Re: Ken, need help with this
From: Androcles (dummy_at_dummy.net)
Date: 10/15/04
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Date: Fri, 15 Oct 2004 19:24:41 GMT
"Paul Draper" <pdraper@yahoo.com> wrote in message
news:74768d2d.0410150955.203bcdef@posting.google.com...
> "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.
That is misguiding statement,
p = mv works perfectly at high speeds too, and is a definition.
In FACT, the problem you face is not assuming an equation is right.
>>
>> 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.
Correct. We can discuss the momentum of a photon, however; it is
hf(c+v/c, but since f = 1/t, this becomes h(c+v)/tc.
>> 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).
This is where you make misleading statements.
For the photon traveling at high speed, in this case c,
E = hf (Planck) = mc^2 (Langevin)
m = hf/c^2.
So p = hf/c if and only if p = mc.
>>
>> so the equation for finding energy (E),
>>
>> E2/c2 = (m02c2) + p2,
No it isn't.
E = Rest Energy + K.E.
= mc^2 + 1/2mv^2.
E^2 = ( mc^2 + 1/2mv^2)^2
= m^2c^4 +m^2v^2c^2 + m^2v^4 /4
Since p = mv,
E^2 = m^2c^4 +p^2c^2 + m^2v^4 / 4.
Dividing through by c^2,
E^2/c^2 = m^2c^2 + p^2 + m^2v^4/c^2 /4
You've left out m^2v^4/c^2 / 4.
For v close to c, this approximates to m^2c^2/4.
>
> 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.
>
I should tip you off that your algebra is incorrect. You didn't complete the
square.
>>
>>
>> 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.
Correct.
>> 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.
Cows are not mammals? Who claims that?
>
>>
>>
>>
>> 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.
That's fine, but the wad has both momemtum and mass equivalence.
E = hf = mc^2
p = hf/c = mc
>>
>>
>> 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.
YOU are incorrect, but I lack time to point it out.
I'll just say
p = h/w , ct = w, p = h/ct, f = 1/t,
p = hf/c.
so w=h/p is correct.
> 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
>>
Yes, Einstein's math was hopeless and nobody ever pointed out his errors:
For quotations following, reference:
http://www.fourmilab.ch/etexts/einstein/specrel/www/
("On the Electrodynamics of Moving Bodies" by Albert Einstein)
1) "light is always propagated in empty space with a definite velocity c
which is independent of the state of motion of the emitting body",
a totally unproven assumption without any evidence to support it.
2) "In agreement with experience we further assume the quantity
2AB/(t'A-tA) = c to be a universal constant- the velocity of light in empty
space.",
an admitted assumption that is quite worthless when there is any
relative motion between A and B, yet essential to the derivation of the
remainder of Einstein's nonsense.
3) The equation
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v)) ,
the ½ of which is derived from 2) above and is tantamount to saying
(1/3 + 2/3)/2 = 1/3.
4) The missing 0' from that equation, since x' = x-vt, hence 0' = 0-vt,
and the equation should be
½[tau(-vt,0,0,t)+tau(-vt,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
at the very least.
5) The further assumption "IF we place x' = x-vt ... " without considering
IF we place x' = x+vt, from which we derive (using Einstein's method)
tau = (t+xv/c^2)/sqrt(1-v^2/c^2)
xi = (x + vt)/sqrt(1-v^2/c^2)" -Paul B. Andersen
6) The statements
"But the ray moves relatively to the initial point of k,
when measured in the stationary system, with the velocity c-v..."
and
"It follows, further, that the velocity of light c cannot be altered by
composition with a velocity less than that of light. For this case we obtain
V = (c+w)/(1+w/c) = c."
which are contradictory, the first being Galilean, the second being
contrary to the vector addition of velocities, an axiom of a vector space.
7) The lack of a check to verify the theory is self-consistent by feeding
the new PoR given in 6) into the equation given in 3) and finding a total
failure.
Check:
(t1-t)/(t2-t)*[tau(-vt,0,0,t)+tau(-vt,0,0,t+x'/V+x'/V)] = tau(x',0,0,t+x'/V)
where V = (c+v)/(1+v/c) as required by the redefined PoR.
Androcles.
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