Re: Ken, need help with this
From: Paul Draper (pdraper_at_yahoo.com)
Date: 10/15/04
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Date: 15 Oct 2004 10:55:43 -0700
"Pax" <pax1@whitesweb.com> wrote in message news:<e_Fbd.4721$Lk3.2906@newssvr12.news.prodigy.com>...
[a bunch of misguided statements, strewn willy-nilly across the
countryside, which I will now attempt to set upright, one by one.]
>
>
> The following math relies on my abilities in algebra, and I'm not
> certain if algebra is up to what really determines the results of it
> all.
It doesn't, and you shouldn't let it. Your intuition should guide your
use of the math, not the other way round. If your intuition is not
sound, the math will just be a meat grinder, churning away on garbage,
and you will still get garbage out.
> As presented by the renowned men responsible for the foundational
> calculations, one would think the reductions into algebra are accurate,
> but have trouble understanding many of the conclusions that spring form
> this math. The algebra seems to say certain assumptions taken so much
> for granted are not warranted.
The math is not so much the problem. In fact, the problem you face is
always assuming an equation is right.
>
>
>
> Below is a section of my "study notes". :)
>
>
>
>
> -------------------------------------------------------------------------
> -------
>
>
>
>
> 1) The law of conservation of momentum
>
> 2) The First Law of Thermodynamics concerning conservation of energy
> (This needs no further explanation it merely needs to be remembered.)
>
>
>
> The law concerning the conservation of momentum (p) results from a
> consideration of both rest mass (m0) and velocity (v)...
No, it doesn't, especially if you're going to talk about nonclassical
things.
>
> p = m0v (classic statement), its checking permutations being
The key disclaimer is "classical statement". In general, this formula
for momentum is WRONG. It is an approximation. It only works at low
speeds.
>
> m0 = p/v,
>
> and
>
> v = p/m0,
>
> but v = p/m0 will not hold true if m0 = 0 unless v = 0 to begin
> with.
Here your math *is* weak. This will not hold true if m0=0 regardless
whether v=0 or not. Something divided by zero is not zero.
>
> For a massless particle if
>
> m0 = 0,
>
> then
>
> p = 0,
>
> since 0*v = 0,
And here is where you go wrong, using an approximate formula in a case
where it doesn't work. "p=mv" doesn't apply for anything that is
massless or traveling at high speed (or both).
>
> so the equation for finding energy (E),
>
> E2/c2 = (m02c2) + p2,
And now you compound the mistake by using a classical approximation
("p=mv") to embed in a relativistic one. You can't do that. It doesn't
follow. You shouldn't be surprised that using an incorrect formula
inside a correct one yields funny results.
>
> that uses for its basic elements both mass (m) and momentum (p), breaks
> down to
>
> E2/c2 = (02c2) + 02
>
> E2/c2 = 0c + 0
>
> reducing to
>
> E = 0+0
>
> (or, classically, E = pc) must result in
>
> E = 0
>
> for a massless object.
>
Which of course should tip you off that you've done something wrong.
Not in the algebra, but in the understanding of what formulas apply
when.
>
>
> 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.
No, that doesn't follow at all. It's not like there's two energy
buckets for light, one for a wave and one for a photon. The SAME
energy in the light must be accounted for in EITHER description.
> 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?
Certainly not. You just concluded, by your own analysis, that photons
have no meaning because they have no energy (when physicists
practically *define* photons as packets of light energy). And yet you
then claim they must mean something else? That's like claiming that,
because cows are not mammals, they must be fish.
>
>
>
> This would seem on the surface to be a ridiculous assertion, since
> photons were first described by Max Planck by their virtue of being
> packets of energy (See Planck's Law). If you examine that you might
> notice the word "packets". A packet is a container, a carrier, something
> that is capable of containing and, one would assume, delivering
> something else. I am asserting nothing different from that.
>
A packet is not just a container, a carrier. If you like, define a
photon as a "wad" of energy, or a "clump" of energy, or a "bit" of
energy, or a "chunk" of energy. Let the concept drive the language,
not the language drive the concept.
>
>
> 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".
Lovely concept. Not what a photon is. Nor is a bucket-brigade
required, any more than a baseball, once struck, needs hands passing
it from one to another for it to reach the outfield bleachers. Why do
you think light needs to be carried in a bucket?
> In normal terms, photons
> pass energy on through waves. This would explain why photons seem so
> commonly to be "destroyed" in tied groups.
I have no idea where you got the impression that photons are destroyed
in tied groups. Nothing could be further from the truth.
>
>
>
> In order for the energy carried in a wave to diminish, gravity or
> friction of the medium must offer resistance, in other words, the medium
> must absorb some of the energy or be inhibited by gravity, and/or both.
> A massless medium is incapable of such absorption or inhibition... or is
> it? What is warped by gravity in the vicinity of a gravitational mass,
> is it only time or is it space as well? Most certainly it must be both.
Yes it is both. And you're right, in a vacuum, the energy in light
does not diminish as long as it's not going uphill. The same is true
for any wave. It might get *spread out* with distance as it moves from
a source, but it doesn't diminish.
>
>
>
> 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]."
Both of the added bracketed statements are incorrect. Like the formula
for momentum above, you have used the low speed approximation, which
is not correct in general. Moreover, the de Broglie wavelength as
expressed above does not apply to all particles, it only applies to
massive particles. Once again, you have led the formulas lead you
astray from the correct understanding.
>
> [http://www.hsphys.com/ibahl quantum summary L.doc]
>
>
>
> 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.
Well, that all assumes that all the formulas you've used will work for
massless or high-speed particles, which they won't.
>
> Planck's Constant is
>
> H = 6.626*10-34J*s,
>
> where J is Joules and s is seconds) and
>
> J = 1 kg m2/s2,
>
> where kg is kilograms and m is meters.
>
>
>
> 2) "In classical mechanics, massless objects are an ill-defined
> concept, since applying any force to one would produce, via Newton's
> second law, an infinite acceleration - a nonsensical result."
> [http://en.wikipedia.org/wiki/Mass#Relation between Mass.2C Energy and Mo
> mentum in relativistic mechanics]
>
Which goes to warn that it's not a good idea to apply a classical idea
(like Newton's 2nd Law) to a non-classical object.
>
>
> Re 2): First, it's far from "nonsensical", but more to the point, one
> must wonder how someone goes about applying force to a massless object.
I'm not sure you do! Applying a force is a classical concept.
> That's like trying to put a circle around "nothing", the minute you
> attempt such an exercise, it's no longer "nothing", it's "something",
> the area within the circle.
>
>
>
> Force is a form of mass.
The heck it is! Where did you get that idea?
> Mass and energy are interchangeable.
That statement doesn't say that mass and energy are equal. What it
says is, that within the bounds of the other laws of physics that work
even non-classically (like conservation of momentum), mass can be
turned into energy and energy into mass.
> Zero mass
> must then equal zero energy.
No, that doesn't follow. Even taking your mass-energy
interchangeability statement literally, take the following case. I
have an object with some mass, and I convert it to energy completely
somehow. Now I have all energy left, and no mass. Are you going to
conclude that because I have no mass left, that there must be no
energy, either?
> If something has no mass, what is there to
> convert to energy or add energy to? Photons could very well be the
> bottom of the particle "food chain", the enablers of force interaction.
> However, this seems to already be assumed, justified, explained, and
> utilized to the point of being more a law than anything else.
What if -- and I'm just supposing here -- what if the photons *are*
the energy that you converted mass into? And if you convert energy
back into mass, it's photons that disappear and massive particles that
appear?
>
>
>
> 3) "The net force on any massless object must be zero. If a string
> is massless, the tension force is the same at either end (and any point
> in the middle)."
>
> [http://scott-yost.baylor.edu/phy1422f04/review2.pdf]
>
This is a blatant case of taking a statement out of context. You are
quoting a classical treatment of an idealized piece of string, not a
photon; moreover the author of that statement makes no pretense that
the statement applies to photons. Moreover, a force acting on a photon
is mixing classical and nonclassical concepts. Again, you are letting
language dictate the concepts, rather than a good conceptual
understanding dictate your language.
>
>
> Re 3): Sense at last.
>
>
>
> 4) "Massless objects such as photons also carry momentum; the
> formula is p=E/c, where E is the energy the photon carries and c is
> the speed of light."
>
> [http://en.wikipedia.org/wiki/Momentum]
>
>
>
> Re 4): How do they justify the above statement? Did someone just decide
> all the numbers were wrong where photons were concerned simply by virtue
> of the fact they found nothing of mass to be the medium for aether?
Not quite. They decided that "p=mv" is a classical approximation to a
better definition, which you have not learned or invoked.
> Did
> no one at any time notice photons have no mass and add the two facts?
Sure they did, but you apparently got left behind, confused.
> An
> undetectable medium + a massless particle = an undetectable, massless
> aethereal medium composed of massless particles. Photons do not have
> energy, but waves are very capable of having it, in fact, all waves are
> a form of energy in motion.
>
>
>
> Let us consider the Photoelectric Effect. How can a massless object like
> a photon be thrown, what can interact with something that has no mass?
Why do you assume that something has to have mass to interact with it?
You interact with waves all the time. A sound wave can cause blood to
come from your ears. A microwave oven can congeal the protein in an
egg. A radio wave can send electrons zinging up and down an antenna.
> Further, how can a massless thing be thrown against something of mass in
> such a manner as to cause parts of that object of mass, a metal plate,
> to dislodge? First, it is electrons (negatively charged leptons) that
> are dislodged from the metal plate. Could the Photoelectric Effect be
> the result of the interaction of charged forces rather than particles?
It could be, but it isn't. Otherwise we'd detect the charge beyond the
electrons liberated. Note, by the way, that the wave description of
light FAILS to account for the behavior of the photoelectric effect.
Read that section again. Waves CANNOT make what we observe in the
photoelectric effect.
> Where do quarks and gluons fit in? The number parade into the tiny seems
> neverending. However, the number of gluons in each of the gluon chains
> binding quarks together is determined by force.
I'm not sure what you're issue here is. Gluons have nothing to do with
light. Or are you just complaining that it's all so complicated?
>
>
>
> Einstein concluded ejected electrons take on a photon and that is what
> causes them to become dislodged, many even call such ejected electrons
> "photoelectrons", denoting the energy added to the electron when it
> absorbs the force of a photon. Is there a difference between force, as
> assigned the name "quanta", and the deliverer of that force, the photon?
> Yes, however the photon is incorporated into the electron, but for
> another reason, the electron is a particle of mass with properties
> immediately sympathetic to such absorption.
>
I'm sorry, I didn't even understand what you were trying to say in
that last paragraph so I can't fix the errors.
>
>
> There must first be mass in order for a reaction to force to occur,
No that's not true, and even Newton didn't say that. Where did you
conclude that?
> however, even if force could be applied, that force would result in
> imparting mass to the former massless object.
No, that's not true, either. Exerting a force does not impart mass.
> but then, what are waves
> if not the results of force? It could only be that they are not waves of
> physical force in the sense commonly assumed (as when a bat hits a
> ball), they are waves composed entirely of electromagnetic force. This
> explains the EM pulse that accompanies an atomic blast.
>
OK, I have to admit I'm getting a little winded.
>
>
> It is the transformation of mass into electromagnetism and heat (the
> highest forms of energy) during thermal reactions that causes
> displacement waves. Though true waves of physical force in the form of
> heat are released with the conversion of mass into its highest order
> manifestation of photons, it is the EM waves that are responsible for
> light. My Lord! Does that mean that Universal expansion is due to the
> transformation of denser forms of energy into higher forms of energy?
>
No, and you've not only wandered into the weeds, but are leaping over
bushes in the weeds.
>
>
> That is what happened at the time of the Big Bang (which is quite
> plausible now since the introduction of the Clashing Branes Theory).
> Then transformation into denser forms of energy would result,
> ultimately, in a return to a state of greatest density, such as in the
> case of Schwarzschild Radii, and this would result in a shrinking of
> space!
>
>
>
> What must follow from that thought is that gravity itself is an
> indication of a shrinking of space, which would explain why spacetime is
> warped in the vicinity of mass.
>
>
>
> Wait! It suddenly occurred to me that a supraforce "enveloped within" a
> supraforce could manifest as multiple dimensions and time, since the
> enveloping supraforce could pull in all directions "outward" from the
> internal supraforce!
OK, I'm standing back here on the road, and I can't even see you
anymore. I hear you whooping and making a lot of animal noises, but I
can't make it out.
>
>
> -------------------------------------------------------------------------
> -------
>
>
> That last just sort of flowed from the other, so left it in.
>
> Be well - Pax
>
> .~*~. .~*~. .~*~. .~*~. .~*~. .~*~. .~*~. .~*~. .~*~.
>
> What are ten years in the history of humanity? Must
> not all those forces that determine the life of a man
> be regarded as constant compared with such a trifling
> interval? - Albert Einstein - Out of My Later Years
>
> As far as the laws of mathematics refer to reality,
> they are not certain; and as far as they are certain,
> they do not refer to reality - Albert Einstein
>
> I don't believe in mathematics. - Albert Einstein
>
> Do not worry about your problems with mathematics,
> I assure you mine are far greater. - Albert Einstein
>
>
> --
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