Re: The massless photon?
From: TomGee (lvlus_at_hotmail.com)
Date: 11/29/04
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Date: 29 Nov 2004 01:04:24 -0800
vergon_enterprises@highstream.net (V ertner Vergon) wrote in message news:<b337f5db.0411281940.67d32c98@posting.google.com>...
> AN ANALYSIS OF THE
> ENERGY-MOMENTUM 4 VECTOR EQUATION
> AND THE MASSLESS PARTICLE
>
>
>
> A wise man once said, regarding theoretical physicists, "They are
> often ?algebraically' correct in their ?proofs', while fumbling on the
> interpretation."
>
>
I think that is so simply because it is hard to distinguish between
fantasy and reality because theoretical physics is not so concerned
with empirical evidence as it is with theory.
> Einstein gives the mass of radiation as m = E/c^2, that of the photon
> being
> h nu/c^2.
>
> The energy of the photon is given as E = pc = h nu.
>
> Then there is the oft misinterpretation of the energy-momentum 4
> vector equation.
>
> One interpretation of the said equation, E^2 = (mc^2)^2 + (pc)^2, is
> as follows:
>
> "If we set the m in the right hand first term to zero, then we get E =
> pc
> which we know is true. This shows that the mass of the photon is
> zero."
But isn't that like saying, "I will toss the die and I predict it will
land zero up", and tossing a die with zeros on every face of it?
>
> THE FALLACY:
>
> Following is a brilliant analysis by Jim Redgewell:
>
>
> EINSTEIN'S ENERGY FORMULA
>
> Page created 9 February 2003 (http://www.daftwat.com)
>
> Many people are familiar with the formula E = mc˛ , but this is a
> simplified version of the following formula:
>
> E^2 = mc^2 + p^2 c^2
>
>
> This formula takes into account the momentum 'p'. When p = 0, then E =
> mc˛.
> -------------------------------------------------------------------------------
> INTERJECTION by vergon:
>
> It is imperative to note here that first and foremost, mathematical
> expressions are required to reflect the physical conditions.
> In this case there are two ways to set p to zero. One is to have the
> object at rest ? then as a result mc^2 is the rest energy at rest. The
> other is to set p to 0 by setting the mass to 0 then both m and p
> would be 0 and E would equal 0. Redgewell has obviously chosen the
> former.
> We must exercise caution and practice astuteness when manipulating
> mathematics.
> ------------------------------------------------------------------------------
There is only one way to have anything visible at rest, and that is to
have it moving at constant velocity with that which measures it. That
is because no visible object can be at rest due to atomic/molecular
motion and/or the effects of an expanding universe.
>
>
> PYTHAGORAS
> The formula can be written differently to show that it is in fact
> based upon the Pythagoras theorem:
>
> E^2= (mc^2)^2 + (pc)^2
>
> ENERGY
> The next observation to be made, is that the above formula is in fact
> the following two formulas applied at right angles to each other.
>
>
> E = hf E = mc^2
>
> Obviously, the next step is to prove that pc = hf. So let's start with
> the formula for wavelength:
>
>
> lambda = h/p
>
>
> THE MATHS
> .
> p = h/lambda c = f lambda pc = h/lambda x f lambda . . pc
> = hf
>
> So the maths has worked out well. So basically, the formula has two
> energy functions which are vectors applied at 90° to each other. One
> of the energy functions is for matter at rest and the other is for
> matter in motion.
> -------------------------------------------------------------------------------
> INTERJECTION:
>
> This author is not sure what Redgewell means by " matter in motion".
> Since the equations above used to develop this result are equations of
> radiation (photon) then we can assume the matter spoken of is the mass
> of the photon. This would be distinct from ponderable mass (physical
> bodies) the kinetic energy given for such being E = (gamma ?1) mc^2.
Why would it be distinct from visible matter? Do you suppose matter
can mean in one case that it is non-ponderable mass, a non-physical
body? Has the meaning of the term changed to include now massless
particles (which I assume are the opposite of "ponderable matter"?
> Therefore, the energy-momentum 4 vector equation is not applicable to
> ponderous mass. This despite the fact it was developed, in part, from
> the equation for ponderable mass.
> -------------------------------------------------------------------------------
> PHOTONS
>
> The rest mass of a photon is zero, so the first part of the equation
> tends to zero. A photon travels at the velocity of light and at this
> speed it has mass.
A photon cannot ever actually have zero rest mass. It is created, it
is said, moving at c. Any consideration of it having rest mass puts
the work into the realm of theoretical physics where not any of its
results can be empirically based and thus they cannot be reported as
fact.
> Therefore, the energy of a photon is given by the
> second part of the equation. So a photon has a momentum of mc and
> therefore, the energy of a photon is pc, and is equal to mc˛. So once
> again E = mc˛.
> -------------------------------------------------------------------------------
>
> COMMENT:
>
> We have here a peculiar coincidence. mc^2 is the kinetic energy of the
> photon ? but also the rest energy of ponderous mass.
>
> Or is it a coincidence? We might draw the conclusion that the rest
> energy of ponderous mass is convertible to the kinetic energy of the
> photon ? both being mc^2, i.e., with no change in absolute dimensions.
> Of course the atomic bomb proves this true.
>
>
The atomic bomb proves that the potential energy inherent in a mass
can be released in certain situations. The rest energy of the mass
would be the potential energy of the mass in that case. When released
it becomes the kinetic energy of the mass.
>
> In further consideration, every equation for momentum contains MASS
> and motion. Therefore, E = pc states that the photon in flight has
> momentum, p = (h nu/c). And we observe (h nu/c^2) = m. Thus
>
> (h nu/c^2)c = mc = p -? and pc = mc^2.
>
> Therefore, we see that whereas mc^2 is the rest energy for a ponderous
> mass it is also the kinetic energy of a photon.
>
>
> We also note that the photon in flight possess momentum and that there
> can be no momentum without mass. Therefore, the photon possess mass.
> This is the same position taken by Einstein when he declared the mass
> of radiation to be m = E/c^2.
>
> When faced with that fact, those who misinterpret the equation try to
> maintain their position of a mass-less photon by declaring a new
> physics whereby there exists momentum with no mass.
>
> The problem is, they cannot substantiate that.
Agreed. They have tried, though, unrelentlessly. Some have decided
to just believe it is reality anyway.
>
> The overall result is that there is no such thing as a mass-less
> particle -- of any kind.
>
> The objective universe consists only of matter, the space between
> matter ? and the motion of matter in that space. All else are concepts
> in the mind of man.
>
> If matter (mass) is removed, there is left only empty space.
>
>
> Vertner Vergon
>
> 1998
But aren't you rejecting the evidence of the indirect effects seen and
thought to be caused by dark matter and energy? Space cannot be
completely empty for if it was, it would be absolute space. Most
scientists lean toward the idea that dark matter/energy comprise up to
70 percent of the universe and that certain effects observed stem from
such matter/energy. Dirac's positron was confirmed long ago and it
proposed the existence of electron particles having negative mass and
negative energy. Empty space could not make those effects since they
require force to make them happen.
TomGee 112904
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