Re: Part 1 Re: Ken, need help with this
From: Pax (pax1_at_whitesweb.com)
Date: 10/21/04
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Date: Thu, 21 Oct 2004 04:37:22 GMT
> "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
> news:2202379a.0410201041.7443fbca@posting.google.com...
> > "Pax" <pax1@whitesweb.com> wrote in message
> > news:<fEqdd.7822$Lk3.3671@newssvr12.news.prodigy.com>...
> > [Pax]
> > Part 1 of "Re: Ken, need help with this"... this sucker's too long to
post in one piece, so cutting it in two.
> [Ken]
> Hi Pax, I'm having a problem with your format, it's hard for anyone to
know who said what!
[Pax]
Sorry for the confusing formatting, trying to fix it here. :)
> > > "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
> > > news:2202379a.0410150154.2fe4ed05@posting.google.com...
> > > > "Pax" <pax1@whitesweb.com> wrote in message
> > > > news:<e Fbd.4721$Lk3.2906@newssvr12.news.prodigy.com>...
> > > > [Pax]
> > > > p = m_0v (classic statement), its checking permutations being
> > > [Ken]
> > > ((never heard it put that way, interesting))
> > [Pax]
> > Because the m_0 assumes it doesn't include kinetic energy? Guess I
colored it with m0 due to my current fixation on massless objects. The
classic statement is actually p = mv, should've just left it at that since
it's all the same in the end. Thanks!
> > > > [Pax]
> > > > so the equation for finding energy (E),
> > > > E^2/c^2 = (m_0^2c^2) + p^2,
> >
> >
> > [Pax]
> > Guess this should just be E^2/c^2 = (m^2c^2) + p^2, although I've seen
it
> > written both ways (with m_0 and with just m).
> [Ken]
> Well what's important is,
>
> E = m_0 c^2 /sqrt(1-v^2/c^2) = m_0 c^2 *gamma
>
> where E includes the contribution of the rest mass m_0 and the kinetic
energy from gamma.
[Pax]
The problem is that mass is a determining factor for finding kinetic energy
as well.
> > > > [Pax]
> > > > Strange, it seems a massless particle can have no velocity, is
incapable of momentum, and has no energy. If the energy carried by an object
is directly proportional to its mass and momentum, then for a massless
object it appears that would calculate to 0 energy, yet we know light has
energy. The energy must therefore be in the wave and not in the photon. Can
we then go the final step and conclude that photons are the stationary
medium of transmission, better known classically as aether, for
electromagnetic waves that propagate through it?
> > > [Ken]
> > > ((That's topical, pardon my rant))...
> > [Pax]
> > "Ranting" is what I asked for. :)
> > > [Ken]
> > > IMO, the rest mass of a photon can only be measured when it leaves a
system K and takes some energy, and the rest mass of the photon appears as a
deficit to the emitting system and/or is then brought to rest by absorption
in system K' adding energy to that system.
> > >
> > Between K and K' a doppler shift can occur or in a g-field an Einstein
shift, so the emission deficit may not be the same as the absorption
increment.
> > >
> > > There is really no point in discussing the "rest mass" of a "photon in
flight", (I call that a floton), because it's not at rest.
> > >
> > > It gets complicated when a "floton" is deflected in a g-field.
Basically what happens is the floton appears to be deaccelerated by the
g-field in the process of deflection. That deacceleration from speed c to a
lesser speed C (C = speed of light in a g-field) can be regarded as part of
the floton being absorbed by the g-field, thus exhibiting a "rest mass"
absorption, and the rest whizzes by but red shifted by momentum exchange.
> > [Pax]
> > It's taken me a long time to post what you're replying to because I knew
I'd have to fight through the "always moving photon" assumption... heck, not
"assumption", more "given". If you could just suspend that assumption for a
moment... I don't mean consider it to be wrong, just suspend the notion that
it *must* be right...
> [Ken]
> ok
> > [Pax]
> > First, every effect that is attributed to photons existing at c is based
> > on four things:
> > 1) photons are massless
> > 2) photons deliver energy
> > 3) photons move in waves
> > 4) photon reception --
> > a) resulting in absorption
> > b) resulting in reflection/deflection/etc.
> [Ken]
> ok
> > [Pax]
> > There are variables, such as interference, but no good explanation for
one photon interfering with itself (double slit) has been discovered. The
phenomenon has been noted and accepted, that's different. Of course, some
have tried to explain it, but the reasoning has more to do with wave
functions.
> >
> > However, on closer examination, considering photons to be moving really
does no harm, I guess. It just precludes there being any sort of actual
"fabric" to spacetime. Quantum foam doesn't seem to be a good alternative,
unless it allows for wave displacement... but I haven't examined it closely
enough yet to actually make that sort of judgment.
> > > > [Pax]
> > > > What I am saying is that, rather than one photon carrying a portion
of energy separately from source to destination, photons work in tandem like
"bucket-brigades", picking up energy from the source and then handing it off
to the next photon in "line". In normal terms, photons pass energy on
through waves.
> [Ken]
> ok, assume everywhere in spacetime is a virtual photon, corresponding to
your empty bucket, then the photon energy fills the bucket...and so
on...that's interesting.
[Pax]
Exactly! That's why I can't get it out of my mind, especially due to all the
special-case exceptions applied for photons. The only things we know about
light are through interception. In vacuo, we can't see it going away from
us, we can't watch it going sideways, we can't detect it until at least some
of it is redirected toward us or a collector.
Visualize a true bucket brigade, but everything is invisible... men,
buckets, and water... until the line is intercepted somewhere. Now, what if,
no matter where you intercepted that line of moving water, a bucket
materialized to deliver the water, but you only knew the bucket and water
were there because, as the bucket delivered its load of water, the bucket
hit at a definite point of impact in such a way that you knew it had to be
an object? Would it be easy to draw the conclusion that the bucket was
actually the water and only one bucket was being used from source to
interceptor?
> > > [Ken]
> > > Sure a photon is a tiny oscillator, it has wave characteristics.
> > [Pax]
> > Every particle has wave characteristics. :) There is no such thing as a
"solid". But examine what you're saying:
> >
> > A tiny particle, unlike every other know thing that waves, is both its
own medium and its actual wave movement. A corresponding alternative would
be to say that each molecule of water drops from the sky and rushes in a
wave outward... not *causes* waves outward, but *is* a wave that moves
outward. Yes molecules are huge in comparison to particles, but the analogy
holds.
> >
> > Waves in the ocean are transferred from molecule to molecule, in other
words: the waves *propagate* through their medium of water molecules.
Water's wave action is a result of force applied at some point to the medium
as an entity. Light waves are also said to *propagate*, in fact, c is their
detectable propagation speed. [Dictionary.com -- propagate - Physics. To
cause (a wave, for example) to move in some direction or through a medium;
transmit.]
> > > > [Pax]
> > > > This would explain why photons seem so commonly to be "destroyed" in
tied groups.
> [Ken]
> Oh I understand, I was not familiar with the word "tied", no need for a
ref, your explanation was good for me.
[Pax]
Thanks. :) Wish it was good enough for me, too. I really hate trusting
completely to memory.
> > > [Ken]
> > > Yes both, recall we can convert mass to sucking 1.5 km of length call
that number m = 1.5 km.We also suck 1.5 km of time, for a total of 2m sucked
in spacetime.
> > [Pax]
> > Was very grateful you pointed that out to me in that other thread. :)
Had no idea until then there was any math that backed up my deductions.
> > > [Ken]
> > > It turns out that the deflection of light by the spacetime "warp",
(maybe we should call it spacetime suction) is
> > >
> > > deflection = 2m / R, R = closest approach to m
> > > (oo => R)
> > >
> > > and that is reported to be confirmed by measuring the deflection and
retarded flight times of photons by the sun.
> > [Pax]
> > Thanks. :) Keeping that too.
> > > > [Pax]
> > > > In my studies I have come across these statements ([bracketed] are
my insertions):
> > > >
> > > > 1) "The de Broglie wavelength of a particle equals Planck's constant
divided by its momentum [w=h/p]. The momentum of a particle equals the
square root of twice its mass times its kinetic energy [p=sqrt(2m0)*Ek]. The
kinetic energy of a particle equals the square of its momentum divided by
twice its mass [Ek=p2/2m0]."
> > > >
> > > > Re 1): All of the above assertions seem to hinge upon rest mass
being greater than zero (m0>0), would they not all otherwise zero out?
Wavelength depends upon momentum, momentum depends upon velocity times mass,
kinetic energy depends upon momentum divided by mass. Is the assertion that
photons always move at c based on anything other than the observation of
light waves? If wavelength equals Planck's Constant divided by momentum,
then photons cannot have a wavelength.
> > > [Ken]
> > > Before discussing those questions, please see if they are altered by
my rant.
> > [Pax]
> > Sorry, but no. Discuss? :)
> [Ken]
> Ha, ok. Let's keep our terms straight, a photon is something that is
measurable, as I ranted above, it is defined by how it interacts with
matter. A "floton" is an "in flight flying photon", that is supposed to be
defined in the absence of interaction.
[Pax]
So you've divided the energy from the particle that delivers it as a matter
of course in the past. :)
> [Ken]
> Well, one can define a floton from some dynamics, and it can possess
frequency, in the form of oscillating EM-fields. Electric charges are
orthogonal to x,y,z,t and so any lengths and motion within the floton
relating electric charge is also perpendicular to x,y,z,t so the lorentz
transform does not apply.
[Pax]
It seems you're describing the transverse nature of lightwaves, but your
terminology is strange to me. The Lorentz Transformation applies to
lightwaves, since Lorentz is based on the velocity of a light emitting
object. But perhaps you mean the Lorentz Transformation doesn't apply to the
strength of the electromagnetism making up the wave.
Need some clarification here. :) Afraid to comment at present.
> > > > [Pax]
> > > > Planck's Constant is
> > > > H = 6.626*(10^-34*J)*s,
> > > > where J is Joules and s is seconds) and
> > > > J = 1 kg m^2/s^2,
> > > > where kg is kilograms and m is meters.
> > > [Ken]
> > > yup
> > On to Part 2. :)
> >
> > Be well - Pax
> You too.
> Ken
Thanks. :)
Be well - Pax
.~*~._.~*~._.~*~._.~*~._.~*~._.~*~._.~*~._.~*~._.~*~.
May people say of you:
"The world is a better place because you are in it."
>From Andromeda:
"Dillon Hunt, there are three kinds of people in this
universe, those who can count, and those who can't."
We sit inside the impossible and say the impossible
is impossible.
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