warren
> >> >> in his treatment of x-ray diffraction.
> >> >>
> >> >
> >> >Warren was considering quantized systems involving electrons trapped in
> >> >atoms. Free electrons in very strong magnetic fields can give off
> >> >x-ray synchrotron emission. There's no quantization involved there -
> >> >until you are use an SSD to measure the radiation.
> >>
> >> You didn't read Warren. But as I'd mentioned, yes there is quantization
> >> involved with free electrons in very strong electromagnetic fields.
> >>
> >
> >I'm very curious to hear more, please educate me further.
>
> He considered a free electron. Since an x-ray frequency is much higher
> than the visible light frequencies of typical electronic transitions, it's
> reasonable to neglect the restoring force of the nucleus.
>
"Typical electronic transistions".. I would have guessed that's UV
range, not visible. There are plenty of transitions in the x-ray band,
especially:
K transitions connecting the K shell (n = 1) to the shells with
principal quantum numbers n = 2 to 4 and L transitions connecting the
L1, L2, and L3 shells (n = 2) to the shells with principal quantum
numbers n = 3 and 4.
I'm still curious about the quanitzation of free electrons in very
strong fields, please say more - the Wiedemann book isn't in my
library, but I see both volumes are available for about $40 each..
THanks again - shevek
Gregory L. Hansen
There are photons associated with that, sure. But they just conserve
energy and quantum numbers. Dropping from one discrete energy state to
another is more about discrete energy states, the way I see it. Photons
come along for the ride.
>
>> >
>> >> And quantum mechanics
>> >> applies perfectly well to free electrons in radio antennas or magnetic
>> >> fields or whatever else you like.
>> >
>> >You can try to apply techniques of quantum mechanics - however you are
>> >applying them to a system which isn't quantized.
>>
>> It turns out that quantum mechanics doesn't just apply to the electronic
>> transitions of atoms. You recall one of the postulates of quantum
>> mechanics-- DeBroglie's relation p=h/lambda. That means, for instance, if
>> you have a pure sinusoidal electromagnetic wave it can transfer a momentum
>> p=h/lambda to a particle. Not p=0.5*h/lambda, not p=1.127*h/lambda. Just
>> p=h/lambda. It can do it twice, p=2h/lambda, or three times, or more.
>> But always in steps of h/lambda. That's quantization. That's what makes
>> an interaction with a field like a collision between particles. You don't
>> need discrete energy states for that, it applies to all interactions.
>>
>> Undergraduate quantum mechanics doesn't quantize the field. In that sense
>> it's really a semiclassical theory-- the quantum mechanics of particles in
>> classical fields. That's a good enough approximation for many purposes.
>> But not, e.g., for the Lamb shift.
>>
>
>I don't know much about the Lamb shift, but I do know it involves bound
>electrons - quantized due to the central potential of the nucleus.
>It's not about free electrons.
It's the effect on spectra when you quantize the field.
They're also produced in synchrotrons. Probably the most important use of
synchrotrons is to generate x-rays.
>
>> >
>> >> >
>> >> >>And that worked well enough for Warren
>> >> >> in his treatment of x-ray diffraction.
>> >> >>
>> >> >
>> >> >Warren was considering quantized systems involving electrons trapped in
>> >> >atoms. Free electrons in very strong magnetic fields can give off
>> >> >x-ray synchrotron emission. There's no quantization involved there -
>> >> >until you are use an SSD to measure the radiation.
>> >>
>> >> You didn't read Warren. But as I'd mentioned, yes there is quantization
>> >> involved with free electrons in very strong electromagnetic fields.
>> >>
>> >
>> >I'm very curious to hear more, please educate me further.
>>
>> He considered a free electron. Since an x-ray frequency is much higher
>> than the visible light frequencies of typical electronic transitions, it's
>> reasonable to neglect the restoring force of the nucleus.
>>
>
>"Typical electronic transistions".. I would have guessed that's UV
>range, not visible. There are plenty of transitions in the x-ray band,
>especially:
>
>K transitions connecting the K shell (n = 1) to the shells with
>principal quantum numbers n = 2 to 4 and L transitions connecting the
>L1, L2, and L3 shells (n = 2) to the shells with principal quantum
>numbers n = 3 and 4.
X-rays cover a larger breadth of wavelengths than visible light does. The
x-rays that materials scientists are interested in are usually a few
Angstroms, comparable to the separation of nuclei in materials.
>
>I'm still curious about the quanitzation of free electrons in very
>strong fields, please say more - the Wiedemann book isn't in my
>library, but I see both volumes are available for about $40 each..
Wiedemann isn't a good book to learn about that. He has a purely
practical interest, which he disposes of by dividing the classical
radiation pattern by hf. The subject is quantum field theory or quantum
electrodynamics. And there's no one book that I can really recommend.
But Bjorken and Drell, despite being old-fashioned (or maybe because of
it) have a pretty clear treatment and some applications.
she...@yahoo.com
They are absorbed or emitted for the ride - a quick ride giving them a
precise quantum of energy in the rest frame of the system. This
(photon) radiation is of a very different character than the other kind
of light we were discussing - slow or continuum emission such as
cyclotron radiation or radio waves from an antenna.
> >> >> >
> >> >> >
> >> >> >> The best classical
> >> >> >> treatment you can do of x-rays is to divide the field by hf and
> >call that
> >> >> >> the number of photons per second.
> >> >> >
> >> >> >You mean that's the best -quantum- treatment.
> >> >>
> >> >> Nope. That's a kludge that allows you to use the classical theory on a
> >> >> problem that's really quantum mechanical. You realize that x-rays aren't
> >> >> detected with a radio antenna, right?
> >> >>
> >> >
> >> >Yes! :)
> >>
> >> What's the difference between an x-ray and a radio wave?
> >>
> >
> >In general, just the frequency/wavelength/energy. However there are
> >lots of other differences, such as x-rays are often produced as
> >products of nuclear decays - those discrete energy states again.
>
> They're also produced in synchrotrons. Probably the most important use of
> synchrotrons is to generate x-rays.
>
So please tell me how this kind of light is quantized! The light is
generated by the acceleration of free electrons by magnetic fields.
The energy of the free electron is not quantized, nor is the magnetic
field strength (those two parameters can easily be adjusted as
experimental parameters). How could the emitted light be quantized?
Thanks, that one's in my library - I'll give it a go. I'll give you my
opinion vs. the Kaku text.
Cheers - shevek
Timo Nieminen
> Gregory L. Hansen wrote:
>>>
>>> An atom emits a photon.. by dropping from one discrete energy state to
>>> another. That contradicts your statement that photons are not about
>>> discrete energy states.
>>
>> There are photons associated with that, sure. But they just conserve
>> energy and quantum numbers. Dropping from one discrete energy state to
>> another is more about discrete energy states, the way I see it. Photons
>> come along for the ride.
>
> They are absorbed or emitted for the ride - a quick ride giving them a
> precise quantum of energy in the rest frame of the system. This
> (photon) radiation is of a very different character than the other kind
> of light we were discussing - slow or continuum emission such as
> cyclotron radiation or radio waves from an antenna.
>> They're also produced in synchrotrons. Probably the most important use of
>> synchrotrons is to generate x-rays.
>
> So please tell me how this kind of light is quantized! The light is
> generated by the acceleration of free electrons by magnetic fields.
> The energy of the free electron is not quantized, nor is the magnetic
> field strength (those two parameters can easily be adjusted as
> experimental parameters). How could the emitted light be quantized?
How indeed? But were not the first definitive photon experiments
(Millikan's photoelectric effect experiments) absorption of photons by
conduction electrons in metals, with unquantised energies both before and
after absorption?
The spectra due to free-free and bound-free atomic transitions matches
theory if you assume quantisation, E=hf. How are free-free atomic
transitions fundamentally different from synchrotron-accelerated
electrons?
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
newedana
medium equals the mass, mainly because of the influence of false
particle physics. This obsolete idea seriously distorted the science of
understanding the light and its propagation characters.
Light is exactly the same as sound wave in terms of propagation
character. However, particle physicists in high school texts deny this
fact, saying that the light is entirely different from sound wave. It
is mainly due to a bad influence of particle physicists who believe
light is corpuscular photons.
Once sound wave is released from its source, it propagates through the
air phase with its own speed that is nothing to do with the speed of
its source. If the source moves at some speeds there occurs the Doppler
effect.
Thus there occurs the Doppler effect when the source moves at some
speeds. One cannot hear the steamwhistle influenced by the Doppler
effect if one is in the same train. It is because the elongated
wavelength of steamwhistle due to Doppler effect turns out to be
restored to its initial dimensions when he receives it in the same
train, because he runs with the same speed as the train does.
Light is exactly the same. If light source and the detector are on the
same coordinate system, or on the earth, one can measure always the
absolute speed of light, without any relation with the orbiting speed
of the earth. Elongated or condensed wavelength of light due to Doppler
effect becomes restored when the detector receives it.
Because the detector has to receive the wave signal as it advances or
in reverse case retreats with the same speed as the source, Albert
Einstein didn't know this simple plain truth. Since Einstein's photon
travels with its momentum given by its source, as though a shut bullet
travels in the space with its momentum given by its rifles. Obviously
the speed of bullet involves the speed of rifle.
So Einstein set forth the famous, but wrong, postulation for his
special theory of relativity: If a number of observer are moving at
uniform velocity in respect to each other and to a source of light, and
if each observer measures the speed of the light emerging from the
source, they will all obtain the same value. The same value in
Einstein's word means the absolute speed of light from which the speed
of source is excluded.
In order to explain the failed experiment of Michelson-Morley, the
speed of source v had to be removed, but Einstein didn't know how to
remove it from additive formula, c=c'+ v, so he made his postulation
c=c' when v approach the speed c.
Thus Einstein proposed the general principle of projectile mechanics
involving the travelling mechanism of light, expressed with a WRONG
equation: v=(v'+u)/(1+v'u/c^2), by borrowing the idea of Lorentz's
space contraction. And he announced that the speed of light is constant
anywhere in the cosmic space, because the light has the fastest speed
of all possible speeds in the nature, based on his also WRONG equation,
m=m'(1-v^2/c^2)^-1/2.
According to Einstein, mass increases its absolute value as its speed
increases, so if the speed of mass approaches the speed of light its
acceleration can no more contribute to its speed increment.
However this is a WRONG that if we believe that the vacant space itself
is only the medium of light propagation. Because the vacant space is
absolutely uniform anywhere in this cosmic space, so the speed of light
has to have a naturally constant speed. Then how can we explain the
light refraction taking place between different materials with
different optical density?
This phenomenon is more scientifically explained in Dr. Hansik Yoon's
book, "Natural Science Founded on A New Atomic Model",
(http://www.yoonsatom.net). Dr. Yoon in the Light Section of his
book(Part IV) clearly explains the refraction phenomenon with a simple
equation built without any postulation, including no speed factor, but
containing
wavelength of incident light, mass density factor, as well as incident
angle of incoming light.
Dr. Yoon also explains in his book that the light refraction can take
place because the atomic nuclei in mass system subdivides the incoming
light wave into numerous micro beams which develope into spherical
waves, so the constructive interference between them build a refractive
light with a different running direction.
Dr. Yoon also shows a number of schematical experiments of light
refraction, utilizing a large number of concentric half circles
representing sequential wave phases, drawn on two transparent films,
and superimposing them.
And Dr. Yoon claims the empty space itself is the only medium of light
propagation, and element particles building material system has nothing
to do with this light propagation. Although this assertion conflicts
critically against the traditional concept, it is quite correct. People
today has been taught that electrons building the material system
serves to transmit the light passing through material system.
Feynman had also the same idea, so Dr. Yoon in his book debates
critically the Feynman's equation representing refractive index, built
based on his QED theory, saying that it is a typical example of
misleading people with a complex mathematical trick.
As a plain truth, electric and magnetic force can act through this
empty vacant space without any aid of mass particles. Dr. Yoon
emphasises in his book that we have to abandon our old prejudice,
action only through a medium and medium equals the mass.
Particle physicists wrongly believe that the electric and magnetic
force acting through this vacant space is due to exchange of their
energy grains traveling with their momentum, and disregard the true
mechanism of how these forces can act without medium.
This incorrect belief, mass system interferes the speed of light is
inherited from our science pioneer such as Fizeau who tried to
investigate in 1845, how does the speed of light change due to moving
speed of its medium, such as water. Fizeau mis-evaluated the light
interference occurred between two light beams, one running along the
flowing water and the other against that, as speed difference between
them. But it is quite WRONG!
Imagine two rockets. One approaches the earth and the other departs
from the earth with the same speed. If they emit lights with the same
wavelength to a detector on the earth, the detector would receive two
signals interfering with one another, exactly the same pattern as that
Fizeau obtained in his experiment. Have these two lights different
speed? Absolutely no. They are exactly the same.
This is the reevaluation of Fizeau's experiment by Dr.Yoon in his book,
and Dr. Yoon says if vacuum empty space itself is only the medium of
light, there CANNOT exist such a moving coordinate system in this
universe as far as concerned to transmission of light, which is the
base of Einstein's two relativity theory.
Dr. Yoon proves in his book that light does not change its speed even
in material system.
He also proves that light is not a flow of photons.
Based on Dr. Hansik Yoon's new physics, "Natural Science Founded on A
New Atomic Model". (http://www.yoonsatom.net and
http://yoonsphysics.blogspot.com/ )
Autymn D. C.
Autymn D. C.
The lossless/lossfree accelerators meseem like magnetic crustals.
Autymn D. C.
Autymn D. C.
> Real electrons are almost infinitely big,
>
> **************
>
> Huh. Can one fit inside the causal horizon?
part of it
Autymn D. C.
> >> >> >> >> In article <1134873846.9...@g49g2000cwa.googlegroups.com>,
> >> >> >> >> chuckles <chuckleb...@gmail.com> wrote:
> >> >> >> >> >If sound travels through the compression and rarefaction of
Gregory, arsehole, trim posts or shut up. Stop cascading or die.
> As far as I know, buckyballs are the largest particles that have been
> diffracted.
Viri are.
Autymn D. C.
Autymn D. C. wrote:
> chuckles wrote:
> > If sound travels through the compression and rarefaction of particles
> > in the air how does light travel through a vacuum?
>
> This is easy:
> <http://groups.google.com/groups?q=liht+OR+fotons+Autymn&start=0&scoring=d&filter=0>.
> Look for the two links I posted in one message to Ross for my
> transactional interpretation of energic radiation. Real electrons are
> almost infinitely big, so the vacuum is no problem. And there is a
> vacuum between the air molecules, so sound does fare in a vacuum.
>
> There /are/ dumb asknesses (questions), those that are too easily or
> already answeren. Frosend (Dismiss) fake wisdom. Fotons are waves,
> but waves are not things. Fotons do not ever will (act) as particles;
> all things are particles, yet they will wave and will as waves. The
> fotons that seem to part are but particle conjugates, such as e+/-,
> about to come out of the field. But the particles were always there.
Read the above carefully, ye, and stop asking about fotons and empty
space. Fotons are not real. They depend on particles to propagate, as
they /are/ the propagating particles. The only particular properties
they have are the line-of-fliht/siht needen by the distance term in
Coulomb's law and the lack of refraction for more energic fotons which
is wreaken only by faster collisions with charges to abide by
momentumal relations. I have redefinate a particle from being a
point-centric bit of matter commensurate with its charge to being the
fieldly region outside the charge dome so that
"interaction-at-a-distance" is no longer a fuddle. The æther is the
field; the particle is the fieldly whit (quantum); space is the set of
all whits; therefore, particles may occupy any region of space, by
definition. Where particles end, at nodes, is where space for that
particle ends. The universe, any universe, has a volume definate by
its energic content--the same is true for any gas. (Note that
"definate" is a past participle of "define", as it should be spelt, and
not a misspelling of "definite".) Existence is solvate by classic
mechanics and alternative algebra.
-Aut
Autymn D. C.
it's -> its, retard
behaviour -> behavred
cloths -> clothes
childs -> child's
colours -> colors, hues
donsto...@hotmail.com
the information.
chuckles
Colours is an acceptable spelling of the word. The US dictionary is NOT
a de facto standard for the internet.
If you weren't implying colour spelt with a u was incorrect, sorry.
Gregory L. Hansen
she...@yahoo.com <she...@yahoo.com> wrote:
>
>Gregory L. Hansen wrote:
>> In article <1135745014....@g43g2000cwa.googlegroups.com>,
>> she...@yahoo.com <she...@yahoo.com> wrote:
>> >
>> >Gregory L. Hansen wrote:
>> >> In article <1135718856....@g14g2000cwa.googlegroups.com>,
>> >> she...@yahoo.com <she...@yahoo.com> wrote:
>> >
>> >An atom emits a photon.. by dropping from one discrete energy state to
>> >another. That contradicts your statement that photons are not about
>> >discrete energy states.
>>
>> There are photons associated with that, sure. But they just conserve
>> energy and quantum numbers. Dropping from one discrete energy state to
>> another is more about discrete energy states, the way I see it. Photons
>> come along for the ride.
>>
>
>They are absorbed or emitted for the ride - a quick ride giving them a
>precise quantum of energy in the rest frame of the system. This
>(photon) radiation is of a very different character than the other kind
>of light we were discussing - slow or continuum emission such as
>cyclotron radiation or radio waves from an antenna.
...
>> >>
>> >> What's the difference between an x-ray and a radio wave?
>> >>
>> >
>> >In general, just the frequency/wavelength/energy. However there are
>> >lots of other differences, such as x-rays are often produced as
>> >products of nuclear decays - those discrete energy states again.
>>
>> They're also produced in synchrotrons. Probably the most important use of
>> synchrotrons is to generate x-rays.
>>
>
>So please tell me how this kind of light is quantized! The light is
>generated by the acceleration of free electrons by magnetic fields.
>The energy of the free electron is not quantized, nor is the magnetic
>field strength (those two parameters can easily be adjusted as
>experimental parameters). How could the emitted light be quantized?
If x-rays are produced in synchrotrons then it must be quantized, mustn't
it? Because x-rays are detected by the recoil energies of the particles
in the detector that they bump in to. You can't mistake an x-ray for a
continuous wave.
It's DeBroglie's relation, E=h/lambda=hf/c. Since the energy of the free
electron is not quantized there's no particular limitation on what the
wavelength is. But whatever the wavelength is, the emitted radiation has
energy h/lambda, or integral multiples thereof.
To give it a semiclassical picture (it's quantum mechanics and so there
will be a superposition of states), that means as the electron goes
through the magnetic field, in any time interval it may or may not
interact with the field, it may or may not swerve, it may or may not
emit a photon.
I don't like to think of the photon as a particle because that brings to
mind little billiard balls knocking into each other, and questions like
"How can virtual particles create an attractive force?" Quantum field
theory is a theory of fields. But when fields interact, when momentum is
transferred from one to another, it occurs in units of p=h/lambda from
DeBroglie's relation. That sudden and finite momentum transfer gives all
the appearance of particles colliding.
I'm going to try a classical and semiclassical picture. Suppose you have
a particle going through an electric field E(x), the force on it is
F=-eE(x), the definition of force is F=dp/dt. Suppose the particle spends
a short enough time in the field that you can describe the interaction as
dp = F dt
= -e E(x) dt
Momentum eigenstates are plane waves, and that means a Fourier transform.
dp = -e dt \int E(q) exp(iqx) dq
with E(q) the transform of E(x),
E(q) = 1/(2*pi)^(3/2) \int E(x) exp(-iqx) dx
and p, q, x are vector quantities, and qx a dot product.
When we remain classical, the resulting momentum transferred, dp, is the
sum of all the possible momentum transfers. If we go to a semiclassical
picture with a classical electron and a quantized field, the momentum
transfered is q*dt with a probability proportional to e*E(q), and the
integral is a superposition of states. The superposition is given by an
integral-- q varies continuously, the energy levels are not quantized. q
can be anything that conserves energy.
If we make the field bigger we can look at possible momentum transfers dp1
in time interval dt1, transfer dp2 in time interval dt2, and so on, and
then integrate. And then we're halfway to a Green's function method of
solving electrodynamics problems. Each time some momentum is transferred
between the particle and the field, we say "A photon has been exchanged."
Gregory L. Hansen
Gregory L. Hansen <glha...@steel.ucs.indiana.edu> wrote:
>In article <1135887555.0...@g14g2000cwa.googlegroups.com>,
>she...@yahoo.com <she...@yahoo.com> wrote:
This is what I get for flying by the seat of my pants.
>It's DeBroglie's relation, E=h/lambda=hf/c. Since the energy of the free
>electron is not quantized there's no particular limitation on what the
>wavelength is. But whatever the wavelength is, the emitted radiation has
>energy h/lambda, or integral multiples thereof.
p=h/lambda=hf/c, energy p=h/lambda
There might be other corrections to be made. Just read what I mean, not
what I say.
--
"Voice or no voice, the people can always be brought to the bidding of
the leaders. This is easy. All you have to do is to tell them they
are being attacked, and denounce the pacifists for lack of patriotism
and exposing the country to danger." -- Hermann Goering
Autymn D. C.
I have no country. No, it's not acceptable: The Latin is color. Do
you write authour or commutatour as well? Stop borrowing French
corruptions and think about what you write.
-Aut
Anonymous
Nonsense. Spelling is what it is and needs no justification except the
sanction of time.
she...@yahoo.com
E=hf does not necessarily imply quantization. If the frequency can
vary continuously, then so can the energy, and you can have any energy
photon you'd like. The formula applies to emission and absorption of
energy. "Free-free" atomic transitions are perhaps somewhat misnamed -
if they are atomic transitions then they are governed by the central
potential of a nucleus, and can act to quantize the energy - only
discrete energy eigenvalues are solutions. Just because the captured
electron can quickly be transferred to neighboring atoms due to its
valence nature doesn't mean that the electron is "free" - it is still
bound by the atoms of the crystal.
A key is your use of the word "absorption" - something that is
absorbed isn't free. Synchrotron accelerated electrons see no central
potential, atomic transition electrons do see a central potential.
srp
Actually they do, kind of. They are forced to move in locally
uniform orthogonal electric and magnetic fields around a very
precise approximately circular path.
The precision and intensity of the fields allow setting very
precise energies to be released as synchrotron photons, that
in my view, just about amount to forced quantization.
I refer you again to that excellent textbook
"Principles of Charged Particle Acceleration" by S. Humphries Jr.
>>--
>>Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
>>E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
>>Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
>
>
--
André Michaud
Service de Recherche Pédagogique http://pages.globetrotter.net/srp/
she...@yahoo.com
To your first question, the answer is no. You hit the nail on the head
however when describing how the radiation is detected - it is detected
in discrete photons - that is the limitation of the detector.
One can easily create continuous x-rays.. just continously send high
energy electrons into a strong magnetic field. Continuously sampling
the signal is difficult however, unless you have a petaHertz
oscilloscope lying around over there.
It's your detector that is doing the quantizing.
> It's DeBroglie's relation, E=h/lambda=hf/c. Since the energy of the free
> electron is not quantized there's no particular limitation on what the
> wavelength is. But whatever the wavelength is, the emitted radiation has
> energy h/lambda, or integral multiples thereof.
>
> To give it a semiclassical picture (it's quantum mechanics and so there
> will be a superposition of states), that means as the electron goes
> through the magnetic field, in any time interval it may or may not
> interact with the field, it may or may not swerve, it may or may not
> emit a photon.
>
That's not a very useful picture. An electron - by definition - has a
charge, meaning a divergence of electric field around it. When you
accelerate the electron, the electric field at a point in space also
changes. These changes in the electromagnetic field propagate outward
at the speed of light. I see no need for a superposition of states in
this case.
> I don't like to think of the photon as a particle because that brings to
> mind little billiard balls knocking into each other, and questions like
> "How can virtual particles create an attractive force?" Quantum field
> theory is a theory of fields. But when fields interact, when momentum is
> transferred from one to another, it occurs in units of p=h/lambda from
> DeBroglie's relation. That sudden and finite momentum transfer gives all
> the appearance of particles colliding.
>
Well, how can virtual particles create an attractive force? :) What
does such a contrived explanation gain us anyway?
One can easily design a large impact parameter scattering experiment in
which arbitrarily small amounts of momentum are exchanged.. the only
problem is that in -detection- of the momentum or energy we rely on
atomic transition states.
> I'm going to try a classical and semiclassical picture. Suppose you have
> a particle going through an electric field E(x), the force on it is
> F=-eE(x), the definition of force is F=dp/dt. Suppose the particle spends
> a short enough time in the field that you can describe the interaction as
>
> dp = F dt
>
> = -e E(x) dt
>
> Momentum eigenstates are plane waves, and that means a Fourier transform.
>
> dp = -e dt \int E(q) exp(iqx) dq
>
> with E(q) the transform of E(x),
>
> E(q) = 1/(2*pi)^(3/2) \int E(x) exp(-iqx) dx
>
> and p, q, x are vector quantities, and qx a dot product.
>
> When we remain classical, the resulting momentum transferred, dp, is the
> sum of all the possible momentum transfers. If we go to a semiclassical
> picture with a classical electron and a quantized field, the momentum
> transfered is q*dt with a probability proportional to e*E(q), and the
> integral is a superposition of states. The superposition is given by an
> integral-- q varies continuously, the energy levels are not quantized. q
> can be anything that conserves energy.
>
> If we make the field bigger we can look at possible momentum transfers dp1
> in time interval dt1, transfer dp2 in time interval dt2, and so on, and
> then integrate. And then we're halfway to a Green's function method of
> solving electrodynamics problems. Each time some momentum is transferred
> between the particle and the field, we say "A photon has been exchanged."
>
I suppose I'd better get used to that terminology, but it seems
obfuscatory to me. If a 'photon' is a placeholder for any
electromagnetic energy, then how does one indicate a quantized amount
of electromagnetic radiation such as that emitted or absorbed by atomic
transitions?
Cheers - shevek
Timo Nieminen
> Timo Nieminen wrote:
>>
>> How indeed? But were not the first definitive photon experiments
>> (Millikan's photoelectric effect experiments) absorption of photons by
>> conduction electrons in metals, with unquantised energies both before and
>> after absorption?
>>
>> The spectra due to free-free and bound-free atomic transitions matches
>> theory if you assume quantisation, E=hf. How are free-free atomic
>> transitions fundamentally different from synchrotron-accelerated
>> electrons?
>
> E=hf does not necessarily imply quantization.
It does. It's the whole point. For a frequency f, you only get energy in
and out of the field in chunks of E=hf.
> If the frequency can
> vary continuously, then so can the energy, and you can have any energy
> photon you'd like.
Sure. Note that the first theoretical stuff done with quantised emission
of electromagnetic radiation was black-body radiation, with a continuous
spectrum. Quantisation doesn't mean that you can only have photons of
certain energies. Even from atomic transitions between two states, even if
the atom is unperturbed by its environment and is stationary, the spectrum
is still Lorentzian, and thus photons of any frequency can be emitted -
the probability distribution is peaked at the "transition wavelength", but
this is not the only permitted wavelength.
> The formula applies to emission and absorption of
> energy. "Free-free" atomic transitions are perhaps somewhat misnamed -
> if they are atomic transitions then they are governed by the central
> potential of a nucleus, and can act to quantize the energy - only
> discrete energy eigenvalues are solutions. Just because the captured
> electron can quickly be transferred to neighboring atoms due to its
> valence nature doesn't mean that the electron is "free" - it is still
> bound by the atoms of the crystal.
Free-free atomic transitions are governed by the central potential of the
nucleus, and the allowed energies are a continuum (because they're
"free", not "bound").
> A key is your use of the word "absorption" - something that is
> absorbed isn't free. Synchrotron accelerated electrons see no central
> potential, atomic transition electrons do see a central potential.
And the electron cares for what reason? Surely all that is important is
that the electron is accelerated? Why should emission by atoms, including
free-free transitions be quantised ie governed by E=hf, while synchrotron
electrons not be?
Timo Nieminen
> I suppose I'd better get used to that terminology, but it seems
> obfuscatory to me. If a 'photon' is a placeholder for any
> electromagnetic energy, then how does one indicate a quantized amount
> of electromagnetic radiation such as that emitted or absorbed by atomic
> transitions?
"Photon", since it's all just EM radiation, and the physics is the same,
the quantisation is the same, regardless of the source of the photon.
Your comments that it's the emission/absorption that is quantised, not the
EM field in any fundamental sense is quite sensible. For some reason that
I never quite grokked, the EM field is not quite suitable as a
wavefunctions for photons, but it's close enough for most purposes. The
photon is the quantum of excitation of its wavefunction, effectively the
EM field, which is continuous, both classically and quantumly. Only the
excitation is quantised.
If you think of the EM field being fundamentally continuous, with only the
exchange of energy between the EM field and matter being quantised (ie
E=hf when exchanging energy), then that is far more correct than the
commonly-held picture of billiard-ball photons.
FrediFizzx
news:2006010105...@emu.uq.edu.au...
I think the botton line for this is the question; can hbar also be due
to quantum "vacuum" processes like c is? If not, then the quantization
of the source free Maxwell equations is wrong to do. Milonni does not
agree with that in his book "The Quantum Vacuum; an Introduction to
QED". Of course if we take the quantum "vacuum" to be a medium, then
hbar can be also due to "vacuum" processes and the source free Maxwell
equations can be quantized. Seems fairly simple to me. Quantum Vacuum
Charge = +,- sqrt(hbar*c). ;-)
FrediFizzx
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
she...@yahoo.com
Thanks for your reply.
You may want to note that a Hydrogen atom, unlike a synchrotron, cannot
be tuned to emit light of any frequency. Of course this is true for
all atomic emitters and this solved the "ultaviolet catastrophe" of
blackbody emission.